# Vanishing of homology for hyperelliptic locus

It is a theorem due to Harer that $H_k(M_{g,n},\mathbb{Q})=0$ for $k>C(g,n)$, where $C(0,n)=n-3, C(g,0)=4g-5$ for $g>0$, and $C(g,n)=4g-4+n$ for $g,n>0$. Here $M_{g,n}$ denotes the coarse moduli space of $n$-pointed genus $g$ curves. I was wondering if an analogous vanishing theorem holds for $H_{g,n}$, and if yes I would love to see a reference.

As is clear and well known, the theorem of Harer would follow from this famous conjecture of Looijenga, claiming that if $2g-2+n>0$, $M_{g,n}$ is the union of $g$ affine open subsets (actually even $g-1$, when $n=0$). Recall that $H_{g,0}$ is affine, so for $n=0$ the homology is zero for $k\geq 2g$. However I don't know if $H_{g,n}$ is affine/Stein in general, nor I am aware of a bound on the number of affine opens required to cover it. Since the maps $H_{g,n+1}\to H_{g,n}$ are affine for $n\geq 1$, it would suffice to find affine covers with few open sets for $n=1$.

The proof of Harer's theorem is (for me) rather intricate, and I don't think it adapts easily to the case of hyperelliptic curves.

• The space $H_g$ has the rational cohomology of a point. And $H_{g,1}$ has the rational cohomology of a circle. I can supply an argument later. – Dan Petersen Dec 23 '17 at 11:54
• The fibration exists, and it is a fiber bundle in the sense of orbifolds. But the fundamental group of the base acts nontrivially on the cohomology of the fiber, so you need to work with twisted coefficients, which kills your suggested argument (the vanishing results for $n=0,1$ are only for constant coefficients). – Dan Petersen Dec 26 '17 at 21:01
• Also, I shouldn't have said "circle" in the comment three days ago, but "2-sphere". – Dan Petersen Dec 26 '17 at 21:03

A cohomological bound can be obtained as follows.

Proposition The affine stratification number of $H_{g,n}$ is $1$ for $n > 0$, $0$ for $n=0$.

Proof Since $H_{g,0}$ is affine we are only interested in the case $n > 0$. Since $H_{g,n+1} \to H_{g,n}$ is an affine morphism for $n>0$, it's enough to do the case $n=1$. Write $H_{g,1}$ as the disjoint union of the locus where the marked point is (resp. is not) a Weierstrass point. Both these loci are affine since the projection down to $H_g$ is an affine morphism on each of them. To see that $H_{g,1}$ is not affine, note that it contains many complete curves - specifically, any fiber of $H_{g,1} \to H_g$ is proper. QED

This implies that not just the rational cohomology, but the cohomology of any constructible sheaf (over $\mathbb Q$, to take care of stacky issues) on $H_{g,n}$ vanishes above degree $2g+n$ for $n > 0$. Also vanishing theorems for constructible sheaves - see the paper of Roth and Vakil.

In general I would expect this bound to be sharp. By this I mean that for a given $g$, there will only be finitely many $n$ for which $H^{2g+n}(H_{g,n},\mathbb Q) = 0$. I have no idea how one would prove this, though.

Now let me prove the claim I made in a comment.

Proposition The space $H_g$ has the rational cohomology of a point for all $g$.

Proof Note that $H_g = M_{0,2g+2}/S_{2g+2}$. So it will be enough to prove that $M_{0,2g+2}/S_{2g+1}$, parametrizing $2g+1$ unordered marked points on $\mathbb P^1$ and a distinguished other point, has the rational cohomology of a point. If we put the distinguished marking at infinity, we get an isomorphism $M_{0,2g+2}/S_{2g+1} = B(\mathbb A^1,2g+1)/G$. Here $B(\mathbb A^1,n)$ denotes the configuration space of $n$ unordered marked points on $\mathbb A^1$, and $G= \mathbb G_m \ltimes \mathbb G_a$ denotes the group of Möbius transformations fixing the point at infinity. In fact the projection from the configuration space to $M_{0,2g+2}/S_{2g+1}$ is a trivial $G$-bundle.

The rational cohomology of $B(\mathbb A^1,n)$ was computed by Arnold in his famous paper on the cohomology of the braid group: it has the rational cohomology groups of a circle. And $G$ is homotopic to a circle. So $M_{0,2g+2}/S_{2g+1}$ has the rational cohomology of a point, as claimed. QED

Proposition $H^0(H_{g,1},\mathbb Q) \cong H^2(H_{g,1},\mathbb Q) \cong \mathbb Q$ and all other cohomology groups vanish.

Proof Apply the Leray spectral sequence for $f:H_{g,1} \to H_g$. We have $R^0 f_\ast \mathbb Q \cong R^2 f_\ast \mathbb Q \cong \mathbb Q$. The locally constant sheaf $R^1 f_\ast \mathbb Q$ has vanishing cohomology, because every point of $H_g$ has the hyperelliptic involution as a nontrivial stabilizer, and the hyperelliptic involution acts as $-1$ on the stalk of the local system. The result follows. QED

The cohomologies of $H_{g,2}$ and $H_{g,3}$ have also been computed for all $g$ by Orsola Tommasi in her Ph.D. thesis.

Let me also mention two results about the cohomology of $H_{g,n}$ that I can prove but haven't written down yet.

Proposition The Leray spectral sequence for $H_{g,n} \to H_g$ degenerates for all $g,n$.

This is in contrast with the Leray spectral sequence for $M_{g,n} \to M_g$, which typically does not degenerate. It is really very special to families of hyperelliptic curves. It means that understanding the cohomology of $H_{g,n}$ is equivalent to understanding the cohomology of $H_g$ with coefficients in a symplectic representation.

Proposition There is an analogue of Harer stability for moduli of hyperelliptic curves: $H^k(H_{g,n},\mathbb Q) \cong H^k(H_{g+1,n},\mathbb Q)$ for $g \gg k$.

This isn't really "my" theorem - it follows from the work of Randal-Williams and Wahl.