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 trivilization of $det(H^1(E))^{\otimes 2}$?

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 $\det(H^1(E))^{\otimes 2}$?

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 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 trivilization of $det(H^1(E))^{\otimes 2}$?

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 trivilization of $det(H^1(E))^{\otimes 2}$?