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Everything is over $\mathbb C$. Let $X\to \mathbb P^1$ be a fiber bundle with fiber some smooth projective variety $F$. If we have another such bundle $Y \to \mathbb P^1$, then, for which $p,q$ we have the Hodge number $h^{p,q}(X)=h^{p,q}(Y)$?

Apparently, the outer Hodge numbers (essentially $h^{p,0}$) are equal since they are birational equivalent. But can we expect more? Is it true that all the Hodge numbers of $X$ and $Y$ are equal? (I know this is true for projective bundles, i.e. $F\cong \mathbb P^n$, but I have no idea for general bundles. I guess it is not.)

Thanks!

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    $\begingroup$ Could you clarify what you mean by a fiber bundle? For instance, do you take a fiber bundle to mean that this fiber bundle is locally trivial in the Zariski topology? Or something else? $\endgroup$
    – KReiser
    Commented Dec 12, 2019 at 10:18
  • $\begingroup$ @KReiser Yes, I mean locally trivial in the Zariski topology. $\endgroup$
    – Akatsuki
    Commented Dec 12, 2019 at 15:05

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It is true in general, the keyword is Grothendieck ring of varieties. I am not sure for a reference, but it's probably in the recent book by Chambert-Loir, Nicaise and Sebag about motivic integration.

There is a ring $\mathrm K_0(\mathrm{Var}/\Bbb C)$ defined as follows : take the free abelian group generated by algebraic varieties over $\Bbb C$. Define a product as $[X] \cdot [Y] = [X \times Y]$, and impose the relations $[X] = [U] + [Z]$ for any open set $U \subset X, Z = X \backslash U$.

Theorem : There is a unique morphism $\mathrm K_0(\mathrm{Var}/\Bbb C) \to \Bbb Z[u,v]$ that coincide with Hodge polynomials $\sum_{p,q}h^{p,q}u^pv^q$ on projective smooth varieties.

It is easy to check that if $X \to B$ is a fiber bundle in the Zariski topology with fiber $F$, then $[X] = [F] \cdot [B]$ in $K_0$. In particular, with your notation $[X] = [Y]$ in $\mathrm{K}_0(\mathrm{Var}/ \Bbb C)$ so their Hodge polynomial coincide.

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  • $\begingroup$ In the past there has been much wailing and gnashing of teeth here on MSE about actually demonstrating that $[X]=[F]\cdot[B]$, see for instance these counterexamples which show that the obvious strategy of "find a Zariski open so the fibration is trivial, cut, repeat" does not work. Can you provide the proof of this statement or a reference to a source where it is proven? $\endgroup$
    – KReiser
    Commented Dec 12, 2019 at 9:57
  • $\begingroup$ @KReiser : Thanks a lot for all your comments it's really helpful as often I forget to check details. I assumed that "fiber bundles" meant locally trivial fiber bundles (for the Zariski topology). If it means "fibration with fiber $F$" not indeed my answer doesn't work (but I don't know any counter-example to the OP question). Maybe the OP can say what is meant by "fiber bundles"? $\endgroup$
    – Nicolas Hemelsoet
    Commented Dec 12, 2019 at 10:10
  • $\begingroup$ @KReiser : I realize now that this notion of "fiber bundles" in the Zariski topology is too strong and probably not so interesting, for example the monodromy should be trivial. So indeed it doesn't really answer the question. $\endgroup$
    – Nicolas Hemelsoet
    Commented Dec 12, 2019 at 10:15
  • $\begingroup$ Oh boy! I had read "fibration" instead of fiber bundle. If the OP does mean a Zariski-locally trivial fiber bundle, the strategy in your answer is correct as you say and I'm sorry to have troubled you as a result of my hasty reading. (The secret about these counterexamples is that I got hit with this issue on a separate post a few years ago - you can get around the issue by using the fact that $K_0$ is invariant under passing from schemes/C to algebraic spaces, where the stated strategy actually works. So your post is on the whole correct, up to possibly one argument being adjusted.) $\endgroup$
    – KReiser
    Commented Dec 12, 2019 at 10:15
  • $\begingroup$ @KReiser : Wonderful !! Do you know a reference about this invariance of $K_0$ ? $\endgroup$
    – Nicolas Hemelsoet
    Commented Dec 12, 2019 at 10:25

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