Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 134 / 1

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 $A_{\Gamma }$.