Plus closure/Two-dimensional/Graded/Torsor/Strongly semistable/Section

We deal first with the situation of a strongly semistable sheaf ${}{\mathcal {S}}$ of degree ${}0$. The following two results are due to Lange and Stuhler. We say that a locally free sheaf is étale trivializable if there exists a finite étale morphism ${}\varphi \colon C'\rightarrow C$ such that ${}\varphi ^{*}({\mathcal {S}})\cong {\mathcal {O}}_{C'}^{r}$. Such bundles are directly related to linear representations of the étale fundamental group.

Lemma

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}X$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a locally free sheaf over ${}X$. Then ${}{\mathcal {S}}$ is étale trivializable if and only if there exists some ${}n$ such that

${}F^{n*}{\mathcal {S}}\cong {\mathcal {S}}$.

Theorem

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}X$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a strongly semistable locally free sheaf over ${}X$ of degree ${}0$. Then there exists a finite morphism

$\varphi \colon Y\longrightarrow X$

such that ${}\varphi ^{*}({\mathcal {S}})$ is trivial.

Proof

We consider the family of locally free sheaves ${}F^{e*}({\mathcal {S}})$, ${}e\in \mathbb {N}$. Because these are all semistable of degree ${}0$, and defined over the same finite field, we must have (by the existence of the moduli space for vector bundles) a repetition, i.e.

${}F^{e*}({\mathcal {S}})\cong F^{e'*}({\mathcal {S}})\,$

for some ${}e'>e$. By fact, the bundle ${}F^{e*}({\mathcal {S}})$ admits an étale trivialization ${}\varphi \colon Y\rightarrow X$. Hence the finite map ${}F^{e}\circ \varphi$ trivializes the bundle.

$\Box$

Theorem

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}X$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a strongly semistable locally free sheaf over ${}X$ of nonnegative degree and let ${}c\in H^{1}(X,{\mathcal {S}})$ denote a cohomology class. Then there exists a finite morphism

$\varphi \colon Y\longrightarrow X$

such that ${}\varphi ^{*}(c)$ is trivial.

Proof

If the degree of ${}{\mathcal {S}}$ is positive, then a Frobenius pull-back ${}F^{e*}({\mathcal {S}})$ has arbitrary large degree and is still semistable. By Serre duality we get that ${}H^{1}(X,F^{e*}({\mathcal {S}}))=0$. So in this case we can annihilate the class by an iteration of the Frobenius alone.

So suppose that the degree is ${}0$. Then there exists by fact a finite morphism which trivializes the bundle. So we may assume that ${}{\mathcal {S}}\cong {\mathcal {O}}_{X}^{r}$. Then the cohomology class has several components ${}c_{i}\in H^{1}(X,{\mathcal {O}}_{X})$ and it is enough to annihilate them separately by finite morphisms. But this is possible by the parameter theorem of K. Smith (or directly using Frobenius and Artin-Schreier extensions).

$\Box$