CAT(0) Actions of Artin Groups

CAT$0$ Actions of Artin Groups

OpenAI's claim

OpenAI presents a mathematical result relevant to this question in the selected manuscript. A specific Artin group on 116 generators is claimed to have no proper cocompact isometric action on a proper CAT$0$ space.

Mathematical statement of the claimed result

Result 1.

For an Artin matrix $M$ on 116 generators with off-diagonal entries only in $\{2,3,\infty\}$, the constructed Artin group $A_M$ is claimed to have no action that is simultaneously isometric, proper, and cocompact on any nonempty proper CAT$0$ metric space. In quantified form, \[ \nexists X,\ A_M\curvearrowright X:\quad X\text{ proper CAT}(0),\quad\text{action proper and cocompact}. \] The metric space is required to be proper, and the exclusion is independent of dimension.

Outline of the argument

Stable translation lengths and centralizer characters impose quantitative braid constraints. Six asymptotic sectors and labeled triangular diagrams yield a local winding count incompatible with those constraints; a finite Artin matrix realizes the contradictory configuration.

Formal verification

For family 254, the scope includes finite-rank Salvetti contractibility, the 116-generator Artin CAT$0$ counterexample, and arbitrary parabolic intersections with closure consequences. Each has its own Comparator statement. The listed Comparator files are ArtinCAT0.lean, ArtinParabolicIntersections.lean, HarmonicArtin.lean. Named declarations include OAI.ArtinCAT0.main, OAI.HarmonicArtin.salvetti_cover_contractible. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

Sources

• OpenAI, An Artin group with no geometric CAT$0$ action, 2026-09-23; citation key OAI:An-Artin-group-with-no-geometric-CAT-0-action-September-23-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main.
• Family 254: Lean scope.
• Comparator ArtinCAT0.json.
• Comparator ArtinCAT0.lean.
• Comparator ArtinParabolicIntersections.json.
• Comparator ArtinParabolicIntersections.lean.
• Comparator HarmonicArtin.json.
• Comparator HarmonicArtin.lean.
• OpenAI catalogue entry.

CAT$0$ Actions of Artin Groups

Definitions and mathematical statement

Let $S$ be a finite set and $(m_{st})$ a Coxeter matrix with $m_{ss}=1$ and $m_{st}\in\{2,3,\ldots,\infty\}$ for $s\ne t$. Its Artin group is \[ A=\langle S\mid \underbrace{sts\cdots}_{m_{st}\text{ letters}}= \underbrace{tst\cdots}_{m_{st}\text{ letters}} \text{ whenever }m_{st}<\infty\rangle. \] Finite matrix means finitely many generators, not that the associated Coxeter group is finite. A geodesic metric space is CAT$0$ when distances within every geodesic triangle are at most the corresponding Euclidean comparison distances.

Does every such Artin group act properly and cocompactly by isometries on a CAT$0$ space, in Haettel's Conjecture 1.1 convention? Properness requires only finitely many group translates of a compact set to meet that set, and cocompactness requires compact quotient. No particular standard complex, cube-complex structure, or faithful linear representation is prescribed as the construction.

Formulation and concept references: [S1].

Short English statement

Does every finitely generated Artin group admit a proper cocompact isometric action on some CAT$0$ space?

Research attempt

Scope, source audit, and strategy (27 September 2026).

The quantifier is over every finite generating set and every permitted matrix; the space and its dimension may depend on that matrix. The empty generating set gives the trivial group acting on a point. Finite rank does not mean spherical type. Throughout, trees have unit-length edges and products carry the Euclidean product metric.

Haettel's Conjecture 1.1 in [S1] is exactly the geometric-action assertion above. His Theorem 1.4 supplies compact locally CAT$0$ models when every finite label is at least five. This is a proved subcase, not an argument covering the labels two, three, and four simultaneously. His Section 2 uses dihedral groups as building blocks; the calculation below independently spells out their tree-times-line action, without claiming novelty for that known subcase. The accessible sources checked here do not establish the all-matrix conjecture. A search for recent work was a scope check, not an exhaustive certification that no further special cases exist.

The main attack is to start with explicit rank-two models, then ask what compatibility would allow them to assemble. A second benchmark is the right-angled cubical construction, whose local curvature can be checked combinatorially. The counterexample track tests the simplest attempted assembly: putting a Euclidean polygon on each defining relation. It fails already for a group which does have a CAT$0$ action. Consequently a failed presentation complex cannot itself disprove the group conjecture.

An action lemma with all properness hypotheses exposed.

Lemma 500.1. Suppose a group $G$ acts without inversions on a locally finite tree $T$, with finite quotient. Suppose there is a homomorphism $h:G\to\mathbb R$ whose restriction to every vertex stabilizer is injective with discrete image. Finally suppose there is $z\in G$ fixing $T$ pointwise with $h(z)\ne0$. Then \[ g\cdot(p,r)=(gp,r+h(g)) \] is a proper cocompact isometric action on the CAT$0$ space $T\times\mathbb R$.

Proof. The action law follows from additivity of $h$. A metric tree is CAT$0$, and the squared distances in a product add; the CAT$0$ comparison inequality therefore passes to its Euclidean product with a line. The product is proper as a metric space because the tree is locally finite.

Fix a vertex $v$. For any bound $R$, there are finitely many vertices $w$ with $d(v,w)\le R$. For each such $w$ in the orbit choose $g_wv=w$. Every $g$ satisfying $gv=w$ is $g_wk$, with $k\in G_v$. The further condition $|h(g)|\le R$ bounds $h(k)$ in an interval of finite length. Discreteness and injectivity on $G_v$ leave finitely many $k$. Thus there are finitely many group elements moving $(v,0)$ by at most any fixed distance. If $gK\cap K\ne\varnothing$ for a compact $K$, choose a ball of radius $R_K$ about $(v,0)$ containing $K$; then $d((v,0),g(v,0))\le2R_K$. This proves the stated compact-set properness.

A finite union $Q$ of closed edges and vertices meets every $G$-orbit in $T$. Move $p$ into $Q$ using an element $g$, and then use an integral power of $z$ to move the second coordinate into $[0,|h(z)|]$. The latter operation leaves the first coordinate unchanged. Hence the compact set $Q\times[0,|h(z)|]$ meets every orbit. This proves cocompactness. $\square$

The discrete-image condition cannot be replaced merely by a nonzero homomorphism somewhere in $G$. Infinite vertex stabilizers must be controlled where they occur. Likewise, the action on $T$ alone is not proper in the examples that follow; adjoining the line removes that exact defect.

Odd dihedral labels: a full algebra-to-geometry calculation.

Let $m=2k+1\ge3$, put $u=st$, and write \[ G_m=\langle s,t\mid (st)^ks=(ts)^kt\rangle. \] Since $ts=s^{-1}us$ and $t=s^{-1}u$, the defining relation becomes $u^ks=s^{-1}u^{k+1}$, equivalently $su^ks=u^{k+1}$. Putting $v=u^ks$ gives \[ v^2=u^k(su^ks)=u^{2k+1}=u^m. \] Conversely, from $\langle u,v\mid v^2=u^m\rangle$ define $s=u^{-k}v$ and $t=v^{-1}u^{k+1}$. Then $st=u$ and $su^ks=u^{-k}v^2=u^{k+1}$, recovering the original relation. These substitutions are inverse, so they prove the presentation isomorphism rather than only a quotient map: \begin{equation}\label{wsref-0-72-61-35-30-30-3a-6f-64-64} G_m\cong\langle u,v\mid u^m=v^2\rangle =\langle u\rangle*_{\langle u^m\rangle=\langle v^2\rangle} \langle v\rangle. \end{equation}

The Bass--Serre tree of this amalgam has vertex stabilizers conjugate to $\langle u\rangle$ and $\langle v\rangle$, and valences $m$ and $2$. It has one edge in the quotient. These facts follow directly from its coset description: vertices are $G_m/\langle u\rangle$ and $G_m/\langle v\rangle$, edges are $G_m/\langle u^m\rangle$, and an edge $g\langle u^m\rangle$ joins the two vertices represented by $g$. The normal form for an amalgam says a reduced alternating product of non-edge-group factors is nontrivial; it rules out a nonbacktracking cycle in this connected coset graph, so the graph is a tree.

The element $z=u^m=v^2$ commutes with both generators and belongs to the edge group, hence fixes every vertex and edge. The homomorphism \[ h(u)=2,\qquad h(v)=m \] is well-defined because both sides of the relation have height $2m$. Its restrictions to the vertex stabilizers are injective discrete maps, also after conjugation, since $h(gag^{-1})=h(a)$. Moreover $h(z)=2m$. Lemma 500.1 now proves a proper cocompact action. Notice that the recovered original generators have $h(s)=h(t)=1$, as required for consistency with the homogeneous Artin relation.

Even dihedral labels, including the commuting boundary case.

For $m=2k\ge2$, again put $u=st$. The equality $(st)^k=(ts)^k$ is precisely $u^k=s^{-1}u^ks$, so \begin{equation}\label{wsref-0-72-61-35-30-30-3a-65-76-65-6e} G_{2k}\cong\langle u,s\mid su^ks^{-1}=u^k\rangle. \end{equation} The inverse substitution is $t=s^{-1}u$. This is an HNN extension of $\langle u\rangle\cong\mathbb Z$ with both associated subgroups $\langle u^k\rangle$, identified by the identity. Its coset tree has vertex stabilizers conjugate to $\langle u\rangle$, valence $2k$, and a single loop in the quotient. The tree property is the HNN normal-form statement: a reduced word containing a stable letter cannot represent the identity unless a cancellable associated-subgroup segment occurs.

Here $z=u^k$ is central and fixes the entire tree. Set $h(u)=2$ and $h(s)=1$. Again the relation preserves height, $h$ is injective discrete on each vertex stabilizer, and $h(z)=2k$. Lemma 500.1 applies. When $k=1$ the tree is a line and this construction produces the expected Euclidean-plane action of $G_2=\mathbb Z^2$; it is not necessary to exclude this degenerate valence-two case. For $m=\infty$ the group is free on $s,t$ and its ordinary Cayley tree already supplies a proper cocompact action. A one-generator Artin group acts on $\mathbb R$ by integral translations. Thus all groups of rank at most two have been covered.

The underlying Bass--Serre normal forms are the only group-action input in these constructions; the corresponding dihedral presentations and geometry are also recorded in [S1, Section 2]. They are not being used to assert that arbitrary unions of these models are nonpositively curved.

A separate all-rank benchmark: right-angled matrices.

Assume every off-diagonal entry is $2$ or $\infty$. Let $\Gamma$ be the graph of the entries equal to $2$. Construct a finite cubical complex $X$ with one vertex, one oriented circle for each generator, and a coordinate $q$-torus for each $q$-clique of $\Gamma$, glued on the corresponding coordinate subtori. Its two-cells impose exactly the commuting relations; cells of dimension at least three do not change the fundamental group. Consequently $\pi_1X$ is the desired Artin group.

Here is the full combinatorial curvature check. A link vertex is a signed generator $s^+$ or $s^-$. A collection spans a simplex precisely when its underlying generators are distinct and form a clique of $\Gamma$. Opposite signs of the same generator are not adjacent. Thus a pairwise adjacent collection has distinct underlying generators which commute pairwise, so the required torus exists and fills its link simplex. Every clique in the link is therefore filled: the link is flag.

The cubical link criterion and the CAT$0$ globalization theorem imply that the universal cover $\widetilde X$ is CAT$0$; the exact statement needed is that a finite Euclidean cube complex with flag links has a CAT$0$ universal cover in the lifted length metric [E1, Section 2]. Its finite cell structure ensures local finiteness and completeness of that lifted metric. Deck transformations act properly, and the quotient $X$ is compact. This rederives the known right-angled subcase with no restriction on rank.

The importance of the higher cubes is visible for three mutually commuting generators. If only the three commutator squares are retained, the link has an unfilled triangle whose edge lengths are $\pi/2$; the loop has length $3\pi/2<2\pi$. Attaching the three-torus provides the missing simplex. Having each rank-two square locally suitable is therefore insufficient even in this particularly simple setting.

A concrete failed construction for a noncommuting relation.

Take the presentation $\langle s,t\mid sts=tst\rangle$. Its standard two-complex has a hexagonal cell with attaching word \[ stst^{-1}s^{-1}t^{-1}. \] If that cell is a regular Euclidean hexagon, every corner contributes a link edge of length $2\pi/3$. A consecutive pair of letters $a,b$ contributes an edge joining the outgoing directions $a^{-1}$ and $b$. The corner $(s,t)$ and the distinct corner $(t^{-1},s^{-1})$ both join $s^-$ to $t^+$. They are different link edges with the same endpoints. Their union is a locally geodesic two-edge loop of length $4\pi/3$. The link fails the CAT$1$ condition, and this polygonal metric is not locally CAT$0$. This is an exact obstruction, not an impression based on a picture.

The same group is $G_3$, for which the tree-times-line construction above works. Thus replacing the group-existence question by the assertion that its most obvious presentation metric is CAT$0$ would change the problem and yield a false stronger assertion. Subdividing the hexagon without changing the metric does not remove this short link loop.

Where assembly stops and what would suffice next.

The Artin presentation is assembled from two-generator relations, but a geometric amalgamation requires more than matching those relations. One would need compact local models with compatible lengths for each shared generator loop and a proof that every link in the assembled space is CAT$1$, including loops that traverse three or more different rank-two blocks. Convex-gluing theorems apply to suitably embedded convex subsets in CAT$0$ spaces; an arbitrary immersed loop in a quotient is not such a subset. The preceding two examples show specific ways in which this distinction matters. Neither abstract presentation gluing nor the existence of a rank-two action verifies these hypotheses.

A precise sufficient next lemma for this route would be: for every finite Artin matrix one can choose finitely many compact piecewise Euclidean blocks, attach them along the generator data to obtain the stated Artin fundamental group, and verify the CAT$1$ inequality for all resulting links. The unsupplied part is the last compatibility assertion, particularly when small and large labels interact. It is not proved by the rank-two calculations. The XXL construction in [S1] gives one carefully controlled successful assembly, which explains why its angle hypotheses cannot simply be dropped.

Verification and outcome.

The algebraic substitutions above were checked in both directions; the height of the central translation is nonzero in each finite-label case. Properness was checked on compact sets, not inferred solely from a trivial point stabilizer. The $m=2$, $m=3$, $m=\infty$, rank-one, and empty-rank cases were separated. A small exact combinatorial script in the local research record enumerates the six corners of the braid hexagon and confirms the duplicated link edges; it also checks the signed-link flag description on all 1,100 graphs with at most five vertices. These finite tests are diagnostics of the constructions, not evidence for all matrices.

Outcome: UNRESOLVED, with complete reconstructions of known subcases. The present work proves the explicit rank-two geometric actions and the right-angled construction, and refutes two tempting shortcuts for gluing. It supplies neither an all-matrix construction nor a non-CAT$0$ Artin group. The first unresolved step is compatible local curvature control when the rank-two pieces are assembled beyond these verified families. No novelty or resolution of the catalogue conjecture is claimed.

Sources

• S1 T. Haettel, XXL type Artin groups are CAT$0$ and acylindrically hyperbolic, AIF 72 $2022$, 2541–2555. https://www.numdam.org/item/10.5802/aif.3524.pdf. Formulation source.
• E1 Ruth Charney, An introduction to right-angled Artin groups, Geometriae Dedicata 125 $2007$, 141--158, Section 2.4 (link criterion), Section 2.6 and Theorem 2.6 (Salvetti complex). https://arxiv.org/pdf/math/0610668. Author manuscript inspected 27 September 2026.