MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is well known, that the intermediate extension functor $j_{!*}$ preserves injections and surjections. However it seems that it is not exact in general!

1) What would be an example which shows that $j_{!*}$ is not exact?

2) It is easy to see, that $j_{!*}$ is exact if $j$ is the inclusion of a cell into a cell stratified variety and we are considering say $\mathbb Q_l$-sheaves. Is $j_{!*}$ still exact if we replace $\mathbb Q_l$ by $\mathbb Z_l$ in this situation?

share|cite|improve this question
For your first question, see Example 2.7.1. in de Cataldo and Migliorini's "The decomposition theorem, perverse sheaves and the topology of algebraic maps". – Dan Petersen Oct 2 '12 at 10:06
Thanks! $ $ – Jan Weidner Oct 2 '12 at 12:19
up vote 2 down vote accepted

I think that the answer for 2) is no.

Consider the affine Grassmannian for $SL_2$ and let $X$ denote the Schubert variety corresponding to the three dimensional representation of $PSL_2$. Then $X$ is stratified by Iwahori orbits into three affine strata of dimensions 2, 1 and 0. Let $U$ denote the open stratum.

Consider the exact sequence of local systems

$0 \to \mathbb{Z}_2 \to \mathbb{Z}_2 \to \mathbb{F_2} \to 0$

on $U$. Shifting by $[2]$ we get an exact sequence of perverse sheaves on $U$ and if we apply $j_{!*}$ we get a sequence

$0 \to IC(X,\mathbb{Z}_2) \stackrel{2}{\to} IC(X,\mathbb{Z}_2) \to IC(X,\mathbb{F}_2) \to 0$

(Here one needs to be careful: because $\mathbb{Z}_2$ is not a field there are two possible $t$-structures (denoted $p$ and $p^+$) which are interchanged by duality. See Daniel Juteau's paper "Decomposition numbers for perverse sheaves" arXiv:0803.2326.) Here I am considering the perversity $p$.

I claim that this sequence cannot be exact. Indeed, the stalks of $IC(X,\mathbb{Z}_2)$ are $\mathbb{Z}_2$ in degree -2, and 0 elsewhere. Whereas the stalks of $IC(X,\mathbb{F}_2)$ are $\mathbb{F}_2$ in degrees $-2$, $-1$ and zero elsewhere.

(These stalks calculations are explained in detail in arXiv:0901.3322.)

share|cite|improve this answer

Your Answer


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

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