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

Given a coherent $\mathcal{D}_X$-Module $M$, one can assign to it its characteristic variety $$ch(M)\subseteq T^*X$$ or one could look at its support $$supp(M)\subseteq X$$ as an $\mathcal {O}_X$ module. Is there a relation between these two spaces?

Edit: Is it true, that the characteristic variety of a holonomic $\mathcal{D}_X$-Module is the conormal space of its support?

share|improve this question

2 Answers 2

up vote 4 down vote accepted

I'm not an expert either but, at least for holonomic D-modules, the relation you're looking for should be $$ supp(M) = ch(M) \cap T^*_X X $$ where $T^*_XX$ is the zero section of the cotangent bundle identified with $X$.

You can check it in the basic examples: $X = \mathbb{C}$, $M$ the trivial bundle or the $\delta$-module at $x\in X$. The corresponding $ch(M)$ is $T^*_XX$ or $T^*_xX$. The general case should be an easy exercise in linear algebra.

share|improve this answer
3  
Yes, this is correct, though it is more simple, even. The characteristic variety is a conical subspace of the cotangent bundle, so any in fiber of the cotangent bundle it intersects, it also contains the zero point. Therefore, supp(M) is the image of ch(M) under the map from the cotangent bundle to the base space. Also, while the proof is indeed not hard (passing to associated graded does not change the O_X-module structure), the phrase 'an easy exercise in linear algebra' reflexively makes me suspicious. –  Greg Muller Aug 23 '10 at 19:03
    
Thanks Greg and YBL. Is it true, that the characteristic variety of a holonomic D-module is the conormal space of its support? At least in the examples this is the case and it makes sense, the conormal space is conic and lagrangian and its projection is of course the support. –  Jan Weidner Aug 24 '10 at 8:54
2  
The support can have arbitray singularities so it's not obvious what you mean by conormal space. What you can always do is find a stratification $X= \amalg X_\alpha$ by smooth subvarieties so that $ch(M) \subset \coprod T^*_{X_\alpha} X$. This is equivalent to the de Rham complex $DR(M)$ being constructible for this stratification. –  YBL Aug 24 '10 at 10:35

I'm not a $D$-module person. I'm hoping someone else can give a slightly more insightful explanation. (It looks like YBL gives a clear picture of the holonomic case.)

By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.

Continuation: Perhaps it's worthwhile making this a little more explicit. The simplest example is $X=\mathbb{A}^n$. Then (the global sections of) $D_X$ is the Weyl algebra with generators $x_1,\ldots, x_n, \partial_1,\ldots, \partial_n$ with commutation relations $[x_i,\partial_j]= \delta_{ij}$. If we force these to commute, by passing to the associated graded with respect to the filtration by order of operators, we obtain the polynomial ring in $2n$ variables or in other words the coordinate ring of $T^*\mathbb{A}^n$. Any finitely generated $D_X$-module $M$, carries a (noncanonical) compatible filtration, so can define $Ch(M)= Supp(GrM)\subset T^*\mathbb{A}^n$ (it is independent of the filtration). Now let $M= D/\sum D\partial_i$, which corresponds to $\mathcal{O}_X$. I'll omit the details, but one can see that $Ch(M) = V(\partial_1,\ldots \partial_n)$ (the zero section), and $\pi^*supp(M) = T^*\mathbb{A}^n$ where $\pi:T^*\mathbb{A}^n\to \mathbb{A}^n$.

For a nonholonomic example, take $M= D_X$. Then $Ch(M)= \pi^{-1}supp(M) = T^*\mathbb{A}^n$.

share|improve this answer

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.