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Let $f\colon X\to \mathbb{A}^1$ be a smooth projective morphism of complex algebraic manifolds, where the target $\mathbb{A}^1$ is the affine line. Are there any restrictions on the Hodge structures on the cohomology groups of fibers of $f$ over different complex points of $\mathbb{A}^1$? (Say are there examples where these Hodge structures are not isomorphic to each other?)

I apologize if this question is not of the research level; I am not an algebraic geometer. If there is a reference, it would be helpful.

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    $\begingroup$ What is a good example of a nontrivial such family? $\endgroup$
    – Piotr Achinger
    Commented Sep 14, 2017 at 12:31
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    $\begingroup$ Such a variation of Hodge structure is trivial, by e.g. the theorem of the fixed part, as $\mathbb{A}^1$ is simply connected, so all the Hodge structures are isomorphic. More is true: by work of Viehweg, Moller, Zuo, and others, "most" moduli spaces are hyperbolic, and so contain no $\mathbb{A}^1$'s... $\endgroup$
    – Daniel Litt
    Commented Sep 14, 2017 at 13:07
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    $\begingroup$ Of course one can make examples; for example, families of del Pezzo surfaces; or one may take $X\times\mathbb{A}^1$ for any $X$ containing a rational curve, so that the map $X\times \mathbb{A}^1\to\mathbb{A}^1$ has a section, and blow up the image of the section. But I don't know any examples with really "interesting" geometry, for the reasons described in my previous comment. $\endgroup$
    – Daniel Litt
    Commented Sep 14, 2017 at 13:09
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    $\begingroup$ @DanielLitt: Thank you for your interesting comment. Could you state explicitly the theorem of the fixed part? This could be a final answer to my question. $\endgroup$
    – asv
    Commented Sep 14, 2017 at 14:31

1 Answer 1

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See Theorem 11 on page 191 of these notes. A special case is as follows.

Theorem of the fixed part. Let $S$ be a smooth quasiprojective variety, and $V$ a variation of $\mathbb{Q}$-Hodge structures on $S$ (for example, $R^i\pi_*\underline{\mathbb{Q}}$, for $\pi: X\to S$ a smooth projective morphism). Then $H^0(S,V)$ naturally admits a Hodge structure, such that the map $H^0(S, V)\to V_s$ is a morphism of Hodge structures for any $s\in S$.

In particular, if $S$ is simply connected (as it is in your case, where we take $S=\mathbb{A}^1_{\mathbb{C}}$), then the map $H^0(S,V)\to V_s$ is an isomorphism for any $S$. Hence all the $V_s$ are isomorphic.

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  • $\begingroup$ Why does the second paragraph follow from the first? I understand all fibres must have isomorphic cohomology if the base is simply connected. But why must the map $H^0(S,V)\rightarrow V_s$ be an isomorphism of Hodge Structures? $\endgroup$
    – Dr. Evil
    Commented Jul 12 at 10:30
  • $\begingroup$ @Dr.Evil This follows from the last clause in the statement of the theorem. An isomorphism of vector spaces which respects Hodge structures is an isomorphism of Hodge structures. $\endgroup$
    – Daniel Litt
    Commented Jul 12 at 12:54

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