Let $X$ be a stratified variety, and let $i:S\hookrightarrow X$ be the inclusion of a stratum. Let $F := H^k(i^!\operatorname{IC}\_X)$. This is a local system on $S$ whose fiber at a point is isomorphic to the degree $k$ compactly supported cohomology of the stalk of $\operatorname{IC}\_X$ at that point.


Now let’s assume that $S \cong \mathbb{C}^\times$, and let $V$ be the fiber of $F$ at $1\in S$.
Let $\sigma\in\operatorname{Aut}(V)$ be the monodromy map. Then the cohomology of $S$ with coefficients in $F$ is equal to the cohomology of the complex
$$H^0(\mathbb{C}^\times) \otimes V\overset{1-f}{\longrightarrow} H^1(\mathbb{C}^\times) \otimes V.$$


Question: Does the map $1-f$ have to be strictly compatible with the weight filtrations on these groups? 


We know that $H^0(\mathbb{C}^\times)$ has weight 0 and $H^1(\mathbb{C}^\times)$ has weight 2. This would mean that $1-f$ can only be nonzero if $\operatorname{gr}(V)$ contains two different summands whose weights differ by exactly 2. In particular, if $V$ is pure of some weight, then $1-f$ would have to vanish. This is the conclusion that I want, but I am suspicious; it sounds too strong!



Cohomology of a local system and Deligne's weight filtration
Consider the variety with equation $y^2=tx^2$. Let $S$ be the locus with $x=0,y=0,t\neq 0$. Then clearly the sheaf at each point of $S$ is a sum of one-dimensional vector spaces from the surfaces $y=\pm \sqrt{t}x$. Monodromy swaps these two, so the action is nontrivial.


What you can get from purity in this case is that the unipotent monodromy is nontrivial. The operator $f-1$ does not behave in a predictable way on weights because it is not Galois-equivariant. But because the monodromy must be quadi-unipotent, we can take a power $f^n$ which is unipotent. Then $N=\log f^n/n$ is a well-defined map $V \to V$ that subtracts $2$ from the weight, showing that the unipotent part is trivial by exactly the argument you give.


