degree 0 vector bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:20:22Z http://mathoverflow.net/feeds/question/85121 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85121/degree-0-vector-bundles degree 0 vector bundles unknown 2012-01-07T09:14:09Z 2012-01-11T14:59:17Z <p>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?</p> http://mathoverflow.net/questions/85121/degree-0-vector-bundles/85122#85122 Answer by a-fortiori for degree 0 vector bundles a-fortiori 2012-01-07T09:22:40Z 2012-01-07T09:22:40Z <p>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 $-1$.</p> http://mathoverflow.net/questions/85121/degree-0-vector-bundles/85422#85422 Answer by inkspot for degree 0 vector bundles inkspot 2012-01-11T14:59:17Z 2012-01-11T14:59:17Z <p>[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.</p>