# Sheaf on abelian variety preserved by tensor product with a translation invariant line bundle

If $A$ is an abelian variety, $\mathscr{F}$ is a coherent sheaf on $A$, and $\mathscr{F}\otimes \mathscr{L}\cong \mathscr{F}$ for all translation invariant line bundles $\mathscr{L}$, why is the support of $\mathscr{F}$ dimension 0?

This is asserted in a book I'm trying to read (I think it's supposed to be obvious), and I don't see why it is true. Before this assertion, I think the book has already proven that the rank of $\mathscr{F}$ is zero (i.e. it is supported on a proper closed subset) by taking a resolution of $\mathscr{F}$ by line bundles, tensoring with $\mathscr{L}$, and taking the determinant bundle.

This argument is taken from Mukai's famous "Duality between $D(X)$ and $D(\hat{X})$ with its application to Picard sheaves" (Lemma 3.3.).
Let $X$ be the abelian variety we're working on. Assume the dimension of the support is $\geq1$ and let $C$ be a curve in $\mbox{supp}(\mathcal{F})$ with $f:\tilde{C}\to C$ its normalization. Then $\mathcal{G}:=f^*\mathcal{F}/(\mbox{torsion subsheaf})$ is a line bundle on $\tilde{C}$ (every torsion-free coherent sheaf on a smooth curve is locally free) and $f^*P\otimes \mathcal{G}\simeq\mathcal{G}$ for all $P\in\mbox{Pic}^0(X)$ ($\mbox{Pic}^0(X)$ is the set of translation-invariant line bundles on $X$). In particular, if we take the determinant on both sides, we obtain that $(f^*P)^{\mbox{rank}(\mathcal{G})}\simeq\mathcal{O}_{\tilde{C}}$. However, we have a non-trivial map $F:J_{\tilde{C}}\to X$ where $J_{\tilde{C}}$ is the Jacobian of $\tilde{C}$, given by the universal property of the Jacobian (and so $F$ restricted to the image of $\tilde{C}$ in $J$ is just $f$). Moreover, the map $F^*=f^*:\mbox{Pic}^0(X)\to\mbox{Pic}^0(J_{\tilde{C}})\simeq\mbox{Pic}^0(\tilde{C})$ is not trivial, and so we obtain a contradiction. Therefore $\mbox{supp}(\mathcal{F})$ is 0-dimensional.