# Induced action on Prym variety

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.)

• 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. Aug 19, 2021 at 5:19
• 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 $\pm 1$. Aug 19, 2021 at 5:53

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 $$(-D)$$.
• 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? Aug 23, 2021 at 4:21
• @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. Aug 23, 2021 at 16:07