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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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up vote 2 down vote accepted

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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I think non-degenerate means $\det(V)\to\det(V^*)\otimes\ell^{n}$ is never zero on $X$, and generically non-degenerate means it can be zero at most along a divisor. Right? (assume $X$ irreducible) – Simplicius Oct 27 '12 at 15:00
(when I say the map is zero at a point $x$, I mean on the fiber at $x$ not at the stalk at $x$) – Simplicius Oct 27 '12 at 15:04
@Simplicius: as this is a map of invertible sheaves, if it is not identically zero, then the kernel is zero. The property is local, so you can suppose you are dealing with an $A$-linear map $A\to A$. Now think about the possibilities for the kernel. – user18119 Oct 27 '12 at 18:57
Wait. I was talking about the map of bundles not the corresponding map of sheaves. Take for example an inclusion $\mathcal{O}(-1)\to\mathcal{O}$ on $\mathbb{P}^1$: it's an inclusion of sheaves (it's injective on stalks), but it is not an injective map of line bundles (it's not injective on fibers). – Simplicius Oct 29 '12 at 12:32
@Simplicius, OK I see. – user18119 Oct 29 '12 at 16:40

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