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Let Mg be the Mapping Class Group for $S_g$, the genus-g orientable surface, and consider the action of Mg on $H_1(S_g,\mathbb Z)$ sending f in Mg to m in $Sp^2(2g,\mathbb Z)$ through the induced map on homology .Why is this action a surjection onto $Sp^2(2g,\mathbb Z)$? ; I cannot see why every m in $Sp^2(2g,\mathbb Z)$ necessarily comes from some f in Mg. My group theory is a bit weak, so I may be missing some basic results. Sp^2 is the kernel of the mod2 reduction map from $H_1(Sg,\mathbb Z)$ to $H_1(Sg.\mathbb Z_2)$ , which I know is natural, but I cannot if the naturality helps. A similar second one: Let SPg be the spin-mapping class group , and consider this time the action on $H_1(Sg,\mathbb Z_2)$, sending g in SPg to m' in Og (orthogonal group, preserving the Rokhlin form, but I'm pretty sure this generalizes to all orthogonal groups), via the induced map on homology again. I cannot see either why this second action is surjective onto Og, tho I suspect that since Og is defined over $Z_2$ (more precisely over $H_1(Sg,\mathbb Z_2)$, this makes it a symplectic group, so that this case may reduce to the one above with $Sp(2g,\mathbb Z)$? ; I cannot see why every m in $Sp(2g,\mathbb Z)$ necessarily comes from some f in Mg. My group theory is a bit weak, so I may be missing some basic results.

Any Ideas?

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  • $\begingroup$ Andy: sorry for my screwup; my post was somehow deleted (possibly by me, unknowingly). If you post your answer again, I'll accept it. $\endgroup$
    – Larry
    Commented Jul 23, 2011 at 21:58

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This (at least the first question) is answered in Farb and Margalit's book

www.math.uchicago.edu/~margalit/mcg/mcgv50.pdf. The basic idea is that the symplectic group is generated by transvections, while the mapping class group is generated by Dehn twists.

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  • $\begingroup$ Igor: Sorry to not be able to give you the answer, but Andy Putman had answered before, though I screwed up by deleting my own post somehow. Thanks, though. $\endgroup$
    – Larry
    Commented Jul 23, 2011 at 22:31

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