# Equivariant Riemann-Hurwitz

The Riemann-Hurwitz formula starts with a genus $g$ algebraic curve $Y$ and a ramified cover $\pi\colon X\to Y$ of degree $N$, with ramification indices $e_P$ and computes invariants of $X$, such as the genus, or more simply the Euler characteristic: $$\chi(X)=N\chi(Y)-\sum_P (e_P-1)$$

This computes the Euler characteristic as a number. Categorifying to the homology suggests generalizations. If the cover is Galois, the quotient by a finite group $G$, then the homology carries an action by that group and I would like to know the representation. An analogue of the Riemann-Hurwitz formula holds in the representation ring $RG$. The number $N=|G|$ is replaced by the regular representation. Thus the first homology is $2(g-1)$ copies of the regular representation, plus a couple more trivial representations, to account for $H_0$ and $H_2$, plus some additional representations from the ramifications. The typical way I have seen this proved is to form an $G$-stable cellular decomposition of $X$ by lifting a cellular decomposition of $Y$ with a vertex for each ramification point. Assuming that there are $r>0$ ramification points, there are $r$ vertices, $(r-1)+2g$ edges, and one face. The edges and faces contribute regular representations, while a vertex with isotropy $I$ contributes $\mathbb C[G/I]$, which is smaller than $\mathbb C[G]$. The chain complex of $X$ is: $\bigoplus_P\mathbb C[G/I_P]\leftarrow \mathbb C[G]^{2g+r-1}\leftarrow \mathbb C[G]$. Together $C_0$ and $C_2$ are generated by $r+1$ elements, so they partially cancel at most $r+1$ regular representations, leaving $2g-2$ regular representations, plus more representations parameterized by the ramifications (specifically $\mathbb C[G]-\mathbb C[G/I_P])$.

But what I really want is to understand the combination of the representation and the intersection form, equivalently to understand a Lagrangian subspace as a representation. This depends only on the topology, but it is convenient to exploit the complex structure of the curve, because it provides Lagrangian subspaces $H^0(X;\Omega^1)$ and $H^1(X;\mathcal O)$. I think that there should be a Riemann-Hurwitz formula that says that each has $g-1$ copies of the regular representations, plus a trivial representation, plus additional contributions from the ramification $R$. Is this true?

$$H^0(X;\Omega^1)\overset{?}{=}\mathbb C\oplus \mathbb C[G]^{g-1}\oplus ?(R,T_R)$$

In the unramified case, $\pi_*\mathcal O$ is a flat vector bundle, the one corresponding to the regular representation. Note that the isotypic decomposition of the this vector bundle induces the isotypic decomposition of its cohomology. Each of the of the isotypic bundles is flat, so it has degree zero. So each has Euler characteristic equal to $1-g$ times its dimension. So each to the Euler characteristic $1-g$ times its corresponding representation. We started with the regular representation, broke it into pieces, multiplied each piece by $1-g$ and added them together again. Thus $\chi(X)=(1-g)\mathbb C[G]$; or $H^1(X;\mathcal O)=H^1(Y;\pi_*\mathcal O)=\mathbb C[G]^{g-1}\oplus\mathbb C$. But in the ramified case, I don’t see how to exploit flatness.

There are two separate questions here. One question is what is the value of $?(R,T_R)$, as a function of the ramification divisor $R$ and representation of the isotropy on the tangent spaces $T_R$. I think that the Woods Hole / Holomorphic Lefschetz / Atiyah-Bott formula says that there is such an equivariant Riemann-Hurwitz formula depending only on those two ingredients; and that $?(R,T_R)$ is linear in the divisor. And it provides a way to check such a formula, if one could write it down. In some sense it even gives a formula.

But the other question, the one that really interests me, is whether the correction term is positive, whether there really are regular representations, regardless of anything else. I don’t think that the contributions from individual fixed points are all positive, or even real, so all we can ask is that sum over all ramification points is positive. Is it plausible to prove this without understanding the ramification term? For the earlier theorem, the one identifying the representation, but not the symplectic form, is there any way to prove such a weak version, finding $2g-2$ copies of the regular representation in $H^1(X)$, without going so far as to compute the correction from the ramification?

The context for my interest in the presence of the regular representations is that Andy Putman and I made a conjecture about the monodromy of symmetric curves acting on the homology of $X$. We conjectured that if $Y$ has genus at least 2 that the homology of $X$ has no vectors fixed by any finite index subgroup of the monodromy of the space of curves with the same symmetry. An upper bound on the monodromy of symmetric curves is the monodromy of symmetric abelian varieties. Indeed, I think it is a standard conjecture that they have the same Zariski closure. The bigger group is an arithmetic group easily described in terms of the representation and symplectic form. The cotangent space to the space of symmetric abelian varieties is $(S^2\;H^0(X;\Omega^1))^F$. Thus a symmetric (real) representation is never fixed; and a skew (quaternionic) representation is fixed only if it has multiplicity 1. Those cases don’t depend on the symplectic form (which doesn’t vary), only multiplicity, so they can be dealt with by the first equivariant Riemann-Hurwitz formula proved by cellular decomposition. The hard part is a non-self-dual (complex) representation. They have several symplectic forms, depending on how the representation and its dual are distributed between $H^{1,0}$ and $H^{0,1}$. A complex representation is fixed if it appears in $H^{1,0}$ and its dual does not appear there, only in $H^{0,1}$. But if the regular representation appears in $H^{1,0}$, it includes both representations, so it handles this case.

• I know a topological / combinatorial way to compute examples - maybe you can push it a little further: remove all the branch points from $Y$, and all their pullbacks from $X$, giving you a Galois covering map $X'\to Y'$. Note that 1. both $X', Y'$ retract to bouquets of circles, 2. the map $\pi_1(X')\to\pi_1(Y')$ is a finite map of free groups, 3. the map $\pi_1(Y')\to\pi_1(Y)$ is killing generators, and 4. the map $\pi_1(X')\to\pi_1(X)$ is either killing generators or powers of them. I.e. on the level of fundamental groups it's all completely explicit. Jun 18, 2017 at 19:55

This is a well known and well understood problem when the base field is $\mathbb C$. It was first studied by Chevalley and Weil (almost a century ago !) who were interested in modular curves (what else ?).

For a modern account, see

Kani, Ernst : The Galois-module structure of the space of holomorphic differentials of a curve. J. reine angew. Math. 367 (1986), 187-206 .

Nakajima, Shoichi : Galois module structure of cohomology groups for tamely ramified coverings of algebraic varieties. J. Number Theory 22, 115-123 (1986).

Here is a brief report. Warning : this problem is well understood for a tame action only, over an algebraically closed field $k$, in particular over $\mathbb C$, but it is still open, as far as I know, in the wild action case. So I assume tameness.

At each point $P\in X$, define the ramification character $\psi_P$ as the character given by the action of the cyclic stabilizer $G_P$ on the cotangent space $\mathcal m_P/\mathcal m_P^2$. Define the ramification module by

$$\Gamma _{G}=\sum_{P\in X}{\rm Ind}_{G_{P}}^{G}\sum_{l=1}^{e_{P}-1}l\psi _{P}^{l}$$

where $e_P$ is the ramification index at $P$. Then the equivariant Hurwitz formula says : there exists a unique $k\left[ G\right]$ -module $\widetilde{\Gamma} _{G}$ such that

$$\widetilde{\Gamma}_{G}^{\oplus \#G}=\Gamma _{G}$$ and in the Grothendieck ring $R_{k}(G)$ holds the following equality :

$$\chi(G,\mathcal{O}_{X})=\chi(\mathcal{O}_{Y})\cdot \left[ k \left[ G\right] \right] -[ \widetilde{\Gamma} _{G}]$$

where, as you can assume, $\chi(G,\cdot)$ is the equivariant Euler characteristic, and $\chi(\cdot)$ is the ordinary Euler characteristic. From this, and Serre duality, which is of course equivariant, you get the structure of $H^0(X,\Omega^1_X)$.

The proof of the equivariant Hurwitz formula is not hard. Since it is an equality between characters, it is enough to prove it when you evaluate at each $g\in G$. For $g=e$, this is the usual Hurwitz formula. For $g\neq e$, one uses a Lefschetz fix point formula for the corresponding automorphism of $X$.

• Chevalley and Weil didn't have holomorphic Lefschetz, did they? What did they do? Jun 20, 2017 at 23:00
• It is more a question for an historian. If you are fluent in German, you can find Chevalley-Weil's article here : doi.org/10.1007/BF02940687 Jun 21, 2017 at 7:28
• You may wonder why Chevalley and Weil, both French, wrote in German. The reason is that they were answering a letter from Hecke (here you can see why modular curves are at the origin of this question). Jun 21, 2017 at 7:30
• The wild action case (for l-adic homology) is also due to Weil: see Serre, Corps locaux, ch. VI, §4, prop. 7. Jun 26, 2020 at 13:28