Six-Dimensional Kissing Configurations with Q₅ or R₅ Cross Sections

1. Introduction

A kissing configuration in ${\mathbb{R}}^n$ is a finite set X of unit vectors with $\langle x,y\rangle \le 1/2$ for all distinct $x,y\in X$. The kissing number is known in dimensions $1,2,3,4,8$ and 24; in dimensions five, six and seven the largest known configurations have 40, 72 and 126 points, while the best known upper bounds are 44, 77 and 134, from the semidefinite programming bounds of Mittelmann and Vallentin [10] in dimensions five and seven and of de Laat, Leijenhorst and de Muinck Keizer [5] in dimension six. In these three dimensions the configurations attaining these lower bounds are not unique: Leech [8,9] found non-lattice examples, Cohn, Jiao, Kumar and Torquato [2] listed two, four and four configurations up to isometry, Szöllosi [11] found a new configuration $Q_5$ in ${\mathbb{R}}^5$, and Cohn and Rajagopal [3] found a fourth one, $R_5$.

A cross section of a configuration $X\subset {\mathbb{R}}^n$ is the intersection $X\cap H$ with a hyperplane H through the origin. In ${\mathbb{R}}^6$ the $E_6$ root system and several other 72-point configurations can be described as a $D_5$ or $L_5$ cross section together with two layers of 16 points placed over deep holes of the section [3]. For $Q_5$ and $R_5$ this construction cannot be carried out, and Cohn and Rajagopal proposed:

Conjecture 1 

([3]). No six-dimensional kissing configuration with $Q_5$ or $R_5$ as a cross section can contain 72 or more points.

The bound of 15 points on each side, and hence 70 in total, is also proved by Iuzhakov and Vetrov [7], whose manuscript is dated 29 September 2026 and whose version 1.0.0 was publicly released on 30 September 2026. Their proof uses the same height and equality mechanism. We use their explicit polar bound and its equality list below, followed by a different final certificate: a partition into 12 incompatible classes. This supplies a hand-checkable proof together with 64-point examples. The classifications and quantitative neighborhoods are separate results. We make no claim of priority for the shared upper bound.

Throughout, the four five-dimensional configurations $D_5,L_5,Q_5,R_5$ are normalized to unit vectors, and “X has C as a cross section” means that $X\cap H$ is isometric to C for some hyperplane H through the origin. An isometry between these spanning sets of unit vectors preserves their Gram matrices, and therefore extends to a linear orthogonal map of their spans.

Theorem 1. 

Let $X\subset {\mathbb{R}}^6$ be a kissing configuration with $Q_5$ or $R_5$ as a cross section. Then each of the two open half-spaces bounded by the section contains at most 15 points of X, and hence

$$\left|X\right|\le 70.$$

Moreover, for each of $Q_5$ and $R_5$ there is a 64-point kissing configuration in ${\mathbb{R}}^6$ having it as a cross section.

The mechanism behind Theorem 1 is saturation of a layer bound. A side has at most 16 points. If it has 16, equality in the positive-kernel bound forces every projected point to have a partner at inner product $1/5$. These pairs force equality in the height estimates for every point, placing all 16 directions at deep holes. But a compatible deep-hole layer has at most 12 points. This excludes 16 and gives 15; it does not give 12, because a side with fewer than 16 points need not lie over deep holes.

In particular Conjecture 1 holds. The same analysis classifies the extensions of the other two five-dimensional configurations.

Theorem 2. 

Let $C\in \{D_5,L_5\}$ and let $X\subset {\mathbb{R}}^6$ be a kissing configuration with cross section $X\cap H=C$. Then $\left|X\right|\le 72$. If $\left|X\right|=72$, then there are two sets $U_{+},U_{-}$ of 16 deep holes of C (Definition 2) each with pairwise inner products at most $1/5$, such that, with n a unit normal of H,

$$X=C\cup \left\{\sqrt{5/8}u+\sqrt{3/8}n:u\in U_{+}\right\}\cup \left\{\sqrt{5/8}u-\sqrt{3/8}n:u\in U_{-}\right\}.$$

There are exactly two such 16-sets for $D_5$ and exactly three for $L_5$, and every ordered choice of $(U_{+},U_{-})$ yields a kissing configuration. Up to isometry, these $4+9$ choices give exactly the four configurations $C_{6--72a},C_{6--72b},C_{6--72c},C_{6--72d}$ of [2]; the first is $E_6$.

The constructions in Theorem 2 are those described by Cohn and Rajagopal [3]. Theorem 2 proves that there are no other extensions attaining 72 points. One dimension higher the same argument applies to $E_6$.

Theorem 3. 

Let $X\subset {\mathbb{R}}^7$ be a kissing configuration with $E_6$ as a cross section. Then $\left|X\right|\le 126$. If $\left|X\right|=126$, then X is isometric to exactly one of two configurations: the $E_7$ root system, or the non-antipodal configuration $C_{7--126c}$ of [2], the kissing configuration of Leech’s packing ${\Lambda }_7^2$. Both consist of $E_6$ and two layers of 27 points at heights $\pm 1/\sqrt{3}$ lying over deep holes of $E_6$.

The identification in Theorem 3 is exact: an explicit bijection between our second configuration and the coordinates of $C_{7--126c}$ published with [2] preserves all ${126}^2$ inner products (Appendix A). In particular the non-uniqueness itself is classical [2,9]; Theorem 3 proves that these two configurations are the only 126-point extensions of an exact $E_6$ cross section.

Method. All three theorems follow from one mechanism, in the spirit of [2], where the analogous bound 272 was proved for nine-dimensional configurations with an $E_8$ cross section. If a cross section C has covering constant $\rho$ (Definition 1) with ${\rho }^2>1/3$, every other point lies at height at least $\sqrt{1-1/\left(4{\rho }^2\right)}$ above or below the section, and the projections of the points on one side form a spherical code with maximal inner product $1-2{\rho }^2$ (Section 3). A positive-kernel bound then limits the size of each side (Section 4), and in the cases treated here its equality case forces every point to sit exactly over a deepest hole of C (Lemmas 5 and 6 guarantee that every point of an extremal layer has a partner at the extreme inner product). The deepest holes of the relevant sections are determined in Section 5, for root systems by a Weyl-chamber argument for $Q_5,R_5$ by the explicit certificate of [7], and for $L_5$ by an exact exhaustive computation. Theorem 1 uses only the printed polar identity, hole partitions and witnesses, while the remaining finite classifications also use exact certificates. Section 6 assembles the proofs. The computations use only integer and rational arithmetic; they are described in Appendix A.

2. Covering Constant, Polar Polytope and Deep Holes

Definition 1. 

Let C be a finite set of unit vectors spanning a Euclidean space V of dimension m. Itssupport functionis $h_C\left(u\right)={max}_{c\in C}\langle c,u\rangle$, and itscovering constantis

$$\rho \left(C\right)=\underset{u\in V,\left|u\right|=1}{min}{h}_C\left(u\right).$$

Since C spans V and consists of unit vectors, $\rho \left(C\right)>0$ holds exactly when the origin is an interior point of $convC$; this is the case for all configurations below. The polar polytope of C is

$$C^{\circ }=\{y\in V:\langle c,y\rangle \le 1\mathrm{for}\mathrm{all}c\in C\}.$$

It is a bounded polytope when $\rho \left(C\right)>0$.

Definition 2. 

A unit vector $u\in V$ with $h_C\left(u\right)=\rho \left(C\right)$ is adeep holeof C.

Lemma 2. 

Assume $\rho \left(C\right)>0$ and let $M={max}_{y\in C^{\circ }}\left|y\right|$. Then $\rho \left(C\right)=1/M$. A unit vector u is a deep hole if and only if $u=y/\left|y\right|$ for a vertex y of $C^{\circ }$ with $\left|y\right|=M$. In particular the set of deep holes is finite.

Proof. 

For a unit vector u the ray $\{\lambda u:\lambda \ge 0\}$ meets $C^{\circ }$ in the segment $0\le \lambda \le 1/h_C\left(u\right)$. Hence ${max}_{y\in C^{\circ }}\left|y\right|={max}_{u}1/h_C\left(u\right)=1/\rho \left(C\right)$, and u is a deep hole exactly when $u/\rho \left(C\right)\in C^{\circ }$ has norm M. The function $y\mapsto {\left|y\right|}^2$ is strictly convex. If $y=\sum {\lambda }_i{v}_i$ is a convex combination of vertices $v_i$ of $C^{\circ }$ with all ${\lambda }_i>0$, then ${\left|y\right|}^2\le \sum {\lambda }_i{\left|v_i\right|}^2\le M^2$, and equality in the first inequality forces all $v_i$ to coincide. Thus $\left|y\right|=M$ forces y to be a vertex of norm M.    □

3. The Slice Lemma

Let $X\subset {\mathbb{R}}^{m+1}$ be a kissing configuration and H a hyperplane through the origin with unit normal n. Put $C=X\cap H$ and assume that C spans H. Write each $x\in X\setminus C$ as

$$x=p+zn,p\in H,z=\langle x,n\rangle \ne 0,$$

and call $\left|z\right|$ the height of x and the sign of z its side.

Lemma 3 

(Heights). Let $\rho =\rho \left(C\right)$ and assume $\rho >1/2$. Every $x\in X\setminus C$ satisfies

$$\left|p\right|\le {\displaystyle \frac{1}{2\rho }},z^2\ge b^2:=1-{\displaystyle \frac{1}{4{\rho }^2}}>0.$$

If $\left|z\right|=b$, then $p/\left|p\right|$ is a deep hole of C.

Proof. 

For $c\in C$ we have $\langle c,n\rangle =0$, so $\langle c,p\rangle =\langle c,x\rangle \le 1/2$. If $p\ne 0$ then $\left|p\right|\rho \le \left|p\right|h_C(p/|p\left|\right)=h_C\left(p\right)\le 1/2$. This gives $\left|p\right|\le 1/\left(2\rho \right)<1$, hence $z\ne 0$ and $z^2=1-{\left|p\right|}^2\ge b^2$. If $\left|z\right|=b$, then $\left|p\right|=1/\left(2\rho \right)$ and $h_C(p/|p\left|\right)\le 1/\left(2\right|p\left|\right)=\rho$, so $p/\left|p\right|$ is a deep hole.    □

Lemma 4 

(One side). Assume ${\rho }^2>1/3$, equivalently $b>1/2$. Let $x=p+zn$ and $x^{\prime }=p^{\prime }+z^{\prime }n$ be distinct points of $X\setminus C$ on the same side.

1.

If $p=0$ then $\langle x,x^{\prime }\rangle =\left|z^{\prime }\right|>1/2$, which is impossible. Hence a side that contains a point with $p=0$ contains no other point.

2.

If $p,p^{\prime }\ne 0$ and $u=p/\left|p\right|$, $u^{\prime }=p^{\prime }/\left|p^{\prime }\right|$, then

$$\langle u,u^{\prime }\rangle \le s:={\displaystyle \frac{1/2-b^2}{1-b^2}}=1-2{\rho }^2,$$

with equality only if $\left|z\right|=|z^{\prime }|=b$.

Proof. (1) If $p=0$ then $x=\pm n$ and $\langle x,x^{\prime }\rangle =\left|z^{\prime }\right|\ge b>1/2$. (2) Write $\alpha =\left|z\right|$, $\beta =|z^{\prime }|$, both in $[b,1)$. Since $\langle x,x^{\prime }\rangle =\langle p,p^{\prime }\rangle +\alpha \beta \le 1/2$ and $\left|p\right|=\sqrt{1-{\alpha }^2}$, $|p^{\prime }|=\sqrt{1-{\beta }^2}$,

$$\langle u,u^{\prime }\rangle \le \phi (\alpha ,\beta ):={\displaystyle \frac{1/2-\alpha \beta }{\sqrt{(1-{\alpha }^2)(1-{\beta }^2)}}}.$$

A direct computation gives

$${\displaystyle \frac{\partial \phi }{\partial \alpha }}={\displaystyle \frac{\alpha /2-\beta }{{(1-{\alpha }^2)}^{3/2}{(1-{\beta }^2)}^{1/2}}},$$

which is negative because $\alpha /2<1/2<b\le \beta$; by symmetry $\phi$ is also strictly decreasing in $\beta$. Hence $\phi (\alpha ,\beta )\le \phi (b,b)=(1/2-b^2)/(1-b^2)$, with equality only at $\alpha =\beta =b$. Finally $1-b^2=1/\left(4{\rho }^2\right)$ gives $\phi (b,b)=1-2{\rho }^2$.    □

For $s<1$ let $A_m\left(s\right)$ denote the largest size of a set of unit vectors in ${\mathbb{R}}^m$ with pairwise inner products at most s.

Corollary 1. 

If $\rho {\left(C\right)}^2>1/3$, then each side contains at most $max\{1,A_m(1-2{\rho }^2)\}$ points of X, and $\left|X\right|\le \left|C\right|+2max\{1,A_m(1-2{\rho }^2)\}$.

Proof. 

By Lemma 4, a side either consists of a single point with $p=0$, or its points have nonzero projections whose normalized directions have pairwise inner products at most $1-2{\rho }^2<1$. The latter are in particular distinct.    □

4. Positive-Kernel Bounds for One Layer

For a unit vector $u\in {\mathbb{R}}^m$ define the symmetric tensors

$$Q\left(u\right)=u{u}^{\mathsf{T}}-\frac{1}{m}I,T{\left(u\right)}_{abc}=u_a{u}_b{u}_c-\frac{1}{m+2}\left(u_a{\delta }_{bc}+u_b{\delta }_{ac}+u_c{\delta }_{ab}\right).$$

Both are trace free: contracting $b=c$ in $T\left(u\right)$ gives $u_a-{\displaystyle \frac{1}{m+2}}(m+2)u_a=0$. A trace-free symmetric tensor is orthogonal to every symmetric tensor containing a Kronecker factor, hence $\langle T\left(u\right),T\left(v\right)\rangle =\langle T\left(u\right),v\otimes v\otimes v\rangle$ and $\langle Q\left(u\right),Q\left(v\right)\rangle =\langle Q\left(u\right),v{v}^{\mathsf{T}}\rangle$. Direct contraction gives, with $t=\langle u,v\rangle$,

$$\langle Q\left(u\right),Q\left(v\right)\rangle =t^2-{\displaystyle \frac{1}{m}},\langle T\left(u\right),T\left(v\right)\rangle =t^3-{\displaystyle \frac{3t}{m+2}}.$$

(1)

Consequently, for any unit vectors $u_1,\dots ,u_N$ and real ${\lambda }_1,{\lambda }_2,{\lambda }_3\ge 0$,

$$\begin{array}{cc}\hfill & \sum _{i,j}\left({\lambda }_1\langle u_i,u_j\rangle +{\lambda }_2\langle Q\left(u_i\right),Q\left(u_j\right)\rangle +{\lambda }_3\langle T\left(u_i\right),T\left(u_j\right)\rangle \right)\hfill \\ \hfill & ={\lambda }_1{\left|\sum u_i\right|}^2+{\lambda }_2{\left|\sum Q\left(u_i\right)\right|}^2+{\lambda }_3{\left|\sum T\left(u_i\right)\right|}^2\ge 0.\hfill \end{array}$$

(2)

Lemma 5. 

Let $u_1,\dots ,u_N$ be distinct unit vectors in ${\mathbb{R}}^5$ with $\langle u_i,u_j\rangle \le 1/5$ for $i\ne j$. Then $N\le 16$. If $N=16$, then ${\sum }_i{u}_i=0$, every inner product $\langle u_i,u_j\rangle$ with $i\ne j$ lies in $\{1/5,-3/5\}$, and every $u_i$ has inner product $1/5$ with exactly 10 of the others.

Proof. 

Let $f\left(t\right)=(t-\frac{1}{5}){(t+\frac{3}{5})}^2=t^3+t^2+\frac{3}{25}t-\frac{9}{125}$. By (1) with $m=5$,

$$f\left(t\right)={\displaystyle \frac{16}{125}}+{\displaystyle \frac{96}{175}}t+\left(t^2-{\displaystyle \frac{1}{5}}\right)+\left(t^3-{\displaystyle \frac{3t}{7}}\right),$$

as one checks by comparing coefficients. By (2), ${\sum }_{i,j}f\left(\langle u_i,u_j\rangle \right)\ge {\displaystyle \frac{16}{125}}{N}^2$. On the other hand $f\le 0$ on $[-1,{\displaystyle \frac{1}{5}}]$, so ${\sum }_{i,j}f\left(\langle u_i,u_j\rangle \right)\le Nf\left(1\right)={\displaystyle \frac{256}{125}}N$. Hence $N\le 16$.

If $N=16$ both inequalities are equalities. Therefore $f\left(\langle u_i,u_j\rangle \right)=0$ for $i\ne j$, i.e. $\langle u_i,u_j\rangle \in \{{\displaystyle \frac{1}{5}},-{\displaystyle \frac{3}{5}}\}$, and each nonnegative square in (2) vanishes; in particular $\sum u_i=0$. Taking the inner product of ${\sum }_j{u}_j=0$ with $u_i$ gives ${\sum }_{j\ne i}\langle u_i,u_j\rangle =-1$. If k of these 15 terms equal ${\displaystyle \frac{1}{5}}$, then ${\displaystyle \frac{k}{5}}-{\displaystyle \frac{3(15-k)}{5}}=-1$, so $k=10$.    □

Lemma 6. 

Let $u_1,\dots ,u_N$ be distinct unit vectors in ${\mathbb{R}}^6$ with $\langle u_i,u_j\rangle \le 1/4$ for $i\ne j$. Then $N\le 27$. If $N=27$, then ${\sum }_i{u}_i=0$, every inner product $\langle u_i,u_j\rangle$ with $i\ne j$ lies in $\{1/4,-1/2\}$, and every $u_i$ has inner product $1/4$ with exactly 16 of the others.

Proof. 

Let $g\left(t\right)=(t-\frac{1}{4}){(t+\frac{1}{2})}^2=t^3+\frac{3}{4}{t}^2-\frac{1}{16}$. By (1) with $m=6$,

$$g\left(t\right)={\displaystyle \frac{1}{16}}+{\displaystyle \frac{3}{8}}t+{\displaystyle \frac{3}{4}}\left(t^2-{\displaystyle \frac{1}{6}}\right)+\left(t^3-{\displaystyle \frac{3t}{8}}\right).$$

As before ${\displaystyle \frac{1}{16}}{N}^2\le {\sum }_{i,j}g\left(\langle u_i,u_j\rangle \right)\le Ng\left(1\right)={\displaystyle \frac{27}{16}}N$, so $N\le 27$. If $N=27$, all off-diagonal values of g vanish, so the inner products lie in $\{{\displaystyle \frac{1}{4}},-{\displaystyle \frac{1}{2}}\}$, and $\sum u_i=0$. Then ${\sum }_{j\ne i}\langle u_i,u_j\rangle =-1$, and ${\displaystyle \frac{k}{4}}-{\displaystyle \frac{26-k}{2}}=-1$ gives $k=16$.    □

Remark 1. 

The bounds of Lemmas 5 and 6 are the Delsarte linear programming bounds for these parameters [4]. Both bounds are attained; the 27-point example is the Schläfli configuration. The tensor form above makes the positivity self-contained.

5. Deep Holes of the Sections

5.1. The Configurations

We use rational coordinates in which all points have squared norm 2, and normalize by $1/\sqrt{2}$ when unit vectors are needed. Let

$$D_5=\{\pm e_i\pm e_j:1\le i<j\le 5\}.$$

Following Leech [8] and [3], $L_5$ is obtained from $D_5$ by replacing its 8 points with fifth coordinate 1 by the 8 points $(\pm {\displaystyle \frac{1}{2}},\pm {\displaystyle \frac{1}{2}},\pm {\displaystyle \frac{1}{2}},\pm {\displaystyle \frac{1}{2}},1)$ with an odd number of minus signs. Let $\sigma$ be the reflection $v\mapsto v-{\displaystyle \frac{2}{5}}\left({\sum }_k{v}_k\right)(1,1,1,1,1)$ in the hyperplane of coordinate sum 0. Both $D_5$ and $L_5$ contain 20 points of coordinate sum 0, 10 of sum 2 and 10 of sum $-2$. Then $Q_5$ (resp. $R_5$) is obtained from $D_5$ (resp. $L_5$) by replacing the 10 points of coordinate sum 2 by the images under $\sigma$ of the 10 points of coordinate sum $-2$ [3,11]. All four are 40-point kissing configurations spanning ${\mathbb{R}}^5$; the inner-product distributions agree with [3] (Appendix A). Each has centroid zero. Thus $0\in intconvC$: otherwise a nonzero supporting functional at 0 would be nonnegative on every point and, since the sum of the points is zero, would vanish on all of C, contradicting the spanning property. In particular all four polar polytopes are bounded.

5.2. Root Systems

For an irreducible simply laced root system $\Phi$ with roots of squared norm 2, simple roots ${\alpha }_1,\dots ,{\alpha }_r$, fundamental weights ${\omega }_i$ ($\langle {\omega }_i,{\alpha }_j\rangle ={\delta }_{ij}$), Cartan matrix A and highest root $\theta =\sum m_i{\alpha }_i$, the Gram matrix of the fundamental weights is $A^{-1}$.

Proposition 1. 

The polar polytope ${\Phi }^{\circ }=\{y:\langle r,y\rangle \le 1\forall r\in \Phi \}$ is W-invariant, and

$$\underset{y\in {\Phi }^{\circ }}{max}{\left|y\right|}^2=\underset{i}{max}{\displaystyle \frac{{\left(A^{-1}\right)}_{ii}}{m_i^2}}.$$

The points of ${\Phi }^{\circ }$ of maximal norm are exactly the W-images of the vertices ${\omega }_i/m_i$ at which this maximum is attained.

Proof. 

${\Phi }^{\circ }$ is W-invariant because $\Phi$ is. Let $h={\sum }_i{\omega }_i$, so $\langle {\alpha }_i,h\rangle =1$. For $y\in {\Phi }^{\circ }$ choose $y^{\prime }\in Wy$ maximizing $\langle y^{\prime },h\rangle$; if $\langle y^{\prime },{\alpha }_i\rangle <0$ for some i, the reflection $s_i$ would increase $\langle ·,h\rangle$ by $-\langle y^{\prime },{\alpha }_i\rangle >0$. Hence $y^{\prime }=\sum b_i{\omega }_i$ with $b_i\ge 0$. For dominant $y^{\prime }$ and a negative root r we have $\langle y^{\prime },r\rangle \le 0\le \langle y^{\prime },\theta \rangle$, and for a positive root r the difference $\theta -r$ is a nonnegative combination of simple roots [6]; hence $\langle y^{\prime },r\rangle \le \langle y^{\prime },\theta \rangle =\sum m_i{b}_i$ for every root r. Therefore

$${\Phi }^{\circ }\cap \left\{\mathrm{dominant}\right\}=\left\{\sum b_i{\omega }_i:b_i\ge 0,\sum m_i{b}_i\le 1\right\}=conv\{0,{\omega }_1/m_1,\dots ,{\omega }_r/m_r\}.$$

By strict convexity of ${\left|y\right|}^2$ (as in Lemma 2) the maximum on this simplex is attained only at vertices ${\omega }_i/m_i$ of maximal norm, and $|{\omega }_i{|}^2={\left(A^{-1}\right)}_{ii}$. Since every point of ${\Phi }^{\circ }$ is W-equivalent to a dominant one of the same norm, the claim follows.    □

For $D_5$ (Bourbaki labelling, $m=(1,2,2,1,1)$) the ratios ${\left(A^{-1}\right)}_{ii}/m_i^2$ are $1,{\displaystyle \frac{1}{2}},{\displaystyle \frac{3}{4}},{\displaystyle \frac{5}{4}},{\displaystyle \frac{5}{4}}$, and for $E_6$ ($m=(1,2,2,3,2,1)$) they are ${\displaystyle \frac{4}{3}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{5}{6}},{\displaystyle \frac{2}{3}},{\displaystyle \frac{5}{6}},{\displaystyle \frac{4}{3}}$. With Lemma 2 and the normalization $c=r/\sqrt{2}$ this gives:

Corollary 2. 

$\rho {\left(D_5\right)}^2={\displaystyle \frac{2}{5}}$, and the deep holes of $D_5$ are the 32 unit vectors in the directions $W{\omega }_4\cup W{\omega }_5$, i.e. ${\displaystyle \frac{1}{\sqrt{5}}}(\pm 1,\dots ,\pm 1)$. $\rho {\left(E_6\right)}^2={\displaystyle \frac{3}{8}}$, and the deep holes of $E_6$ are the 54 unit vectors in the directions $W{\omega }_1\cup W{\omega }_6$ (the weights of the two minuscule representations of dimension 27).

Proof. 

For $D_5$: ${max\left|y\right|}^2={\displaystyle \frac{5}{4}}$, so ${max}_{u}1/h_{\Phi }\left(u\right)=\sqrt{5/4}$ for $h_{\Phi }$ computed with roots of norm $\sqrt{2}$; after normalization $\rho \left(D_5\right)={\displaystyle \frac{2}{\sqrt{5}}}·{\displaystyle \frac{1}{\sqrt{2}}}$, i.e. ${\rho }^2={\displaystyle \frac{2}{5}}$. The orbits $W{\omega }_4$, $W{\omega }_5$ are the 16 half-spin weights each, namely ${\displaystyle \frac{1}{2}}(\pm 1,\dots ,\pm 1)$ with an even, respectively odd, number of minus signs. For $E_6$: ${max\left|y\right|}^2={\displaystyle \frac{4}{3}}$, so $\rho {\left(E_6\right)}^2={\displaystyle \frac{3}{4}}·{\displaystyle \frac{1}{2}}={\displaystyle \frac{3}{8}}$; the orbits $W{\omega }_1$, $W{\omega }_6=-W{\omega }_1$ have $|W\left(E_6\right)|/|W\left(D_5\right)|=27$ elements each.    □

5.3. Explicit Polar Certificates for $Q_5$ and $R_5$

The following argument uses the polar bound and positive identity of Iuzhakov and Vetrov [7]. We give the details to make the proof of Theorem 1 independent of vertex enumeration. In this subsection $K=\sqrt{2}C$ has squared norm 2, and

$$P_K=\{y:\langle y,k\rangle \le 1(k\in K)\},S\left(y\right)={\displaystyle \sum _{i=1}^5}{y}_i.$$

Write $v_A=-\frac{1}{2}(1,1,1,1,1)+{\sum }_{i\in A}{e}_i$ and $u_i=-e_i+\frac{1}{2}{e}_5$ ($1\le i\le 4$). Define

$$\begin{array}{cc}\hfill H_Q^{-}& =\{v_A:A\subseteq \{1,\dots ,5\},|A|\le 2\},\hfill \\ \hfill H_R^{-}& =\{v_A:A\subseteq \{1,\dots ,4\},\left|A\right|\le 2\}\cup \{v_{\left\{5\right\}},u_1,u_2,u_3,u_4\},\hfill \\ \hfill H_K^{+}& =\sigma \left(H_K^{-}\right),\sigma \left(y\right)=y-\frac{2}{5}S\left(y\right)(1,1,1,1,1).\hfill \end{array}$$

Proposition 2 

(Iuzhakov–Vetrov [7]). For $K=\sqrt{2}{Q}_5$ or $\sqrt{2}{R}_5$, every $y\in P_K$ satisfies ${\left|y\right|}^2\le 5/4$. Equality holds precisely at the 32 points of $H_K^{-}\cup H_K^{+}$. Consequently $\rho {\left(C\right)}^2=2/5$, and its deep holes are the vectors $2y/\sqrt{5}$ for these 32 points.

Proof. 

Both carriers are invariant under $\sigma$, so assume $S\left(y\right)\le 0$. To see the polar inequalities directly, put $B=\{-e_i-e_j:i<j\}$. The unnormalized carriers are $B\cup \sigma \left(B\right)\cup M_K$, where

$$\begin{array}{cc}\hfill M_Q& =\{e_i-e_j:i\ne j\},\hfill \\ \hfill M_R& =\{e_i-e_j:1\le i\ne j\le 4\}\cup \{e_i-e_5:1\le i\le 4\}\hfill \\ \hfill & \cup \{\frac{1}{2}(e_i-\sum _{j\ne i,j\le 4}{e}_j)+e_5:1\le i\le 4\}.\hfill \end{array}$$

For $b\in B$, $\langle y,\sigma \left(b\right)\rangle =\langle y,b\rangle +\frac{4}{5}S\left(y\right)\le \langle y,b\rangle$, so the reflected inequalities are redundant on this half-space.

For $Q_5$ the remaining inequalities are $y_i+y_j\ge -1$ and ${max}_i{y}_i-{min}_i{y}_i\le 1$. Put $m={min}_i{y}_i\le 0$. If $m\ge -1/2$, then $y_i\in [m,m+1]$ gives

$${\left|y\right|}^2\le (2m+1)S\left(y\right)-5m(m+1)\le -5m(m+1)\le 5/4.$$

Equality requires $m=-1/2$ and every coordinate to be $-1/2$ or $1/2$; $S\left(y\right)\le 0$ permits at most two positive coordinates. If $m<-1/2$, put $b=1+m$. The other four coordinates lie in $[-b,b]$, with $0\le b<1/2$, whence

$${\left|y\right|}^2\le {(b-1)}^2+4b^2=5/4-(1/2-b)(5b+1/2)<5/4.$$

This proves the assertion for $Q_5$, including equality.

For $R_5$, permute the first four coordinates so that $y_1\ge y_2\ge y_3\ge y_4$. In this chamber the polar inequalities, the ordering and $S\le 0$ are exactly ${\ell }_i\ge 0$ for

$$\begin{array}{cccccc}\hfill {\ell }_1& =y_1-y_2,\hfill & \hfill {\ell }_2& =y_2-y_3,\hfill & \hfill {\ell }_3& =y_3-y_4,\hfill \\ \hfill {\ell }_4& =1+y_3+y_4,\hfill & \hfill {\ell }_5& =1+y_4+y_5,\hfill \\ \hfill {\ell }_6& =1-y_1+y_4,\hfill & \hfill {\ell }_7& =1-y_1+y_5,\hfill \\ \hfill {\ell }_8& =1-\frac{1}{2}{y}_1+\frac{1}{2}{y}_2+\frac{1}{2}{y}_3+\frac{1}{2}{y}_4-y_5,\hfill & \hfill {\ell }_9& =-S\left(y\right).\hfill \end{array}$$

Indeed, $y_3+y_4$ and $y_4+y_5$ are the smallest pair sums of the two types in B, and the three types in $M_R$ give ${\ell }_6,{\ell }_7,{\ell }_8$ by maximizing their inner products in the ordered chamber. The identity of [7] is

$${112(5/4-|y|}^2)=\sum _I{c}_I\prod _{i\in I}{\ell }_i.$$

(3)

All coefficients are positive and are displayed in Table 1. Direct expansion leaves the constant 140 and the five terms $-112y_i^2$; every other coefficient vanishes. Thus (3) proves the bound throughout the chamber.

For equality, each term in (3) must vanish. The quadratic terms alone give an exhaustive four-case calculation.

1.

If ${\ell }_1>0$, the terms $12,13,16,17$ give ${\ell }_2={\ell }_3={\ell }_6={\ell }_7=0$. Thus $y=(a+1,a,a,a,a)$, and ${\ell }_4{\ell }_5=0$ forces $a=-1/2$.

2.

If ${\ell }_1=0<{\ell }_2$, the terms $23,26,27$ give $y=(a+1,a+1,a,a,a)$; again ${\ell }_4{\ell }_5=0$ gives $a=-1/2$.

3.

If ${\ell }_1={\ell }_2=0<{\ell }_3$, the terms $36,38$ give $y=(a,a,a,a-1,a+1/2)$. Here ${\ell }_4=2a\ge 0$ and ${\ell }_5=2a+1/2>0$, so the term 45 gives $a=0$.

4.

If ${\ell }_1={\ell }_2={\ell }_3=0$, write $y=(a,a,a,a,t)$. Were ${\ell }_4>0$, the terms $45,49$ would give ${\ell }_5={\ell }_9=0$, hence $a=1/3$, $t=-4/3$, contradicting ${\ell }_7\ge 0$. Therefore $a=-1/2$, and ${\ell }_5{\ell }_8=0$ gives $t=\pm 1/2$.

These five points satisfy every chamber inequality. Their orbits under permutations of the first four coordinates have sizes $4,6,4,1,1$, and form $H_R^{-}$. All have negative coordinate sum, so reflection supplies 16 distinct positive-sum points. This proves completeness of the equality lists. Finally, normalizing the carrier multiplies its polar by $\sqrt{2}$, giving $\rho {\left(C\right)}^2=1/(2·5/4)=2/5$ by Lemma 2.    □

5.4. Additional Polar Enumeration

Proposition 3 

(exact computation). For each $C\in \{D_5,L_5,Q_5,R_5\}$ one has $\rho {\left(C\right)}^2={\displaystyle \frac{2}{5}}$, and C has exactly 32 deep holes. The polar polytopes $C^{\circ }$ have 42, 50, 92 and 100 vertices respectively.

Proof. 

A vertex of $C^{\circ }\subset {\mathbb{R}}^5$ is the unique solution y of five linearly independent equations $\langle c,y\rangle =1$, $c\in S\subset C$, $\left|S\right|=5$, which satisfies $\langle c,y\rangle \le 1$ for all $c\in C$. For each configuration all $\left({\displaystyle \genfrac{}{}{0pt}{}{40}{5}}\right)=658008$ five-element subsets S were examined in exact integer arithmetic (Appendix A): the determinant and Cramer numerators were computed by Laplace expansion, and feasibility was tested by exact integer comparisons. This yields the complete vertex list. The maximal squared norm is attained at exactly 32 vertices and corresponds to ${\rho }^2=2/5$ by Lemma 2. For $D_5$ the result agrees with Corollary 2. For $Q_5,R_5$, Proposition 2 already proves the radius and the complete deep-hole list without enumeration; only the total vertex counts require this computation.    □

5.5. Admissible Layers of Deep Holes

Definition 3. 

Let C be one of the sections above, with $s=1-2\rho {\left(C\right)}^2$. Anadmissible layeris a set of deep holes of C with pairwise inner products at most s.

For $D_5,L_5,Q_5,R_5$ we have $s={\displaystyle \frac{1}{5}}$; for $E_6$, $s={\displaystyle \frac{1}{4}}$.

Proposition 4 

(exact computation).

1.

$D_5$ has exactly two admissible layers of size 16 (the half-spin directions with an even, respectively odd, number of minus signs), and $L_5$ has exactly three.

2.

$Q_5$ and $R_5$ have no admissible layer of size 16; the largest admissible layers have size 12.

3.

$E_6$ has exactly two admissible layers of size 27, namely the normalized directions of the two minuscule orbits $W{\omega }_1$ and $W{\omega }_6=-W{\omega }_1$.

We use the following elementary fact for the root-system cases.

Lemma 7. 

Let G be a connected d-regular bipartite graph, $d\ge 1$, with parts of size n. Then G has exactly two independent sets of size n, namely its two parts.

Proof. 

An independent set I meets $d\left|I\right|$ distinct edges, and G has $dn$ edges, so $\left|I\right|\le n$. If $\left|I\right|=n$, every edge has exactly one end in I, so the complement of I is independent too, and $(I,V\setminus I)$ is a proper 2-colouring. A connected bipartite graph has a unique proper 2-colouring up to swapping the colours.    □

Proof 

(Proof of Proposition 4). An admissible layer is an independent set in the incompatibility graph on the deep holes, in which two holes are adjacent when their inner product exceeds s.

(3) Let U be the 27 normalized weights of $W{\omega }_1$. Distinct elements of U have inner products ${\displaystyle \frac{1}{4}}$ or $-{\displaystyle \frac{1}{2}}$; these values can be checked directly from the root coordinates in Appendix A. By Lemma 6, each element has 16 partners of the first kind and 10 of the second. No two elements of U, and no two of $-U$, are incompatible, while $u\in U$ and $-u^{\prime }\in -U$ are incompatible exactly when $\langle u,u^{\prime }\rangle =-{\displaystyle \frac{1}{2}}$. The exact coordinate calculation verifies that the graph on U with adjacency $\langle u,u^{\prime }\rangle =-{\displaystyle \frac{1}{2}}$ has 27 vertices of degree 10, with one common neighbor for an adjacent pair and five for a nonadjacent pair. Thus this graph is connected and contains a triangle. Its bipartite double cover is consequently connected: a path lifts to a path with prescribed starting sheet, and an odd closed walk interchanges the two sheets. This double cover is precisely the incompatibility graph on $U\cup (-U)$. It is connected and 10-regular, so Lemma 7 gives exactly the two admissible 27-sets U and $-U$.

(1) For $D_5$ the incompatibility graph on the 32 half-spin directions is 5-regular, connected and bipartite, with parts the directions with an even, respectively odd, number of minus signs (two such directions differing in h signs have inner product $1-{\displaystyle \frac{2h}{5}}$, so they are incompatible exactly when $h=1$; the incompatibility graph is the 5-dimensional hypercube graph). Lemma 7 gives exactly two admissible 16-sets.

(2) For each of $Q_5$ and $R_5$ the 32 deep holes can be partitioned into 12 classes of pairwise incompatible holes (inner products greater than ${\displaystyle \frac{1}{5}}$); the classes and their minimum inner products are printed in Table 2, and checking them requires only 35 inner-product comparisons for each configuration. An admissible layer contains at most one hole of each class, hence at most 12 holes; in particular there is no admissible layer of size 16. The explicit lists in Section 5.6 give admissible layers of size 12, so 12 is the maximum.

(1) for $L_5$, and the additional enumeration counts: admissible layers are cliques in the graph on the deep holes in which two holes are adjacent when their inner product is at most s. All inner products are rational multiples of a common squared norm, so adjacency was decided exactly, and all maximal cliques were enumerated by the Bron–Kerbosch algorithm [1] on graphs with 32 vertices (Appendix A). The computation also reproduces the count for $D_5$, and the counts for $D_5$ and $L_5$ agree with the descriptions in [3].    □

5.6. Printed Partitions and 12-Point Witnesses

Use the labels above, abbreviating $v_{\{i,j\}}$ to $v_{ij}$ and writing $\overline{y}=\sigma \left(y\right)$. Table 2 gives two partitions of the unnormalized outer points. Its last two columns give the minimum inner product among distinct members of each class; a dash denotes a singleton. All these minima exceed $1/4$. Since the corresponding unit directions have inner product ${\displaystyle \frac{4}{5}}\langle y,y^{\prime }\rangle$, at most one point of an admissible layer can belong to each class. This proves the upper bound 12 without a clique search. Every entry can be checked using

$$\begin{array}{cc}\hfill \langle v_A,v_B\rangle & =\frac{5}{4}-\frac{1}{2}\left(\right|A|+|B\left|\right)+|A\cap B|,\hfill \\ \hfill \langle u_i,v_A\rangle & =\frac{1}{4}-{\mathbf{1}}_{i\in A}+\frac{1}{2}{\mathbf{1}}_{5\in A},\langle u_i,u_j\rangle ={\delta }_{ij}+\frac{1}{4},\hfill \\ \hfill \langle y,\sigma \left(z\right)\rangle & =\langle y,z\rangle -\frac{2}{5}S\left(y\right)S\left(z\right),\langle \sigma \left(y\right),\sigma \left(z\right)\rangle =\langle y,z\rangle .\hfill \end{array}$$

The bound is attained by the following two lists. Every pair of distinct vectors in either list has inner product at most $1/4$, as follows directly from the coordinate formulas for $v_A,u_i,\sigma$:

$$\begin{array}{cc}\hfill U_Q^{\mathrm{pol}}& =\{v_{⌀},{\overline{v}}_{⌀}\}\cup \{v_{ij}:1\le i<j\le 5\},\hfill \\ \hfill U_R^{\mathrm{pol}}& =\{v_{⌀},{\overline{v}}_{⌀}\}\cup \{v_{ij}:1\le i<j\le 4\}\cup \{u_1,u_2,u_3,u_4\}.\hfill \end{array}$$

Normalizing these lists gives the admissible 12-sets used in Theorem 1. The exact exhaustive enumeration additionally finds 1024 maximum layers for each carrier; this count is not needed for the bound or the construction.

6. Proofs of the Main Theorems

Proof 

(Proof of Theorem 1). We first bound a side by 16 points, then exclude saturation by forcing all its points over deep holes. Let $C=X\cap H$ be isometric to $Q_5$ or $R_5$. By Proposition 2, $\rho {\left(C\right)}^2={\displaystyle \frac{2}{5}}>{\displaystyle \frac{1}{3}}$, so Lemmas 3 and 4 apply with $b^2=1-{\displaystyle \frac{5}{8}}={\displaystyle \frac{3}{8}}$ and $s={\displaystyle \frac{1}{5}}$. (Since $\rho >1/2$, no point of X other than those of C lies in H.) Fix a side containing $N\ge 2$ points. By Lemma 4, their projections have nonzero norms and unit directions $u_1,\dots ,u_N\in H\cong {\mathbb{R}}^5$ with pairwise inner products at most ${\displaystyle \frac{1}{5}}$. By Lemma 5, $N\le 16$.

Suppose $N=16$, so the positive-kernel bound is saturated. By Lemma 5, every $u_i$ has inner product exactly ${\displaystyle \frac{1}{5}}$ with some $u_j$. By the equality statement of Lemma 4(2), both corresponding points have height exactly b. Hence all 16 points have height b, and by Lemma 3 every $u_i$ is a deep hole of C. Thus $\{u_1,\dots ,u_{16}\}$ is an admissible layer of size 16, contradicting Proposition 4(2). Therefore each side contains at most 15 points, and $\left|X\right|\le 40+15+15=70$.

For the lower bound take an admissible layer U of size 12 (Proposition 4(2)), a unit normal n of H, and

$$X=C\cup \{\sqrt{5/8}u\pm \sqrt{3/8}n:u\in U\}.$$

For $c\in C$ and $u\in U$, $\langle c,\sqrt{5/8}u\rangle \le \sqrt{5/8}\rho =\sqrt{5/8}\sqrt{2/5}={\displaystyle \frac{1}{2}}$. Two points on the same side have inner product ${\displaystyle \frac{5}{8}}\langle u,u^{\prime }\rangle +{\displaystyle \frac{3}{8}}\le {\displaystyle \frac{5}{8}}·{\displaystyle \frac{1}{5}}+{\displaystyle \frac{3}{8}}={\displaystyle \frac{1}{2}}$, and two points on opposite sides have inner product ${\displaystyle \frac{5}{8}}\langle u,u^{\prime }\rangle -{\displaystyle \frac{3}{8}}\le {\displaystyle \frac{1}{4}}$. Hence X is a kissing configuration with $40+24=64$ points.    □

Proof 

(Proof of Theorem 2). Here $\rho {\left(C\right)}^2={\displaystyle \frac{2}{5}}$ again, so the argument above gives at most 16 points on each side and $\left|X\right|\le 72$. If $\left|X\right|=72$, both sides contain 16 points, and as in the proof of Theorem 1 each side consists of the points $\sqrt{5/8}u\pm \sqrt{3/8}n$ with u running through an admissible layer $U_{\pm }$ of size 16. Conversely, for any two admissible layers $U_{+},U_{-}$ of size 16 the three estimates at the end of the proof of Theorem 1 show that the displayed set is a kissing configuration. The number of admissible layers is given by Proposition 4(1).

For the isometry classification, list the 32 deep polar vertices lexicographically in the stored coordinates of Appendix A, and order the 16-sets lexicographically by their increasing index tuples. Label these sets $0,1$ for $D_5$ and $0,1,2$ for $L_5$. The following matrices give the isometry type for every ordered pair, with the upper choice indexing rows and the lower choice indexing columns:

$$D_5:\left(\begin{array}{cc}b& a\\ a& b\end{array}\right),L_5:\left(\begin{array}{ccc}d& d& b\\ d& c& d\\ b& d& d\end{array}\right).$$

Here $a,b,c,d$ denote $C_{6--72a},\dots ,C_{6--72d}$. Each of these 13 identifications is certified by an explicit permutation preserving all ${72}^2$ normalized inner products. Both point sets span ${\mathbb{R}}^6$, so Gram preservation proves an orthogonal isometry. The four types are distinct: their numbers of unordered antipodal pairs are $36,20,12,12$, and the numbers of unordered pairs with inner product $-3/4$ are $0,80,144,128$, respectively. Thus the two types with 12 antipodal pairs are also distinguished. All four occur in the displayed matrices, proving both completeness and attainment.

   □

Proof 

(Proof of Theorem 3). Let $C=X\cap H$ be isometric to $E_6$. By Corollary 2, ${\rho }^2={\displaystyle \frac{3}{8}}>{\displaystyle \frac{1}{3}}$, so $b^2=1-{\displaystyle \frac{2}{3}}={\displaystyle \frac{1}{3}}$ and $s=1-{\displaystyle \frac{3}{4}}={\displaystyle \frac{1}{4}}$. By Lemmas 4 and 6, each side contains at most 27 points, so $\left|X\right|\le 72+54=126$.

If $\left|X\right|=126$, each side contains 27 points. Every $u_i$ has inner product ${\displaystyle \frac{1}{4}}$ with some other $u_j$ (Lemma 6), so as before every point has height $b=1/\sqrt{3}$ and every $u_i$ is a deep hole. Hence each side is $\{\sqrt{2/3}u\pm {\displaystyle \frac{1}{\sqrt{3}}}n:u\in U_{\pm }\}$ with $U_{\pm }\in \{W{\omega }_1,W{\omega }_6\}$ (normalized) by Proposition 4(3). Every such choice is a kissing configuration: for $c\in C$, $\langle c,\sqrt{2/3}u\rangle \le \sqrt{2/3}\sqrt{3/8}={\displaystyle \frac{1}{2}}$; on one side ${\displaystyle \frac{2}{3}}\langle u,u^{\prime }\rangle +{\displaystyle \frac{1}{3}}\le {\displaystyle \frac{1}{2}}$; across sides ${\displaystyle \frac{2}{3}}\langle u,u^{\prime }\rangle -{\displaystyle \frac{1}{3}}\le {\displaystyle \frac{1}{3}}$.

Write $U=W{\omega }_1$, so $W{\omega }_6=-U$. The choice $(U,-U)$ is the $E_7$ root system: relative to the hyperplane spanned by an $E_6$ subsystem, the 126 normalized roots of $E_7$ split into the 72 roots of $E_6$ and two layers of 27 roots at squared height ${\displaystyle \frac{1}{3}}$, whose projections point in the directions of U and of $-U$ (after possibly replacing U by $-U$; verified exactly in Appendix A). The choice $(-U,U)$ is its image under the reflection in H. The choices $(U,U)$ and $(-U,-U)$ are exchanged by $-1$, which preserves $E_6$. The configuration $(U,U)$ is not antipodal, since the antipode of $\sqrt{2/3}u+{\displaystyle \frac{1}{\sqrt{3}}}n$ would be $\sqrt{2/3}(-u)-{\displaystyle \frac{1}{\sqrt{3}}}n$, and $-u\notin U$; as $E_7$ is antipodal, the two configurations are not isometric. This gives exactly two isometry classes. The configuration $(U,U)$ is invariant under $W\left(E_6\right)$ acting on H and under the reflection in H.    □

The two equality cases can also be distinguished by their deep graphs, whose edges are the pairs with inner product strictly less than $-1/2$.

Table 3. The two 126-point extensions of exact $E_6$. Here U is one of the normalized 27-element minuscule orbits. All counts are checked from exact coordinates.

	${\mathit{E}}_7$	${\mathit{C}}_{7--126\mathit{c}}$
Upper and lower hole sets	$(U,-U)$	$(U,U)$
Distinct occupied holes	54	27
Antipodal	yes	no
Deep edges	63	306
Deep-graph degrees	1 at every vertex	1 at 72 vertices; 10 at 54

Table 3. The two 126-point extensions of exact $E_6$. Here U is one of the normalized 27-element minuscule orbits. All counts are checked from exact coordinates.

	${\mathit{E}}_7$	${\mathit{C}}_{7--126\mathit{c}}$
Upper and lower hole sets	$(U,-U)$	$(U,U)$
Distinct occupied holes	54	27
Antipodal	yes	no
Deep edges	63	306
Deep-graph degrees	1 at every vertex	1 at 72 vertices; 10 at 54

In particular, one hole need not hold only one point: the two points $\sqrt{2/3}u\pm n/\sqrt{3}$ over the same hole have inner product $1/3$. The second equality case leaves 27 hole directions unused, yet Theorem 3 precludes adding another point. The capacity is therefore controlled by the joint layer constraints, not by counting unoccupied hole directions.

7. Explicit Robustness of the Carrier Bounds

The exact section hypothesis can be relaxed for cardinality bounds. Let C be a finite unit configuration in ${\mathbb{R}}^m$ and let $X=\{x_1,\dots ,x_N\}\subset {\mathbb{R}}^{m+1}$, where $N\ge \left|C\right|$. Define the positional carrier distance

$${\Delta }_C\left(X\right)=\underset{U,\iota }{min}\underset{c\in C}{max}|x_{\iota \left(c\right)}-Uc|,$$

where $U:{\mathbb{R}}^m\to {\mathbb{R}}^{m+1}$ ranges over linear isometric embeddings and $\iota$ over injections from C into the labels of X. The minimum is attained. This measures Euclidean distances between unit vectors, not discrepancies between Gram entries. The matched points need not lie in a common hyperplane. Here $D_4=\{(\pm e_i\pm e_j)/\sqrt{2}:1\le i<j\le 4\}$.

Theorem 4 

(Explicit carrier neighborhoods). Let $X\subset {\mathbb{R}}^{m+1}$ be a kissing configuration with ${\Delta }_C\left(X\right)\le {\epsilon }_C$, for a carrier in the following table. Then $\left|X\right|$ is at most the indicated bound.

C	$m+1$	${\epsilon }_C$	Bound on $\left|X\right|$
$D_4$	5	$\sqrt{7/27}-1/2$	40
$D_5,L_5,Q_5,R_5$	6	$\sqrt{87/340}-1/2$	72
$E_6$	7	$\sqrt{113/448}-1/2$	126

In particular, the common radius $1/450$, and hence $1/500$, is sufficient in every row. The endpoints of the displayed intervals are included. No optimality of these radii is asserted.

Lemma 8 

(Perturbed slice bound). Suppose ${\Delta }_C\left(X\right)\le \epsilon$, put $a=1/2+\epsilon$ and $\rho =\rho \left(C\right)$, and assume $a^2<3{\rho }^2/4$. After removing the matched $\left|C\right|$ points, every residual point has height at least

$$b=\sqrt{1-a^2/{\rho }^2}>1/2$$

relative to the ideal hyperplane $U{\mathbb{R}}^m$. On each side, either there is just one point or the normalized projections form a code with pairwise inner products at most

$$s=1-{\displaystyle \frac{{\rho }^2}{2a^2}}.$$

Proof. 

For a residual point x and any matched point $\tilde{c}=x_{\iota \left(c\right)}$, the original code constraint and Cauchy–Schwarz give

$$\langle x,Uc\rangle \le \langle x,\tilde{c}\rangle +|Uc-\tilde{c}|\le 1/2+\epsilon =a.$$

Writing $x=p+zn$ relative to $U{\mathbb{R}}^m$ gives $\rho \left|p\right|\le a$, and hence $\left|z\right|\ge b$. If $p=0$, no other point can lie on its side. Otherwise the derivative calculation in Lemma 4 applies without change and bounds the projected product by $(1/2-b^2)/(1-b^2)=s$. The numerator still contains $1/2$, since the constraints between residual points have not been relaxed.    □

Lemma 9 

(Layer bounds including the threshold). For $m=4,5,6$, put respectively

$$(s_{*},L)=(1/28,8),(19/87,16),(29/113,27).$$

A set of unit vectors in ${\mathbb{R}}^m$ with pairwise inner products at most $s_{*}$ has at most L points.

Proof. 

Use the polynomials in the following table. Their decompositions have the form

$$f\left(t\right)=f_0+{\lambda }_{1}t+{\lambda }_2(t^2-1/m)+{\lambda }_3(t^3-3t/(m+2)).$$

m	$f\left(t\right)$	$f_0$	${\lambda }_1$	$({\lambda }_2,{\lambda }_3)$
4	$(t-s)(t+1)$	$1/4-s$	$1-s$	$(1,0)$
5	$(t-s){(t+3/5)}^2$	$(6-14s)/25$	$3/7+9/25-6s/5$	$(6/5-s,1)$
6	$(t-s){(t+1/2)}^2$	$(2-5s)/12$	$5/8-s$	$(1-s,1)$

At the respective $s=s_{*}$, all displayed coefficients except the unused cubic coefficient for $m=4$ are strictly positive. Also $f\left(t\right)\le 0$ for $-1\le t\le s_{*}$. Equations (1)–(2) therefore give

$$N\le {\displaystyle \frac{f\left(1\right)}{f_0}}=L+1.$$

It remains to exclude equality, and this is what allows the threshold itself. If $N=L+1$, equality in the kernel bound forces ${\sum }_i{u}_i=0$ and forces every off-diagonal inner product to be one of the two roots $s_{*}$ and $-r$, where $r=1,3/5,1/2$ respectively. For any fixed $u_i$, if k of the other L vectors have product $s_{*}$, then

$$k{s}_{*}-(L-k)r=-1,k={\displaystyle \frac{Lr-1}{s_{*}+r}}.$$

The three values are $196/29$, $3741/356$ and $2825/171$, none an integer. Thus $N=L+1$ is impossible, proving $N\le L$. A smaller permitted product only restricts the same class further.    □

Proof 

(Proof of Theorem 4). The covering constants are $\rho {\left(D_4\right)}^2=1/2$, $\rho {\left(C\right)}^2=2/5$ for the four five-dimensional carriers, and $\rho {\left(E_6\right)}^2=3/8$. Only the first needs an additional calculation. For a unit vector in ${\mathbb{R}}^4$, let $v\ge w$ be its two largest absolute coordinates and put $h=v+w$. Then

$$1\le {(h-w)}^2+3w^2\le h^2,0\le w\le h/2,$$

since the convex quadratic has value $h^2$ at both endpoints. The support function of normalized $D_4$ is $h/\sqrt{2}$, and a coordinate unit vector attains $1/\sqrt{2}$.

At the three asserted radii, the values of $a^2$ are $7/27$, $87/340$ and $113/448$. Lemma 8 gives respectively

$$b^2=13/27,49/136,55/168,$$

all greater than $1/4$, and $s=s_{*}$ from Lemma 9. Each residual side therefore has at most 8, 16 or 27 points. Adding the matched carrier gives $24+16=40$, $40+32=72$ and $72+54=126$. Finally ${(1/2+1/450)}^2={(113/225)}^2<113/448$, and $113/448<87/340<7/27$, proving the common-radius statement.    □

For comparison, at the original rational tolerance $\epsilon =1/500$ the direct polynomial bounds are as follows. They lie strictly below the next integers, so the equality argument is unnecessary at this smaller tolerance.

Table 4. Exact layer bounds at positional tolerance $1/500$. The middle row applies to all four five-dimensional carriers.

m	Projected product s	$\mathit{f}\left(1\right)/{\mathit{f}}_0$	Integer capacity
4	$501/63001$	$500000/60997<9$	8
5	$13001/63001$	$400000/24499<17$	16
6	$16126/63001$	$421875/15124<28$	27

Table 4. Exact layer bounds at positional tolerance $1/500$. The middle row applies to all four five-dimensional carriers.

m	Projected product s	$\mathit{f}\left(1\right)/{\mathit{f}}_0$	Integer capacity
4	$501/63001$	$500000/60997<9$	8
5	$13001/63001$	$400000/24499<17$	16
6	$16126/63001$	$421875/15124<28$	27

Corollary 3. 

For each $C\in \{Q_5,R_5\}$ there exists ${\eta }_C>0$ such that ${\Delta }_C\left(X\right)\le {\eta }_C$ implies $\left|X\right|\le 70$. No explicit value of ${\eta }_C$ is claimed.

Proof. 

Otherwise there are codes with at least 71 points and carrier distances tending to zero. Retain 71 points including the matched 40, label these consistently, and rotate the ideal carrier to a fixed embedding. Compactness of ${\left(S^5\right)}^{71}$ gives a limiting 71-point kissing code containing exact C; the separation constraints prevent point collisions. Since $\rho \left(C\right)>1/2$, there are no further points in the span of C, so this is a cross section. Theorem 1 gives a contradiction.    □

The explicit radius of Theorem 4 gives the bound 72 near $Q_5$ or $R_5$. The stronger bound 70 in Corollary 3 has a separate, presently nonquantitative radius.

8. Remarks

1.

The general form of the argument is Corollary 1: a cross section with covering constant $\rho$, ${\rho }^2>1/3$, limits a configuration to $\left|C\right|+2A_m(1-2{\rho }^2)$ points. When the layer bound has an equality case in which every point has a partner at inner product exactly $1-2{\rho }^2$, as in Lemmas 5 and 6, every point of an extremal side sits exactly over a deepest hole; without such an equality statement no conclusion about the positions is drawn. For $E_8\subset {\mathbb{R}}^9$ one has $\rho {\left(E_8\right)}^2={\displaystyle \frac{1}{2}}$, $1-2{\rho }^2=0$ and $A_8\left(0\right)=16$, which is [2].

2.

For $Q_5$ and $R_5$ the true maximum lies between 64 and 70. Points of a side need not lie over deep holes unless the side has 16 points, so improving 70 requires information about sides with 13 to 15 points at heights above b.

3.

Stability of the equality classification. Theorem 4 gives an explicit neighborhood with cardinality at most 126. Exact classification of the 126-point codes in a possibly smaller neighborhood is a different conclusion. Equip the space $\Omega$ of 126-point kissing configurations in ${\mathbb{R}}^7$ modulo orthogonal transformations and relabelling with the distance

$$d(X,Y)=\underset{Q\in O\left(7\right),\pi \in S_{126}}{min}\underset{i}{max}|x_i-Q{y}_{\pi \left(i\right)}|.$$

The labelled space is a closed subset of ${\left(S^6\right)}^{126}$, so $\Omega$ is compact. Fix an enumeration $c_1,\dots ,c_{72}$ of normalized $E_6$ and define

$$\Delta \left(X\right)=\underset{Q\in O\left(7\right),\iota }{min}\underset{1\le i\le 72}{max}|x_{\iota \left(i\right)}-Q{c}_i|,$$

where $\iota$ ranges over injections into the 126 labels. This function is continuous (indeed 1-Lipschitz with respect to d). Its zero set consists of precisely the two classes in Theorem 3; an exact $E_6$ subset is a full cross section because $\rho \left(E_6\right)>1/2$ excludes further points in its span. Therefore, for every $\eta >0$ there is $\delta >0$ such that $\Delta \left(X\right)<\delta$ implies $min\{d(X,E_7),d(X,C_{7--126c})\}<\eta$. Otherwise compactness would give a limit with $\Delta =0$ at distance at least $\eta$ from both classes. These two configurations are isolated modulo isometry by the infinitesimal rigidity results of [2]. For the non-root configuration the cited result is an exact rational linear-programming verification, as stated there; it is not inferred merely from the coordinate file. Choosing $\eta$ inside their isolation neighborhoods shows that some ${\delta }_0>0$ even forces exact isometry to one of them. No numerical value of ${\delta }_0$ is asserted; in particular it is not identified with the explicit cardinality radius in Theorem 4.

4.

Theorem 3 implies that any other 126-point kissing configuration in ${\mathbb{R}}^7$, in particular the distinct configurations $C_{7--126b}$ and $C_{7--126d}$ of [2], has no cross section isometric to $E_6$.