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Let $X$ be a smooth Fano threefold with a finite group $G$ action. Assume that the orbit space $X/G$ is smooth. Is it true that $J(X/G)\cong J(X)^G$ As an abelian variety? Here, $J(X)^G$ is the $G$-invariant part of $J(X)$.

I am particular interested in the case that $G$ is $\mathbb{Z}_2=\langle 1,\tau\rangle$ and $\tau$ is an involution.

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No, because there is no reason for $J(X)^G$ to be connected. Here is a silly example: consider a smooth elliptic curve $C\subset \mathbb{P}^3$ given by $\sum x_i^2=\sum a_ix_i^2=0$; take for $X$ $\mathbb{P}^3$ with $C$ blown up, and let $\tau $ be the involution of $\mathbb{P}^3$ which changes the sign of one coordinate. We have a $G$-equivariant isomorphism $J(X) \cong JC $, and $\tau $ acts on $JC$ ($\cong C$) with 4 fixed points, while $J(X/G)=0$.

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  • $\begingroup$ Thanks for such nice example, If we know J(X)^G is connected, is this equality true? I am thinking about a particular example, say X is a smooth cubic threefold with a specific involution $\tau$ and I know the $\tau$-invariant part of J(X) is a Prym variety associated with a discriminant curve from the conic fibration. I was wondering if there is a particular nice situation such that this equality holds. I mean if J(X)^G has nice property, say irreducibility, smoothness, then does this equality hold? $\endgroup$
    – user41650
    Commented Dec 12, 2022 at 16:48
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    $\begingroup$ You must certainly ask for $J(X)^G$ to be connected, but I doubt that it is sufficient. If it holds you get an isogeny $J(X/G)\rightarrow J(X)^G$, but I don't see why it should be an isomorphism. $\endgroup$
    – abx
    Commented Dec 12, 2022 at 17:16

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