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Suppose $\pi:C'\to C$ is a branched cover of compact Riemann surfaces such that the associated extension of function fields is Galois with group $G$ -- so that $\pi$ presents $C$ as the quotient $C'$ by the action of $G = \text{Aut}(C'/C)$. Now, let $\rho:G\to GL(W)$ be a finite dimensional complex representation of $G$. Below, we identify locally free coherent sheaves with holomorphic vector bundles.

Let $\underline W$ be the trivial vector bundle on $C'$ with fibre $W$. Define the vector bundle $W^\rho$ on $C$ to be the subsheaf of $\pi_*\underline W$ whose sections over $U\subset C$ are the $G$-equivariant holomorphic functions $U' = \pi^{-1}(U)\to W$. We want to compute $c_1(W^\rho)$ as follows. We have a natural map $\varphi:\pi^*W^\rho\to\underline W$ on $C'$ (coming from the adjunction counit $\pi^*\pi_*\underline W\to\underline W$) which is an injective map of coherent sheaves. Now, by looking at the zeros of the determinant of $\varphi$ (which occur exactly at the critical points of $\pi$), we can figure out the value of $c_1(W^\rho) = \frac1{|G|}c_1(\pi^*W^\rho) = -\frac1{|G|}\cdot\dim H^0(C',\text{coker }\varphi)$.

Carrying out this computation explicitly, we seem to get the following answer. Given any branch point $p\in C$ of $\pi$, pick a preimage $p'\in C$. Let $G_{p'}\subset G$ be the (necessarily cyclic) stabilizer group of $p'$, of order $n_p$. For $0\le i<n_p$, define the numbers $w_{p,i} := \dim\text{Hom}^{G_{p'}}((T_{p'}C)^{\otimes i},W)$ and set $w_p:=\sum_{0\le i<n_p}\frac{i}{n_p}w_{p,i}\in\mathbb Q$. We then get $c_1(W^\rho) = -\sum_p w_p$, where the sum is over the branch points of $\pi$. Is the result of this computation correct and can I verify it by comparing it to some well-known/basic theorem? (I tried to calculate explicitly in local holomorphic coordinates where the map is given by $z\mapsto z^{n_p}$ and got the above answer.)

I find it quite interesting that the sum of the rational numbers $w_p$ is an integer. But clearly, these numbers can be defined without referring to the vector bundle $W^\rho$ or its Chern class. Is there some direct way in which we could prove this integrality statement?

This question is motivated by trying to understand the index computation (Theorem 4.1) in Chris Wendl's paper on super-rigidity and equivariant transversality (https://arxiv.org/abs/1609.09867).

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This computation is related to the well-known semiorthogonal decomposition of the $G$-equivariant derived category of $C'$, or equivalently, of the quotient stack $C'/G$. The latter can be thought of as the curve $C$ with a root stack of order $n_p$ structure at each of the branch point $p \in C$.

The semiorthogonal decomposition takes the form $$ D(C'/G) = \Big\langle D(p_1),\dots,D(p_1),\dots,D(p_m),\dots,D(p_m), D(C) \Big\rangle, $$ where the component $D(p_i)$ repeats $n_{p_i}-1$ times. The exceptional objects corresponding to the components $D(p_i)$ can be described as follows. Let $X_i = \pi^{-1}(p_i)$ (taken with reduced structure). This is a subscheme of length $|G|/n_{p_i}$ of $C'$. Denote by $E_{i,j}$ the sheaf $\mathcal{O}_{X_i}$ with $G$-equivariant structure corresponding to the action of $G$ on $T_{x_i}C'$, where $x_i \in X_i$ is any point. Then $E_{i,1}, \dots, E_{i,n_{p_i}-1}$ are the required objects.

The projection functor to the component $D(C)$ of the above semiorthogonal decomposition is equal to $$ F \mapsto \pi^{\ast}((\pi_{\ast}F)^G). $$ Therefore, taking $F = W$ we obtain a distingusihed triangle $$ \pi^\ast((\pi_\ast W)^G) \to W \to W' $$ where $W'$ is the projection of $W$ onto the subcategory generated by $E_{i,j}$. Note that $(\pi_\ast W)^G = W^\rho$. Therefore, $$ c_1(\pi^\ast((\pi_\ast W)^G)) = - c_1(W'). $$ It remains to note that $W'$ is an extension of the sheaves $E_{i,j}$ and the integers $w_{i,j}$ encode the multiplicities. This implies that $$ c_1(\pi^\ast(W^\rho) = \sum \left(jw_{i,j} \frac{|G|}{n_i}\right) $$ (the last factor is the length of $E_{i,j}$). Finally, to obtain $c_1(W^\rho)$ the above expression should be divided by the degree $|G|$ of the map $\pi$.

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  • $\begingroup$ Thanks, this seems to confirm the computation in question. I'll wait and see if someone can explain why the sum of the very concretely defined rational numbers $w_p$ is an integer directly (i.e., without referencing vector bundles, coherent sheaves or their Chern classes) before accepting this answer. $\endgroup$ Apr 17, 2020 at 12:40

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