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Dec 20, 2022 at 2:44 vote accept user41650
Dec 12, 2022 at 17:16 comment added abx 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.
Dec 12, 2022 at 16:48 comment added user41650 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?
Dec 12, 2022 at 16:14 history answered abx CC BY-SA 4.0