Six and ten labels

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Supports at equality showed that an admissible map on the lines of $\mathbb R^k$ with $N_k$ labels has a support complex whose facets have $k$ labels and form a cycle modulo two. This lesson turns that information into a contradiction for $k=4$, $N_4=10$, by finite combinatorics. For $k=3$ the triangles of a six-label complex form a rigid configuration, a six-label triangle system (Section 1). In a ten-label complex, the six labels outside each tetrahedron carry such a system, and moving from a tetrahedron to a neighbouring one transports these systems along transpositions of labels; the rigidity of six-label systems then forces the tetrahedra into a block pattern that cannot close up (Section 2). Section 3 assembles the counterexample to Borsuk's conjecture in dimension nine and in every higher dimension.

Notation is that of Supports at equality. A face with three labels is a triangle, one with four labels a tetrahedron. For a finite set $C$, $\binom C3$ is the set of its three-element subsets.

1. Six labels

Definition 1.1. A six-label triangle system on a set $C$ with six elements is a set $\mathcal T\subseteq\binom C3$ such that

for every $T\in\binom C3$, exactly one of $T$ and $C\setminus T$ belongs to $\mathcal T$;
every pair of elements of $C$ lies in exactly two members of $\mathcal T$.

The image of a six-label triangle system under a bijection $C\to C'$ is a six-label triangle system on $C'$. For $a\in C$, the link graph of $\mathcal T$ at $a$ has vertex set $C\setminus\{a\}$ and an edge $\{u,v\}$ whenever $\{a,u,v\}\in\mathcal T$.

Lemma 1.2. Let $f:\mathbb P(\mathbb R^3)\to\Delta^5$ be admissible. Then the triangles of its support complex form a six-label triangle system on $[6]$.

Proof. Here $m=6=N_3$. By Theorem 2.1 and Theorem 3.3 of Supports at equality, the facets are triangles and every pair of labels lies in a positive even number of triangles; by Proposition 1.1 there, every pair is a face, so it lies in at least two triangles. Two complementary triples $T$ and $[6]\setminus T$ cannot both be triangles: if $x$ and $y$ witness them, a line orthogonal to both $x$ and $y$ exists in $\mathbb R^3$, and by admissibility its support is disjoint from $S(x)\cup S(y)=[6]$, which is impossible. So there are at most ten triangles, one from each complementary pair. Counting pairs of labels in triangles, there are at least $2\cdot15/3=10$ triangles. Hence there are exactly ten, one from each complementary pair, and every pair lies in exactly two. $\square$

Lemma 1.3. Let $\mathcal T$ be a six-label triangle system on $C$.

Every link graph is a cycle of length five.
Every four-element subset of $C$ contains a member of $\mathcal T$, but not all four of its three-element subsets belong to $\mathcal T$.
If $a\ne b$ in $C$ and $R\subseteq C\setminus\{a,b\}$ has three elements, then the link graphs at $a$ and at $b$, restricted to $R$, are different. In particular, no transposition of two elements of $C$ maps $\mathcal T$ to itself.
The members of $\mathcal T$ contained in any five-element subset of $C$ determine $\mathcal T$.

Proof. (1) By condition 2, every vertex of a link graph has degree two, so the graph is a disjoint union of cycles, each of length at least three. On five vertices this leaves a single five-cycle.

(2) Let $A\subseteq C$ have four elements and $a\in A$. Three vertices of a five-cycle always include two adjacent ones (otherwise each of the three would be followed by a vertex outside the set, and the cycle would have at least six vertices). So the link at $a$ has an edge $\{u,v\}$ inside $A\setminus\{a\}$, and $\{a,u,v\}\in\mathcal T$. If all four triples of $A$ were in $\mathcal T$, the link at $a$ would contain a triangle on $A\setminus\{a\}$, which a five-cycle does not.

(3) Name the elements $C=\{a,b,c,d,e,g\}$ so that the two members of $\mathcal T$ containing $\{a,b\}$ are $\{a,b,c\}$ and $\{a,b,d\}$. In the five-cycle at $a$, the neighbours of $b$ are $c$ and $d$; deleting $b$ leaves a path through $c,d,e,g$ with ends $c$ and $d$, whose middle edge is $\{e,g\}$. Its two end edges form one of the matchings $\begin{gathered} M_1=\bigl\{\{c,e\},\{d,g\}\bigr\},\\ M_2=\bigl\{\{c,g\},\{d,e\}\bigr\}. \end{gathered}$ The same holds for the link at $b$, whose neighbours of $a$ are also $c$ and $d$. The two links use different matchings: if the link at $a$ uses $M_1$, then $\{a,c,e\},\{a,d,g\}\in\mathcal T$, so their complements $\{b,d,g\},\{b,c,e\}\notin\mathcal T$, and the link at $b$ contains neither $\{d,g\}$ nor $\{c,e\}$; it uses $M_2$. The restrictions of the two links to $\{c,d,e,g\}$ are the paths $\{e,g\}\cup M_1$ and $\{e,g\}\cup M_2$, which differ exactly in the four-cycle $M_1\cup M_2$. Any three vertices of a four-cycle contain one of its edges, so the restrictions to any three-element $R\subseteq\{c,d,e,g\}$ differ. A transposition $(a\ b)$ preserving $\mathcal T$ would map the link at $a$ onto the link at $b$ and fix $R$ pointwise, making the restrictions equal.

(4) Let $z$ be the element outside the five-element subset. A triple containing $z$ has its complement inside the subset, and by condition 1 its membership is the opposite of the complement's. $\square$

2. Ten labels

In this section $f:\mathbb P(\mathbb R^4)\to\Delta^9$ is admissible, and $K$ is its support complex. By Corollary 1.2 of Supports at equality every label is used, so $m=10=N_4$, and Theorems 2.1 and 3.3 there apply: $K$ is pure of dimension three (every face lies in a tetrahedron, and the facets are the tetrahedra), and every triangle lies in a positive even number of tetrahedra. For a tetrahedron $S$ put $C_S=[10]\setminus S$, a set of six labels, and let $\mathcal T_S$ be the set of triangles of $K$ contained in $C_S$.

Lemma 2.1 (Facet complements). For every tetrahedron $S$, $\mathcal T_S$ is a six-label triangle system on $C_S$. No two tetrahedra of $K$ are disjoint, and every four labels of $C_S$ contain a triangle of $K$.

Proof. Let $x$ witness $S$, so $S(x)=S$. Lines of $x^\perp\cong\mathbb R^3$ use only labels in $C_S$, and at least $N_3=6$ labels occur there (Corollary 1.2 there), so all six occur, and the restriction of $f$ is an admissible map $\mathbb P(x^\perp)\to\Delta_{C_S}\cong\Delta^5$. By Lemma 1.2 its triangles form a six-label triangle system $\mathcal T'\subseteq\mathcal T_S$. If some $T\in\mathcal T_S$ were not in $\mathcal T'$, then $C_S\setminus T\in\mathcal T'$, and $T$, $C_S\setminus T$ and $S$ would all be faces of $K$; a line orthogonal to witnesses of all three exists in $\mathbb R^4$, and its support would avoid $S\cup T\cup(C_S\setminus T)=[10]$, which is impossible. So $\mathcal T_S=\mathcal T'$. A tetrahedron disjoint from $S$ would be a four-element subset of $C_S$ all of whose triples are triangles, contradicting Lemma 1.3(2), which also gives the last claim. $\square$

Lemma 2.2 (Transport; two tetrahedra per triangle). (a) If $S$ and $S'=(S\setminus\{a\})\cup\{b\}$ are tetrahedra, then the transposition $(a\ b)$ maps $\mathcal T_S$ onto $\mathcal T_{S'}$.

(b) Every triangle of $K$ lies in exactly two tetrahedra.

Proof. (a) $C_S$ contains $b$, $C_{S'}$ contains $a$, and $Q=C_S\setminus\{b\}=C_{S'}\setminus\{a\}$ has five elements. The transposition maps $C_S$ onto $C_{S'}$, fixes $Q$ pointwise, and maps $\mathcal T_S$ to a six-label triangle system on $C_{S'}$ whose members inside $Q$ are those of $\mathcal T_S$, namely the triangles of $K$ inside $Q$. These are also the members of $\mathcal T_{S'}$ inside $Q$, so the two systems agree by Lemma 1.3(4).

(b) Suppose a triangle $T$ lies in three tetrahedra $T\cup\{a\}$, $T\cup\{b\}$, $T\cup\{c\}$, and put $R=[10]\setminus(T\cup\{a,b,c\})$, a set of four labels. By (a), the transpositions $(a\ b)$, $(b\ c)$, $(c\ a)$ carry $\mathcal T_{T\cup\{a\}}$ to $\mathcal T_{T\cup\{b\}}$, then to $\mathcal T_{T\cup\{c\}}$, then back to $\mathcal T_{T\cup\{a\}}$. The composite $(c\ a)(b\ c)(a\ b)$ fixes $R$ and exchanges $b$ and $c$, so the transposition $(b\ c)$ of $C_{T\cup\{a\}}=R\cup\{b,c\}$ preserves $\mathcal T_{T\cup\{a\}}$, contradicting Lemma 1.3(3). So at most two tetrahedra contain $T$, and the number is positive and even. $\square$

For an edge $D$ of $K$, the link of $D$ is the graph whose vertices are the labels $v\notin D$ with $D\cup\{v\}$ a triangle, and whose edges are the pairs $\{v,w\}$ with $D\cup\{v,w\}$ a tetrahedron.

Lemma 2.3 (Short links). Every connected component of the link of an edge is a cycle of length three or four.

Proof. If $D\cup\{v,w\}$ is a tetrahedron, $D\cup\{w\}$ is a triangle, so edges join vertices of the link. By Lemma 2.2(b) every vertex has degree two, so every component is a cycle. Suppose the link contains a path $v_0,v_1,v_2,v_3,v_4$ on five distinct vertices, and let $R$ be the three labels outside $D\cup\{v_0,\dots,v_4\}$. The tetrahedra $S_j=D\cup\{v_j,v_{j+1}\}$, $j=0,\dots,3$, satisfy $S_{j+1}=(S_j\setminus\{v_j\})\cup\{v_{j+2}\}$, so by Lemma 2.2(a) the composite $\pi=(v_2\ v_4)(v_1\ v_3)(v_0\ v_2)$ maps $\mathcal T_{S_0}$ onto $\mathcal T_{S_3}$. It fixes $R$ and sends $v_4$ to $v_2$. For $r,s\in R$, therefore, $\{v_4,r,s\}\in\mathcal T_{S_0}$ if and only if $\{v_2,r,s\}\in\mathcal T_{S_3}$. The triple $\{v_2,r,s\}$ lies in both $C_{S_0}=R\cup\{v_2,v_3,v_4\}$ and $C_{S_3}=R\cup\{v_0,v_1,v_2\}$, so its membership in either system is its membership in $K$. Hence the link graphs of $\mathcal T_{S_0}$ at $v_4$ and at $v_2$ agree on $R$, contradicting Lemma 1.3(3). A cycle of length at least five contains such a path. $\square$

Lemma 2.4 (Partner blocks). Let $S$ be a tetrahedron. For $v\in S$ let $p(v)$ be the label such that $(S\setminus\{v\})\cup\{p(v)\}$ is the second tetrahedron containing the triangle $S\setminus\{v\}$ (Lemma 2.2(b)). For each value $y$ of $p$ put $B_y=p^{-1}(y)\cup\{y\}$. Then:

(a) $p(v)\notin S$; the blocks $B_y$ are pairwise disjoint, have at least two elements, contain $S$ in their union, and $\sum_y(|B_y|-1)=4$;

(b) for every choice of one element $q_y\in B_y$ in each block, $\bigcup_y(B_y\setminus\{q_y\})$ is a tetrahedron;

Proof. (a) The second tetrahedron is not $S$, so $p(v)\ne v$, and $p(v)\notin S\setminus\{v\}$ because the union has four elements. Disjointness and the sizes are immediate, and $\sum_y|p^{-1}(y)|=|S|=4$.

Exchanges keep the blocks. Let $v\in S$, $a=p(v)$, $S'=(S\setminus\{v\})\cup\{a\}$, and let $p'$ be the partner map of $S'$. The triangle $S'\setminus\{a\}=S\setminus\{v\}$ lies in $S'$ and $S$, so $p'(a)=v$. Let $w\in S\setminus\{v\}$, $D=S\setminus\{v,w\}$ and $b=p(w)$. The link of the edge $D$ contains the edges $\{v,w\}$ (from $S$), $\{w,a\}$ (from $S'$) and $\{v,b\}$ (from $(S\setminus\{w\})\cup\{b\}$). If $a=b$, these three edges form a triangle, which is a whole component because degrees are two; so the tetrahedra containing $D\cup\{a\}=S'\setminus\{w\}$ are $S'$ and $D\cup\{a,v\}$, and $p'(w)=v$. If $a\ne b$, the edges form the path $a,w,v,b$ on four distinct vertices, and by Lemma 2.3 its component is the four-cycle closed by the edge $\{a,b\}$; so $D\cup\{a,b\}$ is a tetrahedron containing $S'\setminus\{w\}$, and $p'(w)=b=p(w)$. Thus $p'(w)=v$ if $p(w)=a$ and $p'(w)=p(w)$ otherwise. The block of $S'$ indexed by $v$ is $\{a\}\cup(p^{-1}(a)\setminus\{v\})\cup\{v\}=B_a$, and every other block is unchanged.

(b) $S$ contains $p^{-1}(y)=B_y\setminus\{y\}$ and not $y$, so $S=\bigcup_y(B_y\setminus\{y\})$: it omits one element from each block. Every tetrahedron $T$ obtained from $S$ by exchanges has the same blocks and omits exactly one element $q$ of each block $B$, its index there. To change the omitted element of $B$ from $q$ to $u\in B\setminus\{q\}$, note that $u\in T$ and the partner of $u$ for $T$ is $q$, so $(T\setminus\{u\})\cup\{q\}$ is a tetrahedron; it omits $u$ from $B$ and the same elements as $T$ from the other blocks. Changing one block at a time reaches every choice.

(c) If a tetrahedron $T$ met every block in at most one element, choose $q_y$ to be that element when it exists and arbitrary otherwise. The tetrahedron $\bigcup_y(B_y\setminus\{q_y\})$ of (b) is disjoint from $T$, contradicting Lemma 2.1. $\square$

Theorem 2.5. There is no admissible map $\mathbb P(\mathbb R^4)\to\Delta^9$.

Proof. Suppose $f$ is one, and fix a tetrahedron $S$ with its blocks. Let $t$ be the number of blocks and $R$ the set of labels outside all blocks. The blocks contain $\sum_y|B_y|=4+t$ labels, so $1\le t\le4$ and $|R|=6-t$.

If $t=4$, each block has two elements and $S$ contains exactly one element of each, contradicting Lemma 2.4(c). So $t\le3$ and $|R|\ge3$.

No triangle lies in $R$: it would lie in a tetrahedron, whose fourth label is the only one that can belong to a block, against Lemma 2.4(c). But $R$ is disjoint from $S$, so $R\subseteq C_S$, and every four labels of $C_S$ contain a triangle (Lemma 2.1). Hence $|R|\le3$, so $t=3$, $|R|=3$, and the block sizes are $3,2,2$.

The six labels of $C_S$ are those of $R$ and the three labels $y$ omitted by $S$, one in each block. Choose distinct $r,s\in R$. In the six-label system $\mathcal T_S$, the pair $\{r,s\}$ lies in two triangles $\{r,s,z_1\}$, $\{r,s,z_2\}$ with $z_1\ne z_2$. Since $R$ contains no triangle, $z_1$ and $z_2$ are omitted labels of two different blocks, and as only one block has three elements, one of them, say $z$, lies in a block $B$ with two elements. A tetrahedron containing the triangle $\{r,s,z\}$ must contain two elements of some block (Lemma 2.4(c)); as $r,s$ lie in no block, its fourth label is the other element of $B$. So at most one tetrahedron contains $\{r,s,z\}$, contradicting Lemma 2.2(b). $\square$

3. The counterexample

Theorem 3.1 (OpenAI 2026). The set $X_4=\{uu^{\mathsf T}:u\in\mathbb R^4,\|u\|=1\}$, in the nine-dimensional Euclidean space of symmetric $4\times4$ matrices of trace one with the Frobenius distance, is compact, has diameter $\sqrt2$, and cannot be covered by ten subsets of diameter less than $\sqrt2$. So Borsuk's conjecture fails in dimension nine.

Proof. Compactness and the diameter are Lemma 1.1 of Diameter covers and admissible maps. A cover by ten subsets of smaller diameter would give an admissible map $\mathbb P(\mathbb R^4)\to\Delta^9$ (Lemma 2.3 there), which does not exist by Theorem 2.5. $\square$

Corollary 3.2. For every $d\ge9$ there is a compact set of diameter $\sqrt2$ in $\mathbb R^d$ that cannot be covered by $d+1$ subsets of smaller diameter.

Proof. For $d=9$ this is Theorem 3.1. Let $d=9+s$ with $s\ge1$. Translate $X_4$ to $Y=\{P-\frac14I:P\in X_4\}$ in the nine-dimensional space of symmetric matrices of trace zero; then $\|y\|_F^2=1-\frac12+\frac14=\frac34$ for $y\in Y$. The matrix $I_s+\frac14\mathbf 1\mathbf 1^{\mathsf T}$ is positive definite, so it is the Gram matrix of vectors $v_1,\dots,v_s\in\mathbb R^s$; they have $|v_i|^2=\frac54$ and $|v_i-v_j|^2=\frac54+\frac54-\frac12=2$ for $i\ne j$. In $\mathbb R^9\oplus\mathbb R^s$, the compact set $Z=(Y\times\{0\})\cup\{(0,v_i):1\le i\le s\}$ has diameter $\sqrt2$: each new point is at distance $\sqrt{\frac34+\frac54}=\sqrt2$ from every point of $Y\times\{0\}$ and from every other new point. A subset of $Z$ of diameter less than $\sqrt2$ containing a new point contains no other point of $Z$. So a cover of $Z$ by $d+1=10+s$ such subsets leaves at most ten of them to cover $Y\times\{0\}$, contradicting Theorem 3.1. $\square$

The proof does not determine the smallest dimension in which Borsuk's conjecture fails, which lies between four and nine, nor the exact number of smaller-diameter pieces needed for $X_4$.

4. Exercises

4.1. Show that the ten triples $\{1,2,3\}$, $\{1,3,4\}$, $\{1,4,5\}$, $\{1,5,6\}$, $\{1,2,6\}$, $\{2,3,5\}$, $\{3,4,6\}$, $\{2,4,5\}$, $\{3,5,6\}$, $\{2,4,6\}$ form a six-label triangle system on $[6]$ (they are the triangles of the six-vertex triangulation of the projective plane), and find the link graph at $1$.

4.2. Show that there are exactly twelve six-label triangle systems on $[6]$. (Show that the system is determined by the link at $1$, a five-cycle on $\{2,\dots,6\}$, and that each of the twelve five-cycles occurs.)

4.3. In Lemma 2.4, describe the blocks when all four partners coincide ($t=1$). Why does the proof of Theorem 2.5 not need to treat this case separately?

4.4. Check the distances in Corollary 3.2 for $s=1$ directly: give $v_1$ and verify that $Z\subset\mathbb R^{10}$ has diameter $\sqrt2$.

5. Solutions

4.1. The complements of the listed triples are $\{4,5,6\}$, $\{2,5,6\}$, $\{2,3,6\}$, $\{2,3,4\}$, $\{3,4,5\}$, $\{1,4,6\}$, $\{1,2,5\}$, $\{1,3,6\}$, $\{1,2,4\}$, $\{1,3,5\}$, none of which is listed, and together the twenty triples are all of $\binom{[6]}3$. Each pair occurs in exactly two listed triples; for instance $\{1,2\}$ in $\{1,2,3\}$ and $\{1,2,6\}$. The link at $1$ has edges $\{2,3\},\{3,4\},\{4,5\},\{5,6\},\{6,2\}$: the five-cycle $2,3,4,5,6$.

4.2. The members containing $1$ are given by the link at $1$, and every other triple is the complement of a triple containing $1$, so by condition 1 the link determines the system. There are $4!/2=12$ five-cycles on the five labels $2,\dots,6$. Relabelling the system of 4.1 by a permutation of $\{2,\dots,6\}$ produces every five-cycle as the link at $1$, and the resulting system satisfies conditions 1 and 2 because relabelling preserves them.

4.3. If $t=1$, all four elements of $S$ have the same partner $y$, and $B_y=S\cup\{y\}$ has five elements. Then $|R|=5\ge4$, and the proof of Theorem 2.5 rules out $|R|\ge4$ in general: a set of four labels of $R\subseteq C_S$ contains a triangle, while $R$ contains none.

4.4. $v_1\in\mathbb R$ with $v_1^2=\frac54$, say $v_1=\frac{\sqrt5}2$. For $y\in Y$, $|(y,0)-(0,v_1)|^2=\frac34+\frac54=2$, and distances within $Y\times\{0\}$ are at most $\sqrt2$ by Lemma 1.1 of Diameter covers and admissible maps.

References

[OpenAI-B9] OpenAI, A nine-dimensional counterexample to Borsuk's covering assertion, OpenAI Math Release preprint, 23 September 2026. https://github.com/openai/math/blob/main/preprints/A-nine-dimensional-counterexample-to-Borsuks-covering-assertion-September-23-2026/paper.pdf
[Bon] A. V. Bondarenko, On Borsuk's conjecture for two-distance sets, Discrete & Computational Geometry 51 (2014). https://arxiv.org/abs/1305.2584
[Kal] G. Kalai, Some old and new problems in combinatorial geometry I: around Borsuk's problem, Surveys in Combinatorics 2015. https://arxiv.org/abs/1505.04952