Polarizations on intermediate Jacobians - MathOverflow

Polarizations on intermediate Jacobians

2010-06-01T15:43:34Z

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?

Answer by Charles Siegel for Polarizations on intermediate Jacobians

2010-06-01T16:06:03Z

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.

Answer by Donu Arapura for Polarizations on intermediate Jacobians

2010-06-01T16:09:04Z

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.