Lowest weight of compactly supported cohomology with coefficients 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.
 A: 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.
