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

(1) Suppose that $X$ is a smooth complex algebraic variety, stratified by some nice smooth stratification $S$. Let $M$ be a $D$-module on $X$, s.t. its shriek-pullback (or star... whatever is comfortable to you) to every stratum is smooth. What can we say then about the singular support of this $D$-module? For example, I hope that every point in the singular support will be zero when evaluated on tangents to stratums. Maybe one will need $M$ to be holonomic R.S. for that.

(2) I want this in order to understand how to restrict this $D$-module to a subvariety, transversal to this stratification (say that upper-star and upper-shriek differ by a shift, and are in one cohomological degree).

So if anyone can explain this, or give some reference, I will be thankful.

Edit: I will be happy with an answer for (2) in the constructible world as well.

Sasha

share|improve this question
1  
Hi, if I'm not mistaken (1) is true if $M$ is reg hol because then $DR(M)$ is $S$-perverse and $SS(M) = SS(DR(M)) \subset T_SX$. (2) If $Z \subset X$ is transverse to the stratification then it is non characteristic for $M$, in which case the upper-shriek inverse image computation has been studied quite extensively see Maisonobe and Torelli's SMF article for a review. –  YBL Jul 27 '13 at 4:51
    
Thank you! So actually what I want is a reference for your claim $SS(DR(M)) \subset T_S X$. Do you know such? Does it also appear in the above SMF article? –  Sasha Jul 27 '13 at 8:26
1  
Because of your assumptions on $M$, $DR(M)$ will be $S$-constructible. Then for the singular support of constructible sheaves the reference is Kashiwara and Schapira's book Sheaves on manifolds. –  YBL Jul 27 '13 at 15:11
1  
I'd like to add that I would very much like to see a purely algebraic proof of (1) (without using the RH correspondance). think I might hold for all holonomic D-modules. –  YBL Jul 27 '13 at 17:40
    
Thank you! . . . . –  Sasha Jul 28 '13 at 9:37

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.