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We know the classical Leray-Hirsch theorem for fibrations. My question is, whether a similar statement also holds for flat, proper morphism? In particular, consider a faithfully flat, proper morphism $f:Y \to X$ with both $X$ and $Y$ non-singular, irreducible varieties over $\mathbb{C}$. Can we say that if for every $x \in X$, $H^q(f^{-1}(x), \mathbb{Q})$ is the same, then $R^qf_*\mathbb{Q}$ is a local system? More generally if for every $q$, $H^q(f^{-1}(x),\mathbb{Q})$ does not depend on the choice of $x$, then can we write $H^q(Y,\mathbb{Q})$ as a direct sum of $H^i(f^{-1}(x),\mathbb{Q}) \otimes H^{q-i}(X,\mathbb{Q})$ as $i$ varies? Any reference will be most welcome.

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  • $\begingroup$ I don't know much Hodge theory, but since $H^q(f^{-1}(x), \mathbf{Q})$ is the stalk of $R^qf_\ast \mathbf{Q}$ at $x$, and an equivalent characterization of local constancy is that all specialization maps are isomorphisms, it should follow that $R^qf_\ast \mathbf{Q}$ is a local system. $\endgroup$ Commented Apr 11, 2021 at 2:50
  • $\begingroup$ This might be overkill, but you might find the discussion in Section 6 of the paper by Brosnan and Chow "Unit interval orders and the dot action on the cohomology of regular semisimple Hessenberg varieties" useful. In particular, the results there do imply that the sheaves $R^q f_* \mathbb{Q}$ are local systems under your assumptions. $\endgroup$
    – naf
    Commented Apr 11, 2021 at 3:33

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Here's a version of Leray-Hirsch for a proper morphism not a priori assumed to be a fibration.

Suppose that:

  • $f \colon X \to Y$ is a proper morphism of smooth varieties,
  • all fibers of $f$ have the same Betti numbers,
  • for a generic point $y$ of $Y$, the restriction map $H^\ast(X,\mathbf Q) \to H^\ast(f^{-1}(y),\mathbf Q)$ is surjective ("Leray-Hirsch").

Then all sheaves $R^qf_\ast \mathbf Q$ are trivial local systems, the Leray spectral sequence degenerates, and $H^\ast(X,\mathbf Q) \cong H^\ast(Y,\mathbf Q) \otimes H^\ast(F,\mathbf Q)$ where $F$ denotes any fiber of $f$.

Proof: By the decomposition theorem, $Rf_\ast\mathbf Q$ is a sum of shifted perverse sheaves. Choose a dense open $U \subset Y$ over which these are local systems. This open subset contains the generic point $y$, so by the usual Leray-Hirsch principle these local systems are all trivial over $U$. The intermediate extension of these local systems gives a summand of $Rf_\ast\mathbf Q$, but the intermediate extensions are again trivial. By the assumption on the Betti numbers of the fibers there can be no further summands inside $Rf_\ast\mathbf Q$, and we are done.

Remark: I do not know an example of a morphism $f$ satisfying the assumptions of the argument, but which is not in fact a fibration.

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For any reasonable interpretation of the terms "$H^q(f^{-1}(x), \mathbb{Q})$ does not depend on $x$" or "...is the same", $R^qf_*\mathbb{Q}$ would be a local system. So then by Deligne, the Leray spectral sequence degenerates to give isomorphisms $$ H^q(Y,\mathbb{Q}) = \bigoplus_i H^i(Y, R^{q-i}f_*\mathbb{Q})$$ If the monodromies of the all the above local systems are trivial, then you would get the sort of isomorphisms that you are asking for, but not otherwise.

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  • $\begingroup$ Could you give a reference or idea on how to prove that $R^qf_*\mathbb{Q}$ is a local system? $\endgroup$
    – user43198
    Commented Apr 10, 2021 at 23:35
  • $\begingroup$ I interpreted "...is the same" as saying that all fibers have the same Betti numbers. It's not clear to me what assumption to add to deduce that the higher direct images are trivial local systems. One observation is that by the decomposition theorem it would be enough to prove it on a dense open subset of X. $\endgroup$ Commented Apr 11, 2021 at 0:24
  • $\begingroup$ @DanPetersen Yes, by "... is the same" I mean the same Betti numbers. Also, on an open dense subset of X it is going to be a local system (using the theorem on generic smoothness). Why does same Betti numbers imply local system? And could you elaborate a bit more on what application of the decomposition theorem you have in mind. $\endgroup$
    – user43198
    Commented Apr 11, 2021 at 0:28
  • $\begingroup$ If it's a trivial local system on a dense open it's really easy to compute the intermediate extension! $\endgroup$ Commented Apr 11, 2021 at 0:31

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