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

Consider a regular holonomic D-module (or a perverse sheaf) $M$ on a smooth variety $X$. Let $f:X\to A^1$ be a polynomial (or holomorphic) function.

Question: Is it true that the $\lambda \in A^1$ such that the vanishing cycles $\phi_{f-\lambda}(M) = 0$ is a dense open set?

Here are my thoughts:

If $M = O_X$ (or the constant perverse sheaf $A[dim X]$), this is just the fact that the critical values of $f$ are isolated.

In the general case, we can factor $f$ through its graph $X\to X\times A^1$, $x\mapsto (x,f(x))$, reducing to the case where $f$ is the (smooth) projection $t:X\times A^1 \to A^1$. Our sheaf $M$ on $X\times A^1$ has a characteristic variety $\bigcup_\alpha T^*_{S_\alpha}(X\times A^1)$ for a stratification $X\times A^1 = \bigcup_\alpha S_\alpha$. My guess is that $\phi_{t-\lambda}(M) = 0$ when $\{t-\lambda = 0 \}$ is transverse to all the $S_\alpha$ and that this is a generic condition but I'm having trouble making this intuition precise.

share|improve this question
    
I think you can fix the jsMath rendering problems in your post by escaping the problematic math with backticks. See the box at the bottom of the "Related" list to the right. –  Ian Shipman Feb 22 '10 at 20:55
    
I tried but it doesn't seem to help. –  YBL Feb 22 '10 at 23:25
    
Just refreshing page works. –  YBL Mar 2 '10 at 12:27
add comment

1 Answer 1

An answer to this question was given to me by Pierre Schapira. This is known as the microlocal Bertini-Sard theorem (cf. Sheaves on manifolds cor. 8.3.12).

Consider a map $f:X\to A^1$. It induces $f_\pi : X\times_{A^1} T^*A^1 \to T^*A^1$ and $f_d : X\times_{A^1} T^*A^1 \to T^*X$. Set $\Lambda = SS(M)$ the characteristic variety of $M$. This is a closed conic isotropic subset of $T^*X$. Now $$ supp(\phi_{f-t}(M)) \subset [ x ~|~ f(x) = t,~(x,df(x)) \in \Lambda ] $$ so $$ [ t\in A^1 ~|~ \phi_{f-\lambda}(M) \neq 0 ] \subset [ t\in A^1 ~|~ (t,dt)\in f_\pi f_d^{-1}(\Lambda) ] $$

Now assume that $f$ is compactifiable as $X\overset{j}{\to} \bar{X} \overset{\bar{f}}{\to} A^1$, $j$ an open immersion and $\bar{f}$ proper. The closure $\bar{\Lambda}$ of $\Lambda$ is $T^*\bar{X}$ is a closed conic isotropic subset and since $\bar{f}$ is proper, $\bar{f}_\pi \bar{f}_d^{-1}(\bar{\Lambda})$ is a closed conic isotropic subset of $T^*A^1$. So its intersection with the nowhere vanishing section

$$ [t \in A^1 ~|~ (t,dt) \in \bar{f}_\pi \bar{f}_d^{-1}(\bar{\Lambda})] $$

has dimension 0. Since $f_\pi f_d^{-1}(\Lambda) \subset \bar{f}_\pi \bar{f}_d^{-1}(\bar{\Lambda})$ the same is true for

$$ [ t\in A^1 ~|~ \phi_{f-\lambda}(M) \neq 0 ] \subset [ t\in A^1 ~|~ (t,dt)\in f_\pi f_d^{-1}(\Lambda) ] $$

and the theorem is proved.

If $f$ is algebraic it is always compactifiable. If $f$ is analytic, I don't know if the theorem still holds in general.

PS: If $f(x) = \lambda $, the condition $(x,df(x)) \in T^*_Z X$ just says that the fiber $\{f = \lambda\}$ is tranverse to $Z$ at $x$. So when $SS(M) \subset \bigcup T^*_{S_\alpha} X$ this gives the geometric interpretation that the vanishing cycles are 0 whenever the fibers are transverse to the strata.

share|improve this answer
add comment

Your Answer

 
discard

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.