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).
Best Answer
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.