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Let $X$ be a scheme over an algebraically closed field (I'm mainly interested in the characteristic zero case), $V$ a rank $n$ vector bundle and $\ell$ a line bundle on $X$, and

$\phi:V\otimes V\to\ell$

an $\ell$-valued quadratic form on $V$, which can be seen as a map $\phi:V\to V^{*}\otimes\ell$. The form $\phi$ is generically non degenerate if it induces an isomorphism

$\det(\phi):\det (V)\otimes\mathcal{O}(D)\to\det(V^{*})\otimes\ell^{\;n}$

for some divisor $D$ on $X$.

My question is: is it possible for a form $\phi$ to be generically non degenerate yet not to induce an isomorphism $V\otimes\mathcal{O}(D)\to V^{*}\otimes\ell$ for some divisor $D$?

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Take $X=\mathbb P^1_{\mathbb C}$, $V=O_X\oplus O_X(-1)\oplus O_X(-m)$ with $m\ge 3$, $\ell=O_X$ and consider the inclusion $V\to V^*=O_X\oplus O_X(1) \oplus O_X(m).$ It is generically non-degenerated, but $V\otimes O_X(k)$ is not isomorphic to $V^*$ for any $k$ (edit by Grothendieck's theorem on the classification of vector bundles on $\mathbb P^1$).

When $X$ is irreducible, non-degenerated just means that the map $\det(V)\to \det(V^*)\otimes \ell^n$ is not the zero map. This is not a strong condition.

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    @Simplicius, OK I see.2012-10-29