5
$\begingroup$

What can one say about the relation between the Hodge decompositions of $H^\*(X,C)$ and $H^{*}(X_\sigma,C)$ for a complex algebraic smooth projective variety $X$ and $\sigma$ an automorphism of the field of complex numbers ? I am also interested in conjectural results, e.g. whether Standard Conjectures imply something about the relationship of the Hodge structure of conjugate varieties $X$ and $X_\sigma$.

For example, it appears that the Hodge numbers $h^{ab}=dim_C H^a(X,\Omega_X^b)$ and $h_\sigma ^{ab}=dim_C H^a(X_\sigma,\Omega_{X_\sigma}^b)$ are necessarily the same (being defined via de Rham cohomology); is this right (cf. discussion in the comments)? what other invariants/subspaces etc are the same or known to be possibly different ? Can one define a related Galois (or $Aut(C/Q)$) action on the Hodge structures, something like where an automorphism $\sigma\in Aut(C/Q)$ takes the Hodge structure of $X$ into that of $X_\sigma$'s ?

I know of several references that show that conjugate varieties may be rather different topologically. F.Charles, Conjugate varieties with distinct real cohomology algebras lists most references I know of and well, proves what's in the title. Serre constructed two two non-homotopic conjugated varieties with different fundamental groups; his proof exploits the fact that for a CM elliptic curve, its fundamental group has different $EndE$-module structures for different embeddings. Abelson constructs two non-homotopic conjugated varieties with the same finite fundamental groups.

$\endgroup$
7
  • $\begingroup$ I think you should reconsider your first question, "For example ..." (the answer is very easy). $\endgroup$ Commented Aug 12, 2012 at 14:33
  • $\begingroup$ Jason Starr: what is the answer to that part ? Do you mean to say that $dim_C H^a(X,Ω^b_X)$ is a definition of these numbers in algebraic de Rham cohomology ? $\endgroup$
    – o a
    Commented Aug 12, 2012 at 15:04
  • $\begingroup$ maybe this is just a confusion/silliness on my part, I admit... $\endgroup$
    – o a
    Commented Aug 12, 2012 at 15:04
  • 1
    $\begingroup$ Will Sawin: what does it mean 'the structure factors through the automorphism' $\sigma$ ; in what sense ? Then I am still confused whether Hodge numbers are preserved... $\endgroup$
    – o a
    Commented Aug 12, 2012 at 15:29
  • $\begingroup$ They are because they are the dimension of algebraically defined vector spaces. $\endgroup$
    – Will Sawin
    Commented Aug 12, 2012 at 15:55

1 Answer 1

9
$\begingroup$

As Jason & Will have already commented, the Hodge numbers are the same for conjugate pairs because of GAGA and the fact algebraic coherent cohomology behaves well with respect to field extensions. In fact, there would be an isomorphism of filtered vector spaces $$(\mathbb{H}^i(X,\Omega^\bullet), F^\bullet)\otimes_\mathbb{C} \mathbb{C}_\sigma \cong (\mathbb{H}^i(X_\sigma,\Omega^\bullet), F^\bullet)$$ Nevertheless, the Hodge structures need not be the same because this may not be compatible with integral structures. For example, if $X$ is an elliptic curve with transcendental $j$-invariant, we can find an automorphism with $j(X)\not= j(X_\sigma)$ so that $H^1(X)\not\cong H^1(X_\sigma)$ as Hodge structures. (So assuming I've understood your last question, that would be a no.)

$\endgroup$
2
  • 2
    $\begingroup$ This is true for non-rational $j$-invariant. Only for transcendental ones are the possible Hodge structures dense, though. $\endgroup$
    – Will Sawin
    Commented Aug 12, 2012 at 16:57
  • $\begingroup$ Thank you! Are there any references where people discuss this sort of questions in detail? $\endgroup$
    – o a
    Commented Aug 12, 2012 at 20:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .