Foundations of Recursive Interval Algebras

1. Introduction

An ordered pair carries two pieces of data. Repeating this construction gives objects whose components are themselves ordered pairs, until a scalar layer is reached. Recursive interval geometry begins with this elementary structure. Its foundational algebra is developed here over

$$R=\mathbb{Z}\left[{\Omega }^{-1}\right],\Omega \in \mathbb{Z},\Omega >1.$$

(1)

Set ${\mathcal{I}}_0=R$ and ${\mathcal{I}}_{n+1}={\mathcal{I}}_n\times {\mathcal{I}}_n$. The components of a pair are pos and len. At the scalar layer they satisfy $pos\left(c\right)=c$ and $len\left(c\right)=0$. These boundary values allow the same recursive formulas to act on inputs of different depths. In particular, the product is

$$xy=\left(pos\left(x\right)pos\left(y\right),pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)\right)$$

(2)

whenever at least one input has positive depth. Each term is evaluated by the same rule, and every branch ends with multiplication and addition in R.

The main questions concern what this recursion determines. Which elements are invertible or nilpotent? How does multiplication change a norm built from the same components? What happens when one recursive tree is grafted onto another in either order? Which scalar functions admit evaluation by recursive descent, and which estimates survive as depth increases? These questions connect the algebraic and metric parts of the construction directly.

At each fixed depth, choosing pos or len at every stage gives a path coordinate. The positions at which len is chosen form a subset of $\{1,\dots ,n\}$. Multiplication combines disjoint subsets and annihilates overlapping ones. Thus the finite algebra is the square-free quotient

$${\mathcal{I}}_n\cong R[t_1,\dots ,t_n]/(t_1^2,\dots ,t_n^2).$$

(3)

These quotients and their square-zero extensions belong to standard commutative algebra [1,2,3,4]. After real scalar extension, they are the zeon algebras generated by commuting square-zero elements. Their subset basis and Boolean-lattice representations are studied by Feinsilver [5], and their role in derivative and combinatorial identities by Neto and dos Anjos [6]. In the recursive realization, the coordinates are extracted by pos and len and remain compatible with arithmetic between different depths. The coefficient ring is a principal ideal domain, and is dense in the real numbers with their usual topology. Divisibility therefore controls which exact constructions stay in R, whereas density identifies sharp real norm bounds with suprema over the original recursive objects.

The weighted form gives a path with r occurrences of len squared weight ${\Omega }^{2r}$. Normalizing the path basis gives the Euclidean coefficient norm with the same disjoint-support multiplication. Norm inequalities for zeon algebras were developed by Lindell and Staples [7]. At depth one, Cook and Nguyen determined the optimal constant for any number of factors in the dual-number algebra [8]. Applying this estimate at each recursive position gives a constant depending on the full list of input depths. For k inputs of depths $m_1,\dots ,m_k$, put $r_j=\#\{i:m_i\ge j\}$. The exact multiplication constant is

$${\displaystyle \prod _{j=1}^{{\max }_i{m}_i}}{\left({\displaystyle \frac{r_j^{r_j}}{{(r_j+1)}^{r_j-1}}}\right)}^{1/2}.$$

(4)

In particular, two inputs have constant ${(2/\sqrt{3})}^{\min (m_1,m_2)}$. Theorem 15 also classifies every extremizing tuple, including the unrestricted tail of a uniquely deepest input, and determines when the real bound is attained over R. Homogeneous inputs satisfy additional estimates depending on the number of len positions rather than on recursive depth. These estimates distinguish a single product from a sum of products.

Directed tensor products concatenate two blocks of recursive positions. The left orientation places the first input outside the second, while the right orientation places the second input outside the first. They have exact norm factorization, but four-input inner-product factorization depends on the alignment of the blocks. This distinction persists when equal data at successive depths are identified. The tensor operation then defines a faithful one-sided action; the global nilpotent ideal is a free module of rank one for that action. Keeping the typed depths separately gives a free associative algebra on the two level-one component directions. Its Lie and Jordan structures connect recursive concatenation with the classical theories of free Lie algebras and special Jordan algebras [9,10,11].

Function evaluation supplies another connection between the operations. Polynomial and regular rational evaluation have the recursive rule

$$f\left(x\right)=\left(f\left(pos\left(x\right)\right),f^{\prime }\left(pos\left(x\right)\right)len\left(x\right)\right).$$

(5)

The same formula applies to coherent scalar derivative data and to smooth functions. Elementary functions and factorizations in zeon algebras are treated by Staples and Weygandt [12]; continuity and differentiability of zeon functions are studied by Staples [13]. Here finite path expansions give necessary and sufficient conditions for recursive evaluation to remain in R. Exponential and logarithmic operations on nilpotent elements and normalized units have integral coefficients even when factorial denominators are not units in R. For prescribed consecutive blocks, logarithmic supports determine all separating cuts and the unique finest normalized tensor factorization in Theorem 26.

Several structural consequences sharpen this foundation. Complete len descent gives a perfect multiplication pairing and exact annihilator formulas. Elements of maximal nilpotence index have a common multiplication Jordan type after rational scalar extension, including arbitrary higher-order terms. The rank calculation is related to the classical square-free multiplication theory [14,15,16]; a triangular recursive normalization handles the higher-order terms explicitly. On the analytic side, derivative data determine sharp function bounds on first-homogeneous inputs. Allowing arbitrary finite depths leads to a stricter stability question. Theorem 36 proves that, on a fixed-scalar fiber, recursive evaluation is locally bounded at some point exactly when every derivative datum of order at least two vanishes; otherwise it is locally unbounded at every point. In Theorem 35, degree scaling with real parameters $a_1,\dots ,a_k$ makes products contractive at every depth exactly when ${\sum }_i{a}_i^2\le 1$. Beyond this threshold no finite constant independent of depth exists. For tensor concatenation, Theorem 45 determines block-length-dependent commutator and anticommutator bounds together with the difference between real extremizers and original-ring attainability. These results relate finite recursive calculation to the norm behavior of operations across all depths.

Section 2 constructs the arithmetic and establishes the finite and global algebraic structure. Section 3 develops the weighted form, duality, sharp multiplication estimates, and completion. Section 4 treats the two tensor orientations and their block structure. Section 5 develops recursive function evaluation and its bounds. Section 6 studies the associative, Lie, and Jordan structures associated with concatenation. The final section summarizes the resulting foundation.

2. Recursive Arithmetic and Algebraic Structure

The arithmetic outlined in the introduction begins with ordered pairs and the two component operations pos and len. Their scalar boundary rules allow inputs of different depths to be combined by the same finite recursion. Recording the successive component choices gives path coordinates, which make the ring laws and the passage between depths explicit. The scalar coefficient then determines units and nilpotents, while further descent gives logarithmic coordinates, multiplication pairings, and the structure of nilpotent multiplication operators. These algebraic descriptions provide the basis for the weighted estimates and tensor factorizations that follow.

2.1. Scalar Descent and Path Coordinates

Definition 1 

(Recursive objects and scalar boundary). Let ${\mathcal{I}}_0=R$ and ${\mathcal{I}}_{n+1}={\mathcal{I}}_n\times {\mathcal{I}}_n$. An element specified in ${\mathcal{I}}_n$ has typed depth n. For $x=(a,b)$ of positive depth, set

$$pos\left(x\right)=a,len\left(x\right)=b.$$

(6)

For $c\in R$, set $pos\left(c\right)=c$ and $len\left(c\right)=0$. The zero and unit at depth n are given by

$$0_0=0,1_0=1,0_{n+1}=(0_n,0_n),1_{n+1}=(1_n,0_n).$$

(7)

Within a specified level, 0 and 1 denote its zero and ordinary multiplicative identity. In a recursive pair, their level is determined by that of the other component. Subscripts are retained when the depths themselves enter a calculation or comparison. The notation ${pos}^{\left(r\right)}$ and ${len}^{\left(r\right)}$ means r repeated applications; the zeroth iteration is the identity.

Definition 2 

(Arithmetic by scalar descent). When all inputs are scalars, use the arithmetic of R. Otherwise addition and multiplication are defined by

$$\begin{array}{cc}\hfill x+y& =\left(pos\left(x\right)+pos\left(y\right),len\left(x\right)+len\left(y\right)\right),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill xy& =\left(pos\left(x\right)pos\left(y\right),pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)\right).\hfill \end{array}$$

(9)

For a positive-depth input, define

$$-x=\left(-pos\left(x\right),-len\left(x\right)\right),cx=\left(cpos\left(x\right),clen\left(x\right)\right)(c\in R).$$

(10)

Every operation on the right is evaluated by the same recursion. Subtraction is $x-y=x+(-y)$.

Example 1 

(Subtraction at different depths). For $x=(1,2)\in {\mathcal{I}}_1$ and $y=((a,b),(c,d))\in {\mathcal{I}}_2$, descent gives

$$\begin{array}{cc}\hfill x-y& =\left(pos\left(x\right)-pos\left(y\right),len\left(x\right)-len\left(y\right)\right)\hfill \\ \hfill & =\left(1-(a,b),2-(c,d)\right)\hfill \\ \hfill & =\left(\left(pos\left(1\right)-a,len\left(1\right)-b\right),\left(pos\left(2\right)-c,len\left(2\right)-d\right)\right)\hfill \\ \hfill & =\left((1-a,-b),(2-c,-d)\right).\hfill \end{array}$$

The scalar boundary supplies a zero whenever a scalar meets a second component. The same calculation applies repeatedly across a larger depth gap.

Theorem 1 

(Termination and ring laws). For inputs of depths m and n, addition, subtraction, and multiplication terminate in ${\mathcal{I}}_{\max (m,n)}$. Negation and scalar multiplication preserve the input depth. Mixed multiplication by a scalar agrees with the scalar action. The commutative, associative, distributive, and unit laws hold in their common output levels. In particular, for arbitrary finite-depth $x,y,z$,

$$\begin{array}{cccc}\hfill x+y& =y+x,\hfill & \hfill (x+y)+z& =x+(y+z),\hfill \\ \hfill xy& =yx,\hfill & \hfill \left(xy\right)z& =x\left(yz\right),\hfill \\ \hfill x(y+z)& =xy+xz,\hfill & \hfill x+0& =x,1x=x.\hfill \end{array}$$

(11)

Here the identities hold with the scalar $0,1$ or with the zero and unit at the depth of x. For $x\in {\mathcal{I}}_n$, one also has $x+(-x)=0$ in ${\mathcal{I}}_n$.

Proof. 

Induct on the largest input depth d. The case $d=0$ is arithmetic in R. For $d\ge 1$, each input $x\in {\mathcal{I}}_j$ with $0\le j\le d$ satisfies

$$pos\left(x\right),len\left(x\right)\in {\mathcal{I}}_{\max (j-1,0)}.$$

Every recursive call therefore has depth at most $d-1$; pairing its outputs gives an element of ${\mathcal{I}}_d={\mathcal{I}}_{d-1}\times {\mathcal{I}}_{d-1}$. This proves termination and the depth assertions.

The same induction proves the ring laws by equality of the two components. For additive associativity,

$$\begin{array}{cc}\hfill pos\left((x+y)+z\right)& =\left(pos\left(x\right)+pos\left(y\right)\right)+pos\left(z\right)\hfill \\ \hfill & =pos\left(x\right)+\left(pos\left(y\right)+pos\left(z\right)\right)=pos\left(x+(y+z)\right),\hfill \\ \hfill len\left((x+y)+z\right)& =\left(len\left(x\right)+len\left(y\right)\right)+len\left(z\right)\hfill \\ \hfill & =len\left(x\right)+\left(len\left(y\right)+len\left(z\right)\right)=len\left(x+(y+z)\right).\hfill \end{array}$$

For additive commutativity, the lower-depth equalities give

$$\begin{array}{cc}\hfill pos(x+y)& =pos\left(x\right)+pos\left(y\right)=pos\left(y\right)+pos\left(x\right)=pos(y+x),\hfill \\ \hfill len(x+y)& =len\left(x\right)+len\left(y\right)=len\left(y\right)+len\left(x\right)=len(y+x).\hfill \end{array}$$

At a positive prescribed depth n, the zero and inverse calculations are

$$\begin{array}{cc}\hfill x+0_n& =\left(pos\left(x\right)+0_{n-1},len\left(x\right)+0_{n-1}\right)=x,\hfill \\ \hfill x+(-x)& =\left(pos\left(x\right)-pos\left(x\right),len\left(x\right)-len\left(x\right)\right)=(0_{n-1},0_{n-1})=0_n.\hfill \end{array}$$

The same zero calculation with scalar boundary values proves $x+0_0=x$. For $c\in R$ and positive-depth x,

$$cx=\left(cpos\left(x\right),clen\left(x\right)+0pos\left(x\right)\right)=\left(cpos\left(x\right),clen\left(x\right)\right).$$

This is the scalar action; taking $c=1$ gives $1_{0}x=x$. The typed unit $1_n=(1_{n-1},0_{n-1})$ gives the same pair of components and hence $1_{n}x=x$. For commutativity of multiplication,

$$\begin{array}{cc}\hfill pos\left(xy\right)& =pos\left(x\right)pos\left(y\right)=pos\left(y\right)pos\left(x\right)=pos\left(yx\right),\hfill \\ \hfill len\left(xy\right)& =pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)\hfill \\ \hfill & =pos\left(y\right)len\left(x\right)+len\left(y\right)pos\left(x\right)=len\left(yx\right).\hfill \end{array}$$

Finally, scalar distributivity and compatibility of two scalar actions hold in each component by the induction hypothesis. Pairing those component equalities proves the corresponding R-module laws.

The lower-depth ring laws give multiplicative associativity:

$$\begin{array}{cc}\hfill pos\left(\left(xy\right)z\right)& =pos\left(x\right)pos\left(y\right)pos\left(z\right)=pos\left(x\left(yz\right)\right),\hfill \\ \hfill len\left(\left(xy\right)z\right)& =pos\left(x\right)pos\left(y\right)len\left(z\right)\hfill \\ \hfill & +\left(pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)\right)pos\left(z\right)\hfill \\ \hfill & =pos\left(x\right)\left(pos\left(y\right)len\left(z\right)+len\left(y\right)pos\left(z\right)\right)\hfill \\ \hfill & +len\left(x\right)pos\left(y\right)pos\left(z\right)\hfill \\ \hfill & =len\left(x\left(yz\right)\right).\hfill \end{array}$$

Similarly, distributivity follows from

$$\begin{array}{cc}\hfill pos\left(x(y+z)\right)& =pos\left(x\right)\left(pos\left(y\right)+pos\left(z\right)\right)=pos(xy+xz),\hfill \\ \hfill len\left(x(y+z)\right)& =pos\left(x\right)\left(len\left(y\right)+len\left(z\right)\right)\hfill \\ \hfill & +len\left(x\right)\left(pos\left(y\right)+pos\left(z\right)\right)\hfill \\ \hfill & =len(xy+xz).\hfill \end{array}$$

These calculations include scalar inputs through $pos\left(c\right)=c$ and $len\left(c\right)=0$, so the induction applies to mixed depths. □

The component product identities, valid also at the scalar boundary, are

$$\begin{array}{cc}\hfill pos\left(xy\right)& =pos\left(x\right)pos\left(y\right),\hfill \\ \hfill len\left(xy\right)& =pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right).\hfill \end{array}$$

(12)

Thus pos is multiplicative and len satisfies a product rule relative to pos.

Definition 3 

(Path coordinates). Write $\left[n\right]=\{1,\dots ,n\}$ and $\left[0\right]=⌀$. For $S\subseteq \left[n\right]$, the path coefficient $x_S$ of $x\in {\mathcal{I}}_n$ is obtained by choosing len at positions in S and pos at the other positions, counting from the outside. The element ${\mathbf{b}}_S$ has coefficient 1 at this path and zero at every other path. In particular ${\mathbf{b}}_{⌀}=1$, and ${\mathbf{b}}_j$ means ${\mathbf{b}}_{\left\{j\right\}}$. Every element has the unique expansion

$$x=\sum _{S\subseteq \left[n\right]}{x}_S{\mathbf{b}}_S.$$

(13)

The homogeneous part of order r consists of the terms with $\left|S\right|=r$. Recursive depth records the available positions, while homogeneous order counts the len choices in a path.

Theorem 2 

(Path multiplication and polynomial model). The path elements form an R-basis of ${\mathcal{I}}_n$, and

$${\mathbf{b}}_A{\mathbf{b}}_B=\left\{\begin{array}{cc}{\mathbf{b}}_{A\cup B},\hfill & A\cap B=⌀,\hfill \\ 0,\hfill & A\cap B\ne ⌀.\hfill \end{array}\right.$$

(14)

Consequently

$${\left(xy\right)}_S=\sum _{A\bigsqcup B=S}{x}_A{y}_B,{\mathcal{I}}_n\cong R[t_1,\dots ,t_n]/(t_1^2,\dots ,t_n^2),$$

(15)

where $t_j$ corresponds to ${\mathbf{b}}_j$. Here $A\bigsqcup B=S$ means that A and B are disjoint and their union is S.

Proof. 

Path basis. At depth zero the only path is empty, ${\mathbf{b}}_{⌀}=1$, and each $c\in R$ has the unique expansion $c=c{\mathbf{b}}_{⌀}$. Suppose that the path expansion has been established for every object of depth n. If $x\in {\mathcal{I}}_{n+1}$, then $pos\left(x\right),len\left(x\right)\in {\mathcal{I}}_n$, so the induction hypothesis applies separately to these two objects:

$$\begin{array}{cc}\hfill pos\left(x\right)& =\sum _{A\subseteq \left[n\right]}{\left(pos\left(x\right)\right)}_A{\mathbf{b}}_A,\hfill \\ \hfill len\left(x\right)& =\sum _{A\subseteq \left[n\right]}{\left(len\left(x\right)\right)}_A{\mathbf{b}}_A.\hfill \end{array}$$

Here ${\left(pos\left(x\right)\right)}_A$ is the scalar obtained by following path A within the object $pos\left(x\right)$; the analogous interpretation applies to ${\left(len\left(x\right)\right)}_A$.

For $A\subseteq \left[n\right]$, write $1+A=\{1+j:j\in A\}$. An initial component choice shifts all subsequent positions by one. Hence

$$\begin{array}{cccc}\hfill {\left(pos\left(x\right)\right)}_A& =x_{1+A},\hfill & \hfill {\mathbf{b}}_{1+A}& =({\mathbf{b}}_A,0_n),\hfill \\ \hfill {\left(len\left(x\right)\right)}_A& =x_{\left\{1\right\}\cup (1+A)},\hfill & \hfill {\mathbf{b}}_{\left\{1\right\}\cup (1+A)}& =(0_n,{\mathbf{b}}_A).\hfill \end{array}$$

In each pair the entries ${\mathbf{b}}_A$ and $0_n$ belong to ${\mathcal{I}}_n$; the pair itself is a path element in ${\mathcal{I}}_{n+1}$. The first pair begins with a pos choice and the second with a len choice. Componentwise addition and scalar multiplication give, for $c\in R$,

$$\begin{array}{cc}\hfill c({\mathbf{b}}_A,0_n)& =(c{\mathbf{b}}_A,0_n),\hfill \\ \hfill c(0_n,{\mathbf{b}}_A)& =(0_n,c{\mathbf{b}}_A),\hfill \\ \hfill \left(pos\left(x\right),0_n\right)+\left(0_n,len\left(x\right)\right)& =\left(pos\left(x\right),len\left(x\right)\right)=x.\hfill \end{array}$$

Applying these identities to the two component expansions yields

$$\begin{array}{cc}\hfill x& =\sum _{A\subseteq \left[n\right]}{\left(pos\left(x\right)\right)}_A({\mathbf{b}}_A,0_n)+\sum _{A\subseteq \left[n\right]}{\left(len\left(x\right)\right)}_A(0_n,{\mathbf{b}}_A)\hfill \\ \hfill & =\sum _{A\subseteq \left[n\right]}{x}_{1+A}{\mathbf{b}}_{1+A}+\sum _{A\subseteq \left[n\right]}{x}_{\left\{1\right\}\cup (1+A)}{\mathbf{b}}_{\left\{1\right\}\cup (1+A)}\hfill \\ \hfill & =\sum _{S\subseteq [n+1]}{x}_S{\mathbf{b}}_S.\hfill \end{array}$$

Indeed, every subset S of $[n+1]$ has exactly one of the forms $1+A$ or $\left\{1\right\}\cup (1+A)$, according to whether 1 is absent or present. This proves existence of the expansion.

Successive component choices respect addition and scalar multiplication by Definition 2, so taking a path coefficient is R-linear. Moreover, ${\left({\mathbf{b}}_S\right)}_U$ equals 1 when $S=U$ and 0 otherwise. If a linear combination of the depth-$n+1$ paths is zero, taking its coefficient at any $U\subseteq [n+1]$ gives

$$0={\left(\sum _{S\subseteq [n+1]}{c}_S{\mathbf{b}}_S\right)}_U=\sum _{S\subseteq [n+1]}{c}_S{\left({\mathbf{b}}_S\right)}_U=c_U.$$

Thus all coefficients vanish. The expansion is unique and the path elements form an R-basis, completing the induction.

Products of path elements. We again induct on depth. At depth zero the rule is $1·1=1$. For the induction step, let $A,B\subseteq \left[n\right]$, and use the four products obtained from the recursive definition:

$$\begin{array}{cc}\hfill ({\mathbf{b}}_A,0_n)({\mathbf{b}}_B,0_n)& =({\mathbf{b}}_A{\mathbf{b}}_B,0_n),\hfill \\ \hfill ({\mathbf{b}}_A,0_n)(0_n,{\mathbf{b}}_B)& =(0_n,{\mathbf{b}}_A{\mathbf{b}}_B),\hfill \\ \hfill (0_n,{\mathbf{b}}_A)({\mathbf{b}}_B,0_n)& =(0_n,{\mathbf{b}}_A{\mathbf{b}}_B),\hfill \\ \hfill (0_n,{\mathbf{b}}_A)(0_n,{\mathbf{b}}_B)& =\left(0_n{0}_n,0_n{\mathbf{b}}_B+{\mathbf{b}}_A{0}_n\right)=0_{n+1}.\hfill \end{array}$$

The last line is zero because both first choices are len. In the other three lines, the only remaining multiplication is the depth-n product ${\mathbf{b}}_A{\mathbf{b}}_B$.

More explicitly, for $U,V\subseteq [n+1]$, let

$$A=\{j\in [n]:j+1\in U\},B=\{j\in [n]:j+1\in V\}.$$

If $1\in U\cap V$, the fourth product above gives ${\mathbf{b}}_U{\mathbf{b}}_V=0$. Otherwise one of the first three products applies. If $A\cap B\ne ⌀$, the lower-depth induction hypothesis gives ${\mathbf{b}}_A{\mathbf{b}}_B=0$, so the upper-depth product is again zero. These are exactly the cases in which $U\cap V\ne ⌀$. If $U\cap V=⌀$, then $A\cap B=⌀$ and ${\mathbf{b}}_A{\mathbf{b}}_B={\mathbf{b}}_{A\cup B}$. The corresponding product is

$${\mathbf{b}}_U{\mathbf{b}}_V=\left\{\begin{array}{cc}({\mathbf{b}}_{A\cup B},0_n)={\mathbf{b}}_{1+(A\cup B)},\hfill & 1\notin U\cup V,\hfill \\ (0_n,{\mathbf{b}}_{A\cup B})={\mathbf{b}}_{\left\{1\right\}\cup (1+(A\cup B\left)\right)},\hfill & 1\in U\cup V.\hfill \end{array}\right.$$

Both cases give ${\mathbf{b}}_{U\cup V}$. This proves (14) at depth $n+1$ and completes the induction.

Coefficients of an arbitrary product. For $x,y\in {\mathcal{I}}_n$, expand both inputs in their path bases and use distributivity, scalar multiplication, and (14):

$$\begin{array}{cc}\hfill xy& =\left(\sum _{A\subseteq \left[n\right]}{x}_A{\mathbf{b}}_A\right)\left(\sum _{B\subseteq \left[n\right]}{y}_B{\mathbf{b}}_B\right)\hfill \\ \hfill & =\sum _{A,B\subseteq \left[n\right]}{x}_A{y}_B{\mathbf{b}}_A{\mathbf{b}}_B\hfill \\ \hfill & =\sum _{\begin{array}{c}A,B\subseteq \left[n\right]\\ A\cap B=⌀\end{array}}{x}_A{y}_B{\mathbf{b}}_{A\cup B}\hfill \\ \hfill & =\sum _{S\subseteq \left[n\right]}\left(\sum _{A\bigsqcup B=S}{x}_A{y}_B\right){\mathbf{b}}_S.\hfill \end{array}$$

The sums are finite, and the last line groups terms according to their output support $S=A\cup B$. Uniqueness of the path expansion gives the coefficient formula in (15). The partitions in this sum are ordered: A belongs to the first input and B to the second. Equivalently,

$${\left(xy\right)}_S=\sum _{A\subseteq S}{x}_A{y}_{S\setminus A}.$$

For the empty path there is just the term ${\left(xy\right)}_{⌀}=x_{⌀}{y}_{⌀}$.

Polynomial model. Every ordinary monomial in $R[t_1,\dots ,t_n]$ either has all exponents at most one or is divisible by some $t_j^2$. Consequently each polynomial p can be written as

$$p(t_1,\dots ,t_n)=\sum _{S\subseteq \left[n\right]}{c}_S\prod _{j\in S}{t}_j+{\displaystyle \sum _{j=1}^n}{t}_j^2{q}_j(t_1,\dots ,t_n),$$

with $c_S\in R$ and ordinary polynomials $q_j$. Thus every class in the quotient has a square-free representative. This representative is unique: the difference of two such representatives contains only square-free monomials, whereas every nonzero monomial of a polynomial in $(t_1^2,\dots ,t_n^2)$ has a square factor. Equality of ordinary polynomial coefficients therefore forces their square-free coefficients to agree.

The path product rule also gives

$${\mathbf{b}}_j^2=0,\prod _{j\in S}{\mathbf{b}}_j={\mathbf{b}}_S,$$

where the empty product is 1. Substituting the commuting objects ${\mathbf{b}}_1,\dots ,{\mathbf{b}}_n$ into the preceding polynomial decomposition gives

$$\begin{array}{cc}\hfill p({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)& =\sum _{S\subseteq \left[n\right]}{c}_S\prod _{j\in S}{\mathbf{b}}_j+{\displaystyle \sum _{j=1}^n}{\mathbf{b}}_j^2{q}_j({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)\hfill \\ \hfill & =\sum _{S\subseteq \left[n\right]}{c}_S{\mathbf{b}}_S.\hfill \end{array}$$

By linear independence of the path basis, this value is zero precisely when all $c_S$ vanish, or equivalently when $p\in (t_1^2,\dots ,t_n^2)$. Hence two polynomials have the same value on these path generators precisely when they represent the same quotient class. Substitution therefore defines a well-defined injective correspondence from the quotient to ${\mathcal{I}}_n$.

For any $x\in {\mathcal{I}}_n$, the polynomial ${\sum }_{S\subseteq \left[n\right]}{x}_S{\prod }_{j\in S}{t}_j$ has value ${\sum }_{S\subseteq \left[n\right]}{x}_S{\mathbf{b}}_S=x$, so this correspondence is also surjective. Finally, the ring laws already proved by scalar descent give

$$\begin{array}{cc}\hfill (p+q)({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)& =p({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)+q({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n),\hfill \\ \hfill \left(pq\right)({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)& =p({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)q({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n).\hfill \end{array}$$

Constants $c\in R$ have value $c1$, so substitution preserves the scalar action and the unit as well. It is therefore the stated unital R-algebra isomorphism, with the class of $t_j$ corresponding to ${\mathbf{b}}_j$. At depth zero the polynomial ring is R and the ideal is zero, giving the same conclusion. □

This is the subset multiplication of the zeon algebra [5,6]. The recursive path description also records the order of the component choices, which determines how refinements and directed tensor blocks interact with the same finite algebra.

Definition 4 

(Refinement and global objects). Define a one-step refinement recursively by

$$\iota \left(c\right)=(c,0)(c\in R),\iota \left(x\right)=\left(\iota \left(pos\left(x\right)\right),\iota \left(len\left(x\right)\right)\right)$$

(16)

for positive-depth x. Identify two typed elements when finitely many refinements give equal elements at a common depth. The resulting set of classes is denoted $\mathcal{I}$, and $\left[x\right]$ is the class represented by x. The minimal depth of a class is the smallest depth of one of its representatives.

Proposition 1 

(Refinement and one-sided compatibility). Refinement is injective, preserves the scalar path coefficients, and introduces an unused last position. It preserves addition and multiplication, zero and unit. For $x\in {\mathcal{I}}_m$, $y\in {\mathcal{I}}_n$, and $m<n$, one has in ${\mathcal{I}}_n$

$$\iota \left(x\right)+y=x+y,\iota \left(x\right)y=xy.$$

(17)

Hence the recursively defined operations induce a commutative unital R-algebra on $\mathcal{I}$, with

$$\mathcal{I}\cong R[t_1,t_2,\dots ]/(t_1^2,t_2^2,\dots ),$$

(18)

whose elements involve finitely many variables.

Proof. 

Path coefficients and injectivity. At depth zero, $\iota \left(c\right)=(c,0)$ has coefficients c at the empty path and zero at the path $\left\{1\right\}$. For the induction step, take $n\ge 1$ and $x\in {\mathcal{I}}_n$. Both $pos\left(x\right)$ and $len\left(x\right)$ belong to ${\mathcal{I}}_{n-1}$. For $A\subseteq \left[n\right]$, the recursive definition and the path convention give

$$\begin{array}{cc}\hfill {\left(\iota \left(x\right)\right)}_{1+A}& ={\left(\iota \left(pos\left(x\right)\right)\right)}_A,\hfill \\ \hfill {\left(\iota \left(x\right)\right)}_{\left\{1\right\}\cup (1+A)}& ={\left(\iota \left(len\left(x\right)\right)\right)}_A.\hfill \end{array}$$

Here $1+A=\{1+j:j\in A\}$, as in the preceding proof. If $A\subseteq [n-1]$, the lower-depth induction hypothesis identifies these two coefficients with

$${\left(pos\left(x\right)\right)}_A=x_{1+A},{\left(len\left(x\right)\right)}_A=x_{\left\{1\right\}\cup (1+A)}.$$

If $n\in A$, that hypothesis makes both coefficients zero. These two cases exhaust the paths at depth $n+1$, and prove

$${\left(\iota \left(x\right)\right)}_S=\left\{\begin{array}{cc}{x}_S,\hfill & S\subseteq \left[n\right],\hfill \\ 0,\hfill & n+1\in S.\hfill \end{array}\right.$$

Thus the old positions retain their numbers, while a len choice at the new last position always has coefficient zero. In particular, if $x,y\in {\mathcal{I}}_n$ and $\iota \left(x\right)=\iota \left(y\right)$, then

$$x_S={\left(\iota \left(x\right)\right)}_S={\left(\iota \left(y\right)\right)}_S=y_S(S\subseteq \left[n\right]).$$

Theorem 2 now gives $x=y$.

Arithmetic at equal depths. Let $x,y\in {\mathcal{I}}_n$. For a path $S\subseteq \left[n\right]$, the coefficient formulas give

$$\begin{array}{cc}\hfill {\left(\iota \left(x\right)+\iota \left(y\right)\right)}_S& =x_S+y_S={(x+y)}_S={\left(\iota (x+y)\right)}_S,\hfill \\ \hfill {\left(\iota \left(x\right)\iota \left(y\right)\right)}_S& =\sum _{A\bigsqcup B=S}{\left(\iota \left(x\right)\right)}_A{\left(\iota \left(y\right)\right)}_B\hfill \\ \hfill & =\sum _{A\bigsqcup B=S}{x}_A{y}_B={\left(xy\right)}_S={\left(\iota \left(xy\right)\right)}_S.\hfill \end{array}$$

If $n+1\in S$, both refined inputs have zero coefficient at S. In each ordered partition $A\bigsqcup B=S$, exactly one of $A,B$ contains $n+1$, so every summand in the product formula is zero as well. The refined sum and product also have zero coefficients on these paths. Equality at every path therefore proves

$$\iota (x+y)=\iota \left(x\right)+\iota \left(y\right),\iota \left(xy\right)=\iota \left(x\right)\iota \left(y\right)$$

at equal input depths. Scalar multiplication is coefficientwise, so $\iota \left(cx\right)=c\iota \left(x\right)$ for $c\in R$. The zero has all coefficients zero; the unit has coefficient one only at the empty path. Their refinements have exactly the same respective descriptions at the next depth. Hence

$$\iota \left(0_n\right)=0_{n+1},\iota \left(1_n\right)=1_{n+1},\iota \left(c{1}_n\right)=c{1}_{n+1}.$$

One-sided refinement and common depth. We prove (17) by induction on the smaller depth m, allowing every $n>m$. At $m=0$, write $x=c\in R$. Since y has positive depth, the scalar boundary and Theorem 1 give

$$\begin{array}{cc}\hfill \iota \left(c\right)+y& =\left(c+pos\left(y\right),0_0+len\left(y\right)\right)\hfill \\ \hfill & =\left(c+pos\left(y\right),len\left(y\right)\right)=c+y,\hfill \\ \hfill \iota \left(c\right)y& =\left(cpos\left(y\right),clen\left(y\right)+0_{0}pos\left(y\right)\right)\hfill \\ \hfill & =\left(cpos\left(y\right),clen\left(y\right)\right)=cy.\hfill \end{array}$$

These equalities hold in ${\mathcal{I}}_n$, even when $n>1$; the operations in each component are themselves evaluated by scalar descent.

For $m\ge 1$, the two components of x belong to ${\mathcal{I}}_{m-1}$, and those of y belong to ${\mathcal{I}}_{n-1}$. Since $m-1<n-1$, the induction hypothesis applies to each of the following component operations:

$$\begin{array}{cc}\hfill pos\left(\iota \left(x\right)+y\right)& =\iota \left(pos\left(x\right)\right)+pos\left(y\right)\hfill \\ \hfill & =pos\left(x\right)+pos\left(y\right)=pos(x+y),\hfill \\ \hfill len\left(\iota \left(x\right)+y\right)& =\iota \left(len\left(x\right)\right)+len\left(y\right)\hfill \\ \hfill & =len\left(x\right)+len\left(y\right)=len(x+y),\hfill \\ \hfill pos\left(\iota \left(x\right)y\right)& =\iota \left(pos\left(x\right)\right)pos\left(y\right)\hfill \\ \hfill & =pos\left(x\right)pos\left(y\right)=pos\left(xy\right),\hfill \\ \hfill len\left(\iota \left(x\right)y\right)& =\iota \left(pos\left(x\right)\right)len\left(y\right)+\iota \left(len\left(x\right)\right)pos\left(y\right)\hfill \\ \hfill & =pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)=len\left(xy\right).\hfill \end{array}$$

Both sides have depth n. Equality of their two components completes the induction.

Write ${\iota }^{\left(r\right)}$ for r successive refinements, with ${\iota }^{\left(0\right)}\left(x\right)=x$. Repeating the one-step result gives, for $0\le r\le n-m$,

$${\iota }^{\left(r\right)}\left(x\right)+y=x+y,{\iota }^{\left(r\right)}\left(x\right)y=xy.$$

At each step before the last, the refined first input still has depth strictly less than n. More generally, take arbitrary input depths $m,n$ and put $d=\max (m,n)$. At depth d, repeated one-sided refinement aligns the shallower input; commutativity allows either input to be the shallower one. Applying equal-depth preservation a further $N-d$ times then gives, for every $N\ge d$,

$$\begin{array}{cc}\hfill {\iota }^{(N-d)}(x+y)& ={\iota }^{(N-m)}\left(x\right)+{\iota }^{(N-n)}\left(y\right),\hfill \\ \hfill {\iota }^{(N-d)}\left(xy\right)& ={\iota }^{(N-m)}\left(x\right){\iota }^{(N-n)}\left(y\right).\hfill \end{array}$$

In particular, aligning $\iota \left(x\right),\iota \left(y\right)$ at depth $d+1$ yields

$$\begin{array}{cc}\hfill \iota \left(x\right)+\iota \left(y\right)& ={\iota }^{(d+1-m)}\left(x\right)+{\iota }^{(d+1-n)}\left(y\right)=\iota (x+y),\hfill \\ \hfill \iota \left(x\right)\iota \left(y\right)& ={\iota }^{(d+1-m)}\left(x\right){\iota }^{(d+1-n)}\left(y\right)=\iota \left(xy\right).\hfill \end{array}$$

Thus preservation of arithmetic also holds for mixed input depths.

Operations on classes and the polynomial model. Equality after refinement is reflexive and symmetric. It is transitive because two equalities witnessed at different depths can both be refined to the larger witnessing depth. Thus Definition 4 does give an equivalence relation.

Suppose $\left[x\right]=\left[x^{\prime }\right]$ and $\left[y\right]=\left[y^{\prime }\right]$, with their typed depths denoted by $m,m^{\prime },n,n^{\prime }$, respectively. Choose N large enough that both comparisons hold at depth N:

$$\begin{array}{cc}\hfill {\iota }^{(N-m)}\left(x\right)& ={\iota }^{(N-m^{\prime })}\left(x^{\prime }\right),\hfill \\ \hfill {\iota }^{(N-n)}\left(y\right)& ={\iota }^{(N-n^{\prime })}\left(y^{\prime }\right).\hfill \end{array}$$

With $d=\max (m,n)$ and $d^{\prime }=\max (m^{\prime },n^{\prime })$, substituting these equalities into the common-depth formulas proves

$$\begin{array}{cc}\hfill {\iota }^{(N-d)}(x+y)& ={\iota }^{(N-d^{\prime })}(x^{\prime }+y^{\prime }),\hfill \\ \hfill {\iota }^{(N-d)}\left(xy\right)& ={\iota }^{(N-d^{\prime })}\left(x^{\prime }{y}^{\prime }\right).\hfill \end{array}$$

Scalar multiplication commutes with repeated refinement as well. Consequently

$$\left[x\right]+\left[y\right]=[x+y],\left[x\right]\left[y\right]=\left[xy\right],c\left[x\right]=\left[cx\right]$$

are independent of the chosen representatives. All ring and module laws follow by placing the finitely many inputs in a common level and applying Theorem 1 there. The zero and unit are $\left[0_0\right]$ and $\left[1_0\right]$, and the additive inverse of $\left[x\right]$ is $[-x]$. This proves the commutative unital R-algebra assertion, with the same arithmetic as direct scalar descent.

For $x\in {\mathcal{I}}_n$, form its square-free polynomial

$$p_x(t_1,\dots ,t_n)=\sum _{S\subseteq \left[n\right]}{x}_S\prod _{j\in S}{t}_j.$$

The coefficient formula for refinement gives the exact polynomial identity

$$p_{\iota \left(x\right)}(t_1,\dots ,t_{n+1})=p_x(t_1,\dots ,t_n).$$

Hence the class of $p_x$ in the quotient in (18) depends only on $\left[x\right]$. At a common finite depth, Theorem 2 shows that this correspondence preserves addition, multiplication, scalars, and the unit.

To check injectivity, represent two objects at a common depth N. The difference of their square-free polynomials is square-free. Every polynomial in $(t_1^2,t_2^2,\dots )$ is a finite sum of multiples of generator squares, so all its square-free coefficients are zero. If the two quotient classes agree, their path coefficients therefore agree, and their depth-N representatives are equal. Their global classes are consequently equal as well. Finally, every ordinary polynomial uses only finitely many variables, say $t_1,\dots ,t_N$. The finite-level isomorphism realizes its quotient class by an element of ${\mathcal{I}}_N$. This proves surjectivity and completes (18). □

Lemma 1 

(Zeros, units, and prescribed depths). The recursive zeros and units satisfy

$$\begin{array}{cccc}\hfill \iota \left(0_n\right)& =0_{n+1},\hfill & \hfill \iota \left(1_n\right)& =1_{n+1},\hfill \\ \hfill [0_n]& =\left[0_0\right],\hfill & \hfill \left[1_n\right]& =\left[1_0\right].\hfill \end{array}$$

Their common classes are the zero 0 and the unit 1 of $\mathcal{I}$. Each class has exactly one representative at each prescribed depth, namely $0_n$ and $1_n$, respectively. For $x\in {\mathcal{I}}_m$ and $n\ge 0$,

$$\begin{array}{cccc}\hfill 0_m+0_n& =0_{\max (m,n)},\hfill & \hfill x{0}_n& =0_{\max (m,n)},\hfill \\ \hfill 1_m{1}_n& =1_{\max (m,n)},\hfill & \hfill x+0_n=x{1}_n& =\left\{\begin{array}{cc}x,\hfill & n\le m,\hfill \\ {\iota }^{(n-m)}\left(x\right),\hfill & n>m.\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

At depth zero the refinement identities are $\iota \left(0\right)=(0,0)$ and $\iota \left(1\right)=(1,0)$. At the next depth,

$$\begin{array}{cc}\hfill \iota \left(0_{n+1}\right)& =\left(\iota \left(0_n\right),\iota \left(0_n\right)\right)=0_{n+2},\hfill \\ \hfill \iota \left(1_{n+1}\right)& =\left(\iota \left(1_n\right),\iota \left(0_n\right)\right)=1_{n+2}.\hfill \end{array}$$

This proves the refinement identities by induction, and iteration gives $\left[0_n\right]=\left[0_0\right]$ and $\left[1_n\right]=\left[1_0\right]$. If $x\in {\mathcal{I}}_n$ represents the zero class, choose $N\ge n$ at which its refinement equals $0_N$. Then

$${\iota }^{(N-n)}\left(x\right)=0_N={\iota }^{(N-n)}\left(0_n\right).$$

Injectivity of repeated refinement gives $x=0_n$. Replacing $0_N,0_n$ by $1_N,1_n$ proves uniqueness for the unit class.

For the mixed-depth identities, put $d=\max (m,n)$. Proposition 1 and the just-proved refinement laws give

$$\begin{array}{cc}\hfill 0_m+0_n& =0_d+0_d=0_d,\hfill \\ \hfill x{0}_n& ={\iota }^{(d-m)}\left(x\right)0_d=0_d,\hfill \\ \hfill 1_m{1}_n& =1_d{1}_d=1_d,\hfill \\ \hfill x+0_n& ={\iota }^{(d-m)}\left(x\right)+0_d={\iota }^{(d-m)}\left(x\right),\hfill \\ \hfill x{1}_n& ={\iota }^{(d-m)}\left(x\right)1_d={\iota }^{(d-m)}\left(x\right).\hfill \end{array}$$

When $n\le m$, the exponent $d-m$ is zero and the last two outputs are x. When $n>m$, that exponent is $n-m$. These are precisely the two cases in the statement. □

Thus adding a deeper zero, or multiplying by a deeper ordinary unit, changes only the prescribed representation. The recursive operations compute these values directly. A scalar c used as a constant in ${\mathcal{I}}_n$ denotes $c{1}_n$; globally its class is simply c.

Theorem 3 

(Global components and reconstruction). The expressions $[pos(x\left)\right]$ and $[len(x\left)\right]$ depend only on $\left[x\right]$. They define pos and len on $\mathcal{I}$. Conversely, two global objects $a,b$ determine an ordered pair by taking representatives at a common depth. These constructions are inverse, so

$$x=\left(pos\left(x\right),len\left(x\right)\right)(x\in \mathcal{I}).$$

(19)

The corresponding algebra structure on $\mathcal{I}\times \mathcal{I}$ is $(a,b)(c,d)=(ac,ad+bc)$.

Proof. 

For positive depth, the two components of a refined object are refinements of its components:

$$pos\left(\iota \left(x\right)\right)=\iota \left(pos\left(x\right)\right),len\left(\iota \left(x\right)\right)=\iota \left(len\left(x\right)\right).$$

For a scalar c, the corresponding class identities are $[pos((c,0)\left)\right]=\left[c\right]$ and $[len((c,0)\left)\right]=\left[0\right]$. If two representatives agree after refinement, applying either component to that equality and using these identities makes their component classes equal. This proves independence of representatives for both operations, including a scalar representative.

For reconstruction, choose representatives of $a,b$ in one depth n. For every $r\ge 0$, the pair recursion gives

$${\iota }^{\left(r\right)}\left((a,b)\right)=\left({\iota }^{\left(r\right)}\left(a\right),{\iota }^{\left(r\right)}\left(b\right)\right).$$

Indeed, the case $r=0$ is the identity and one more refinement applies $\iota$ separately to the two displayed entries. Two different choices of representatives can be compared at a still larger common depth; the displayed equality then makes their reconstructed pairs equivalent. At a common depth,

$$pos\left((a,b)\right)=a,len\left((a,b)\right)=b,\left(pos\left(x\right),len\left(x\right)\right)=x.$$

For a scalar the last equality holds on classes since $\left[\right(c,0\left)\right]=\left[c\right]$. Thus the constructions are inverse. For two pairs at a common level, the arithmetic formulas are explicitly

$$(a,b)+(c,d)=(a+c,b+d),(a,b)(c,d)=(ac,ad+bc).$$

Independence of representatives carries both identities to global pairs. □

In particular the global zero and unit have the component values

$$pos\left(0\right)=len\left(0\right)=0,pos\left(1\right)=1,len\left(1\right)=0.$$

2.2. Nilpotents, Units, and Logarithms

In path coordinates the components separate the first position and then renumber the remaining positions. The coefficient of the empty path is

$$x_{⌀}={pos}^{\left(n\right)}\left(x\right)(x\in {\mathcal{I}}_n).$$

(20)

It is independent of the representative and is additive and multiplicative. Let ${\mathcal{N}}_n$ consist of the elements of ${\mathcal{I}}_n$ with this coefficient zero, and let $\mathcal{N}$ be the corresponding ideal of $\mathcal{I}$.

Theorem 4 

(Classification by scalar descent). An element with scalar coefficient c is nilpotent exactly when $c=0$, is a non-zero-divisor exactly when $c\ne 0$, and is a unit exactly when $c\in R^{\times }$. The only idempotents are 0 and 1. At finite depth,

$${\mathcal{N}}_n^{n+1}=0,{\mathcal{N}}_n^n=R{\mathbf{b}}_{\left[n\right]}(n\ge 1).$$

(21)

Globally, $\mathcal{N}$ is locally nilpotent but is not nilpotent. It is both the nilradical and the Jacobson radical. Every prime ideal is the inverse image of a prime ideal of R under scalar-coefficient evaluation. Thus the global algebra has Krull dimension one and is not Noetherian.

Proof. 

Finite nilpotence and scalar classification. A product of r nonempty paths can survive only when their supports are pairwise disjoint, and its resulting support then has size at least r. Conversely, every support of size at least r can be partitioned into r nonempty parts. Consequently, for $1\le r\le n$,

$${\mathcal{N}}_n^r=\sum _{\begin{array}{c}S\subseteq \left[n\right]\\ \left|S\right|\ge r\end{array}}R{\mathbf{b}}_S.$$

There are no supports of size $n+1$, whereas at size n there is exactly $\left[n\right]$. This proves (21). Write $x=c+u$ with $u\in {\mathcal{N}}_n$. If a power of x vanishes, its scalar coefficient c vanishes because R is a domain. If $c\ne 0$ and $xy=0$, place both inputs in a common finite level and choose the lowest nonzero homogeneous part $y_d$ of a nonzero y. The degree-d part of $xy$ is $c{y}_d$: multiplying the positive-degree part of x by any nonzero part of y raises the degree above d. Thus

$${\left(xy\right)}_S=c{y}_S\left(\right|S|=d),$$

and at least one of these coefficients is nonzero in the domain R. This contradicts $xy=0$. If $c=0$ and $x\ne 0$, let r be its least positive vanishing power. Then $r\ge 2$, $x^{r-1}\ne 0$, and $x{x}^{r-1}=0$. The zero element itself annihilates 1. A unit must have unit scalar coefficient, since ${\left(xy\right)}_{⌀}=1$ gives $x_{⌀}{y}_{⌀}=1$ in R. Conversely, when c is a unit, put

$$y=c^{-1}{\displaystyle \sum _{j=0}^n}{(-c^{-1}u)}^j.$$

Then cancellation of consecutive powers gives

$$xy=(1+c^{-1}u){\displaystyle \sum _{j=0}^n}{(-c^{-1}u)}^j=1-{(-c^{-1}u)}^{n+1}=1.$$

An idempotent satisfies $c^2=c$, so $c=0$ or $c=1$. In the first case $x^2=x$ gives $x=x^{n+1}=0$. In the second, ${(1-x)}^2=1-x$ and $1-x$ has scalar coefficient zero, so $1-x=0$.

Global ideals. A finite family in $\mathcal{N}$ uses only finitely many positions, say $\left[n\right]$. In any product of $n+1$ members of the ideal it generates, expansion contains $n+1$ nonempty supports drawn from $\left[n\right]$. At least two overlap, even if their other factors use later positions. Thus that generated ideal is nilpotent. On the other hand,

$${\mathbf{b}}_1\dots {\mathbf{b}}_n={\mathbf{b}}_{\left[n\right]}\ne 0$$

for every n, so $\mathcal{N}$ is not nilpotent.

Scalar-coefficient evaluation is surjective onto R and has kernel $\mathcal{N}$, giving $\mathcal{I}/\mathcal{N}\cong R$. If a prime ideal contains a power of an element, primality applied repeatedly makes it contain the element. Every prime therefore contains the nilradical $\mathcal{N}$. Ideals containing $\mathcal{N}$ correspond, by taking scalar coefficients and inverse images, to ideals of R; this correspondence preserves primality and maximality. The nonzero prime ideals of $R=\mathbb{Z}\left[{\Omega }^{-1}\right]$ are exactly $pR$ for primes p not dividing $\Omega$, and each is maximal. Their intersection is zero: for $c=a/{\Omega }^q\ne 0$, choose a prime p dividing neither a nor $\Omega$, so $c\notin pR$. Hence

$$\bigcap _{\mathfrak{m}\mathrm{maximal}\mathrm{in}\mathcal{I}}\mathfrak{m}=\mathcal{N}.$$

The prime chains correspond to the chains $0⊊pR$ in R, so the Krull dimension is one. Finally, every monomial in $({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)$ meets $\left[n\right]$, whereas ${\mathbf{b}}_{n+1}$ does not. Therefore

$$({\mathbf{b}}_1,\dots ,{\mathbf{b}}_n)⊊({\mathbf{b}}_1,\dots ,{\mathbf{b}}_{n+1})$$

is a strictly increasing chain, proving failure of the Noetherian property. □

The units of the coefficient ring are explicitly

$$R^{\times }=\left\{\pm \prod _{p\mid \Omega }{p}^{k_p}:k_p\in \mathbb{Z}\right\},$$

(22)

where p ranges over the prime divisors of $\Omega$. In particular, 2 is a unit exactly when $\Omega$ is even.

Proposition 2 

(Inverse, quotient, and recursive unit coordinates). A positive-depth x is a unit exactly when $pos\left(x\right)$ is a unit, and then

$$x^{-1}=\left(pos{\left(x\right)}^{-1},-pos{\left(x\right)}^{-2}len\left(x\right)\right).$$

(23)

For a unit y, division at non-scalar depth is

$${\displaystyle \frac{x}{y}}=\left({\displaystyle \frac{pos\left(x\right)}{pos\left(y\right)}},{\displaystyle \frac{len\left(x\right)pos\left(y\right)-pos\left(x\right)len\left(y\right)}{pos{\left(y\right)}^2}}\right).$$

(24)

For units $x,y$,

$$pos{\left(xy\right)}^{-1}len\left(xy\right)=pos{\left(x\right)}^{-1}len\left(x\right)+pos{\left(y\right)}^{-1}len\left(y\right).$$

(25)

The coordinates $(pos\left(x\right),pos{\left(x\right)}^{-1}len\left(x\right))$ give group isomorphisms

$${\mathcal{I}}_{n+1}^{\times }\cong {\mathcal{I}}_n^{\times }\times ({\mathcal{I}}_n,+),{\mathcal{I}}^{\times }\cong {\mathcal{I}}^{\times }\times (\mathcal{I},+).$$

(26)

In particular ${\mathcal{I}}_n^{\times }\cong R^{\times }\times {(R,+)}^{2^n-1}$.

Proof. 

If $xy=1$, the first component gives $pos\left(x\right)pos\left(y\right)=1$. Conversely, if $pos\left(x\right)$ is a unit, take y to be the pair in (23). Then

$$\begin{array}{cc}\hfill pos\left(xy\right)& =pos\left(x\right)pos{\left(x\right)}^{-1}=1,\hfill \\ \hfill len\left(xy\right)& =-pos\left(x\right)pos{\left(x\right)}^{-2}len\left(x\right)+len\left(x\right)pos{\left(x\right)}^{-1}=0.\hfill \end{array}$$

Hence $xy=1$. Substituting that inverse for a unit denominator gives

$$\begin{array}{cc}\hfill pos\left(x{y}^{-1}\right)& =pos\left(x\right)pos{\left(y\right)}^{-1},\hfill \\ \hfill len\left(x{y}^{-1}\right)& =-pos\left(x\right)pos{\left(y\right)}^{-2}len\left(y\right)+len\left(x\right)pos{\left(y\right)}^{-1}\hfill \\ \hfill & ={\displaystyle \frac{len\left(x\right)pos\left(y\right)-pos\left(x\right)len\left(y\right)}{pos{\left(y\right)}^2}}.\hfill \end{array}$$

These are (24). For units $x,y$,

$$\begin{array}{cc}\hfill pos{\left(xy\right)}^{-1}len\left(xy\right)& ={\displaystyle \frac{pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)}{pos\left(x\right)pos\left(y\right)}}\hfill \\ \hfill & =pos{\left(y\right)}^{-1}len\left(y\right)+pos{\left(x\right)}^{-1}len\left(x\right).\hfill \end{array}$$

Thus the first coordinate multiplies and the second adds. The inverse coordinate construction is $(a,v)$ reconstructed as $(a,av)$, since $pos\left(\right(a,av\left)\right)=a$ and $a^{-1}len\left((a,av)\right)=v$. Thus the coordinate construction and its inverse are both defined over R, and the first-coordinate product and second-coordinate sum make them mutually inverse group homomorphisms. The same formulas hold globally by Theorem 3. Since the additive group of ${\mathcal{I}}_j$ is ${(R,+)}^{2^j}$, iteration gives

$${\mathcal{I}}_n^{\times }\cong R^{\times }\times {\displaystyle \prod _{j=0}^{n-1}}({\mathcal{I}}_j,+)\cong R^{\times }\times {(R,+)}^{{\sum }_{j=0}^{n-1}{2}^j}=R^{\times }\times {(R,+)}^{2^n-1}.$$

At $n=0$ the product is empty and only $R^{\times }$ remains. □

Example 2 

(Iteration and powers). Take $x=\left(\right(1,b),(c,d\left)\right)$. Then

$${len}^{\left(2\right)}\left(x\right)=d,len{\left(x\right)}^2=(c^2,2cd),$$

while $pos{\left(x\right)}^{-1}=(1,-b)$. Consequently

$${len}^{\left(2\right)}\left(x\right)pos{\left(x\right)}^{-2}=d{(1,-b)}^2=(d,-2db).$$

An iteration extracts successive components; an ordinary power uses the recursive multiplication.

Exponential and logarithmic functions on finite zeon algebras, together with factorizations of invertible elements, are studied in [12]. The component recursions below express these operations over R; the path expansion proves that their coefficients belong to R even when the displayed factorial denominators are not units.

Theorem 5 

(Integral recursive exponential and logarithm). On $\mathcal{N}$ and $1+\mathcal{N}$, respectively, define the scalar values $\exp \left(0\right)=1$, $\log \left(1\right)=0$, and the positive-depth recursions

$$\begin{array}{cc}\hfill \exp \left(u\right)& =\left(\exp \left(pos\left(u\right)\right),\exp \left(pos\left(u\right)\right)len\left(u\right)\right),\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill \log \left(v\right)& =\left(\log \left(pos\left(v\right)\right),pos{\left(v\right)}^{-1}len\left(v\right)\right).\hfill \end{array}$$

(28)

These operations are well defined over R, are inverse, and satisfy

$$\exp (u+w)=\exp \left(u\right)\exp \left(w\right),\log \left(vy\right)=\log \left(v\right)+\log \left(y\right).$$

(29)

For $u\in {\mathcal{N}}_n$, they have the finite expansions

$$\exp \left(u\right)={\displaystyle \sum _{r=0}^n}{\displaystyle \frac{u^r}{r!}},\log (1+u)={\displaystyle \sum _{r=1}^n}{\displaystyle \frac{{(-1)}^{r-1}{u}^r}{r}}.$$

(30)

Every indicated division in these finite expressions has a unique result in ${\mathcal{I}}_n$. Thus

$$(\mathcal{N},+)\cong (1+\mathcal{N},\times ),{\mathcal{I}}^{\times }\cong R^{\times }\times (\mathcal{N},+).$$

(31)

Proof. 

The scalar coefficient of $pos\left(u\right)$ is zero, and that of $pos\left(v\right)$ is one. The recursions therefore continue within their domains and end at $\exp \left(0\right)=1$ and $\log \left(1\right)=0$. Every first component inverted in the logarithm has scalar coefficient one and is a unit by Theorem 4. Thus both recursions are defined over R. For $u,w$ at a common positive depth, lower-depth induction gives

$$\begin{array}{cc}\hfill pos\left(\exp (u+w)\right)& =\exp \left(pos\left(u\right)+pos\left(w\right)\right)\hfill \\ \hfill & =\exp \left(pos\left(u\right)\right)\exp \left(pos\left(w\right)\right),\hfill \\ \hfill len\left(\exp (u+w)\right)& =\exp \left(pos\left(u\right)\right)\exp \left(pos\left(w\right)\right)\left(len\left(u\right)+len\left(w\right)\right)\hfill \\ \hfill & =\exp \left(pos\left(u\right)\right)\left(\exp \left(pos\left(w\right)\right)len\left(w\right)\right)\hfill \\ \hfill & +\left(\exp \left(pos\left(u\right)\right)len\left(u\right)\right)\exp \left(pos\left(w\right)\right).\hfill \end{array}$$

These are exactly the two components of $\exp \left(u\right)\exp \left(w\right)$. For the two compositions, induction and the component rules give

$$\begin{array}{cc}\hfill pos\left(\log \left(\exp \left(u\right)\right)\right)& =\log \left(\exp \left(pos\left(u\right)\right)\right)=pos\left(u\right),\hfill \\ \hfill len\left(\log \left(\exp \left(u\right)\right)\right)& =\exp {\left(pos\left(u\right)\right)}^{-1}\exp \left(pos\left(u\right)\right)len\left(u\right)=len\left(u\right),\hfill \\ \hfill pos\left(\exp \left(\log \left(v\right)\right)\right)& =\exp \left(\log \left(pos\left(v\right)\right)\right)=pos\left(v\right),\hfill \\ \hfill len\left(\exp \left(\log \left(v\right)\right)\right)& =pos\left(v\right)pos{\left(v\right)}^{-1}len\left(v\right)=len\left(v\right).\hfill \end{array}$$

This proves mutual inversion. Taking $u=\log \left(v\right)$ and $w=\log \left(y\right)$ in the exponential product law gives

$$\exp \left(\log \left(v\right)+\log \left(y\right)\right)=vy,\log \left(vy\right)=\log \left(v\right)+\log \left(y\right).$$

At the scalar boundary, $\exp \left(\right(0,0\left)\right)=(1,0)=\iota \left(1\right)$ and $\log \left(\right(1,0\left)\right)=(0,0)=\iota \left(0\right)$. At positive depth, preservation of products and inverses by refinement, followed by the lower-depth induction hypothesis, gives

$$\begin{array}{cc}\hfill \exp \left(\iota \left(u\right)\right)& =\left(\iota \left(\exp \left(pos\left(u\right)\right)\right),\iota \left(\exp \left(pos\left(u\right)\right)len\left(u\right)\right)\right)=\iota \left(\exp \left(u\right)\right),\hfill \\ \hfill \log \left(\iota \left(v\right)\right)& =\left(\iota \left(\log \left(pos\left(v\right)\right)\right),\iota \left(pos{\left(v\right)}^{-1}len\left(v\right)\right)\right)=\iota \left(\log \left(v\right)\right).\hfill \end{array}$$

Iteration proves representative independence. In particular the equal-depth product laws already proved hold for arbitrary global inputs.

In a nonzero coefficient of $u^r$, the contributing ordered supports are r pairwise disjoint nonempty sets. Permuting them is a free action with $r!$ elements in each orbit, all with the same coefficient product. Every coefficient is therefore divisible by $r!$ in R. More explicitly, for a nonempty path S,

$${\left(u^r\right)}_S=r!\sum _{\begin{array}{c}\pi \mathrm{a}\mathrm{partition}\mathrm{of}S\\ \left|\pi \right|=r\end{array}}\prod _{B\in \pi }{u}_B.$$

The quotient is unique because R has no integer torsion. Division by r is consequently also valid. To identify the finite expressions with the recursions, first work over $\mathbb{Q}$. The recursive multiplication gives, for $r\ge 1$,

$$u^r=\left(pos{\left(u\right)}^r,rpos{\left(u\right)}^{r-1}len\left(u\right)\right).$$

Since $pos{\left(u\right)}^n=0$ at depth $n-1$, summing the two components gives

$$\begin{array}{cc}\hfill pos\left({\displaystyle \sum _{r=0}^n}{\displaystyle \frac{u^r}{r!}}\right)& ={\displaystyle \sum _{r=0}^{n-1}}{\displaystyle \frac{pos{\left(u\right)}^r}{r!}},\hfill \\ \hfill len\left({\displaystyle \sum _{r=0}^n}{\displaystyle \frac{u^r}{r!}}\right)& =\left({\displaystyle \sum _{r=0}^{n-1}}{\displaystyle \frac{pos{\left(u\right)}^r}{r!}}\right)len\left(u\right),\hfill \\ \hfill pos\left({\displaystyle \sum _{r=1}^n}{\displaystyle \frac{{(-1)}^{r-1}{u}^r}{r}}\right)& ={\displaystyle \sum _{r=1}^{n-1}}{\displaystyle \frac{{(-1)}^{r-1}pos{\left(u\right)}^r}{r}},\hfill \\ \hfill len\left({\displaystyle \sum _{r=1}^n}{\displaystyle \frac{{(-1)}^{r-1}{u}^r}{r}}\right)& =\left({\displaystyle \sum _{r=0}^{n-1}}{(-pos\left(u\right))}^r\right)len\left(u\right)\hfill \\ \hfill & ={(1+pos\left(u\right))}^{-1}len\left(u\right).\hfill \end{array}$$

Induction identifies these pairs with the recursive exponential and logarithm. All divided coefficients belong to R by the partition calculation, so the identities also hold over R. Finally a unit with scalar coefficient c has coordinates $(c,\log \left(c^{-1}x\right))$, with inverse $c\exp \left(u\right)$. □

For a nonempty path S, the explicit scalar formulas are

$$\begin{array}{cc}\hfill {\left(\exp u\right)}_S& =\sum _{\pi \mathrm{a}\mathrm{partition}\mathrm{of}S}\prod _{B\in \pi }{u}_B,\hfill \\ \hfill {\left(\log (1+u)\right)}_S& =\sum _{\pi \mathrm{a}\mathrm{partition}\mathrm{of}S}{(-1)}^{\left|\pi \right|-1}\left(\right|\pi |-1)!\prod _{B\in \pi }{u}_B.\hfill \end{array}$$

(32)

They express the finite calculations without denominators. The divisibility used here is also an instance of the usual divided-power identities [17].

Corollary 1 

(Unique factorization of normalized units). Every $v\in 1+{\mathcal{N}}_n$ has a unique expression

$$v=\prod _{⌀\ne S\subseteq \left[n\right]}(1+c_S{\mathbf{b}}_S).$$

(33)

The coefficients $c_S$ are precisely the path coefficients of $\log \left(v\right)$.

Proof. 

At depth zero the only normalized unit is 1 and the empty product is 1. At positive depth, $\log \left(v\right)$ has scalar coefficient zero, so its unique path expansion uses exactly the nonempty supports. For each such S, one has ${\mathbf{b}}_S^2=0$ and hence $\exp \left(c_S{\mathbf{b}}_S\right)=1+c_S{\mathbf{b}}_S$. Writing $\log \left(v\right)={\sum }_{S\ne ⌀}{c}_S{\mathbf{b}}_S$ gives

$$v=\exp \left(\sum _{S\ne ⌀}{c}_S{\mathbf{b}}_S\right)=\prod _{S\ne ⌀}\exp \left(c_S{\mathbf{b}}_S\right)=\prod _{S\ne ⌀}(1+c_S{\mathbf{b}}_S).$$

Conversely, logarithmic additivity gives

$$\log \left(\prod _{S\ne ⌀}(1+d_S{\mathbf{b}}_S)\right)=\sum _{S\ne ⌀}{d}_S{\mathbf{b}}_S.$$

Uniqueness of path coefficients forces $d_S=c_S$ for every S. □

2.3. Multiplication Duality and Recursive Rigidity

After scalar extension to a field, the finite square-free quotient is an Artinian complete intersection. Its pairing between complementary degrees is the corresponding Gorenstein duality [3,18]. Complete len descent gives this pairing directly and keeps its matrix invertible over the original ring R.

Theorem 6 

(Perfect multiplication pairing). At depth n, define

$${\beta }_n(x,y)={len}^{\left(n\right)}\left(xy\right).$$

(34)

This symmetric R-bilinear pairing is perfect: its matrix in the path basis is invertible over R. It satisfies

$$\begin{array}{cc}\hfill {\beta }_n(xy,z)& ={\beta }_n(x,yz),\hfill \\ \hfill {\beta }_{n+1}(x,y)& ={\beta }_n\left(pos\left(x\right),len\left(y\right)\right)+{\beta }_n\left(len\left(x\right),pos\left(y\right)\right),\hfill \\ \hfill {\beta }_n({\mathbf{b}}_S,{\mathbf{b}}_A)& =\left\{\begin{array}{cc}1,\hfill & A=\left[n\right]\setminus S,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

(35)

In particular $x_S={\beta }_n(x,{\mathbf{b}}_{\left[n\right]\setminus S})$.

Proof. 

Complete len descent reads the coefficient of ${\mathbf{b}}_{\left[n\right]}$. The path product gives

$${\beta }_n(x,y)={\left(xy\right)}_{\left[n\right]}=\sum _{S\subseteq \left[n\right]}{x}_S{y}_{\left[n\right]\setminus S}.$$

In particular, ${\beta }_n({\mathbf{b}}_S,{\mathbf{b}}_A)$ is one exactly when $A=\left[n\right]\setminus S$, and zero otherwise. The matrix is a permutation matrix with determinant 1 or $-1$, so it is invertible over R. Explicitly, for any R-linear functional ℓ on ${\mathcal{I}}_n$, the element

$$y=\sum _{S\subseteq \left[n\right]}\ell \left({\mathbf{b}}_S\right){\mathbf{b}}_{\left[n\right]\setminus S}$$

satisfies ${\beta }_n(x,y)={\sum }_S{x}_S\ell \left({\mathbf{b}}_S\right)=\ell \left(x\right)$. The complementary coefficient formula also makes this y unique. This proves perfectness without scalar extension. Associativity gives

$${\beta }_n(xy,z)={len}^{\left(n\right)}\left(\left(xy\right)z\right)={len}^{\left(n\right)}\left(x\left(yz\right)\right)={\beta }_n(x,yz).$$

At depth $n+1$, taking the outer second component gives

$$\begin{array}{cc}\hfill {\beta }_{n+1}(x,y)& ={len}^{\left(n\right)}\left(pos\left(x\right)len\left(y\right)+len\left(x\right)pos\left(y\right)\right)\hfill \\ \hfill & ={\beta }_n\left(pos\left(x\right),len\left(y\right)\right)+{\beta }_n\left(len\left(x\right),pos\left(y\right)\right).\hfill \end{array}$$

□

Theorem 7 

(Exact annihilators). For an ideal $J\subseteq {\mathcal{I}}_n$, let $J^{{\perp }_{{\beta }_n}}$ denote its orthogonal complement for ${\beta }_n$. Then

$$Ann\left(J\right)=J^{{\perp }_{{\beta }_n}},Ann\left({\mathcal{N}}_n^r\right)={\mathcal{N}}_n^{n+1-r}(1\le r\le n+1).$$

(36)

The double annihilator of J is its R-saturation:

$$Ann\left(Ann\left(J\right)\right)=\{x\in {\mathcal{I}}_n:ax\in J\mathrm{for}\mathrm{some}\mathrm{nonzero}a\in R\}.$$

(37)

Globally, for $k\ge 1$ put $J_k=\ker \left({pos}^{\left(k\right)}\right)$. Then

$$\begin{array}{cc}\hfill J_k& =({\mathbf{b}}_1,\dots ,{\mathbf{b}}_k)\mathcal{I},\hfill \\ \hfill {Ann}_{\mathcal{I}}\left(J_k^r\right)& =J_k^{k+1-r}(1\le r\le k+1),\hfill \\ \hfill \mathcal{N}& =\bigcup _{k\ge 1}{J}_k,{Ann}_{\mathcal{I}}\left({\mathcal{N}}^r\right)=0(r\ge 1).\hfill \end{array}$$

(38)

Proof. 

An annihilator is orthogonal by definition. If ${\beta }_n(x,J)=0$, then for $y\in J$ and any z one has

$${\beta }_n(xy,z)={\beta }_n(x,yz)=0,$$

because $yz\in J$. Taking $z={\mathbf{b}}_{\left[n\right]\setminus S}$ gives ${\left(xy\right)}_S=0$ for every path S, hence $xy=0$. The ideal ${\mathcal{N}}_n^r$ is spanned by paths of cardinality at least r. For a support A of such a path, orthogonality to ${\mathbf{b}}_A$ requires $x_{\left[n\right]\setminus A}=0$. These are exactly the coefficients with size at most $n-r$. The unrestricted coefficients have size at least $n-r+1$, so

$${\left({\mathcal{N}}_n^r\right)}^{{\perp }_{{\beta }_n}}=\sum _{\left|S\right|\ge n+1-r}R{\mathbf{b}}_S={\mathcal{N}}_n^{n+1-r}.$$

For $r=n+1$, both sides are ${\mathcal{I}}_n$, with ${\mathcal{N}}_n^0={\mathcal{I}}_n$.

Over $\mathbb{Q}$, put $J_{\mathbb{Q}}={span}_{\mathbb{Q}}J$. Finite-dimensional nondegenerate duality gives ${\left(J_{\mathbb{Q}}^{{\perp }_{{\beta }_n}}\right)}^{{\perp }_{{\beta }_n}}=J_{\mathbb{Q}}$. To justify the passage back to R, take a rational vector v orthogonal to $J_{\mathbb{Q}}$. A common nonzero integer denominator a gives $av\in {\mathcal{I}}_n$, and ${\beta }_n(av,y)=a{\beta }_n(v,y)=0$ for $y\in J$. Thus $av\in J^{{\perp }_{{\beta }_n}}$. The converse inclusion follows by bilinearity. Hence

$${span}_{\mathbb{Q}}\left(J^{{\perp }_{{\beta }_n}}\right)=J_{\mathbb{Q}}^{{\perp }_{{\beta }_n}}.$$

Applying the same observation to the second complement, and using rational nondegeneracy, gives

$$\begin{array}{cc}\hfill Ann\left(Ann\left(J\right)\right)& =J_{\mathbb{Q}}\cap {\mathcal{I}}_n\hfill \\ \hfill & =\left\{x\in {\mathcal{I}}_n:ax\in J\mathrm{for}\mathrm{some}a\in R\setminus \left\{0\right\}\right\}.\hfill \end{array}$$

Indeed, a finite expression $x={\sum }_j{q}_j{y}_j$ with $q_j\in \mathbb{Q}$ and $y_j\in J$ becomes $ax\in J$ after a common nonzero denominator is chosen; conversely $ax\in J$ gives $x=a^{-1}\left(ax\right)\in J_{\mathbb{Q}}$. This proves (37). For example, if p is prime and does not divide $\Omega$, then $J=p{\mathcal{I}}_n$ has zero annihilator and double annihilator ${\mathcal{I}}_n$.

Applying posk times discards precisely the paths meeting $\left[k\right]$, then shifts the remaining positions. Thus $J_k$ has the displayed generators. Regard the later positions as coefficients and apply the same complementary path calculation on $\left[k\right]$. It remains valid over this coefficient algebra: its pairing matrix is still a permutation matrix. In these first k positions, $J_k^r$ consists of paths using at least r positions, and its annihilator consists of paths using at least $k+1-r$ positions. This proves (38). Every finite family of nonempty supports lies in some $\left[k\right]$, giving $\mathcal{N}={\bigcup }_k{J}_k$. Finally, for nonzero x supported in $\left[n\right]$, choose a set B of r new positions. Then ${\mathbf{b}}_B\in {\mathcal{N}}^r$ and

$$x{\mathbf{b}}_B=\sum _{A\subseteq \left[n\right]}{x}_A{\mathbf{b}}_{A\cup B}\ne 0.$$

The unions $A\cup B$ are distinct, so no nonzero coefficient of x is cancelled. Hence no nonzero x annihilates ${\mathcal{N}}^r$. □

Theorem 8 

(Maximal nilpotence and multiplication Jordan type). For $u={\sum }_{⌀\ne S\subseteq \left[n\right]}{a}_S{\mathbf{b}}_S$,

$$u^n=n!\left({\displaystyle \prod _{j=1}^n}{a}_{\left\{j\right\}}\right){\mathbf{b}}_{\left[n\right]}.$$

(39)

Its nilpotence index is $n+1$ exactly when every first-homogeneous coefficient $a_{\left\{j\right\}}$ is nonzero. Under this condition, multiplication by u on ${\mathcal{I}}_n{\otimes }_R\mathbb{Q}$ has Jordan blocks of lengths

$$n-2d+1\mathrm{with}\mathrm{multiplicity}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d-1}}\right),0\le d\le \lfloor n/2\rfloor .$$

(40)

Here $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{-1}}\right)=0$. For every integer $r\ge 0$, its rth power has rank

$${\displaystyle \sum _{d=0}^{\lfloor n/2\rfloor }}\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d-1}}\right)\right)\max \{n-2d+1-r,0\}.$$

(41)

Proof. 

A nonzero contribution to $u^n$ requires n disjoint nonempty subsets of $\left[n\right]$. They must be its singletons. Their $n!$ orders give (39) and the nilpotence assertion.

Assign each nonempty support to its smallest position. This writes

$$u={\displaystyle \sum _{j=1}^n}{\mathbf{b}}_j{v}_j({\mathbf{b}}_{j+1},\dots ,{\mathbf{b}}_n),$$

where the scalar coefficient of $v_j$ is $a_{\left\{j\right\}}$. Over $\mathbb{Q}$ every $v_j$ is a unit. Define $\theta \left({\mathbf{b}}_j\right)={\mathbf{b}}_j{v}_j$. Its square is zero, so this prescription defines an algebra endomorphism. Starting at $j=n$ and then decreasing j, define

$$\eta \left({\mathbf{b}}_j\right)={\mathbf{b}}_j\eta {\left(v_j\right)}^{-1}.$$

Here $\eta \left(v_j\right)$ is already determined by the later positions and has nonzero scalar coefficient $a_{\left\{j\right\}}$. Its inverse exists. Induction gives

$$\begin{array}{cc}\hfill \eta \left(\theta \left({\mathbf{b}}_j\right)\right)& ={\mathbf{b}}_j\eta {\left(v_j\right)}^{-1}\eta \left(v_j\right)={\mathbf{b}}_j,\hfill \\ \hfill \theta \left(\eta \left({\mathbf{b}}_j\right)\right)& ={\mathbf{b}}_j{v}_j{\left(\theta \left(\eta \left(v_j\right)\right)\right)}^{-1}={\mathbf{b}}_j.\hfill \end{array}$$

Thus $\eta ={\theta }^{-1}$. With $s={\mathbf{b}}_1+\dots +{\mathbf{b}}_n$,

$$\theta \left(s\right)=u,M_u=\theta M_s{\theta }^{-1}.$$

If all $a_{\left\{j\right\}}$ are units in R, the same normalization is defined over R.

It remains to compute the type of multiplication by ${\mathbf{b}}_1+\dots +{\mathbf{b}}_n$. Give the path basis an auxiliary orthonormal real inner product and define

$$U{\mathbf{b}}_S=\sum _{j\notin S}{\mathbf{b}}_{S\cup \left\{j\right\}},D{\mathbf{b}}_S=\sum _{j\in S}{\mathbf{b}}_{S\setminus \left\{j\right\}}.$$

Then U is that multiplication operator and D is its adjoint. On the homogeneous space of cardinality d, cancellation of the terms that add and remove different positions can be seen on a single path:

$$\begin{array}{cc}\hfill DU{\mathbf{b}}_S& =(n-d){\mathbf{b}}_S+\sum _{\begin{array}{c}i\in S\\ j\notin S\end{array}}{\mathbf{b}}_{(S\setminus \{i\left\}\right)\cup \left\{j\right\}},\hfill \\ \hfill UD{\mathbf{b}}_S& =d{\mathbf{b}}_S+\sum _{\begin{array}{c}i\in S\\ j\notin S\end{array}}{\mathbf{b}}_{(S\setminus \{i\left\}\right)\cup \left\{j\right\}}.\hfill \end{array}$$

Subtraction and linearity give

$$(DU-UD)v=(n-2d)v.$$

For $0\le d\le \lfloor n/2\rfloor$ and a homogeneous v of order d with $Dv=0$, induction gives

$$D{U}^{r}v=r(n-2d-r+1)U^{r-1}v.$$

(42)

The induction step is explicit: $U^{r}v$ has order $d+r$, so

$$\begin{array}{cc}\hfill D{U}^{r+1}v& =UD{U}^{r}v+\left(n-2(d+r)\right)U^{r}v\hfill \\ \hfill & =\left(r(n-2d-r+1)+n-2d-2r\right)U^{r}v\hfill \\ \hfill & =(r+1)(n-2d-r)U^{r}v.\hfill \end{array}$$

Taking the auxiliary inner product with $U^{r-1}v$ gives

$$\begin{array}{cc}\hfill {\parallel U^{r}v\parallel }^2& =r(n-2d-r+1){\parallel U^{r-1}v\parallel }^2\hfill \\ \hfill & ={\displaystyle \frac{r!(n-2d)!}{(n-2d-r)!}}{\parallel v\parallel }^2(0\le r\le n-2d).\hfill \end{array}$$

These values are positive for $v\ne 0$. At $r=n-2d+1$ the same recurrence gives ${\parallel U^{n-2d+1}v\parallel }^2=0$. For a homogeneous w of cardinality $d-1$,

$${\parallel Uw\parallel }^2={\parallel Dw\parallel }^2+(n-2d+2){\parallel w\parallel }^2.$$

Thus U is injective from cardinality $d-1$ to d when $d\le n/2$. Its adjoint is surjective in the reverse direction, and the kernel of D at cardinality d has dimension $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d-1}}\right)$. Choose an orthogonal basis of each such kernel and generate its U-chains. For primitive vectors $v,w$ of orders $d<e$, two chain vectors can pair nontrivially only when $d+r=e+s$, hence $r>s$. Repeated use of (42) gives a scalar c such that

$$\langle U^{r}v,U^{s}w\rangle =\langle D^s{U}^{r}v,w\rangle =c\langle v,D^{r-s}w\rangle =0.$$

For primitive vectors of the same order, the corresponding expression is

$$\langle U^{r}v,U^{r}w\rangle ={\displaystyle \frac{r!(n-2d)!}{(n-2d-r)!}}\langle v,w\rangle .$$

Thus the chosen orthogonal kernel bases generate independent chains. The chain dimensions sum to $2^n$ by the binomial identity

$${\displaystyle \sum _{d=0}^{\lfloor n/2\rfloor }}(n-2d+1)\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d-1}}\right)\right)=2^n.$$

To verify the dimension count, put $h=\lfloor n/2\rfloor$ and telescope:

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{d=0}^h}(n-2d+1)\left(\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d-1}}\right)\right)\hfill \\ \hfill & =2{\displaystyle \sum _{d=0}^{h-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{d}}\right)+(n-2h+1)\left({\displaystyle \genfrac{}{}{0pt}{}{n}{h}}\right)\hfill \\ \hfill & =2^n.\hfill \end{array}$$

The last equality uses symmetry of the binomial coefficients, separately for even and odd n. Thus these independent chains fill the space. A chain of length l contributes $\max \{l-r,0\}$ to the rank of $U^r$: its first $l-r$ vectors have independent nonzero images and its last r vectors, or all its vectors if $r\ge l$, are killed. Adding over the chains proves (41) over the reals. A rational matrix has the same rank over $\mathbb{Q}$ and $\mathbb{R}$, since rank is the largest size of a nonzero minor. The ranks of all powers determine the nilpotent Jordan block sizes, so the same block list and rank formula hold over $\mathbb{Q}$. □

The square-free multiplication underlying this chain calculation is part of classical Lefschetz and Sperner theory [14,15]. Feinsilver develops the corresponding Boolean-lattice representation and its irreducible chains directly in zeon coordinates [5]. Its formulation through the Jordan type of a nilpotent multiplication operator is developed more generally in [16]. The explicit normalization above also shows that all higher path coefficients leave the type unchanged when the first-homogeneous coefficients remain nonzero.

Proposition 3 

(Rigidity of both component operations). A unital R-algebra endomorphism of $\mathcal{I}$ that commutes with both pos and len is the identity.

Proof. 

Let $\theta$ be the endomorphism. Since it is a unital R-algebra endomorphism, $\theta \left(c\right)=c$ for every scalar c. Suppose it fixes the classes represented at depths at most $n-1$. For $x\in {\mathcal{I}}_n$,

$$\begin{array}{cc}\hfill pos\left(\theta \left(\left[x\right]\right)\right)& =\theta \left(\left[pos\left(x\right)\right]\right)=\left[pos\left(x\right)\right],\hfill \\ \hfill len\left(\theta \left(\left[x\right]\right)\right)& =\theta \left(\left[len\left(x\right)\right]\right)=\left[len\left(x\right)\right].\hfill \end{array}$$

Both equalities use the induction hypothesis on the corresponding component. Global reconstruction gives

$$\theta \left(\left[x\right]\right)=\left(\left[pos\left(x\right)\right],\left[len\left(x\right)\right]\right)=\left[x\right].$$

Induction covers every finite representative. □

3. Weighted Forms and Sharp Inequalities

The component recursion also gives a notion of size: the pos and len branches contribute orthogonally, with their relative weight determined by $\Omega$. In path coordinates, this form connects the multiplication pairing of Section 2 with finite-level duality and a Hilbert completion. Ordinary multiplication is governed by the depths of the inputs and by the overlap of their nonzero paths. Tracking these features through the recursion yields sharp constants and equality conditions. The resulting estimates also distinguish the products that extend continuously to the completion from the general coefficient product, whose domain is only partial.

3.1. The Recursive Form and Finite-Level Duality

Throughout this section, scalar coefficients are viewed in $\mathbb{R}$ when inequalities are used. The real scalar extension of a finite level consists of the same recursive pairs with real leaves.

Definition 5 

(Weighted recursive form). For scalar inputs set ${〈c,d〉}_{\Omega }=cd$. If at least one input has positive depth, set

$${〈x,y〉}_{\Omega }={〈pos\left(x\right),pos\left(y\right)〉}_{\Omega }+{\Omega }^2{〈len\left(x\right),len\left(y\right)〉}_{\Omega }.$$

(43)

The scalar rules $pos\left(c\right)=c$ and $len\left(c\right)=0$ apply whenever one branch has already reached the scalar layer.

Theorem 9 

(Coordinates and independent refinement). The recursion terminates for inputs of arbitrary finite depths and defines a symmetric R-bilinear form. In a common level,

$${〈x,y〉}_{\Omega }=\sum _{S\subseteq \left[n\right]}{\Omega }^{2\left|S\right|}{x}_S{y}_S,{〈{\mathbf{b}}_S,{\mathbf{b}}_U〉}_{\Omega }=\left\{\begin{array}{cc}{\Omega }^{2\left|S\right|},\hfill & S=U,\hfill \\ 0,\hfill & S\ne U.\hfill \end{array}\right.$$

(44)

Either input may be independently replaced by any of its recursive refinements without changing the pairing. The resulting form on $\mathcal{I}$ is positive definite, and its real scalar extension is an inner product. In particular,

$${∥x∥}_{\Omega }=\sqrt{{〈x,x〉}_{\Omega }},d_{\Omega }(x,y)={∥x-y∥}_{\Omega }$$

(45)

define a norm and its metric.

Proof. 

At each non-scalar step the maximum depth decreases. Symmetry follows at the scalar boundary and is preserved by the recursion. For bilinearity, descend simultaneously in the inputs of ${〈x+y,z〉}_{\Omega }$, using the recursive addition rule and its scalar boundary. At a positive largest depth the induction hypothesis gives

$$\begin{array}{cc}\hfill {〈x+y,z〉}_{\Omega }& ={〈pos\left(x\right)+pos\left(y\right),pos\left(z\right)〉}_{\Omega }+{\Omega }^2{〈len\left(x\right)+len\left(y\right),len\left(z\right)〉}_{\Omega }\hfill \\ \hfill & ={〈pos\left(x\right),pos\left(z\right)〉}_{\Omega }+{〈pos\left(y\right),pos\left(z\right)〉}_{\Omega }\hfill \\ \hfill & +{\Omega }^2{〈len\left(x\right),len\left(z\right)〉}_{\Omega }+{\Omega }^2{〈len\left(y\right),len\left(z\right)〉}_{\Omega }\hfill \\ \hfill & ={〈x,z〉}_{\Omega }+{〈y,z〉}_{\Omega }.\hfill \end{array}$$

The same descent gives ${〈cx,y〉}_{\Omega }=c{〈x,y〉}_{\Omega }$ for $c\in R$; symmetry supplies linearity in the second input.

For the coordinate formula, the scalar case is $cd$. At depth $n+1$, the lower-depth formula on the two branches gives

$$\begin{array}{cc}\hfill {〈x,y〉}_{\Omega }& =\sum _{A\subseteq \left[n\right]}{\Omega }^{2\left|A\right|}{\left(pos\left(x\right)\right)}_A{\left(pos\left(y\right)\right)}_A\hfill \\ \hfill & +\sum _{A\subseteq \left[n\right]}{\Omega }^{2\left(\right|A|+1)}{\left(len\left(x\right)\right)}_A{\left(len\left(y\right)\right)}_A\hfill \\ \hfill & =\sum _{A\subseteq \left[n\right]}{\Omega }^{2\left|A\right|}{x}_{1+A}{y}_{1+A}\hfill \\ \hfill & +\sum _{A\subseteq \left[n\right]}{\Omega }^{2\left(\right|A|+1)}{x}_{\left\{1\right\}\cup (1+A)}{y}_{\left\{1\right\}\cup (1+A)}\hfill \\ \hfill & =\sum _{S\subseteq [n+1]}{\Omega }^{2\left|S\right|}{x}_S{y}_S.\hfill \end{array}$$

The two sums exhaust the supports not containing and containing the first position, respectively. Substitution of two path basis elements gives their stated diagonal pairing. For different depths, regard missing later positions as absent: a scalar branch contributes only through further pos choices, since every further len choice gives zero. Thus the same formula holds after placing the coefficient lists in any common finite level. By Proposition 1, refinement preserves every existing path coefficient and adds zero coefficients at paths involving the new last position. It follows from the coordinate formula that refining just one input leaves the pairing unchanged. The pairing is therefore independent of both representatives.

More explicitly, extend an input coefficient list by zero on every path using a position beyond its original depth. At any common depth N, the pairing is the same finite sum

$$\sum _{S\subseteq \left[N\right]}{\Omega }^{2\left|S\right|}{x}_S{y}_S.$$

Refining either input changes only the available zero entries in this sum, not any nonzero summand. This proves independent refinement invariance.

For equal inputs the formula is a sum of strictly positive weights times squares. It vanishes precisely when all coefficients vanish, that is, when the input is the zero of its prescribed level. The same argument applies to real coefficients. □

The zero and unit of any finite level therefore satisfy

$${〈x,0〉}_{\Omega }=0,{〈x,1〉}_{\Omega }=x_{⌀},{∥0∥}_{\Omega }=0,{∥1∥}_{\Omega }=1.$$

These values are unchanged by the choice of their representative depths.

The normalized path basis ${\Omega }^{-\left|S\right|}{\mathbf{b}}_S$ is orthonormal and retains the disjoint-support multiplication rule:

$$\left({\Omega }^{-\left|A\right|}{\mathbf{b}}_A\right)\left({\Omega }^{-\left|B\right|}{\mathbf{b}}_B\right)=\left\{\begin{array}{cc}{\Omega }^{-|A\cup B|}{\mathbf{b}}_{A\cup B},\hfill & A\cap B=⌀,\hfill \\ 0,\hfill & A\cap B\ne ⌀.\hfill \end{array}\right.$$

For disjoint supports this follows from $|A\cup B|=\left|A\right|+\left|B\right|$. Thus the finite-level weighted multiplication problem is isometrically identified with the Euclidean coefficient problem for a zeon algebra. The basis change is already defined over R, since $\Omega$ is a unit. Lindell and Staples study coefficient p-norms and multiplicative inequalities in these algebras [7]. Here the recursive form also fixes the comparison between inputs of different depths.

Theorem 10 

(Orthogonal recursion and path norms). For every global element,

$${∥x∥}_{\Omega }^2={∥pos\left(x\right)∥}_{\Omega }^2+{\Omega }^2{∥len\left(x\right)∥}_{\Omega }^2,{〈(u,0),(0,v)〉}_{\Omega }=0.$$

(46)

A composition of r component operations, containing exactly h occurrences of len and $r-h$ occurrences of pos, satisfies

$${∥{\tau }_r\circ \dots \circ {\tau }_1\left(x\right)∥}_{\Omega }\le {\Omega }^{-h}{∥x∥}_{\Omega },{\tau }_j\in \{pos,len\}.$$

(47)

Its operator norm on the real scalar extension and on the Hilbert completion is exactly ${\Omega }^{-h}$. This value is already attained by an element with integral path coefficients.

Proof. 

The first identity is the defining recursion on equal inputs. The cross term is

$${〈(u,0),(0,v)〉}_{\Omega }={〈u,0〉}_{\Omega }+{\Omega }^2{〈0,v〉}_{\Omega }=0.$$

Dropping either nonnegative term of the squared norm identity gives

$${∥pos\left(x\right)∥}_{\Omega }\le {∥x∥}_{\Omega },{∥len\left(x\right)∥}_{\Omega }\le {\Omega }^{-1}{∥x∥}_{\Omega }.$$

Apply these bounds successively in the order ${\tau }_1,\dots ,{\tau }_r$. Every pos step has factor one and each of the hlen steps has factor ${\Omega }^{-1}$, proving (47). For equality start with $x_r=1$ and, for $j=r,r-1,\dots ,1$, set

$$x_{j-1}=\left\{\begin{array}{cc}(x_j,0),\hfill & {\tau }_j=pos,\hfill \\ (0,x_j),\hfill & {\tau }_j=len.\hfill \end{array}\right.$$

At the jth step, ${\tau }_j\left(x_{j-1}\right)=x_j$, and the norm recursion gives ${∥x_{j-1}∥}_{\Omega }={∥x_j∥}_{\Omega }$ or $\Omega {∥x_j∥}_{\Omega }$ according to the chosen branch. Consequently

$${\tau }_r\circ \dots \circ {\tau }_1\left(x_0\right)=1,{∥x_0∥}_{\Omega }={\Omega }^h,{\displaystyle \frac{{∥{\tau }_r\circ \dots \circ {\tau }_1\left(x_0\right)∥}_{\Omega }}{{∥x_0∥}_{\Omega }}}={\Omega }^{-h}.$$

All leaves of $x_0$ are zero or one, so this extremizer is integral. The norm-preserving value passes to the continuous extension to the completion, which is identified below. □

The classical Cauchy–Schwarz and triangle inequalities [19] apply to the weighted path coordinates. The component recursion also identifies their equality conditions and the proportionality that is available over R.

Theorem 11 

(Cauchy–Schwarz and metric inequalities). For finite or completed real elements,

$$|{〈x,y〉}_{\Omega }|\le {∥x∥}_{\Omega }{∥y∥}_{\Omega }.$$

(48)

For nonzero real elements, equality holds precisely when they are real scalar multiples of each other. For nonzero original-ring elements, this is equivalent to proportionality over $\mathbb{Q}$, or to the existence of nonzero $a,b\in R$ with $ay=bx$. Moreover,

$$\begin{array}{cc}\hfill {∥x+y∥}_{\Omega }& \le {∥x∥}_{\Omega }+{∥y∥}_{\Omega },\hfill \\ \hfill |{∥x∥}_{\Omega }-{∥y∥}_{\Omega }|& \le {∥x-y∥}_{\Omega },\hfill \\ \hfill {∥x+y∥}_{\Omega }^2+{∥x-y∥}_{\Omega }^2& =2{∥x∥}_{\Omega }^2+2{∥y∥}_{\Omega }^2.\hfill \end{array}$$

(49)

For nonzero inputs, equality in the first inequality holds precisely for positive scalar proportionality; equality in the second holds precisely for positive scalar proportionality as well.

Proof. 

First consider finite inputs and induct on their largest prescribed depth. At the scalar boundary,

$$|{〈c,d〉}_{\Omega }|=|cd|=|c\left|\right|d|={∥c∥}_{\Omega }{∥d∥}_{\Omega }.$$

At positive depth, both component pairs have smaller largest depth. The scalar triangle inequality and the induction hypothesis give

$$\begin{array}{cc}\hfill |{〈x,y〉}_{\Omega }|& =\left|{〈pos\left(x\right),pos\left(y\right)〉}_{\Omega }+{\Omega }^2{〈len\left(x\right),len\left(y\right)〉}_{\Omega }\right|\hfill \\ \hfill & \le \left|{〈pos\left(x\right),pos\left(y\right)〉}_{\Omega }\right|+{\Omega }^2\left|{〈len\left(x\right),len\left(y\right)〉}_{\Omega }\right|\hfill \\ \hfill & \le {∥pos\left(x\right)∥}_{\Omega }{∥pos\left(y\right)∥}_{\Omega }+{\Omega }^2{∥len\left(x\right)∥}_{\Omega }{∥len\left(y\right)∥}_{\Omega }.\hfill \end{array}$$

For nonnegative real $a,b,c,d$, the identity

$$(a^2+b^2)(c^2+d^2)-{(ac+bd)}^2={(ad-bc)}^2\ge 0$$

bounds a two-term scalar product by the product of the two lengths. Apply it to the two component-norm pairs to obtain

$$\begin{array}{cc}\hfill & {∥pos\left(x\right)∥}_{\Omega }{∥pos\left(y\right)∥}_{\Omega }+{\Omega }^2{∥len\left(x\right)∥}_{\Omega }{∥len\left(y\right)∥}_{\Omega }\hfill \\ \hfill & \le \sqrt{{∥pos\left(x\right)∥}_{\Omega }^2+{\Omega }^2{∥len\left(x\right)∥}_{\Omega }^2}\sqrt{{∥pos\left(y\right)∥}_{\Omega }^2+{\Omega }^2{∥len\left(y\right)∥}_{\Omega }^2}\hfill \\ \hfill & ={∥x∥}_{\Omega }{∥y∥}_{\Omega }.\hfill \end{array}$$

This proves (48) by scalar descent, including unequal input depths through the scalar component rules.

To determine equality, suppose $x\ne 0$ and work in the real scalar extension. Set $\lambda ={〈x,y〉}_{\Omega }/{∥x∥}_{\Omega }^2$. Bilinearity gives

$$\begin{array}{cc}\hfill {∥y-\lambda x∥}_{\Omega }^2& ={∥y∥}_{\Omega }^2-2\lambda {〈x,y〉}_{\Omega }+{\lambda }^2{∥x∥}_{\Omega }^2\hfill \\ \hfill & ={∥y∥}_{\Omega }^2-{\displaystyle \frac{{〈x,y〉}_{\Omega }^2}{{∥x∥}_{\Omega }^2}}.\hfill \end{array}$$

Thus equality in (48) holds exactly when ${∥y-\lambda x∥}_{\Omega }=0$, or $y=\lambda x$. Conversely, that relation gives

$$|{〈x,y〉}_{\Omega }|=|\lambda |{∥x∥}_{\Omega }^2={∥x∥}_{\Omega }{∥y∥}_{\Omega }.$$

For nonzero original-ring inputs attaining equality, choose a common finite level and a path S with $x_S\ne 0$. Then

$$\lambda ={\displaystyle \frac{y_S}{x_S}}\in \mathbb{Q}\setminus \left\{0\right\},x_{S}y=y_{S}x.$$

Taking $a=x_S$ and $b=y_S$ gives the claimed relation in R. Conversely, $ay=bx$ with $a,b\ne 0$ gives $y=(b/a)x$ over the reals and hence equality. A zero input gives equality directly.

For finite inputs, expand the squared norms:

$$\begin{array}{cc}\hfill {∥x+y∥}_{\Omega }^2& ={∥x∥}_{\Omega }^2+2{〈x,y〉}_{\Omega }+{∥y∥}_{\Omega }^2\hfill \\ \hfill & \le {∥x∥}_{\Omega }^2+2{∥x∥}_{\Omega }{∥y∥}_{\Omega }+{∥y∥}_{\Omega }^2={\left({∥x∥}_{\Omega }+{∥y∥}_{\Omega }\right)}^2,\hfill \\ \hfill {∥x-y∥}_{\Omega }^2& ={∥x∥}_{\Omega }^2-2{〈x,y〉}_{\Omega }+{∥y∥}_{\Omega }^2\hfill \\ \hfill & \ge {∥x∥}_{\Omega }^2-2{∥x∥}_{\Omega }{∥y∥}_{\Omega }+{∥y∥}_{\Omega }^2={\left({∥x∥}_{\Omega }-{∥y∥}_{\Omega }\right)}^2.\hfill \end{array}$$

Taking nonnegative square roots proves both metric inequalities. For nonzero inputs, either equality requires ${〈x,y〉}_{\Omega }={∥x∥}_{\Omega }{∥y∥}_{\Omega }$, which is precisely $y=\lambda x$ with $\lambda >0$. Adding the two exact expansions cancels the cross terms and gives the parallelogram identity.

The form extends to the completion without a new inequality. Indeed, for Cauchy sequences of finite real inputs $\left(x_j\right)$ and $\left(y_j\right)$, the finite bound gives

$$\begin{array}{cc}\hfill \left|{〈x_j,y_j〉}_{\Omega }-{〈x_k,y_k〉}_{\Omega }\right|& \le \left|{〈x_j-x_k,y_j〉}_{\Omega }\right|+\left|{〈x_k,y_j-y_k〉}_{\Omega }\right|\hfill \\ \hfill & \le {∥x_j-x_k∥}_{\Omega }{∥y_j∥}_{\Omega }+{∥x_k∥}_{\Omega }{∥y_j-y_k∥}_{\Omega }.\hfill \end{array}$$

The norms of the Cauchy sequences are bounded, so the right side has limit zero as $j,k$ increase. The limit pairing is independent of the approximating sequences and satisfies

$$|{〈x,y〉}_{\Omega }|=\underset{j}{\lim }\left|{〈x_j,y_j〉}_{\Omega }\right|\le \underset{j}{\lim }{∥x_j∥}_{\Omega }{∥y_j∥}_{\Omega }={∥x∥}_{\Omega }{∥y∥}_{\Omega }.$$

The completed-square calculation and the squared-norm expansions remain valid for the extended form. They prove the same equality conditions, metric inequalities, and parallelogram identity. □

Proposition 4 

(Arithmetic norm levels). Let q be a nonnegative integer and suppose that $X={\Omega }^{q}x$ has integral path coefficients. Then

$${∥x∥}_{\Omega }^2\in {\Omega }^{-2q}{\mathbb{Z}}_{\ge 0}.$$

(50)

If ${\Omega }^{q}x$ and ${\Omega }^{q}y$ both have integral path coefficients and their norms differ, then

$$|{∥x∥}_{\Omega }^2-{∥y∥}_{\Omega }^2|\ge {\Omega }^{-2q}.$$

(51)

The constant is attained by $x=0$ and $y={\Omega }^{-q}1$.

When both components are nonzero, their weighted norm ratio is unchanged by nonzero scalar multiplication. For the integral element X above,

$${\displaystyle \frac{\Omega {∥len\left(x\right)∥}_{\Omega }}{{∥pos\left(x\right)∥}_{\Omega }}}={\displaystyle \frac{\Omega {∥len\left(X\right)∥}_{\Omega }}{{∥pos\left(X\right)∥}_{\Omega }}}\ge {\displaystyle \frac{\Omega }{{∥pos\left(X\right)∥}_{\Omega }}}\ge {\displaystyle \frac{\Omega }{{∥X∥}_{\Omega }}}.$$

(52)

The ratio is undefined when $pos\left(x\right)=0$; when $len\left(x\right)=0$ and $pos\left(x\right)\ne 0$, it is zero.

The set $\mathcal{I}$ is countable, dense in its Hilbert completion, and totally disconnected in the norm metric. Each ${\mathcal{I}}_n$ is a Lebesgue-null subset of its finite-dimensional real scalar extension.

Proof. 

The integral path coefficients of X give

$${\Omega }^{2q}{∥x∥}_{\Omega }^2={∥X∥}_{\Omega }^2=\sum _S{\Omega }^{2\left|S\right|}{X}_S^2\in {\mathbb{Z}}_{\ge 0}.$$

For unequal squared norms, their scaled difference is a nonzero integer, so its absolute value is at least one. This proves the gap. For a nonzero scalar c, scalar-linearity gives

$${\displaystyle \frac{\Omega {∥len\left(cx\right)∥}_{\Omega }}{{∥pos\left(cx\right)∥}_{\Omega }}}={\displaystyle \frac{\Omega \left|c\right|{∥len\left(x\right)∥}_{\Omega }}{\left|c\right|{∥pos\left(x\right)∥}_{\Omega }}}={\displaystyle \frac{\Omega {∥len\left(x\right)∥}_{\Omega }}{{∥pos\left(x\right)∥}_{\Omega }}}$$

whenever the denominator is nonzero. A nonzero integral path coefficient contributes at least one to the squared norm. Hence ${∥len\left(X\right)∥}_{\Omega }\ge 1$ when that component is nonzero, while ${∥pos\left(X\right)∥}_{\Omega }\le {∥X∥}_{\Omega }$. Substitution proves every step of (52). A zero second component gives ratio zero; a zero first component leaves the denominator zero.

Each finite level is a finite Cartesian power of the countable ring R, and there are countably many levels. Thus $\mathcal{I}$ and each ${\mathcal{I}}_n$ are countable; the latter has Lebesgue measure zero. The numbers ${\Omega }^{-j}$ have limit zero, so R is dense in $\mathbb{R}$. Coefficientwise approximation at a finite level, followed by truncation, gives density in the completion described in Theorem 14. Finally, two different original-ring elements differ in some path coordinate. Choose an irrational real number strictly between those two coordinate values. The corresponding coordinate cut separates $\mathcal{I}$ into two disjoint relatively open sets. Indeed, all its coordinate values are rational, and

$$|u_S-v_S|\le {\Omega }^{-\left|S\right|}{∥u-v∥}_{\Omega }$$

proves continuity of coordinate evaluation. Every subset containing the two points is therefore disconnected. □

Theorem 12 

(Weighted complementation). At a specified level define

$${★}_0\left(c\right)=c,{★}_{n+1}\left(x\right)=\left(\Omega {★}_n\left(len\left(x\right)\right),{\Omega }^{-1}{★}_n\left(pos\left(x\right)\right)\right).$$

(53)

This is an R-linear isometric involution, with

$$\begin{array}{cc}\hfill {★}_n\left({\mathbf{b}}_S\right)& ={\Omega }^{2\left|S\right|-n}{\mathbf{b}}_{\left[n\right]\setminus S},\hfill \\ \hfill {〈x,y〉}_{\Omega }& ={\Omega }^n{len}^{\left(n\right)}\left(x{★}_n\left(y\right)\right).\hfill \end{array}$$

(54)

For ordinary multiplication $M_x\left(y\right)=xy$ on the real extension of ${\mathcal{I}}_n$, its adjoint is

$$M_x^{*}={★}_n{M}_x{★}_n.$$

(55)

Proof. 

At each recursive position, a first-component path is exchanged with a second-component path and receives weight ${\Omega }^{-1}$; a second-component path receives weight $\Omega$. Thus a path with $\left|S\right|$ second choices has total weight ${\Omega }^{\left|S\right|}{\Omega }^{-(n-|S\left|\right)}$, proving (54). Consequently

$$\begin{array}{cc}\hfill {★}_n^2\left({\mathbf{b}}_S\right)& ={\Omega }^{2\left|S\right|-n}{\Omega }^{2(n-|S\left|\right)-n}{\mathbf{b}}_S={\mathbf{b}}_S,\hfill \\ \hfill {∥{★}_n\left({\mathbf{b}}_S\right)∥}_{\Omega }^2& ={\Omega }^{4\left|S\right|-2n}{\Omega }^{2(n-|S\left|\right)}={\Omega }^{2\left|S\right|}.\hfill \end{array}$$

Distinct paths still have distinct, orthogonal complements, so linearity proves the isometric involution assertion.

The perfect multiplication pairing of Theorem 6 pairs a path with its complement. Using its formula (34) gives

$${\Omega }^n{\beta }_n\left(x,{★}_n\left(y\right)\right)={\Omega }^n\sum _S{x}_S{\Omega }^{2\left|S\right|-n}{y}_S={〈x,y〉}_{\Omega }.$$

Associativity of that pairing then gives

$${〈xy,z〉}_{\Omega }={\Omega }^n{\beta }_n\left(y,x{★}_n\left(z\right)\right)={〈y,{★}_n\left(x{★}_n\left(z\right)\right)〉}_{\Omega },$$

which is the adjoint identity. □

Remark 1. 

Complementation depends on the specified level. For example,

$${★}_n\left(0\right)=0,{★}_n\left(1_n\right)={\Omega }^{-n}{\mathbf{b}}_{\left[n\right]},$$

and in depths zero and one,

$${★}_1\left(1_1\right)=(0,{\Omega }^{-1}),\iota \left({★}_0\left(1_0\right)\right)=(1,0).$$

These distinct global classes show that complementation is not invariant under refinement.

Theorem 13 

(Algebra isometries). Every unital R-algebra isometric automorphism of ${\mathcal{I}}_n$ is given by

$$\theta \left({\mathbf{b}}_j\right)={\sigma }_j{\mathbf{b}}_{\pi \left(j\right)},{\sigma }_j\in \{-1,1\},$$

(56)

where π is a permutation of $\left[n\right]$. Conversely every such choice defines an algebra isometry. The same classification on $\mathcal{I}$ uses permutations of the positive integers and an independent sign at each position. A global algebra isometry commuting with both pos and len is the identity.

Proof. 

An algebra automorphism preserves the nilradical $\mathcal{N}$ and its square. The ideal ${\mathcal{N}}^2$ is spanned by the paths of size at least two. Path orthogonality therefore identifies its orthogonal complement inside $\mathcal{N}$ with exactly the singleton-path span. A nonzero higher-order coefficient cannot belong to that complement, as pairing with its own path detects it. Polarization of the norm shows that an R-linear isometry preserves the inner product. The automorphism and its inverse consequently preserve this span. Write a singleton image as $\theta \left({\mathbf{b}}_k\right)={\sum }_j{c}_j{\mathbf{b}}_j$. Its square gives

$$0=\theta \left({\mathbf{b}}_k^2\right)={\left(\sum _j{c}_j{\mathbf{b}}_j\right)}^2=2\sum _{i<j}{c}_i{c}_j{\mathbf{b}}_{\{i,j\}}.$$

The coefficient ring has characteristic zero and no zero divisors, so at most one $c_j$ is nonzero. Its norm is $\Omega$, forcing that coefficient to be 1 or $-1$. Bijectivity gives a permutation, and multiplication determines every path image:

$$\theta \left({\mathbf{b}}_S\right)=\prod _{j\in S}\theta \left({\mathbf{b}}_j\right)=\left(\prod _{j\in S}{\sigma }_j\right){\mathbf{b}}_{\pi \left(S\right)}.$$

Signed permutations preserve path products and the weights ${\Omega }^{2\left|S\right|}$, which proves the converse. These arguments apply unchanged to finite-support global elements. The last assertion also follows from Proposition 3: commuting with both components and fixing scalars determines every finite pair recursively. □

The weighted coordinate norm identifies the completion with a sequence Hilbert space, as in the standard theory of square-summable families [20].

Theorem 14 

(Hilbert completion). The metric completion of $\mathcal{I}$, equivalently the Hilbert completion of its real scalar extension, consists of the real coefficient families

$${\mathcal{H}}_{\Omega }=\left\{{\left(x_S\right)}_{S\subseteq {\mathbb{N}}_{>0},\left|S\right|<\infty }:\sum _S{\Omega }^{2\left|S\right|}{x}_S^2<\infty \right\},$$

(57)

with inner product ${〈x,y〉}_{\Omega }={\sum }_S{\Omega }^{2\left|S\right|}{x}_S{y}_S$. Both component operations extend continuously, and the assignment of the ordered pair $(pos(x),\Omega len(x\left)\right)$ to x is an isometric bijection between ${\mathcal{H}}_{\Omega }$ and ${\mathcal{H}}_{\Omega }\oplus {\mathcal{H}}_{\Omega }$.

Proof. 

The vectors ${\Omega }^{-\left|S\right|}{\mathbf{b}}_S$ form an orthonormal basis of the finite-support real space. Multiplying each coordinate $x_S$ by ${\Omega }^{\left|S\right|}$ identifies its norm with the usual square-sum norm, whose complete space is (57). For a family in that space, let $P_{n}x={\sum }_{S\subseteq \left[n\right]}{x}_S{\mathbf{b}}_S$. Then

$${∥x-P_{n}x∥}_{\Omega }^2=\sum _{S\neg \subseteq \left[n\right]}{\Omega }^{2\left|S\right|}{x}_S^2,\underset{n}{\lim }{∥x-P_{n}x∥}_{\Omega }=0.$$

For a prescribed error $\delta >0$, first choose n with ${∥x-P_{n}x∥}_{\Omega }<\delta /2$. Then choose $c_S\in R$ for $S\subseteq \left[n\right]$ such that

$$|c_S-x_S|<{\displaystyle \frac{\delta }{2\sqrt{2^n}{\Omega }^{\left|S\right|}}}.$$

The element $y={\sum }_{S\subseteq \left[n\right]}{c}_S{\mathbf{b}}_S$ satisfies

$${∥P_{n}x-y∥}_{\Omega }^2=\sum _{S\subseteq \left[n\right]}{\Omega }^{2\left|S\right|}{|x_S-c_S|}^2<{\delta }^2/4,{∥x-y∥}_{\Omega }<\delta .$$

This proves density of the original-ring elements, and therefore the same completion for $\mathcal{I}$ as for its real scalar extension. The component bounds in Theorem 10 give their unique continuous extensions. For the weighted pair assignment, the exact identity is

$${∥pos\left(x\right)∥}_{\Omega }^2+{∥\Omega len\left(x\right)∥}_{\Omega }^2={∥x∥}_{\Omega }^2.$$

A pair $(u,v)$ of finite elements has preimage $(u,{\Omega }^{-1}v)$: its first component is u and its weighted second component is v. Thus its range is dense; as the range of an isometry from a complete space it is closed, and hence it is the entire direct sum. □

3.2. Sharp Multiplication at Prescribed Depths

In orthogonal component coordinates, multiplication becomes a block matrix. Its largest singular value measures the exact amplification of the norm; the matrix-norm facts used below are standard [21].

Proposition 5 

(A fixed multiplier and its recursive bound). Write ${\mu }_n\left(x\right)$ for the operator norm of multiplication by $x\in {\mathcal{I}}_n$ on its real scalar extension. At the scalar level ${\mu }_0\left(c\right)=\left|c\right|$. Under the orthogonal coordinates $(pos(y),\Omega len(y\left)\right)$, the exact multiplication matrix is

$$M_x=\left(\begin{array}{cc}{M}_{pos\left(x\right)}& 0\\ \Omega M_{len\left(x\right)}& M_{pos\left(x\right)}\end{array}\right).$$

(58)

In particular, if $a={\mu }_{n-1}(pos\left(x\right))$ and $b=\Omega {\mu }_{n-1}(len\left(x\right))$, then

$${\mu }_n\left(x\right)\le {\displaystyle \frac{\sqrt{4a^2+b^2}+b}{2}}.$$

(59)

This is a recursive upper bound, with the exact norm determined by (58). For a separated element $x={\prod }_{j=1}^n(c_j+d_j{\mathbf{b}}_j)$ there is the exact formula

$${\mu }_n\left(x\right)={\displaystyle \prod _{j=1}^n}{\displaystyle \frac{\sqrt{4c_j^2+{\Omega }^2{d}_j^2}+\Omega \left|d_j\right|}{2}}.$$

(60)

Proof. 

The two components of $xy$ are $pos\left(x\right)pos\left(y\right)$ and $len\left(x\right)pos\left(y\right)+pos\left(x\right)len\left(y\right)$, giving the block matrix. Taking the norms of these two expressions gives

$${∥xy∥}_{\Omega }^2\le a^2{∥pos\left(y\right)∥}_{\Omega }^2+{\left(b{∥pos\left(y\right)∥}_{\Omega }+a\Omega {∥len\left(y\right)∥}_{\Omega }\right)}^2.$$

Thus it suffices to compute the norm of $C=\left(\begin{array}{cc}a& 0\\ b& a\end{array}\right)$. Since

$$C^{\mathsf{t}}C=\left(\begin{array}{cc}{a}^2+b^2& ab\\ ab& a^2\end{array}\right),\det (\lambda I-C^{\mathsf{t}}C)={\lambda }^2-(2a^2+b^2)\lambda +a^4,$$

its larger eigenvalue is

$${\lambda }_{+}={\displaystyle \frac{2a^2+b^2+b\sqrt{4a^2+b^2}}{2}}={\left({\displaystyle \frac{\sqrt{4a^2+b^2}+b}{2}}\right)}^2.$$

Taking its square root proves the recursive bound. The comparison uses the nonnegative vector $\left({∥pos\left(y\right)∥}_{\Omega },\Omega {∥len\left(y\right)∥}_{\Omega }\right)$, whose Euclidean norm is exactly ${∥y∥}_{\Omega }$. Thus it applies even when a multiplier component vanishes. For separated elements, normalized path coordinates identify multiplication with the tensor product of the matrices $\left(\begin{array}{cc}{c}_j& 0\\ \Omega d_j& c_j\end{array}\right)$. The positive matrix of this tensor operator is the tensor product of the individual matrices $C_j^{\mathsf{t}}{C}_j$. An orthonormal eigenbasis for each factor gives a tensor eigenbasis, with eigenvalues the products of the individual nonnegative eigenvalues. The largest is the product of their largest eigenvalues. Taking square roots and using the preceding two-dimensional calculation, with $a=|c_j|$ and $b=\Omega |d_j|$, proves (60). □

The next inequality is the dual-number multiplication estimate of Cook and Nguyen [8], written for nonnegative component norms. Their optimal constant covers every number of inputs, including $2/\sqrt{3}$ for two inputs. The proof records the maximizing ratios used at successive recursive positions.

Lemma 2 

(The sharp recursive multiplication step). For $r\ge 2$ and nonnegative real numbers $a_i,b_i$,

$${\left({\displaystyle \prod _{i=1}^r}{a}_i\right)}^2+{\left({\displaystyle \sum _{i=1}^r}{b}_i\prod _{h\ne i}{a}_h\right)}^2\le {\displaystyle \frac{r^r}{{(r+1)}^{r-1}}}{\displaystyle \prod _{i=1}^r}(a_i^2+b_i^2).$$

(61)

If every pair $(a_i,b_i)$ is nonzero, equality holds precisely when every $a_i>0$ and $b_i/a_i=1/\sqrt{r}$. For $r=1$ the corresponding expression is an identity.

Proof. 

Normalize every nonzero pair to length one; zero pairs give zero on both sides. The normalized domain is compact. If some $a_i=0$, at most one term in the second sum survives, and the objective is at most one. If all $a_i>0$ and all $b_i=0$, its value is one. If $b_i=0$ and another $b_h>0$, put $t_j=b_j/a_j$ in the region where all $a_j>0$. The normalized objective and its one-sided derivative at $t_i=0$, with $s={\sum }_j{t}_j>0$, are

$$F(t_1,\dots ,t_r)={\displaystyle \frac{1+{\left({\sum }_j{t}_j\right)}^2}{{\prod }_j(1+t_j^2)}},{\left.{\displaystyle \frac{\partial F}{\partial t_i}}\right|}_{t_i=0}={\displaystyle \frac{2s}{{\prod }_{j\ne i}(1+t_j^2)}}>0.$$

Thus this point is not a maximum. On the other hand, choosing $b_i/a_i=t>0$ gives

$${\displaystyle \frac{1+r^2{t}^2}{{(1+t^2)}^r}}=1+r(r-1)t^2+O\left(t^4\right)>1$$

for sufficiently small t. The maximum is therefore an interior point.

Put $t_i=b_i/a_i$ and $s={\sum }_i{t}_i$. Logarithmic differentiation of $(1+s^2)/{\prod }_i(1+t_i^2)$ at an interior maximum gives

$${\displaystyle \frac{t_i}{1+t_i^2}}={\displaystyle \frac{s}{1+s^2}}.$$

The two possible roots are s and $1/s$. Since $r\ge 2$ and $0<t_i<s$, every $t_i=1/s$. Explicitly, the stationary equation is

$$s{t}_i^2-(1+s^2)t_i+s=(t_i-s)(s{t}_i-1)=0.$$

Summing $t_i=1/s$ gives $s=r/s$, so $s^2=r$ and $t_i=1/\sqrt{r}$. The value there is

$${\displaystyle \frac{1+s^2}{{\prod }_i(1+t_i^2)}}={\displaystyle \frac{1+r}{{(1+1/r)}^r}}={\displaystyle \frac{r^r}{{(r+1)}^{r-1}}}.$$

The boundary values are strictly smaller, so these are all maximizers. □

At a position shared by r inputs, Lemma 2 gives the corresponding amplification factor. Positions belonging to only one input introduce no amplification. Following equality through the same descent determines the common factors and any remaining unshared tail.

Theorem 15 

(Mixed-depth multiplication and all extremizers). Fix $k\ge 2$ input depths $m_1,\dots ,m_k$, put $N={\max }_i{m}_i$, and let

$$r_j=\#\{i:m_i\ge j\}(1\le j\le N).$$

For inputs $x_i\in {\mathcal{I}}_{m_i}$, and also for their real scalar extensions,

$${∥x_1\dots x_k∥}_{\Omega }\le {\displaystyle \prod _{j=1}^N}{\left({\displaystyle \frac{r_j^{r_j}}{{(r_j+1)}^{r_j-1}}}\right)}^{1/2}{\displaystyle \prod _{i=1}^k}{∥x_i∥}_{\Omega }.$$

(62)

The factor for $r_j=1$ is one, and an empty product is one. This constant is the exact real maximum and the exact original-ring supremum.

For the equality classification suppose that all inputs are nonzero, and let s be the second largest input depth, counting multiplicity. If $s=0$, every such tuple gives equality. If $s>0$, all real equality tuples have precisely the following form:

$$x_i={\lambda }_i{\displaystyle \prod _{j=1}^{\min (m_i,s)}}\left(1+{\displaystyle \frac{{\sigma }_j}{\Omega \sqrt{r_j}}}{\mathbf{b}}_j\right)h_i.$$

(63)

Here ${\sigma }_j\in \{-1,1\}$ is common to all inputs using position j, each ${\lambda }_i$ is a nonzero real number, and $h_i=1$ except possibly for the unique input with $m_i>s$. For that input, $h_i$ is any nonzero element supported on positions $s+1,\dots ,m_i$. In particular its scalar coefficient need not be nonzero. If the largest depth is repeated, there is no free tail.

A nonzero equality tuple exists over R precisely when each $r_j\ge 2$ is a perfect square. This includes the vacuous condition when $s=0$.

Proof. 

At a recursive step,

$$pos(x_1\dots x_k)=\prod _{i}pos\left(x_i\right),len(x_1\dots x_k)=\sum _{i}len\left(x_i\right)\prod _{h\ne i}pos\left(x_h\right).$$

An input which has reached a scalar has zero second component and acts simply as a scalar factor. After removing these scalar factors, let r be the number of remaining inputs and let C be their lower-depth multiplication bound. Put $a_i={∥pos\left(x_i\right)∥}_{\Omega }$ and $b_i=\Omega {∥len\left(x_i\right)∥}_{\Omega }$. The recursive norm identity, the triangle inequality, and Lemma 2 give

$$\begin{array}{cc}\hfill {∥x_1\dots x_r∥}_{\Omega }^2& \le C^2\left[{\left(\prod _i{a}_i\right)}^2+{\left(\sum _i{b}_i\prod _{h\ne i}{a}_h\right)}^2\right]\hfill \\ \hfill & \le C^2{\displaystyle \frac{r^r}{{(r+1)}^{r-1}}}\prod _i(a_i^2+b_i^2)\hfill \\ \hfill & =C^2{\displaystyle \frac{r^r}{{(r+1)}^{r-1}}}\prod _i{∥x_i∥}_{\Omega }^2.\hfill \end{array}$$

The case $r=1$ is exact scalar multiplication. At successive positions the active counts are $r_1,\dots ,r_N$, so iterating this estimate proves (62).

More precisely, let $C_j$ be the asserted bound for the tails beginning after position j, with $C_N=1$. The calculation gives

$$C_{j-1}={\left({\displaystyle \frac{r_j^{r_j}}{{(r_j+1)}^{r_j-1}}}\right)}^{1/2}{C}_j(1\le j\le N).$$

Every component product at a given position has the same remaining typed depths. It therefore has the same applicable tail bound $C_j$, including when one component is the zero object.

In the proposed tuple, different positions are independent. At a shared position j, square-zero multiplication gives

$${\left(1+{\displaystyle \frac{{\sigma }_j}{\Omega \sqrt{r_j}}}{\mathbf{b}}_j\right)}^{r_j}=1+{\displaystyle \frac{{\sigma }_j\sqrt{r_j}}{\Omega }}{\mathbf{b}}_j,{\displaystyle \frac{1+r_j}{{(1+1/r_j)}^{r_j}}}={\displaystyle \frac{r_j^{r_j}}{{(r_j+1)}^{r_j-1}}}.$$

The ratio is the squared norm of the product divided by the product of the squared input norms. Positions used by only one input, including any free tail, give exact norm factorization. Thus (63) attains the bound. Density of R in finite real coordinate spaces gives the same supremum over R.

For completeness, the equality conditions can also be read at each recursive step. When at least two inputs remain active, the strict boundary exclusions in Lemma 2 force all their first and second components to be nonzero and

$$\Omega {∥len\left(x_i\right)∥}_{\Omega }={\displaystyle \frac{{∥pos\left(x_i\right)∥}_{\Omega }}{\sqrt{r}}}.$$

(64)

Every lower-level all-first-component product and every product having exactly one second component must itself attain its lower-level bound; the latter products must also lie on the same positive ray. Apply induction to this collection of products. At every lower shared position their common signs agree: two such products retain the same nonzero component in every unchanged slot, and the inductive factor at that position is determined by that component. Comparing the all-first-component product with the product in which one slot is changed therefore identifies the lower shared factors of that slot’s first and second components.

After these common factors have been removed, each non-deepest slot has only a scalar left. If a unique deepest slot remains, its first and second components may at this stage have two tail factors. The positive-ray equality for the single-second-component products makes these two tails proportional: the other factors in these products are nonzero scalars and units supported on the shared positions, so they may be cancelled. This is also the argument when only one active input remains at the next level, where the lower-level multiplication bound imposes no tail restriction. For a possible deepest slot with index d, the cancellation is explicit. After the common units on shared positions are removed, write its two remaining tails as $H,K$, and the nonzero scalar factors of the other slots as $a_i,b_i$. Two single-second-component products are scalar multiples of

$$K\prod _{i\ne d}{a}_i,H{b}_l\prod _{i\ne d,l}{a}_i(l\ne d).$$

Their positive-ray relation forces $K,H$ to be proportional. In the other slots the remaining factors are already scalars. Consequently every active slot has its second component equal to a scalar multiple of its first component at the present position. The positive-ray condition makes all these scalar multiples have one common sign, and (64) fixes their absolute value as $1/\left(\Omega \sqrt{r}\right)$. Thus, for one common $\sigma \in \{-1,1\}$,

$$len\left(x_i\right)={\displaystyle \frac{\sigma }{\Omega \sqrt{r}}}pos\left(x_i\right),x_i=\left(pos\left(x_i\right),{\displaystyle \frac{\sigma }{\Omega \sqrt{r}}}pos\left(x_i\right)\right)$$

at the present recursive position. This proves its factorization and completes the induction. Below the last shared position there is at most one active input and no further equality restriction.

For original-ring equality, at each shared position choose an active slot other than a possible unique deepest slot. Its scalar coefficient is nonzero, and its singleton-to-scalar coefficient ratio is ${\sigma }_j/\left(\Omega \sqrt{r_j}\right)$. That ratio must be rational, so $r_j$ is a square. Conversely, writing $r_j=q_j^2$ with positive integers $q_j$, take in each active slot the product of $q_j\Omega +{\sigma }_j{\mathbf{b}}_j$ over its shared positions and choose any nonzero original-ring tail when there is one. These are original-ring extremizers. □

Corollary 2 

(Two inputs and equal depths). For depths $m,n$, the exact bilinear multiplication constant is ${(2/\sqrt{3})}^{\min (m,n)}$. For k inputs of depth n it is

$${\left({\displaystyle \frac{k^k}{{(k+1)}^{k-1}}}\right)}^{n/2}.$$

(65)

When $n\ge 1$, nonzero equality tuples at equal depth are exactly

$$x_i={\lambda }_i{\displaystyle \prod _{j=1}^n}\left(1+{\displaystyle \frac{{\sigma }_j}{\Omega \sqrt{k}}}{\mathbf{b}}_j\right).$$

They exist over R precisely when k is a perfect square. In particular, no original-ring nonzero tuple attains the two-input constant when both prescribed depths are positive. There is no depth-independent bound for ordinary multiplication on $\mathcal{I}$, and no continuous bilinear extension of this multiplication to all of ${\mathcal{H}}_{\Omega }$.

Proof. 

For two inputs, the active count is $r_j=2$ through position $\min (m,n)$ and is one afterwards. For k equal depths it is $r_j=k$ at all n positions. Substitution into (62) gives both constants and their equality conditions. A continuous bilinear extension would have a finite bound C with ${∥xy∥}_{\Omega }\le C{∥x∥}_{\Omega }{∥y∥}_{\Omega }$ on all finite inputs. To obtain such a uniform bound from continuity at the origin, choose $\delta >0$ for which both input norms less than $\delta$ give output norm at most one. Rescaling each nonzero input to norm $\delta /2$ gives $C=4/{\delta }^2$. Sharpness on equal depths would then require

$$C\ge {\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^n\mathrm{for}\mathrm{every}n\ge 0,$$

which no finite C satisfies. □

3.3. Homogeneous Multiplication and Support Overlap

An element supported on singleton paths is homogeneous of degree one at every recursive depth containing those positions.

Lemma 3 

(Symmetric multilinear forms on a real Hilbert space). Let B be a symmetric scalar-valued k-linear form on a finite-dimensional real Hilbert space, with $k\ge 2$. Its maximum absolute value on unit inputs is attained with all inputs equal.

Proof. 

This is the real Hilbert-space form of Banach’s multilinear theorem [22]; a tensor spectral-norm formulation appears in [23]. The following argument supplies the finite-dimensional proof used here. Write M for the maximum. The case $M=0$ is immediate. Among extremizing unit tuples choose one maximizing the squared norm of the sum of its inputs; compactness gives such a tuple. Fix two entries $u,v$, let s be the sum of the others, and represent the resulting symmetric bilinear form by a self-adjoint matrix C. Its operator norm is at most M. Equality gives

$$Cv=\delta Mu,Cu=\delta Mv,\delta \in \{-1,1\}.$$

Indeed, $|\langle u,Cv\rangle |=M$ while $\parallel Cv\parallel \le M$ and $\parallel u\parallel =1$. Cauchy–Schwarz equality gives the first vector identity. Symmetry gives the second one with the same sign. If $u+v\ne 0$, put $a=\parallel u+v\parallel$ and $w=(u+v)/a$. Then $Cw=\delta Mw$, so replacing $u,v$ by $w,w$ preserves the absolute maximum. Replacing both $u,v$ by their negatives also preserves it; maximality of the sum norm therefore implies $\langle s,u+v\rangle \ge 0$. If $u\ne v$, the first replacement increases the sum norm squared by

$$(2-a)\left(2+a+2\langle s,w\rangle \right)>0,$$

a contradiction. If $u=-v$, then $Cu=-\delta Mu$. Replacing the pair by $u,u$ or by $-u,-u$ preserves the absolute maximum; choosing the sign having nonnegative inner product with s strictly increases the sum norm. Thus every pair of inputs in the selected tuple is equal. □

Lemma 4 

(The sharp power of a degree-one element). For $u={\sum }_{j=1}^n{a}_j{\mathbf{b}}_j$ and $1\le k\le n$,

$$\begin{array}{cc}\hfill u^k& =k!\sum _{\begin{array}{c}S\subseteq \left[n\right]\\ \left|S\right|=k\end{array}}\left(\prod _{j\in S}{a}_j\right){\mathbf{b}}_S,\hfill \\ \hfill {∥u^k∥}_{\Omega }^2& ={(k!)}^2{\Omega }^{2k}\sum _{\begin{array}{c}S\subseteq \left[n\right]\\ \left|S\right|=k\end{array}}\prod _{j\in S}{a}_j^2\le {(k!)}^2{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}{n^k}}{∥u∥}_{\Omega }^{2k}.\hfill \end{array}$$

(66)

For $2\le k\le n$ and $u\ne 0$, equality holds precisely when all $|a_j|$ are equal. For $k=1$ the formula is an identity; for $k>n$ the power is zero.

Proof. 

A repeated singleton position gives a zero product; each choice of k distinct positions has $k!$ orders. Path orthogonality then gives the norm identity. For nonnegative $p_1,\dots ,p_n$ with fixed positive sum, the sum of their square-free products of degree k has a positive maximum on the compact simplex. At least k coordinates of a maximizing point are positive. With all but $p_i,p_j$ and their sum fixed, that expression has the form

$$p_i{p}_j{e}_{k-2}+(p_i+p_j)e_{k-1}+e_k,$$

where $e_l$ is the degree-l elementary symmetric sum in the remaining coordinates, and $e_0=1$. If $p_i\ne p_j$, at least $k-2$ of the other coordinates are positive, so $e_{k-2}>0$. Replacing both entries by $(p_i+p_j)/2$ increases the expression by

$$\left({\displaystyle \frac{{(p_i+p_j)}^2}{4}}-p_i{p}_j\right)e_{k-2}={\displaystyle \frac{{(p_i-p_j)}^2}{4}}{e}_{k-2}>0,$$

a contradiction. For $k\ge 2$ the maximizing coordinates are therefore all equal. Apply this to $p_j={\Omega }^2{a}_j^2$. □

The elementary symmetric-sum estimate in the preceding lemma is Maclaurin’s inequality, with the equality condition obtained by averaging two coordinates [19,24].

Theorem 16 

(Degree-one multiplication: exact norm and equality). Suppose $x_i={\sum }_{j=1}^n{a}_{ij}{\mathbf{b}}_j$, with $2\le k\le n$. Then

$${∥x_1\dots x_k∥}_{\Omega }\le k!\sqrt{{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}{n^k}}}{\displaystyle \prod _{i=1}^k}{∥x_i∥}_{\Omega }.$$

(67)

The constant is best possible and nonzero original-ring tuples attain it. For $n\ge 3$, all nonzero real equality tuples are

$$x_i={\lambda }_i{\displaystyle \sum _{j=1}^n}{\sigma }_j{\mathbf{b}}_j,{\lambda }_i\ne 0,{\sigma }_j\in \{-1,1\},$$

(68)

with the same signs for every input. Original-ring inputs have ${\lambda }_i\in R$. In the exceptional case $n=k=2$, equality is instead equivalent to

$$a_{11}{a}_{21}=a_{12}{a}_{22}.$$

(69)

For $k=1$ every input gives norm equality; for $k>n$ the product is zero.

Proof. 

Let M be the maximum product norm on unit inputs in the degree-one real subspace. Pair an extremizing product with its unit output direction v. The resulting scalar form $B(u_1,\dots ,u_k)={〈u_1\dots u_k,v〉}_{\Omega }$ is symmetric and has maximum absolute value M. By Lemma 3 this maximum occurs at a diagonal tuple. Lemma 4 computes all diagonal maxima, proving the bound. Explicitly, the product at a maximizing unit tuple is nonzero, since an equal-magnitude diagonal tuple already has positive product norm. For its unit direction v, the scalar maximum is at least M and at most M by Cauchy–Schwarz. Its diagonal maximizer w satisfies

$$M=|{〈w^k,v〉}_{\Omega }|\le {∥w^k∥}_{\Omega }\le k!\sqrt{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)/n^k}.$$

The equal-magnitude diagonal example attains the right side, proving the exact value of M.

For the equality classification assume $n\ge 3$. Replacing input coordinates by their absolute values cannot decrease any absolute output coordinate. Hence an extremizer gives a nonnegative extremizing tuple. Normalize its inputs, and let v be their nonnegative unit product direction. The scalar-form argument gives a unit w with $|{〈w^k,v〉}_{\Omega }|=M$ and ${∥w^k∥}_{\Omega }\le M$. Cauchy–Schwarz forces $w^k=\delta Mv$. The power equality condition makes the singleton coefficients of w equal in absolute value. Consequently v is the normalized sum of all degree-k path vectors in orthonormal coordinates.

Fix this v and the scalar form B. In nonnegative orthonormal singleton coordinates, repeatedly select a pair $u,v_0$ of maximum distance in an extremizing unit tuple and replace both by $(u+v_0)/\parallel u+v_0\parallel$. The sum is nonzero. With other slots fixed, the matrix calculation of Lemma 3, now with positive extremal value M, shows that the new tuple still has value M. Since all other slots are nonnegative, the squared norm of the input sum increases by at least

$$4-{\parallel u+v_0\parallel }^2={\parallel u-v_0\parallel }^2.$$

That squared norm is at most $k^2$, so the largest pairwise distance has limit zero. Compactness shows that every limit tuple is diagonal and extremizing. The unique nonnegative diagonal extremizer is $w_0=(1,\dots ,1)/\sqrt{n}$, by Lemma 4. The entire sequence therefore approaches $(w_0,\dots ,w_0)$.

At this tuple, fixing any $k-2$ slots gives a symmetric matrix $C_0$ with zero diagonal and one positive value at every off-diagonal entry. Since $C_0{w}_0=M{w}_0$, each off-diagonal entry is $M/(n-1)$, and

$$C_{0}u={\displaystyle \frac{M}{n-1}}\left(\left({\displaystyle \sum _{j=1}^n}{u}_j\right)(1,\dots ,1)-u\right).$$

Thus $C_0{w}_0=M{w}_0$, whereas $C_{0}u=-Mu/(n-1)$ when ${\sum }_j{u}_j=0$. For $n\ge 3$ its least eigenvalue is strictly greater than $-M$. By continuity this remains true when the fixed slots are sufficiently close to $w_0$. At an extremizing pair,

$$C{v}_0=Mu,Cu=M{v}_0,C(u-v_0)=-M(u-v_0),$$

which in this neighborhood forces $u=v_0$. There are only finitely many pairs. Thus after finitely many averaging steps all slots are equal and the tuple is exactly $(w_0,\dots ,w_0)$. Reverse the steps: if a replacement produced that tuple, the unchanged slots are $w_0$, so the same spectral argument forces the original two slots to be equal already. They must have been $w_0,w_0$. Induction backwards proves that the original nonnegative extremizer was flat in every slot.

Return to signed inputs. Their absolute values are now strictly positive and flat. Equality under taking absolute values requires all permutation contributions to each output support to have the same sign. Given two rows and two distinct positions, complete them by any $k-2$ other distinct positions and compare the assignments interchanging only those two rows. Their two sign products are equal. Writing ${\sigma }_{ij}=sgn\left(a_{ij}\right)$, this gives

$${\sigma }_{ij}{\sigma }_{\ell r}={\sigma }_{ir}{\sigma }_{\ell j},{\sigma }_{ij}={\sigma }_{i1}{\sigma }_{1j}{\sigma }_{11}.$$

The second identity follows by setting $\ell =r=1$; coincident row or column indices give it directly. Each row is therefore the same sign vector times an overall row sign. Absorb that sign and amplitude into ${\lambda }_i$ to obtain (68).

For $n=k=2$, direct expansion gives

$${∥x_1∥}_{\Omega }^2{∥x_2∥}_{\Omega }^2-{∥x_1{x}_2∥}_{\Omega }^2={\Omega }^4{(a_{11}{a}_{21}-a_{12}{a}_{22})}^2.$$

Finally, taking all $x_i={\sum }_{j=1}^n{\mathbf{b}}_j$ proves sharpness over R. □

For $k=n$, the coefficient of $x_1\dots x_n$ at $\left[n\right]$ is the permanent of the matrix $\left(a_{ij}\right)$. This case of Theorem 16 is therefore exactly the real form of the sharp permanent inequality of Carlen, Lieb, and Loss [25]. For $k<n$, the product norm combines the coefficients of all supports of size k.

Example 3 

(A recursive original-ring extremizer). The elements $s_n\in {\mathcal{I}}_n$ defined by $s_0=0$ and $s_{n+1}=(s_n,1)$ satisfy $s_n={\sum }_{j=1}^n{\mathbf{b}}_j$. Every input equal to $s_n$ attains (67). For example, ${∥s_3∥}_{\Omega }=\sqrt{3}\Omega$ and ${∥s_3^3∥}_{\Omega }=6{\Omega }^3$, so the three-input ratio is $2/\sqrt{3}$.

Corollary 3 

(The highest possible nilpotent power). For $u\in {\mathcal{N}}_n$ and $n\ge 1$,

$${∥u^n∥}_{\Omega }\le {\displaystyle \frac{n!}{n^{n/2}}}{∥u∥}_{\Omega }^n.$$

(70)

For $u\ne 0$, equality holds precisely when u has only singleton coefficients and their absolute values are all equal. The bound is sharp over R.

Proof. 

Write $u={\sum }_{S\ne ⌀}{a}_S{\mathbf{b}}_S$. A nonzero contribution to $u^n$ must use n disjoint nonempty supports among n positions, necessarily all singletons. Hence

$${∥u^n∥}_{\Omega }=n!{\displaystyle \prod _{j=1}^n}\left|\Omega a_{\left\{j\right\}}\right|\le n!{\left({\displaystyle \frac{{\sum }_j{\Omega }^2{a}_{\left\{j\right\}}^2}{n}}\right)}^{n/2}\le {\displaystyle \frac{n!}{n^{n/2}}}{∥u∥}_{\Omega }^n.$$

Equality in the arithmetic–geometric mean inequality gives equal singleton magnitudes; equality in the last step excludes every higher-degree coefficient. Indeed, path orthogonality gives

$${∥u∥}_{\Omega }^2={\Omega }^2{\displaystyle \sum _{j=1}^n}{a}_{\left\{j\right\}}^2+\sum _{\begin{array}{c}S\subseteq \left[n\right]\\ \left|S\right|\ge 2\end{array}}{\Omega }^{2\left|S\right|}{a}_S^2.$$

The second sum consists of nonnegative terms with positive weights. It vanishes exactly when all higher-degree coefficients vanish. For a nonzero equality input the common singleton magnitude is positive, so the arithmetic–geometric mean equality is also sufficient. Example 3 gives sharpness. □

Proposition 6 

(Exact overlap of degree-one inputs). For $x={\sum }_j{a}_j{\mathbf{b}}_j$ and $y={\sum }_j{c}_j{\mathbf{b}}_j$,

$${∥xy∥}_{\Omega }^2={∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2+{〈x,y〉}_{\Omega }^2-2{\Omega }^4\sum _j{a}_j^2{c}_j^2.$$

(71)

Let D be the set of common nonzero positions, let $r=\left|D\right|$, and let $x_D,y_D$ retain just those coordinates. If $r=0$, then ${∥xy∥}_{\Omega }={∥x∥}_{\Omega }{∥y∥}_{\Omega }$. If $D=\left\{j\right\}$, then ${∥xy∥}_{\Omega }^2={∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-{\Omega }^4{a}_j^2{c}_j^2$. For $r\ge 2$ there is the sharp estimate

$${∥xy∥}_{\Omega }^2\le {∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2+\left(1-{\displaystyle \frac{2}{r}}\right){∥x_D∥}_{\Omega }^2{∥y_D∥}_{\Omega }^2.$$

(72)

For $r\ge 3$, equality means that the restricted vectors are proportional and each has constant coordinate magnitude. For $r=2$, equality means that the two products $a_j{c}_j$ are equal. There is no further restriction outside D, and all these equality cases occur over R.

Proof. 

Expand the squares of the coefficients $a_i{c}_j+a_j{c}_i$ at ${\mathbf{b}}_{\{i,j\}}$:

$$\begin{array}{cc}\hfill {\Omega }^{-4}{∥xy∥}_{\Omega }^2& =\sum _{i<j}{(a_i{c}_j+a_j{c}_i)}^2\hfill \\ \hfill & =\sum _{i\ne j}{a}_i^2{c}_j^2+2\sum _{i<j}{a}_i{c}_i{a}_j{c}_j\hfill \\ \hfill & =\left(\sum _i{a}_i^2\right)\left(\sum _i{c}_i^2\right)+{\left(\sum _i{a}_i{c}_i\right)}^2-2\sum _i{a}_i^2{c}_i^2.\hfill \end{array}$$

Multiplication by ${\Omega }^4$ gives the stated identity. Put $s={\sum }_D{a}_j{c}_j$ and $q={\sum }_D{a}_j^2{c}_j^2$. Two applications of Cauchy–Schwarz give

$$s^2-2q\le \left(1-{\displaystyle \frac{2}{r}}\right)s^2\le \left(1-{\displaystyle \frac{2}{r}}\right)\left(\sum _D{a}_j^2\right)\left(\sum _D{c}_j^2\right).$$

For the first step the exact scalar deficit is

$$rq-s^2=\sum _{\begin{array}{c}i,j\in D\\ i<j\end{array}}{(a_i{c}_i-a_j{c}_j)}^2.$$

It vanishes precisely when every coordinate product on D is equal. For the second step the coefficient $1-2/r$ is positive when $r\ge 3$ and zero when $r=2$. Thus, for $r\ge 3$, equality additionally requires equality in the scalar Cauchy–Schwarz inequality for ${\left(a_j\right)}_{j\in D}$ and ${\left(c_j\right)}_{j\in D}$. For $r\ge 3$, equality requires proportionality of the restricted vectors and equality of all their coordinate products. Since none is zero on D, this is precisely flat proportionality. For $r=2$, only the first inequality is relevant, and its equality condition is equality of the two products. All contributions outside D remained exact throughout. Equal integral amplitudes give original-ring sharpness. □

3.4. Uniform Homogeneous Bounds and Completed Multiplication

Theorem 17 

(Uniform bounds at prescribed homogeneous degrees). Suppose each $x_i$ is homogeneous of path degree $p_i\ge 0$, and put $P=p_1+\dots +p_k$. There is the depth-independent bound

$${∥x_1\dots x_k∥}_{\Omega }\le \sqrt{{\displaystyle \frac{P!}{p_1!\dots p_k!}}}\prod _i{∥x_i∥}_{\Omega }.$$

(73)

Its constant is optimal over all finite depths, both over R and over $\mathbb{R}$. If at least two of the $p_i$ are positive, no nonzero finite tuple attains it. If at most one is positive, the formula is an equality for every tuple. On the closed homogeneous subspaces of ${\mathcal{H}}_{\Omega }$, ordinary multiplication has a unique bounded multilinear extension with the same norm.

Proof. 

The coefficient at an output support S of size P is

$$\sum _{\begin{array}{c}{S}_1\bigsqcup \dots \bigsqcup S_k=S\\ |S_i|=p_i\end{array}}\prod _i{\left(x_i\right)}_{S_i}.$$

There are exactly $C=P!/(p_1!\dots p_k!)$ ordered partitions in this sum. Each input tuple has total weight ${\Omega }^{2P}$. Cauchy–Schwarz on the partition sum, followed by summation, gives

$$\begin{array}{cc}\hfill {∥x_1\dots x_k∥}_{\Omega }^2& \le C{\Omega }^{2P}\sum _{\begin{array}{c}{S}_1,\dots ,S_k\mathrm{pairwise}\mathrm{disjoint}\\ |S_i|=p_i\end{array}}\prod _i{\left(x_i\right)}_{S_i}^2\hfill \\ \hfill & \le C\sum _{\begin{array}{c}{S}_1,\dots ,S_k\\ |S_i|=p_i\end{array}}\prod _i{\Omega }^{2p_i}{\left(x_i\right)}_{S_i}^2\hfill \\ \hfill & =C\prod _i\left(\sum _{|S_i|=p_i}{\Omega }^{2p_i}{\left(x_i\right)}_{S_i}^2\right)=C\prod _i{∥x_i∥}_{\Omega }^2.\hfill \end{array}$$

For sharpness, at depth $n\ge P$ take original-ring inputs

$$x_i=\sum _{\begin{array}{c}S\subseteq \left[n\right]\\ \left|S\right|=p_i\end{array}}{\mathbf{b}}_S.$$

Their squared norm ratio is

$${\left({\displaystyle \frac{P!}{p_1!\dots p_k!}}\right)}^2{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{P}\right)}{{\prod }_i\left(\genfrac{}{}{0pt}{}{n}{p_i}\right)}},$$

whose limit as n increases is $P!/(p_1!\dots p_k!)$. Indeed, for each fixed integer r,

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)={\displaystyle \frac{n^r}{r!}}{\displaystyle \prod _{j=0}^{r-1}}\left(1-{\displaystyle \frac{j}{n}}\right).$$

The powers of n cancel because ${\sum }_i{p}_i=P$, and each remaining finite product has limit one. These integral tuples therefore establish the asserted original-ring supremum. If a nonzero tuple attained equality when the partition number exceeds one, dropping disjointness could lose no nonzero term. Every nonzero support from one input would then be disjoint from every nonzero support of each other input. Thus the unions of the supports used by distinct inputs are disjoint. For an output support with nonzero coefficient, at most one ordered partition term could be nonzero. The Cauchy–Schwarz step, with more than one term, would be strict, a contradiction. With at most one positive degree, all other inputs are scalars and norm factorization is exact.

Finite elements are dense in each homogeneous subspace. To see independence of approximation, use the telescoping identity

$${\displaystyle \prod _{i=1}^k}{x}_i-{\displaystyle \prod _{i=1}^k}{y}_i={\displaystyle \sum _{j=1}^k}\left(\prod _{i<j}{y}_i\right)(x_j-y_j)\left(\prod _{i>j}{x}_i\right).$$

When $x_i,y_i$ have degree $p_i$, the bound already proved gives

$${∥\prod _i{x}_i-\prod _i{y}_i∥}_{\Omega }\le \sqrt{C}{\displaystyle \sum _{j=1}^k}{∥x_j-y_j∥}_{\Omega }\prod _{i<j}{∥y_i∥}_{\Omega }\prod _{i>j}{∥x_i∥}_{\Omega }.$$

For Cauchy approximations the factors are bounded and each difference has limit zero. This gives the unique bounded multilinear extension; its coefficient formula is the same finite partition sum. Its norm equals the finite-input supremum already computed. □

Proposition 7 

(Disjoint positions). If the positions used by x and y are disjoint, then

$${∥xy∥}_{\Omega }={∥x∥}_{\Omega }{∥y∥}_{\Omega }.$$

(74)

This holds for finite elements and for square-summable coefficient families supported on disjoint sets of positions.

Proof. 

Let $E,F$ be the disjoint position sets supporting the two inputs. Every output support has the unique decomposition $S=(S\cap E)\bigsqcup (S\cap F)$. Thus

$$\begin{array}{cc}\hfill {∥xy∥}_{\Omega }^2& =\sum _{A\subseteq E}\sum _{B\subseteq F}{\Omega }^{2\left(\right|A|+|B\left|\right)}{x}_A^2{y}_B^2\hfill \\ \hfill & =\left(\sum _{A\subseteq E}{\Omega }^{2\left|A\right|}{x}_A^2\right)\left(\sum _{B\subseteq F}{\Omega }^{2\left|B\right|}{y}_B^2\right)={∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2.\hfill \end{array}$$

For infinite position sets the sums range over finite subsets. All summands are nonnegative, so the same factorization follows by increasing finite partial sums. Their finite product proves that the coefficient product belongs to the completion. □

Theorem 18 

(The closed coefficient product). For $x,y\in {\mathcal{H}}_{\Omega }$, prescribe

$${\left(xy\right)}_S=\sum _{A\bigsqcup B=S}{x}_A{y}_B.$$

(75)

The product is defined in ${\mathcal{H}}_{\Omega }$ precisely when this coefficient family is square-summable with the weights ${\Omega }^{2\left|S\right|}$. This partial product has a closed graph. Its graph is exactly the closure of the graph of multiplication on finite real elements.

For fixed $x\in {\mathcal{H}}_{\Omega }$, the maximal coefficient multiplier $M_x\left(y\right)=xy$ is a densely defined closed linear operator. Let $P_n$ retain the coefficients with support in $\left[n\right]$. With the finite-level norms of Proposition 5, $M_x$ is bounded and defined everywhere precisely when

$$\underset{n}{\sup }{\mu }_n\left(P_{n}x\right)<\infty .$$

(76)

Its operator norm then equals this supremum. For a finite element $x\in {\mathcal{I}}_m$, the completed multiplier is bounded with exact norm ${\mu }_m\left(x\right)$.

Proof. 

Each sum in (75) is finite. Coordinate evaluation is continuous, since $|x_S|\le {\Omega }^{-\left|S\right|}{∥x∥}_{\Omega }$. Suppose $x_l,y_l,x_l{y}_l$ converge in norm to $x,y,z$. Taking limits in each coefficient sum gives $z_S={\sum }_{A\bigsqcup B=S}{x}_A{y}_B$. The pair $(x,y)$ therefore belongs to the stated domain and has product z, proving closedness. Conversely, for a pair in that domain,

$$\left(P_{n}x\right)\left(P_{n}y\right)=P_n\left(xy\right).$$

All three truncations converge in norm, so every graph point is a limit of finite-product graph points.

For fixed x, the coefficient formula is linear in y, and closedness follows from the same argument. Every path belongs to its domain, because

$${∥x{\mathbf{b}}_A∥}_{\Omega }^2={\Omega }^{2\left|A\right|}\sum _{B\cap A=⌀}{\Omega }^{2\left|B\right|}{x}_B^2\le {\Omega }^{2\left|A\right|}{∥x∥}_{\Omega }^2.$$

All finite elements therefore lie in its domain, proving density.

Suppose the supremum in (76) is $C<\infty$. For arbitrary $y\in {\mathcal{H}}_{\Omega }$, the finite sums of weighted squared product coefficients satisfy

$${∥P_n\left(xy\right)∥}_{\Omega }={∥\left(P_{n}x\right)\left(P_{n}y\right)∥}_{\Omega }\le C{∥P_{n}y∥}_{\Omega }\le C{∥y∥}_{\Omega }.$$

These nonnegative sums exhaust all coefficients, so the product lies in ${\mathcal{H}}_{\Omega }$ and has norm at most $C{∥y∥}_{\Omega }$. Conversely, a bounded full multiplier compresses to multiplication by $P_{n}x$ at level n. Thus ${\mu }_n\left(P_{n}x\right)\le \parallel M_x\parallel$, proving the criterion and the exact norm.

If x uses only the first m positions, write a finite input as

$$y=\sum _{B\subseteq \{m+1,\dots ,n\}}{y}^{\left[B\right]}{\mathbf{b}}_B,y^{\left[B\right]}=\sum _{A\subseteq \left[m\right]}{y}_{A\cup B}{\mathbf{b}}_A.$$

Different B give orthogonal components, and multiplication by x does not change B. Hence

$$\begin{array}{cc}\hfill {∥xy∥}_{\Omega }^2& =\sum _B{\Omega }^{2\left|B\right|}{∥x{y}^{\left[B\right]}∥}_{\Omega }^2\hfill \\ \hfill & \le {\mu }_m{\left(x\right)}^2\sum _B{\Omega }^{2\left|B\right|}{∥y^{\left[B\right]}∥}_{\Omega }^2={\mu }_m{\left(x\right)}^2{∥y∥}_{\Omega }^2.\hfill \end{array}$$

Density extends this bound to the completion. Inputs supported in the first m positions attain its real operator norm, so this value is exact on the completion. □

Example 4 

(A square outside the completion). Choose disjoint sets $B_j$ with $|B_j|=10j$, and distinct additional positions $p_j,q_j$ disjoint from all these sets and from each other. Put

$$u_j=\prod _{r\in B_j}\left(\sqrt{{\displaystyle \frac{2}{3}}}+\sqrt{{\displaystyle \frac{1}{3}}}{\displaystyle \frac{{\mathbf{b}}_r}{\Omega }}\right),v_j={\displaystyle \frac{{\mathbf{b}}_{p_j}+{\mathbf{b}}_{q_j}}{\sqrt{2}\Omega }}{u}_j,x=\sum _{j\ge 1}{2}^{-j}{v}_j.$$

The $v_j$ are orthogonal and have norm one, so ${∥x∥}_{\Omega }^2={\sum }_{j\ge 1}{4}^{-j}=1/3$. Direct multiplication gives

$$v_j^2={\displaystyle \frac{{\mathbf{b}}_{\{p_j,q_j\}}}{{\Omega }^2}}{u}_j^2,{∥v_j^2∥}_{\Omega }={\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^{10j}.$$

The terms with both markers $p_j,q_j$ are orthogonal to every other same-block square and every cross-block product. The weighted squared coefficient sum of the formal square is therefore at least

$$\sum _{j\ge 1}{2}^{-4j}{∥v_j^2∥}_{\Omega }^2=\sum _{j\ge 1}{\left({\displaystyle \frac{{(4/3)}^{10}}{16}}\right)}^j=\infty .$$

Thus $x\in {\mathcal{H}}_{\Omega }$ but its coefficient square does not belong to that space. The closed product in Theorem 18 is genuinely partial.

4. Directed Tensors and Recursive Block Factorization

Ordinary multiplication combines path supports within a common level; a recursive tensor instead places the two inputs in consecutive blocks. This separation explains the exact norm factorization, in contrast to the depth-dependent multiplication bounds of Section 3. The order and lengths of the blocks remain essential when comparing inner products, refining inputs, or distributing over sums of different depths. Once these compatibility conditions are established, path coefficients give tests for tensor decomposability and commuting blocks. For units, the recursive logarithm turns separation into a condition on supports and determines the finest consecutive factorization.

4.1. Recursive Grafting and Fixed-Block Structure

The two tensor operations read the same scalar data in two orders: the left tensor first reads the components of its first input, whereas the right tensor first reads those of its second input.

Definition 6 

(Left and right recursive tensors). For $x\in {\mathcal{I}}_m$ and $y\in {\mathcal{I}}_n$, both $x\otimes y$ and $x{\otimes }^{\mathrm{op}}y$ belong to ${\mathcal{I}}_{m+n}$. The left tensor is defined by

$$\begin{array}{cc}\hfill s\otimes y& =sy(s\in R),\hfill \\ \hfill x\otimes y& =\left(pos\left(x\right)\otimes y,len\left(x\right)\otimes y\right)(m\ge 1).\hfill \end{array}$$

(77)

The right tensor is defined by

$$\begin{array}{cc}\hfill x{\otimes }^{\mathrm{op}}s& =sx(s\in R),\hfill \\ \hfill x{\otimes }^{\mathrm{op}}y& =\left(x{\otimes }^{\mathrm{op}}pos\left(y\right),x{\otimes }^{\mathrm{op}}len\left(y\right)\right)(n\ge 1).\hfill \end{array}$$

(78)

Each recursion removes one level from the input through which it descends. It therefore terminates at scalar multiplication.

Example 5 

(The two component blocks). For $x=(a,b)$ and $y=(c,d)$ in ${\mathcal{I}}_1$,

$$x\otimes y=\left((ac,ad),(bc,bd)\right),x{\otimes }^{\mathrm{op}}y=\left((ca,cb),(da,db)\right).$$

For a full depth-two calculation, let

$$x=\left((a,b),(c,d)\right),y=\left((r,s),(t,u)\right),$$

where the eight displayed entries belong to R. The first recursive step is

$$x\otimes y=\left((a,b)\otimes y,(c,d)\otimes y\right).$$

Continuing to the scalar layer gives all sixteen output coefficients:

$$x\otimes y=\left(\begin{array}{cc}\hfill & \left(\left((ar,as),(at,au)\right),\left((br,bs),(bt,bu)\right)\right),\hfill \\ \hfill & \left(\left((cr,cs),(ct,cu)\right),\left((dr,ds),(dt,du)\right)\right)\hfill \end{array}\right).$$

The right recursion instead begins with

$$x{\otimes }^{\mathrm{op}}y=\left(x{\otimes }^{\mathrm{op}}(r,s),x{\otimes }^{\mathrm{op}}(t,u)\right),$$

and yields

$$x{\otimes }^{\mathrm{op}}y=\left(\begin{array}{cc}\hfill & \left(\left((ra,rb),(rc,rd)\right),\left((sa,sb),(sc,sd)\right)\right),\hfill \\ \hfill & \left(\left((ta,tb),(tc,td)\right),\left((ua,ub),(uc,ud)\right)\right)\hfill \end{array}\right).$$

In both cases the result has depth four. The difference is the order of the two blocks of component choices.

At fixed depths, the recursive construction realizes the usual algebraic tensor product by an explicit pairing of path bases. Over real scalars, the same identification will carry the Hilbert tensor norm; the general tensor-product and crossnorm constructions are described in [26].

Theorem 19 

(Block paths and tensor algebra laws). For $A\subseteq \left[m\right]$ and $B\subseteq \left[n\right]$, with $m+B=\{m+j:j\in B\}$,

$$\begin{array}{cc}\hfill {\mathbf{b}}_A\otimes {\mathbf{b}}_B& ={\mathbf{b}}_{A\cup (m+B)},\hfill \\ \hfill {\mathbf{b}}_A{\otimes }^{\mathrm{op}}{\mathbf{b}}_B& ={\mathbf{b}}_{B\cup (n+A)}.\hfill \end{array}$$

(79)

Both tensors are R-bilinear on each fixed pair of input levels, and

$$\begin{array}{cc}\hfill x{\otimes }^{\mathrm{op}}y& =y\otimes x,\hfill \\ \hfill (x\otimes y)\otimes z& =x\otimes (y\otimes z),\hfill \\ \hfill \left(x{\otimes }^{\mathrm{op}}y\right){\otimes }^{\mathrm{op}}z& =x{\otimes }^{\mathrm{op}}\left(y{\otimes }^{\mathrm{op}}z\right),\hfill \\ \hfill 1_0\otimes x& =x=x\otimes 1_0.\hfill \end{array}$$

(80)

At every fixed pair of depths there is a unital R-algebra isomorphism

$${\mathcal{I}}_m{\otimes }_R{\mathcal{I}}_n\cong {\mathcal{I}}_{m+n},$$

(81)

under which the elementary module tensor $x{\otimes }_{R}y$ corresponds to $x\otimes y$. Here the algebra structure on the module tensor product is given by

$$\left(x{\otimes }_{R}y\right)\left(u{\otimes }_{R}v\right)=xu{\otimes }_{R}yv,1_{{\mathcal{I}}_m{\otimes }_R{\mathcal{I}}_n}=1_m{\otimes }_R{1}_n.$$

(82)

In particular, for $x,u\in {\mathcal{I}}_m$ and $y,v\in {\mathcal{I}}_n$,

$$(x\otimes y)(u\otimes v)=\left(xu\right)\otimes \left(yv\right).$$

(83)

The same identity holds with every tensor replaced by the right tensor.

Proof. 

The path formulas follow by applying the defining recursion once at each position: the first m choices come from the first input of the left tensor, and the last n come from its second input. Scalar coefficients multiply at the last step. Consequently, for arbitrary inputs at the fixed depths,

$$x\otimes y=\sum _{A\subseteq \left[m\right]}\sum _{B\subseteq \left[n\right]}{x}_A{y}_B{\mathbf{b}}_{A\cup (m+B)}.$$

Every output support S has the unique decomposition $A=S\cap \left[m\right]$ and $B=\{j\in [n]:m+j\in S\}$. The displayed coefficient is linear in each input list, proving fixed-level bilinearity.

The opposite identity has a direct recursive proof. When y is scalar, both sides are its scalar multiple of x. For positive-depth y, the two components of the right tensor, by lower-depth induction, are $pos\left(y\right)\otimes x$ and $len\left(y\right)\otimes x$. Their pair is $y\otimes x$. For left associativity, descend through x. The scalar case follows from scalar homogeneity. At positive depth the induction hypothesis gives

$$\begin{array}{cc}\hfill pos\left((x\otimes y)\otimes z\right)& =\left(pos\left(x\right)\otimes y\right)\otimes z\hfill \\ \hfill & =pos\left(x\right)\otimes (y\otimes z)=pos\left(x\otimes (y\otimes z)\right),\hfill \\ \hfill len\left((x\otimes y)\otimes z\right)& =\left(len\left(x\right)\otimes y\right)\otimes z\hfill \\ \hfill & =len\left(x\right)\otimes (y\otimes z)=len\left(x\otimes (y\otimes z)\right).\hfill \end{array}$$

For the right orientation, left associativity gives the explicit chain

$$\begin{array}{cc}\hfill \left(x{\otimes }^{\mathrm{op}}y\right){\otimes }^{\mathrm{op}}z& =z\otimes (y\otimes x)\hfill \\ \hfill & =(z\otimes y)\otimes x=x{\otimes }^{\mathrm{op}}\left(y{\otimes }^{\mathrm{op}}z\right).\hfill \end{array}$$

The unit formulas and the further scalar boundary $x\otimes s=sx$ follow from the same recursion.

The module tensors of the path bases form a basis of ${\mathcal{I}}_m{\otimes }_R{\mathcal{I}}_n$. Their images in (79) are exactly the path basis of ${\mathcal{I}}_{m+n}$, each occurring once. This proves the module isomorphism. Multiplying two such paths gives zero exactly when their supports intersect in either block; otherwise it gives their union. For supports $A,C\subseteq \left[m\right]$ and $B,D\subseteq \left[n\right]$, their output supports have intersection

$$\left(A\cup (m+B)\right)\cap \left(C\cup (m+D)\right)=(A\cap C)\cup \left(m+(B\cap D)\right).$$

Thus both sides of interchange vanish if either block intersection is nonempty. Otherwise both sides are the path ${\mathbf{b}}_{(A\cup C)\cup (m+(B\cup D\left)\right)}$. Bilinear expansion in all four inputs proves (83). The empty supports give $1_m\otimes 1_n=1_{m+n}$, so the module isomorphism also preserves the algebra unit. Equivalently, a recursive proof of (83) first multiplies the two pos components, then adds the two products containing exactly one len component. Lower-depth interchange combines these as $pos\left(xu\right)\otimes \left(yv\right)$ and $len\left(xu\right)\otimes \left(yv\right)$. Exchanging the inputs proves the right-oriented statement. □

The prescribed levels in (83) determine its common block boundary. For example, with $x=1_0$, $u=1_1$, $y={\mathbf{b}}_{\left\{1\right\}}\in {\mathcal{I}}_1$, and $v=1_0$, its left side is ${\mathbf{b}}_{\left\{1\right\}}$, whereas its right side is ${\mathbf{b}}_{\left\{2\right\}}$. These remain different after comparison at depth two.

Proposition 8 

(Zero factors and unit blocks). For $x\in {\mathcal{I}}_m$ and $n\ge 0$, zero factors satisfy

$$\begin{array}{cc}\hfill x\otimes 0_n=0_n\otimes x& =0_{m+n},\hfill \\ \hfill x{\otimes }^{\mathrm{op}}{0}_n=0_n{\otimes }^{\mathrm{op}}x& =0_{m+n}.\hfill \end{array}$$

The unit blocks satisfy

$$\begin{array}{cc}\hfill 1_m\otimes 1_n=1_m{\otimes }^{\mathrm{op}}{1}_n& =1_{m+n},\hfill \\ \hfill x\otimes 1_n=1_n{\otimes }^{\mathrm{op}}x& ={\iota }^{\left(n\right)}\left(x\right),\hfill \\ \hfill 1_{n+1}\otimes x=x{\otimes }^{\mathrm{op}}{1}_{n+1}& =\left(1_n\otimes x,0\right).\hfill \end{array}$$

(84)

For $n\ge 1$, the global equality $[1_n\otimes x]=\left[x\right]$ holds exactly when $\left[x\right]$ is a scalar. The same criterion holds for $\left[x{\otimes }^{\mathrm{op}}{1}_n\right]=\left[x\right]$.

Proof. 

At the scalar boundary a zero factor gives zero. Each further recursive step pairs two lower-depth zeros, giving $0_{m+n}$. For a scalar c, one has $c\otimes (1,0)=(c,0)=\iota \left(c\right)$. At positive depth, induction gives

$$\begin{array}{cc}\hfill x\otimes (1,0)& =\left(pos\left(x\right)\otimes (1,0),len\left(x\right)\otimes (1,0)\right)\hfill \\ \hfill & =\left(\iota \left(pos\left(x\right)\right),\iota \left(len\left(x\right)\right)\right)=\iota \left(x\right),\hfill \\ \hfill 1_{n+1}\otimes x& =\left(1_n\otimes x,0_n\otimes x\right)=\left(1_n\otimes x,0_{m+n}\right).\hfill \end{array}$$

Associativity and $1_n=1_1^{\otimes n}$ give

$$x\otimes 1_n={\iota }^{\left(n\right)}\left(x\right),1_m\otimes 1_n={\iota }^{\left(n\right)}\left(1_m\right)=1_{m+n}.$$

Exchanging the two inputs gives every right-oriented identity in (84).

By the path formula, $1_n\otimes x$ moves each position j used by x to position $n+j$. For a nonscalar $\left[x\right]$, let r be the largest position in a nonzero nonempty path coefficient. Its minimal depth is r, whereas that of $[1_n\otimes x]$ is $n+r$. Thus the two classes differ. If $\left[x\right]=c$ is scalar, then $x=c{1}_m$ and $1_n\otimes x=c{1}_{m+n}$, whose class is c. The opposite tensor identity gives the right-oriented criterion. □

At specified input levels, the zero-factor identities may be written $x\otimes 0=0=0\otimes x$, and likewise for the right tensor. The scalar $1_0$ is the two-sided tensor identity; positive-depth $1_n$ specifies a block length in (84).

Proposition 9 

(Minimal depth of a nonzero tensor). Let $x\in {\mathcal{I}}_m$ and $y\in {\mathcal{I}}_n$ be nonzero minimal representatives of their global classes. Both $[x\otimes y]$ and $\left[x{\otimes }^{\mathrm{op}}y\right]$ have minimal depth $m+n$.

Proof. 

An element of positive depth n has a representative at depth $n-1$ exactly when all path coefficients whose support contains n vanish. This follows from the refinement description in Proposition 1. If $n>0$, choose a nonzero coefficient of y whose support contains n, and any nonzero coefficient of x. If these supports are $B\subseteq \left[n\right]$ and $A\subseteq \left[m\right]$, the coefficient is

$${(x\otimes y)}_{A\cup (m+B)}=x_A{y}_B\ne 0,m+n\in A\cup (m+B).$$

The block decomposition is unique, so this coefficient has no other contributions which could cancel it. It is nonzero because R is a domain. Thus $x\otimes y$ has minimal depth $m+n$. If $n=0$, multiplication by a nonzero scalar preserves the nonzero last-position coefficients of x. Interchanging the inputs proves the assertion for the right tensor. □

Theorem 20 

(Fixed-block metric factorization). For $x_1,x_2\in {\mathcal{I}}_m$ and $y_1,y_2\in {\mathcal{I}}_n$,

$$\begin{array}{cc}\hfill {〈x_1\otimes y_1,x_2\otimes y_2〉}_{\Omega }& ={〈x_1,x_2〉}_{\Omega }{〈y_1,y_2〉}_{\Omega },\hfill \\ \hfill {〈x_1{\otimes }^{\mathrm{op}}{y}_1,x_2{\otimes }^{\mathrm{op}}{y}_2〉}_{\Omega }& ={〈x_1,x_2〉}_{\Omega }{〈y_1,y_2〉}_{\Omega }.\hfill \end{array}$$

(85)

Consequently, at arbitrary prescribed depths,

$${∥x\otimes y∥}_{\Omega }={∥x∥}_{\Omega }{∥y∥}_{\Omega }={∥x{\otimes }^{\mathrm{op}}y∥}_{\Omega }.$$

(86)

For any finite group of pairs in the same two levels,

$${∥\sum _i{x}_i\otimes y_i∥}_{\Omega }={∥\sum _i{x}_i{\otimes }^{\mathrm{op}}{y}_i∥}_{\Omega }.$$

(87)

The analogous equality holds when the order of any fixed collection of input blocks is reversed or permuted, provided that the same permutation is applied to every summand.

Proof. 

Induct on the common first-input depth m. At $m=0$, scalar homogeneity gives

$${〈x_1\otimes y_1,x_2\otimes y_2〉}_{\Omega }=x_1{x}_2{〈y_1,y_2〉}_{\Omega }={〈x_1,x_2〉}_{\Omega }{〈y_1,y_2〉}_{\Omega }.$$

For $m\ge 1$, the two tensor outputs have the same depth $m+n$. Their component recursions and the induction hypothesis give

$$\begin{array}{cc}\hfill {〈x_1\otimes y_1,x_2\otimes y_2〉}_{\Omega }& ={〈pos\left(x_1\right)\otimes y_1,pos\left(x_2\right)\otimes y_2〉}_{\Omega }\hfill \\ \hfill & +{\Omega }^2{〈len\left(x_1\right)\otimes y_1,len\left(x_2\right)\otimes y_2〉}_{\Omega }\hfill \\ \hfill & ={〈pos\left(x_1\right),pos\left(x_2\right)〉}_{\Omega }{〈y_1,y_2〉}_{\Omega }\hfill \\ \hfill & +{\Omega }^2{〈len\left(x_1\right),len\left(x_2\right)〉}_{\Omega }{〈y_1,y_2〉}_{\Omega }\hfill \\ \hfill & ={〈x_1,x_2〉}_{\Omega }{〈y_1,y_2〉}_{\Omega }.\hfill \end{array}$$

For the right tensor, the same left-oriented result, now with first depth n and second depth m, gives

$$\begin{array}{cc}\hfill {〈x_1{\otimes }^{\mathrm{op}}{y}_1,x_2{\otimes }^{\mathrm{op}}{y}_2〉}_{\Omega }& ={〈y_1\otimes x_1,y_2\otimes x_2〉}_{\Omega }\hfill \\ \hfill & ={〈y_1,y_2〉}_{\Omega }{〈x_1,x_2〉}_{\Omega }.\hfill \end{array}$$

Taking $x_1=x_2=x$ and $y_1=y_2=y$ and then nonnegative square roots proves (86). For a finite sum at the same two depths,

$${∥\sum _i{x}_i\otimes y_i∥}_{\Omega }^2=\sum _{i,j}{〈x_i,x_j〉}_{\Omega }{〈y_i,y_j〉}_{\Omega }={∥\sum _i{x}_i{\otimes }^{\mathrm{op}}{y}_i∥}_{\Omega }^2.$$

More generally, a common permutation of entire blocks permutes the output path coefficients and preserves each path cardinality. It is therefore an isometry, proving the finite-sum assertions over R and over the real scalar extension. In particular it carries every equality example of a fixed-block inequality to an equality example over the same ring. □

Since ${∥1∥}_{\Omega }=1$, insertion of a unit block on either side is isometric:

$${∥x\otimes 1_n∥}_{\Omega }={∥1_n\otimes x∥}_{\Omega }={∥x∥}_{\Omega }.$$

The same holds for the right tensor. By Proposition 8, only insertion on the right for the left tensor, or on the left for the right tensor, always preserves the global object itself.

4.2. Metric Factorization and Refinement

Theorem 21 

(Exact mixed-depth factorization criterion). Fix nonnegative depths $m,n,p,q$. The identity

$${〈x_1\otimes y_1,x_2\otimes y_2〉}_{\Omega }={〈x_1,x_2〉}_{\Omega }{〈y_1,y_2〉}_{\Omega }$$

(88)

holds for every $x_1\in {\mathcal{I}}_m$, $y_1\in {\mathcal{I}}_n$, $x_2\in {\mathcal{I}}_p$, $y_2\in {\mathcal{I}}_q$ precisely under one of the following conditions:

$$m=p;m<p\mathrm{and}n=0;p<m\mathrm{and}q=0.$$

(89)

For the right tensor the corresponding complete list is

$$n=q;n<q\mathrm{and}m=0;q<n\mathrm{and}p=0.$$

(90)

Proof. 

By fixed-level bilinearity it suffices to take paths with supports $A\subseteq \left[m\right]$, $B\subseteq \left[n\right]$, $C\subseteq \left[p\right]$, and $D\subseteq \left[q\right]$. The two output supports are

$$A\cup (m+B),C\cup (p+D).$$

Their pairing is nonzero exactly when these sets are equal. The product of the separate pairings is nonzero exactly when $A=C$ and $B=D$, where empty last positions are understood as in the mixed-depth form. The corresponding values are explicitly

$$\begin{array}{cc}\hfill & {〈{\mathbf{b}}_A\otimes {\mathbf{b}}_B,{\mathbf{b}}_C\otimes {\mathbf{b}}_D〉}_{\Omega }\hfill \\ \hfill & =\left\{\begin{array}{cc}{\Omega }^{2\left(\right|A|+|B\left|\right)},\hfill & A\cup (m+B)=C\cup (p+D),\hfill \\ 0,\hfill & \mathrm{otherwise},\hfill \end{array}\right.\hfill \\ \hfill & {〈{\mathbf{b}}_A,{\mathbf{b}}_C〉}_{\Omega }{〈{\mathbf{b}}_B,{\mathbf{b}}_D〉}_{\Omega }=\left\{\begin{array}{cc}{\Omega }^{2\left(\right|A|+|B\left|\right)},\hfill & A=C\mathrm{and}B=D,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

Thus it suffices to determine when the two stated support conditions agree for every choice of paths.

If $m=p$, the common boundary separates the equality of the two unions into equality of their two parts. If $m<p$ and $n=0$, equality of the unions forces D to be empty and forces C to have no position beyond m; it then says $C=A$. This is exactly the separate pairing condition. The case $p<m$, $q=0$ is symmetric.

Conversely, suppose $m<p$ and $n>0$. Choose

$$A=D=⌀,B=\left\{1\right\},C=\{m+1\}.$$

Both output supports equal $\{m+1\}$, so the left side of (88) is ${\Omega }^2$, while the right side is zero. The symmetric example deals with $p<m$ and $q>0$. This exhausts the excluded cases. Applying the same result after interchanging the two inputs proves (90). □

For a simple failure of four-input factorization, take $x_1=y_2=1_0$ and $y_1=x_2=(0,1)$. Both left tensor outputs are $(0,1)$, so their pairing is ${\Omega }^2$, but both separate pairings are zero.

Example 6 

(Varying block boundaries). Let

$$x_1=1_1,y_1={\mathbf{b}}_{\left\{1\right\}}\in {\mathcal{I}}_1,x_2={\mathbf{b}}_{\left\{2\right\}}\in {\mathcal{I}}_2,y_2=-1_0.$$

Both tensor terms have output depth two, but their input depths are $(1,1)$ and $(2,0)$. Direct calculation gives

$$\begin{array}{cc}\hfill x_1\otimes y_1+x_2\otimes y_2& =0,\hfill \\ \hfill x_1{\otimes }^{\mathrm{op}}{y}_1+x_2{\otimes }^{\mathrm{op}}{y}_2& ={\mathbf{b}}_{\left\{1\right\}}-{\mathbf{b}}_{\left\{2\right\}}.\hfill \end{array}$$

The second sum has norm $\sqrt{2}\Omega$. The ordinary product sum $x_1{y}_1+x_2{y}_2$ is also ${\mathbf{b}}_{\left\{1\right\}}-{\mathbf{b}}_{\left\{2\right\}}$. Thus a common output depth does not suffice either for (87) or for a finite-sum comparison of ordinary products with left tensors. The common input depths in such comparisons are essential.

Proposition 10 

(Refinement and mixed-depth distributivity). Put $v_{pos}=(1,0)$ and $v_{len}=(0,1)$ in ${\mathcal{I}}_1$. Then

$$\begin{array}{cc}\hfill \iota \left(x\right)& =x\otimes v_{pos},\hfill \\ \hfill x\otimes \iota \left(y\right)& =\iota (x\otimes y),\hfill \\ \hfill \iota \left(x\right){\otimes }^{\mathrm{op}}y& =\iota \left(x{\otimes }^{\mathrm{op}}y\right).\hfill \end{array}$$

(91)

The left tensor distributes over mixed-depth addition in its second input. The right tensor distributes over mixed-depth addition in its first input. In the other slot neither refinement compatibility nor mixed-depth distributivity holds in general. Consequently neither orientation defines a two-input operation on global refinement classes.

Proof. 

For a scalar c, $c\otimes (1,0)=(c,0)=\iota \left(c\right)$. At positive depth the two components of $x\otimes (1,0)$ agree recursively with the defining components of $\iota \left(x\right)$. Associativity now gives

$$x\otimes \iota \left(y\right)=x\otimes (y\otimes v_{pos})=(x\otimes y)\otimes v_{pos}=\iota (x\otimes y).$$

The opposite identity proves (91). For $x\in {\mathcal{I}}_m$, $y\in {\mathcal{I}}_n$, $z\in {\mathcal{I}}_p$, put $N=\max (n,p)$. Iterating covariance and using fixed-level bilinearity gives

$$\begin{array}{cc}\hfill x\otimes (y+z)& =x\otimes \left({\iota }^{(N-n)}\left(y\right)+{\iota }^{(N-p)}\left(z\right)\right)\hfill \\ \hfill & ={\iota }^{(N-n)}(x\otimes y)+{\iota }^{(N-p)}(x\otimes z)\hfill \\ \hfill & =x\otimes y+x\otimes z.\hfill \end{array}$$

Both sides belong to ${\mathcal{I}}_{m+N}$, and the last equality is the recursive mixed-depth sum. Exchanging the tensor inputs proves the right-oriented assertion. Explicitly, its distributive law is

$$\begin{array}{cc}\hfill (x+y){\otimes }^{\mathrm{op}}z& =z\otimes (x+y)=z\otimes x+z\otimes y\hfill \\ \hfill & =x{\otimes }^{\mathrm{op}}z+y{\otimes }^{\mathrm{op}}z.\hfill \end{array}$$

The mixed depths here belong to $x,y$; the typed depth of z fixes the unchanged first block of the left-oriented expression.

For the other slot, let $y=v_{len}$. In depth two,

$$\iota \left(1_0\right)\otimes y=v_{pos}\otimes v_{len}={\mathbf{b}}_{\left\{2\right\}},\iota (1_0\otimes y)=v_{len}\otimes v_{pos}={\mathbf{b}}_{\left\{1\right\}}.$$

(92)

The distinct path coefficients remain distinct under further refinements. Explicitly, the two depth-two values are

$$1_1\otimes y=\left((0,1),(0,0)\right),\iota \left(y\right)=\left((0,0),(1,0)\right),$$

and ${∥1_1\otimes y-\iota \left(y\right)∥}_{\Omega }=\sqrt{2}\Omega$. For mixed-depth distributivity, $1_0+0_1=1_1=v_{pos}$, so

$$(1_0+0_1)\otimes y={\mathbf{b}}_{\left\{2\right\}},1_0\otimes y+0_1\otimes y={\mathbf{b}}_{\left\{1\right\}}.$$

The right tensor has the exchanged counterexamples. Since $1_0$ and $\iota \left(1_0\right)$ represent the same global object, the first counterexample also proves the failure of a two-input operation on classes. □

Theorem 22 

(Finite sums: from directed tensors to multiplication). Fix depths $m_1,\dots ,m_k$, put $N={\max }_h{m}_h$, and define $r_j=\#\{h:m_h\ge j\}$ and $q={\prod }_{j=1}^N{r}_j$. For a finite family with $x_{i,h}\in {\mathcal{I}}_{m_h}$,

$${∥\sum _i{\displaystyle \prod _{h=1}^k}{x}_{i,h}∥}_{\Omega }\le \sqrt{q}{∥\sum _i{x}_{i,1}\otimes \dots \otimes x_{i,k}∥}_{\Omega }.$$

(93)

Every tensor factor retains its specified depth and block division. The constant is optimal and is attained over R. For two fixed depths $m,n$ it is $2^{\min (m,n)/2}$.

Write $c_{A_1,\dots ,A_k}$ for the path coefficients of the tensor sum, with $A_h\subseteq \left[m_h\right]$, and put $J=\{j:r_j>1\}$. Equality holds precisely under these three conditions:

1.

Every coefficient with overlapping $A_h$ is zero.

2.

Every coefficient with pairwise disjoint $A_h$ whose union does not contain J is zero.

3.

For each S containing J, all coefficients with $A_1\bigsqcup \dots \bigsqcup A_k=S$ have one common value, which may depend on S.

These conditions include the zero tensor sum. If all inputs are scalar, the empty product defining q is one.

Proof. 

The path-block description identifies the squared norm of the tensor sum with

$$\sum _{A_1,\dots ,A_k}{\Omega }^{2{\sum }_h\left|A_h\right|}{c}_{A_1,\dots ,A_k}^2.$$

This is also the fixed-block metric identity of Theorem 20; no input depth changes here. Ordinary multiplication sends overlapping paths to zero and disjoint paths to their union. Its output coefficient at S is therefore

$$\sum _{A_1\bigsqcup \dots \bigsqcup A_k=S}{c}_{A_1,\dots ,A_k}.$$

Each position $j\in S$ can be assigned to exactly $r_j$ blocks. The number of terms is $q_S={\prod }_{j\in S}{r}_j$. Cauchy–Schwarz gives

$${\left(\sum _{A_1\bigsqcup \dots \bigsqcup A_k=S}{c}_{A_1,\dots ,A_k}\right)}^2\le q_S\sum _{A_1\bigsqcup \dots \bigsqcup A_k=S}{c}_{A_1,\dots ,A_k}^2.$$

Multiply by ${\Omega }^{2\left|S\right|}$ and sum. For disjoint supports this is the original tensor weight. Use $q_S\le q$ and include the nonnegative terms from overlapping supports to prove the bound.

Equality in that last inclusion requires condition (i). Equality in replacing $q_S$ by q excludes all coefficients unless $q_S=q$, which is exactly $J\subseteq S$. Equality in each Cauchy–Schwarz inequality requires all its entries to be equal. These are conditions (ii) and (iii), and together they are also sufficient. To see the equality test directly, list the $q_S$ coefficients at a fixed disjoint output support as $c_1,\dots ,c_{q_S}$. The scalar deficit is

$$q{\displaystyle \sum _{l=1}^{q_S}}{c}_l^2-{\left({\displaystyle \sum _{l=1}^{q_S}}{c}_l\right)}^2=(q-q_S){\displaystyle \sum _{l=1}^{q_S}}{c}_l^2+\sum _{l<h}{(c_l-c_h)}^2.$$

Its two nonnegative terms vanish exactly under conditions (ii) and (iii) at that support. The omitted overlapping coefficients contribute q times their original positive weighted squares, giving condition (i). Thus no cancellation between different supports can create an additional equality case.

For original-ring sharpness, sum over all allowed assignments $A_1\bigsqcup \dots \bigsqcup A_k=\left[N\right]$, using the paths ${\mathbf{b}}_{A_h}$ at depths $m_h$. There are q assignments. The ordinary product sum is $q{\mathbf{b}}_{\left[N\right]}$, of norm $q{\Omega }^N$. The tensor sum consists of q different orthogonal paths, each of norm ${\Omega }^N$, and has norm $\sqrt{q}{\Omega }^N$. Its ratio is $\sqrt{q}$. □

Example 7 

(The two-block equality space). For $m\le n$, all tensor sums attaining the two-input bound have the form

$$\sum _{U\subseteq \{m+1,\dots ,n\}}{d}_U\sum _{A\subseteq \left[m\right]}{\mathbf{b}}_A\otimes {\mathbf{b}}_{\left(\right[m]\setminus A)\cup U},$$

with every factor placed in its prescribed level. For $m=n=1$ and $d_{⌀}=1$,

$${∥(1,0)(0,1)+(0,1)(1,0)∥}_{\Omega }=2\Omega ,$$

$${∥(1,0)\otimes (0,1)+(0,1)\otimes (1,0)∥}_{\Omega }=\sqrt{2}\Omega .$$

Fixed total depth alone does not suffice. Take $x_1=1_1$, $y_1={\mathbf{b}}_1\in {\mathcal{I}}_1$, $x_2={\mathbf{b}}_2\in {\mathcal{I}}_2$, and $y_2=-1_0$. Both tensor terms have total depth two and their sum is zero, whereas $x_1{y}_1+x_2{y}_2={\mathbf{b}}_1-{\mathbf{b}}_2$ has norm $\sqrt{2}\Omega$.

4.3. Decomposability and Functional Block Structure

The two fixed blocks also make tensor decomposability an exact coefficient question over the original ring, without division by arbitrary scalars.

Theorem 23 

(Fixed-block decomposability over R). For $w\in {\mathcal{I}}_{m+n}$, arrange its coefficients as the matrix

$$c_{A,B}=w_{A\cup (m+B)},A\subseteq \left[m\right],B\subseteq \left[n\right].$$

(94)

The element w can be written $x\otimes y$ with $x\in {\mathcal{I}}_m$, $y\in {\mathcal{I}}_n$ exactly when all two-by-two minors of this matrix vanish. If $w\ne 0$, it has a factorization with the coefficients of x generating the unit ideal of R. Such a primitive first factor is unique up to multiplication by a unit, and the second factor changes by the inverse unit.

More generally, if $x\otimes y=x^{\prime }\otimes y^{\prime }\ne 0$ at the same block boundary, there is a nonzero rational number a such that

$$x^{\prime }=ax,y^{\prime }=a^{-1}y.$$

(95)

The rational number is required to keep the displayed factors over R. The nonzero support of a pure tensor consists exactly of all pairs of nonzero input supports in the prescribed two blocks.

Proof. 

A pure tensor has $c_{A,B}=x_A{y}_B$, so its minors vanish. Conversely, the zero matrix is a tensor with a zero factor. For a nonzero matrix whose minors vanish, every column is a rational multiple of any chosen nonzero column: choose a nonzero entry in that column and apply the minor equations to each row. Specifically, if $c_{A_0,B_0}\ne 0$, then

$$c_{A,B}{c}_{A_0,B_0}=c_{A,B_0}{c}_{A_0,B},c_{A,B}={\displaystyle \frac{c_{A_0,B}}{c_{A_0,B_0}}}{c}_{A,B_0}.$$

The ring R is a localization of the principal ideal domain $\mathbb{Z}$, and is itself a principal ideal domain. Divide the chosen column by a generator of its coefficient ideal to obtain a primitive column p over R. If another column is $ap$ with $a\in \mathbb{Q}$, choose R-coefficients whose combination of the entries of p equals 1. The same combination of the entries of $ap$ equals a, proving $a\in R$. In formulas, choose $r_A\in R$ with

$$\sum _A{r}_A{p}_A=1;a=\sum _A{r}_A\left(a{p}_A\right)\in R.$$

These column multipliers give the second factor over R.

For uniqueness over $\mathbb{Q}$, choose $A_0,B_0$ with $x_{A_0}{y}_{B_0}\ne 0$. Equality of the tensor coefficients also gives $x_{A_0}^{\prime }{y}_{B_0}^{\prime }\ne 0$. Set $a=x_{A_0}^{\prime }/x_{A_0}$. Comparing the $A_0$ row and then the $B_0$ column gives

$$\begin{array}{cccc}\hfill x_{A_0}{y}_B& =x_{A_0}^{\prime }{y}_B^{\prime },\hfill & \hfill y_B^{\prime }& =a^{-1}{y}_B,\hfill \\ \hfill x_A{y}_{B_0}& =x_A^{\prime }{y}_{B_0}^{\prime },\hfill & \hfill x_A^{\prime }& =a{x}_A.\hfill \end{array}$$

These coefficient identities are (95). If the first factors are both primitive, their proportionality number and its inverse belong to R by the same coefficient-combination argument, so that number is a unit. Finally, $x_A{y}_B$ is nonzero precisely when both coefficients are nonzero, since R is a domain. This proves the support statement. □

For instance, a depth-two element $\left(\right(a,b),(c,d\left)\right)$ is a tensor of two depth-one elements precisely when $ad=bc$. A rectangular nonzero support alone does not suffice: the four coefficients $(1,1,1,2)$ have that support but fail the determinant condition.

Theorem 24 

(Commuting tensors have a common primitive block). Let $m,n>0$, let $x\in {\mathcal{I}}_m$ and $y\in {\mathcal{I}}_n$ be nonzero, and put $d=\gcd (m,n)$. The identity $x\otimes y=y\otimes x$ holds precisely when there are a primitive $w\in {\mathcal{I}}_d$ and nonzero $a,b\in R$ such that

$$x=a{w}^{\otimes (m/d)},y=b{w}^{\otimes (n/d)}.$$

(96)

The primitive block w is unique up to multiplication by a unit of R. A zero input or an input of depth zero commutes with every other input.

Proof. 

First extend coefficients to $\mathbb{Q}$. The successive block identifications of Theorem 19 identify depth r with the rth module tensor power of the two-dimensional space $V={\mathcal{I}}_1{\otimes }_R\mathbb{Q}$. If $m=n$, choose a linear functional on that common space taking value 1 on y. Calling it ℓ and applying it to the first block gives

$$\ell \left(x\right)y=\ell \left(y\right)x=x.$$

Thus x is a nonzero scalar multiple of y.

Suppose $m>n$. Choose ℓ on the depth-n coefficient space with $\ell \left(y\right)=1$, and contract the first n positions of x to obtain $u=(\ell {\otimes }_{\mathbb{Q}}\mathrm{id})\left(x\right)$ of depth $m-n$. Contracting the first n positions in the commuting equality gives

$$u\otimes y=x.$$

In particular $u\ne 0$. Substitution gives

$$u\otimes y\otimes y=y\otimes u\otimes y.$$

Contracting the final block by ℓ cancels the common y and yields $u\otimes y=y\otimes u$. Induction on $m+n$, with the symmetric step when $n>m$, is the Euclidean algorithm for the two depths. It produces a common block of depth d whose appropriate powers are scalar multiples of x and y. Explicitly, the smaller pair $(u,y)$ has depths $m-n,n$ and greatest common divisor d. If its induction conclusion is

$$u=a{w}^{\otimes \left(\right(m-n)/d)},y=b{w}^{\otimes (n/d)},$$

then $x=u\otimes y=ab{w}^{\otimes (m/d)}$. The sum of the two positive depths decreases at each unequal-depth step. At equal depths the preceding proportionality calculation is the induction base, so the process terminates.

Return to R. Scale the rational common block to a primitive block w over R. For two finite coefficient lists the ideal generated by all their pairwise products is the product of their coefficient ideals. By (79), every tensor power of a primitive block is therefore primitive. If a rational multiple of a primitive vector has all coefficients in R, a coefficient combination equal to 1 shows that the multiple belongs to R. Hence the two scalars in (96) belong to R.

If two primitive blocks work, compare their nonzero tensor powers in the first depth-d block of x. Pure-tensor uniqueness over $\mathbb{Q}$ shows that the blocks are proportional. Their primitiveness makes the proportion a unit of R. Conversely, powers of a common block commute by associativity. The scalar and zero cases follow from the scalar boundary and bilinearity. □

Over a field, Bergman’s centralizer theorem places every commuting family containing a nonscalar element of a free associative algebra inside a polynomial algebra in one element [27]. The homogeneous block calculation above gives a direct common-power description with depth $\gcd (m,n)$ and retains primitive factors over R.

Proposition 11 

(Polynomial and rational evaluation in two blocks). Fix $x\in {\mathcal{I}}_m$, $y\in {\mathcal{I}}_n$. All powers and evaluated values below retain these levels, including $x^0=1_m$ and $y^0=1_n$. For polynomials $f,g\in R\left[t\right]$,

$$f(x\otimes 1_n)g(1_m\otimes y)=f\left(x\right)\otimes g\left(y\right).$$

(97)

If $F(s,t)={\sum }_{p,q}{a}_{pq}{s}^p{t}^q$, then

$$F(x\otimes 1_n,1_m\otimes y)=\sum _{p,q}{a}_{pq}{x}^p\otimes y^q,$$

(98)

and its exact squared norm is

$$\sum _{p,q,r,s}{a}_{pq}{a}_{rs}{〈x^p,x^r〉}_{\Omega }{〈y^q,y^s〉}_{\Omega }.$$

(99)

Every tensor in these assertions may be replaced by the right tensor, without changing the norm. In particular, for $f\left(t\right)={\sum }_j{c}_j{t}^j$,

$$f(x\otimes y)=\sum _j{c}_j{x}^j\otimes y^j.$$

(100)

The polynomials satisfying $f(x\otimes y)=f\left(x\right)\otimes f\left(y\right)$ for all inputs are exactly the zero polynomial and the monomials $t^j$ with $j\ge 0$.

Formula (97) also holds for rational functions whose denominators take unit values at x and y. For units one has

$${(x\otimes y)}^{-1}=x^{-1}\otimes y^{-1},$$

(101)

and the same formula in the opposite direction.

Proof. 

The block copies $x\otimes 1_n$ and $1_m\otimes y$ preserve ordinary powers and scalars by (83). For $p,q\ge 0$,

$$\begin{array}{cc}\hfill {(x\otimes 1_n)}^p{(1_m\otimes y)}^q& =(x^p\otimes 1_n)(1_m\otimes y^q)\hfill \\ \hfill & =x^p\otimes y^q.\hfill \end{array}$$

Thus, for $f\left(t\right)={\sum }_p{a}_p{t}^p$ and $g\left(t\right)={\sum }_q{b}_q{t}^q$,

$$\begin{array}{cc}\hfill f(x\otimes 1_n)g(1_m\otimes y)& =\sum _{p,q}{a}_p{b}_q(x^p\otimes y^q)\hfill \\ \hfill & =\left(\sum _p{a}_p{x}^p\right)\otimes \left(\sum _q{b}_q{y}^q\right)=f\left(x\right)\otimes g\left(y\right).\hfill \end{array}$$

The same monomial calculation proves (98). Expanding its squared norm and using fixed-block factorization gives

$$\begin{array}{cc}\hfill {∥\sum _{p,q}{a}_{pq}{x}^p\otimes y^q∥}_{\Omega }^2& =\sum _{p,q,r,s}{a}_{pq}{a}_{rs}{〈x^p\otimes y^q,x^r\otimes y^s〉}_{\Omega }\hfill \\ \hfill & =\sum _{p,q,r,s}{a}_{pq}{a}_{rs}{〈x^p,x^r〉}_{\Omega }{〈y^q,y^s〉}_{\Omega }.\hfill \end{array}$$

Taking $F(s,t)=f\left(st\right)$ proves (100). These arguments are unchanged in the right orientation.

For the classification, scalar inputs already impose $f\left(ab\right)=f\left(a\right)f\left(b\right)$ for every $a,b\in R$. Since R is infinite, this is a polynomial identity in two indeterminates. If f is nonzero of degree j and leading coefficient c, comparison of the coefficient of $a^j$ gives $c{b}^j=cf\left(b\right)$. Cancellation in R yields $f\left(t\right)=t^j$. Conversely, these polynomials and the zero polynomial have the claimed property by (83).

For units, multiply the proposed inverse by $x\otimes y$ and use (83):

$$(x\otimes y)(x^{-1}\otimes y^{-1})=\left(x{x}^{-1}\right)\otimes \left(y{y}^{-1}\right)=1_m\otimes 1_n=1_{m+n}.$$

The same calculation applied to unit denominator values proves the rational-function assertion. In particular, if $q\left(x\right)$ is a unit, then

$${\left(q\left(x\right)\otimes 1_n\right)}^{-1}=q{\left(x\right)}^{-1}\otimes 1_n.$$

The corresponding formula holds in the second block. Writing each rational function as numerator times inverse denominator therefore reduces (97) to the polynomial identity and this inverse identity. □

Theorem 25 

(Exponential block sums and orthogonal logarithms). For $u\in {\mathcal{N}}_m$ and $v\in {\mathcal{N}}_n$,

$$\exp (u\otimes 1_n+1_m\otimes v)=\exp \left(u\right)\otimes \exp \left(v\right).$$

(102)

For $x\in 1+{\mathcal{N}}_m$ and $y\in 1+{\mathcal{N}}_n$,

$$\log (x\otimes y)=\log \left(x\right)\otimes 1_n+1_m\otimes \log \left(y\right).$$

(103)

Its two terms are orthogonal. Thus

$$\begin{array}{cc}\hfill {∥\log (x\otimes y)∥}_{\Omega }^2& ={∥\log \left(x\right)∥}_{\Omega }^2+{∥\log \left(y\right)∥}_{\Omega }^2,\hfill \\ \hfill {∥\exp (u\otimes 1_n+1_m\otimes v)∥}_{\Omega }& ={∥\exp \left(u\right)∥}_{\Omega }{∥\exp \left(v\right)∥}_{\Omega }.\hfill \end{array}$$

(104)

All four formulas hold in the right orientation as well. For more than two prescribed blocks, the logarithmic squared norms add and the exponential norms multiply.

Proof. 

Apply the finite formulas of Theorem 5 to a block copy. Interchange preserves each power, so $\exp (u\otimes 1_n)=\exp \left(u\right)\otimes 1_n$ and $\exp (1_m\otimes v)=1_m\otimes \exp \left(v\right)$. These identities hold over R even when a factorial is not a unit: the block copy is R-linear, and if $r!a=u^r$, then $r!(a\otimes 1_n)=u^r\otimes 1_n$. Uniqueness of exact integer division in the torsion-free coefficient module identifies this with ${(u\otimes 1_n)}^r/r!$. The additive exponential law and interchange give

$$\begin{array}{cc}\hfill \exp (u\otimes 1_n+1_m\otimes v)& =\exp (u\otimes 1_n)\exp (1_m\otimes v)\hfill \\ \hfill & =\left(\exp \left(u\right)\otimes 1_n\right)\left(1_m\otimes \exp \left(v\right)\right)\hfill \\ \hfill & =\exp \left(u\right)\otimes \exp \left(v\right).\hfill \end{array}$$

Taking $u=\log \left(x\right)$ and $v=\log \left(y\right)$, and then applying the inverse logarithm, proves (103).

Both logarithms have scalar coefficient zero. Fixed-block factorization therefore gives

$${〈\log \left(x\right)\otimes 1_n,1_m\otimes \log \left(y\right)〉}_{\Omega }={〈\log \left(x\right),1〉}_{\Omega }{〈1,\log \left(y\right)〉}_{\Omega }=0.$$

Their individual norms are unchanged by unit blocks, so

$$\begin{array}{cc}\hfill {∥\log (x\otimes y)∥}_{\Omega }^2& ={∥\log \left(x\right)\otimes 1_n∥}_{\Omega }^2+{∥1_m\otimes \log \left(y\right)∥}_{\Omega }^2\hfill \\ \hfill & ={∥\log \left(x\right)∥}_{\Omega }^2+{∥\log \left(y\right)∥}_{\Omega }^2.\hfill \end{array}$$

Equation (104) follows from tensor norm factorization. Block reversal proves the right-oriented formulas, and induction on the number of blocks proves the final assertion. □

Factorizations of invertible zeons into elementary and homogeneous factors are studied in [12]. For the directed products here, the factors occupy prescribed consecutive blocks of recursive positions. The logarithmic identity above determines which cuts separate such blocks and how all admissible cuts combine.

Theorem 26 

(Unit cuts and the finest consecutive factorization). Let $w\in {\mathcal{I}}_{m+n}$ be a unit with scalar coefficient $c\in R^{\times }$. It is a tensor $x\otimes y$ with $x\in {\mathcal{I}}_m$, $y\in {\mathcal{I}}_n$ exactly when every nonzero path coefficient of $\log \left(c^{-1}w\right)$ has its support entirely in $\left[m\right]$ or entirely in $m+\left[n\right]$. In this case there are unique $x_0\in 1+{\mathcal{N}}_m$ and $y_0\in 1+{\mathcal{N}}_n$ such that

$$w=c(x_0\otimes y_0).$$

(105)

All factorizations with these two depths have the form $x=a{x}_0$, $y=b{y}_0$, where $a,b\in R^{\times }$ and $ab=c$.

For a unit of prescribed depth $N\ge 1$, test each cut between positions r and $r+1$, with $1\le r<N$, by the same logarithmic condition. All admissible cuts can be used simultaneously. They give a unique finest factorization into consecutive blocks, with the total scalar coefficient outside the tensor and every block factor normalized to scalar coefficient one. No resulting block admits a further internal cut. Reversing the order of the prescribed blocks gives the corresponding right-oriented assertion.

Proof. 

If $w=x\otimes y$, its scalar coefficient is the product of the scalar coefficients of x and y. Since that product is a unit, both coefficients are units. Normalize them and apply (103); it has no path crossing the cut.

Conversely, the support condition gives the unique separation

$$\log \left(c^{-1}w\right)=u\otimes 1_n+1_m\otimes v,u\in {\mathcal{N}}_m,v\in {\mathcal{N}}_n.$$

The support blocks determine u and v coefficient by coefficient. Writing $l_S$ for the coefficients of $\log \left(c^{-1}w\right)$, they are explicitly

$$u=\sum _{⌀\ne A\subseteq \left[m\right]}{l}_A{\mathbf{b}}_A,v=\sum _{⌀\ne B\subseteq \left[n\right]}{l}_{m+B}{\mathbf{b}}_B.$$

The empty coefficient is zero, so it is assigned to neither block. The hypothesis excludes every remaining support, namely those meeting both blocks. Thus these formulas recover the entire logarithm. Taking $x_0=\exp \left(u\right)$ and $y_0=\exp \left(v\right)$ and applying (102) gives (105). Uniqueness of the logarithmic separation gives uniqueness of the normalized factors and the stated description of all other factors.

For several cuts, a nonzero logarithmic path crosses none of the admissible cuts. It must therefore lie wholly in one of the consecutive blocks determined by all those cuts. Group the logarithm by these blocks and exponentiate. This supplies their simultaneous factorization and proves uniqueness by uniqueness of the path coefficients. More explicitly, for consecutive blocks $B_1,\dots ,B_s$, each nonempty logarithmic support belongs to exactly one $B_j$. Extracting those coefficients and renumbering within $B_j$ gives $u_j$ with scalar coefficient zero. The exponential block law then gives

$$w=c\exp \left(u_1\right)\otimes \dots \otimes \exp \left(u_s\right).$$

Each normalized factor is forced by its own logarithmic coefficients. A further cut inside a resulting block would have no crossing logarithmic path, so would already have been among the admissible cuts. This proves fineness. When the normalized unit is $1_N$, all cuts are admissible and each depth-one factor is $1_1$. Depth zero consists only of the outside scalar. □

For a depth-two normalized unit,

$$w=1+a{\mathbf{b}}_{\left\{1\right\}}+b{\mathbf{b}}_{\left\{2\right\}}+c{\mathbf{b}}_{\{1,2\}},$$

the finite logarithm is

$$\log \left(w\right)=a{\mathbf{b}}_{\left\{1\right\}}+b{\mathbf{b}}_{\left\{2\right\}}+(c-ab){\mathbf{b}}_{\{1,2\}}.$$

Thus the cut $1+1$ is admissible precisely when $c=ab$; in that case $w=(1,a)\otimes (1,b)$. This coefficient calculation is the first case of the general no-crossing criterion.

5. Recursive Functions and Sharp Function Estimates

The polynomial and logarithmic block identities of Section 4 are instances of a more general recursive evaluation. A function value is determined by scalar data and their successive use along the pos and len branches. Nilpotence makes the resulting expansions finite, so preservation of R, composition, and the solution of equations can be decided coefficient by coefficient. The weighted norm then measures how evaluation depends on its inputs. Sharp bounds on specified homogeneous classes, together with explicit unbounded families at increasing depth, determine the scope of uniform estimates and continuity for these functions.

5.1. Scalar Data and Recursive Evaluation

Polynomial evaluation uses the ordinary multiplication of Section 2. Throughout this section a scalar coefficient is read by (20). Real differentiation and prescribed real radii are understood after extension of scalars to $\mathbb{R}$; preservation of R will be stated separately.

Theorem 27 

(Polynomial and rational evaluation). For $f\in R\left[t\right]$ and a positive-depth input,

$$f\left(x\right)=\left(f\left(pos\left(x\right)\right),f^{\prime }\left(pos\left(x\right)\right)len\left(x\right)\right).$$

(106)

Each evaluation on the right continues to the scalar layer. If $x=c+u$ and $u^{q+1}=0$, then

$$f\left(x\right)={\displaystyle \sum _{r=0}^q}{f}^{\left[r\right]}\left(c\right)u^r,f^{\left[r\right]}\left(t\right)=\sum _{j\ge r}\left({\displaystyle \genfrac{}{}{0pt}{}{j}{r}}\right)a_j{t}^{j-r}\mathrm{for}f\left(t\right)=\sum _j{a}_j{t}^j.$$

(107)

These Hasse derivatives belong to $R\left[t\right]$.

For polynomials $p,q$, the expression $p\left(x\right)/q\left(x\right)$ is defined whenever $q\left(c\right)$ is a unit of R. It is independent of the representation of the fraction among fractions with admissible denominators, and satisfies (106) with

$${(p/q)}^{\prime }=(p^{\prime }q-p{q}^{\prime })/q^2.$$

(108)

Equivalently, evaluation is defined on the localization of $R\left[t\right]$ at the polynomials whose values at c are units.

Proof. 

For $j=1$, the power formula below is simply $x=\left(pos\left(x\right),len\left(x\right)\right)$. If it holds at j, multiplication by x gives its two components at $j+1$:

$$\begin{array}{cc}\hfill x^j& =\left(pos{\left(x\right)}^j,jpos{\left(x\right)}^{j-1}len\left(x\right)\right),\hfill \\ \hfill x^{j+1}& =\left(pos{\left(x\right)}^{j+1},pos{\left(x\right)}^{j}len\left(x\right)+jpos{\left(x\right)}^{j}len\left(x\right)\right)\hfill \\ \hfill & =\left(pos{\left(x\right)}^{j+1},(j+1)pos{\left(x\right)}^{j}len\left(x\right)\right).\hfill \end{array}$$

The constant term has second component zero. Hence, for $f\left(t\right)={\sum }_j{a}_j{t}^j$,

$$\begin{array}{cc}\hfill pos\left(f\left(x\right)\right)& =\sum _j{a}_{j}pos{\left(x\right)}^j=f\left(pos\left(x\right)\right),\hfill \\ \hfill len\left(f\left(x\right)\right)& =\sum _{j\ge 1}j{a}_{j}pos{\left(x\right)}^{j-1}len\left(x\right)=f^{\prime }\left(pos\left(x\right)\right)len\left(x\right).\hfill \end{array}$$

For $x=c+u$ and $u^{q+1}=0$, the finite binomial calculation is

$$\begin{array}{cc}\hfill f(c+u)& =\sum _j{a}_j{\displaystyle \sum _{r=0}^{\min (j,q)}}\left({\displaystyle \genfrac{}{}{0pt}{}{j}{r}}\right)c^{j-r}{u}^r\hfill \\ \hfill & ={\displaystyle \sum _{r=0}^q}\left(\sum _{j\ge r}{a}_j\left({\displaystyle \genfrac{}{}{0pt}{}{j}{r}}\right)c^{j-r}\right)u^r={\displaystyle \sum _{r=0}^q}{f}^{\left[r\right]}\left(c\right)u^r.\hfill \end{array}$$

The scalar coefficient of $q\left(x\right)$ is $q\left(c\right)$, so Theorem 4 gives exactly the asserted unit condition. The quotient formula (24), applied to $p\left(x\right)$ and $q\left(x\right)$, gives (108) and the two recursive components. In particular, the second component is

$$\begin{array}{cc}\hfill len\left(p\left(x\right)q{\left(x\right)}^{-1}\right)& ={\displaystyle \frac{p^{\prime }\left(pos\left(x\right)\right)q\left(pos\left(x\right)\right)-p\left(pos\left(x\right)\right)q^{\prime }\left(pos\left(x\right)\right)}{q{\left(pos\left(x\right)\right)}^2}}len\left(x\right)\hfill \\ \hfill & ={\left({\displaystyle \frac{p}{q}}\right)}^{\prime }\left(pos\left(x\right)\right)len\left(x\right).\hfill \end{array}$$

The first component is $p(pos(x\left)\right)/q(pos(x\left)\right)$. Its denominator is a unit because it has the same scalar coefficient $q\left(c\right)$. If two admissible fractions represent the same rational function, cross multiplication in $R\left[t\right]$ gives equal numerators after evaluation. Multiplication by the two available inverses proves equality of their values. Sums and products of admissible fractions remain admissible, which proves the localization statement. □

Corollary 4 

(Square-zero displacement). If $u^2=0$, then for every polynomial f and every finite input x,

$$f(x+u)=f\left(x\right)+f^{\prime }\left(x\right)u.$$

(109)

The same formula holds for an admissible rational function: admissibility at x and at $x+u$ is the same scalar unit condition.

Proof. 

Since $u^2=0$, the binomial formula gives

$${(x+u)}^j=x^j+j{x}^{j-1}u(j\ge 1),f(x+u)=f\left(x\right)+f^{\prime }\left(x\right)u.$$

The scalar coefficient of u is zero, so $q\left(x\right)$ and $q(x+u)$ are units under the same condition. Their inverses satisfy

$$q{(x+u)}^{-1}=q{\left(x\right)}^{-1}-q{\left(x\right)}^{-2}{q}^{\prime }\left(x\right)u,$$

because the product with $q\left(x\right)+q^{\prime }\left(x\right)u$ is

$$\begin{array}{cc}\hfill & \left(q\left(x\right)+q^{\prime }\left(x\right)u\right)\left(q{\left(x\right)}^{-1}-q{\left(x\right)}^{-2}{q}^{\prime }\left(x\right)u\right)\hfill \\ \hfill & =1-q{\left(x\right)}^{-1}{q}^{\prime }\left(x\right)u+q^{\prime }\left(x\right)q{\left(x\right)}^{-1}u-q{\left(x\right)}^{-2}{q}^{\prime }{\left(x\right)}^2{u}^2=1.\hfill \end{array}$$

The linear terms cancel and the quadratic term vanishes. Thus

$$\begin{array}{cc}\hfill {\displaystyle \frac{p(x+u)}{q(x+u)}}& =\left(p\left(x\right)+p^{\prime }\left(x\right)u\right)\left(q{\left(x\right)}^{-1}-q{\left(x\right)}^{-2}{q}^{\prime }\left(x\right)u\right)\hfill \\ \hfill & ={\displaystyle \frac{p\left(x\right)}{q\left(x\right)}}+{\displaystyle \frac{p^{\prime }\left(x\right)q\left(x\right)-p\left(x\right)q^{\prime }\left(x\right)}{q{\left(x\right)}^2}}u.\hfill \end{array}$$

□

Definition 7 

(Evaluation from scalar data). Let $f_0,f_1,\dots$ be real-valued scalar functions on a set of allowed scalar coefficients. Define their recursive evaluations by

$$\begin{array}{cc}\hfill {\widehat{f}}_j\left(c\right)& =f_j\left(c\right),\hfill \\ \hfill {\widehat{f}}_j\left(x\right)& =\left({\widehat{f}}_j\left(pos\left(x\right)\right),{\widehat{f}}_{j+1}\left(pos\left(x\right)\right)len\left(x\right)\right)\mathrm{at}\mathrm{positive}\mathrm{depth}.\hfill \end{array}$$

(110)

The data need not initially be derivatives of one function.

Theorem 28 

(Finite evaluation and its exact arithmetic domain). For $x=c+u$ with $u\in {\mathcal{N}}_n$,

$${\widehat{f}}_j\left(x\right)={\displaystyle \sum _{r=0}^n}{f}_{j+r}\left(c\right){\displaystyle \frac{u^r}{r!}}.$$

(111)

Every $u^r/r!$ belongs to ${\mathcal{I}}_n$ when u does. If S is nonempty,

$${\left({\widehat{f}}_j\left(x\right)\right)}_S=\sum _{\pi \mathrm{a}\mathrm{partition}\mathrm{of}S}{f}_{j+\left|\pi \right|}\left(c\right)\prod _{B\in \pi }{u}_B.$$

(112)

The scalar coefficient is $f_j\left(c\right)$. Evaluation commutes with refinement and hence is well defined on finite global objects.

For a fixed $c\in R$, the operation ${\widehat{f}}_0$ takes $c+{\mathcal{N}}_n$ into ${\mathcal{I}}_n$ exactly when $f_0\left(c\right),\dots ,f_n\left(c\right)$ all belong to R. It preserves the original coefficient ring at every finite depth exactly when all $f_r\left(c\right)$ belong to R.

Proof. 

The scalar case starts the induction. At depth $n\ge 1$, let y be the finite sum in (111). The power formula and $pos{\left(u\right)}^n=0$ give

$$\begin{array}{cc}\hfill pos\left(y\right)& ={\displaystyle \sum _{r=0}^{n-1}}{f}_{j+r}\left(c\right){\displaystyle \frac{pos{\left(u\right)}^r}{r!}}={\widehat{f}}_j\left(pos\left(x\right)\right),\hfill \\ \hfill len\left(y\right)& ={\displaystyle \sum _{r=1}^n}{f}_{j+r}\left(c\right){\displaystyle \frac{pos{\left(u\right)}^{r-1}}{(r-1)!}}len\left(u\right)\hfill \\ \hfill & ={\widehat{f}}_{j+1}\left(pos\left(x\right)\right)len\left(x\right).\hfill \end{array}$$

Here $pos\left(x\right)=c+pos\left(u\right)$ and $len\left(x\right)=len\left(u\right)$. The two components are exactly those of (110), proving the finite formula by induction.

In each coefficient of $u^r$, surviving terms use r pairwise disjoint nonempty supports. An unordered family contributes once for each of its $r!$ orders, with the same coefficient product. Division by $r!$ is thus coefficientwise division of an exact multiple in R, uniquely determined because R has no integer torsion. Summing the resulting unordered families gives (112).

At the scalar boundary, ${\widehat{f}}_j\left((c,0)\right)=(f_j\left(c\right),0)=\iota \left(f_j\left(c\right)\right)$. At arbitrary depth, refinement fixes the scalar coefficient and preserves powers and their unique exact divisions. Thus the finite formula just established gives

$$\begin{array}{cc}\hfill {\widehat{f}}_j\left(\iota \left(x\right)\right)& ={\displaystyle \sum _{r=0}^n}{f}_{j+r}\left(c\right){\displaystyle \frac{\iota {\left(u\right)}^r}{r!}}\hfill \\ \hfill & =\iota \left({\displaystyle \sum _{r=0}^n}{f}_{j+r}\left(c\right){\displaystyle \frac{u^r}{r!}}\right)=\iota \left({\widehat{f}}_j\left(x\right)\right).\hfill \end{array}$$

The possible extra order $n+1$ is zero because $\iota {\left(u\right)}^{n+1}=0$. The calculation also applies after real scalar extension. Repeating it proves representative independence.

Formula (112) proves the sufficiency of the stated arithmetic condition. For necessity use $u={\mathbf{b}}_1+\dots +{\mathbf{b}}_n$. The coefficient at any support of size r is exactly $f_r\left(c\right)$, including $r=0$ at the empty support. This recovers every required scalar value from an original-ring input. □

On a scalar constant in a specified level, the finite expansion reduces to

$${\widehat{f}}_j\left(c{1}_n\right)=f_j\left(c\right)1_n.$$

In particular, whenever the scalar data are defined at 0 and 1,

$${\widehat{f}}_j\left(0\right)=f_j\left(0\right)1,{\widehat{f}}_j\left(1\right)=f_j\left(1\right)1.$$

The outputs use the same level as the inputs. The recursive exponential and logarithm accordingly satisfy $\exp \left(0\right)=1$ and $\log \left(1\right)=0$ at every level, as well as on global objects.

Set partitions record how scalar data combine under composition. When the data are derivatives, the resulting sum is the classical Faà di Bruno formula; Bell-polynomial and multivariate treatments can be found in [28,29]. Neto and dos Anjos obtain derivative and combinatorial identities, including Faà di Bruno and Bell-number formulas, within zeon algebra [6]. The disjoint path supports below give the set partitions directly from recursive multiplication.

Proposition 12 

(Function algebra and composition). For two scalar data families, addition is termwise. The data for the product of their evaluated functions are

$$h_r\left(c\right)={\displaystyle \sum _{j=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{j}}\right)f_j\left(c\right)g_{r-j}\left(c\right).$$

(113)

When the scalar domains permit composition, its data are

$$\begin{array}{cc}\hfill h_0\left(c\right)& =f_0\left(g_0\left(c\right)\right),\hfill \\ \hfill h_r\left(c\right)& =\sum _{\pi \mathrm{a}\mathrm{partition}\mathrm{of}\left[r\right]}{f}_{\left|\pi \right|}\left(g_0\left(c\right)\right)\prod _{B\in \pi }{g}_{\left|B\right|}\left(c\right)(r\ge 1).\hfill \end{array}$$

(114)

Consequently these rules give, respectively, ${\widehat{h}}_0\left(x\right)={\widehat{f}}_0\left(x\right){\widehat{g}}_0\left(x\right)$ and ${\widehat{h}}_0\left(x\right)={\widehat{f}}_0\left({\widehat{g}}_0\left(x\right)\right)$. They are also necessary if these identities hold at all finite depths.

Proof. 

For $x=c+u$ in depth n, multiplication of the two finite sums gives

$$\begin{array}{cc}\hfill {\widehat{f}}_0\left(x\right){\widehat{g}}_0\left(x\right)& =\sum _{p+q\le n}{f}_p\left(c\right)g_q\left(c\right){\displaystyle \frac{u^{p+q}}{p!q!}}\hfill \\ \hfill & ={\displaystyle \sum _{r=0}^n}\left({\displaystyle \sum _{p=0}^r}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{p}}\right)f_p\left(c\right)g_{r-p}\left(c\right)\right){\displaystyle \frac{u^r}{r!}}.\hfill \end{array}$$

This is (113). For composition, the inner value has scalar coefficient $g_0\left(c\right)$, so

$${\widehat{f}}_0\left({\widehat{g}}_0(c+u)\right)={\displaystyle \sum _{\ell =0}^n}{\displaystyle \frac{f_{\ell }\left(g_0\left(c\right)\right)}{\ell !}}{\left({\displaystyle \sum _{j=1}^n}{g}_j\left(c\right){\displaystyle \frac{u^j}{j!}}\right)}^{\ell }.$$

For $r\ge 1$, the coefficient of $u^r/r!$ is

$${\displaystyle \sum _{\ell =1}^r}{f}_{\ell }\left(g_0\left(c\right)\right)\sum _{\begin{array}{c}{j}_1+\dots +j_{\ell }=r\\ j_i\ge 1\end{array}}{\displaystyle \frac{r!}{\ell !j_1!\dots j_{\ell }!}}{\displaystyle \prod _{i=1}^{\ell }}{g}_{j_i}\left(c\right).$$

An ordered division of the r labelled positions into blocks of sizes $j_1,\dots ,j_{\ell }$ has $r!/(j_1!\dots j_{\ell }!)$ choices; division by $\ell !$ removes the block order. Grouping by the resulting set partitions gives exactly (114). For instance, in the first two nontrivial orders this reads

$$\begin{array}{cc}\hfill h_2\left(c\right)& =f_1\left(g_0\left(c\right)\right)g_2\left(c\right)+f_2\left(g_0\left(c\right)\right)g_1{\left(c\right)}^2,\hfill \\ \hfill h_3\left(c\right)& =f_1\left(g_0\left(c\right)\right)g_3\left(c\right)+3f_2\left(g_0\left(c\right)\right)g_1\left(c\right)g_2\left(c\right)\hfill \\ \hfill & +f_3\left(g_0\left(c\right)\right)g_1{\left(c\right)}^3.\hfill \end{array}$$

The coefficient three counts the partitions of three positions into one singleton and one two-element block. Thus the formula uses only the prescribed scalar data, without assuming they are derivatives. This expression has values in R when the scalar data do, and Theorem 28 gives original-ring evaluations. Finally, evaluation at $c+{\mathbf{b}}_1+\dots +{\mathbf{b}}_r$ and extraction of the full support recovers the rth datum. Thus the rules are necessary as well. □

Proposition 13 

(Determination by path substitutions). Suppose an operation on every finite level commutes with every unital R-algebra substitution between such levels. On inputs with a given scalar coefficient c, it is uniquely of the form (112) for a sequence of scalar values $f_r\left(c\right)\in R$. Conversely, each such sequence defines operations with this substitution property.

Proof. 

Evaluate the operation at $c+{\mathbf{b}}_1+\dots +{\mathbf{b}}_n$. Permuting the positions shows that a coefficient depends only on its support size. Setting unwanted positions equal to zero shows that the coefficient of size r is independent of $n\ge r$. Denote it by $f_r\left(c\right)$, and write F for the operation. Then

$$F\left(c+{\displaystyle \sum _{j=1}^n}{\mathbf{b}}_j\right)=\sum _{A\subseteq \left[n\right]}{f}_{\left|A\right|}\left(c\right){\mathbf{b}}_A.$$

For an arbitrary $x=c+{\sum }_{A\ne ⌀}{u}_A{\mathbf{b}}_A$, introduce one new position for each nonzero term. Substituting that new path by $u_A{\mathbf{b}}_A$ is a valid algebra substitution, since each image has square zero. To make this calculation explicit, list the nonzero supports as $S_1,\dots ,S_q$, and let $\theta$ be the substitution with $\theta \left({\mathbf{b}}_i\right)=u_{S_i}{\mathbf{b}}_{S_i}$. Then

$$\begin{array}{cc}\hfill F\left(x\right)& =\theta \left(F\left(c+{\displaystyle \sum _{i=1}^q}{\mathbf{b}}_i\right)\right)\hfill \\ \hfill & =\sum _{A\subseteq \left[q\right]}{f}_{\left|A\right|}\left(c\right)\prod _{i\in A}{u}_{S_i}{\mathbf{b}}_{S_i}.\hfill \end{array}$$

The product is zero unless the selected supports are pairwise disjoint. For a nonempty output support, the surviving selections are exactly its partitions into nonzero input supports, giving (112). If there are no nonzero supports, the scalar-level value gives the same formula with only its empty term.

Conversely, a unital R-algebra substitution fixes scalars and takes nilpotents to nilpotents, so it preserves the scalar coefficient. It preserves every power. If $r!v=u^r$, then

$$r!\theta \left(v\right)=\theta \left(u^r\right)=\theta {\left(u\right)}^r.$$

Uniqueness of exact integer division in the target identifies $\theta \left(v\right)$ with $\theta {\left(u\right)}^r/r!$. Applying this term by term to (111) proves the converse. □

After real scalar extension, a finite level is a real unital commutative algebra with a nilpotent ideal and scalar quotient $\mathbb{R}$. Finite Taylor evaluation on such algebras is classical; see the treatment of Weil algebras in [30]. Elementary zeon functions and their differential calculus are developed in [12,13]. Here the evaluation is expressed by the same two component operations as the arithmetic, with scalar-data conditions that specify when its coefficients remain in R.

Theorem 29 

(Smooth functional calculus). Let f be smooth on a real open interval. On elements whose scalar coefficient belongs to that interval, take $f_j=f^{\left(j\right)}$. Then

$$\widehat{f}\left(x\right)=\left(\widehat{f}\left(pos\left(x\right)\right),\widehat{f^{\prime }}\left(pos\left(x\right)\right)len\left(x\right)\right)$$

(115)

at positive depth, with scalar value $f\left(c\right)$. It preserves sums, products, and compositions wherever the scalar domains allow composition:

$$\widehat{f+g}\left(x\right)=\widehat{f}\left(x\right)+\widehat{g}\left(x\right),\widehat{fg}\left(x\right)=\widehat{f}\left(x\right)\widehat{g}\left(x\right),\widehat{f\circ g}\left(x\right)=\widehat{f}\left(\widehat{g}\left(x\right)\right).$$

(116)

At a fixed real input x, this is the unique evaluation on the algebra of smooth functions that preserves constants, addition, and multiplication and evaluates the coordinate function at x.

Proof. 

The ordinary derivative rules give the data in Proposition 12; alternatively their finite expansions give the same identities directly. For uniqueness, let c be the scalar coefficient of $x\in {\mathcal{I}}_n$. Smooth Taylor division gives

$$f\left(t\right)={\displaystyle \sum _{r=0}^n}{\displaystyle \frac{f^{\left(r\right)}\left(c\right)}{r!}}{(t-c)}^r+{(t-c)}^{n+1}g\left(t\right)$$

with g smooth on the interval. For example its value at c and its derivatives are given by the smooth integral formula

$$g\left(t\right)={\displaystyle \frac{1}{n!}}{\int }_0^1{(1-s)}^n{f}^{(n+1)}\left(c+s(t-c)\right)ds.$$

For any evaluation E with the stated algebra properties,

$$E\left({(t-c)}^{n+1}g\left(t\right)\right)={(x-c)}^{n+1}E\left(g\right)=0,E\left({(t-c)}^r\right)={(x-c)}^r.$$

The remaining terms force (111). Existence has already been proved, so the forced formula is the unique one. The remainder vanishes by the exact algebraic equality ${(x-c)}^{n+1}=0$. Thus this conclusion applies to smooth functions without requiring an infinite power-series expansion. □

The value at depth n uses only the first n derivatives at the scalar coefficient. A $C^n$ function therefore suffices for that finite formula and its sum, product, and composition identities. Smoothness supplies all levels simultaneously; analyticity is not needed. In particular, the exponential on $\mathcal{N}$, the logarithm on $1+\mathcal{N}$, and the inverse on units agree with the operations of Section 2.

In the finite regular representation, the same terminating expansion is the scalar-plus-nilpotent case of matrix functional calculus [31]. At depth one, the pair $(f\left(c\right),f^{\prime }\left(c\right)d)$ is the value-and-derivative rule of forward differentiation [32]; successive component descent applies this rule to the higher recursive levels.

Example 8 

(An arithmetic domain beyond polynomial coefficients). A smooth scalar function preserves the entire original-ring tower at the allowed scalar coefficients exactly when all its derivatives there, including the function itself, belong to R. This class contains functions that are polynomials over $\mathbb{Q}$ but not over R. Choose a prime p not dividing Ω and set

$$f\left(t\right)={\displaystyle \frac{{(t^p-t)}^p}{p}}.$$

Reduction of R modulo p is the field with p elements, so $c^p-c\in pR$ for $c\in R$ and hence $f\left(c\right)\in R$. Moreover

$$f^{\prime }\left(t\right)={(t^p-t)}^{p-1}(p{t}^{p-1}-1)\in R\left[t\right].$$

All higher derivatives consequently take values in R, whereas the leading coefficient $1/p$ of f does not belong to R. Theorem 28 supplies an original-ring value at every finite input. This extends the arithmetic domain beyond the unit-denominator localization in Theorem 27.

Proposition 14 

(Obstruction for all scalar set functions). At a nonscalar input $x=c+u$ with nonzero nilpotent u, no evaluation of all scalar set functions can simultaneously preserve constants, addition, multiplication, the coordinate function, and scalar coefficients.

Proof. 

Let e be the indicator function of the singleton $\left\{c\right\}$. Its value would satisfy

$$\widehat{e}{\left(x\right)}^2=\widehat{e^2}\left(x\right)=\widehat{e}\left(x\right),{\left(\widehat{e}\left(x\right)\right)}_{⌀}=e\left(c\right)=1.$$

Theorem 4 forces $\widehat{e}\left(x\right)=1$. But the scalar identity $(t-c)e\left(t\right)=0$ would then give

$$0=(x-c)\widehat{e}\left(x\right)=u·1=u,$$

contrary to the choice of the input. □

Theorem 30 

(Several inputs). For a smooth function F of d real variables, evaluate at the scalar tuple when every input is scalar. Otherwise define

$$\begin{array}{cc}\hfill \widehat{F}(x_1,\dots ,x_d)=(& \widehat{F}\left(pos\left(x_1\right),\dots ,pos\left(x_d\right)\right),\hfill \\ \hfill & {\displaystyle \sum _{j=1}^d}\widehat{{\partial }_{j}F}\left(pos\left(x_1\right),\dots ,pos\left(x_d\right)\right)len\left(x_j\right)).\hfill \end{array}$$

(117)

The scalar boundary of pos and len permits inputs of different depths. At common depth n, write $x_j=c_j+u_j$. For a tuple $\alpha =({\alpha }_1,\dots ,{\alpha }_d)$ of nonnegative integers, put $\left|\alpha \right|={\sum }_j{\alpha }_j$ and $\alpha !={\prod }_j{\alpha }_j!$. Then

$$\widehat{F}(x_1,\dots ,x_d)=\sum _{\left|\alpha \right|\le n}\left({\partial }^{\alpha }F\right)(c_1,\dots ,c_d){\displaystyle \prod _{j=1}^d}{\displaystyle \frac{u_j^{{\alpha }_j}}{{\alpha }_j!}}.$$

(118)

Preservation of R on this scalar tuple through depth n is equivalent to all the displayed mixed partial derivatives belonging to R. At every finite depth it is equivalent to the same condition for all orders. The evaluation preserves the multivariate sum, product, and composition rules.

Proof. 

We first check the components of the finite expression (118). Recursive multiplication gives

$$\begin{array}{cc}\hfill pos\left({\displaystyle \prod _{j=1}^d}{u}_j^{{\alpha }_j}\right)& ={\displaystyle \prod _{j=1}^d}pos{\left(u_j\right)}^{{\alpha }_j},\hfill \\ \hfill len\left({\displaystyle \prod _{j=1}^d}{u}_j^{{\alpha }_j}\right)& =\sum _{\begin{array}{c}1\le l\le d\\ {\alpha }_l>0\end{array}}{\alpha }_{l}pos{\left(u_l\right)}^{{\alpha }_l-1}len\left(u_l\right)\prod _{j\ne l}pos{\left(u_j\right)}^{{\alpha }_j}.\hfill \end{array}$$

The first identity is multiplicativity of pos. The second follows by applying the component product rule to one factor at a time; the power rule supplies the coefficient ${\alpha }_l$. In the outer pair decomposition every second component contains the same first square-zero path. In particular,

$$\left(0,len\left(x_i\right)\right)\left(0,len\left(x_j\right)\right)=0(1\le i,j\le d).$$

Insert the two monomial component formulas into the finite sum. For pos, every term of total order n vanishes at depth $n-1$. The remaining sum is the lower-depth evaluation of F at the first components. For len, replace ${\alpha }_l$ by ${\alpha }_l-1$ in the lth summand and use ${\alpha }_l/\alpha !=1/(\alpha -e_l)!$, where $e_l$ is the tuple with one in position l and zeros elsewhere. The resulting sum is the lower-depth evaluation of ${\partial }_{l}F$, multiplied by $len\left(x_l\right)=len\left(u_l\right)$. Summing over l gives exactly (117). The scalar case is its induction base. A scalar input has zero second component, so the calculation also includes mixed depths through the scalar boundary. Terms of total order greater than n vanish because they belong to ${\mathcal{N}}_n^{n+1}$. Each divided power in the product is integral by Theorem 28, proving sufficiency. For necessity, choose disjoint position sets of sizes ${\alpha }_1,\dots ,{\alpha }_d$, and let $u_j$ be the sum of the singleton paths in its set. Only the specified multi-index contributes to the complete support coefficient. If those sets are $A_1,\dots ,A_d$, then

$${\displaystyle \frac{u_j^{{\alpha }_j}}{{\alpha }_j!}}={\mathbf{b}}_{A_j},\prod _j{\displaystyle \frac{u_j^{{\alpha }_j}}{{\alpha }_j!}}={\mathbf{b}}_{A_1\bigsqcup \dots \bigsqcup A_d}.$$

That coefficient is therefore $\left({\partial }^{\alpha }F\right)(c_1,\dots ,c_d)$. The finite expansions also prove the sum, product, and composition rules by ordinary polynomial substitution after discarding vanishing total orders. □

5.2. Recursive Equations and Sharp Function Bounds

Theorem 31 

(Equations with prescribed scalar coefficient). Fix $c\in R$ and scalar data $f_j\left(c\right)\in R$, and put $d=f_0\left(c\right)$, $a=f_1\left(c\right)$. For $n\ge 1$, evaluation on $c+{\mathcal{N}}_n$ is injective exactly when $a\ne 0$. It is surjective onto $d+{\mathcal{N}}_n$, and hence bijective, exactly when a is a unit of R. In the bijective case its inverse is a polynomial in the finitely many path coefficients on this whole set.

Proof. 

For a nonempty support S, the output coefficient is

$$a{u}_S+\sum _{\begin{array}{c}\pi \mathrm{a}\mathrm{partition}\mathrm{of}S\\ \left|\pi \right|\ge 2\end{array}}{f}_{\left|\pi \right|}\left(c\right)\prod _{B\in \pi }{u}_B.$$

Every coefficient in the sum has strictly smaller support. If $a\ne 0$, comparison in increasing support size proves injectivity over the domain R: at a singleton support, equality of two outputs gives $a{u}_S=a{u}_S^{\prime }$ and hence $u_S=u_S^{\prime }$. If the coefficients agree on all proper supports of S, the partition sums agree, leaving $a(u_S-u_S^{\prime })=0$. Cancellation in the domain again gives equality. Induction on $\left|S\right|$ identifies the entire input. If a is a unit and the target is $d+v$, the solution is

$$u_S=a^{-1}\left(v_S-\sum _{\begin{array}{c}\pi \mathrm{a}\mathrm{partition}\mathrm{of}S\\ \left|\pi \right|\ge 2\end{array}}{f}_{\left|\pi \right|}\left(c\right)\prod _{B\in \pi }{u}_B\right).$$

For singleton S the sum is empty. Each later coefficient depends only on previously computed proper supports and division by the fixed unit a. Thus every coefficient is polynomial in the target coefficients. Substitution into the displayed output formula makes each coefficient exactly $v_S$, so the construction proves both existence and uniqueness for every target. If $a=0$, the distinct inputs c and $c+{\mathbf{b}}_{\left[n\right]}$ have the same value d, since ${\mathbf{b}}_{\left[n\right]}^2=0$ and the linear datum is zero. This proves noninjectivity. If a is not a unit, a target with a first-path coefficient equal to one would require $a{u}_{\left\{j\right\}}=1$, proving nonsurjectivity.

The solution can equally be obtained by scalar descent. For a target y, solve for the first component at lower depth, then compute the second:

$${\widehat{f}}_0\left(pos\left(x\right)\right)=pos\left(y\right),len\left(x\right)={\widehat{f}}_1{\left(pos\left(x\right)\right)}^{-1}len\left(y\right).$$

The inverse exists because its scalar coefficient is the unit a. Pairing the two components gives the same unique solution as the triangular calculation. □

The theorem requires no bound on the nilpotent coordinates. On a selected set of scalar coefficients, evaluation is bijective at all levels exactly when its scalar function is bijective onto the selected target set and its first datum at each scalar is a unit. Necessity follows already from depths zero and one. Over the reals, the unit condition is simply $f^{\prime }\left(c\right)\ne 0$, so each regular scalar solution of $f\left(c\right)=d$ gives exactly one recursive solution with that scalar coefficient to any target with scalar coefficient d.

Proposition 15 

(Derivatives and inverse data). For smooth f on a finite real level,

$$\begin{array}{cc}\hfill D^r\widehat{f}\left(x\right)[h_1,\dots ,h_r]& =\widehat{f^{\left(r\right)}}\left(x\right)h_1\dots h_r,\hfill \\ \hfill \det D\widehat{f}\left(x\right)& ={\left(f^{\prime }\left(c\right)\right)}^{2^n}(x\in {\mathcal{I}}_n).\hfill \end{array}$$

(119)

For scalar inverse data at $d=f_0\left(c\right)$ with $a=f_1\left(c\right)$ a unit, set $g_0\left(d\right)=c$, $g_1\left(d\right)=a^{-1}$. The remaining data are determined within R by

$$g_r\left(d\right)=-a^{-1}\sum _{\begin{array}{c}\pi \mathrm{a}\mathrm{partition}\mathrm{of}\left[r\right]\\ \left|\pi \right|\ge 2\end{array}}{f}_{\left|\pi \right|}\left(c\right)\prod _{B\in \pi }{g}_{\left|B\right|}\left(d\right),r\ge 2.$$

(120)

Their recursive evaluation is the inverse on the fixed scalar fiber.

Proof. 

Differentiate the finite formula with $x=c+u$. If h has scalar coefficient b, differentiating c gives b and differentiating u gives $h-b$. Therefore

$$\begin{array}{cc}\hfill D\widehat{f}\left(x\right)\left[h\right]& =b{\displaystyle \sum _{r=0}^n}{f}^{(r+1)}\left(c\right){\displaystyle \frac{u^r}{r!}}+(h-b){\displaystyle \sum _{r=1}^n}{f}^{\left(r\right)}\left(c\right){\displaystyle \frac{u^{r-1}}{(r-1)!}}\hfill \\ \hfill & =h{\displaystyle \sum _{r=0}^n}{f}^{(r+1)}\left(c\right){\displaystyle \frac{u^r}{r!}}-(h-b)f^{(n+1)}\left(c\right){\displaystyle \frac{u^n}{n!}}\hfill \\ \hfill & =\widehat{f^{\prime }}\left(x\right)h.\hfill \end{array}$$

The last correction is zero because $h-b\in {\mathcal{N}}_n$ and $u^n\in {\mathcal{N}}_n^n$. Apply this first-derivative identity successively to $f^{\prime },f^{\prime \prime },\dots$ while keeping the directions $h_j$ fixed. This gives $D^r\widehat{f}\left(x\right)[h_1,\dots ,h_r]=\widehat{f^{\left(r\right)}}\left(x\right)h_1\dots h_r$. Multiplication by $\widehat{f^{\prime }}\left(x\right)$ has a triangular matrix in the path basis ordered by support size, with every diagonal entry $f^{\prime }\left(c\right)$. This proves the determinant formula.

In the composition formula (114), the one-block partition contributes $a{g}_r\left(d\right)$. Every other block has size less than r, giving (120) recursively. The first composition therefore has the data of the coordinate function. Uniqueness in Theorem 31 makes this evaluation the inverse. For the first two nontrivial orders, the equations read

$$\begin{array}{cc}\hfill a{g}_2\left(d\right)+f_2\left(c\right)g_1{\left(d\right)}^2& =0,\hfill \\ \hfill a{g}_3\left(d\right)+3f_2\left(c\right)g_1\left(d\right)g_2\left(d\right)+f_3\left(c\right)g_1{\left(d\right)}^3& =0.\hfill \end{array}$$

Using $g_1\left(d\right)=a^{-1}$ gives

$$g_2\left(d\right)=-{\displaystyle \frac{f_2\left(c\right)}{a^3}},g_3\left(d\right)={\displaystyle \frac{3f_2{\left(c\right)}^2-a{f}_3\left(c\right)}{a^5}}.$$

At every later order, the same equation solves for $g_r\left(d\right)$ using only earlier data and division by the fixed unit a. □

Corollary 5 

(Arithmetic criterion for roots). For $v\in 1+\mathcal{N}$ and a positive integer m, a normalized unit solution of $x^m=v$ exists exactly when every path coefficient of $\log \left(v\right)$ belongs to $mR$. It is unique when it exists, and equals

$$x=\exp \left(\log \left(v\right)/m\right).$$

(121)

In particular, every normalized unit has a unique mth root if m is a unit of R. More generally, a unit target of scalar coefficient d has one root for each scalar root $c^m=d$ in R, provided that the coefficients of $\log \left(d^{-1}v\right)$ belong to $mR$; otherwise it has none.

Proof. 

For a normalized unit, logarithmic additivity gives

$$\log \left(x^m\right)=m\log \left(x\right),m{\left(\log \left(x\right)\right)}_S={\left(\log \left(v\right)\right)}_S.$$

The coordinate equations are soluble in R precisely when the right sides belong to $mR$. Their solutions are unique because R has no integer torsion. Exponentiating these coordinates gives

$${\left(\exp \left(\log \left(v\right)/m\right)\right)}^m=\exp \left(\log \left(v\right)\right)=v.$$

For a general unit write $v=d{v}_0$ and $x=c{x}_0$, where $v_0,x_0$ are normalized. Then $x^m=v$ consists of the two equations $c^m=d$ and $x_0^m=v_0$. A scalar root of a unit is a unit, so this normalization is defined over R and proves the general case. Explicitly, $c^{-1}=c^{m-1}{d}^{-1}\in R$. Each allowed scalar root therefore gives exactly one normalized solution, and roots with different scalar coefficients are distinct. If $m\in R^{\times }$, division of every logarithmic coefficient by m stays in R, so there is no further obstruction. □

For example, if a prime p does not divide $\Omega$, the scalar function $t^p$ has nonzero derivative at 1, but the normalized equation $x^p=1+{\mathbf{b}}_1$ has no solution over R: its first-path equation would be $p{x}_{\left\{1\right\}}=1$.

Function estimates have a particularly precise form when the input displacement is first-homogeneous. The following statements first use real coefficients; original-ring conclusions apply when the scalar coefficient and the required derivative values satisfy Theorem 28.

Theorem 32 

(Exact first-homogeneous function norms). Let $n\ge 1$, $u={\sum }_{j=1}^n{a}_j{\mathbf{b}}_j$, and $t={∥u∥}_{\Omega }$. For a smooth function defined near c,

$${∥\widehat{f}(c+u)∥}_{\Omega }^2=\sum _{S\subseteq \left[n\right]}{\left|f^{\left(\right|S\left|\right)}\left(c\right)\right|}^2\prod _{j\in S}{\Omega }^2{a}_j^2.$$

(122)

Consequently the sharp bounds at fixed depth and radius are

$${\left|f\left(c\right)\right|}^2+|f^{\prime }{\left(c\right)|}^2{t}^2\le {∥\widehat{f}(c+u)∥}_{\Omega }^2\le {\displaystyle \sum _{r=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right){\left|f^{\left(r\right)}\left(c\right)\right|}^2{(t^2/n)}^r.$$

(123)

The lower bound is attained by a single nonzero position. If u has s nonzero positions, lower equality holds exactly when the derivatives of orders $2,\dots ,s$ at c vanish. For $t>0$, upper equality requires all $|a_j|=t/\left(\Omega \sqrt{n}\right)$ as soon as at least one derivative of order $2,\dots ,n$ is nonzero. If all these derivatives vanish, every input of that radius attains the upper bound.

Proof. 

The coefficient at a support of size r is $f^{\left(r\right)}\left(c\right){\prod }_{j\in S}{a}_j$. Path orthogonality proves (122). Put $p_j={\Omega }^2{a}_j^2$, so that ${\sum }_j{p}_j=t^2$. Every elementary symmetric sum of order r satisfies

$$\sum _{\left|S\right|=r}\prod _{j\in S}{p}_j\le \left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right){(t^2/n)}^r.$$

(124)

For $2\le r\le n$ and $t>0$, equality holds only when the $p_j$ are equal. To see this, maximize on the compact simplex. A positive maximum has at least r positive coordinates. With the sum of two coordinates fixed, the expression has their product multiplied by the nonnegative elementary symmetric sum of order $r-2$ on the remaining coordinates. At a maximizer this multiplier is positive for every unequal pair. Averaging that pair would strictly increase the value. Thus all coordinates are equal. The cases $r=0,1$ are identities.

All terms in (122) are nonnegative. Dropping orders at least two proves the lower bound, and these orders are positive precisely when $r\le s$ and the corresponding derivative is nonzero. Applying (124) order by order proves the upper bound. The same equal-magnitude input maximizes every order, proving sharpness and all stated equality conditions. For $t=0$, the input is the scalar c and both bounds equal ${\left|f\left(c\right)\right|}^2$. For $t>0$, a single coefficient $a_1=t/\Omega$ gives the lower endpoint, while $a_j=t/\left(\Omega \sqrt{n}\right)$ for every j gives the upper endpoint over the reals. These constructions attain the prescribed radius; original-ring attainment additionally requires the chosen coefficients to lie in R. □

Over R, equal-magnitude coefficient families give genuine equality examples at their actual radii. An arbitrary prescribed real radius need not occur on an original-ring sphere. On an original-ring ball, approximation from its interior gives the same upper supremum as in the real extension, since R is dense and finite-coordinate evaluation is continuous.

Theorem 33 

(Uniform radius and optimal Lipschitz constant). Fix c and $t>0$, and allow all finite first-homogeneous inputs with ${∥u∥}_{\Omega }\le t$. Then

$$\sup {∥\widehat{f}(c+u)∥}_{\Omega }^2={\displaystyle \sum _{r=0}^{\infty }}{\displaystyle \frac{|f^{\left(r\right)}{\left(c\right)|}^2}{r!}}{t}^{2r}.$$

(125)

The optimal Lipschitz constant on this ball, with the scalar coefficient fixed at c, has square

$$L_c{\left(t\right)}^2={\displaystyle \sum _{r=1}^{\infty }}{\displaystyle \frac{r|f^{\left(r\right)}{\left(c\right)|}^2}{(r-1)!}}{t}^{2r-2}.$$

(126)

Either right-hand side may be infinite. The same suprema hold on original-ring balls whenever the evaluation preserves R there.

Proof. 

Expanding ${\left({\sum }_j{p}_j\right)}^r$ and retaining only distinct positions gives the upper estimate for the order-r elementary symmetric sum by $t^{2r}/r!$. Equation (122) proves the upper bound in (125). Equal-magnitude inputs give the sum in (123); for each fixed r,

$$\underset{n}{\lim }\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)n^{-r}=1/r!.$$

Fix finitely many orders before increasing n. Their limits prove sharpness whether the full nonnegative series converges or diverges.

Let h be first-homogeneous. Within this calculation write the normalized singleton coefficients as $a_j=\Omega u_{\left\{j\right\}}$ and $b_j=\Omega h_{\left\{j\right\}}$. At a support S of size r, differentiation of the finite formula gives the normalized coefficient

$$f^{\left(r\right)}\left(c\right)\sum _{j\in S}{b}_j\prod _{i\in S\setminus \left\{j\right\}}{a}_i.$$

Cauchy–Schwarz for its r terms, followed by summation at the common finite level n, gives

$$\begin{array}{cc}\hfill {∥f^{\left(r\right)}\left(c\right)u^{r-1}h/(r-1)!∥}_{\Omega }^2& =|f^{\left(r\right)}{\left(c\right)|}^2\sum _{\left|S\right|=r}{\left(\sum _{j\in S}{b}_j\prod _{i\in S\setminus \left\{j\right\}}{a}_i\right)}^2\hfill \\ \hfill & \le r|f^{\left(r\right)}{\left(c\right)|}^2{\displaystyle \sum _{j=1}^n}{b}_j^2\sum _{\begin{array}{c}A\subseteq \left[n\right]\setminus \left\{j\right\}\\ \left|A\right|=r-1\end{array}}\prod _{i\in A}{a}_i^2\hfill \\ \hfill & \le {\displaystyle \frac{r|f^{\left(r\right)}{\left(c\right)|}^2}{(r-1)!}}{∥u∥}_{\Omega }^{2r-2}{∥h∥}_{\Omega }^2.\hfill \end{array}$$

Indeed the sum over the other $r-1$ distinct positions is at most ${∥u∥}_{\Omega }^{2r-2}/(r-1)!$. Different output orders are orthogonal. Thus their squared bounds add to $L_c{\left(t\right)}^2{∥h∥}_{\Omega }^2$ when ${∥u∥}_{\Omega }\le t$. For two inputs in that ball, the segment remains in the same finite-dimensional ball, and

$$\begin{array}{cc}\hfill \widehat{f}(c+v)-\widehat{f}(c+u)& ={\int }_0^{1}D\widehat{f}\left(c+(1-s)u+sv\right)[v-u]ds,\hfill \\ \hfill {∥\widehat{f}(c+v)-\widehat{f}(c+u)∥}_{\Omega }& \le {\int }_0^1{L}_c\left(t\right){∥v-u∥}_{\Omega }ds=L_c\left(t\right){∥v-u∥}_{\Omega }.\hfill \end{array}$$

This proves the Lipschitz bound when the series is finite.

At an equal-magnitude input of norm t, differentiation in its unit radial direction has squared norm

$${\displaystyle \sum _{r=1}^n}{\left|f^{\left(r\right)}\left(c\right)\right|}^2{r}^2\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)n^{-r}{t}^{2r-2}.$$

Interior radial difference quotients approach this value. Increasing the number of positions and then the number of retained orders proves the lower bound in (126), including the infinite case. Explicitly, take a flat input $u_n$ with norm t and compare it with $(1-s)u_n$ for $0<s<1$. Both lie in the ball. Taking $s=1/l$ for integers $l\ge 2$ gives

$$\underset{l}{\lim }{\displaystyle \frac{{∥\widehat{f}(c+u_n)-\widehat{f}\left(c+(1-l^{-1})u_n\right)∥}_{\Omega }}{t/l}}={∥D\widehat{f}(c+u_n)[u_n/t]∥}_{\Omega }.$$

For any fixed number of orders, the squared right side has limit the corresponding partial sum of (126) as n increases. Every Lipschitz constant is therefore at least every partial-sum square root. This excludes a finite constant when the series diverges. For original-ring inputs, approximate each finite pair used in these difference quotients from inside the ball. Continuity of the finite formula preserves the limit of the ratios. □

For the exponential, (125) equals $\exp (2c+t^2)$; at $c=0$ its optimal Lipschitz constant is $\exp (t^2/2)\sqrt{1+t^2}$. In contrast, $f\left(s\right)={(1-s)}^{-1}$ at $c=0$ gives the divergent series ${\sum }_{r\ge 0}r!t^{2r}$ for every $t>0$. Each finite value still exists over R, since $1-u$ is a unit.

Corollary 6 

(Separable exponential, logarithm, and inverse). For $n\ge 1$, $u={\sum }_{j=1}^n{a}_j{\mathbf{b}}_j$, and $x=\exp \left(u\right)$,

$$\begin{array}{cc}\hfill x& ={\displaystyle \prod _{j=1}^n}(1+a_j{\mathbf{b}}_j),\hfill \\ \hfill {∥x∥}_{\Omega }^2& ={∥x^{-1}∥}_{\Omega }^2={\displaystyle \prod _{j=1}^n}(1+{\Omega }^2{a}_j^2),\hfill \end{array}$$

(127)

$$\begin{array}{cc}\hfill 1+{∥u∥}_{\Omega }^2& \le {∥\exp \left(u\right)∥}_{\Omega }^2\le {(1+{∥u∥}_{\Omega }^2/n)}^n,\hfill \end{array}$$

(128)

$$\begin{array}{cc}\hfill n({∥x∥}_{\Omega }^{2/n}-1)& \le {∥\log \left(x\right)∥}_{\Omega }^2\le {∥x∥}_{\Omega }^2-1.\hfill \end{array}$$

(129)

In (), lower equality means at most one nonzero coefficient, and upper equality for nonzero u means all $|a_j|$ are equal. These conditions give original-ring equality examples and the corresponding equalities in (). More generally, if u is the sum of $s\ge 1$ nonzero path terms with pairwise disjoint nonempty supports, the same statements hold with s in place of n and the norms of those terms in place of $|\Omega a_j|$.

Proof. 

Each path term is square zero, and the additive exponential law in Theorem 5 gives (127). All products of distinct terms have distinct supports. Therefore

$${∥x∥}_{\Omega }^2=\sum _{S\subseteq \left[n\right]}\prod _{j\in S}{\Omega }^2{a}_j^2={\displaystyle \prod _{j=1}^n}(1+{\Omega }^2{a}_j^2).$$

Replacing every $a_j$ by $-a_j$ gives $x^{-1}$ with the same norm. For $p_j={\Omega }^2{a}_j^2$, product expansion and arithmetic–geometric mean give

$$1+\sum _j{p}_j\le \prod _j(1+p_j)\le {\left({\displaystyle \frac{{\sum }_j(1+p_j)}{n}}\right)}^n={\left(1+{\displaystyle \frac{{\sum }_j{p}_j}{n}}\right)}^n.$$

Lower equality requires every product $p_i{p}_j$ with $i\ne j$ to vanish. Upper equality requires all $1+p_j$ to agree. These are exactly the stated support and magnitude conditions. Substitution of $u=\log \left(x\right)$ gives (), since the two rearrangements are

$${∥x∥}_{\Omega }^{2/n}\le 1+{∥u∥}_{\Omega }^2/n,{∥u∥}_{\Omega }^2\le {∥x∥}_{\Omega }^2-1.$$

For disjoint nonempty supports, distinct collections of supports have distinct unions. Therefore the product expansion and all orthogonality steps remain valid, with the norm of each path term providing its weight. □

The uniform exponential supremum on the first-homogeneous ball of radius t is $\exp \left(t^2\right)$. For $t>0$ it is not attained at any finite depth, because ${(1+t^2/n)}^n<\exp \left(t^2\right)$. The lower bound in () depends on the input form. For example, in the original ring take

$$u={\Omega }^{-1}({\mathbf{b}}_1+{\mathbf{b}}_2)-{\Omega }^{-2}{\mathbf{b}}_{\{1,2\}}.$$

Then ${∥u∥}_{\Omega }^2=3$, whereas $\exp \left(u\right)=1+{\Omega }^{-1}({\mathbf{b}}_1+{\mathbf{b}}_2)$ has squared norm 3, strictly less than $1+{∥u∥}_{\Omega }^2$.

5.3. Degree Scaling and Global Domains

Theorem 34 

(Square-zero inputs and a cubic-nilpotent family). If $u^2=0$, then

$${∥\exp \left(u\right)∥}_{\Omega }={∥{(1+u)}^{-1}∥}_{\Omega }=\sqrt{1+{∥u∥}_{\Omega }^2},{∥\log (1+u)∥}_{\Omega }={∥u∥}_{\Omega }.$$

(130)

For $h\in {\mathcal{I}}_n$, consider instead

$$u={\Omega }^{-1}({\mathbf{b}}_{n+1}+{\mathbf{b}}_{n+2})h\mathrm{in}{\mathcal{I}}_{n+2}.$$

(131)

Then $u^3=0$ and

$$\begin{array}{cccc}\hfill {∥u∥}_{\Omega }^2& =2{∥h∥}_{\Omega }^2,\hfill & \hfill {∥u^2∥}_{\Omega }^2& =4{∥h^2∥}_{\Omega }^2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {∥\exp \left(u\right)∥}_{\Omega }^2& =1+{∥u∥}_{\Omega }^2+\frac{1}{4}{∥u^2∥}_{\Omega }^2,\hfill \\ \hfill {∥\log (1+u)∥}_{\Omega }^2& ={∥u∥}_{\Omega }^2+\frac{1}{4}{∥u^2∥}_{\Omega }^2,\hfill \end{array}$$

(132)

$$\begin{array}{cc}\hfill {∥{(1+u)}^{-1}∥}_{\Omega }^2& =1+{∥u∥}_{\Omega }^2+{∥u^2∥}_{\Omega }^2.\hfill \end{array}$$

(133)

For $n\ge 1$ and $t>0$, the suprema of the last three quantities over the family (131) with ${∥u∥}_{\Omega }\le t$ are, respectively,

$$1+t^2+\frac{1}{4}{(4/3)}^n{t}^4,t^2+\frac{1}{4}{(4/3)}^n{t}^4,1+t^2+{(4/3)}^n{t}^4.$$

(134)

They are attained over the reals. Their nonzero equality inputs have ${∥h∥}_{\Omega }=t/\sqrt{2}$ and

$$h=\lambda {\displaystyle \prod _{j=1}^n}\left(1+{\displaystyle \frac{{\sigma }_j}{\Omega \sqrt{2}}}{\mathbf{b}}_j\right),{\sigma }_j\in \{-1,1\},\lambda \ne 0.$$

(135)

Over R the suprema are the same but are not attained for $n\ge 1$.

Proof. 

A square-zero element has scalar coefficient zero. Its exponential, logarithm, and normalized inverse reduce to $1+u$, u, and $1-u$, respectively; the scalar and nonempty paths are orthogonal.

For (131), multiplication of the two marker positions gives

$$u^2=2{\Omega }^{-2}{\mathbf{b}}_{\{n+1,n+2\}}{h}^2,u^3=0.$$

The positions of h are disjoint from these markers. The two terms of u are orthogonal, each with norm ${∥h∥}_{\Omega }$; the displayed square has norm $2{∥h^2∥}_{\Omega }$. Moreover $1,u,u^2$ use respectively zero, one, and two markers and hence are pairwise orthogonal. Substitute the finite formulas $\exp \left(u\right)=1+u+u^2/2$, $\log (1+u)=u-u^2/2$, and ${(1+u)}^{-1}=1-u+u^2$. The division by two is integral because of the explicit factor two in $u^2$. For example, the exponential calculation is

$$\begin{array}{cc}\hfill {∥1+u+u^2/2∥}_{\Omega }^2& =1+{∥u∥}_{\Omega }^2+\frac{1}{4}{∥u^2∥}_{\Omega }^2\hfill \\ \hfill & +2{〈1,u〉}_{\Omega }+{〈1,u^2〉}_{\Omega }+{〈u,u^2〉}_{\Omega }\hfill \\ \hfill & =1+{∥u∥}_{\Omega }^2+\frac{1}{4}{∥u^2∥}_{\Omega }^2.\hfill \end{array}$$

The logarithm and inverse use the same three orthogonal components with coefficients $(0,1,-1/2)$ and $(1,-1,1)$, respectively.

Corollary 2 gives ${∥h^2∥}_{\Omega }\le {(2/\sqrt{3})}^n{∥h∥}_{\Omega }^2$, with precisely the nonzero equality form (135). Since ${∥u∥}_{\Omega }^2=2{∥h∥}_{\Omega }^2$, substitution in the three increasing expressions uses the exact bound

$${∥u^2∥}_{\Omega }^2=4{∥h^2∥}_{\Omega }^2\le 4{(4/3)}^n{∥h∥}_{\Omega }^4={(4/3)}^n{∥u∥}_{\Omega }^4\le {(4/3)}^n{t}^4.$$

This proves (134); equality requires both the maximal radius and the multiplication equality form already stated. Their original-ring suprema follow by finite-dimensional approximation from the ball interior. A nonzero original-ring extremizer would require the ratio of a first-path coefficient of h to its scalar coefficient to be ${\sigma }_j/\left(\Omega \sqrt{2}\right)$, which is irrational. Thus none exists. □

Example 9 

(An explicit original-ring sequence). The cubic family permits arbitrarily small inputs with arbitrarily large exponential, logarithmic, and inverse norms. Set $a=10/7$, so that $\sqrt{2}<a<5^{1/4}$, and define

$${\lambda }_n={\Omega }^{-\lceil n{\log }_{\Omega }a\rceil },h_n={\lambda }_n{(1,{\Omega }^{-1})}^{\otimes n},u_n={\Omega }^{-1}({\mathbf{b}}_{n+1}+{\mathbf{b}}_{n+2})h_n.$$

The actual coefficients are in R, and

$${∥h_n∥}_{\Omega }={\lambda }_n{2}^{n/2},{∥h_n^2∥}_{\Omega }={\lambda }_n^2{5}^{n/2},{\Omega }^{-1}{a}^{-n}\le {\lambda }_n\le a^{-n}.$$

Consequently

$${∥u_n∥}_{\Omega }\le \sqrt{2}{(\sqrt{2}/a)}^n,{∥u_n^2∥}_{\Omega }\ge 2{\Omega }^{-2}{(\sqrt{5}/a^2)}^n.$$

The first quantity tends to zero and the second to infinity. Equations (132)–() show that all three function norms tend to infinity, even though every $u_n^3$ is zero. Thus the stable identities on square-zero inputs do not extend to cubic-nilpotent inputs with a bound independent of depth.

A second type of recursive function acts directly on the path coefficients and interacts with both tensor orientations.

Definition 8 

(Recursive coefficient scaling). For a real number a, define

$${\mathcal{D}}_a\left(c\right)=c,{\mathcal{D}}_a\left(x\right)=\left({\mathcal{D}}_a\left(pos\left(x\right)\right),a{\mathcal{D}}_a\left(len\left(x\right)\right)\right)\mathrm{at}\mathrm{positive}\mathrm{depth}.$$

(136)

When $a\in R$, this is an operation over the original coefficient ring.

Proposition 16 

(Algebraic compatibility and exact norm). The scaling satisfies

$${\mathcal{D}}_a\left({\mathbf{b}}_S\right)=a^{\left|S\right|}{\mathbf{b}}_S,{\mathcal{D}}_a{\mathcal{D}}_b={\mathcal{D}}_{ab}.$$

(137)

It commutes with refinement and preserves sums, ordinary products, the unit, and both directed tensor products at their specified depths. It consequently preserves inverses of units, powers, and the Lie and Jordan operations where defined. For every admissible scalar-data evaluation,

$${\mathcal{D}}_a\left({\widehat{f}}_j\left(x\right)\right)={\widehat{f}}_j\left({\mathcal{D}}_a\left(x\right)\right).$$

(138)

Its relations with the components are

$$pos\left({\mathcal{D}}_a\left(x\right)\right)={\mathcal{D}}_a\left(pos\left(x\right)\right),len\left({\mathcal{D}}_a\left(x\right)\right)=a{\mathcal{D}}_a\left(len\left(x\right)\right).$$

(139)

The norm and the positive-depth operator norm are exactly

$${∥{\mathcal{D}}_a\left(x\right)∥}_{\Omega }^2=\sum _{S\subseteq \left[n\right]}{\left|a\right|}^{2\left|S\right|}{\Omega }^{2\left|S\right|}{x}_S^2,{\parallel {\mathcal{D}}_a\parallel }_{{\mathcal{I}}_n}={\max \{1,|a|}^n\}(n\ge 1).$$

(140)

The scalar-level operator norm is one. It has a finite operator norm independent of depth exactly when $\left|a\right|\le 1$, in which case that norm is one.

Proof. 

Every occurrence of len contributes one factor a, proving (137). Disjoint path multiplication adds support sizes, so

$$\begin{array}{cc}\hfill {\left({\mathcal{D}}_a\left(x\right){\mathcal{D}}_a\left(y\right)\right)}_S& =\sum _{A\bigsqcup B=S}{a}^{\left|A\right|+\left|B\right|}{x}_A{y}_B\hfill \\ \hfill & =a^{\left|S\right|}{\left(xy\right)}_S={\left({\mathcal{D}}_a\left(xy\right)\right)}_S.\hfill \end{array}$$

Overlapping products are zero before and after scaling. For a concatenated path, its two blocks have cardinalities $\left|A\right|,\left|B\right|$, so both sides of tensor compatibility have coefficient $a^{\left|A\right|+\left|B\right|}{x}_A{y}_B$. The argument is the same in either order. Addition is coefficientwise. Refinement adds an unused position and hence leaves all factors unchanged. Preservation of the unit and product implies preservation of unit inverses and therefore of all stated derived operations.

The scalar coefficient is fixed. Apply product preservation and preservation of exact integer division to (111) to obtain, for $x=c+u$ in depth n,

$$\begin{array}{cc}\hfill {\mathcal{D}}_a\left({\widehat{f}}_j\left(x\right)\right)& ={\displaystyle \sum _{r=0}^n}{f}_{j+r}\left(c\right){\displaystyle \frac{{\mathcal{D}}_a{\left(u\right)}^r}{r!}}\hfill \\ \hfill & ={\widehat{f}}_j\left(c+{\mathcal{D}}_a\left(u\right)\right)={\widehat{f}}_j\left({\mathcal{D}}_a\left(x\right)\right).\hfill \end{array}$$

The component identities follow directly from the defining recursion, including scalar inputs. Path orthogonality proves the first norm formula. Its diagonal factors have maximum $\max \{1,|a{|}^n\}$, attained respectively on the scalar path or the full path. For $\left|a\right|\le 1$, each diagonal factor is at most one, and the scalar path attains one. If $\left|a\right|>1$, the full path gives

$${\displaystyle \frac{{∥{\mathcal{D}}_a\left({\mathbf{b}}_{\left[n\right]}\right)∥}_{\Omega }}{{∥{\mathbf{b}}_{\left[n\right]}∥}_{\Omega }}}={\left|a\right|}^n.$$

These ratios have no finite common bound as n increases. This proves both directions of the depth-independent assertion. □

Theorem 35 

(Sharp depth-independent product threshold). For $k\ge 1$ and real parameters $a_1,\dots ,a_k$, the inequality

$${∥{\displaystyle \prod _{i=1}^k}{\mathcal{D}}_{a_i}\left(x_i\right)∥}_{\Omega }\le {\displaystyle \prod _{i=1}^k}{∥x_i∥}_{\Omega }$$

(141)

holds for all finite inputs of arbitrary depths exactly when

$${\displaystyle \sum _{i=1}^k}{a}_i^2\le 1.$$

(142)

Under this condition the constant one is best possible and is attained by nonzero original-ring scalar inputs. If the sum exceeds one, no finite constant independent of depth can replace one in (141). When the parameters belong to R, the latter assertion already holds for original-ring inputs.

For nonzero inputs under (142), equality is classified as follows. If exactly one parameter has absolute value one, all the other inputs must be scalar and that input is unrestricted. In every other case all inputs must be scalar. Here an input is scalar when all its nonempty path coefficients vanish, regardless of its typed depth.

Proof. 

Induct on the largest input depth. To record the real component norms inside this calculation, put $p_i={∥pos\left(x_i\right)∥}_{\Omega }$ and $q_i=\Omega {∥len\left(x_i\right)∥}_{\Omega }$. The lower-depth inequality, applied to each ordinary product in the two components, and the triangle inequality bound the squared output by

$$\prod _i{p}_i^2+{\left(\sum _i\left|a_i\right|q_i\prod _{j\ne i}{p}_j\right)}^2.$$

(143)

Cauchy–Schwarz and (142) bound the second term as follows:

$${\left(\sum _i\left|a_i\right|q_i\prod _{j\ne i}{p}_j\right)}^2\le \left(\sum _i{a}_i^2\right)\sum _i{q}_i^2\prod _{j\ne i}{p}_j^2\le \sum _i{q}_i^2\prod _{j\ne i}{p}_j^2.$$

The difference between the full input product and the resulting bound is

$$\begin{array}{cc}\hfill & \prod _i(p_i^2+q_i^2)-\prod _i{p}_i^2-\sum _i{q}_i^2\prod _{j\ne i}{p}_j^2\hfill \\ \hfill & =\sum _{\begin{array}{c}A\subseteq \left[k\right]\\ \left|A\right|\ge 2\end{array}}\prod _{i\in A}{q}_i^2\prod _{j\notin A}{p}_j^2\ge 0.\hfill \end{array}$$

A scalar input has $q_i=0$; thus the same induction handles different depths. Nonzero scalar inputs prove sharpness of the constant.

If $s={\sum }_i{a}_i^2>1$, choose a small nonzero real t and depth-one inputs $x_i=(1,t{a}_i/\Omega )$. Their squared norm ratio is

$${\displaystyle \frac{1+t^2{s}^2}{{\prod }_i(1+t^2{a}_i^2)}}=1+s(s-1)t^2+O\left(t^4\right)>1$$

for sufficiently small t. Replace each input by its repeated directed tensor power. Product compatibility within fixed blocks and tensor norm factorization raise this ratio to the number of blocks. It is therefore unbounded. For parameters in R, choose $t\in R$ small; every coefficient of this construction then remains in R.

For equality, first suppose every $|a_i|<1$. If two or more of the $p_i$ vanish while every input is nonzero, (143) is zero and equality is impossible. If exactly one $p_i$ vanishes, its ratio to ${\prod }_i(p_i^2+q_i^2)$ is at most $a_i^2<1$. When all $p_i>0$, equality in the expansion of the product allows at most one nonzero $q_i/p_i$. Equality in Cauchy–Schwarz then requires either every $q_i=0$, or a single parameter of absolute value one. The latter is excluded in this case. The strictness with a single nonzero $q_l$ can also be read directly: after cancelling the other positive $p_i^2$, the squared ratio in the component estimate is

$${\displaystyle \frac{p_l^2+a_l^2{q}_l^2}{p_l^2+q_l^2}}=1-{\displaystyle \frac{(1-a_l^2)q_l^2}{p_l^2+q_l^2}}<1.$$

Hence all second components vanish. Repeating on the first components shows that all inputs are scalar. In the remaining parameter case, one parameter is 1 or $-1$ and every other parameter is zero. Its operation is an isometry, while the other operations take scalar coefficients. Their norms equal the input norms exactly for scalar inputs, proving the asserted equality conditions. □

The first-homogeneous radius estimates and the damping threshold give different forms of uniform control. On the whole set of nilpotent displacements, the finite functional calculus has a stricter continuity criterion. Continuity and differentiability on zeon algebras are studied in [13]. The question here concerns one normed space of finite recursive objects with no common depth bound. On each fixed finite level a fixed-scalar evaluation is a polynomial in finitely many path coefficients; the following result determines what survives when perturbations may use arbitrarily many further positions.

Theorem 36 

(Continuity on a fixed scalar fiber). Fix a scalar coefficient c and admissible scalar data $f_j\left(c\right)$. On $c+\mathcal{N}$, equipped with the Ω norm, the following conditions are equivalent: evaluation is locally bounded at some point; it is continuous at some point; and $f_j\left(c\right)=0$ for every integer $j\ge 2$. In this case

$${\widehat{f}}_0(c+u)=f_0\left(c\right)+f_1\left(c\right)u$$

(144)

at every finite input, and its optimal Lipschitz constant is $|f_1\left(c\right)|$. If any higher datum is nonzero, evaluation is locally unbounded at every point of this fiber. These statements hold both for admissible original-ring data and for real scalar extension.

Proof. 

If all higher data vanish, the finite formula reduces to (144). Scalar multiplication has norm $|f_1\left(c\right)|$, which is the exact difference ratio on every nonzero displacement. Continuity at a point always implies local boundedness there.

For the converse choose an integer $d\ge 2$ with $f_d\left(c\right)\ne 0$ and fix any finite input x of scalar coefficient c. Choose $b>0$ in R sufficiently small that

$${(1+b^2)}^d<1+d^2{b}^2.$$

Such a choice exists: as a polynomial in $b^2$, the difference of the right and left sides has positive linear coefficient $d^2-d$. Choose a real number a strictly between $\sqrt{1+b^2}$ and ${(1+d^2{b}^2)}^{1/\left(2d\right)}$, and put

$${\lambda }_n={\Omega }^{-\lceil n{\log }_{\Omega }a\rceil },h_n={\lambda }_n{(1,b/\Omega )}^{\otimes n}.$$

Place the n positions of $h_n$ after every position used by x; inserting an initial scalar unit block does this without changing its norms. The path product and fixed-block norm factorization give

$${∥h_n∥}_{\Omega }={\lambda }_n{(1+b^2)}^{n/2},{∥h_n^d∥}_{\Omega }={\lambda }_n^d{(1+d^2{b}^2)}^{n/2}.$$

Since ${\Omega }^{-1}{a}^{-n}\le {\lambda }_n\le a^{-n}$, the two bounds are

$$\begin{array}{cc}\hfill {∥h_n∥}_{\Omega }& \le {\left({\displaystyle \frac{\sqrt{1+b^2}}{a}}\right)}^n,\hfill \\ \hfill {∥h_n^d∥}_{\Omega }& \ge {\Omega }^{-d}{\left({\displaystyle \frac{\sqrt{1+d^2{b}^2}}{a^d}}\right)}^n.\hfill \end{array}$$

The first base is less than one and the second is greater than one, so the first norm has limit zero and the second is unbounded.

Choose d further positions $p_1,\dots ,p_d$, disjoint from all those already used, and define

$$u_n={\Omega }^{-1}({\mathbf{b}}_{p_1}+\dots +{\mathbf{b}}_{p_d})h_n.$$

Then $u_n^{d+1}=0$, ${∥u_n∥}_{\Omega }=\sqrt{d}{∥h_n∥}_{\Omega }$ tends to zero, and ${∥u_n^d∥}_{\Omega }=d!{∥h_n^d∥}_{\Omega }$ tends to infinity. The finite translation identity for the scalar data is

$${\widehat{f}}_0(x+u_n)={\displaystyle \sum _{j=0}^d}{\widehat{f}}_j\left(x\right){\displaystyle \frac{u_n^j}{j!}}.$$

To verify it, expand (111) at the common scalar coefficient c and apply the binomial theorem to $(x-c)+u_n$; collection of the $u_n$ powers gives the displayed sum. Its terms contain different numbers of the new marker positions and are orthogonal. In addition, ${\widehat{f}}_j\left(x\right)$ uses only the old positions, whereas $u_n$ uses only new ones, so their product norm factorizes. Therefore

$${∥{\widehat{f}}_0(x+u_n)∥}_{\Omega }\ge {\displaystyle \frac{{∥{\widehat{f}}_d\left(x\right)∥}_{\Omega }}{d!}}{∥u_n^d∥}_{\Omega }.$$

The first factor is positive because its scalar coefficient is $f_d\left(c\right)\ne 0$. This proves unbounded values in every neighborhood of x. All constructed coefficients are in R when the original ring is under consideration. □

Corollary 7 

(Varying the real scalar coefficient). Let f be smooth on a real open interval and let x be a finite real input of scalar coefficient c in that interval. On all finite real inputs with allowed scalar coefficients, the following are equivalent: evaluation is locally bounded at x; it is continuous at x; and f is affine on a neighborhood of c.

Proof. 

If $f\left(t\right)=at+b$ on such a neighborhood, every nearby input evaluates to the corresponding affine expression. The scalar coefficient is a continuous norm-one functional, and for nearby y one has

$$|y_{⌀}-c|\le {∥y-x∥}_{\Omega },{∥\widehat{f}\left(y\right)-\widehat{f}\left(x\right)∥}_{\Omega }=\left|a\right|{∥y-x∥}_{\Omega }.$$

The first inequality keeps the scalar coefficient in the affine neighborhood; the second proves continuity.

If f is not affine on any neighborhood of c, arbitrarily close scalar coefficients $c^{\prime }$ satisfy $f^{\prime \prime }\left(c^{\prime }\right)\ne 0$. Otherwise its second derivative would vanish on a neighborhood, making it affine there. Given a radius $r>0$, choose such a $c^{\prime }$ with $|c^{\prime }-c|<r/2$ and set $x^{\prime }=x+(c^{\prime }-c)$. Theorem 36, with $d=2$, supplies inputs $y_j$ of scalar coefficient $c^{\prime }$ such that

$${∥y_j-x^{\prime }∥}_{\Omega }<r/2,\underset{j}{\sup }{∥\widehat{f}\left(y_j\right)∥}_{\Omega }=+\infty .$$

They remain in the original ball, since

$${∥y_j-x∥}_{\Omega }\le {∥y_j-x^{\prime }∥}_{\Omega }+{∥x^{\prime }-x∥}_{\Omega }<r/2+|c^{\prime }-c|<r.$$

Thus evaluation is unbounded in every ball about x; local boundedness and continuity both fail. □

The fixed-coefficient condition involves derivatives at one point, whereas Corollary 7 involves an actual scalar neighborhood. A smooth function can have every higher derivative zero at one point without being affine near it. On original-ring rational domains restricted by unit denominators, the allowed scalar coefficients can be locally discrete. The fixed-scalar criterion still applies, while the neighborhood criterion requires an open real scalar domain.

6. Free Concatenation, Lie Brackets, and Jordan Products

The two tensor orders developed above also determine operations through their difference and sum. Keeping the prescribed depths gives a free associative algebra whose commutator is a Lie product and whose symmetrization is a Jordan product when 2 is invertible. Its faithful action on global recursive objects connects this graded construction with scalar descent. Fixed-block metric factorization then gives exact norm identities and sharp bounds for commutators and associators. Reversal of the recursive paths organizes the symmetric and antisymmetric components, while the dependence on block lengths specifies which conclusions belong to typed objects.

6.1. Free Concatenation and Its Global Action

In the graded construction, addition keeps different levels as separate summands. It is the direct-sum addition used to extend concatenation bilinearly, distinct from mixed-depth addition on global refinement classes.

Definition 9 

(The graded tensor algebra). Let

$$\mathcal{A}=\underset{n\ge 0}{⨁}{\mathcal{I}}_n.$$

(145)

An element of $\mathcal{A}$ is a finite sum of typed components. Extend the left tensor by R-bilinearity with respect to this direct-sum addition. Its zero is denoted by 0, and its identity by $1=1_0$. Write

$$v_{pos}=(1,0),v_{len}=(0,1)$$

(146)

for the two basis elements of its degree-one component. Each summand has the same zero in $\mathcal{A}$. Its ordinary unit, however, is the homogeneous word $1_n=v_{pos}^{\otimes n}$ for $n\ge 1$, distinct from the tensor identity $1_0$.

Theorem 37 

(Free associative structure). The algebra $(\mathcal{A},\otimes )$ is the free associative unital R-algebra on $v_{pos}$ and $v_{len}$. Its degree-n component has the $2^n$ words of length n in these two elements as a basis. The right tensor is the multiplication of the opposite algebra.

Proof. 

Successive use of (81) identifies

$${\mathcal{I}}_n\cong \underset{n\mathrm{factors}}{\underbrace{{\mathcal{I}}_1{\otimes }_R\dots {\otimes }_R{\mathcal{I}}_1}}.$$

In each position, $v_{pos}$ chooses the pos component and $v_{len}$ chooses the len component. The resulting words are exactly the path basis ${\mathbf{b}}_S$ of depth n. Associativity from Theorem 19 makes the product of two words their concatenation. Given two elements of any associative unital R-algebra, evaluate each word by replacing the two letters with these elements and extend linearly. The empty word has value the target unit. For two words $u,v$, evaluation satisfies

$$\theta (u\otimes v)=\theta \left(u\right)\theta \left(v\right).$$

Distributing over two finite word expansions proves this identity for arbitrary elements of $\mathcal{A}$. Conversely, any homomorphism with the prescribed generator values must have these values on every word and hence on every finite linear combination. This proves existence and uniqueness in the universal property of the free algebra. The identity $x{\otimes }^{\mathrm{op}}y=y\otimes x$ proves the last assertion. □

Theorem 38 

(Faithful action and the nilradical as a free module). For a fixed $x\in {\mathcal{I}}_m$, the rule

$$L_x\left(\left[y\right]\right)=[x\otimes y]$$

(147)

defines an R-linear operator on $\mathcal{I}$. Its real scalar extension satisfies

$${∥L_x\left(u\right)∥}_{\Omega }={∥x∥}_{\Omega }{∥u∥}_{\Omega },\parallel L_x\parallel ={∥x∥}_{\Omega },L_{x\otimes y}=L_x{L}_y.$$

(148)

It extends to the weighted completion ${\mathcal{H}}_{\Omega }$. For a fixed y, assigning $\left[x{\otimes }^{\mathrm{op}}y\right]$ to the class $\left[x\right]$ gives the same operator $L_y$.

Extending x linearly across the graded summands gives a faithful unital action of $\mathcal{A}$ on $\mathcal{I}$ by bounded operators. The nilradical $\mathcal{N}$ is invariant under this action, and

$$\psi \left(a\right)=L_a\left(\left[v_{len}\right]\right)(a\in \mathcal{A})$$

(149)

is an isomorphism of left $\mathcal{A}$-modules between the regular module $\mathcal{A}$ and $\mathcal{N}$. In particular, $\mathcal{N}$ is free of rank one for this directed tensor action. If $a={\sum }_{n=0}^N{a}_n$ with $a_n\in {\mathcal{I}}_n$, then

$${∥\psi \left(a\right)∥}_{\Omega }^2={\Omega }^2{\displaystyle \sum _{n=0}^N}{∥a_n∥}_{\Omega }^2.$$

(150)

Proof. 

Second-input covariance in Proposition 10 shows that (147) is independent of the representative of $\left[y\right]$. Indeed, for every $r\ge 0$,

$$x\otimes {\iota }^{\left(r\right)}\left(y\right)={\iota }^{\left(r\right)}(x\otimes y).$$

If two representatives of $\left[y\right]$ agree after refinement, their outputs therefore agree after the corresponding refinements as well. The depth of the fixed action parameter x stays unchanged throughout. Mixed-depth distributivity in that input proves linearity. Tensor norm factorization proves the first identity in (148), and evaluation at the global unit 1 proves the exact operator norm. Associativity gives, for a third input z,

$$L_{x\otimes y}\left(\left[z\right]\right)=\left[(x\otimes y)\otimes z\right]=\left[x\otimes (y\otimes z)\right]=L_x\left(L_y\left(\left[z\right]\right)\right).$$

Boundedness gives the continuous extension to the completion. The right-oriented rule follows from the opposite identity. For a finite graded sum, its action is the sum of the individual bounded operators, with norm at most ${\sum }_n{∥a_n∥}_{\Omega }$.

An input with zero scalar coefficient remains of zero scalar coefficient after tensoring on the left, so $\mathcal{N}$ is invariant. By the block path formula,

$$\psi \left({\mathcal{I}}_n\right)={span}_R\{{\mathbf{b}}_S:\max S=n+1\}.$$

Indeed, the last position of $a_n\otimes v_{len}$ is occupied and its earlier positions have exactly the arbitrary coefficients of $a_n$. Every nonempty finite support has a unique last position. Explicitly, for $u\in \mathcal{N}$ define

$$a_n=\sum _{S\subseteq \left[n\right]}{u}_{S\cup \{n+1\}}{\mathbf{b}}_S\in {\mathcal{I}}_n.$$

Only finitely many $a_n$ are nonzero, and the block path formula gives

$$\psi \left(\sum _n{a}_n\right)=\sum _n\sum _{S\subseteq \left[n\right]}{u}_{S\cup \{n+1\}}{\mathbf{b}}_{S\cup \{n+1\}}=u.$$

The same coefficients are forced by u, so $\psi$ is bijective. Associativity gives

$$\psi (b\otimes a)=L_b\psi \left(a\right),$$

which proves the module assertion. These submodules are also mutually orthogonal for the global weighted form. Therefore

$${∥\psi \left(a\right)∥}_{\Omega }^2=\sum _n\sum _{S\subseteq \left[n\right]}{\Omega }^{2\left(\right|S|+1)}{\left(a_n\right)}_S^2={\Omega }^2\sum _n{∥a_n∥}_{\Omega }^2.$$

Finally, if $a\ne 0$, then $L_a\left(\left[v_{len}\right]\right)=\psi \left(a\right)\ne 0$, proving faithfulness. □

The free-module assertion concerns the directed action. Ordinary commutative multiplication has the different nilpotent and annihilator properties established in Section 2. The action parameter retains its depth: for instance, $L_{1_0}$ and $L_{1_1}$ take different values on $\left[v_{len}\right]$ by (92).

Proposition 17 

(The refinement quotient as a module). Let

$$\mathcal{J}={span}_R\{x\otimes v_{pos}-x:x\in {\mathcal{I}}_n,n\ge 0\}.$$

(151)

There is an R-module identification $\mathcal{I}\cong \mathcal{A}/\mathcal{J}$. The submodule $\mathcal{J}$ is a left ideal of $(\mathcal{A},\otimes )$, but is not a right ideal.

Proof. 

Assign to a finite graded sum the sum of its global classes. The generators of $\mathcal{J}$ have value zero because $x\otimes v_{pos}=\iota \left(x\right)$. Conversely, for a kernel element $a={\sum }_{n=0}^N{a}_n$, the common-depth sum ${\sum }_n{\iota }^{(N-n)}\left(a_n\right)$ is zero in ${\mathcal{I}}_N$. In $\mathcal{A}$, each difference is the telescoping sum

$${\iota }^{(N-n)}\left(a_n\right)-a_n={\displaystyle \sum _{j=0}^{N-n-1}}\left({\iota }^{\left(j\right)}\left(a_n\right)\otimes v_{pos}-{\iota }^{\left(j\right)}\left(a_n\right)\right)\in \mathcal{J}.$$

The common-depth sum is the zero element of its typed level, by Lemma 1. Hence, in the graded module,

$$a={\displaystyle \sum _{n=0}^N}\left(a_n-{\iota }^{(N-n)}\left(a_n\right)\right)\in \mathcal{J}.$$

This proves equality with the kernel. Surjectivity holds because each global object has a finite typed representative, which is itself one homogeneous element of $\mathcal{A}$. The quotient identification follows. Left multiplication preserves each generator because

$$y\otimes (x\otimes v_{pos}-x)=(y\otimes x)\otimes v_{pos}-y\otimes x\in \mathcal{J}.$$

However, $v_{pos}-1_0$ belongs to $\mathcal{J}$, while

$$(v_{pos}-1_0)\otimes v_{len}=v_{pos}\otimes v_{len}-v_{len}$$

has global value ${\mathbf{b}}_{\left\{2\right\}}-{\mathbf{b}}_{\left\{1\right\}}\ne 0$. It does not belong to $\mathcal{J}$. This proves the right-ideal failure. □

Definition 10 

(Tensor commutator). For homogeneous elements, and then by bilinear extension in $\mathcal{A}$, define

$${[x,y]}_{\otimes }=x\otimes y-y\otimes x.$$

(152)

Theorem 39 

(The commutator Lie algebra). The bracket is an ordinary $\mathbb{N}$-graded Lie bracket over R. It satisfies

$$\begin{array}{cc}\hfill {[{\mathcal{I}}_m,{\mathcal{I}}_n]}_{\otimes }& \subseteq {\mathcal{I}}_{m+n},\hfill \\ \hfill {[x,x]}_{\otimes }& =0,{[x,y]}_{\otimes }=-{[y,x]}_{\otimes },\hfill \\ \hfill {\left[x,{[y,z]}_{\otimes }\right]}_{\otimes }+{\left[y,{[z,x]}_{\otimes }\right]}_{\otimes }+{\left[z,{[x,y]}_{\otimes }\right]}_{\otimes }& =0,\hfill \\ \hfill {[x,y\otimes z]}_{\otimes }& ={[x,y]}_{\otimes }\otimes z+y\otimes {[x,z]}_{\otimes }.\hfill \end{array}$$

(153)

Its action is faithful and respects operator commutators:

$$L_{{[x,y]}_{\otimes }}=L_x{L}_y-L_y{L}_x.$$

(154)

Proof. 

The tensor grading and bilinearity give the degree and bilinearity statements. Subtraction proves alternation and antisymmetry. For Jacobi, associativity gives the three expansions

$$\begin{array}{cc}\hfill {\left[x,{[y,z]}_{\otimes }\right]}_{\otimes }& =x\otimes y\otimes z-x\otimes z\otimes y\hfill \\ \hfill & -y\otimes z\otimes x+z\otimes y\otimes x,\hfill \\ \hfill {\left[y,{[z,x]}_{\otimes }\right]}_{\otimes }& =y\otimes z\otimes x-y\otimes x\otimes z\hfill \\ \hfill & -z\otimes x\otimes y+x\otimes z\otimes y,\hfill \\ \hfill {\left[z,{[x,y]}_{\otimes }\right]}_{\otimes }& =z\otimes x\otimes y-z\otimes y\otimes x\hfill \\ \hfill & -x\otimes y\otimes z+y\otimes x\otimes z.\hfill \end{array}$$

Adding cancels every ordered product with its negative. Similarly,

$$\begin{array}{cc}\hfill {[x,y]}_{\otimes }\otimes z+y\otimes {[x,z]}_{\otimes }& =(x\otimes y-y\otimes x)\otimes z+y\otimes (x\otimes z-z\otimes x)\hfill \\ \hfill & =x\otimes y\otimes z-y\otimes z\otimes x={[x,y\otimes z]}_{\otimes }.\hfill \end{array}$$

Linearity and the composition law in Theorem 38 give

$$L_{{[x,y]}_{\otimes }}=L_{x\otimes y}-L_{y\otimes x}=L_x{L}_y-L_y{L}_x.$$

If an element of the Lie algebra acted by zero, its action on $\left[v_{len}\right]$ would be zero. The injectivity of $\psi$ in that theorem then makes the element zero. Thus the Lie action is faithful. The grading records depth; these identities use the ordinary signs of a Lie algebra. □

Theorem 40 

(Free Lie algebra and its ranks). Let $\mathfrak{l}$ be the Lie subalgebra generated by $v_{pos}$ and $v_{len}$. It is the free Lie algebra over R on these two generators. Its degree-n component is free of rank

$${\ell }_n={\displaystyle \frac{1}{n}}\sum _{d\mid n}\mu \left(d\right)2^{n/d}(n\ge 1),$$

(155)

where μ is the Möbius function. More precisely, the component with a occurrences of $v_{pos}$ and b occurrences of $v_{len}$ has rank

$${\ell }_{a,b}={\displaystyle \frac{1}{a+b}}\sum _{d\mid \gcd (a,b)}\mu \left(d\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(a+b)/d}{a/d}}\right),a,b\ge 0,a+b>0.$$

(156)

Both expressions are integers. In particular,

$${\ell }_1=2,{\ell }_2=1,{\ell }_3=2,{\ell }_4=3.$$

(157)

Proof. 

The integral Hall-basis theorem realizes the free Lie algebra on two letters as the Lie subalgebra they generate in the integral free associative algebra. Its homogeneous components have integral bases; equivalently, one can use the standard bracketings of Lyndon words, whose associative expansions have the corresponding word with coefficient one as their first term. The triangularity gives linear independence, and the Hall straightening identities, obtained from antisymmetry and Jacobi, give spanning. These facts hold integrally [9,33]. Localization from $\mathbb{Z}$ to R preserves the basis and its injection into the word algebra. By Theorem 37, its image is exactly $\mathfrak{l}$.

The Poincaré–Birkhoff–Witt basis of the enveloping algebra, which here is the word algebra on two letters, gives the formal identity

$${\displaystyle \frac{1}{1-2t}}=\prod _{n\ge 1}{(1-t^n)}^{-{\ell }_n}.$$

Take formal logarithms over $\mathbb{Q}$ and compare the coefficient of $t^n$:

$$\sum _{r\ge 1}{\displaystyle \frac{2^r{t}^r}{r}}=\sum _{d\ge 1}{\ell }_d\sum _{j\ge 1}{\displaystyle \frac{t^{dj}}{j}},{\displaystyle \frac{2^n}{n}}=\sum _{d\mid n}{\displaystyle \frac{d{\ell }_d}{n}}.$$

Multiplying by n gives

$$2^n=\sum _{d\mid n}d{\ell }_d.$$

The inversion can be written directly:

$$\begin{array}{cc}\hfill \sum _{d\mid n}\mu \left(d\right)2^{n/d}& =\sum _{d\mid n}\mu \left(d\right)\sum _{r\mid n/d}r{\ell }_r\hfill \\ \hfill & =\sum _{r\mid n}r{\ell }_r\sum _{d\mid n/r}\mu \left(d\right)=n{\ell }_n.\hfill \end{array}$$

The inner divisor sum is one when $r=n$ and zero otherwise. Dividing by n proves (155). Since ${\ell }_n$ is the cardinality of a basis, this also proves the divisibility implicit in that expression.

Keeping the two letter counts in the same basis argument gives

$${\displaystyle \frac{1}{1-s-t}}=\prod _{\begin{array}{c}a,b\ge 0\\ a+b>0\end{array}}{(1-s^a{t}^b)}^{-{\ell }_{a,b}}.$$

The coefficient of $s^a{t}^b$ in its logarithm, multiplied by $a+b$, is

$$\left({\displaystyle \genfrac{}{}{0pt}{}{a+b}{a}}\right)=\sum _{d\mid \gcd (a,b)}{\displaystyle \frac{a+b}{d}}{\ell }_{a/d,b/d}.$$

Möbius inversion along the common divisors of a and b proves (156). The same integral basis proves its integrality, including the cases in which one letter count is zero. □

For example, the first nonzero bracket and two degree-three brackets have the path expansions

$$\begin{array}{cc}\hfill {[v_{pos},v_{len}]}_{\otimes }& ={\mathbf{b}}_{\left\{2\right\}}-{\mathbf{b}}_{\left\{1\right\}},\hfill \\ \hfill {\left[v_{pos},{[v_{pos},v_{len}]}_{\otimes }\right]}_{\otimes }& ={\mathbf{b}}_{\left\{3\right\}}-2{\mathbf{b}}_{\left\{2\right\}}+{\mathbf{b}}_{\left\{1\right\}},\hfill \\ \hfill {\left[v_{len},{[v_{pos},v_{len}]}_{\otimes }\right]}_{\otimes }& =2{\mathbf{b}}_{\{1,3\}}-{\mathbf{b}}_{\{1,2\}}-{\mathbf{b}}_{\{2,3\}}.\hfill \end{array}$$

The two latter elements have different letter counts and are independent. The commuting criterion for arbitrary positive depths is stronger than the same-depth proportionality test: by Theorem 24, two nonzero homogeneous elements commute exactly when they are scalar multiples of powers of one primitive block of depth equal to the greatest common divisor of their two depths.

6.2. Lie and Jordan Norm Identities

Definition 11 

(Symmetric tensor product). When $2\in R^{\times }$, define

$$x\circ y={\displaystyle \frac{1}{2}}(x\otimes y+y\otimes x)$$

(158)

on $\mathcal{A}$. For $R=\mathbb{Z}\left[{\Omega }^{-1}\right]$, the condition $2\in R^{\times }$ holds precisely when Ω is even. If Ω is odd, the same normalized product is considered over $R[1/2]$ or over the real scalar extension.

Theorem 41 

(Jordan identity and exact associator). Under the coefficient condition in Definition 11, ∘ is a unital special Jordan product. Its unit is 1, and

$$\begin{array}{cc}\hfill (x\circ x)\circ (y\circ x)& =\left((x\circ x)\circ y\right)\circ x,\hfill \\ \hfill (x\circ y)\circ z-x\circ (y\circ z)& ={\displaystyle \frac{1}{4}}{\left[y,{[x,z]}_{\otimes }\right]}_{\otimes },\hfill \\ \hfill {\circ z]}_{\otimes }& ={[x,y]}_{\otimes }\circ z+y\circ {[x,z]}_{\otimes }.\hfill \end{array}$$

(159)

The two orientations decompose as

$$x\otimes y=x\circ y+\frac{1}{2}{[x,y]}_{\otimes },x{\otimes }^{\mathrm{op}}y=x\circ y-\frac{1}{2}{[x,y]}_{\otimes }.$$

(160)

In particular, the right tensor induces the same Jordan product and the negative Lie bracket.

Proof. 

Commutativity follows by exchanging the two summands, and $x\circ 1=(x\otimes 1+1\otimes x)/2=x$. The two three-input products expand separately as

$$\begin{array}{cc}\hfill 4(x\circ y)\circ z& =x\otimes y\otimes z+y\otimes x\otimes z\hfill \\ \hfill & +z\otimes x\otimes y+z\otimes y\otimes x,\hfill \\ \hfill 4x\circ (y\circ z)& =x\otimes y\otimes z+x\otimes z\otimes y\hfill \\ \hfill & +y\otimes z\otimes x+z\otimes y\otimes x.\hfill \end{array}$$

Subtracting the two common ordered products gives

$$\begin{array}{cc}\hfill 4\left((x\circ y)\circ z-x\circ (y\circ z)\right)=& y\otimes x\otimes z+z\otimes x\otimes y\hfill \\ \hfill & -x\otimes z\otimes y-y\otimes z\otimes x\hfill \\ \hfill =& {\left[y,{[x,z]}_{\otimes }\right]}_{\otimes }.\hfill \end{array}$$

Since $x\circ x=x\otimes x$, associativity gives

$$\begin{array}{cc}\hfill {[x\circ x,x]}_{\otimes }& =(x\otimes x)\otimes x-x\otimes (x\otimes x)=0,\hfill \\ \hfill \left((x\circ x)\circ y\right)\circ x-(x\circ x)\circ (y\circ x)& =\frac{1}{4}{\left[y,{[x\circ x,x]}_{\otimes }\right]}_{\otimes }=0.\hfill \end{array}$$

For the derivation identity, apply (153) to both orders of $y,z$:

$$\begin{array}{cc}\hfill 2{[x,y\circ z]}_{\otimes }& ={[x,y]}_{\otimes }\otimes z+y\otimes {[x,z]}_{\otimes }\hfill \\ \hfill & +{[x,z]}_{\otimes }\otimes y+z\otimes {[x,y]}_{\otimes }\hfill \\ \hfill & =2\left({[x,y]}_{\otimes }\circ z+y\circ {[x,z]}_{\otimes }\right).\hfill \end{array}$$

Thus this is the special Jordan algebra obtained from an associative algebra by symmetrization [10,11,34]. Adding and subtracting the two tensor orders proves (160) and its uniqueness. □

Without inverting 2, the unnormalized product $x\otimes y+y\otimes x$ still satisfies the Jordan identity. In its two sides the associative expansion is, in both cases,

$$2\left(x\otimes x\otimes y\otimes x+x\otimes x\otimes x\otimes y+y\otimes x\otimes x\otimes x+x\otimes y\otimes x\otimes x\right).$$

Its product with 1 is $2x$, which explains the coefficient condition for the normalized unital product.

Theorem 42 

(Orthogonality of the two orientation components). For $x\in {\mathcal{I}}_m$ and $y\in {\mathcal{I}}_n$,

$$\begin{array}{cc}\hfill {〈x\otimes y+y\otimes x,x\otimes y-y\otimes x〉}_{\Omega }& =0,\hfill \\ \hfill {∥x\otimes y+y\otimes x∥}_{\Omega }^2+{∥x\otimes y-y\otimes x∥}_{\Omega }^2& =4{∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2.\hfill \end{array}$$

(161)

When the normalized Jordan product is defined,

$${〈x\circ y,{[x,y]}_{\otimes }〉}_{\Omega }=0,{∥x\otimes y∥}_{\Omega }^2={∥x\circ y∥}_{\Omega }^2+\frac{1}{4}{∥{[x,y]}_{\otimes }∥}_{\Omega }^2.$$

(162)

For two inputs in the same level ${\mathcal{I}}_m$, the exact formulas are

$$\begin{array}{cc}\hfill {∥{[x,y]}_{\otimes }∥}_{\Omega }^2& =2\left({∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-{〈x,y〉}_{\Omega }^2\right),\hfill \\ \hfill {∥x\circ y∥}_{\Omega }^2& =\frac{1}{2}\left({∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2+{〈x,y〉}_{\Omega }^2\right).\hfill \end{array}$$

(163)

For nonzero same-depth inputs over R, the bracket vanishes precisely when they are linearly dependent over $\mathbb{Q}$.

Proof. 

The two orders have the same norm by (86). Expanding their sum-difference pairing gives

$$\begin{array}{cc}\hfill {〈x\otimes y+y\otimes x,x\otimes y-y\otimes x〉}_{\Omega }& ={∥x\otimes y∥}_{\Omega }^2-{∥y\otimes x∥}_{\Omega }^2\hfill \\ \hfill & ={∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-{∥y∥}_{\Omega }^2{∥x∥}_{\Omega }^2=0.\hfill \end{array}$$

The two mixed terms cancel by symmetry. The parallelogram identity proves (161). Division by the indicated powers of two gives (162). At equal input depths, (85) gives

$${〈x\otimes y,y\otimes x〉}_{\Omega }={〈x,y〉}_{\Omega }^2.$$

Consequently

$$\begin{array}{cc}\hfill {∥{[x,y]}_{\otimes }∥}_{\Omega }^2& ={∥x\otimes y∥}_{\Omega }^2+{∥y\otimes x∥}_{\Omega }^2-2{〈x\otimes y,y\otimes x〉}_{\Omega }\hfill \\ \hfill & =2{∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-2{〈x,y〉}_{\Omega }^2,\hfill \\ \hfill 4{∥x\circ y∥}_{\Omega }^2& ={∥x\otimes y∥}_{\Omega }^2+{∥y\otimes x∥}_{\Omega }^2+2{〈x\otimes y,y\otimes x〉}_{\Omega }\hfill \\ \hfill & =2{∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2+2{〈x,y〉}_{\Omega }^2.\hfill \end{array}$$

The final assertion is the equality condition in the weighted Cauchy–Schwarz inequality. A proportionality relation between two nonzero vectors with rational path coefficients can be expressed over $\mathbb{Q}$. Explicitly, if $y=\lambda x$ and $x_S\ne 0$ at some path, then

$$\lambda ={\displaystyle \frac{y_S}{x_S}}\in \mathbb{Q},x_{S}y=y_{S}x.$$

Conversely, any such rational proportionality makes the two tensor orders equal by fixed-level bilinearity, so the bracket vanishes. □

For $m\ge 1$, the same-depth bracket bound ${∥{[x,y]}_{\otimes }∥}_{\Omega }\le \sqrt{2}{∥x∥}_{\Omega }{∥y∥}_{\Omega }$ is sharp, with nonzero equality exactly when ${〈x,y〉}_{\Omega }=0$. The Jordan norm lies between ${∥x∥}_{\Omega }{∥y∥}_{\Omega }/\sqrt{2}$ and ${∥x∥}_{\Omega }{∥y∥}_{\Omega }$; orthogonal inputs attain the lower endpoint and proportional inputs attain the upper endpoint. Path basis inputs realize these endpoints over R, with the stated coefficient condition for normalized Jordan products. The equal-depth hypothesis in the Gram formulas cannot be removed. For $x=1_0$ and $y=v_{len}$, the bracket is zero while ${〈x,y〉}_{\Omega }=0$ and ${∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2={\Omega }^2$.

Theorem 43 

(The three-input Jordan norm). Assume the normalized Jordan product is defined, and take $x,y,z\in {\mathcal{I}}_m$. Its associator has the exact squared norm

$$\begin{array}{cc}\hfill {∥(x\circ y)\circ z-x\circ (y\circ z)∥}_{\Omega }^2=& {\displaystyle \frac{1}{4}}{∥y∥}_{\Omega }^2\left({∥x∥}_{\Omega }^2{∥z∥}_{\Omega }^2-{〈x,z〉}_{\Omega }^2\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}{∥{〈x,y〉}_{\Omega }z-{〈y,z〉}_{\Omega }x∥}_{\Omega }^2.\hfill \end{array}$$

(164)

For $m\ge 1$, its sharp trilinear bound is

$${∥(x\circ y)\circ z-x\circ (y\circ z)∥}_{\Omega }\le {\displaystyle \frac{\sqrt{6}}{4}}{∥x∥}_{\Omega }{∥y∥}_{\Omega }{∥z∥}_{\Omega }.$$

(165)

For nonzero real inputs, equality holds exactly when x and z are orthogonal and y belongs to their real linear span. These conditions have examples over R whenever $2\in R^{\times }$. For nonzero y, the associator vanishes exactly when x and z are linearly dependent. At depth zero the associator is identically zero.

Proof. 

Use the four-word expression in the proof of (159). Put

$$\begin{array}{cc}\hfill \alpha & ={∥x∥}_{\Omega }^2,\beta ={∥y∥}_{\Omega }^2,\gamma ={∥z∥}_{\Omega }^2,\hfill \\ \hfill a& ={〈x,y〉}_{\Omega },b={〈x,z〉}_{\Omega },c={〈y,z〉}_{\Omega }.\hfill \end{array}$$

All four words have the same three depth-m blocks. Their pairings can be recorded explicitly. Order the words as

$$\begin{array}{cccc}\hfill w_1& =y\otimes x\otimes z,\hfill & \hfill w_2& =z\otimes x\otimes y,\hfill \\ \hfill w_3& =x\otimes z\otimes y,\hfill & \hfill w_4& =y\otimes z\otimes x.\hfill \end{array}$$

Then four times the associator is $w_1+w_2-w_3-w_4$, and fixed-block factorization gives the Gram matrix

$${\left({〈w_i,w_j〉}_{\Omega }\right)}_{i,j=1}^4=\left(\begin{array}{cccc}\alpha \beta \gamma & \alpha c^2& abc& \beta b^2\\ \alpha c^2& \alpha \beta \gamma & \beta b^2& abc\\ abc& \beta b^2& \alpha \beta \gamma & \gamma a^2\\ \beta b^2& abc& \gamma a^2& \alpha \beta \gamma \end{array}\right).$$

The diagonal contribution is $4\alpha \beta \gamma$. Multiplying the off-diagonal entries by their signs and counting both orders gives $2\alpha c^2+2\gamma a^2-4abc-4\beta b^2$. After division by 16, the result is

$${\displaystyle \frac{1}{4}}\beta (\alpha \gamma -b^2)+{\displaystyle \frac{1}{8}}(\gamma a^2+\alpha c^2-2abc),$$

which is (164).

If x and z are independent, let P be the orthogonal projection onto their span. Choose an orthonormal basis $e_1,e_2$ of that plane and real scalars $s,r>0$, $t,p,q$ such that

$$x=s{e}_1,z=t{e}_1+r{e}_2,y=p{e}_1+q{e}_2+y_{\perp },y_{\perp }\perp e_1,e_2.$$

Then $a=sp$, $c=tp+rq$, and $\alpha \gamma -b^2=s^2{r}^2$. Hence

$${〈x,y〉}_{\Omega }z-{〈y,z〉}_{\Omega }x=sp(t{e}_1+r{e}_2)-(tp+rq)s{e}_1=sr(p{e}_2-q{e}_1).$$

Taking squared norms gives

$${∥{〈x,y〉}_{\Omega }z-{〈y,z〉}_{\Omega }x∥}_{\Omega }^2=(\alpha \gamma -b^2){∥Py∥}_{\Omega }^2\le (\alpha \gamma -b^2)\beta .$$

Here ${∥Py∥}_{\Omega }^2=p^2+q^2$ and $\beta =p^2+q^2+{∥y_{\perp }∥}_{\Omega }^2$, so equality holds exactly when $y_{\perp }=0$, or $y=Py$. If $x,z$ are dependent, both sides of the identity are zero. Substitution into (164) gives

$${∥(x\circ y)\circ z-x\circ (y\circ z)∥}_{\Omega }^2\le {\displaystyle \frac{3}{8}}\beta (\alpha \gamma -b^2)\le {\displaystyle \frac{3}{8}}\alpha \beta \gamma .$$

For nonzero inputs, both inequalities are equalities precisely under the stated orthogonality and span conditions. The first nonnegative term in (164) proves the vanishing criterion. To attain the bound over R, take two distinct path basis vectors as $x,z$ and take $y=x$. They are orthogonal, and all the normalized operations remain over R when 2 is a unit. For a unit-norm example at depth $m\ge 1$, take

$$x=y=1_m,z={\Omega }^{-1}{\mathbf{b}}_{\left\{1\right\}}.$$

Here $\alpha =\beta =\gamma =1$, $a=1$, and $b=c=0$, so the exact squared associator norm is $1/4+1/8=3/8$. This gives the stated constant $\sqrt{6}/4$ over the original ring. □

Theorem 44 

(Exact norms of iterated commutators). Let $x,y\in {\mathcal{I}}_m$, where $m\ge 1$ and $x\ne 0$, and define recursively

$$h_0=y,h_{k+1}={[x,h_k]}_{\otimes }.$$

(166)

For every $k\ge 1$,

$${∥h_k∥}_{\Omega }^2=\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){∥x∥}_{\Omega }^{2k-2}\left({∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-{〈x,y〉}_{\Omega }^2\right).$$

(167)

Thus the vanishing of one positive iterate, the vanishing of the first bracket, and rational linear dependence of $x,y$ over R are equivalent. A nonzero first bracket produces a nonzero element at every later iteration. When 2 is invertible, one also has

$${∥(x\circ x)\circ y-x\circ (x\circ y)∥}_{\Omega }^2={\displaystyle \frac{3}{8}}{∥x∥}_{\Omega }^2\left({∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-{〈x,y〉}_{\Omega }^2\right).$$

(168)

Proof. 

Associativity and induction give the finite expansion

$$h_k={\displaystyle \sum _{j=0}^k}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)x^{\otimes (k-j)}\otimes y\otimes x^{\otimes j}.$$

Here $x^{\otimes 0}=1_0$. To verify the induction, apply ${[x,·]}_{\otimes }$ to the displayed sum. Prefixing x leaves the index j unchanged; subtracting the term with an appended x replaces it by $j+1$. Thus the coefficient at index j becomes

$${(-1)}^j\left(\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j-1}}\right)\right)={(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{j}}\right),$$

where binomial coefficients outside $0,\dots ,k$ are zero. The case $k=0$ is $h_0=y$, completing the induction. Every term has $k+1$ blocks of the same length m. A term paired with itself contributes ${∥x∥}_{\Omega }^{2k}{∥y∥}_{\Omega }^2$. Two distinct terms place y in different positions, and their pairing is ${∥x∥}_{\Omega }^{2k-2}{〈x,y〉}_{\Omega }^2$. Therefore

$${\displaystyle \sum _{j=0}^k}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)=0,{\displaystyle \sum _{j=0}^k}{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)}^2=\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)$$

combine the diagonal and off-diagonal contributions into the following squared norm:

$$\begin{array}{cc}\hfill {∥h_k∥}_{\Omega }^2& ={\displaystyle \sum _{j=0}^k}{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)}^2{∥x∥}_{\Omega }^{2k}{∥y∥}_{\Omega }^2\hfill \\ \hfill & +\left[{\left({\displaystyle \sum _{j=0}^k}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\right)}^2-{\displaystyle \sum _{j=0}^k}{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)}^2\right]{∥x∥}_{\Omega }^{2k-2}{〈x,y〉}_{\Omega }^2\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){∥x∥}_{\Omega }^{2k-2}\left({∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2-{〈x,y〉}_{\Omega }^2\right).\hfill \end{array}$$

The last binomial identity follows by comparing the coefficient of $t^k$ in ${(1+t)}^k{(1+t)}^k$. Positivity and the Cauchy–Schwarz equality condition prove the vanishing statements, including $y=0$. Equation (159) identifies the associator in (168) with $h_2/4$. Taking $k=2$ proves the formula, which is also a special case of (164). □

6.3. Sharp Block Estimates and Reversal

Theorem 45 

(Sharp bounds for different tensor block lengths). Fix a nonzero $x\in {\mathcal{I}}_m$ with $m\ge 1$, and let $k\ge 1$. For $y\in {\mathcal{I}}_{km}$, put $\theta =\pi /\left(2\right(k+1\left)\right)$. On the real scalar extension,

$$\begin{array}{cc}\hfill {∥x\otimes y-y\otimes x∥}_{\Omega }& \le 2\cos \theta {∥x∥}_{\Omega }{∥y∥}_{\Omega },\hfill \end{array}$$

(169)

$$\begin{array}{cc}\hfill 2\sin \theta {∥x∥}_{\Omega }{∥y∥}_{\Omega }& \le {∥x\otimes y+y\otimes x∥}_{\Omega }\le 2{∥x∥}_{\Omega }{∥y∥}_{\Omega }.\hfill \end{array}$$

(170)

All three constants are exact for each fixed nonzero x. The nonzero equality vectors for (169) and the lower bound in () are exactly

$$y={\displaystyle \sum _{j=0}^{k-1}}{(-1)}^j\sin \left({\displaystyle \frac{(j+1)\pi }{k+1}}\right)x^{\otimes j}\otimes h\otimes x^{\otimes (k-1-j)},$$

(171)

where $h\in {\mathcal{I}}_m{\otimes }_R\mathbb{R}$ is nonzero and ${〈h,x〉}_{\Omega }=0$. The nonzero equality vectors for the upper bound in () are exactly $y=t{x}^{\otimes k}$ with a nonzero real scalar t.

For fixed nonzero x over R, the commutator upper bound and the anticommutator lower bound have nonzero equality vectors over R exactly when $k=1$ or $k=2$. For $k\ge 3$ the same supremum and infimum hold over R, but neither endpoint is attained by a nonzero y over R. The anticommutator upper bound is attained over R for every k. All assertions remain valid after replacing every tensor with the right tensor. When the normalized Jordan product is defined, its corresponding bounds are obtained by dividing () by two.

Proof. 

Let $W={\mathcal{I}}_m{\otimes }_R\mathbb{R}$ with its weighted inner product, and put $e=x/{∥x∥}_{\Omega }$. Fixed-block metric factorization identifies the real extension of ${\mathcal{I}}_{km}$ isometrically with the k-fold tensor power of W. In this proof define two isometric operators by

$$A\left(y\right)=e\otimes y,B\left(y\right)=y\otimes e.$$

Their adjoints satisfy

$${(A-B)}^{*}(A-B)=2id-A^{*}B-B^{*}A,{(A+B)}^{*}(A+B)=2id+A^{*}B+B^{*}A.$$

(172)

Choose an orthonormal basis of W containing e. For a word $u_1\otimes \dots \otimes u_k$ in this basis,

$$\begin{array}{cc}\hfill A^{*}B(u_1\otimes \dots \otimes u_k)& ={〈e,u_1〉}_{\Omega }{u}_2\otimes \dots \otimes u_k\otimes e,\hfill \\ \hfill B^{*}A(u_1\otimes \dots \otimes u_k)& ={〈e,u_k〉}_{\Omega }e\otimes u_1\otimes \dots \otimes u_{k-1}.\hfill \end{array}$$

Thus $A^{*}B$ removes an initial e and appends it at the end, and $B^{*}A$ performs the reverse operation. An endpoint orthogonal to e makes the corresponding term zero.

The all-e word $e^{\otimes k}$ forms a one-dimensional subspace on which $A^{*}B+B^{*}A$ equals 2. Any other basis word has a unique core obtained by removing its initial and final strings of e. Its core has some length $r\ge 1$, and its first and last letters are orthogonal to e. For this fixed core h, the words

$$e^{\otimes j}\otimes h\otimes e^{\otimes (k-r-j)},0\le j\le k-r,$$

span an invariant subspace. On that orthonormal basis, $A^{*}B+B^{*}A$ is the adjacency matrix of a path with $\ell =k-r+1$ vertices: its only nonzero entries are the ones just above and below the diagonal. These subspaces, together with the all-e word, are an orthogonal decomposition of the entire space.

The spectrum of this path matrix is the standard one from algebraic graph theory [35]: its eigenvalues are

$$2\cos \left({\displaystyle \frac{s\pi }{\ell +1}}\right),1\le s\le \ell .$$

Indeed, put $\alpha =s\pi /(\ell +1)$ and $c_j=\sin \left((j+1)\alpha \right)$. The endpoint values and the three-term relation are

$$\begin{array}{cc}\hfill c_{-1}& =0,c_{\ell }=\sin \left(s\pi \right)=0,\hfill \\ \hfill c_{j-1}+c_{j+1}& =\sin \left(j\alpha \right)+\sin \left((j+2)\alpha \right)=2\cos \alpha c_j(0\le j<\ell ).\hfill \end{array}$$

The eigenvalues are distinct, so these ℓ nonzero vectors account for the whole path matrix. The smallest eigenvalue among all blocks is $-2\cos (\pi /(k+1\left)\right)$, attained only when $r=1$ and $s=k$. The largest is 2, attained only on $e^{\otimes k}$. Consequently (172) gives

$$\begin{array}{cc}\hfill {∥(A-B)y∥}_{\Omega }^2& \le \left(2+2\cos \left(2\theta \right)\right){∥y∥}_{\Omega }^2=4{\cos }^2\theta {∥y∥}_{\Omega }^2,\hfill \\ \hfill 4{\sin }^2\theta {∥y∥}_{\Omega }^2& =\left(2-2\cos \left(2\theta \right)\right){∥y∥}_{\Omega }^2\le {∥(A+B)y∥}_{\Omega }^2\le 4{∥y∥}_{\Omega }^2.\hfill \end{array}$$

Multiplication by ${∥x∥}_{\Omega }^2$ proves the three bounds.

The endpoint eigenvector of a longest path has coordinates

$${(-1)}^j\sin \left({\displaystyle \frac{(j+1)\pi }{k+1}}\right),0\le j<k.$$

Here the core is one basis vector orthogonal to e. Summing over all such cores permits an arbitrary nonzero $h\perp e$, giving (171). Replacing e by x changes every term by the same scalar factor, which is absorbed into h. No shorter path has this endpoint eigenvalue. Moreover, in an orthogonal eigen-decomposition $y={\sum }_{\lambda }{y}_{\lambda }$ of $A^{*}B+B^{*}A$, equality at its lower endpoint is equivalent to

$$\sum _{\lambda }(\lambda -{\lambda }_{\min }){∥y_{\lambda }∥}_{\Omega }^2=0.$$

All terms are nonnegative, so every component outside that eigenspace vanishes. This proves completeness of the equality list. Since $\dim W=2^m\ge 2$, its orthogonal complement to e contains a nonzero vector, so the longest-path eigenspace really occurs. The largest sum norm comes only from the all-e word, giving its stated equality classification.

Suppose now that x has coefficients in R. Since $m\ge 1$, the weighted orthogonal complement of x contains a nonzero vector over R: solve its one rational linear equation and clear denominators. For $k=1$ take $y=h$. For $k=2$ take $y=x\otimes h-h\otimes x$; the two sine coefficients in (171) have equal absolute value, so this vector attains both indicated endpoints. For $k\ge 3$, an equality vector over R would make the rational number

$${\displaystyle \frac{{∥x\otimes y\mp y\otimes x∥}_{\Omega }^2}{{∥x∥}_{\Omega }^2{∥y∥}_{\Omega }^2}}$$

equal to $2\pm 2\cos (\pi /(k+1\left)\right)$. These values are irrational. To verify this directly, set $t=2\cos (\pi /(k+1\left)\right)$ and define integer polynomials by $q_0\left(t\right)=2$, $q_1\left(t\right)=t$, and $q_{j+1}\left(t\right)=t{q}_j\left(t\right)-q_{j-1}\left(t\right)$. The cosine recurrence gives

$$q_j\left(2\cos \alpha \right)=2\cos \left(j\alpha \right),q_{k+1}\left(2\cos {\displaystyle \frac{\pi }{k+1}}\right)+2=0.$$

Thus $q_{k+1}\left(t\right)+2$ is monic and vanishes at this value of t. A rational root of a monic integer polynomial is an integer, whereas $1<t<2$ for $k\ge 3$. This proves the irrationality and the nonattainment. Density of R in $\mathbb{R}$, applied to the finitely many coordinates of an endpoint vector, gives the same sharp constants over R. The upper sum norm is attained by $y=x^{\otimes k}$ over R. Finally, replacing the tensor by the right tensor negates the commutator and leaves the sum unchanged. It preserves every bound and every equality condition. □

In particular, the successive sharp commutator constants for block ratios $1:1$, $1:2$, and $1:3$ are $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{2+\sqrt{2}}$. If $m=1$, the theorem covers every positive depth of the second input. All these estimates concern typed blocks, as does their complete equality classification.

Definition 12 

(Reversal). On a word in $v_{pos},v_{len}$, let ρ reverse the order of its letters, and extend it linearly on $\mathcal{A}$. Thus $\rho \left(1\right)=1$ and, on ${\mathcal{I}}_n$,

$$\rho \left({\mathbf{b}}_S\right)={\mathbf{b}}_{\{n+1-j:j\in S\}}.$$

(173)

Proposition 18 

(Reversal, ranks, and parity). The operation ρ is an involutive, degreewise isometric anti-automorphism of the tensor algebra:

$${\rho }^2=id,\rho (x\otimes y)=\rho \left(y\right)\otimes \rho \left(x\right).$$

(174)

It satisfies

$$\rho \left({[x,y]}_{\otimes }\right)=-{\left[\rho \left(x\right),\rho \left(y\right)\right]}_{\otimes },\rho (x\circ y)=\rho \left(x\right)\circ \rho \left(y\right)$$

(175)

whenever the normalized Jordan product is defined. At each fixed level it also preserves ordinary multiplication.

If $2\in R^{\times }$, put

$${\mathcal{A}}^{+}=\left\{x\in \mathcal{A}:\rho \left(x\right)=x\right\},{\mathcal{A}}^{-}=\left\{x\in \mathcal{A}:\rho \left(x\right)=-x\right\}.$$

(176)

Then

$$\mathcal{A}={\mathcal{A}}^{+}\oplus {\mathcal{A}}^{-},x=x^{+}+x^{-},x^{\pm }=\frac{1}{2}\left(x\pm \rho \left(x\right)\right).$$

(177)

The submodules ${\mathcal{A}}_n^{\pm }={\mathcal{A}}^{\pm }\cap {\mathcal{I}}_n$ are orthogonal within ${\mathcal{I}}_n$ and are free of ranks

$${rank}_R{\mathcal{A}}_n^{\pm }={\displaystyle \frac{2^n\pm 2^{\lceil n/2\rceil }}{2}}(n\ge 0).$$

(178)

Their products obey

$$\begin{array}{cccccc}\hfill {\mathcal{A}}^{+}\circ {\mathcal{A}}^{+}& \subseteq {\mathcal{A}}^{+},\hfill & \hfill {\mathcal{A}}^{-}\circ {\mathcal{A}}^{-}& \subseteq {\mathcal{A}}^{+},\hfill & \hfill {\mathcal{A}}^{+}\circ {\mathcal{A}}^{-}& \subseteq {\mathcal{A}}^{-},\hfill \\ \hfill {\mathcal{A}}^{+},{\mathcal{A}}^{+}{]}_{\otimes }& \subseteq {\mathcal{A}}^{-},\hfill & \hfill {[{\mathcal{A}}^{+},{\mathcal{A}}^{-}]}_{\otimes }& \subseteq {\mathcal{A}}^{+},\hfill & \hfill {[{\mathcal{A}}^{-},{\mathcal{A}}^{-}]}_{\otimes }& \subseteq {\mathcal{A}}^{-}.\hfill \end{array}$$

(179)

For a homogeneous $u\in \mathfrak{l}\cap {\mathcal{I}}_n$, even without inverting 2,

$$\rho \left(u\right)={(-1)}^{n-1}u(n\ge 1).$$

(180)

Proof. 

Reversing a word twice restores it, and reversing a concatenation reverses the order of its two blocks. A reversed path has the same cardinality and therefore the same weight. Thus reversal is an isometry on each level. Permuting the positions also preserves intersection and union of path supports, so it preserves the ordinary path multiplication at that level. Taking the difference or average of the two orders proves (175).

When 2 is invertible, the projections satisfy

$$\rho \left(x^{\pm }\right)=\frac{1}{2}\left(\rho \left(x\right)\pm x\right)=\pm x^{\pm },x^{+}+x^{-}=x.$$

An element u in their two images satisfies $u=\rho \left(u\right)=-u$, so $2u=0$ and $u=0$. This proves the direct sum. For $u\in {\mathcal{A}}_n^{+}$ and $v\in {\mathcal{A}}_n^{-}$, degreewise isometry gives

$${〈u,v〉}_{\Omega }={〈\rho \left(u\right),\rho \left(v\right)〉}_{\Omega }=-{〈u,v〉}_{\Omega },$$

so the pairing is zero. There are $2^{\lceil n/2\rceil }$ palindromic words of length n: the first half, including the middle letter when present, determines the rest. Every remaining word belongs to a two-element reversal orbit. An orbit $\{w,\rho (w\left)\right\}$ has the basis $w+\rho \left(w\right),w-\rho \left(w\right)$, since

$$w=\frac{1}{2}\left(\left(w+\rho \left(w\right)\right)+\left(w-\rho \left(w\right)\right)\right),\rho \left(w\right)=\frac{1}{2}\left(\left(w+\rho \left(w\right)\right)-\left(w-\rho \left(w\right)\right)\right).$$

It contributes one vector to each sign component, while each palindrome contributes one positive vector. Hence

$${rank}_R{\mathcal{A}}_n^{-}={\displaystyle \frac{2^n-2^{\lceil n/2\rceil }}{2}},{rank}_R{\mathcal{A}}_n^{+}=2^{\lceil n/2\rceil }+{rank}_R{\mathcal{A}}_n^{-}.$$

For $\rho \left(x\right)=\sigma x$ and $\rho \left(y\right)=\tau y$, with $\sigma ,\tau \in \{-1,1\}$, (175) gives

$$\rho (x\circ y)=\sigma \tau (x\circ y),\rho \left({[x,y]}_{\otimes }\right)=-\sigma \tau {[x,y]}_{\otimes }.$$

The choices of signs prove all six containments.

Both generators of $\mathfrak{l}$ have reversal sign 1. If two bracket expressions of degrees $p,q$ have signs ${(-1)}^{p-1}$ and ${(-1)}^{q-1}$, the additional minus sign in (175) gives their bracket sign ${(-1)}^{p+q-1}$. Induction on bracket length and linearity prove (180). □

The free Lie component therefore lies in the positive reversal component at odd degree and in the negative one at even degree. These inclusions can be strict: at degree three the free Lie rank is 2, whereas the positive reversal rank is 6. The positive and negative reversal spaces are orthogonal at each fixed degree; different degrees need not be orthogonal for the mixed-depth global form.

Proposition 19 

(Dependence on typed representatives). Reversal, the tensor Lie bracket, and the symmetric tensor product depend on the chosen depths. They do not define corresponding operations on the global refinement classes. This includes the unnormalized symmetric product over R and, when defined, the normalized Jordan product.

Proof. 

The equivalent representatives $v_{len}\in {\mathcal{I}}_1$ and $v_{len}\otimes v_{pos}\in {\mathcal{I}}_2$ have reversals $v_{len}$ and $v_{pos}\otimes v_{len}$. The refinement of the first is $v_{len}\otimes v_{pos}$, different from the second. At depth two their difference is

$$v_{pos}\otimes v_{len}-v_{len}\otimes v_{pos}={\mathbf{b}}_{\left\{2\right\}}-{\mathbf{b}}_{\left\{1\right\}},{∥{\mathbf{b}}_{\left\{2\right\}}-{\mathbf{b}}_{\left\{1\right\}}∥}_{\Omega }=\sqrt{2}\Omega >0.$$

Refinement preserves these coefficients and their norm, so later refinements cannot identify the two outputs.

Likewise, $1_0$ and $v_{pos}=\iota \left(1_0\right)$ are equivalent, but

$${[1_0,v_{len}]}_{\otimes }=0,{[v_{pos},v_{len}]}_{\otimes }={\mathbf{b}}_{\left\{2\right\}}-{\mathbf{b}}_{\left\{1\right\}}\ne 0.$$

The unnormalized symmetric products with $v_{len}$ are $2v_{len}$ and $v_{pos}\otimes v_{len}+v_{len}\otimes v_{pos}$. At depth two these are $2{\mathbf{b}}_{\left\{1\right\}}$ and ${\mathbf{b}}_{\left\{1\right\}}+{\mathbf{b}}_{\left\{2\right\}}$, which are different. Dividing by two gives the same obstruction for the normalized product. For the direct-sum addition of $\mathcal{A}$, one has $1_0+0_1=1_0$. For recursive mixed-depth addition, instead $1_0+0_1=v_{pos}$, giving

$${[1_0+0_1,v_{len}]}_{\otimes }\ne {[1_0,v_{len}]}_{\otimes }+{[0_1,v_{len}]}_{\otimes }.$$

Thus the bilinear Lie and Jordan laws use the direct-sum addition of $\mathcal{A}$, in accord with their constructions. □

7. Conclusions

The two component operations realize commutative square-zero algebras as recursive objects with arithmetic evaluated by scalar descent. Path coordinates describe the finite zeon multiplication and remain compatible with refinement, giving the global algebra and its recursive reconstruction. They also give explicit unit and nilpotent classifications, multiplication duality, annihilators, and finite exponential and logarithmic operations over the original coefficient ring.

The weighted form connects this arithmetic with exact norm estimates. The one-level dual-number bound assembles recursively into mixed-depth multiplication constants, while equality at each position determines all extremizers and their original-ring attainability. Directed tensor products preserve norms multiplicatively and retain the prescribed input blocks. Their alignment determines mixed-depth inner-product factorization, and logarithmic supports determine the finest consecutive-block decomposition of units. Typed concatenation equips the same blocks with associative, Lie, and Jordan operations; their norm estimates include exact block-length-dependent constants and arithmetic restrictions on endpoint equality.

Finite function evaluation is determined by scalar derivative data, with explicit conditions for coefficient-ring preservation and sharp bounds on first-homogeneous inputs. Across unrestricted depths, the same evaluation has a precise boundary: on a fixed-scalar fiber, any nonzero datum of order at least two forces local unboundedness at every point. Degree scaling provides uniform product control exactly when the squared parameters sum to at most one; outside this region no finite depth-independent constant exists. These positive estimates and obstructions describe how the finite recursive arithmetic behaves in the global normed space and its completion. The resulting foundation links scalar computation, recursive depth, and tensor-block structure within a single system of operations.