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Let $X$ be an algebraic curve over $\mathbb C$ and let $E$ be a symplectic local system on $X$. For simplicity let us assume that $H^0(E)=H^2(E)=0$. Then $H^1(E)$ has a symmetric bilinear form and thus the square of its determinant is canonically trivialized.

$\mathbf{Question:}$ Is there a canonical trivialization of the determinant of $H^1(E)$ whose square is equal to the above trivialization of $\det(H^1(E))^{\otimes 2}$?

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  • $\begingroup$ Take a cell decomposition of $X$ which gives a chain complex computing the cohomology of $E$. The alternating product of determinants of this chain complex is the same before and after you take cohomology, and is a power of $\det(E)$, so can be trivialized by the canonical trivialization of $\det(E)$. Maybe this one works? $\endgroup$
    – Will Sawin
    Commented Nov 11 at 14:06
  • $\begingroup$ Where exactly are you using the symplectic structure on E? $\endgroup$
    – Alexander Braverman
    Commented Nov 11 at 14:38
  • $\begingroup$ To trivialize the determinant of $E$. $\endgroup$
    – Will Sawin
    Commented Nov 11 at 15:22
  • $\begingroup$ I actually think that it is important that the form on E is skew-symmetric (but I might be wrong) $\endgroup$
    – Alexander Braverman
    Commented Nov 11 at 16:41

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