2
$\begingroup$

Let $X$ be a smooth complex algebraic variety. From Deligne's work, we know that the have a Mixed Hodge structure over its (rational) compactly supported cohomology $H^{*}_c(X,\mathbb{Q})$. With this, one can define the Euler-Hodge polynomial $$E(X,x,y)=\sum_{p,q}e_{p,q}x^py^q$$ where $$e_{p,q}=\sum_i (-1)^{i} dim Gr^W_{p+q} Gr_p^F H^i_c(X,\mathbb{Q}) $$.

Let us suppose we have another complex smooth algebraic variety $P$ with an algebraic map $f:P \to X$ which is a fiber bundle with fiber $F$. I would like to know if there is any result that ensures us that $$E(X,x,y)E(F,x,y)=E(P,x,y).$$

This should somehow be thought of as a generalization of the similar equation which is true for Euler characteristic.

$\endgroup$
6
  • $\begingroup$ If it's a Zariski fibre-bundle, then it's true. You can use the fact that the $E$-polynomial only depends on the class in the Grothendieck ring of varieties. $\endgroup$ Apr 12, 2021 at 15:22
  • $\begingroup$ If it just a fibre bundle in the etale topology? Can we still say it's true? $\endgroup$ Apr 12, 2021 at 15:27
  • $\begingroup$ I don't know. I remember discussing it here math.stackexchange.com/questions/3472329/… but we couldn't come to a conclusion. $\endgroup$ Apr 12, 2021 at 16:16
  • $\begingroup$ If that could help I'm looking for a statement with both $X,P$ assumed to be affine. But thank you for the reference and the other answer! $\endgroup$ Apr 12, 2021 at 16:55
  • 3
    $\begingroup$ @Tommaso, perhaps I'm being silly but isn't the squaring map from punctured affine line to itself a counterexample if u relax the Zar loc trivial condition to étale loc triv? $\endgroup$
    – EBz
    Apr 12, 2021 at 17:39

2 Answers 2

4
$\begingroup$

If $\pi_1(X)=0$, or more generally if the fundamental group of $X$ acts trivially on the compact support cohomology of the fiber, this is true because of the Leray spectral sequence. Otherwise it is almost never true (a simple counterexample was given by EBz in the comments: the squaring map $\mathbb G_m \to \mathbb G_m$).

$\endgroup$
5
  • $\begingroup$ Is there like an intuitive way of see why this should fall? Or like a geometric explanation? I thought this was true also because of Katz theorem which relates E polynomial to number of points over finite fields. So connectedness hypothesis would not be useful at all? $\endgroup$ Apr 12, 2021 at 20:39
  • $\begingroup$ No, connectedness wouldn't help, either, though examples become more complicated. You may for example think of the projection from the Legendre family of elliptic curves down to $\mathbb P^1 \setminus \{0,1,\infty\}$. $\endgroup$ Apr 12, 2021 at 21:41
  • $\begingroup$ Okok really thank you for the counterexamples :) $\endgroup$ Apr 12, 2021 at 21:54
  • $\begingroup$ What if $E(F, x, y)=1$? $\endgroup$
    – user178279
    Apr 15, 2021 at 13:43
  • $\begingroup$ @virkkunen This happens only if $F$ is a point, which is not so interesting. $\endgroup$ Apr 16, 2021 at 8:21
2
$\begingroup$

There has been much research on computing $E$-polynomials of character varieties. You can find a lot of general theory by reading those papers (just do a search for key words).

In particular, the theorem you want is Proposition 2.1 here:

Hodge polynomials of the SL(2,C)-character variety of an elliptic curve with two marked points by Marina Logares, Vicente Muñoz.

I quote:

Suppose that $B$ is connected and $\pi : Z \longrightarrow B$ is an algebraic fibre bundle with fibre $F$ (not necessarily locally trivial in the Zariski topology) and that the action of $\pi_1(B)$ on $H^∗_c(F)$ is trivial. Suppose that $Z, F, B$ are smooth. Then $e(Z) = e(F)e(B).$

The hypotheses hold in particular in the following cases:

  • $B$ is irreducible and $\pi$ is locally trivial in the Zariski topology.
  • $\pi$ is a principal $G$-bundle with $G$ a connected algebraic group.
$\endgroup$
1
  • 1
    $\begingroup$ The result is true without assuming that the spaces are smooth. It follows from the Leray spectral sequence for compact support cohomology. $\endgroup$ Apr 13, 2021 at 8:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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