Let $X\_0/\mathbb F\_q$ be a variety, and let $\mathcal F$ be a Weil sheaf on $X := (X\_0)\_{\overline{\mathbb{F}\_q}}$ that is pure of weight $n$. If $j < n$, does the weight $j$ piece of $H^i\_c(X,\mathcal F)$ necessarily vanish for all $i$?


I thought this was true, but I have made some computations that seem to contradict it.



Lowest weight of compactly supported cohomology with coefficients
This is not true.


If you take an open curve $U \subset C$ and a pure sheaf $\mathcal F$ on $U$ of weight $w$, with $j$ the open immersion $U \to C$ and $i$ the complementary closed immersion, then the exact sequence $$ 0 \to j\_! \mathcal F \to j\_\* \mathcal F \to i\_\* i^\* j\_\* \mathcal F \to 0$$ induces a long exact sequence on cohoomology


$$ H^\*\_c( U, \mathcal F) \to H^\* (C, j\_\* \mathcal F) \to H^\* (C- U, i^\* j\_\* \mathcal F ) $$


The first term is what we want to compute and the middle term is pure of weight $w+d$ in degree $d$ by Deligne. So low weight cohomology can only come from the third term.


However, come from the third term it does when $\mathcal F$ has unipotent local monodromy. For each Jordan block of size $n$ in the local monodromy around a point of $C - U$, $i^\* j\_\* \mathcal F$ has a single Frobenius eigenvalue of weight $w+ 1-n$, and that eigenvalue will show up in cohomology as $C-U$ is finite and thus cohomology is simply taking the sums of the stalks at its points.


The sheaves on $M\_{1,1}$ you get from $M\_{1,n}$ can be constructed from symmetric powers of the Tate module of the universal elliptic curve. Because the universal elliptic curve has semistable reduction at $\infty$, the Tate module has unipotent local monodromy there, and its symmetric powers do as well, explaining why the phenomenon occurs in this case.


