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Let $\pi:X\rightarrow Y$ be a double cover of curves (smooth and projective), and $E$ a vector bundle on $X$ with a non-degenerate symmetric bilinear form $\phi:E\otimes E\rightarrow \mathcal O_X$. With this given, I can associate to it a non-degenerate symmetric bilinear form $$\pi_*\phi:\pi_*E\otimes\pi_*E\rightarrow \pi_*\mathcal O_X$$

My question is , how could one construct $\phi$ starting from $\pi_*\phi$? I mean if given $\psi:\pi_*E\otimes\pi_*E\rightarrow \pi_*\mathcal O_X$, could one construct a $\phi$ s.t $$\pi_*\phi=\psi$$

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    $\begingroup$ If $\psi$ is of the form $\pi_*\phi$, then $\psi$ is $\pi_*\mathcal{O}_X$-bilinear, not merely $\mathcal{O}_Y$-bilinear. $\endgroup$
    – Jason Starr
    Sep 8, 2015 at 11:35
  • $\begingroup$ So I think in general it is not true that any $\psi$ is of the form $\pi_*\phi$? $\endgroup$
    – Gest2015
    Sep 9, 2015 at 9:14
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    $\begingroup$ That is correct. There is a necessary and sufficient condition for existence of $\phi$: $\pi_*\mathcal{O}_X$-bilinearity of $\psi$. $\endgroup$
    – Jason Starr
    Sep 9, 2015 at 9:57
  • $\begingroup$ @JasonStarr, could you please detail a little bit your ansewer, I didn't see why this is sufficient condition! $\endgroup$
    – Gest2015
    Sep 9, 2015 at 13:42
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    $\begingroup$ Please read Exercise II.5.17(e) on p. 128 of Hartshorne's "Algebraic Geometry". $\endgroup$
    – Jason Starr
    Sep 9, 2015 at 15:27

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