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Let $f : X \to Y$ be projective and smooth morphism of complex algebraic varieties. Here we care about the algebraic topology of $X$ and $Y$, so use classical topology for simplicity.

I can take the constant sheaf $\mathbb{Q}_X$ and (derived) push it forward to get $f_* \mathbb{Q}_X \in D^b_c(Y,\mathbb{Q})$. There is a celebrated theorem of Deligne that $f_* \mathbb{Q}_X$ is semi-simple, i.e. isomorphic to a direct sum of its cohomology sheaves. The argument uses hard Lefschetz along the fibres. (It is also true that each summand is a semi-simple local system, as a polarizable VHS, however I want to ignore that extra piece of information below.)

Suppose I replace $\mathbb{Q}$ with $k := \mathbb{F}_p$.

Question: Is it true that $f_* k_X \in D^b_c(Y,k)$ is always semi-simple? That is, does it always split as a direct sum of its (not-necessarily semi-simple) cohomology sheaves.

I had always assumed the answer was no, but woke up this morning feeling unusually optimistic. (I have tried several times to produce a counter-example.) I understand that this is deep water, and I am happy with a heuristic answer either way (potentially using motives).

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    $\begingroup$ Here's a candidate example: If $f:X \to Y$ is a smooth proper genus $0$ curve, then I think $R^i f_* k$ looks the same as that for $\mathbf{P}^1_Y \to Y$ (as $\mathrm{PGL}_2$ is connected). So the obstruction to splitting with $\mathbf{Z}/2$-coefficients is an element of $H^2(Y, \mathbf{Z}/2(1))$ that I think is the Brauer class of $f$. So it will be non-trivial in general. $\endgroup$
    – Anonymous
    Commented Sep 7, 2021 at 12:06
  • $\begingroup$ @Anonymous: This looks like a very promising counter-example, thank you. I'll try to work through the details. $\endgroup$ Commented Sep 8, 2021 at 1:26
  • $\begingroup$ Geordie: I dont know a counterexample off the top of my head. However, the first thing that came to my mind was that this would get you into trouble with Steenrod operations (over $\mathbb{F}_2$ say). So perhaps look at approximations to the map $E\mathbf{Z}/2 \to B\mathbf{Z}/2$. However, it has been a long time since I thought about these things and my current intuition might be way off. $\endgroup$
    – rvk
    Commented Oct 7, 2021 at 20:43
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    $\begingroup$ @rvk: thanks a lot. I'll look at this, but I suspect that the example you give (approximations to EZ/Z -> BZ/2) won't work as any example should have fibres of dimension $> 1$. (Btw, I basically convinced myself that Anonymous' proposal works. There is one very natural looking identification that I couldn't convince myself of, but I'm hoping to still do so and post the answer.) Welcome back by the way! $\endgroup$ Commented Oct 13, 2021 at 4:34
  • $\begingroup$ @GeordieWilliamson : Doh! You are too kind. That was a completely dumb idea on my part. You at least need the fibers to have dim > 0 (I dont immediately see right now why you need dim > 1) otherwise your hope holds rather trivially. This is what I get for not thinking a bit carefully. I am tempted to delete to hide my shame! $\endgroup$
    – rvk
    Commented Oct 13, 2021 at 13:58

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We can give many counterexamples to semisimplicity in positive characteristic using the observation that if the canonical morphism $k_Y\rightarrow f_*k_X$ in $D^b_c(Y,k)$ is split, then the induced morphism in $k$ cohomology must be injective: $$f^*:H^*(Y,k)\rightarrow H^*(X,k).$$

So we just need to cook up nontrivial elements in this kernel. For instance, take any nontrivial abelian cover $f:X\rightarrow Y$ of degree $p$. This is classified by an element $\alpha$ of $H^1(Y,\mathbb{Z}/p\mathbb{Z})$, and since this cover is the pullback of the universal degree $p$ cover along a morphism $Y\rightarrow B(\mathbb{Z}/p\mathbb{Z})$, we see that $\alpha$ vanishes in $H^1(X,\mathbb{Z}/p\mathbb{Z})$, since $E(\mathbb{Z}/p\mathbb{Z})$ is contractible.

For a more geometric example, let $E$ be an elliptic curve over $Y$, and $A$ an $E$ torsor over $Y$. Then this is classified up to homotopy by a map to $B(S^1\times S^1)\cong K(\mathbb{Z},2)\times K(\mathbb{Z},2)$. Since the total space $E(S^1\times S^1)$ is contractible, the associated classes $(\alpha,\beta)$ in $H^2(Y,\mathbb{Z})$ will lie in the kernel of $H^2(Y,\mathbb{Z})\rightarrow H^2(A,\mathbb{Z})$. Then one can reduce this mod $p$ and get counterexamples over a field if the orders of these elements aren't divisible by $p$.

Finally, one can use nonliftable projective bundles to give counterexamples, in line with Anonymous' suggestion. If $f:X\rightarrow Y$ is a (complex) dimension $n-1$ projective bundle, then its obstruction to being the projectivisation of a complex vector bundle is a class $\beta_f\in H^2(X,\mathbb{C}^*)\cong H^3(X,\mathbb{Z})$. Then one can check in the universal case of $u:U\rightarrow BPGL_n(\mathbb{C})$, we have $u^*(\beta_u)=0$, so this class $\beta_f$ is always in the kernel of the map: $$H^3(Y,\mathbb{Z})\rightarrow H^3(X,\mathbb{Z}).$$

These examples have been written up in full detail in an upcoming paper, which should be coming to the arXiv soon.

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