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If $X$ is a smooth projective curve over a field $k$, $V$ is a subbundle of some finite direct sum of $O_X$, then is it true that $V$ is of degree 0?

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2 Answers 2

up vote 6 down vote accepted

No. Counterexample: On $\mathbf P^1$, take a (two-element) basis of $\Gamma(\mathbf P^1, \mathcal O(1))$, so that the corresponding homomorphism $\mathcal O^2\to\mathcal O(1)$ is surjective. Its kernel is a subbundle of degree $-1$.

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[Really a comment sed hac marginis ... .] a-fortiori's answer disposes of the question completely, but it is possible to go a little further: if $V$ is a degree $0$ sub-bundle of a free bundle on a curve, then $V$ is free (and is a direct summand). Proof: Write $W=V^\vee$, the dual bundle. Say it is of rank $n$. There is then a surjection $\pi:F=\mathcal O_X^{\oplus r}\to W$. Note that $r\ge n$. Choose general sections $s_1,\ldots,s_n$ in $H^0(X,F)$; then $\pi(s_1),\ldots,\pi(s_n)$ will generate a rank $n$ subsheaf $G$ of $W$ [because $\pi(s_1)$ is a nowhere vanishing section of $W$, and then argue inductively on the sheaves $F/(s_1)$ and $W/(\pi(s_1))$]. Then $G$ generated by $n$ sections and is a rank $n$ subsheaf of $W$. Since $\deg W=0$ it follows that $G=W$ and the homomorphism $\mathcal O_X^{\oplus n}\to W$ generated by the $s_i$ is an isomorphism. Dualizing gives the result.

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Yes, what you said is right. But it is easier to see it using Tannkian. If $V$ is of degree 0 and a sub of a free bundle, then it is an essentially finite vector bundle (see Nori's article for Nori's fundamental group). But the category of essentially finite vector bundles is tensor equivalent to representations of some group scheme over $k$. Since free bundle corresponds to trivial representation. $V$ as a suboject also corresponds to the trivial rep. The splitting assertion is even clearer. –  Lei Jan 11 '12 at 16:44
More generally, if $F$ is a finite direct sum of simple objects $S_i$ in some abelian category, any subobject of $F$ is a direct summand and isomorphic to a direct sum of some of the $S_i$. Apply this to the abelian category of semistable degree 0 bundles. –  user2035 Jan 29 '12 at 13:42

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