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Let $X$ be a Kahler variety of dimension $n$. For each odd number $2k-1 \leq n$ one can consider the $k$-th intermediate Jacobian, that is, the complex torus $$J^{k}X := \frac{H^{2k+1}(X, \mathbb{C})}{F^k H^{2k+1}(X, \mathbb{C}) + H^{2k+1}(X, \mathbb{Z})},$$ where $F^{\cdot}$ is the Hodge filtration. In general $J^k X$ is not an abelian variety, except for the extreme cases of the Picard and Albanese varieties (when $X$ itself is algebraic).

Are there any criteria to determine whether $J^K X$ is polarized? Or have some nontrivial cases where $J^k X$ is polarized been worked out?

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    $\begingroup$ If $X$ is a Kahler manifold which is projective, then any choice of an ample line bundle on $X$ induces a natural polarization on $J^{k}X$. This is a $(1,1)$ integral class on $J^{k}X$ which as a Hermitian pairing is non-degenerate but need not be definite. The corresponding holomorphic line bundle is non-degenerate and its powers have only one non-trivial cohomology group, namely the cohomology of degree the number of negative eigenvalues. The corresponding cohomology classes can be viewed as forms with coefficients in the line bundle. These are the analogues of the theta functions. $\endgroup$
    – Tony Pantev
    Commented Jun 2, 2010 at 14:01

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If enough Hodge numbers vanish so that the Hodge structure $H^{2k+1}(X)$ has level one, then $J^kX$ should be an abelian variety. This applies to Fano (e.g. cubic) 3-folds for example.

Later that day: Partly in response to Charles Siegel's comment/question, let me sketch a proof of a slightly more general statement. Suppose X is a projective rather than just Kaehler (which I forgot to say before), so $H$ has a polarization $Q$. Assume further that $$H^{2k+1}(X) = H= H^{pq}\oplus H^{qp}$$ has only two terms. Let $G$ be the same thing as $H$ viewed as a weight one structure. More precisely, the lattices $G_Z=H_Z$ are the same, and $G^{10}=H^{pq}$.

Then one sees that $J^kH= G^{01}/G_Z$, and that $\pm Q$ gives a polarization on $G$. So this is abelian variety.

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  • $\begingroup$ Something related that I've been wondering...if every other Hodge number is zero, so that, say, you only have the terms where the cup product (say, in the middle cohomology) has the same sign? Like, you have $H^{5,0}\oplus H^{3,2}$ but $H^{4,1}=0$. Is that enough, or still, not quite? $\endgroup$
    – Charles Siegel
    Commented Jun 1, 2010 at 17:16
  • $\begingroup$ @Donu, sorry for this very late comment to this question. I was wondering if you could explain me why in your argument you really need that the Hodge structure has level 1. Thank you for that. $\endgroup$
    – User3773
    Commented Jul 7, 2016 at 15:29
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The best known examples are the cubic threefolds and the quartic double-solids (that is, double covers of $\mathbb{P}^3$ branched along a quartic). In general, this works for quadric bundles, see Beauville's paper.

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