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 We will discuss S. Kim’s (unpublished) account of why the last one admits a hyperbolic surface subgroup.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.


About berstein

berstein is the name under which participants in the Berstein Seminar - a mathematics seminar at Cornell - are blogging.
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