Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

On a smooth complex projective variety of $\dim X=n$, we have $n$ complex tori associated to it via $J^k(X)=F^kH^{2k-1}(X,\mathbb{C})/H_k(X,\mathbb{Z})$ (assuming I've got all the indices right) called the $k$th intermediate Jacobian.

If $k=1$, we have $J^1(X)=H^{1,0}/H_1$, and so $J^1(X)\cong Jac(X)$ is an abelian variety (the bilinear form is a polarization because it has to be definite on each piece of the Hodge decomposition (I think) ) and is in fact isomorphic as PPAV's to the Jacobian of the variety.

If $k=n$, we have $H^{2n-1,1}/H_{2n-1}$, which is also a PPAV, and is in fact the Albanese of $X$.

The ones in the middle, however, the "true" intermediate Jacobians, are generally only complex tori. One example of an application is that Clemens and Griffiths proved that cubic threefolds are unirational but not rational using $J^2(X)$ for $X$ a cubic threefold.

So, what information do the intermediate Jacobians contain? I've been told that we don't really know much about that, but what is known, beyond Clemens/Griffiths?

share|improve this question
    
To elaborate on Emerton's answer below : though $J^k(X)$ is a complex torus in general, there is a subtorus $J_alg^k(X)$ which is polarized (hence an abelian variety). The AJ map restricted to cycles algebraicly equivalent to 0 factors through this, and sometimes this is surjective (ex : smooth projective curves !). But the terrifying and intriguing fact is that the image of AJ:Griffiths group=cycles hom. trivial/cycles alg.$\right arrow J^k/J^k_alg$ is countable ! And no one really knows how to caracterize this countable image inside a big complex torus. –  Simon Pepin Lehalleur Aug 8 '10 at 20:38
    
Sorry : the error stands for $\rightarrow J^k/J_alg^k$ –  Simon Pepin Lehalleur Aug 8 '10 at 20:39
    
Sorry (bis) : the error stands for $\rightarrow J^k/J_{alg}^k$ –  Simon Pepin Lehalleur Aug 8 '10 at 20:40

2 Answers 2

up vote 10 down vote accepted

Probably you know this, but just to be sure: they receive cycle class maps from codimension $k$ cycles. More precisely, if $Z$ is a cycle on $X$ of codimension $k$ that is cohomologically trivial, then it gives an element in $J^k(X)$. This map, from cohomologically trivial cycles to $J^k(X)$, is a (generalized) Abel-Jacobia map.

If we let $CH^k(X)$ be the Chow group of codimension $k$ cycles on $X$, then we get a filtration on $CH^k(X)$, namely $CH^k(X) \supset \text{ cohomologically trivial cycles} \supset \text{ Albanese trivial cycles}.$ (Albanese trivial means "has trivial image in $J^k(X)$.) It is conjectured that this filtration can be further extended; this is part of the Bloch--Beilinson series of conjectures on Chow groups and motives.

On the other other hand, if $X$ is defined over a number field $K$, and we look just at cycles defined over $K$, then it is conjectured that Albanese trivial implies rationally trivial (i.e. any cohomologically trivial cycle that is also is in the kernel of themap to $J^k(X)$ (and is defined over a number field ) is in fact rationally equivalent to zero). I believe this conjecture is extremely wide open at the moment.

share|improve this answer
    
I did know the first bit (though not the stuff about number fields), but I didn't really specify, so +1 for a good answer, going to leave this open a few more days to see if I get anything else. –  Charles Siegel Feb 2 '10 at 20:37

Recently I learned from a talk of Nick Addington one beautiful classical example where intermediate Jacobians contain all information about the variety. Namely, if we consider an intersection of two quadrics in $\mathbb CP^n$ then to such a variety we can associate a hyperelliptic curve. To do this we consider the corresponding pencil of quadrics. Singular quadrics in the pencil define a finite number of points on $CP^1$ and we take the double cover of $CP^1$ with the ramification at these points. The Jacobian of the hyperelliptic curve is isomorphic to the intermediate Jacobian of the intersection of to quadrics (this is the PhD of Miles Reid). The consruction was first considered by Weil. All this is very well explained in the introduction of the article of Nick http://arxiv.org/abs/0904.1764. You can reconstruct the hyperelliptic curve from its Jacobian and you can reconstructs the intersection of two quadrics from the curve.

Intermediate Jacobians of the intersection of 3 quadrics were considered by Turin. He proved the corresponding Torelli theorem: On intersections of quadrics. Russ. Math. Surv., 30, 51–105, 1975.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.