Picard Groups of Moduli Stacks update

A tiny update on this post. I was chatting with Benson and realized that I may as well ask him directly for a presentation of the mapping class group of a genus two surface. Perhaps unsurprisingly, it can be found in his book with Dan Margalit (see page 122 of their book which might be downloadable from a Russian website) and is given as follows:

\(G \simeq \langle a_1,a_2,a_3,a_4,a_5 | \ [a_i,a_j] \ \text{for $|i-j|>1$}, a_i a_{i+1} a_i = a_{i+1} a_i a_{i+1},\)
\( (a_1 a_2 a_3)^4 = a^2_5, [(a_5 a_4 a_3 a_2 a_1 a_1 a_2 a_3 a_4 a_5),a_1], (a_5 a_4 a_3 a_2 a_1 a_1 a_2 a_3 a_4 a_5)^2, \rangle.\)

The next task is to find the representation

\(G \rightarrow \mathrm{Sp}_4(\mathbf{Z}) \rightarrow \mathrm{Sp}_4(\mathbf{F}_2) \simeq S_6\)

and then take the index \(6\) preimage \(\Gamma \subset G\) of the \(S_5 \subset S_6\) corresponding to fixing a Weierstrass point. Note there are two conjugacy classes of \(S_5\), the correct one is the one whose restriction to \(A_5\) still acts absolutely irreducibly on \((\mathbf{F}_2)^4\). Then one can use Reidemeister-Schreier to compute a presentation of \(\Gamma\) and then compute \(H_1(\Gamma,\mathbf{Z})\). This is all good in theory, and Farb-Margalit does have a chapter on the symplectic representation, but actually having to read the book in detail to extract the precise symplectic representation sounded like too much work, especially since all of this is ultimately just for a two sentence comment in a paper that might be removed for space reasons anyway. So instead I just fired up magma with the representation \(G\) and asked it to find *all* index six subgroups. It turns out that there are only two of them (up to conjugation), which must come exactly from the two subgroups of \(S_5 \subset S_6\). The abelianization of one is \(\mathbf{Z}/10 \mathbf{Z} \simeq G^{\mathrm{ab}}\), but the other group is

\(\Gamma = \langle a_1,a_2,a_3,a_4 \rangle\), and one finds that \(H_1(\Gamma,\mathbf{Z}) \simeq \mathbf{Z}/40 \mathbf{Z}.\)

Hence this is (in light of the previous dicussion) the correct subgroup, and this (unsurprisingly although not entirely independently) confirms the analysis of naf in the comments. Now I suspect that if you think a little harder than I am prepared to do (or if you just know a little bit more than me), you might be able to see directly from the definition of the \(a_i\) that \(a_1,a_2,a_3,a_4\) fix a Weierstrass point; if you are such a person please make a comment!

This entry was posted in Mathematics and tagged , , , , . Bookmark the permalink.

2 Responses to Picard Groups of Moduli Stacks update

  1. naf says:

    It looks to me that the map from \(G\) to \(S_6\) simply sends \(a_i\) to the transposition \((i,i+1)\).

    • naf says:

      I realised that I didn’t say why that map from \(G\) to \(S_6\) is the one you want, i.e., so that the \(S_5\) is the obvious one. This follows from the fact that a Dehn twist around a simple closed (non-separating) curve acts as a transposition on the Weierstrass point. This is most easily seen from the complex analytic point of view by thinking of the Dehn twist as the monodromy of a one parameter family of genus \(2\) curves degenerating to a geometric genus \(1\) curve with a node. This corresponds to (via the double cover) the “degeneration” of \(6\) distinct points in \(\mathbb{P}^1\) to a union of a set of \(4\) distinct points and a distinct point of multiplicity \(2\), from which the claim follows.

Leave a Reply

Your email address will not be published. Required fields are marked *