## Surface subgroups of non-positively curved groups II: Artin groups

This is the second of three posts based on a guest lecture by Sam Kim.

We now consider Gromov’s question about the existence of hyperbolic surface subgroups in the context of Artin groups — a family closely related to Coxeter groups.

Define ${\theta(x,y,n)}$ to be the word ${xyxy\cdots}$ that alternates between the letters ${x}$ and ${y}$ and has length ${n}$. Suppose ${\Gamma}$ is a simplicial graph with a labeling of its edges ${m:E(\Gamma)\rightarrow {\mathbb Z}_{>1}}$. The associated Artin group is

$\displaystyle A(\Gamma,m) = \langle v\in V(\Gamma) | \theta(v,w,m_e)=\theta(w,v,m_e) \mbox{\ if } e=\{v,w\} \in E(\Gamma) \rangle.$

For example, the braid group on ${3}$ strands ${\mbox{BRAID}_3}$ has presentation ${\langle a,b | aba=bab \rangle}$ and so is $A(\Gamma,m)$ when $\Gamma$ is a two vertices connected by a singe edge and $m$ assigns the label ${3}$ to that edge.

The Coxeter group ${W(\Gamma,m)}$ is the quotient of ${A(\Gamma,m)}$ arising on adding the relations ${v^2=1}$ for all ${v\in V(\Gamma)}$.

Gordon, Long and Reid were the first to consider the question of which Artin groups contain hyperbolic surface subgroups. Let ${\Delta_{l,m,n}}$ be the triangle with edges labeled by ${(l,m,n)}$. They found that most Artin groups of finite and Euclidean type admit hyperbolic surface subgroups. To be precise, they left the question unresolved for Artin groups associated to ${\Delta_{2,3,5}}$, ${\Delta_{2,3,6}}$, and ${\Delta_{2,4,4}}$, and amongst all others of finite and Euclidean type only the groups in the families ${I_2(m)}$, ${A_1}$ and ${\tilde A_1}$ fail to contain hyperbolic surface groups.

${A(\Delta_{2,3,5})}$ is type ${H_3}$ and is of finite type, ${A(\Delta_{2,3,6})}$ is of type ${\tilde B_2}$, and ${A(\Delta_{2,4,4})}$ is of type ${\tilde G_2}$ and so of Euclidean type. For the first two, the problem remains open. We will discuss S. Kim’s (unpublished) account of why the last one admits a hyperbolic surface subgroup.

The standard presentation of a braid group (which you can find here) gives us that

$\displaystyle \mbox{BRAID}_4 \ = \ A(\Delta_{3,3,2}) \ = \ \langle \, a,b,c \, | \, aba=bab, bcb=cbc, ac=ca \, \rangle.$

First we show that this group contains a hyperbolic surface group. Let ${G=\pi_1(S^3-K)}$ where ${K}$ is the figure-eight knot. ${G}$ is known to contain a hyperbolic surface group (see [Cooper-Long-Reid 1997]), and to admit the presentation

$\displaystyle G \ = \ \langle \, x,y,t \, | \, x^t=xy, \, y^t = yxy \, \rangle.$

Using a similar argument to Mangum and Shanahan’s, one can embed ${G}$ into ${\mbox{BRAID}_4}$ via ${x\mapsto ac^{-1}}$, ${y\mapsto bac^{-1}b^{-1}}$ and ${t\mapsto abc^{-1}b^{-1}}$. So ${\mbox{BRAID}_4}$ also contains a surface group.

Now consider the subgroup ${\mbox{BRAID}_4^{(4)}}$ that fixes the ${4}$th string. It is generated by ${a,b,c^2}$, and it admits the presentation of ${A(\Delta_{3,4,2})}$ in these generators. It is an index ${4}$ subgroup of ${\mbox{BRAID}_4}$, so it also contains a surface group.

On the other hand, we can get an isomorphism ${\mbox{BRAID}_4^{(4)} \cong \mbox{BRAID}_3(S^1\times [0,1])}$, by fattening the ${4}$th string (see Kent-Peifer2002]). We can visualize the ${4}$th (fattened) puncture as the inner hole of the annulus, and the punctures ${1}$, ${2}$ and ${3}$ arranged in cyclic order around the annulus, and invariant under a rotation of ${2\pi/3}$. Then the element ${t=c^2ba}$ corresponds, as a braid in ${\mbox{BRAID}_3(S^1\times [0,1])}$, to a rotation of angle ${2\pi/3}$ on the target annulus. This gives a decomposition of ${\mbox{BRAID}_4^{(4)}}$ as a semi-direct product

$\displaystyle \mbox{BRAID}_4^{(4)} = A(\Delta_{3,3,3}) \rtimes \langle t \rangle$

where the first factor is generated by ${a,a^t,a^{t^2}}$. Consider the index ${3}$ subgroup

$\displaystyle A(\Delta_{3,3,3}) \times \langle t^3 \rangle.$

It contains a surface subgroup, since it has finite index in ${\mbox{BRAID}_4^{(4)}}$. But since it is a direct product, the surface subgroup must project isomorphically into the first factor (since a hyperbolic surface group cannot contain ${{\mathbb Z}^2}$).

Next we will consider a different index–3 subgroup of ${\mbox{BRAID}_4^{(4)}}$; namely, we take the subgroup ${\mbox{BRAID}_4^{(1)(4)}}$ of braids that fix the ${1}$st and ${4}$th punctures.
Let ${\phi:\mbox{BRAID}_4^{(1)(4)} \rightarrow {\mathbb Z}}$ be the morphism given by the winding number of the ${1}$st string around the ${4}$th string. Then ${\ker\phi \cong \mbox{BRAID}_2(\Sigma)}$, where ${\Sigma}$ is a pair of pants (a sphere minus three disks). This can be seen by fattening the ${1}$st and ${4}$th strings, and observing that in a braid in ${\ker\phi}$, they do not wind around each other.

In his thesis, Davide Moroni proved that this exact sequence splits, giving ${\mbox{BRAID}_4^{(1)(4)} \cong \ker\phi\times{\mathbb Z}}$. Thus, by the previous argument, ${\ker\phi \cong \mbox{BRAID}_2(\Sigma)}$ contains a surface group.

Moroni also proved that ${\mbox{BRAID}_2(\Sigma) = A(\Delta_{2,4,4})}$, and this finishes the proof.