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.)



Induced action on Prym variety
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 $(-D)$.


