Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 134 (Cubulating Spaces and Groups)

For a finite graph $\Gamma$, its Right-angled Artin group isIts Salvetti complex $S_{\Gamma }$ is the one-vertex cube complex with an oriented loop labelled $v$ for each vertex of $\Gamma$, a square torus for every edge, and more generally one cubulated $k$-torus for every $k$-clique, attached compatibly along coordinate subtori. Thus ${\pi }_1(S_{\Gamma })\cong A_{\Gamma }$. The vertex link of a Salvetti complex is flag, so $S_{\Gamma }$ is a nonpositively curved cube complex.

If $\Gamma ={\Gamma }_1\bigsqcup {\Gamma }_2$ is a nontrivial disjoint union of graphs, no defining relation mixes the two vertex sets, and henceOn the topological side, every clique lies in one component, soa one-point union of Salvetti complexes.

If $\Gamma ={\Gamma }_1\ast {\Gamma }_2$ is a nontrivial join of graphs, every generator from the first part commutes with every generator from the second. ThereforeEvery clique of the join is the union of a clique in each factor, which gives the cubical identity

For each $v\in V(\Gamma )$, the vertex link of a Salvetti complex has two vertices $v^{+}$ and $v^{-}$. If $v$ has degree $d_{\Gamma }(v)$, then each of $v^{+}$ and $v^{-}$ is adjacent to both signed vertices $w^{+},w^{-}$ for every neighbour $w$, and hence has link degree $2d_{\Gamma }(v)$.

Suppose the whole link is a cycle. Every link vertex has degree two, so every vertex of $\Gamma$ has degree one. The link is connected, which forces $\Gamma$ to be connected. A connected one-regular graph consists of one edge. Conversely, if $\Gamma$ is one edge, then $S_{\Gamma }$ is the square torus and its unique vertex has link the four-cycleThus the vertex link is a cycle exactly when $\Gamma$ is an edge.

The hyperplane of a Salvetti complex dual to the loop labelled $v$ is itself the Salvetti complex of the induced subgraph on the neighbours of $v$. If this hyperplane were homeomorphic to a closed surface, its vertex link would be a cycle. Part (c), applied to that induced graph, says that the graph must be one edge. Its Salvetti complex is therefore the two-dimensional torus, a surface of genus one. It cannot be a closed orientable surface of genus two.

The inclusion of a full subgraph ${\Gamma }^{\prime }\subseteq \Gamma$ sends each generator of $A_{{\Gamma }^{\prime }}$ to the equally named generator of $A_{\Gamma }$. Defineby fixing the generators in ${\Gamma }^{\prime }$ and sending every other generator to the identity. Every commutator relation of $A_{\Gamma }$ maps to a valid relation, so $r$ is a group homomorphism. Its composite with the natural map $A_{{\Gamma }^{\prime }}\to A_{\Gamma }$ is the identity. The natural map has a left inverse and is therefore injective, proving that $A_{{\Gamma }^{\prime }}$ is isomorphic to a subgroup of $A_{\Gamma }$.

A group is cubulated when it admits a cubulation of a group, namely a metrically proper group action by cubical automorphisms on a CAT(0) cube complex. It is cocompactly cubulated when that action is also a cocompact group action. Thus a cocompact cubulation is a proper, cocompact cubical action.

Fix $x\in S$ and let ${\sigma }_x$ be its principal vertex of a dual cube complex. The combinatorial distance in the dual complex is the wall metric:Consequently the wall-metric properness criterion says that the action on $C$ is metrically proper provided this number tends to infinity as $g$ leaves every finite subset of $G$. Equivalently, for every $R$, only finitely many $g$ separate $x$ from $gx$ by at most $R$ walls.

The necessary and sufficient condition is the transverse-wall cocompactness criterion: there must be finitely many $G$-orbits of finite transverse wall collections. Equivalently, their cardinalities must be uniformly bounded and, for every cardinality, there must be only finitely many orbits.

Indeed, an $n$-cube of the dual cube complex of a wallspace is dual to an $n$-element collection of pairwise crossing walls, and this correspondence respects the $G$-action and passage to faces. If there are finitely many orbits of transverse collections, there are finitely many cube orbits, so the quotient $G\backslash C$ is a finite cube complex and is compact. Conversely, if the action is cocompact, a compact fundamental set meets only finitely many open unit cubes. Hence there are finitely many cube orbits and therefore finitely many orbits of their dual transverse wall collections.

The walls occur in three families of parallel lines, and a choice of halfspaces in one family is determined by an integer cut. Lines from different families cross, so the three cuts can be chosen independently. It follows directly from the dual cube complex of a wallspace construction thatand that $C$ is the standard cubulation of ${\mathbb{R}}^3$.

Choose affine coordinates $x_1,x_2,x_3$ for the three wall families so that the original Euclidean plane is $x_1+x_2+x_3=0$ and the walls are the integer level sets. On the cut coordinates $(n_1,n_2,n_3)\in {\mathbb{Z}}^3$, the translation subgroup of the Affine Coxeter group $W$ adds vectors $(m_1,m_2,m_3)$ satisfying $m_1+m_2+m_3=0$, while its finite reflection subgroup permutes the three coordinates. Thereforeis constant on every $W$-orbit. Since $h(n,0,0)=n$ is unbounded, there are infinitely many vertex orbits. A cocompact cubical action on this locally finite cube complex would have only finitely many cube orbits, so the action of $W$ on $C$ is not cocompact.

A group $G$ is residually finite if, for every $1\ne g\in G$, there are a finite group $Q$ and a homomorphism $\varphi :G\to Q$ such that $\varphi (g)\ne 1$.

Let $F(S)$ be a free group and let $1\ne w\in F(S)$ be a reduced word. In the rose with one oriented loop for each $s\in S$, the word $w$ determines a nonclosed reduced path from a chosen vertex in the universal covering tree. Its finite image path can be completed to a finite covering graph of the rose: for each label, pair the still unmatched incoming and outgoing edge germs, adding finitely many vertices if necessary. The lift of $w$ remains nonclosed in this finite cover.

Let $H\le F(S)$ be the finite-index subgroup represented by this based cover. Then $w\notin H$. The action of $F(S)$ on the finite set of right cosets $H\backslash F(S)$ gives a homomorphism to a finite symmetric group, and $w$ does not fix the coset $H$. Thus $w$ survives in a finite quotient. Since $w$ was arbitrary, every free group is residually finite.

Let $X$ be a subcomplex of a product of graphs $A\times B$. Orient every edge of each factor. An edge of $X$ is horizontal or vertical according to its factor, and this type is preserved across opposite sides of every square.

A hyperplane of a cube complex of horizontal type retains one fixed edge of $A$ while moving through edges of $B$; the analogous statement holds vertically. The factor orientation makes every hyperplane two-sided. A square has one horizontal and one vertical direction, so no hyperplane self-intersects. At a vertex there is at most one incident edge with a fixed factor edge and orientation, so no hyperplane self-osculates. Finally, a horizontal hyperplane and a vertical hyperplane can cross only in the unique product square determined by their two factor edges. If that square belongs to $X$, it fills every corner at which those two dual edges meet; if it does not, the hyperplanes never cross. Thus no pair interosculates. All four hyperplane pathologies are absent, so $X$ is a special cube complex.

Call the two horizontal edge classes indicated by one and two arrowheads $a$ and $b$. All vertices in the displayed quotient are identified. In the third displayed square, an $a$-edge and a $b$-edge are opposite, so they are dual to the same hyperplane of a cube complex $H$. At the unique vertex, the distinct edges $a$ and $b$ are therefore dual to $H$, but no square has them as adjacent sides. With the orientations shown, they have the same initial vertex. Hence $H$ is a self-osculating hyperplane, one of the forbidden pathologies of a special cube complex. The displayed cube complex is not special.

Because $X$ is special, the fundamental group of a special cube complex embeds in a Right-angled Artin group. Right-angled Artin groups are residually finite by the residual finiteness of a right-angled Artin group, and a subgroup of a residually finite group is residually finite. Hence ${\pi }_{1}X$ is residually finite.

If ${\pi }_{1}X$ were simple, choose $1\ne g\in {\pi }_{1}X$. A finite quotient in which $g$ survives has a proper normal kernel. Simplicity would force that kernel to be trivial, embedding ${\pi }_{1}X$ into a finite group, contrary to the assumption that ${\pi }_{1}X$ is infinite. This is precisely the obstruction that an infinite residually finite group is not simple.