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Assume $V$ be a genus larger than 1 handlebody, $S=\partial_{+} V$.

Denote $N$ be the normal closure of $MCG(V)$ in $MCG(S)$.

Is there any material related to the quotient group $MCG(S)/N$ ? Thanks!

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@l.Mosher:sorry I have it edited. – yanqing Jun 28 '12 at 13:26
@Lee: He probably means normal closure. – Misha Jun 28 '12 at 13:27
Thanks, it makes sense now. – Lee Mosher Jun 28 '12 at 13:43
up vote 6 down vote accepted

The normal closure of $MCG(V)$ in $MCG(S)$ is all of $MCG(S)$. To see why, we know that $MCG(S)$ is generated by Dehn twists, so it suffices to prove that each Dehn twist is in the normal closure. Consider the Dehn twist $\tau_c$ about an essential simple closed curve $c$. There exists a homeomorphism $f : S \to S$ such that $c' = f(c)$ bounds a properly embedded disc $D \subset V$, and $\tau_{c'}$ extends to a twist about the disc $D$, so $\tau_c = f^{-1} \circ \tau_{c'} \circ f$ is in the normal closure of $MCG(V)$. To find this homeomorphism, observe that the there is one $MCG(S)$-orbit of nonseparating curves, and for each partition $genus(S)=m+n$ there is one $MCG(S)$ orbit of separating curves whose complements have genus $m,n$. Each such orbit clearly has a representative bounding a disc in $V$.

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nice,thanks a lot! – yanqing Jun 28 '12 at 13:46
@L.Mosher: Assign $MCG(S)$ a clay graph $G$. Obviously there is a subgraph $G^{*}$ with vertices corresponding to $MCG(V)$. Is there a vertex in $G$ which is arbitrarily far from $G^{*}$ ? – yanqing Jun 28 '12 at 14:00
@yanqing: I don't know what a clay graph is. – Lee Mosher Jun 28 '12 at 14:38
Cayley graph. And the answer to your question is "Yes." Said another way, the inclusion of $MCG(V)$ into $MCG(S)$ is not "coarsely dense." I don't know a slick proof of this. – Sam Nead Jun 28 '12 at 17:38
This follows from the fact that $MCG(V)$ has infinite index in $MCG(S)$. For any subgroup of any finitely generated group $G$, if the subgroup is coarsely dense then it has finite index in $G$. There might be a very slick proof somewhere, but there is a pretty quick proof of a more general result in my paper with Sageev and Whyte "Quasi-actions on trees I", which says that two subgroups of $G$ at finite Hausdorff distance from each other are commensurable in $G$. – Lee Mosher Jun 28 '12 at 17:56

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