# Abel-Jacobi map over a field

Suppose $$X$$ is a smooth projective variety over a field $$k\subset \mathbb{C}$$. Let $$CH^r(X)_{hom}$$ be the Chow group of codimension $$r$$ cycles defined over $$k$$ and homologous to zero. The usual Abel-Jacobi map is defined on $$CH^r(X(\mathbb{C}))_{hom}$$ taking values in the Griffiths intermediate Jacobian $$IJ(H^*(X))$$.

Question: If this is an abelian variety, is it defined over the field $$k$$?

• Barry Mazur has asked this question: math.harvard.edu/~mazur/papers/page37.pdf There is work on this question by Achter, Casalaina-Martin, and Vial: arxiv.org/abs/1410.5376 Jun 22 '19 at 9:33
• The Griffiths intermediate jacobian is not usually an abelian variety. For example for a general abelian variety X, in codimension 1<k<d-1, for d the dimension of X Even if it is, it is not usually defined over the field of definition of the variety. Jun 22 '19 at 13:54
• See also arxiv.org/abs/1903.08015. Jun 23 '19 at 7:31
• Since this seems to be developing into a "big list", let me mention the earliest results in this direction (that I know of) are due to Welters, "Abel-Jacobi isogenies for certain types of Fano threefolds": mathscinet-ams-org.proxy.library.stonybrook.edu/mathscinet/… I recall that, at least conjecturally, there is a characterization of the isogeny class of the Griffiths intermediate Jacobian of a Fano manifolds in terms of motivic cohomology. So that would imply that the Jacobian is defined over an algebraic extension of $k$. Jun 23 '19 at 11:59
• Also, for one class of projective varieties that are not rationally connected yet have projective Griffiths intermediate Jacobian, Yi Zhu has given a purely algebraic construction (also working in positive characteristic) in Chapter 2 of his thesis: math.stonybrook.edu/alumni/2012-Yi-Zhu.pdf Jun 23 '19 at 12:03

## 1 Answer

Let me just point out that a necessary condition for the $$r$$th intermediate jacobian can descent to an abelian variety over $$k$$, for $$k$$ a number field (or even a finitely generated over $$\mathbb{Q}$$, is that its only non zero $$2r+1$$-dimensional Hodge numbers are the middle two: $$h^{r,r+1}$$ and $$h^{r+1,r}$$. Is under this hypothesis that Mazur asked for the existence of such descent, as Jason Starr points out in the comments. This is verified for example in he case of Fano threefolds (see for example this question and the answer ).