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.



Intermediate Jacobian under group action
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 $J(X/G)=0$.


