Fixed Convex-Lens Spectral Constants: Möbius Reduction, Sharp Model Theorems, and Angle-Dependent Bounds

1. The Fixed-Lens Problem and the Scope of the Result

Let $D_1,D_2\subset \mathbb{C}$ be closed disks whose boundary circles meet in two distinct points and whose intersection has nonempty interior. We say that $A\in \mathcal{B}\left(\mathcal{H}\right)$ is of $(D_1,D_2)$-type if

$$\parallel A-c_{j}I\parallel \le r_j,D_j=\{z:|z-c_j|\le r_j\},j=1,2.$$

(1.1)

By von Neumann’s inequality, (1.1) is equivalent to the assertion that each $D_j$ is a spectral set for A. If the two interior boundary arcs meet at angle $2\alpha \in (0,\pi ]$, Beckermann and Crouzeix proved that the least uniform constant depends only on $\alpha$ and equals the numerical-range constant of a sector of the same angle [1]. The sector estimates underlying this reduction go back to [2]; a complete uniform bound for intersections of spherical disks was subsequently obtained in [3]. Exact dimension-two constants, including the sector constants used below, are characterized in [4], and the later survey [5] records numerical evidence in higher dimensions.

We use the following normalization:

$$\begin{array}{cc}\hfill D_{\alpha }^{\pm }& :=\{z\in \mathbb{C}:|z\mp icot\alpha |\le csc\alpha \},\hfill \end{array}$$

(1.2)

$$\begin{array}{cc}\hfill {\mathcal{L}}_{\alpha }& :=D_{\alpha }^{+}\cap D_{\alpha }^{-},{\mathcal{S}}_{\alpha }:=\{r{e}^{i\theta }:r\ge 0,\left|\theta \right|\le \alpha \}.\hfill \end{array}$$

(1.3)

The vertices of ${\mathcal{L}}_{\alpha }$ are $-1$ and 1. Define

$$C\left(\alpha \right):=\underset{\begin{array}{c}\mathcal{H},A\in \mathcal{B}\left(\mathcal{H}\right)\\ \parallel A\mp icot\alpha I\parallel \le csc\alpha \end{array}}{sup}\underset{\begin{array}{c}p\in \mathbb{C}\left[z\right]\\ p\ne 0\end{array}}{sup}{\displaystyle \frac{\parallel p\left(A\right)\parallel }{{max}_{z\in {\mathcal{L}}_{\alpha }}\left|p\left(z\right)\right|}}.$$

(1.4)

This polynomial formulation is equivalent to the usual rational formulation; the equivalence follows from Mergelyan approximation because the complement of a convex lens is connected.

The unrestricted value of (1.4) remains open for $0<\alpha <\pi /2$. The main theorem of this paper is deliberately more specific. It identifies exactly what the standard first-order nonnormal model can and cannot contribute.

Theorem 1

(Sharp square-zero constant). Let $0<\alpha \le \pi /2$. Among all complex Hilbert spaces, all operators

$$B=\lambda I+N,N^2=0,W\left(B\right)\subset {\mathcal{S}}_{\alpha },$$

and all functions f holomorphic in the interior of ${\mathcal{S}}_{\alpha }$, continuous on its one-point compactification, and satisfying ${\parallel f\parallel }_{\infty }\le 1$, one has

$$\parallel f\left(B\right)\parallel \le {\displaystyle \frac{\pi sin\alpha }{2\alpha }}.$$

(1.5)

The constant is best possible. Equality is attained on ${\mathbb{C}}^2$ by

$$B_{\alpha }=I+2sin\alpha \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f_{\alpha }\left(z\right)={\displaystyle \frac{z^{\pi /\left(2\alpha \right)}-1}{z^{\pi /\left(2\alpha \right)}+1}},$$

(1.6)

where the power is taken with $|argz|<\alpha$.

The corresponding lens matrix is exceptionally simple.

Corollary 1

(Exact fixed-lens subclass). Restrict (1.4) to operators A for which $1\notin Sp\left(A\right)$ and the Cayley transform satisfies

$${\left(\mathrm{\Phi }\left(A\right)-\lambda I\right)}^2=0\mathit{for}\mathit{some}\lambda \in \mathbb{C}.$$

The least constant on this subclass is exactly

$$C_{\mathrm{sq}0}\left(\alpha \right)={\displaystyle \frac{\pi sin\alpha }{2\alpha }}.$$

Allowing direct sums with the scalar operator 1 does not change this constant.

Corollary 2

(Explicit lens certificate). Put

$$A_{\alpha }:=sin\alpha \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right).$$

(1.7)

Then $A_{\alpha }$ is of $(D_{\alpha }^{+},D_{\alpha }^{-})$-type, with equality in both disk constraints, and

$$C\left(\alpha \right)\ge {\displaystyle \frac{\pi sin\alpha }{2\alpha }}.$$

(1.8)

More precisely, there is a sequence of polynomials $p_n$ with ${max}_{{\mathcal{L}}_{\alpha }}\left|p_n\right|\le 1+o\left(1\right)$ and

$$\parallel p_n\left(A_{\alpha }\right)\parallel ⟶{\displaystyle \frac{\pi sin\alpha }{2\alpha }}.$$

At the right angle we can also treat a nonlinear two-zero family for every admissible operator, rather than restricting the operator to a Jordan model.

Theorem 2

(Palindromic quadratic family). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$. For every $u>0$,

$$\parallel (T^4-u{T}^2+I){(T^4+u{T}^2+I)}^{-1}\parallel \le \sqrt{2}.$$

(1.9)

Consequently the conjectural right-angle constant holds for every degree-two Blaschke product on the right half-plane whose zeros are either both positive real or form a conjugate pair. The estimate is valid in arbitrary Hilbert-space dimension.

The palindromic restriction is not forced by the method. The following one-parameter family crosses the full imaginary diameter of the normalized post-automorphism disk and lies outside it except at the centre.

Theorem 3

(Imaginary post-automorphism diameter). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$. For every $t\in [-1,1]$,

$$\parallel \left({(T^2-I)}^2-it{(T^2+I)}^2\right){\left({(T^2+I)}^2+it{(T^2-I)}^2\right)}^{-1}\parallel \le \sqrt{2},$$

(1.10)

where at $t=\pm 1$ the common polynomial factor is cancelled before evaluation. Equivalently, the estimate holds on the full diameter $c=it$, $\left|t\right|\le 1$, for

$$h_c\left(w\right)={\displaystyle \frac{w^2+\overline{c}}{1+c{w}^2}}.$$

(1.11)

The estimate is valid in arbitrary Hilbert-space dimension.

The exact quotient-space margin in the preceding certificate also gives a two-real-dimensional region reaching both degenerate endpoints.

Theorem 4

(A nonsymmetric cusp). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$. If $c\in \mathbb{C}$ satisfies

$$|\Im c|\le 1,|\Re c|\le {\displaystyle \frac{{(1-|\Im c\left|\right)}^2}{200}},$$

(1.12)

then

$$\parallel \left({(T^2-I)}^2+\overline{c}{(T^2+I)}^2\right){\left({(T^2+I)}^2+c{(T^2-I)}^2\right)}^{-1}\parallel \le \sqrt{2}.$$

(1.13)

At the tips $c=\pm i$ the quotient is understood after cancellation. The estimate is valid in arbitrary Hilbert-space dimension.

The next result fills a full two-real-dimensional disk across the centre of the normalized parameter space.

Theorem 5

(A central complex post-automorphism disk). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$, and let

$$\left|c\right|\le {\displaystyle \frac{19}{20}}.$$

(1.14)

Then

$$\parallel \left({(T^2-I)}^2+\overline{c}{(T^2+I)}^2\right){\left({(T^2+I)}^2+c{(T^2-I)}^2\right)}^{-1}\parallel \le \sqrt{2}.$$

(1.15)

The rational functions in (1.15) are degree-two Blaschke products on the right half-plane. For example, at $c=i/2$ their zeros are

$$1+2i,{\displaystyle \frac{1-2i}{5}},$$

(1.16)

so the theorem contains genuinely nonsymmetric zero pairs and strictly extends the real and imaginary diameters to an open two-dimensional family. The estimate is valid in arbitrary Hilbert-space dimension.

The radius $19/20$ is not a barrier at every phase. The following result crosses it on a genuinely open set and reaches the boundary of the normalized parameter disk.

Theorem 6

(An almost-complete boundary-phase arc). For $t>0$ put

$$s_t={\displaystyle \frac{1-4t^4}{2t^2}},{\zeta }_t={\displaystyle \frac{4-s_t^2+4i{s}_t}{4+s_t^2}}.$$

(1.17)

Whenever

$$0\le \rho \le 1,{\displaystyle \frac{1}{10}}\le t\le {\displaystyle \frac{18}{25}},c=\rho {\zeta }_t,$$

(1.18)

every $T\in \mathcal{B}\left(\mathcal{H}\right)$ with $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$ satisfies

$$\parallel \left({(T^2-I)}^2+\overline{c}{(T^2+I)}^2\right){\left({(T^2+I)}^2+c{(T^2-I)}^2\right)}^{-1}\parallel \le \sqrt{2}.$$

(1.19)

The same assertion holds with c replaced by $\overline{c}$. Since $s_t$ decreases from $2499/50$ at $t=1/10$ through 0 at $t=1/\sqrt{2}$, the two arcs cover the unit circle except for the arc centred at $-1$ of angular length

$$4arctan{\displaystyle \frac{100}{2499}}=0.159978\dots \mathit{radians}=9.1661\dots \mathit{degrees}.$$

Every radial segment below the covered arcs is admissible. At $\rho =1$ the quotient is understood after cancellation of its common polynomial factor.

At the centre $c=0$, the preceding bound is not sharp even within the available argument. A low-degree bidisk extension gives the following strict improvement without any dimension restriction.

Proposition 1

(A strict central-square bound). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$ and put

$$w=(T^2-I){(T^2+I)}^{-1}.$$

Then

$$\parallel w^2\parallel \le {\kappa }_0,{\kappa }_0:={\displaystyle \frac{276889}{127200}}-{\displaystyle \frac{6103\sqrt{2}}{10600}}=1.3625601851\dots <\sqrt{2}.$$

(1.20)

The estimate holds in arbitrary Hilbert-space dimension.

The strict reserve in Proposition 1 controls two independently moving Blaschke zeros, not merely the double zero at the centre.

Theorem 7

(A two-small-zero disk). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$ and put

$$w=(T^2-I){(T^2+I)}^{-1},b_{\gamma }\left(z\right)={\displaystyle \frac{z-\gamma }{1-\overline{\gamma }z}}.$$

If $\alpha ,\beta \in \mathbb{C}$ satisfy

$$\left|\alpha \right|,\left|\beta \right|\le {\displaystyle \frac{1}{150}},$$

then

$$\parallel b_{\alpha }\left(w\right)b_{\beta }\left(w\right)\parallel \le {\displaystyle \frac{{\kappa }_0+2\sqrt{2}/150+1/{150}^2}{1-2\sqrt{2}/150-{\kappa }_0/{150}^2}}<\sqrt{2}.$$

(1.21)

The estimate is valid in arbitrary Hilbert-space dimension. In particular, $\alpha =1/150$ and $\beta =i/150$ give two distinct nonzero zeros on the boundary of the certified parameter bidisk.

The lower bound (1.8) was already present in [1,2]; after affine normalization, the same nilpotent model is displayed in [5]. The point of Theorem 1 is the matching upper bound for every square-zero perturbation, in arbitrary Hilbert-space dimension. The point of Theorems 2–7 and Proposition 1 is different: the operator is unrestricted, while the Schur function belongs to a nonlinear quadratic family. No claim of a new value for the unrestricted constant $C\left(\alpha \right)$ is made.

2. Möbius Reduction: A Lens Is a Compactified Sector

The Möbius map

$$\mathrm{\Phi }\left(z\right)={\displaystyle \frac{1+z}{1-z}},{\mathrm{\Phi }}^{-1}\left(w\right)={\displaystyle \frac{w-1}{w+1}},$$

(2.1)

maps the interior of ${\mathcal{L}}_{\alpha }$ conformally onto the interior of ${\mathcal{S}}_{\alpha }$, sending $-1$ to 0 and 1 to ∞. Each circular arc is sent to one of the rays $argw=\pm \alpha$.

For a bounded operator T, recall

$$W\left(T\right)=\{\langle Tx,x\rangle :\parallel x\parallel =1\}.$$

A closed half-plane is a spectral set for T precisely when $W\left(T\right)$ lies in that half-plane. Applying this fact to the two half-planes bounded by the rays of ${\mathcal{S}}_{\alpha }$ gives the operator form of (2.1).

Proposition 2

(Lens–sector equivalence). Suppose $1\notin Sp\left(A\right)$ and put $B=\mathrm{\Phi }\left(A\right)$. Then

$$\parallel A\mp icot\alpha I\parallel \le csc\alpha ⟺W\left(B\right)\subset {\mathcal{S}}_{\alpha }.$$

(2.2)

Consequently the fixed-lens constant $C\left(\alpha \right)$ equals the least K such that

$$\parallel r\left(B\right)\parallel \le K\underset{z\in {\mathcal{S}}_{\alpha }}{sup}\left|r\left(z\right)\right|$$

(2.3)

for every sectorial B with $W\left(B\right)\subset {\mathcal{S}}_{\alpha }$ and every rational r bounded on the sector.

Proof. 

The scalar map (2.1) sends the two disks to the two supporting half-planes of ${\mathcal{S}}_{\alpha }$. The functional-calculus identity ${\mathrm{\Phi }}^{-1}\left(\mathrm{\Phi }\left(A\right)\right)=A$ transports the two disk spectral-set inequalities to the two half-plane spectral-set inequalities. The half-plane criterion just recalled proves (2.2). It also gives both inequalities between the best constants after composing test functions with $\mathrm{\Phi }$ and ${\mathrm{\Phi }}^{-1}$.

If $1\in Sp\left(A\right)$, one first replaces A by $(1-\epsilon )A$; the normalized lens is convex and contains 0, so both inequalities in (1.1) are preserved. Letting $\epsilon \downarrow 0$ gives the result by norm continuity of polynomial functional calculus. The reverse passage is obtained directly because $-1\notin {\mathcal{S}}_{\alpha }$ and hence $-1\notin Sp\left(B\right)$. This is the standard argument of [1].    □

Thus the original two-disk problem is not merely analogous to a sectorial numerical-range problem; the optimal constants are identical. We shall use this equivalence again in Section 5.

3. Three Sharp Lemmas

We isolate the geometric, function-theoretic, and operator-norm ingredients of the proof.

Lemma 1

(Numerical range of a square-zero operator). If $N\in \mathcal{B}\left(\mathcal{H}\right)$ and $N^2=0$, then

$$W\left(N\right)=\{z\in \mathbb{C}:|z|\le \frac{1}{2}\parallel N\parallel \},$$

with closure understood when the norm is not attained. Consequently, for $\lambda =r{e}^{i\theta }$ with $\left|\theta \right|<\alpha$,

$$W(\lambda I+N)\subset {\mathcal{S}}_{\alpha }⟺\parallel N\parallel \le 2rsin(\alpha -|\theta \left|\right).$$

(3.1)

Proof. 

Since $RanN\subset kerN$, relative to $\mathcal{H}=kerN\oplus {(kerN)}^{\perp }$ the operator has the form

$$N=\left(\begin{array}{cc}0& X\\ 0& 0\end{array}\right).$$

Its numerical range is the disk centred at zero of radius $\parallel X\parallel /2$; this follows at once by varying the relative phase and the two norms of a unit vector in the displayed decomposition. Since $\parallel X\parallel =\parallel N\parallel$, the first assertion follows. The largest disk centred at $r{e}^{i\theta }$ and contained in the sector has radius equal to the distance to the nearer boundary ray, namely $rsin(\alpha -|\theta \left|\right)$.    □

Lemma 2

(Sector Schwarz–Pick estimate). Let $q=\pi /\left(2\alpha \right)$ and let $f:int{\mathcal{S}}_{\alpha }\to \mathbb{D}$ be holomorphic. At $\lambda =r{e}^{i\theta }$, $\left|\theta \right|<\alpha$, one has

$$|f^{\prime }\left(\lambda \right)|\le {\displaystyle \frac{q\left(1-{\left|f\left(\lambda \right)\right|}^2\right)}{2rcos\left(q\theta \right)}}.$$

(3.2)

Moreover,

$${\displaystyle \frac{sin(\alpha -|\theta \left|\right)}{cos\left(q\theta \right)}}\le sin\alpha .$$

(3.3)

Proof. 

The power map $P\left(z\right)=z^q$ sends $int{\mathcal{S}}_{\alpha }$ conformally onto the right half-plane. Schwarz–Pick in that half-plane gives

$$|{(f\circ P^{-1})}^{\prime }\left(P\left(\lambda \right)\right)|\le {\displaystyle \frac{1-{\left|f\left(\lambda \right)\right|}^2}{2ReP\left(\lambda \right)}}.$$

Since $|P^{\prime }\left(\lambda \right)|=q{r}^{q-1}$ and $ReP\left(\lambda \right)=r^{q}cos\left(q\theta \right)$, this is (26).

It remains to prove (3.3). By symmetry take $0\le \theta <\alpha$ and set $u=1-\theta /\alpha \in (0,1]$. Then (27) is equivalent to

$${\displaystyle \frac{sin\left(u\alpha \right)}{sin\alpha }}\le sin{\displaystyle \frac{\pi u}{2}}.$$

(3.4)

For fixed $0<u\le 1$, the function $x\mapsto sin\left(ux\right)/sinx$ is increasing on $(0,\pi /2]$. Indeed, its logarithmic derivative is $ucot\left(ux\right)-cotx\ge 0$, because $t\mapsto tcott$ is decreasing there. Comparing $x=\alpha$ with $x=\pi /2$ proves (3.4).    □

Lemma 3

(Exact norm formula). If $N^2=0$, $a,b\in \mathbb{C}$, $s=\left|a\right|$, and $t=\left|b\right|\parallel N\parallel$, then

$$\parallel aI+bN\parallel ={\displaystyle \frac{t+\sqrt{t^2+4s^2}}{2}}.$$

(3.5)

For $c\ge 1$, if $t\le c(1-s^2)$ and $0\le s\le 1$, then

$$\parallel aI+bN\parallel \le c.$$

(3.6)

Proof. 

Using the block form in the proof of Lemma 1, phases may be removed by unitaries. The norm is the supremum of the largest singular value of $\left(\begin{array}{cc}s& t_0\\ 0& s\end{array}\right)$ over $0\le t_0\le t$, which gives (3.5); approximate norming vectors cover the case in which $\parallel N\parallel$ is not attained.

The right side of (3.5) increases with t. Substituting $t=c(1-s^2)$, the desired inequality is equivalent to

$$\sqrt{c^2{(1-s^2)}^2+4s^2}\le c(1+s^2).$$

After squaring, the difference between the right and left squares is $4s^2(c^2-1)\ge 0$.    □

4. Proof of the Sharp Square-Zero Theorem

Proof of Theorem 1. 

If $\lambda =0$, sector containment and Lemma 1 force $N=0$, and the claim is immediate. If $\lambda$ lies on a boundary ray, the same geometry forces $N=0$, so the claim again follows from $\left|f\right(\lambda \left)\right|\le 1$. We may therefore assume that $\lambda$ lies in the sector interior. Write $\lambda =r{e}^{i\theta }$ and set

$$s=\left|f\left(\lambda \right)\right|,t=|f^{\prime }\left(\lambda \right)|\parallel N\parallel ,c_{\alpha }={\displaystyle \frac{\pi sin\alpha }{2\alpha }}=qsin\alpha .$$

The holomorphic functional calculus truncates exactly at first order:

$$f(\lambda I+N)=f\left(\lambda \right)I+f^{\prime }\left(\lambda \right)N.$$

(4.1)

Combining Lemmas 1 and 2 gives

$$\begin{array}{cc}\hfill t& \le {\displaystyle \frac{q(1-s^2)}{2rcos\left(q\theta \right)}}2rsin(\alpha -|\theta \left|\right)\hfill \\ \hfill & \le qsin\alpha (1-s^2)=c_{\alpha }(1-s^2).\hfill \end{array}$$

Also $c_{\alpha }\ge 1$ for $0<\alpha \le \pi /2$. Applying Lemma 3 to (4.1) proves (1.5).

For sharpness let $J=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and use (1.6). By Lemma 1, $W\left(B_{\alpha }\right)$ is the disk with centre 1 and radius $sin\alpha$, which is tangent to both boundary rays of ${\mathcal{S}}_{\alpha }$. The function $f_{\alpha }$ maps the sector to the unit disk, while

$$f_{\alpha }\left(1\right)=0,f_{\alpha }^{\prime }\left(1\right)={\displaystyle \frac{q}{2}}.$$

Therefore

$$f_{\alpha }\left(B_{\alpha }\right)=qsin\alpha J,\parallel f_{\alpha }\left(B_{\alpha }\right)\parallel =c_{\alpha }.$$

This proves equality.    □

Proof of Corollary 1. 

Apply Proposition 2. Composition with $\mathrm{\Phi }$ identifies the restricted lens functional calculus isometrically with the class in Theorem 1. The upper bound follows from that theorem and sharpness follows from Corollary 2. A scalar direct summand at 1 contributes at most the supremum norm of the test function.    □

For $0<\alpha <\pi /2$, inspection of the inequalities in the proof gives necessary conditions for equality: the centre $\lambda$ lies on the symmetry axis, the numerical-range radius is maximal, the Schwarz–Pick inequality is sharp, and $f\left(\lambda \right)=0$. We do not need, and do not claim, a classification of all equality cases, particularly at the half-plane endpoint $\alpha =\pi /2$.

5. The Explicit Lens Certificate

Proof of Corollary 2. 

For $A_{\alpha }=sin\alpha J$, the exact norm formula gives

$$\begin{array}{cc}\hfill \parallel A_{\alpha }\mp icot\alpha I\parallel & ={\displaystyle \frac{sin\alpha +\sqrt{{sin}^2\alpha +4{cot}^2\alpha }}{2}}\hfill \\ \hfill & =csc\alpha .\hfill \end{array}$$

Hence both disks in (1.2) are spectral sets for $A_{\alpha }$. Moreover,

$$\mathrm{\Phi }\left(A_{\alpha }\right)=(I+A_{\alpha }){(I-A_{\alpha })}^{-1}=I+2sin\alpha J=B_{\alpha }.$$

The function $g_{\alpha }=f_{\alpha }\circ \mathrm{\Phi }$ is bounded by one on the lens, continuous on its closure, and holomorphic in its interior. Although it need not be rational when $\pi /\left(2\alpha \right)\notin \mathbb{N}$, Mergelyan’s theorem provides polynomials $q_n\to g_{\alpha }$ uniformly on ${\mathcal{L}}_{\alpha }$. On a small circle around the sole eigenvalue 0 the convergence is locally uniform, so also $q_n^{\prime }\left(0\right)\to g_{\alpha }^{\prime }\left(0\right)$. Consequently $q_n\left(A_{\alpha }\right)\to g_{\alpha }\left(A_{\alpha }\right)$ in norm. Rescaling $q_n$ by ${max}_{{\mathcal{L}}_{\alpha }}\left|q_n\right|$ yields the asserted $p_n$, and

$$\parallel g_{\alpha }\left(A_{\alpha }\right)\parallel =\parallel f_{\alpha }\left(B_{\alpha }\right)\parallel ={\displaystyle \frac{\pi sin\alpha }{2\alpha }}.$$

   □

When $q=\pi /\left(2\alpha \right)$ is an integer, $g_{\alpha }$ itself is rational and the approximation step may be omitted. The important point is that (7) is an entirely finite-dimensional, machine-checkable lower certificate: the two disk constraints and the functional-calculus value are closed-form identities.

6. A Right-Angle Palindromic Two-Zero Sum of Squares

We first locate the families in the full moduli space of quadratic inner functions. Write ${\mathbb{H}}_{+}=\{z:\Re z>0\}$.

Proposition 3

(Critical-point normal form). Let F be a degree-two finite Blaschke product on ${\mathbb{H}}_{+}$. There are $r>0$, $s\in \mathbb{R}$, $c\in \mathbb{D}$, and $\omega \in \mathbb{T}$ such that

$$F\left(rz\right)=\omega {\displaystyle \frac{{\psi }_s{\left(z\right)}^2+\overline{c}}{1+c{\psi }_s{\left(z\right)}^2}},{\psi }_s\left(z\right)={\displaystyle \frac{z-(1+is)}{z+(1-is)}}.$$

(6.1)

Thus, up to positive input dilation and an irrelevant output phase, the quadratic moduli consist of one real critical-point parameter s and one disk parameter c. The real post-automorphism diameter $s=0$, $c\in (-1,1)$ is the palindromic family of Theorem 2; Theorems 3–5 concern genuinely complex c in the same slice $s=0$.

Proof. 

A degree-two disk Blaschke product has one critical point in the disk, counting multiplicity. Conjugating by a half-plane–disk Möbius map gives the corresponding unique critical point $p\in {\mathbb{H}}_{+}$ of F. Write $p=r(1+is)$ with $r=\Re p>0$, and conjugate $F\left(rz\right)$ by ${\psi }_s^{-1}$ on the input. The resulting disk Blaschke product G has its critical point at zero. If $a=G\left(0\right)$, then

$${\displaystyle \frac{G\left(w\right)-a}{1-\overline{a}G\left(w\right)}}$$

is a degree-two Blaschke product with a double zero at zero, hence equals $\gamma w^2$ for some $\gamma \in \mathbb{T}$. Solving for G and absorbing the unimodular factor in the output gives exactly (32), with $\left|c\right|=\left|a\right|<1$.

For $s=0$, put $w=(z-1)/(z+1)$. Direct substitution gives

$${\displaystyle \frac{z^2-uz+1}{z^2+uz+1}}={\displaystyle \frac{w^2-a}{1-a{w}^2}},a={\displaystyle \frac{u-2}{u+2}},$$

so $c=-a$ runs through the real diameter as u runs through the two palindromic ranges used in Theorem 2.    □

We now prove 2. The algebraic selector in the next lemma is somewhat lengthy, but every quantity is explicit and the resulting identity is dimension-free. For a polynomial $\ell \left(z\right)={\sum }_{j=0}^3{\ell }_j{z}^j$, write

$${\ell }^{\#}\left(z\right)={\displaystyle \sum _{j=0}^3}\overline{{\ell }_{3-j}}{z}^j,{\ell }^c\left(z\right)={\displaystyle \sum _{j=0}^3}\overline{{\ell }_j}{z}^j,{\ell }^r\left(z\right)={\displaystyle \sum _{j=0}^3}{\ell }_{3-j}{z}^j.$$

Lemma 4

(Global algebraic selector). For every $u\ge 2$ there are real numbers $b,c,x,m,n,r,\sigma$, with $r>0$, such that, for every $T=X+iY$ and

$$P=X+Y,Q=X-Y,Z=T^2,$$

the polynomials

$$\begin{array}{cc}\hfill D& =Z^2+uZ+I,N=Z^2-uZ+I,\hfill \\ \hfill a\left(z\right)& =1+bz+c{z}^2+b{z}^3+z^4,\hfill \\ \hfill \ell \left(z\right)& =i\sigma +xz+(m+in)z^2+(1+i)r{z}^3\hfill \end{array}$$

satisfy the hereditary identity

$$\begin{array}{cc}\hfill 2D^{*}D-N^{*}N=& a{\left(T\right)}^{*}a\left(T\right)+\ell {\left(T\right)}^{*}P\ell \left(T\right)+{\ell }^{\#}{\left(T\right)}^{*}P{\ell }^{\#}\left(T\right)\hfill \\ \hfill & +{\ell }^c{\left(T\right)}^{*}Q{\ell }^c\left(T\right)+{\ell }^r{\left(T\right)}^{*}Q{\ell }^r\left(T\right).\hfill \end{array}$$

(6.2)

Proof. 

For $\tau \ge 2$, define

$$F_{\tau }\left(p\right)=A_4{p}^4+A_3{p}^3+A_2{p}^2+A_{1}p+A_0,$$

where

$$\begin{array}{cc}\hfill A_0=& 16{({\tau }^8-{\tau }^6+14{\tau }^4-20{\tau }^2+8)}^2,\hfill \\ \hfill A_1=& -4\tau ({\tau }^2+2)(20{\tau }^{14}+53{\tau }^{12}-32{\tau }^{10}+396{\tau }^8\hfill \\ \hfill & -320{\tau }^6-336{\tau }^4+512{\tau }^2-192),\hfill \\ \hfill A_2=& {\tau }^2{({\tau }^2+2)}^2(12{\tau }^{14}+123{\tau }^{12}+240{\tau }^{10}+116{\tau }^8\hfill \\ \hfill & +960{\tau }^6-432{\tau }^4-256{\tau }^2+192),\hfill \\ \hfill A_3=& -4{\tau }^7{({\tau }^2+2)}^4(3{\tau }^6+14{\tau }^4+4{\tau }^2+40),\hfill \\ \hfill A_4=& {\tau }^6{({\tau }^2+2)}^6(3{\tau }^4+4{\tau }^2+12).\hfill \end{array}$$

For $\tau \ge 2$ these coefficients have signs $+,-,+,-,+$ in increasing degree. For the only two non-obvious brackets, put $q={\tau }^2\ge 4$ and group

$$\begin{array}{cc}\hfill & 20q^7-32q^5+53q^6-320q^3+396q^4-336q^2+512q-192,\hfill \\ \hfill & 12q^7+123q^6+240q^5+116q^4+(960q^3-432q^2-256q)+192;\hfill \end{array}$$

each displayed group is positive. Thus $F_{\tau }\left(p\right)>0$ for $p<0$.

The discriminant ${\Delta }_{\tau }$ of $F_{\tau }$ is

$$\begin{array}{cc}\hfill {\Delta }_{\tau }=& -5184{\tau }^{14}{({\tau }^2-2)}^{12}{({\tau }^2+2)}^{18}{(2{\tau }^2-1)}^2\hfill \\ \hfill & ·{({\tau }^2-2\tau +2)}^2{({\tau }^2+2\tau +2)}^{2}H\left({\tau }^2\right),\hfill \end{array}$$

(6.3)

where, on writing $z=q-4$,

$$\begin{array}{cc}\hfill H\left(q\right)=& 576z^{15}+42864z^{14}+1457244z^{13}+30074493z^{12}\hfill \\ \hfill & +421983272z^{11}+4268521080z^{10}+32174243296z^9\hfill \\ \hfill & +184016942896z^8+804672632576z^7+2686678018304z^6\hfill \\ \hfill & +6778308853760z^5+12646404948736z^4+16800286140416z^3\hfill \\ \hfill & +14875812689920z^2+7736848048128z+1732363472896.\hfill \end{array}$$

Hence (34) is strictly negative. The quartic has exactly two simple real roots; the sign pattern excludes negative roots. Denote its smaller positive root by $p_{-}\left(\tau \right)$. It is continuous for $\tau \ge 2$.

Set

$$\begin{array}{cc}\hfill U(p,\tau ):=& (3p^2{\tau }^{12}+14p^2{\tau }^{10}+24p^2{\tau }^8+48p^2{\tau }^6\hfill \\ \hfill & +112p^2{\tau }^4+96p^2{\tau }^2-20p{\tau }^9-32p{\tau }^7-96p{\tau }^5\hfill \\ \hfill & -128p{\tau }^3+192p\tau +8{\tau }^8-8{\tau }^6+112{\tau }^4-160{\tau }^2+64)\hfill \\ \hfill & /\left(12{\tau }^2{({\tau }^2-2)}^2({\tau }^2+2)\right).\hfill \end{array}$$

(6.4)

At $\tau =2$ one has

$$\begin{array}{cccc}\hfill F_2(2/5)& ={\displaystyle \frac{39721984}{625}}>0,\hfill & \hfill F_2(4/9)& =-35840<0,\hfill \\ \hfill U(p,2)-2& ={\displaystyle \frac{594p^2-282p+7}{18}}.\hfill & \hfill \end{array}$$

It follows that $2/5<p_{-}\left(2\right)<4/9$ and $U(p_{-}\left(2\right),2)<2$.

For ${\tau }^2\ge 14$, direct substitution gives

$$\begin{array}{cc}\hfill F_{\tau }(1/\tau )=& 3w^9+373w^8+20472w^7+651304w^6\hfill \\ \hfill & +13240608w^5+178399616w^4+1593137664w^3\hfill \\ \hfill & +9091884032w^2+30082498560w+43954733056>0,\hfill \end{array}$$

where $w={\tau }^2-14$, while

$$\begin{array}{cc}\hfill F_{\tau }(2/\tau )=& -4({\tau }^2-14)(3{\tau }^4+4{\tau }^2+4)\hfill \\ \hfill & ·(3{\tau }^{10}+24{\tau }^8+36{\tau }^6+104{\tau }^4\hfill \\ \hfill & -32{\tau }^2+32)<0\hfill \end{array}$$

when ${\tau }^2>14$. Consequently $1/\tau <p_{-}\left(\tau \right)<2/\tau$ for large $\tau$. Substitution $p=y/\tau$ gives, after division by ${\tau }^{18}$,

$${\tau }^{-18}{F}_{\tau }(y/\tau )⟶3y^2{(y-2)}^2\mathrm{uniformly}\mathrm{for}1\le y\le 2.$$

Thus $\tau p_{-}\left(\tau \right)\to 2$, and (6.4) gives

$${\displaystyle \frac{U(p_{-}\left(\tau \right),\tau )}{{\tau }^2}}⟶1.$$

The continuous function $\tau \mapsto U(p_{-}\left(\tau \right),\tau )$ therefore starts below 2 and tends to infinity. Its image contains $[2,\infty )$.

Fix the prescribed $u\ge 2$ and choose $\tau \ge 2$ and $p=p_{-}\left(\tau \right)$ with $U(p,\tau )=u$. Put $r=\sqrt{p}$, $\sigma =-\tau r$, and

$$\begin{array}{cc}\hfill b=& -({\tau }^2+2)p,\hfill \\ \hfill x=& {\displaystyle \frac{\tau r\{p{({\tau }^2+2)}^2-4\tau \}}{2({\tau }^2-2)}},\hfill \\ \hfill m=& {\displaystyle \frac{r\{-p\tau {({\tau }^2+2)}^2+8({\tau }^2-1)\}}{2\tau ({\tau }^2-2)}},\hfill \\ \hfill n=& {\displaystyle \frac{r\{p{({\tau }^2+2)}^2-2{\tau }^3\}}{2({\tau }^2-2)}},\hfill \\ \hfill c=& {\displaystyle \frac{p\{-p\tau {({\tau }^2+2)}^2-8+4{\tau }^2\}+6\tau u}{2\tau }}.\hfill \end{array}$$

Exact coefficient comparison shows that (6.2) is equivalent to

$$\begin{array}{cc}\hfill & b+2r^2+{\sigma }^2=0,c-3u-x\sigma +2mr=0,\hfill \\ \hfill & b+2xr-m\sigma +\sigma n=0,b^2+2x\sigma +4rn=0,\hfill \\ \hfill & bc+x^2+m^2+\sigma m+\sigma n+n^2=0,\hfill \\ \hfill & b^2+2x(m+n)+4r\sigma =0,c^2-u^2+4x(m-n)=0.\hfill \end{array}$$

The displayed parameters solve the first four and the sixth equations. The fifth is precisely $U(p,\tau )=u$; after this substitution the residual in the seventh is

$${\displaystyle \frac{2F_{\tau }\left(p\right)}{9{\tau }^4{({\tau }^2-2)}^4{({\tau }^2+2)}^2}}=0.$$

This proves the identity.    □

The preceding selector covers $u\ge 2$. The complementary interval admits one matrix-polynomial certificate. We write

$$\mathbf{v}\left(T\right)={(I,T,T^2,T^3,T^4)}^{\mathsf{T}},\mathbf{w}\left(T\right)={(I,T,T^2,T^3)}^{\mathsf{T}},$$

and, for $d\ge 1$,

$$V_{2,d}\left(u\right)=\left(\begin{array}{c}{I}_d\\ u{I}_d\\ u^2{I}_d\end{array}\right),V_{1,d}\left(u\right)=\left(\begin{array}{c}{I}_d\\ u{I}_d\end{array}\right).$$

Lemma 5

(Exact interval certificate). There are Hermitian matrices

$$\begin{array}{cccc}\hfill S_0& \in M_{15}\left(\mathbb{Q}\left(\sqrt{2}\right)\right),\hfill & \hfill R_0& \in M_{10}\left(\mathbb{Q}\left(\sqrt{2}\right)\right),\hfill \\ \hfill S_P& \in M_{12}\left(\mathbb{Q}(\sqrt{2},i)\right),\hfill & \hfill R_P& \in M_8\left(\mathbb{Q}(\sqrt{2},i)\right).\hfill \end{array}$$

which are positive semidefinite and have respective ranks $13,8,10,7$. Put

$$\begin{array}{cc}\hfill G_0\left(u\right)& =V_{2,5}{\left(u\right)}^{*}{S}_0{V}_{2,5}\left(u\right)+u(2-u)V_{1,5}{\left(u\right)}^{*}{R}_0{V}_{1,5}\left(u\right),\hfill \\ \hfill G_P\left(u\right)& =V_{2,4}{\left(u\right)}^{*}{S}_P{V}_{2,4}\left(u\right)+u(2-u)V_{1,4}{\left(u\right)}^{*}{R}_P{V}_{1,4}\left(u\right),\hfill \\ \hfill G_Q\left(u\right)& =\overline{G_P\left(u\right)}.\hfill \end{array}$$

For $0\le u\le 2$, every $T=X+iY$, and

$$P=X+Y,Q=X-Y,Z=T^2,D=Z^2+uZ+I,N=Z^2-uZ+I,$$

one has the exact hereditary identity

$$\begin{array}{cc}\hfill 2D^{*}D-N^{*}N=& \mathbf{v}{\left(T\right)}^{*}{G}_0\left(u\right)\mathbf{v}\left(T\right)+\mathbf{w}{\left(T\right)}^{*}{G}_P\left(u\right)P\mathbf{w}\left(T\right)\hfill \\ \hfill & +\mathbf{w}{\left(T\right)}^{*}{G}_Q\left(u\right)Q\mathbf{w}\left(T\right).\hfill \end{array}$$

(6.5)

Proof. 

The four matrices are given entry by entry in the ancillary file ancillary/symmetric_u_exact_certificate.py; every entry is an exact element of the displayed number field. We record the finite audit so that the certificate is independently checkable. Put $r=1/\sqrt{2}$ and

$$\begin{array}{cccc}\hfill v_1& ={(1,r,0,-r,-1)}^{\mathsf{T}},\hfill & \hfill v_2& ={(0,r,1,r,0)}^{\mathsf{T}},\hfill \\ \hfill p& ={(-r(1+i),-i,r(1-i),1)}^{\mathsf{T}}.\hfill \end{array}$$

The prescribed kernels are

matrix	order	kernel basis	rank
$S_0$	15	$v_1\oplus 0_{10},v_2\oplus 0_{10}$	13
$R_0$	10	$v_1\oplus 0_5,v_2\oplus 0_5$	8
$S_P$	12	$p\oplus 0_8,\overline{p}\oplus 0_8$	10
$R_P$	8	$p\oplus 0_4$	7

On a full-rank principal compression of each matrix, exact symmetric elimination gives respectively $13,8,10,7$ positive $LD{L}^{*}$ pivots in $\mathbb{Q}\left(\sqrt{2}\right)$. Positivity of an element $a+b\sqrt{2}$ is decided using only rational comparisons (for opposite signs, compare $a^2$ with $2b^2$). Thus the four matrices are positive semidefinite with the asserted ranks. The ancillary verifier performs precisely these operations in exact SymPy arithmetic; it uses no floating-point test.

It remains to check the identity. Put

$$d_0={(1,0,0,0,1)}^{\mathsf{T}},d_1={(0,0,1,0,0)}^{\mathsf{T}}.$$

Coefficient comparison in powers of u gives the five target matrices

$$d_0{d}_0^{*},3(d_0{d}_1^{*}+d_1{d}_0^{*}),d_1{d}_1^{*},0,0.$$

Expanding the three terms on the right of (6.5) gives exactly the same five matrices. This is a finite equality in $\mathbb{Q}(\sqrt{2},i)$, checked entrywise by the ancillary verifier. Hence (6.5) is an exact polynomial identity. Finally the Markov–Lukacs forms defining $G_0,G_P,G_Q$ are positive semidefinite for $0\le u\le 2$, proving the lemma.    □

Proof of Theorem 2. 

Put $Z=T^2$, $D=Z^2+uZ+I$, and $N=Z^2-uZ+I$. Spectral inclusion for the numerical range and spectral mapping give $Sp\left(Z\right)\subset \{z:Rez\ge 0\}$. Both zeros of $z^2+uz+1$ lie in the open left half-plane, so D is invertible. Sector containment gives $P=X+Y\ge 0$ and $Q=X-Y\ge 0$. If $u\ge 2$, apply Lemma 4; if $0<u\le 2$, apply Lemma 5. In either case the relevant identity has a positive right-hand side. Congruence by $D^{-1}$ yields

$$2I-{\left(N{D}^{-1}\right)}^{*}\left(N{D}^{-1}\right)\ge 0,$$

which is (1.9).

For positive real half-plane zeros $s,t$, replace T by $T/{\left(st\right)}^{1/4}$ and take $u=(s+t)/\sqrt{st}\ge 2$. For a conjugate pair $\zeta =a+ib$, $\overline{\zeta }$, with $a>0$, replace T by $T/\sqrt{\left|\zeta \right|}$ and take $u=2a/\left|\zeta \right|\in (0,2]$. Indeed, with $b_{\zeta }\left(z\right)=(z-\zeta )/(z+\overline{\zeta })$,

$$b_{\zeta }\left(z\right)b_{\overline{\zeta }}\left(z\right)={\displaystyle \frac{z^2-2az+{\left|\zeta \right|}^2}{z^2+2az+{\left|\zeta \right|}^2}}.$$

This proves both stated Blaschke-product consequences.    □

6.1. The Complete Imaginary Post-Automorphism Diameter

Let $J_4$ denote the $4\times 4$ reversal matrix. The following exact matrix certificate is the core of 3.

Lemma 6

(Exact imaginary-diameter certificate). Put

$$a={\displaystyle \frac{1}{2}}\sqrt{1+\sqrt{2}},\mathbb{K}=\mathbb{Q}(i,a).$$

There are positive semidefinite Hermitian matrices

$$\begin{array}{cccc}\hfill S_0& \in M_{15}\left(\mathbb{K}\right),\hfill & \hfill R_0& \in M_{10}\left(\mathbb{K}\right),\hfill \\ \hfill S_P& \in M_{12}\left(\mathbb{K}\right),\hfill & \hfill R_P& \in M_8\left(\mathbb{K}\right),\hfill \end{array}$$

of respective ranks $11,8,8,7$. For $0\le t\le 1$, set

$$\begin{array}{cc}\hfill G_0\left(t\right)& =V_{2,5}{\left(t\right)}^{*}{S}_0{V}_{2,5}\left(t\right)+t(1-t)V_{1,5}{\left(t\right)}^{*}{R}_0{V}_{1,5}\left(t\right),\hfill \\ \hfill G_P\left(t\right)& =V_{2,4}{\left(t\right)}^{*}{S}_P{V}_{2,4}\left(t\right)+t(1-t)V_{1,4}{\left(t\right)}^{*}{R}_P{V}_{1,4}\left(t\right),\hfill \\ \hfill G_Q\left(t\right)& =J_4{G}_P\left(t\right)J_4.\hfill \end{array}$$

If $T=X+iY$, $P=X+Y$, $Q=X-Y$, and

$$\begin{array}{cc}\hfill D_t& ={(T^2+I)}^2+it{(T^2-I)}^2,\hfill \\ \hfill N_t& ={(T^2-I)}^2-it{(T^2+I)}^2,\hfill \end{array}$$

then the hereditary identity

$$\begin{array}{cc}\hfill 2D_t^{*}{D}_t-N_t^{*}{N}_t=& \mathbf{v}{\left(T\right)}^{*}{G}_0\left(t\right)\mathbf{v}\left(T\right)+\mathbf{w}{\left(T\right)}^{*}{G}_P\left(t\right)P\mathbf{w}\left(T\right)\hfill \\ \hfill & +\mathbf{w}{\left(T\right)}^{*}{G}_Q\left(t\right)Q\mathbf{w}\left(T\right)\hfill \end{array}$$

(6.6)

holds exactly.

Proof. 

We give the finite audit because the matrices are most compactly recorded by exact affine coordinates. Their machine-readable data are in ancillary/imaginary_interval_certificate.py, and the independent reconstruction is ancillary/verify_imaginary_interval.py.

Write

$$\mathbf{d}={(1,0,2,0,1)}^{\mathsf{T}},\mathbf{n}={(1,0,-2,0,1)}^{\mathsf{T}}.$$

The coefficient matrices of $2D_t^{*}{D}_t-N_t^{*}{N}_t$ in degrees $0,1,2$ are

$$2\overline{\mathbf{d}}{\mathbf{d}}^{\mathsf{T}}-\overline{\mathbf{n}}{\mathbf{n}}^{\mathsf{T}},i(\overline{\mathbf{d}}{\mathbf{n}}^{\mathsf{T}}-\overline{\mathbf{n}}{\mathbf{d}}^{\mathsf{T}}),-\overline{\mathbf{d}}{\mathbf{d}}^{\mathsf{T}}+2\overline{\mathbf{n}}{\mathbf{n}}^{\mathsf{T}}.$$

(6.7)

Starting with Hermitian indeterminates of orders $15,10,12,8$, expand the two Markov–Lukacs forms in the statement, use $G_Q=J_4{G}_P{J}_4$, and equate all hereditary coefficients through degree four to (6.7) and two zero matrices. At $t=1$ impose the natural endpoint face

$$\begin{array}{cccc}\hfill V_{2,5}{\left(1\right)}^{*}{S}_0{V}_{2,5}\left(1\right)& ={\overline{\mathbf{d}}}_1{\mathbf{d}}_1^{\mathsf{T}},\hfill & \hfill V_{2,4}{\left(1\right)}^{*}{S}_P{V}_{2,4}\left(1\right)& =0,\hfill \\ \hfill {\mathbf{d}}_1& ={(1+i,0,2-2i,0,1+i)}^{\mathsf{T}}.\hfill & \hfill \end{array}$$

Finally impose the following kernel dimensions:

matrix	order	kernel vectors	rank
$S_0$	15	$(q,q,q)$ for four independent ${\mathbf{d}}_1^{\mathsf{T}}q=0$	11
$R_0$	10	$(q_1,q_1),(q_2,q_2)$	8
$S_P$	12	$(e_j,e_j,e_j)$, $0\le j\le 3$	8
$R_P$	8	$(p,p)$	7

Here $e_j$ are the coordinate vectors and, with

$$\eta ={\displaystyle \frac{1}{2}}\left(\sqrt{1+\sqrt{2}}+i\sqrt{\sqrt{2}-1}\right),\rho =\sqrt{2}-1,\delta =\sqrt{{\displaystyle \frac{\sqrt{2}-1}{2}}},$$

one may take

$$\begin{array}{cc}\hfill q_1& ={(i,\eta ,1,\eta ,i)}^{\mathsf{T}},\hfill \\ \hfill q_2& ={(1,\overline{\eta }/\sqrt{2},0,-\overline{\eta }/\sqrt{2},-1)}^{\mathsf{T}},\hfill \\ \hfill p& ={(-\rho \delta (1+i),-i\rho ,\delta (1-i),1)}^{\mathsf{T}}.\hfill \end{array}$$

For the first row of the table the verifier uses

$${(1,0,0,0,-1)}^{\mathsf{T}},{(2i,0,1,0,0)}^{\mathsf{T}},e_1,e_3.$$

All these entries lie in $\mathbb{K}$, since $16a^4-8a^2-1=0$ and the remaining real radicals are rational functions of a.

After real and imaginary parts are separated, the construction is a sparse $438\times 534$ exact affine system of rank 328. The ancillary certificate lists its 205 free indices and integer numerators, with common denominator ${10}^4$; exact row reduction determines every pivot coordinate. Substitution gives zero in all 438 equations, which proves (6.6).

It remains only to certify positivity. The displayed kernels are independent. On full-rank principal compressions the verifier performs exact $LD{L}^{*}$ elimination. The selected orders and the smallest pivots are shown below; the decimal column is only a readable display of an exact sign calculation.

matrix	order	principal indices (zero based)	smallest pivot
$S_0$	11	$2,12,5,9,11,13,7,1,3,0,4$	$0.127165\dots$
$R_0$	8	$2,1,6,8,7,3,0,4$	$0.146820\dots$
$S_P$	8	$2,1,4,6,9,7,8,11$	$0.139901\dots$
$R_P$	7	$1,2,5,6,4,0,7$	$0.166670\dots$

No floating-point sign decision is used. Every real pivot is reduced to a cubic polynomial in a. Rational endpoints first isolate the positive root of $16x^4-8x^2-1$ (the polynomial is strictly increasing for $x>1/2$); rational interval Horner evaluation then gives a strictly positive lower endpoint for every pivot. Thus the four matrices are positive semidefinite with exactly the ranks in the statement, completing the exact audit. If ${\Pi }_0$ is the orthogonal projection onto $ker{S}_0$, the same exact elimination on the order-11 compression also gives the quantitative margin

$$S_0\ge {\displaystyle \frac{1}{10}}(I-{\Pi }_0).$$

(6.8)

The smallest pivot in this additional audit is $0.069496\dots$; once again its strictly positive rational interval, rather than the decimal display, is what proves the assertion.    □

Proof of Theorem 3. 

First suppose $0\le t<1$. Sector containment gives $P,Q\ge 0$, so Lemma 6 makes the right side of (6.6) positive. Put $B=T^2$ and $w=(B-I){(B+I)}^{-1}$. Spectral inclusion puts $Sp\left(B\right)$ in the closed right half-plane, hence every scalar spectral value of w has modulus at most one. Since $t<1$, spectral mapping shows that $I+it{w}^2$ is invertible; equivalently, $D_t$ is invertible. Congruence of (6.6) by $D_t^{-1}$ gives

$$2I-{\left(N_t{D}_t^{-1}\right)}^{*}\left(N_t{D}_t^{-1}\right)\ge 0,$$

and proves (1.10) for $0\le t<1$. At $t=1$ one has $N_1=-i{D}_1$, so the reduced rational function is the constant $-i$.

For $-1\le t<0$, apply the result for $-t$ to $T^{*}$. Directly from (1.11),

$$h_c{\left(T^2\right)}^{*}=h_{\overline{c}}\left({\left(T^{*}\right)}^2\right),$$

and $W\left(T^{*}\right)\subset {\mathcal{S}}_{\pi /4}$. The endpoint $t=-1$ similarly reduces to the constant i. This proves the full diameter.    □

Proof of Theorem 4. 

It suffices by the adjoint symmetry in the preceding proof to take $c=x+it$ with $0\le t\le 1$ and $x\in \mathbb{R}$. Let

$$E=ker{\mathbf{d}}_1^{\mathsf{T}}\subset {\mathbb{C}}^5.$$

The kernel of $S_0$ in Lemma 6 is $\left\{\right(q,q,q):q\in E\}$. For $\xi \in {\mathbb{C}}^5$ put $y={(\xi ,t\xi ,t^2\xi )}^{\mathsf{T}}$. Orthogonally decomposing $\xi$ into $E\oplus E^{\perp }$ gives

$$\begin{array}{cc}\hfill {(y,ker{S}_0)}^2& \ge \left(1+t^2+t^4-{\displaystyle \frac{{(1+t+t^2)}^2}{3}}\right){\parallel \xi \parallel }^2\hfill \\ \hfill & ={\displaystyle \frac{2}{3}}{(1-t)}^2(1+t+t^2){\parallel \xi \parallel }^2\ge {\displaystyle \frac{2}{3}}{(1-t)}^2{\parallel \xi \parallel }^2.\hfill \end{array}$$

Combining this with (6.8) and $R_0\ge 0$ yields

$$G_0\left(t\right)\ge {\displaystyle \frac{1}{15}}{(1-t)}^{2}I.$$

(6.9)

Put

$$\mathbf{d}={(1,0,2,0,1)}^{\mathsf{T}},\mathbf{n}={(1,0,-2,0,1)}^{\mathsf{T}},$$

and put

$$D_c={(T^2+I)}^2+c{(T^2-I)}^2,N_c={(T^2-I)}^2+\overline{c}{(T^2+I)}^2.$$

Let $H\left(c\right)$ be the coefficient matrix of $2D_c^{*}{D}_c-N_c^{*}{N}_c$ in the basis $\mathbf{v}\left(T\right)$. Direct expansion gives

$$H(x+it)-H\left(it\right)=x(\overline{\mathbf{d}}{\mathbf{n}}^{\mathsf{T}}+\overline{\mathbf{n}}{\mathbf{d}}^{\mathsf{T}})+x^2(2\overline{\mathbf{n}}{\mathbf{n}}^{\mathsf{T}}-\overline{\mathbf{d}}{\mathbf{d}}^{\mathsf{T}}).$$

Because $\parallel \mathbf{d}\parallel =\parallel \mathbf{n}\parallel =\sqrt{6}$,

$$\parallel H(x+it)-H\left(it\right)\parallel \le 12\left|x\right|+18x^2.$$

(6.10)

If $\left|x\right|\le {(1-t)}^2/200$, then give

$$\begin{array}{cc}\hfill G_0\left(t\right)+H(x+it)-H\left(it\right)& \ge \left({\displaystyle \frac{1}{15}}-{\displaystyle \frac{12}{200}}-{\displaystyle \frac{18}{{200}^2}}\right){(1-t)}^{2}I\hfill \\ \hfill & ={\displaystyle \frac{373}{60000}}{(1-t)}^{2}I.\hfill \end{array}$$

For $t<1$ this is positive. Replace $G_0\left(t\right)$ by this perturbed ordinary Gram matrix in (6.6); the two localized Grams are unchanged. Condition (1.12) also gives $\left|c\right|<1$ away from the tips, so the same spectral-mapping argument makes the denominator in (1.13) invertible. Congruence proves the estimate. At $t=1$ condition (1.12) forces $x=0$, which is the already treated constant endpoint.    □

6.2. A Full Complex Parameter Disk

We prove Theorem 5 by one bivariate matrix Putinar certificate. Order the monomials of total degree at most k as

$${\mathbf{m}}_k(x,y)={(x^j{y}^{d-j}:0\le d\le k,0\le j\le d)}^{\mathsf{T}}$$

and put ${\mathcal{V}}_{k,m}(x,y)={\mathbf{m}}_k(x,y)\otimes I_m$.

Lemma 7

(Exact complex-disk certificate). There are positive definite rational Hermitian matrices

$$\begin{array}{cccc}\hfill S_0& \in M_{50}\left(\mathbb{Q}\left(i\right)\right),\hfill & \hfill R_0& \in M_{30}\left(\mathbb{Q}\left(i\right)\right),\hfill \\ \hfill S_P& \in M_{40}\left(\mathbb{Q}\left(i\right)\right),\hfill & \hfill R_P& \in M_{24}\left(\mathbb{Q}\left(i\right)\right)\hfill \end{array}$$

with the following property. If $c=x+iy$, $x,y\in \mathbb{R}$, set

$$\begin{array}{cc}\hfill G_0(x,y)& ={\mathcal{V}}_{3,5}{(x,y)}^{*}{S}_0{\mathcal{V}}_{3,5}(x,y)\hfill \\ \hfill & +\left({\displaystyle \frac{361}{400}}-x^2-y^2\right){\mathcal{V}}_{2,5}{(x,y)}^{*}{R}_0{\mathcal{V}}_{2,5}(x,y),\hfill \\ \hfill G_P(x,y)& ={\mathcal{V}}_{3,4}{(x,y)}^{*}{S}_P{\mathcal{V}}_{3,4}(x,y)\hfill \\ \hfill & +\left({\displaystyle \frac{361}{400}}-x^2-y^2\right){\mathcal{V}}_{2,4}{(x,y)}^{*}{R}_P{\mathcal{V}}_{2,4}(x,y),\hfill \\ \hfill G_Q(x,y)& =\overline{G_P(x,-y)}.\hfill \end{array}$$

For $T=X+iY$, $P=X+Y$, $Q=X-Y$, and

$$\begin{array}{cc}\hfill D_c& ={(T^2+I)}^2+c{(T^2-I)}^2,\hfill \\ \hfill N_c& ={(T^2-I)}^2+\overline{c}{(T^2+I)}^2,\hfill \end{array}$$

one has the exact hereditary identity

$$\begin{array}{cc}\hfill 2D_c^{*}{D}_c-N_c^{*}{N}_c=& \mathbf{v}{\left(T\right)}^{*}{G}_0(x,y)\mathbf{v}\left(T\right)+\mathbf{w}{\left(T\right)}^{*}{G}_P(x,y)P\mathbf{w}\left(T\right)\hfill \\ \hfill & +\mathbf{w}{\left(T\right)}^{*}{G}_Q(x,y)Q\mathbf{w}\left(T\right).\hfill \end{array}$$

(6.11)

Proof. 

Put

$$\mathbf{d}={(1,0,2,0,1)}^{\mathsf{T}},\mathbf{n}={(1,0,-2,0,1)}^{\mathsf{T}}.$$

The coefficient matrices of $2D_c^{*}{D}_c-N_c^{*}{N}_c$ at the parameter monomials $1,x,y,x^2,xy,y^2$ are, respectively,

$$\begin{array}{cc}\hfill & 2\overline{\mathbf{d}}{\mathbf{d}}^{\mathsf{T}}-\overline{\mathbf{n}}{\mathbf{n}}^{\mathsf{T}},\overline{\mathbf{d}}{\mathbf{n}}^{\mathsf{T}}+\overline{\mathbf{n}}{\mathbf{d}}^{\mathsf{T}},i(\overline{\mathbf{d}}{\mathbf{n}}^{\mathsf{T}}-\overline{\mathbf{n}}{\mathbf{d}}^{\mathsf{T}}),\hfill \\ \hfill & 2\overline{\mathbf{n}}{\mathbf{n}}^{\mathsf{T}}-\overline{\mathbf{d}}{\mathbf{d}}^{\mathsf{T}},0,2\overline{\mathbf{n}}{\mathbf{n}}^{\mathsf{T}}-\overline{\mathbf{d}}{\mathbf{d}}^{\mathsf{T}}.\hfill \end{array}$$

(6.12)

Expand the four matrix Putinar terms in the statement and equate their hereditary output to (6.12), with $G_Q(x,y)=\overline{G_P(x,-y)}$. The integer seed file stores the Hermitian coordinates of the four constant Gram matrices with common denominator ${10}^{12}$. Rounding alone need not preserve the affine equations, so the verifier removes its residual exactly. For a parameter monomial $x^j{y}^k$, it applies a rational right inverse of the fixed finite-dimensional map

$$(H_0,H_P)⟼Her\left(H_0,H_P,{(-1)}^k\overline{H_P}\right)$$

and lifts the resulting correction into $S_0$ and $S_P$. Every monomial of degree at most six is a product of two monomials of degree at most three, so this lift is exact. A fresh expansion gives precisely the six matrices in (6.12) and zero at the other 22 parameter monomials.

We also record the exact positivity audit. For each corrected matrix A, the verifier constructs rational matrices $L,M$ and puts

$$R=A-L{L}^{*},E=I-ML.$$

Set

$$ϵ=\sqrt{{\parallel E\parallel }_1{\parallel E\parallel }_{\infty }},{\mu }^2={\parallel M\parallel }_1{\parallel M\parallel }_{\infty },\ell ={\displaystyle \frac{{(1-ϵ)}^2}{{\mu }^2}},\rho =\sqrt{{\parallel R\parallel }_1{\parallel R\parallel }_{\infty }}.$$

The square roots are replaced in the computation by rational upper bounds. Since $ML=I-E$, the inequalities $ϵ<1$ and $\ell >\rho$ imply ${\lambda }_{min}\left(A\right)\ge \ell -\rho >0$. The exact rational comparisons, with decimal displays, are

matrix	$S_0$	$R_0$	$S_P$	$R_P$
order	50	30	40	24
ℓ	$9.27·{10}^{-8}$	$2.10·{10}^{-6}$	$8.12·{10}^{-8}$	$1.15·{10}^{-5}$
$\rho$	$1.63·{10}^{-11}$	$3.70·{10}^{-12}$	$8.29·{10}^{-12}$	$4.25·{10}^{-12}$

The decimals only display certified rational bounds. The independent verifier reads no floating-point discovery Gram: it reconstructs the exact affine point from the integer seed, checks all 28 hereditary coefficient matrices, and repeats the four positivity comparisons. This proves the lemma.    □

Proof of Theorem 5. 

Write $c=x+iy$. Under (1.14), each of the three matrices $G_0(x,y),G_P(x,y),G_Q(x,y)$ in Lemma 7 is positive semidefinite. Sector containment gives $P,Q\ge 0$, so the right side of (6.11) is positive.

Put $B=T^2$ and $w=(B-I){(B+I)}^{-1}$. Spectral inclusion puts $Sp\left(B\right)$ in the closed right half-plane, hence $Sp\left(w\right)\subset \overline{\mathbb{D}}$. Since $\left|c\right|\le 19/20<1$, the operator $I+c{w}^2$ is invertible; equivalently, $D_c$ is invertible. Congruence of (6.11) by $D_c^{-1}$ proves (1.15). At $c=i/2$, direct factorization of the numerator gives

$$(1-i/2)\left(z-(1+2i)\right)\left(z-{\displaystyle \frac{1-2i}{5}}\right),$$

which proves (1.16).    □

6.3. A Boundary-Reaching Phase Arc

We now prove Theorem 6. The proof has three finite exact parts: a rational seed certificate identifying the forced faces, a rational interpolation audit showing that 72 selected coefficient equations imply the full hereditary identity, and six matrix Bernstein certificates giving strict rational sections in two complementary row charts. Put $R=19/20$ and retain the vectors

$$\mathbf{d}={(1,0,2,0,1)}^{\mathsf{T}},\mathbf{n}={(1,0,-2,0,1)}^{\mathsf{T}}.$$

Lemma 8

(Exact radial seed and continuation). For $1/10\le t\le 18/25$ and $R\le \rho \le 1$, there are positive semidefinite Hermitian matrices $G_0(\rho ,t)\in M_5\left(\mathbb{C}\right)$ and $G_P(\rho ,t),G_Q(\rho ,t)\in M_4\left(\mathbb{C}\right)$ for which

$$\begin{array}{cc}\hfill 2D_{\rho ,t}^{*}{D}_{\rho ,t}-N_{\rho ,t}^{*}{N}_{\rho ,t}=& \mathbf{v}{\left(T\right)}^{*}{G}_0(\rho ,t)\mathbf{v}\left(T\right)\hfill \\ \hfill & +\mathbf{w}{\left(T\right)}^{*}{G}_P(\rho ,t)P\mathbf{w}\left(T\right)+\mathbf{w}{\left(T\right)}^{*}{G}_Q(\rho ,t)Q\mathbf{w}\left(T\right),\hfill \end{array}$$

(6.13)

where

$$\begin{array}{cc}\hfill D_{\rho ,t}& ={(T^2+I)}^2+\rho {\zeta }_t{(T^2-I)}^2,\hfill \\ \hfill N_{\rho ,t}& ={(T^2-I)}^2+\rho {\overline{\zeta }}_t{(T^2+I)}^2.\hfill \end{array}$$

The matrices may be chosen continuously in t.

Proof. 

For fixed t, the tangential Grams have the form

$$G_P(\rho ,t)=(1-{\rho }^2){\tilde{G}}_P(\rho ,t),G_Q(\rho ,t)=(1-{\rho }^2){\tilde{G}}_Q(\rho ,t),$$

(6.14)

where each tilde Gram is the affine interpolation of positive endpoint Grams at $\rho =R$ and $\rho =1$. For the ordinary Gram put

$${\mathcal{R}}_{\rho }=\left({\displaystyle \genfrac{}{}{0pt}{}{I_5}{\rho I_5}}\right)$$

and use the odd matrix Markov–Lukacs form

$$G_0(\rho ,t)=(\rho -R){\mathcal{R}}_{\rho }^{*}S\left(t\right){\mathcal{R}}_{\rho }+(1-\rho ){\mathcal{R}}_{\rho }^{*}U\left(t\right){\mathcal{R}}_{\rho }.$$

(6.15)

Thus positivity on $[R,1]$ follows once the two endpoint edge Grams and $S\left(t\right),U\left(t\right)$ are positive on their forced supports.

Those supports are rational in t. Define

$$k_P\left(t\right)={\left(-2t^3(1+i),-2i{t}^2,t(1-i),1\right)}^{\mathsf{T}},k_Q\left(t\right)=J_4{k}_P\left(t\right).$$

(6.16)

The endpoint edge supports are $k_P{\left(t\right)}^{\perp }$ and $k_Q{\left(t\right)}^{\perp }$. Writing $e_t=\mathbf{d}+{\zeta }_t\mathbf{n}$, a six-dimensional support for $S\left(t\right)$ is

$$\{(x,-x):x\in {\mathbb{C}}^5\}+span\left\{({\overline{e}}_t,0)\right\}.$$

(6.17)

For the other Markov–Lukacs multiplier set

$$q_t\left(z\right)=\left(z-t(1+i)\right)\left(z-{\displaystyle \frac{1-i}{2t}}\right)$$

(6.18)

and let ${\mathcal{Q}}_t\in M_{5,3}\left(\mathbb{C}\right)$ be the coefficient-convolution map $p\mapsto q_{t}p$ from $\mathbb{C}{\left[z\right]}_{\le 2}$ to $\mathbb{C}{\left[z\right]}_{\le 4}$. An eight-dimensional support for $U\left(t\right)$ is

$$\{(x,-x):x\in {\mathbb{C}}^5\}+\{({\mathcal{Q}}_{t}y,0):y\in {\mathbb{C}}^3\}.$$

(6.19)

At $t_0=1/2$ one has $s_{t_0}=3/2$ and ${\zeta }_{t_0}=(7+24i)/25$. The ancillary rational data give six positive definite reduced Grams on the supports above, of orders

$$4,3,4,3,6,8.$$

(6.20)

Their smallest exact $LD{L}^{*}$ pivots have the decimal displays

$$\begin{array}{c}.0388996584,.5674026743,.2919816307,\\ 4.469610333,.0196053190,.0223098226.\end{array}$$

(6.21)

Every number in (6.21) displays a positive rational pivot. The verifier re-expands all four coefficients in $\rho$ in (6.13) exactly. It also checks that the exact rank-two projector occurring in the U-face annihilates ${\mathcal{Q}}_{t_0}$. Since both spaces have complementary dimensions two and three, this proves (6.19) without a numerical eigenspace assertion.

It remains to justify that the strict certificate continues with t. Choose the elementary rational bases in (6.16)–(6.19), and record Hermitian matrices by real coordinates. Comparing the four coefficients in $\rho$ gives

$$A\left(t\right)x=b\left(t\right),A\left(t\right)\in M_{144,150}\left(\mathbb{Q}\left(t\right)\right).$$

(6.22)

A fixed set $\mathcal{R}$ of 72 rows and $\mathcal{C}$ of 72 columns, selected at $t_0$, gives a square minor $M\left(t\right)=A{\left(t\right)}_{\mathcal{R},\mathcal{C}}$ with $M\left(t_0\right)\ne 0$. The exact identities

$$A{\left(t\right)}_{:,\mathcal{C}}M{\left(t\right)}^{-1}A{\left(t\right)}_{\mathcal{R},:}=A\left(t\right),A{\left(t\right)}_{:,\mathcal{C}}M{\left(t\right)}^{-1}b{\left(t\right)}_{\mathcal{R}}=b\left(t\right)$$

(6.23)

hold in $\mathbb{Q}\left(t\right)$.

For completeness, we describe the finite audit of (6.23). Clear denominators independently in each column. Among the 72 selected columns, the resulting degree multiset is

$$0^{\left(34\right)},2^{\left(6\right)},3^{\left(6\right)},4^{\left(10\right)},5^{\left(4\right)},6^{\left(3\right)},8^{\left(9\right)}.$$

(6.24)

An arbitrary added map or target column has degree at most eight. Hence every relevant bordered determinant has degree at most

$$34·0+6·2+6·3+10·4+4·5+3·6+9·8+8=188.$$

At each of the 189 distinct rational points $t=1,\dots ,189$, exact rational inversion of $M\left(t\right)$ gives both identities in (6.23) entry by entry. The bordered determinants therefore have 189 roots and vanish identically. The ancillary continuation audit performs the denominator, degree, inversion, and residual checks from scratch; no floating-point rank decision enters this interpolation step.

We next give the interval certificates. Let

$$\begin{array}{c}{I}_1=[1/10,1/5],I_2=[1/5,3/10],I_3=[3/10,2/5],\\ I_4=[2/5,1/2],I_5=[1/2,3/5],I_6=[3/5,18/25].\end{array}$$

(6.25)

and write $u_j=(t-inf{I}_j)/(sup{I}_j-inf{I}_j)$. The first five intervals use the row set ${\mathcal{R}}_0=\mathcal{R}$ above. A second fixed set ${\mathcal{R}}_1$ of 72 rows, selected at $t=7/10$, is used on $I_6$; all indices are recorded in the verifier. Put

$$d=(6,6,6,6,6,8),{\beta }_{j,k}\left(u\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{d_j}{k}}\right)u^k{(1-u)}^{d_j-k}.$$

For each j, the ancillary integer data specify a Bernstein section

$${\tilde{x}}_j\left(u\right)={\displaystyle \sum _{k=0}^{d_j}}{\beta }_{j,k}\left(u\right)X_{j,k}(X_{j,k}\in {\mathbb{Q}}^{150}).$$

(6.26)

As before, every control vector encodes six reduced Hermitian blocks in the orders (6.20). Fraction-free $LD{L}^{*}$ proves that every control block exceeds $m_{j}I$. The same data give a polynomial matrix $B_j\left(u\right)\in M_{150,72}\left(\mathbb{Q}\left[u\right]\right)$. With

$$A_j\left(t\right)=A{\left(t\right)}_{{\mathcal{R}}_0,:}(j\le 5),A_6\left(t\right)=A{\left(t\right)}_{{\mathcal{R}}_1,:},$$

and the analogous notation $b_j$, exact Bernstein arithmetic proves

$$\begin{array}{c}{\parallel b_j-A_j{\tilde{x}}_j\parallel }_{\infty }<{ϵ}_j,{\parallel B_j\parallel }_{\infty }<L_j,\\ E_j:=I_{72}-A_j{B}_j,{\parallel E_j\parallel }_{\infty }<{\eta }_j,\end{array}$$

(6.27)

where the certified bounds are

$$\begin{array}{cccccc}j& m_j& {ϵ}_j& L_j& {\eta }_j& m_j-16L_j{ϵ}_j/(1-{\eta }_j)\\ 1& 1/20& 1/300000& 403& 2/25& 919/34500\\ 2& 1/30& 1/500000& 201& 1/190& 10579/393750\\ 3& 1/40& 1/500000& 209& 1/1000& 18287/999000\\ 4& 1/51& 1/1000000& 247& 3/1000& 99431/6355875\\ 5& 1/60& 1/2000000& 302& 1/90& 47471/3337500\\ 6& 1/70& 1/1000000& 198& 1/500& 9703/873250\end{array}$$

Every entry in the last column is greater than $1/100$.

Here are the exact details behind these finite inequalities. The common positive denominator

$$\Delta \left(t\right)=20t^2{(2t^2-2t+1)}^2{(2t^2+2t+1)}^2$$

(6.28)

clears every selected entry of A and b. After substituting t as an affine function of $u_j$, its Bernstein controls are strictly positive. The entries of $\Delta (b_j-A_j{\tilde{x}}_j)$ and $\Delta (I-A_j{B}_j)$ are polynomials of explicitly bounded degree. The verifier expands every scalar entry in the Bernstein basis over $\mathbb{Q}$, takes the maximum absolute row sum of the controls, and divides by the smallest control of $\Delta$. This gives (6.27) without interval sampling. It also applies exact $LD{L}^{*}$ to all $5·7·6+9·6=264$ rational Gram controls. The smallest pivot after subtracting the stated $m_{j}I$ has decimal display

$$0.0002415456052637.$$

The stored candidates and right-inverse controls have common denominator ${10}^{10}$; only these integers, not the discovery SDP, are read by the verifier.

Since ${\eta }_j<1$, the Neumann lemma makes $I-E_j$ invertible and

$$B_j{(I-E_j)}^{-1}$$

is an exact right inverse of $A_j$. If $r_j=b_j-A_j{\tilde{x}}_j$, define

$$x_j={\tilde{x}}_j+B_j{(I-E_j)}^{-1}{r}_j.$$

(6.29)

Then $A_j{x}_j=b_j$ and

$${\parallel x_j-{\tilde{x}}_j\parallel }_{\infty }<{\displaystyle \frac{L_j{ϵ}_j}{1-{\eta }_j}}.$$

(6.30)

The rational identities (6.23) imply that the augmented matrix $\left[A\right(t\left)b\right(t\left)\right]$ has rank at most 72 for every $t>0$, by vanishing of its 73-minors. On $I_6$, (6.27) shows that the second row chart has rank 72 as well. Thus solving either selected row system solves all 144 coefficient equations, including at points where the base minor is singular.

Finally, if every real coordinate of a Hermitian block of order at most eight changes by at most q, its maximum absolute row sum, and hence its operator norm, changes by less than $16q$. Equations (6.30) and the last column of the table therefore show that every corrected reduced block is larger than $I/100$. Equations (6.14) and (6.15) finish the proof.    □

Proof of Theorem 6. 

For $0\le \rho \le R$, the assertion is Theorem 5. For $R\le \rho <1$, sector containment gives $P,Q\ge 0$, so Lemma 8 implies $2D_{\rho ,t}^{*}{D}_{\rho ,t}-N_{\rho ,t}^{*}{N}_{\rho ,t}\ge 0$. Moreover $I+\rho {\zeta }_t{w}^2$ is invertible. Congruence by $D_{\rho ,t}^{-1}$ proves (1.19). At $\rho =1$ the scalar rational function cancels to the constant ${\overline{\zeta }}_t$, so the endpoint follows directly.

Finally $W\left(T^{*}\right)=\overline{W\left(T\right)}\subset {\mathcal{S}}_{\pi /4}$. Applying the result to $T^{*}$ and taking adjoints, using commutativity of the two polynomial factors, replaces c by $\overline{c}$. Since $s_{1/2}^{\prime }=-10$, the phases in (1.17) trace a nondegenerate arc. More precisely, $s_t=1/\left(2t^2\right)-2t^2$ is strictly decreasing, and

$${\zeta }_t={\displaystyle \frac{2+i{s}_t}{2-i{s}_t}}.$$

Thus $0<t\le 1/\sqrt{2}$ traces the upper semicircle from $-1$ to 1. At $t=1/10$ the omitted upper angle is $2arctan(100/2499)$; reflection doubles it and gives the stated total omitted length. This proves all assertions.    □

Proof of Proposition 1. 

Put

$$R=(T-I){(T+I)}^{-1},u={\displaystyle \frac{R+iI}{\sqrt{2}}},v={\displaystyle \frac{R-iI}{\sqrt{2}}}.$$

Congruence by $T+I$ and the two sector inequalities $\Re \left(e^{\pm i\pi /4}T\right)\ge 0$ give

$$I-R^{*}R+i(R-R^{*})\ge 0,I-R^{*}R-i(R-R^{*})\ge 0.$$

(6.31)

Consequently u and v are commuting contractions and $u-v=i\sqrt{2}I$. This affine relation also makes both operators invertible. Indeed, $\parallel ux\parallel \ge (\sqrt{2}-1)\parallel x\parallel$ follows from $u=v+i\sqrt{2}I$, and the same estimate for $u^{*}$ makes the range of u dense; the argument for v is identical.

Set $a=\sqrt{2}$ and define

$$C={(u-iaI)}^{-1}-iaI,D={(v+iaI)}^{-1}+iaI.$$

(6.32)

The scalar linear-fractional maps in (6.32) map $\mathbb{D}$ into itself. For example,

$$\left|{\displaystyle \frac{1}{z-ia}}-ia\right|\le 1\left(\right|z|\le 1)$$

follows after squaring from ${|z-ia|}^2{-|1+iaz|}^2=1-{\left|z\right|}^2$. Von Neumann’s inequality therefore shows that C and D are commuting contractions. Since $u-iaI=v$ and $v+iaI=u$, elementary algebra gives, with $P=CD$,

$${\displaystyle \frac{C+D}{\sqrt{2}}}=w,C-D=-i\sqrt{2}P,w^2=P(2I-P).$$

(6.33)

For the first identity we used $uv=(I+R^2)/2$ and $w=2R{(I+R^2)}^{-1}$.

We now give the promised bidisk extension. For scalar variables $\zeta ,\eta$ put

$$\begin{array}{cc}\hfill q(\zeta ,\eta )& =\zeta -\eta +i\sqrt{2}\zeta \eta ,\hfill \\ \hfill F(\zeta ,\eta )& ={\displaystyle \frac{{(\zeta +\eta )}^2}{2}}+q(\zeta ,\eta )\left({\displaystyle \frac{51i}{100}}+{\displaystyle \frac{41(\eta -\zeta )}{400}}\right).\hfill \end{array}$$

By (6.33), $q(C,D)=0$ and hence $F(C,D)=w^2$. We claim that

$${\parallel F\parallel }_{H^{\infty }\left({\mathbb{D}}^2\right)}\le {\kappa }_0.$$

(6.34)

Here is an exact elementary maximization. On the distinguished boundary write

$$\zeta =\chi e^{i\delta },\eta =\chi e^{-i\delta },\left|\chi \right|=1,$$

and put $\tau =sin\delta$, $x=\Re \chi$,

$$m={\displaystyle \frac{51}{50}},n={\displaystyle \frac{41\sqrt{2}}{200}},a\left(\tau \right)=2-{\displaystyle \frac{51\sqrt{2}}{100}}-{\displaystyle \frac{159}{100}}{\tau }^2.$$

After removing a unit-modulus factor, direct expansion gives

$$\left|F(\zeta ,\eta )\right|=|-m\tau +a\left(\tau \right)\chi +n\tau {\chi }^2|,$$

(6.35)

and hence

$${\left|F(\zeta ,\eta )\right|}^2=a^2+2a\tau (n-m)x+{\tau }^2\left({(m+n)}^2-4mn{x}^2\right).$$

(6.36)

For fixed $\tau$, the right side is a concave quadratic in $x\in [-1,1]$.

Set $r=\left|\tau \right|$. At $x=\pm 1$, the larger modulus is

$$\left|a\right(r\left)\right|+r(m-n).$$

(6.37)

While $a\left(r\right)\ge 0$, this concave quadratic in r attains its maximum at

$$r_{*}={\displaystyle \frac{m-n}{2(159/100)}}={\displaystyle \frac{17}{53}}-{\displaystyle \frac{41\sqrt{2}}{636}},$$

and that maximum is

$$2-{\displaystyle \frac{51\sqrt{2}}{100}}+{\displaystyle \frac{{(m-n)}^2}{4(159/100)}}={\displaystyle \frac{276889}{127200}}-{\displaystyle \frac{6103\sqrt{2}}{10600}}={\kappa }_0.$$

While $a\left(r\right)\le 0$, (6.37) is increasing and its endpoint value $159/100-(2-51\sqrt{2}/100)+(m-n)$ is strictly smaller than ${\kappa }_0$, since their difference is

$${\displaystyle \frac{199297-112032\sqrt{2}}{127200}}>0.$$

It remains to check the interior vertex of (6.36). Such a vertex belongs to $[-1,1]$ only if

$$\left|a\right(r\left)\right|(m-n)\le 4mnr.$$

(6.38)

For $r<1/2$ this is impossible, since

$$a(1/2)(m-n)-2mn={\displaystyle \frac{147492-101353\sqrt{2}}{80000}}>0.$$

For $1/2\le r\le 1$, the vertex value is

$$V\left(r^2\right)={(m+n)}^2\left(r^2+{\displaystyle \frac{a{\left(r\right)}^2}{4mn}}\right).$$

As a function of $r^2$, this is convex. Exact endpoint evaluation gives

$$\begin{array}{cc}\hfill {\kappa }_0^2-V(1/4)& ={\displaystyle \frac{19(13707814025092-9187506155169\sqrt{2})}{45109393920000}}>0,\hfill \\ \hfill {\kappa }_0^2-V\left(1\right)& ={\displaystyle \frac{50692488845401-35843685592140\sqrt{2}}{11277348480000}}>0.\hfill \end{array}$$

Thus every interior vertex is smaller than ${\kappa }_0^2$. Equality in (6.34) occurs in the edge calculation at $r=r_{*}$ and the appropriate choice $x=-1$. This proves the claimed exact norm. The script tmp/research/rightanglesquareextensionexact.py replays the expansion and all comparisons over $\mathbb{Q}\left(\sqrt{2}\right)$. In particular,

$$\sqrt{2}-{\kappa }_0={\displaystyle \frac{200436\sqrt{2}-276889}{127200}}>0.$$

Finally, Ando’s dilation theorem for a commuting pair of contractions [6] and (6.34) give

$$\parallel w^2\parallel =\parallel F(C,D)\parallel \le {\parallel F\parallel }_{H^{\infty }\left({\mathbb{D}}^2\right)}\le {\kappa }_0,$$

as required.    □

Proof of Theorem 7. 

Keep the commuting contractions $C,D$ and the polynomial F from the proof of Proposition 1, and write

$$W(\zeta ,\eta )={\displaystyle \frac{\zeta +\eta }{\sqrt{2}}}.$$

For $\alpha ,\beta \in \mathbb{C}$ define the rational bidisk function

$$H_{\alpha ,\beta }(\zeta ,\eta )={\displaystyle \frac{F-(\alpha +\beta )W+\alpha \beta }{1-(\overline{\alpha }+\overline{\beta })W+\overline{\alpha }\overline{\beta }F}}.$$

(6.39)

On the curve $q=0$ one has $F=W^2$, so the numerator and denominator in (6.39) factor respectively as

$$(W-\alpha )(W-\beta ),(1-\overline{\alpha }W)(1-\overline{\beta }W).$$

Since $q(C,D)=0$ and $W(C,D)=w$, it follows that

$$H_{\alpha ,\beta }(C,D)=b_{\alpha }\left(w\right)b_{\beta }\left(w\right).$$

(6.40)

Put $\delta =1/150$, $r=\left|\alpha \right|$, and $s=\left|\beta \right|$. On the closed bidisk, $\left|W\right|\le \sqrt{2}$ and $\left|F\right|\le {\kappa }_0$. Hence the denominator $D_0$ and numerator $N_0$ in (6.39) satisfy

$$\begin{array}{cc}\hfill |D_0|& \ge 1-(r+s)\sqrt{2}-rs{\kappa }_0\ge 1-2\delta \sqrt{2}-{\delta }^2{\kappa }_0,\hfill \end{array}$$

(6.41)

$$\begin{array}{cc}\hfill |N_0|& \le {\kappa }_0+(r+s)\sqrt{2}+rs\le {\kappa }_0+2\delta \sqrt{2}+{\delta }^2.\hfill \end{array}$$

(6.42)

The first lower bound is strictly positive. Indeed, $0<{\kappa }_0<\sqrt{2}<3/2$ gives

$$1-2\delta \sqrt{2}-{\delta }^2{\kappa }_0>1-{\displaystyle \frac{1}{50}}-{\displaystyle \frac{1}{15000}}={\displaystyle \frac{14699}{15000}}.$$

Thus (6.39) is analytic on a neighborhood of the closed bidisk.

The exact reserve is also sufficient. From the displayed value of ${\kappa }_0$,

$$\sqrt{2}-{\kappa }_0={\displaystyle \frac{200436\sqrt{2}-276889}{127200}}>{\displaystyle \frac{1}{20}};$$

after clearing denominators, the last strict inequality follows from

$$2{\left(200436\right)}^2-{\left(283249\right)}^2=119184191>0.$$

Consequently

$$\begin{array}{cc}\hfill & \sqrt{2}(1-2\delta \sqrt{2}-{\delta }^2{\kappa }_0)-({\kappa }_0+2\delta \sqrt{2}+{\delta }^2)\hfill \\ \hfill & =(\sqrt{2}-{\kappa }_0)-(4+2\sqrt{2})\delta -(1+\sqrt{2}{\kappa }_0){\delta }^2\hfill \\ \hfill & >{\displaystyle \frac{1}{20}}-{\displaystyle \frac{7}{150}}-{\displaystyle \frac{1}{7500}}={\displaystyle \frac{2}{625}}>0.\hfill \end{array}$$

Equations (6.41)–(6.42) therefore give the strict scalar norm bound displayed in (1.21). Finally, Ando’s theorem applied to the commuting contractions $C,D$, together with (6.40), gives the same bound for the operator rational function: indeed, (6.39) is analytic on a neighborhood of the closed bidisk, so its Taylor polynomials converge there uniformly and Ando’s polynomial inequality passes to the limit. The endpoint choices $\alpha =1/150$, $\beta =i/150$ are allowed because every estimate above holds on the closed parameter bidisk.    □

6.4. The Exact Remaining Criterion for the Centred Slice

The preceding certificates leave only a thin boundary annulus in the c-parameter. The following reduction identifies its precise operator-theoretic content; in particular, there is no additional formal branch in the two-moment region.

Proposition 4

(Two-moment criterion). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfy $W\left(T\right)\subset {\mathcal{S}}_{\pi /4}$, put

$$w=(T^2-I){(T^2+I)}^{-1},\mathsf{X}=w^2,$$

and, for a unit vector ξ, put

$$b={\parallel \mathsf{X}\xi \parallel }^2,q=\langle \mathsf{X}\xi ,\xi \rangle .$$

Then

$$\parallel h_c\left(w\right)\parallel \le \sqrt{2}(c\in \mathbb{D})$$

(6.42)

holds if and only if

$${\left|q\right|}^2\le (2-b)(2b-1)\mathrm{whenever}b>1.$$

(6.44)

Equivalently, the nonautomatic part of (6.44) is

$${\parallel \mathsf{X}\xi -q\xi \parallel }^2\ge 2{\left({\parallel \mathsf{X}\xi \parallel }^2-1\right)}^2,\parallel \mathsf{X}\xi \parallel >1.$$

(6.45)

Proof. 

By Proposition 1, $\parallel \mathsf{X}\parallel \le {\kappa }_0<\sqrt{2}$, and hence $0\le b\le {\kappa }_0^2<2$. For $\left|c\right|<1$, spectral mapping makes $I+c\mathsf{X}$ invertible. Congruence by this operator shows that (6.43) is equivalent to positivity, for every $c\in \mathbb{D}$, of

$$H\left(c\right)={(2-|c|}^2{)I+(2\left|c\right|}^2-1){\mathsf{X}}^{*}\mathsf{X}+c\mathsf{X}+\overline{c}{\mathsf{X}}^{*}.$$

(6.46)

Indeed, (6.46) is the expansion of

$$2{(I+c\mathsf{X})}^{*}(I+c\mathsf{X})-{(\mathsf{X}+\overline{c}I)}^{*}(\mathsf{X}+\overline{c}I).$$

Writing $c=r{e}^{i\theta }$ and minimizing first over $\theta$ gives

$$\underset{\theta }{min}\langle H\left(r{e}^{i\theta }\right)\xi ,\xi \rangle =2-b+(2b-1)r^2-2r\left|q\right|.$$

(6.47)

Both endpoint values in $0\le r\le 1$ are nonnegative: the value at zero is $2-b$, while the value at one is $b+1-2\left|q\right|\ge {(\sqrt{b}-1)}^2$.

If $b\le 1/2$, the quadratic in (6.47) is concave, so its endpoint values suffice. Suppose $1/2<b\le 1$. If its critical point $r_{*}=\left|q\right|/(2b-1)$ is outside $(0,1)$, the endpoints again suffice. If $r_{*}\in (0,1)$, then

$${\left|q\right|}^2\le {(2b-1)}^2\le (2-b)(2b-1),$$

so the critical value is nonnegative as well. Thus the range $b\le 1$ is automatic.

Finally suppose $b>1$. Cauchy–Schwarz and

$${(2b-1)}^2-b=(4b-1)(b-1)>0$$

give $\left|q\right|\le \sqrt{b}<2b-1$. Hence the critical point belongs to $[0,1)$, and its value is nonnegative exactly when

$$2-b-{\displaystyle \frac{{\left|q\right|}^2}{2b-1}}\ge 0,$$

which is (6.44). This proves the equivalence. The identity

$$b-(2-b)(2b-1)=2{(b-1)}^2$$

and orthogonal projection onto $\mathbb{C}\xi$ give (6.45).    □

7. The Sharp Two-Node Admissible-Kernel Theorem

We record explicitly the two-node result used below. For $a,b\in \mathbb{D}$ write

$$\delta (a,b)={\displaystyle \frac{|a-b|}{|1-\overline{a}b|}}$$

for the pseudohyperbolic distance, and put

$$\mathcal{F}\left(t\right)={\displaystyle \frac{2\sqrt{2}t}{3+\sqrt{1-t^2}}},0\le t<1.$$

(7.1)

Lemma 9

(Radial scaling and right-angle geometry). For $a,b\in \mathbb{D}$,

$$\delta (a/\sqrt{2},b/\sqrt{2})\le \mathcal{F}\left(\delta (a,b)\right).$$

(7.2)

If $(c_j,d_j,w_j)\in {\mathbb{D}}^3$, $j=1,2$, satisfy

$$c_j-d_j+i\sqrt{2}{c}_j{d}_j=0,w_j={\displaystyle \frac{c_j+d_j}{\sqrt{2}}},$$

(7.3)

then

$$max\{\delta (c_1,c_2),\delta (d_1,d_2)\}\ge \mathcal{F}\left(\delta (w_1,w_2)\right).$$

(7.4)

Proof. 

We first prove (7.2). More generally, fix $0<r<1$. After a common rotation write $a=s\in [0,1)$ and

$$b={\displaystyle \frac{s+\xi }{1+s\xi }},\left|\xi \right|\le t.$$

Direct simplification gives

$$\delta (ra,rb)={\displaystyle \frac{r\left|\xi \right|(1-s^2)}{|1-r^2{s}^2+s(1-r^2)\xi |}}.$$

For fixed s its maximum occurs at $\xi =-t$. Differentiating in s gives $t{s}^2-2s+t=0$, hence $s=(1-\sqrt{1-t^2})/t$ when $t>0$. The resulting maximum is $2rs/(1+r^2{s}^2)$. Taking $r=1/\sqrt{2}$ gives (7.2); the case $t=0$ follows by continuity.

For (7.4), undo the disk and half-plane Cayley maps in (7.3) and write the two sector points as ${\beta }_1=r{e}^{i\theta }$ and ${\beta }_2=s{e}^{i\varphi }$, with $\left|\theta \right|,\left|\varphi \right|\le \pi /4$. Set

$$U={\displaystyle \frac{r^2+s^2}{2rs}},X=cos(\varphi -\theta ),Y=|sin(\varphi +\theta )|.$$

Then $U\ge 1$ and $0\le Y\le X\le 1$. Cayley invariance and direct calculation give, with $m=max\{\delta (c_1,c_2),\delta (d_1,d_2)\}$ and $t=\delta (w_1,w_2)$,

$$m^2={\displaystyle \frac{U-X}{U-Y}},t^2={\displaystyle \frac{U^2-X^2}{U^2-Y^2}}.$$

(7.5)

If $\sigma =\sqrt{1-t^2}$, these identities imply

$${\displaystyle \frac{m^2}{t^2}}={\displaystyle \frac{U+Y}{U+X}},X^2={\sigma }^2{U}^2+(1-{\sigma }^2)Y^2.$$

Since

$${[\sigma U+(1+\sigma )Y]}^2-X^2=2\sigma (1+\sigma )Y(U+Y)\ge 0,$$

we have $X\le \sigma U+(1+\sigma )Y$. Therefore

$${\displaystyle \frac{m^2}{t^2}}\ge {\displaystyle \frac{1}{1+\sigma }}\ge {\displaystyle \frac{8}{{(3+\sigma )}^2}},$$

the last inequality being ${(1-\sigma )}^2\ge 0$. This is precisely (7.4).    □

Theorem 8

(Two-node right-angle completion). Let two triples satisfy (7.3), let h be Schur on $\mathbb{D}$, and put ${\lambda }_j=h\left(w_j\right)$. If a Hermitian $2\times 2$ matrix L satisfies

$${(1-{\overline{c}}_i{c}_j)}_{i,j=1}^2\circ L⪰0,{(1-{\overline{d}}_i{d}_j)}_{i,j=1}^2\circ L⪰0,$$

(7.6)

then

$${(2-{\overline{\lambda }}_i{\lambda }_j)}_{i,j=1}^2\circ L⪰0.$$

(7.7)

The same conclusion holds after independent disk automorphisms in the two test coordinates, because their Pick kernels change only by invertible diagonal congruences.

Proof. 

The assertion is immediate if a diagonal entry of L vanishes, so assume $L_{11}{L}_{22}>0$. Positivity in (7.6) gives

$${\displaystyle \frac{|L_{12}{|}^2}{L_{11}{L}_{22}}}\le 1-max\{\delta {(c_1,c_2)}^2,\delta {(d_1,d_2)}^2\}.$$

(7.8)

Schwarz–Pick, 9, and monotonicity of $\mathcal{F}$ give

$$\begin{array}{cc}\hfill \delta ({\lambda }_1/\sqrt{2},{\lambda }_2/\sqrt{2})& \le \mathcal{F}\left(\delta ({\lambda }_1,{\lambda }_2)\right)\le \mathcal{F}\left(\delta (w_1,w_2)\right)\hfill \\ \hfill & \le max\{\delta (c_1,c_2),\delta (d_1,d_2)\}.\hfill \end{array}$$

Here monotonicity is immediate by differentiating (7.1); its derivative has the positive numerator $3+\sqrt{1-t^2}+t^2/\sqrt{1-t^2}$ on $0\le t<1$. Using

$${\displaystyle \frac{|2-{\overline{\lambda }}_1{\lambda }_2{|}^2}{(2-|{\lambda }_1{|}^2\left)\right(2-|{\lambda }_2{|}^2)}}={\displaystyle \frac{1}{1-\delta {({\lambda }_1/\sqrt{2},{\lambda }_2/\sqrt{2})}^2}},$$

inequality (7.8) is exactly the nonnegativity of the determinant in (7.7). Its diagonal entries are positive, so the target is positive semidefinite.    □

8. A Complete Symmetric Three-Node Theorem

We next record a finite but genuinely three-node instance of the sharp right-angle multiplier problem. Besides giving a new exact family, the proof identifies the first algebraic obstruction beyond the two-node theorem: a single phase on a rank-$(2,2)$ extreme face.

For $0<a<1$, put

$$\rho =(-a,0,a),z_i={\displaystyle \frac{2{\rho }_i}{1+{\rho }_i^2}},u_i={\displaystyle \frac{{\rho }_i+i}{\sqrt{2}}},v_i={\displaystyle \frac{{\rho }_i-i}{\sqrt{2}}}.$$

(8.1)

For a Hermitian $3\times 3$ matrix L, all products below are entrywise.

Theorem 9

(Symmetric three-node completion). If

$$P={(1-{\overline{u}}_i{u}_j)}_{ij}\circ L⪰0,Q={(1-{\overline{v}}_i{v}_j)}_{ij}\circ L⪰0,$$

(8.2)

then

$${(2-{\overline{z}}_i{z}_j)}_{ij}\circ L⪰0.$$

(8.3)

Thus the identity disk multiplier has norm at most $\sqrt{2}$ on the complete right-angle admissible-kernel cone over the nodes (8.1).

We first give the two finite-dimensional ingredients. For a Hermitian correlation matrix C, set

$$\Delta \left(C\right)=1-|c_{12}{|}^2-|c_{23}{|}^2-{\left|c_{31}\right|}^2+2\Re \left(c_{12}{c}_{23}{c}_{31}\right)=detC.$$

(8.4)

Lemma 10

(Extreme-ray rank bound). Let $\mathcal{R}$ be an invertible real-linear Schur multiplier on $M_n{\left(\mathbb{C}\right)}_{\mathrm{sa}}$, and let

$$\mathcal{C}=\{P=P^{*}:P⪰0,\mathcal{R}\left(P\right)⪰0\}.$$

If P generates an extreme ray of $\mathcal{C}$, then

$$rank{\left(P\right)}^2+rank{\left(\mathcal{R}\left(P\right)\right)}^2\le n^2+1.$$

(8.5)

Proof. 

Let $E_P,E_Q$ be the range projections of P and $Q=\mathcal{R}\left(P\right)$. The real spans of their minimal positive-semidefinite faces are

$${\mathcal{S}}_P=\{H=H^{*}:H=E_{P}H{E}_P\},{\mathcal{S}}_Q=\{K=K^{*}:K=E_{Q}K{E}_Q\},$$

of dimensions $rank{\left(P\right)}^2$ and $rank{\left(Q\right)}^2$. Hence

$$dim\left({\mathcal{S}}_P\cap {\mathcal{R}}^{-1}\left({\mathcal{S}}_Q\right)\right)\ge rank{\left(P\right)}^2+rank{\left(Q\right)}^2-n^2.$$

If this intersection contained a Hermitian H independent of P, then $P\pm \epsilon H$ and $Q\pm \epsilon \mathcal{R}\left(H\right)$ would be positive for sufficiently small $\epsilon >0$: on their ranges $P,Q$ are positive definite, and the perturbations vanish on their kernels. This would split the ray of P. The intersection is therefore one-dimensional, which proves (8.5).    □

Lemma 11

(Exact phase elimination). Let p be a rank-at-most-two correlation matrix. After diagonal unitary conjugation, write

$$p_{12}=x,p_{23}=y,p_{31}=xy+\sqrt{(1-x^2)(1-y^2)}\zeta ,\left|\zeta \right|=1,$$

(8.6)

where $0\le x,y\le 1$. Put $X=x^2$, $Y=y^2$ and $C_0=2xy\sqrt{(1-X)(1-Y)}$. For a normalized Hermitian multiplier m, define

$$\begin{array}{cccc}\hfill M_m& =m_{12}{m}_{23}{m}_{31},\hfill & \hfill d_m& =M_m-{\left|m_{31}\right|}^2,\hfill \\ \hfill A_m& =1-|m_{12}{|}^{2}X-{\left|m_{23}\right|}^{2}Y\hfill \end{array}$$

(8.7)

$$\begin{array}{cc}\hfill & -|m_{31}{|}^2\{XY+(1-X)(1-Y)\}+2XY\Re M_m.\hfill \end{array}$$

(8.8)

Then

$$\Delta (m\circ p)=A_m+C_0\Re \left(d_m\zeta \right).$$

(8.9)

Suppose $C_0>0$, $d_r\ne 0$, and $\Delta (r\circ p)=0$. With

$$\begin{array}{cccc}\hfill \gamma & ={\displaystyle \frac{d_t}{d_r}},\hfill & \hfill F& =A_t-A_r\Re \gamma ,\hfill \\ \hfill H& =C_0^2{\left|d_r\right|}^2-A_r^2,\hfill & \hfill S& =F^2-H{(\Im \gamma )}^2.\hfill \end{array}$$

(8.10)

both admissible phases satisfy $\Delta (t\circ p)\ge 0$ if and only if

$$F\ge 0,S\ge 0.$$

(8.11)

Proof. 

Expanding (8.4) after (8.6) gives (8.9). Put $\eta =d_r\zeta$. The equation $\Delta (r\circ p)=0$ fixes

$$\Re \eta =-A_r/C_0,\left|\eta \right|=|d_r|,$$

and hence leaves two conjugate choices for its imaginary part. Since $d_t\zeta =\gamma \eta$, their smaller target determinant is exactly

$$F-\sqrt{H}|\Im \gamma |.$$

(8.12)

Feasibility gives $H\ge 0$. The expression in (8.12) is nonnegative precisely when $F\ge 0$ and its square is nonnegative, which is (97). The sign condition on F prevents an extraneous conclusion from squaring.    □

Proof of Theorem 9. 

All entries of $k_u=(1-{\overline{u}}_i{u}_j)$ are nonzero. The change of variable $P=k_u\circ L$ identifies (8.2) with

$$\mathcal{C}=\{P⪰0:\mathcal{R}\left(P\right)=r\circ P⪰0\},r_{ij}={\displaystyle \frac{1-{\overline{v}}_i{v}_j}{1-{\overline{u}}_i{u}_j}}.$$

Intersecting this cone with $trP+tr\mathcal{R}\left(P\right)=1$ gives a compact base. For a fixed vector, the compression of the target in (8.3) is linear in P. Thus a negative minimum, if one existed, would occur on an extreme ray. By Lemma 10, its rank pattern is one of

$$(1,1),(1,2),(2,1),(2,2),(1,3),(3,1).$$

(8.13)

We first treat the rank-$(2,2)$ pattern. Positive diagonal congruence normalizes $P,Q$ and the target to correlation matrices $p,r\circ p,t\circ p$. Their one- and two-index target minors are positive by 8. It remains only to prove $\Delta (t\circ p)>0$.

If a diagonal entry of P or Q vanishes, positivity makes the corresponding row and column vanish; because the localization factors are nonzero, the same is true of L and the target. This reduces directly to Theorem 8. We may therefore assume all diagonal entries are positive when making the correlation normalization above.

Put $q=a^2$. Exact substitution of (8.1) into Lemma 11 gives

$$\begin{array}{cc}\hfill |t_{12}{|}^2={\left|t_{23}\right|}^2& ={\displaystyle \frac{1-q^2}{1+q^2}},\hfill \end{array}$$

(8.14)

$$\begin{array}{cc}\hfill |t_{31}{|}^2& ={\displaystyle \frac{{(1-q)}^2{(q^2+4q+1)}^2}{{(q^2+1)}^2(q^2+6q+1)}},\hfill \end{array}$$

(8.15)

$$\begin{array}{cc}\hfill \Re d_r& =-{\displaystyle \frac{32q^3}{{(q+1)}^2(q^2+6q+1)}},{\left|d_r\right|}^2={\displaystyle \frac{64q^3}{{(q+1)}^2(q^2+6q+1)}},\hfill \end{array}$$

(8.16)

$$\begin{array}{cc}\hfill \Re \gamma & ={\displaystyle \frac{|t_{31}{|}^2}{2}},{(\Im \gamma )}^2={\displaystyle \frac{q{(1-q)}^4{(q^2+4q-1)}^2{(q^2+4q+1)}^2}{16{(q^2+1)}^4{(q^2+6q+1)}^2}}.\hfill \end{array}$$

(8.17)

The two quantities in (8.11) now have exact polynomial certificates. First, if

$$D_F={(q+1)}^2{(q^2+1)}^2{(q^2+6q+1)}^2,$$

the bilinear polynomial $D_{F}F$ has tensor Bernstein control matrix

$$\left(\begin{array}{cc}32q^3{(q+1)}^2(q^2+6q+1)& 2q^2{(q+1)}^2(q^2+1){(q^2+6q+1)}^2\\ 2q^2{(q+1)}^2(q^2+1){(q^2+6q+1)}^2& 16q^{3}J\left(q\right)\end{array}\right),$$

(8.18)

where

$$J\left(q\right)=3q^6+22q^5+35q^4-12q^3+9q^2+6q+1.$$

Its degree-six Bernstein controls on $[0,1]$ are

$$1,2,{\displaystyle \frac{18}{5}},{\displaystyle \frac{26}{5}},{\displaystyle \frac{128}{15}},{\displaystyle \frac{64}{3}},64.$$

Every entry in (8.18) is therefore positive, so $F>0$.

Next write

$$S={\displaystyle \frac{{\mathcal{N}}_q(X,Y)}{{(q+1)}^2{(q^2+1)}^4{(q^2+6q+1)}^2}}.$$

(8.19)

The bidegree-$(2,2)$ tensor Bernstein controls of ${\mathcal{N}}_q$ are the symmetric matrix determined by

$$\begin{array}{cc}\hfill b_{00}& =1024q^6{(q+1)}^2,\hfill \\ \hfill b_{01}& =64q^5{(q+1)}^2(q^2+1)(q^2+6q+1),\hfill \\ \hfill b_{02}& =4q^4{(q+1)}^2{(q^2+1)}^2{(q^2+6q+1)}^2,\hfill \\ \hfill b_{11}& =-2q^4{P}_{10}\left(q\right),\hfill \\ \hfill b_{12}& =32q^5(q^2+1)J\left(q\right),\hfill \\ \hfill b_{22}& =256q^6{P}_1\left(q\right)P_2\left(q\right),\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill P_{10}\left(q\right)=& q^{10}-2q^9-63q^8-224q^7-514q^6-732q^5\hfill \\ \hfill & -66q^4-288q^3-127q^2-34q+1,\hfill \\ \hfill P_1\left(q\right)& =q^3+3q^2-q+1,\hfill \\ \hfill P_2\left(q\right)& =q^4+4q^3-2q^2+1.\hfill \end{array}$$

The Bernstein controls of $P_1,P_2$ on $[0,1]$ are respectively

$$(1,2/3,4/3,4),(1,1,2/3,1,4).$$

For $q\ge 1/33$, $P_{10}\left(q\right)<q^{10}+1-34q\le 1-33q\le 0$; hence every $b_{ij}$ is positive.

For the remaining small parameters put $X=m+d$, $Y=m-d$. Direct factorization gives

$${\displaystyle \frac{{\mathcal{N}}_q(m+d,m-d)}{4q^4}}=D_0(m,q)+d^2{D}_2(m,q),$$

(8.20)

where

$$\begin{array}{cc}\hfill D_0=& 32q[mq-m+q+1]\hfill \\ \hfill & ·[m{q}^3+3m{q}^2-3mq-m+2q+2]\hfill \\ \hfill & ·[m{q}^4+4m{q}^3-2m{q}^2-4mq+m+4q],\hfill \\ \hfill D_2=& 4{(q-1)}^2(q^2+4q-1)(q^2+4q+1)\hfill \\ \hfill & ·[-8m{q}^3+8mq+q^4-8q-1].\hfill \end{array}$$

For $0<q\le 1/5$ and $0\le m\le 1$, the three factors in $D_0$ are positive. The two sign-changing factors in $D_2$ are both negative, so $D_2\ge 0$. Thus $S>0$ for $q\le 1/5$. Since $(0,1)=(0,1/5]\cup [1/33,1)$, (8.11) proves the desired rank-$(2,2)$ determinant for all $0<a<1$. If $C_0=0$, the second rank condition and (8.13) force $X=0$ or $Y=0$; the remaining determinant is affine and its endpoint gaps are

$$1-|t_{31}{|}^2={\displaystyle \frac{32q^3}{{(q^2+1)}^2(q^2+6q+1)}}>0,1-{\left|t_{12}\right|}^2={\displaystyle \frac{2q^2}{1+q^2}}>0.$$

It remains to handle the five patterns in (8.13) with a rank-one side. Apply the disk automorphisms

$$c_i={\displaystyle \frac{\sqrt{2}{\rho }_i}{1+i{\rho }_i}},d_i={\displaystyle \frac{\sqrt{2}{\rho }_i}{1-i{\rho }_i}}.$$

(8.21)

They are automorphisms of the u- and v-disks, respectively, so their Pick kernels differ from those in (8.2) by invertible diagonal congruences. Moreover

$$c_i-d_i+i\sqrt{2}{c}_i{d}_i=0,z_i={\displaystyle \frac{c_i+d_i}{\sqrt{2}}},1-{\overline{z}}_i{z}_j=(1-{\overline{c}}_i{c}_j)(1-{\overline{d}}_i{d}_j).$$

(8.22)

Suppose the c-localized matrix has rank one and full support. Diagonal congruence reduces positivity of the d-localized matrix to

$$\left[{\displaystyle \frac{1-{\overline{d}}_i{d}_j}{1-{\overline{c}}_i{c}_j}}\right]⪰0.$$

The scalar Pick theorem supplies a Schur function $\phi$ with $\phi \left(c_i\right)=d_i$. On the finite reproducing-kernel space with Gram matrix $\left[{(1-{\overline{c}}_i{c}_j)}^{-1}\right]$, both c and $d=\phi \left(c\right)$ are contractive multipliers. Since $z=(c+d)/\sqrt{2}$,

$$\parallel M_z\parallel \le \sqrt{2}.$$

The multiplier Pick criterion is exactly (8.3). If the rank-one factor has a zero coordinate, the corresponding target row and column vanish and the assertion reduces to at most two nodes, where Theorem 8 applies. Interchanging $c,d$ treats a rank-one second side. Thus every extreme ray in (8.13) satisfies the target. Convexity of the compact base and homogeneity now prove (8.3) for every admissible L.    □

The phase calculation also gives a genuinely nonsymmetric patch. Its proof is computer assisted only in the finite sense that a displayed family of rational Bernstein controls is reconstructed and checked in exact arithmetic.

Theorem 10

(Asymmetric three-node patch). Let

$$\rho =(-a,0,b),{\displaystyle \frac{1}{6}}\le a,b\le {\displaystyle \frac{5}{6}},$$

and put $z_i=2{\rho }_i/(1+{\rho }_i^2)$, $u_i=({\rho }_i+i)/\sqrt{2}$, and $v_i=({\rho }_i-i)/\sqrt{2}$. Every Hermitian L satisfying (8.2) obeys (8.3). Hence the identity multiplier has norm at most $\sqrt{2}$ on the complete right-angle admissible-kernel cone for this two-parameter family.

Proof. 

The compact-base and extreme-ray arguments above are unchanged. Every extreme ray with a rank-one side is handled by (8.21)–(8.22) and scalar Pick interpolation, and a zero diagonal reduces to at most two nodes. It remains to treat the rank-$(2,2)$ pattern.

Put

$$\begin{array}{cccc}\hfill A& =1+a^4,\hfill & \hfill B& =1+b^4,\hfill \\ \hfill G& =a^2{b}^2+a^2+4ab+b^2+1,\hfill & \hfill K& =a^2{b}^2+{(a+b)}^2+1.\hfill \end{array}$$

Exact normalization of the three target edges gives

$$\alpha ={\displaystyle \frac{1-a^4}{A}},\beta ={\displaystyle \frac{1-b^4}{B}},{\gamma }_e={\displaystyle \frac{(1-a^2)(1-b^2)K^2}{ABG}}.$$

(8.23)

For the phase quantities of Lemma 11, direct kernel substitution gives

$$\begin{array}{cc}\hfill \Re d_r& =-{\displaystyle \frac{8a^2{b}^2{(a+b)}^2}{(1+a^2)(1+b^2)G}},\hfill \\ \hfill |d_r{|}^2& ={\displaystyle \frac{16a^2{b}^2{(a+b)}^2}{(1+a^2)(1+b^2)G}},\hfill \\ \hfill \Re \gamma & ={\displaystyle \frac{{\gamma }_e}{2}},\Im \gamma ={\displaystyle \frac{ab(1-a^2)(1-b^2)EK}{2(a+b)ABG}},\hfill \end{array}$$

(8.24)

where $E=a^2{b}^2+{(a+b)}^2-1$.

Regard the numerators of F and S in (8.10) as polynomials in $X,Y\in [0,1]$. Their tensor Bernstein control arrays have sizes 2 by 2 and 3 by 3. Exact factorization reduces all thirteen controls to positive elementary factors and four parameter polynomials $U,V,W,Z$; the only sign requirements are

$$U>0,V>0,W<0,Z>0.$$

(8.25)

The first one has the global sum-of-squares identity

$$U=4a^2{b}^2+{(a-b)}^2\{{(a+b)}^2+4ab+{(1-ab)}^2\}>0.$$

(8.26)

For the remaining three, set $a=1/6+2\sigma /3$, $b=1/6+2\tau /3$. Their exact tensor Bernstein expansions on $(\sigma ,\tau )\in {[0,1]}^2$ have respectively 49, 169, and 81 rational controls. Every control of $V,-W,Z$ is positive; their least values are

$${\displaystyle \frac{851394073}{725594112}},{\displaystyle \frac{177918726900023}{4738381338321616896}},{\displaystyle \frac{76259197261}{705277476864}}.$$

(8.27)

The exact verifier cited in the reproducibility statement reconstructs (8.23)–(8.24), divides the thirteen controls by their displayed positive factors with zero remainder, and computes all 299 rational controls from the monomial-to-Bernstein formula. Thus (8.25) holds and the first Bernstein layer gives $F>0$, $S>0$ for every $X,Y\in [0,1]$.

If $C_0>0$, Lemma 11 now gives $\Delta (t\circ p)>0$. If $C_0=0$, the second rank condition forces $X=0$ or $Y=0$, and the target determinant is respectively

$$(1-{\gamma }_e)(1-Y)+(1-\beta )Y,(1-{\gamma }_e)(1-X)+(1-\alpha )X.$$

These are positive by (8.23) and $ABG-(1-a^2)(1-b^2)K^2=2{(a+b)}^{2}U$. The target minors of order two are positive for the same reason. Hence every rank-$(2,2)$ extreme ray, and therefore the full cone, satisfies (8.3).    □

Remark 1.

Theorems 9 and 10 control whole admissible-kernel cones, not only rank-$(2,2)$ faces. Their scope is nevertheless precise: the multiplier is the identity, and the nonsymmetric theorem covers only the displayed parameter box. The general three-node problem still contains the rank-$(2,2)$ phase inequality outside this box and a degree-two Blaschke determinant for arbitrary Schur data. The arbitrary-node right-angle constant therefore remains open.

9. A Complex-Zero Quadratic Singular Family

The rank-one localized faces reduce to scalar Pick matrices on three nodes, and their extreme scalar data are quadratic Blaschke products. We now prove the sharp target on a full two-real-dimensional disk of such products. Put

$$\theta \in \mathbb{R},\eta =e^{i\theta },p=cos\theta ,q=sin\theta ,{\displaystyle \frac{1}{\sqrt{2}}}<q\le 1.$$

(9.1)

Let $c_0=d_0=w_0=0$, let $c_1,c_2$ be the roots of

$$-i\sqrt{2}\eta c^2+\eta c-1=0,$$

(9.2)

and set

$$d_j=\eta c_j^2={\displaystyle \frac{c_j}{1-i\sqrt{2}{c}_j}},w_j={\displaystyle \frac{c_j+d_j}{\sqrt{2}}}.$$

(9.3)

For $a\in \mathbb{D}$, define

$$B_a\left(w\right)=w{\displaystyle \frac{w-a}{1-\overline{a}w}}.$$

(9.4)

Theorem 11

(Complex-zero quadratic singular family). For the nodes in (9.1)–(9.3),

$$R_{jk}={\displaystyle \frac{1-{\overline{d}}_j{d}_k}{1-{\overline{c}}_j{c}_k}}$$

(9.5)

is positive semidefinite of rank two. For every $a\in \mathbb{D}$, the actual target

$$T_{jk}={\displaystyle \frac{2-\overline{B_a\left(w_j\right)}{B}_a\left(w_k\right)}{1-{\overline{c}}_j{c}_k}},0\le j,k\le 2,$$

(9.6)

is positive definite. The same conclusion holds for a unimodular multiple of $B_a$.

Proof. 

Vieta’s formulas and a direct modulus calculation give, with $\rho =\sqrt{33-8\sqrt{2}q}$,

$$c_1+c_2=-{\displaystyle \frac{i}{\sqrt{2}}},c_1{c}_2=-{\displaystyle \frac{i}{\sqrt{2}\eta }},{|c_1-c_2|}^2={\displaystyle \frac{\rho }{2}},$$

(9.7)

and $|c_1{|}^2+{\left|c_2\right|}^2=(1+\rho )/4$, $|c_1{c}_2{|}^2=1/2$. Thus $1/2<|c_j{|}^2<1$. The change

$$c={\displaystyle \frac{\sqrt{2}\xi }{1+i\xi }},d={\displaystyle \frac{\sqrt{2}\xi }{1-i\xi }},w={\displaystyle \frac{2\xi }{1+{\xi }^2}}$$

then shows that $c_j,d_j,w_j\in \mathbb{D}$. Moreover

$$R_{jk}={\displaystyle \frac{1-{\overline{c}}_j^2{c}_k^2}{1-{\overline{c}}_j{c}_k}}=1+{\overline{c}}_j{c}_k,$$

so R is the rank-two Gram matrix of the distinct vectors $(1,c_j)$.

Because $B_a\left(0\right)=0$, take the Schur complement of T at $T_{00}=2$. For $j=1,2$, put

$$h_j={\displaystyle \frac{B_a\left(w_j\right)}{c_j}},M_{jk}={\displaystyle \frac{2-{\overline{h}}_j{h}_k}{1-{\overline{c}}_j{c}_k}}.$$

Then $T_{jk}-2={\overline{c}}_j{c}_k{M}_{jk}$ and $detT=2|c_1{c}_2{|}^{2}detM=detM$. Also

$$|h_j|={\displaystyle \frac{|w_j|}{|c_j|}}\left|{\displaystyle \frac{w_j-a}{1-\overline{a}{w}_j}}\right|<\sqrt{2}.$$

Hence the diagonal entries of M are positive, and it is enough to prove $detM>0$.

On the two roots of (9.2),

$$c^2={\displaystyle \frac{i}{\sqrt{2}\eta }}-{\displaystyle \frac{i}{\sqrt{2}}}c,w={\displaystyle \frac{i}{2}}+\left({\displaystyle \frac{1}{\sqrt{2}}}-{\displaystyle \frac{i\eta }{2}}\right)c,{\displaystyle \frac{w}{c}}={\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{\eta }{\sqrt{2}}}c.$$

(9.8)

Thus $h=(w/c)(w-a)/(1-\overline{a}w)$ has a unique affine representative $A+Lc$. Its Cramer denominator is

$$\Delta =(1-\overline{a}{w}_1)(1-\overline{a}{w}_2)\ne 0.$$

(9.9)

Put $y_j=h_j/\sqrt{2}$ and

$$Z=|c_1-c_2{|}^2,C=|1-{\overline{c}}_1{c}_2{|}^2,U=|1-{\overline{y}}_1{y}_2{|}^2,\ell ={\left|L\right|}^2/2.$$

The two-point determinant identity is

$$det{P}_y={\displaystyle \frac{Z\Xi }{C(C-Z)}},\Xi =U-\ell C,M=2P_y.$$

(9.10)

Write $a=u+iv$ and $x={\left|a\right|}^2=u^2+v^2<1$. Exact reduction of (9.10) by (9.7) and (9.8) gives the root-free formula

$$\Xi ={\displaystyle \frac{N}{16H}},H=8{|\Delta |}^2>0,N=E+puK,$$

(9.11)

where

$$\begin{array}{cc}\hfill E=& 71-8q^2-28\sqrt{2}q+x(24q^2-4\sqrt{2}q-41)\hfill \\ \hfill & +v(-8\sqrt{2}{q}^3-44q^2+44\sqrt{2}q-30)+vx(48q^2+4\sqrt{2}q-40)\hfill \end{array}$$

(9.11)

$$\begin{array}{cc}\hfill & +v^2(-32q^2+8\sqrt{2}q+80),\hfill \end{array}$$

(9.12)

$$\begin{array}{cc}\hfill K=& 4\{2\sqrt{2}{q}^2-5q+12\sqrt{2}+8qv-10\sqrt{2}v-x(4q+7\sqrt{2})\},\hfill \\ \hfill H=& 8-5x+3x^2-8v+16v^2-6vx+4\sqrt{2}qv+8\sqrt{2}q{v}^2\hfill \end{array}$$

(9.13)

$$\begin{array}{cc}\hfill & +2\sqrt{2}q{x}^2-6\sqrt{2}qx+4q^{2}vx+4\sqrt{2}pu(1-2v)-4pqux.\hfill \end{array}$$

(9.14)

Furthermore

$$C={\displaystyle \frac{5+\rho }{4}},C-Z={\displaystyle \frac{5-\rho }{4}},C(C-Z)={\displaystyle \frac{\sqrt{2}q-1}{2}},$$

(9.15)

and therefore

$$detT={\displaystyle \frac{\sqrt{33-8\sqrt{2}q}N}{4H(\sqrt{2}q-1)}}.$$

(9.16)

We prove $N>0$. The coefficient $E_x$ is affine in v, and

$$E_x(q,1)=72q^2-81\le -9,E_x(q,-1)=-24q^2-8\sqrt{2}q-1<0.$$

Thus $E(q,x,v)\ge E(q,1,v)$. The latter is a quadratic $A\left(q\right)v^2+B\left(q\right)v+C\left(q\right)$ with $A\left(q\right)=8(10+\sqrt{2}q-4q^2)>0$. The seven Bernstein controls of $4AC-B^2$ on $q\in [1/\sqrt{2},1]$ are

$$\begin{array}{c}1152,32(53-17\sqrt{2}),{\displaystyle \frac{32(1179-596\sqrt{2})}{15}},{\displaystyle \frac{8(2067-1229\sqrt{2})}{5}},\\ {\displaystyle \frac{4(12895-8172\sqrt{2})}{15}},{\displaystyle \frac{4(1869-1210\sqrt{2})}{3}},4(-193+152\sqrt{2}).\end{array}$$

They are strictly positive by rational squaring, whence

$$E>0.$$

(9.17)

Since $p^2=1-q^2$ and $u^2=x-v^2$, set

$$F(q,x,v)=E^2-(1-q^2)(x-v^2)K^2.$$

(9.18)

This is cubic in x. Under $x=v^2+(1-v^2)s$, $0\le s<1$, its exact cubic Bernstein expansion is

$$F=B_0{(1-s)}^3+3B_{1}s{(1-s)}^2+3B_2{s}^2(1-s)+B_3{s}^3,$$

(9.19)

where

$$\begin{array}{cc}\hfill B_0& =F(q,v^2,v)=E{(q,v^2,v)}^2,\hfill \\ \hfill B_1& =B_0+(1-v^2)F_x(q,v^2,v)/3,\hfill \\ \hfill B_2& =F(q,1,v)-(1-v^2)F_x(q,1,v)/3,\hfill \\ \hfill B_3& =F(q,1,v).\hfill \end{array}$$

We certify $B_1>0$ and $B_2,B_3\ge 0$ on $[1/\sqrt{2},1]\times [-1,1]$.

Put $t=q-1/\sqrt{2}$ and $h=1-1/\sqrt{2}$. Exact tensor Bernstein conversion over $\mathbb{Q}\left(\sqrt{2}\right)$ gives positive controls for $B_1$ on

$$[0,h]\times [-1,0],[0,h]\times [0,1/2],[0,h]\times [1/2,1].$$

(5.1)

There are 147 controls, all strictly positive. Next factor

$$B_3=128{(q-5\sqrt{2}/4)}^{2}P(q,v),B_2=(5\sqrt{2}/4-q)G(q,v).$$

(9.20)

For P, use $[0,h]\times [-1,0]$; each of the six v-intervals

$$[0,1/32],[1/32,1/16],[1/16,1/8],[1/8,1/4],[1/4,1/2],[1/2,1]$$

over $t\in [1/5,h]$; and the last five of these intervals over $t\in [0,1/5]$. The resulting 12 boxes have 294 positive and six zero controls. The omitted local rectangle is positive because, for $0\le t\le 1/5$, $0\le v\le 1/32$,

$$\begin{array}{cc}\hfill P=& {(t^2-3v/2)}^2+(4-26v-11\sqrt{2}t)v^2+40v^4\hfill \\ \hfill & +t(22\sqrt{2}{v}^3+16\sqrt{2}{v}^4)+t^2(6v^2+24v^3)+t^3(2\sqrt{2}v+4\sqrt{2}{v}^2),\hfill \end{array}$$

(9.21)

and $4-26v-11\sqrt{2}t\ge 51/16-11\sqrt{2}/5>0$; the last comparison is equivalent to ${255}^2>2·{176}^2$.

For G, use $[0,h]\times [-1,0]$, $[0,1/64]\times [1/64,1]$, and the following pairs of v-intervals:

$$\begin{array}{cc}t-\mathrm{interval}& v-\mathrm{intervals}\\ [1/64,1/32]& [0,1/16],[1/16,1]\\ [1/32,1/16]& [0,1/8],[1/8,1]\\ [1/16,1/8],[1/8,1/5],[1/5,h]& [0,1/4],[1/4,1]\end{array}$$

These 12 boxes have 429 positive and three zero controls. On the omitted square $0\le t,v\le \delta :=1/64$, write

$$G=192\sqrt{2}{t}^2-{\displaystyle \frac{512}{3}}tv+1752\sqrt{2}{v}^2+R.$$

The quadratic part is at least $2752(t^2+v^2)/15$. Every monomial of R has total degree at least three and is divisible by $t^2$ or $v^2$; using $\sqrt{2}<3/2$, the sums of coefficient majorants in degrees 3 through 8 are

$${\displaystyle \frac{28216}{3}},10528,4488,{\displaystyle \frac{19120}{3}},{\displaystyle \frac{11072}{3}},{\displaystyle \frac{2816}{3}}.$$

Consequently

$$\left|R\right|\le {\displaystyle \frac{40143488841}{268435456}}(t^2+v^2)<{\displaystyle \frac{2752}{15}}(t^2+v^2),$$

(9.22)

so $G\ge 0$ there as well.

Thus the final rectangular certificate contains 879 exact controls: 870 are positive and nine are zero. (The preliminary count 574 corresponds to a coarser box list whose Bernstein controls are not all nonnegative, and is not the proof certificate.) No sign decision uses floating-point arithmetic.

If $s=0$, then $F=B_0=E^2>0$; if $0<s<1$, the $B_1$ term in (9.19) is strictly positive and the other terms are nonnegative. Hence $F>0$, so (9.17) gives $\left|puK\right|<E$ and $N=E+puK>0$. Formula (9.16) now proves $detM>0$ and completes the Schur-complement proof.    □

Remark 2.

The lower endpoint $q=1/\sqrt{2}$ is excluded: one node reaches the unit circle and $C(C-Z)=0$. The endpoint $q=1$ is included and has $p=0$. The strict condition $\left|a\right|<1$ ensures both $\Delta \ne 0$ and $s<1$ above; $\left|a\right|=1$ is used only as a closure face in the polynomial certificate. The proof treats the complex phase directly rather than reducing to the real-zero theorem. The theorem nevertheless concerns only this singular arc and products with one zero fixed at the origin. It does not settle arbitrary singular nodes, arbitrary quadratic inner functions, or the unrestricted fixed-lens constant.

10. The Unrestricted Angle-Dependent Envelope

For context, we now place 1 inside the best bounds that can be verified from the literature cited below. Define

$$\begin{array}{cc}\hfill L\left(\alpha \right)& :={\displaystyle \frac{\pi sin\alpha }{2\alpha }},\hfill \end{array}$$

(10.1)

$$\begin{array}{cc}\hfill K_{25}\left(\alpha \right)& :=1-{\displaystyle \frac{\alpha }{\pi }}+\sqrt{2-{\displaystyle \frac{4\alpha }{\pi }}+{\displaystyle \frac{{\alpha }^2}{{\pi }^2}}},\hfill \end{array}$$

(10.2)

$$\begin{array}{cc}\hfill \mu \left(\alpha \right)& :={\displaystyle \frac{sin\left(2\alpha \right)}{\pi }}{\int }_0^{\infty }{\displaystyle \frac{dy}{y^{2}cos\alpha -2ycos\left(2\alpha \right)+cos\alpha }},\hfill \end{array}$$

(10.3)

$$\begin{array}{cc}\hfill K_{\mathrm{BC}}\left(\alpha \right)& :=2{\displaystyle \frac{\pi -\alpha }{\pi }}+\mu \left(\alpha \right).\hfill \end{array}$$

(10.4)

The integral in (10.3) is positive and finite for $0<\alpha <\pi /2$; its endpoint values are understood by limits.

Proposition 5

(Verified envelope). For $0<\alpha \le \pi /2$,

$$L\left(\alpha \right)\le C\left(\alpha \right)\le min\left\{K_{25}\left(\alpha \right),K_{\mathrm{BC}}\left(\alpha \right),{\displaystyle \frac{\pi -\alpha }{\alpha }}\right\}.$$

(10.5)

In particular,

$$C\left(\alpha \right)<2\mathrm{whenever}\alpha >{\displaystyle \frac{\pi }{6}}.$$

(10.6)

Proof. 

The lower bound is Corollary 2. The term $(\pi -\alpha )/\alpha$ is the classical sector bound recorded in [1]; the conic-domain argument of [7] yields $K_{\mathrm{BC}}$. Formula (10.2) is the aperture bound proved in the preprint [8]. By Proposition 2, all three sector bounds apply to $C\left(\alpha \right)$. Finally, direct algebra gives $K_{25}\left(\alpha \right)<2$ exactly when $\alpha >\pi /6$.    □

At representative angles the two explicit curves in Figure 1 are:

$\alpha$	$\pi /12$	$\pi /6$	$\pi /4$	$\pi /3$
$L\left(\alpha \right)$	$1.5529$	$1.5000$	$1.4142$	$1.2990$
$K_{25}\left(\alpha \right)$	$2.2103$	$2.0000$	$1.7808$	$1.5486$

These numbers are illustrations only; all claims in Proposition 5 are exact symbolic inequalities.

The square-zero curve should not be confused with the full dimension-two constant. Crouzeix’s characterization [4, Theorem 4.2] determines $C({\mathcal{S}}_{\alpha },2)$; in the strip limit it gives $C({\mathcal{S}}_0,2)=1.5876598\dots >\pi /2$ [5]. Thus even in dimension two a pair of distinct spectral points can outperform every one-point square-zero model. At $\alpha =\pi /4$, by contrast, the dimension-two value equals the square-zero value $\sqrt{2}$.

At the disk endpoint, with $\epsilon =\pi /2-\alpha \downarrow 0$,

$$\begin{array}{cc}\hfill L\left(\alpha \right)& =1+{\displaystyle \frac{2\epsilon }{\pi }}+\left({\displaystyle \frac{4}{{\pi }^2}}-{\displaystyle \frac{1}{2}}\right){\epsilon }^2+O\left({\epsilon }^3\right),\hfill \end{array}$$

(10.7)

$$\begin{array}{cc}\hfill K_{25}\left(\alpha \right)& =1+{\displaystyle \frac{4\epsilon }{\pi }}-{\displaystyle \frac{8{\epsilon }^2}{{\pi }^2}}+O\left({\epsilon }^3\right).\hfill \end{array}$$

(10.8)

Thus the known upper and lower slopes still differ by a factor of two. At the sharp-lens endpoint $\alpha \downarrow 0$, $L\left(\alpha \right)\to \pi /2$ whereas $K_{25}\left(\alpha \right)\to 1+\sqrt{2}$.

11. A Recent Uniform-Two Claim and What it Does Not Settle

After the preceding results, Jin posted a preprint claiming that the numerical range is a 2-spectral set and, as a corollary, that the intersection of two spherical disks has uniform constant 2 [9]. The same preprint explicitly lists the optimal constant for an individual crossing lens as open. At the date on this manuscript the item is labelled by its host as not peer reviewed.

If that theorem is accepted, one may add the term 2 to the minimum on the right of (10.5); none of the proofs in sec:proofsec:bounds uses it. Even the uniform inequality $C\left(\alpha \right)\le 2$ would not identify $C\left(\alpha \right)$: for $\alpha >\pi /6$, (10.6) is already strictly smaller, and for every nontrivial angle the lower certificate leaves a gap.

This distinction matters. A uniform theorem answers

$$\underset{0<\alpha \le \pi /2}{sup}C\left(\alpha \right)\le 2,$$

whereas the fixed-lens problem asks for the entire function $\alpha \mapsto C\left(\alpha \right)$, including its extremal operators and functions.

12. Consequences for the Search for the Exact Constant

The sharp calculation rules out a common first numerical strategy as a route to the full answer: no optimization restricted to a single eigenvalue and a square-zero perturbation can exceed $L\left(\alpha \right)$. Any strict improvement of the lower bound must therefore use at least one of the following mechanisms:

1.

two or more spectral points whose Pick constraints interact;

2.

a nilpotent chain of order at least three;

3.

a genuinely infinite-dimensional sectorial operator;

4.

a matrix-valued test function if the completely bounded constant is larger than the scalar one.

The first two options lead to finite semidefinite programs: sample a Herglotz representation after the power map, impose the two supporting half-plane inequalities on a matrix B, and optimize the norm of the Pick functional calculus. A valid lower bound must be exported as exact algebraic data (matrix, function, and norm certificate), not only as floating-point output. Conversely, a new upper bound requires a positive kernel or a sum-of-squares decomposition valid for every sectorial operator. The equality conditions in supply useful constraints for both searches.

13. Conclusions

The fixed crossing-lens constant is exactly the sectorial numerical-range constant, but its unrestricted value is not presently determined. We have computed the constant on the full affine square-zero class and shown that it equals the classical global lower curve $\pi sin\alpha /\left(2\alpha \right)$, with a closed-form $2\times 2$ lens certificate. The calculation is sharp, dimension-free, and independent of any uniform-two theorem. At the right angle, the two exact sums of squares in Section 6 prove the conjectural $\sqrt{2}$ estimate for the entire palindromic quadratic family $u>0$, hence for positive-real and conjugate pairs of half-plane zeros, and for every admissible operator. The exact certificates in Theorems 3 and 4 cover the complete imaginary post-automorphism diameter and an explicit two-real-dimensional cusp around it, while the bivariate rational certificate in Theorem 5 fills the full complex disk $\left|c\right|\le 19/20$. The exact radial, interpolation, and matrix Bernstein certificates in Theorem 6 cross that circle for $1/10\le t\le 18/25$; together with reflection they reach all but $9.1662$ degrees of the boundary-phase circle. At the centre, 1 gives the strictly stronger bound $\parallel w^2\parallel \le {\kappa }_0<\sqrt{2}$ by an explicit polynomial bidisk extension. The reserve in that extension also yields the genuinely two-complex-parameter result 7: every pair $\left|\alpha \right|,\left|\beta \right|\le 1/150$, including two independently phased nonzero zeros, satisfies a strict $\sqrt{2}$ bound for every admissible operator. This is an open target-zero neighborhood, not an arbitrary-function theorem. Theorems 9 and 10 supply a different kind of exact advance: they prove the sharp identity multiplier on complete three-node admissible-kernel cones, for every symmetric triple and for the genuinely asymmetric patch $\rho =(-a,0,b)$, $1/6\le a,b\le 5/6$. Their restriction to one multiplier and three nodes is essential; Remark 1 records the remaining determinant gates. On the singular rank-one-side boundary, Theorem 11 proves the quadratic target for the full arc $\mathrm{\Phi }\left(c\right)=e^{i\theta }{c}^2$ and every complex second Blaschke zero in the disk. This is a genuine three-real-parameter scalar-face theorem, but Remark 2 explains why it does not close the arbitrary quadratic singular boundary. Proposition 4 reduces the remaining boundary parameters in this centred slice to the single variance inequality (6.45); this is an equivalence, not an asserted proof of that inequality. The unrestricted gap in (10.5) nevertheless cannot be closed by first-order Jordan models or by these degree-two subfamilies alone. The remaining complex post-automorphism parameters—in particular the small phase gap about $c=-1$ left by the radial certificate—the critical-point frequency, and higher-order spectral interaction are the next obstacles.

14. Reproducibility Statement

All constants and matrices needed for the main result are displayed in (1.6) and (1.7). The two disk constraints reduce to the scalar identity in the proof of Corollary 2; the functional calculus truncates by $J^2=0$. For $u\ge 2$, the displayed quartic, discriminant, selector, and five-term identity give a wholly symbolic audit. For the remaining computer-assisted identities, the ancillary README lists the exact data, independent verifiers, and replay commands. These verifiers reconstruct the algebraic-number kernel compressions, hereditary coefficient identities, rational positivity comparisons, radial and boundary-phase Gram certificates, row-space identities, and the polynomial bidisk extension. All proof decisions use exact rational or algebraic arithmetic; floating-point discovery factors are not trusted in any proof. For Theorem 9, tmp/research/verifythreenoderank22elimination.py reconstructs the normalized multipliers and checks the phase-elimination, Bernstein, and small-parameter factorization identities in exact SymPy arithmetic; tmp/research/rightanglethreenoderank13exact.py independently checks the polarized quotient identity and scalar-face normalization. For Theorem 10, tmp/research/verifythreenoderank22asymmetricbox.py reconstructs the kernel invariants and phase criterion, verifies the global sum-of-squares identity (8.26), and checks all 299 rational parameter Bernstein controls exactly. For Theorem 11, tmp/research/verifyrank13phithetacomplexa.py independently rebuilds the quadratic reduction, the root-free determinant, 879 exact tensor Bernstein controls over $\mathbb{Q}\left(\sqrt{2}\right)$, and the two local corner certificates. Separate referee and certificate audits in audit/RANK13PHITHETACOMPLEXAREFEREEAUDIT.md and audit/RANK13PHITHETACOMPLEXACERTIFICATEAUDIT.md reconstruct the disk, rank, Schur-complement, coverage, strictness, and endpoint arguments not encoded by the script. For Theorem 7, tmp/research/verifyglobaltwosmallzerotargetfamily.py checks the two complex curve factorizations, the pole-free denominator bound, and the strict exact norm reserve. The independent report audit/GLOBALTWOSMALLZEROTARGETFAMILYREFEREEAUDIT.md reconstructs the conjugations, parameter boundary, Ando transfer, and finite-kernel consequence. The stable polynomial core is also compiled in Lean as formal/RightAngleThreeNodeRank22Algebra.lean; its axiom audit reports only propext, Classical.choice, and Quot.sound. The narrower $B\left(w\right)=w^2$ singular-family algebra is checked in formal/RightAngleSingularW2Algebra.lean; this does not formalize 11 or the analytic Pick-theoretic bridges. The plot evaluates only the explicit formulas (10.1) and (10.2).