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Let $X_{n}$ be the (unordered) configuration space of $n$ distinct points in $\mathbb{P}_{\mathbb{C}}^{1}$. The fundamental group of $X_{n}$ is the braid group on $n$ strands on the Riemann sphere, which I denote by $\pi_{1}X_{n}$, and which is generated by $\sigma_{1}, ... , \sigma_{n - 1}$, where $\sigma_{i}$ wraps the $i$th strand over the $(i + 1)$th strand. Consider the family over $X_{2g + 2}$ of hyperelliptic curves of genus $g$ such that for each element of $X_{2g + 2}$, the fiber is the hyperelliptic curve branched over $\mathbb{P}_{\mathbb{C}}^{1}$ at the corresponding set of $2g + 2$ points. After choosing a fiber $C$, this induces a monodromy representation $R : \pi_{1}X_{n} \to \mathrm{Sp}(H_{1}(C, \mathbb{Z}) \cong \mathrm{Sp}_{2g}(\mathbb{Z})$.

I need to show that the element $(\sigma_{1} ... \sigma_{2g + 1})^{2g + 2}$ lies in the kernel of $R$. I know I could do this through a tedious, elementary proof involving induction on rows and columns of $2g \times 2g$ matrices, but I feel that there should be a more enlightening proof using algebraic topology.

For instance, in Braids, Links, and Mapping Class Groups, Birman describes a map taking $\pi_{1}X_{n}$ to the mapping class group $M(0, 2g + 2)$ whose kernel is exactly the center of $\pi_{1}X_{n}$, which in fact is generated by the element $(\sigma_{1} ... \sigma_{2g + 1})^{2g + 2}$. How do I show that the monodromy action $R : \pi_{1}X_{n} \to \mathrm{Sp}_{2g}(\mathbb{Z})$ factors through this map $\pi_{1}X_{n} \to M(0, 2g + 2)$? A number of people (for instance, Mumford in Tata Lectures on Theta II) seem to assume this; can anyone provide a source that proves it, or advise me on what would be a sufficiently rigorous topological argument?

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This question appears to be a duplicate of: – oxeimon Mar 25 '14 at 15:24

See the discussion in the introduction to my paper "Generators for the hyperelliptic Torelli group and the kernel of the Burau representation at $t=-1$" with Brendle and Margalit, available here. The main result of this paper is a generating set for the kernel of the representation you wrote down.

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Am I missing something? The group $M(0,2g+2)$ is the fundamental group of the quotient $X_{2g+2}/\operatorname{PGL}_2$ (which exists as a manifold/algebraic variety and can be identified with the configuration space of $2g-1$ distinct ordered points in $\mathbf P^1\setminus \{0,1,\infty\}$). Then $\pi_1(X_n) \to \operatorname{Sp}_{2g}(\mathbf Z)$ factors through $M(0,2g+2)$ because the universal family of hyperelliptic curves is $\operatorname{PGL}_2$-invariant.

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Thanks, I hadn't realized that $M(0, 2g + 2)$ is the fundamental group of the ordered configuration space modulo $\mathrm{PGL}_{2}$, although now I see where this is mentioned in Mumford. (Actually, the space $X_{2g + 2}$ which I defined above was unordered configuration space, but I don't think it affects your argument.) I'm not sure I get the last part of your argument, though. You mean $\mathrm{PGL}_{2}$-invariant in what way? Moving one fiber to another via an element of $\mathrm{PGL}_{2}$ identifies the homology of one with the homology of the other in some "nice" way? – Jeff Yelton Mar 26 '14 at 20:47
I mean that the universal family is pulled back from a family over $X_{2g+2}/\mathrm{PGL}_2$. – Dan Petersen Mar 27 '14 at 10:33
I'm still confused on several points. For one thing, $X_{2g + 2} \to X_{2g + 2} / \mathrm{PGL}_{2}$ is a covering, correct? So $\pi_{1}X_{2g + 2}$ should inject into $M(0, 2g + 2)$. But the map of Birman $\pi_{1}X_{2g + 2} \to M(0, 2g + 2)$ has the nontrivial element $(\sigma_{1} ... \sigma_{2g + 1})^{2g + 2}$ in its kernel (this is the very fact I'm trying to exploit). Thus, it would seem that Birman's map isn't the map induced by the covering. – Jeff Yelton Apr 7 '14 at 20:31
@JeffYelton No, it's not a covering, since $\mathrm{PGL}_2$ is positive-dimensional. – Dan Petersen Apr 8 '14 at 5:21
@JeffYelton This is a special case of the following more general claim. Let $\mathscr M_{g,n}$ be the moduli space of compact genus $g$ Riemann surfaces with $n$ marked points, thought of as a complex orbifold. Then its fundamental group is the mapping class group $M(g,n)$. This fact should be in any reference on mapping class groups of surfaces. When $g=0$ (and $n \geq 3$) $\mathscr M_{0,n}$ is just the manifold $X_n/\mathrm{PGL}_2$. (Any genus zero Riemann surface is $\mathbf P^1$, so we are just considering configurations of $n$ points on the Riemann sphere modulo automorphisms.) – Dan Petersen Apr 20 '14 at 8:40

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