Let $G$ be a finite group acting freely on a smooth projective complex variety $X$. This induces an action on $\mathrm{Coh}(X)$ by pullback. Let $\pi\colon X \rightarrow Y=X/G$ denote the quotient.
How do I prove the following claim?
Claim $\pi_\ast \mathcal{O}_X$ splits as a direct sum according to the irreducible representations of $G$.
I was given a hint to first show that:
- $\pi_\ast \mathcal{O}_X = \mathcal{O}_Y\otimes \mathbb{C}[G]$
Then we could conclude with Maschke's/Wedderburn-Artin's theorem, i.e. $\mathbb{C}[G]=\oplus_{i=1}^{k}\dim(\rho_i)\cdot \rho_i$, where $\rho_i$ are the irreducible representations of $G$.
But I'm not sure how to make sense of or prove 1.
Under the equivalence $\mathrm{Coh}(Y)\cong \mathrm{Coh}_G(X)$, I can understand what it means to tensor with a representation of $G$, but I'm not sure how to see this on $\mathrm{Coh}(Y)$. In particular, objects of $\mathrm{Coh}_G(X)$ are pairs $(E,\{\lambda_g\}_{g\in G})$ where $E\in\mathrm{Coh}(X)$ is $G$-invariant, and $\lambda_g\colon E \rightarrow g^\ast E$ is a choice of isomorphism for each $g\in G$. Let $\rho\colon G \rightarrow \mathrm{GL}(V)$ be a representation. Then \begin{equation*} (E,\{\lambda_g\}_{g\in G})\otimes\rho = (E^{\oplus \dim V}, \{\lambda_g\otimes\rho(g)\}_{g\in G}). \end{equation*}
Your claim 1 is wrong. For instance, let $Y$ be a smooth curve of positive genus and let $L$ be a 2-torsion line bundle on $Y$. Set $$ X = \operatorname{Spec}_Y(\mathcal{O}_Y \oplus L), $$ where the algebra structure is induced by an isomorphism $L^2 \cong \mathcal{O}_Y$. Then $\pi \colon X \to Y$ is an etale double covering and $Y \cong X / G$, where $G$ is the group of order 2 acting by the deck transformation. In this case $$ \pi_*\mathcal{O}_X \cong \mathcal{O}_Y \oplus L, $$ and the right-hand side is not a trivial vector bundle.