2
$\begingroup$

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!

$\endgroup$
8
  • $\begingroup$ In what sense is that a "local system"? Did you mean to write "constructible sheaf"? $\endgroup$ Apr 4, 2015 at 16:42
  • $\begingroup$ It is a constructible sheaf that is locally constant (because $S$ is a stratum). $\endgroup$
    – user54343
    Apr 4, 2015 at 17:45
  • $\begingroup$ I see now. I thought you were looking at the extension over $X$ of a sheaf on $S$. $\endgroup$ Apr 4, 2015 at 22:17
  • $\begingroup$ You are right to be suspicious. If one takes the direct image of the constant sheaf under $z \mapsto z^2$ one gets a local system which splits into two pieces, both of which are pure, and one of which has no cohomology. The non-trivial summand gives a counter-example to your hopes. Making $1 - f$ compatible with weight filtrations is a tricky business, in the mixed Hodge world it is given by the limit mixed Hodge structure... $\endgroup$ Apr 4, 2015 at 23:12
  • $\begingroup$ @GeordieWilliamson: I agree that $f \neq 1$ in your example, but how is this a counterexample to my hopes? Is there a way to obtain the sheaf that you describe via my construction? $\endgroup$
    – user54343
    Apr 5, 2015 at 1:06

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.