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Let $f\colon X\to Y$ be a surjective morphism of smooth projective varieties. If the decomposition theorem for $f$ is given by $$Rf_*\mathbb{C} \simeq \bigoplus_i R^if_*\mathbb{C}[-i],$$ what are the necessary conditions the morphism $f$ must satisfy? Is there an example where such a morphism is not smooth but the decomposition theorem nonetheless looks like the above?

Edit: If I additionally assume that $R^if_*\mathbb{C}$ are local systems for all $i$, can one conclude that $f$ is smooth? I understand that the limit mixed Hodge structure is pure as there is no monodromy around the singular fibers.

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    $\begingroup$ One can find examples where $f$ is not smooth, even with the extra condition in the edit, e.g., this can happen when there are nonreduced fibres. For a specific example, take $X$ to be a hyperelliptic surface and $f$ the natural map to $\mathbb{P}^1$. $\endgroup$
    – naf
    Commented Sep 2, 2020 at 10:47
  • $\begingroup$ Thank you very much for the comment. Unfortunately I am not sure how to see that the decomposition theorem for your example satisfies all the properties in my question. If you could please expand a bit, I will really appreciate it. $\endgroup$
    – guest0803
    Commented Sep 3, 2020 at 7:10
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    $\begingroup$ OK, here are some details. To make this short, although it is not necessary, I will assume the result in Donu Arapura's answer. It is obvious that $R^0f_*\mathbb{C}$ and $R^2f_*\mathbb{C}$ are constant sheaves, so it suffices to consider $R^1f_*\mathbb{C}$. All the fibres of the map are elliptic curves, but some are not reduced. Using proper base change, the stalks of $L :=R^1f_*\mathbb{C}$ are all isomorphic. To see that it is a local system it suffices to show that $L$ has no summand supported on any point corresponding to a nonreduced fibre. $\endgroup$
    – naf
    Commented Sep 3, 2020 at 14:44
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    $\begingroup$ cont. In the simplest kind of hyperelliptic surface, there are 4 singular fibres and the local behaviour around each of them is the same. If there are a skyscraper sheaf corresponding to these points, using the Leray spectral sequence to compute $H^1(X,\mathbb{C})$, you get that this is at least 4 dimensional . However, it is easy to see from the description of $X$ as a quotient of a product of elliptic curves that this cohomology group is actually 2 dimensional. (There should, of course, be a purely local proof, but I have not worked that out.) $\endgroup$
    – naf
    Commented Sep 3, 2020 at 14:53
  • $\begingroup$ Dear @ulrich, thank you so much for the beautiful explanation. This seems to be an answer and I would happily accept it if you could please post it as an answer instead of a comment. Perhaps I am being slow, but the only point I still do not understand is why $R^1f_*\mathbb{C}$ could not be of the form $IC(\mathbb{L})$ for some non-trivial rank 2 local system $\mathbb{L}$ outside the 4 points. I agree with all the other statements you made. $\endgroup$
    – guest0803
    Commented Sep 5, 2020 at 14:38

2 Answers 2

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Here is an example where $f$ is not smooth but $Rf_* \mathbb{C}$ behaves as if it were:

Let $X$ be a hyperelliptic surface and $f$ the natural morphism to $Y \cong\mathbb{P}^1$. All reduced fibres of $f$ are elliptic curves, but there is a nonzero number of nonreduced fibres, the number dependending on $X$.

The singular cohomology of $X$ is given by $H^0(X, \mathbb{C}) \cong H^4(X,\mathbb{C}) \cong \mathbb{C}$ and $H^1(X, \mathbb{C}) \cong H^3(X, \mathbb{C}) \cong \mathbb{C}^2$. Furthermore, the restriction map $H^1(X, \mathbb{C}) \to H^1(F, \mathbb{C})$ is an isomorphism for any fibre $F$ of $f$.

It is clear that $R^0 f_* \mathbb{C}_X \cong R^2f_* \mathbb{C}_X \cong \mathbb{C}_Y$, so let us consider $R^1f_* \mathbb{C}_X$. Since $H^1(X, \mathbb{C}) \cong \mathbb{C}^2$, we get a natural map $\mathbb{C}^2_Y \to R^1f_* \mathbb{C}$. By evaluating this on stalks and using the proper base change theorem, we see that this is an isomorphism.

Finally, since we know exactly what each sheaf $R^i f_* \mathbb{C}_X$ is, the same proof as in the case $f$ smooth can be used to show that $Rf_*\mathbb{C}_X$ decomposes as a direct sum of its (shifted) cohomology sheaves.

One may ask if a similar statement holds whenever all reduced fibres are smooth (and say $f$ is flat); I did not think about this. It would also be interesting to know if there are examples with non-smooth reduced fibres. Also, note that in the example $R^1 f_* \mathbb{Z}_X$ is not a local system.

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  • $\begingroup$ Oh, it's you... $\endgroup$
    – Donu Arapura
    Commented Oct 11, 2020 at 21:17
  • $\begingroup$ @DonuArapura: Yes, Hi Donu! $\endgroup$
    – naf
    Commented Oct 12, 2020 at 2:33
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I don't know how to characterize such morphisms, which I think is your first question. However, this can certainly happen, even if $f$ is not smooth. (By the way, your comment about absence of local monodromy, and purity of limit MHS isn't correct.)

Prop (Zucker). If $Y$ is a curve, then $$R f_*\mathbb{C} = \bigoplus_i R^if_* \mathbb{C}[-i]$$

Since Zucker in section of 15 of his 1979 Annals paper proves a slightly weaker statement. Let me sketch a proof using stuff that appeared since. I can flesh it out if needed.

Sketch. Let $D\subset Y$ be the discriminant, $j:U\to Y$ the complement. By the decomposition theorem of BBDG, the object above decomposes as a sum $\bigoplus L_i$, where $L_i$ are translates of pure perverse sheaves. We can assume the $L_j$ are translates of minimal extensions. By restricting to $Y-D$ and applying Deligne (Théoremes de Lefschetz...), we can identify $L_i|_U=R^if_*\mathbb{C}|_{U}[-i]$, after reindexing. It follows that $L_i=j_*R^if_*\mathbb{C}|_{U}[-i]$ for sheaves supported on $Y$. There may be other summands supported on $D$ which need to accounted for. Use the local invariant cycle theorem to get a surjection $R^if_*\mathbb{C}\to L_i$. By purity (in the sense of Hodge modules, say) we can split this. So that we can absorb all $L_k$ with support on $D$ into some $R^if_*\mathbb{C}$

Added Comment: Regarding the latest question, I think I was too hasty in my comment. The example I had in mind doesn't satisfy all your requirements, but it may still be interesting to describe. One has a pencil of genus 2 curves degenerating to a union of 2 elliptic curves at each singular fibre. Contracting one of the elliptic curves from each pair results in singular surface mapping to a curve such that the higher direct images are constant. This is probably similar to what Ulrich Naf was suggesting. (Rmk, Oct 11/20: in fact it's different.)

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  • $\begingroup$ Dear Prof. Arapura, thanks so much for the beautiful answer. And of course, I meant to also add that $R^if_*\mathbb{C}$ are all local systems (if I understand correctly the comment about local monodromy is true in this case.). Under this additional assumption, can one conclude smoothness of $f$? Since the answer is really helpful and explicit, I will leave the question as it is and add the part about the local system assumption at the end. $\endgroup$
    – guest0803
    Commented Sep 1, 2020 at 20:48
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    $\begingroup$ You're welcome. OK, with added assumptions about direct images, I agree that the local monodromy is trivial. Surprisingly, this is not enough to guarantee smoothness. I'll expand my answer later. $\endgroup$
    – Donu Arapura
    Commented Sep 1, 2020 at 21:10
  • $\begingroup$ Thanks! I will look forward to it. Somehow I have the feeling that such $f$ (additional assumption inclusive) cannot have divisorial discriminant locus. But I don't have an argument or a counter-example. $\endgroup$
    – guest0803
    Commented Sep 1, 2020 at 21:33
  • $\begingroup$ Isn't the discriminant always of pure codimension one (if non-empty)? $\endgroup$
    – Chris
    Commented Sep 2, 2020 at 6:56
  • $\begingroup$ @Chris That is unfortunately not correct. Blow-ups already give you examples where discriminant is not of pure codim 1. $\endgroup$
    – guest0803
    Commented Sep 3, 2020 at 7:11

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