Let $n \geq 2$ be an integer, $X$ a smooth variety over a field $k$ containing $\mu_n$ and $G$ a cyclic group of order $n$ acting on it. Assume that the action is free. Then the morphism $\pi: X \to Y=X/G$ is etale. I'm trying to understand why $\pi_\ast \mathcal{O}_X$ decomposes as

$$ \pi_\ast \mathcal{O}_X=\mathcal{O}_Y \oplus L \oplus L^2 \oplus \cdots \oplus L^{n-1} $$

where $n$ is the order of $G$ and $L$ is a line bundle on $Y$ such that $L^n \simeq \mathcal{O}_X$.

Can anybody help me please?

  • 2
    $\begingroup$ That is not true without further hypotheses. For instance, let $X$ be $\mathbb{P}^1_{\mathbb{R}}$. Let $G=\mathbb{Z}/2\mathbb{Z}$ act via the involution $[x,y]\mapsto [y,-x]$. Then the quotient $Y$ does not have any invertible sheaf $L$ of degree $-1$, a necessary condition for your assertion to hold. You either need to assume that your field contains appropriate roots of unity, or you should replace $G$ by a group scheme whose Cartier dual is split, e.g., $G=\mathbf{\mu}_n$. $\endgroup$ – Jason Starr Jul 9 '13 at 8:41
  • $\begingroup$ Hi Jason, thanks for the counterexample! I'm actually interested in the situation when the base field contains the roots of unity of order $|G|$. How do you prove the statement in that case? $\endgroup$ – user36795 Jul 9 '13 at 8:48

By construction the sheaf $\pi_* \mathcal{O}_X$ is a $G$-equivariant vector bundle of rank $n$ on $Y$ and, since $G$ is a cyclic group, the representation of $\pi_* \mathcal{O}_X$ as a $G$-module splits into direct summands which are all line bundles.

Now take any isomorphism $G \cong \mathbb{Z}/ n \mathbb{Z}$, and call $L$ the eigensheaf of $\pi_* \mathcal{O}_X$ corresponding to the generator $\bar{1} \in G$. Clearly $L$ is a $n$-torsion line bundle on $X$ and moreover for any $k \in \mathbb Z$ the eigensheaf corresponding to $\bar{k} \in G$ is precisely $L^k$.

So you obtain the desired splitting.

  • $\begingroup$ The fact that each of the $L^k$ appears exactly once may be derived from the Lefschetz fixed point formula (with empty fixed point set !). $\endgroup$ – Damian Rössler Jul 9 '13 at 21:39

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