# Moduli of hyperelliptic curves: odd vs even genus

I'm stumped by Exercise 2.3 in Harris-Morrison, which says:

"Show that there does not exist a universal family of curves of genus 2 over any open subset $U \subset M_2$. In general, if $H_g \subset M_g$ is the locus of hyperelliptic curves, for which $g$ does there exist a universal family over some open subset $U \subset H_g$? Answer: for $g$ odd."

It's not 100% clear what is meant by "universal family"; possibly it just means some family each of whose (closed) fiber is the right curve''.

Question: what precisely is the question, and then what is the answer?

In any case the "exercise" seems to suggest a fundamental difference between hyperelliptic curves with odd vs even genus. I'd be happy to see an explanation of what this difference is.

• There is a Galois cover $M_{0,2g+2}\to H_g$ with Galois group $\mathfrak{S}_{2g+2}$. There is a curve (generically) over $M_{0,2g+2}$. The quotient by the hyperelliptic involution is the universal genus $0$ curve over $M_{0,2g+2}$. Moreover, the dualizing sheaf on the hyperelliptic curve is the pullback from this genus $0$ curve of an invertible sheaf of degree $g-1$. The genus $0$ curve descends to $H_g$. The question is whether the invertible sheaf of degree $g-1$ descends. It does descend if $g-1$ is even, since then it is a (negative) tensor power of the dualizing sheaf. – Jason Starr Jul 14 '16 at 19:07
• Sorry, is it clear that it doesn't descend if $g-1$ is odd? – user84144 Jul 20 '16 at 16:06
• There is a simpler model than $M_{0,2g+2}$, namely $(\mathbb{P}^1)^{2g+2} = \mathbb{P}^1 \times \dots \times \mathbb{P}^1$. This is, generically, a $\textbf{PGL}_2\times \mathfrak{S}_{2g+2}$-torsor over $H_g$. Using the fact that the $\textbf{PGL}_2$-action on $\mathbb{P}^1$ linearizes to $\mathcal{O}_{\mathbb{P}^1}(d)$ if and only if $d$ is even, I did convince myself once that the class does not descend when $g-1$ is odd. The details look a little messy. – Jason Starr Jul 20 '16 at 16:46