Clemens-Griffiths criterion for 3-fold says that if a smooth projective threefold $X/\mathbb C$ is rational, then the intermediate Jacobian $J(X)$ is isomorphic to product of Jacobians $J(C_1)\times \cdots \times J(C_k)$ for some curves $C_i$ as principally polarized abelian varieties.

I'm wondering if the condition is also sufficient (particularly for Fano varieties). In other words, is there a Fano threefold whose intermediate Jacobian is product of Jacobians of curves but it is irrational?

Edited: As abx pointed out, Voisin's example and Artin-Mumford's example are not strictly speaking Fano varieties, as they arise from desingularization of nodal quartic double solids, so anti-canonical bundle is not ample on exceptional divisors. It is still interesting to ask if there is such a (smooth) Fano example.

  • $\begingroup$ You should take a look at the book "Abel-Jacobi Isogenies" by Gerd Welters. Welters describes the intermediate Jacobians for most Fano threefolds. $\endgroup$ Jul 11, 2023 at 10:44
  • $\begingroup$ @JasonStarr Thanks for the reference. I took a quick look at it but did not find the solution to the question. There is good stuff in it though. I'm sure it will be useful for me at some point. $\endgroup$
    – AG learner
    Jul 12, 2023 at 14:54

1 Answer 1


I believe this is answered in Claire Voisin's paper, Cor. 3.7 https://webusers.imj-prg.fr/~claire.voisin/Articlesweb/voisingargnanovrai.pdf

If I understood correctly, she shows that (the desingularization of) a very general quartic double solid with at most 7 nodes is not even stably rational. The case of exactly 7 nodes has a 3 dimensional intermediate Jacobian, hence a Jacobian of a curve of genus 3.

Edit: As pointed out in a comment by abx, this example is not Fano.

  • $\begingroup$ Thanks! That’s a good catch! Maybe Artin-Mumford’s example also works, it is quartic double solid with 10 nodes, its intermediate Jacobian is probably trivial. $\endgroup$
    – AG learner
    Jul 11, 2023 at 23:45
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    $\begingroup$ Good point. I think Claire was trying to give an example where both other tools failed. I.e. for her example it seems the Artin - Mumford invariant vanishes. $\endgroup$
    – roy smith
    Jul 12, 2023 at 1:09
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    $\begingroup$ The nodal quartic solid and the Artin-Mumford exmple are not Fano. $\endgroup$
    – abx
    Jul 12, 2023 at 4:28
  • $\begingroup$ @abx Thanks for pointing it out. I think Clemens-Griffiths criterion works in general for threefolds with $h^{3,0}=h^{1,0}=0$. When I wrote "Fano" in the original post, I was thinking about these conditions that are a bit more general that $K_X^*$ being ample. But now it seems the original question for Fano 3-folds is still interesting. I edited the question accordingly. $\endgroup$
    – AG learner
    Jul 12, 2023 at 14:45
  • $\begingroup$ @roysmith Let me unaccept your solution temporarily to promote future answers for Fano examples. Thanks for your answer though! $\endgroup$
    – AG learner
    Jul 13, 2023 at 15:46

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