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Let $C$ be a smooth projective curve of genus $g$ with an involution $\iota: C \to C$. We have the quotient map $\pi: C \to C/\iota$, with $C/\iota$ a smooth curve of genus $h$. The pullback map $\pi^{*}: \text{Jac}(C/\iota) \to \text{Jac}(C)$ in this case is injective, and we define the Prym variety as the cokernel:

$$0 \to \text{Jac}(C/\iota) \to \text{Jac}(C) \to \text{Prym}(\pi) \to 0$$

Note that $\text{Prym}(\pi)$ is an Abelian variety of genus $g-h$. Now consider the action by $\iota^{*}$ on $\text{Jac}(C)$. The induced action on $\text{Jac}(C/\iota)$ is trivial, i.e. line bundles pulled back from the quotient are $\iota$-invariant. I am nearly certain that the induced action on $\text{Prym}(\pi)$ is by $\pm 1$ and therefore has $2^{2g-2h}$ fixed points. But I am struggling to prove this. Anyone have any tips? I'm unsure how to prove that a particular involution on an Abelian variety is simply the standard $\pm 1$ action.

(I think an equivalent statement is that the only deformations of invariant degree $0$ line bundles on $C$ come from deformations on $C/\iota$. But there must be a more direct way to argue.)

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    $\begingroup$ Note that the tangent space at identity of $J(C)$ is $H^0(C,\Omega_C)^{*}=H^1(C,\mathcal{O}_C)$ on which $\iota$ operates. $\endgroup$
    – Kapil
    Aug 19, 2021 at 5:19
  • $\begingroup$ I see; I was over-complicating this. So the idea is that $H^{0}(C, \Omega_{C})^{*}$ decomposes as an $\iota$-representation. The subspace where the action is trivial corresponds to forms pulled back from the quotient, and the subspace where $\iota$ acts non-trivially is the tangent space of the Prym. This is what you're getting at, correct? So this implies the global action on the Prym is by $\pm 1$. $\endgroup$
    – Benighted
    Aug 19, 2021 at 5:53

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Kapil's suggestion to express the tangent space as $H^1(C,\mathcal O_C)$ is great in characteristic $0$. In characteristic $p$, and in particular characteristic $2$, the statement is still true, but you can't detecet it from the tangent space.

Instead, note that the statement is equivalent to the claim that for $L$ a line bundle of degree $0$ on $C$, $L \otimes \iota^* L$ is isomorphic to the pullback of a line bundle from $C/\iota$.

One can check explicitly that $L \otimes \iota^* L$ is isomorphic to the pullback of $\det (\pi^* L )\otimes \det (\pi^* \mathcal O)^{-1}$. To do this, note that $\pi^* \pi_* L$ is an extension of $\iota^* L (-D)$ by $L$, for $D$ the branch divisor, so its determinant is $L \otimes \iota^* L(-D)$, and the $ \det (\pi^* \mathcal O)^{-1}$ term cancels the $(-D)$.

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  • $\begingroup$ I think I understand the last paragraph (although I believe you meant $\pi_{*}L$ and $\pi_{*} \mathcal{O}$, if I'm not mistaken) but I'm not seeing how your reformulation of the statement is equivalent. Can you explain this briefly? $\endgroup$
    – Benighted
    Aug 23, 2021 at 4:21
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    $\begingroup$ @Benighted By definition, points of the Prym variety correspond to equivalence classes of line bundles of degree $0$ modulo pullbacks from $C/\iota$. To check that $\iota$ and the inverse map act the same on each point, we consider an arbitrary point, take a line bundle $L$ representing that class, and check that $L^{-1}$ is isomorphic to $\iota^* L$ modulo pullbacks, i..e that $\iota^* L \otimes L$ is isomorphic to a pullback. $\endgroup$
    – Will Sawin
    Aug 23, 2021 at 16:07

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