This is a request for a reference to a proof of a result. The result is not very hard, but I'd rather cite than reprove.

I'm looking for a generalization of the following result (Farb and Margalit, A Primer on Mapping Class Groups) to surfaces with boundary.

Let $Y$ be a closed surface of genus $g$, and let $\Psi$ be the representation $$\Psi: \operatorname{Mod}(Y) \rightarrow \operatorname{Sp}(H_1(Y)) \cong \operatorname{Sp}(2g, \mathbb{Z})$$ given by the action of the mapping class group on homology. Then $\Psi$ is surjective.

My question is the following.

Suppose $Y$ is a surface of genus $g$, with $r > 0$ boundary components. What is the image of $$\Psi: \operatorname{Mod}(Y) \rightarrow \operatorname{SL}(H_1(Y))?$$

Again, I'm pretty sure I know what the answer is, and I could prove it myself, but this sounds like something that is well-known and I'd rather be able to cite it.

Here's what I think the answer is: First, 'cap off' the boundary components of $Y$ to embed $Y$ in a closed genus-$g$ surface $Y'$. We get an exact sequence $$0 \rightarrow V \rightarrow H_1(Y) \rightarrow H_1(Y') \rightarrow 0,$$ where $V$ is generated by the classes of the boundary components.

The MCG action must respect this exact sequence, restrict to the identity on $V$, and induce an automorphism of $H_1(Y')$ that respects the symplectic form.

I expect that every such automorphism of $H_1(Y)$ will lie in the image of the MCG action -- and that it's not very hard to produce Dehn twists on $Y$ to prove it.

Does anyone know of a reference where this is proven?

  • $\begingroup$ I don't think you need to produce explicit Dehn twists. The map from MCG(Y) to MCG(Y') is surjective (the capping homomorphism) and since the latter surjects onto Sp(2g,Z) you get all symplectic automorphisms of H_1(Y'). the rest of your question seems to follow in a straightforward way. $\endgroup$ – Dan Margalit Oct 31 '17 at 13:02
  • $\begingroup$ I don't see how to finish without doing something explicit. For lack of better notation, let Sp(H_1(Y)) be the set of automorphisms which preserve the (degenerate) pairing on H_1(Y). The image of MCG(Y) in Sp(H_1(Y)) is a subgroup which surjects onto Sp(H_1(Y')), but this is not enough to specify the subgroup. (For example, choosing a splitting H_1(Y') --> H_1(Y) gives a section Sp(H_1(Y')) --> Sp(H_1(Y)), whose image is a subgroup surjecting onto Sp(H_1(Y')).) $\endgroup$ – Brian Lawrence Nov 1 '17 at 20:24

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