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Suppose $Y$ is a smooth projective variety of dimension $2p-1$ over $\mathbb{C}$. I have a few questions about the $p^{~\text{ th}}$ intermediate Jacobian $J^p(Y)$ of $Y$.

  1. Does it come from (i.e. is analytification of) an abelian variety? The natural map $H^{2p-1}(Y,\mathbb{R})\rightarrow\frac{H^{2p-1}(Y, \mathbb{C})}{F^pH^{2p-1}(Y,\mathbb{C})}~$ is an isomorphism of real vector spaces, so I tried to check if the Pioncare duality pairing $H^{2p-1}(Y,\mathbb{R})\otimes H^{2p-1}(Y,\mathbb{R})\rightarrow\mathbb{R}$ induces a Riemann form on $\frac{H^{2p-1}(Y, \mathbb{C})}{F^pH^{2p-1}(Y,\mathbb{C})}~$, but I'm not sure if it satisfies the positive-definiteness (of the corresponding Hermitian form) requirement.

  2. Let $X$ be a smooth projective curve over $\mathbb{C}$. Does it matter at all for the situation in 1 if we assume $Y=X^{2p-1}$?

  3. What I'm really interested in is $J^p\left(H^1(X)^{\otimes 2p-1}\right)\subset J^p(X^{2p-1})$, where the first $J^p$ is in the sense of Carlson. (For a mixed Hodge structure $H$, $J^p(H)$ is by definition $\frac{H_{\mathbb{C}}}{F^pH_{\mathbb{C}}+H_{\mathbb{Z}}}~$. I'm not sure if the definition was first given by Carlson, sorry if I should say in someone else's sense.) I wanted to know if $J^p\left(H^1(X)^{\otimes 2p-1}\right)$ comes from an abelian variety, and the reason I thought about 1 and 2 is that I was (maybe naively) hoping to get a Riemann form from the Poincare pairing on $H^{2p-1}(X^{2p-1},\mathbb{R})$. Any thought on whether or not $J^p\left(H^1(X)^{\otimes 2p-1}\right)$ comes from an abelian variety? Also, does this object naturally appear anywhere? Any thing interesting about it?

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    $\begingroup$ Unfortunately no, the Griffiths intermediate Jacobian is not abelian in general (no Riemann form), although it is for level one Hodge structures. I doubt that it would work even in your case 3. $\endgroup$ May 9, 2014 at 0:08
  • $\begingroup$ Hi Donu. Can you clarify on the level one portion? Isn't $J^p$ of the maximal level one sub-HS of $H^{2p-1}$ abelian? Also, have you by any chance seen the object in Question 3 above appearing somewhere interesting? $\endgroup$
    – P.E.
    May 9, 2014 at 1:00
  • $\begingroup$ P.E., Hodge level one means that the Hodge decomposition reduces to $H^{p,p-1}\oplus H^{p-1,p}$ which is a very strong assumption. Unfortunately, this fails in your example 3, except in the trivial cases. I hope that this is not something you really needed. $\endgroup$ May 9, 2014 at 8:49
  • $\begingroup$ Thanks. I had misread the first comment, as if it was saying the Hodge structures giving rise to Griffiths intermediate Jacobians are of level one and yet the resulting torus is not an abelian variety. My bad. $\endgroup$
    – P.E.
    May 9, 2014 at 17:05

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