Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am trying to understand Proposition 3.4.2 in `Koszul duality patterns in representation theory' by Beilinson-Ginzburg-Soergel [BGS]. A copy of the paper can be found at http://home.mathematik.uni-freiburg.de/soergel/

I outline the setup below. Text in bold are my own commentary. I have taken the liberty to change some confusing notation in the paper. However, it is entirely possible that the reason I am confused is that one of my "changes of notation" isn't correct. I have tried to be careful in my changes, but apologies in advance for any additional confusion this may contribute to.

My question is that indicated by the last bold face text below.

Let $X$ be a complex variety with an algebraic stratification by affine linear spaces $X = \sqcup_{w\in W} X_w$. Let $IC_w$ denote the intersection cohomology complex on $X$ corresponding to the constant sheaf on $X_w$.

Let $X = Y_0 \supset Y_1 \supset \cdots \supset Y_r = \emptyset$ be the corresponding filtration by closed subvarieties so that $Y_{p}- Y_{p+1} = X_p$ for some strata $X_p$.

Let

$j_w\colon X_w \to X$

be the inclusion of the strata. Assume parity vanishing, i.e., assume

$H^ij_v^*IC_w = 0$ unless $i = dim(X_v) + dim(X_w) \mod 2$, for all $v,w\in W$.

Here $H^*$ denotes perverse cohomology.

Proposition: Under the assumption of parity vanishing, hypercohomology induces an injection

$Hom^{\bullet}_{D^b(X)}(IC_x, IC_y) \to Hom_{\mathbb{C}}(\mathbb{H}^{\bullet}IC_x, \mathbb{H}^{\bullet}IC_y)$.

Proof: By parity vanishing the spectral sequence $\mathbb{H}^{p+q}j_p^!IC_x \implies \mathbb{H}^n IC_x$ is degenerate (the spectral sequence is defined via the filtration by local hypercohomology along the strata, for details see Section 3.4 of [BGS]). So if $f\in Hom^{\bullet}_{D^b(X)}(IC_x, IC_y)$ is given such that $\mathbb{H}^{\bullet}f = 0$, then necessarily $0 = j_p^!f \in Hom^{\bullet}_{D^b(X)}(j_p^!IC_x, j_p^!IC_y)$ for all $p$. Let

$a_p\colon Y_p \to X$

be the closed inclusion. We have a decomposition

$u\colon X_p = Y_p - Y_{p+1} \to Y_p$,

$i\colon Y_{p+1}\to Y_p$

in an open and a closed subset and a distinguished triangle

Edit: the original distinguished triangle (as stated in [BGS]) wasn't correct, I have now made the fix

$i_*i^!a_p^! \to a_p^! \to u_*u^!a_p^!$

(so this distinguished triangle is the same as $i_*a_{p+1}^! \to a_p^! \to u_*j_p^!$ )

which shows that $a_{p+1}^!f = 0 = j_{p}^!f$ implies $a_p^!f = 0$ (it is this implication that I don't understand, related to my confusion is an earlier question of mine Showing morphism of sheaves is zero )

Hence by induction $j_p^!f = 0$ for all $p$ implies $f = a_0^!f = 0$.

Any comments that would clarify the above would be most appreciated!

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.