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This is a follow up to my question about D-modules supported on the nilpotent cone. I got some good answers there but now I have a more basic question.

Consider an affine algebraic variety X, a closed subvariety i:Y-->X, and the intermediate extension of the structure sheaf on Y to all of X (do I denote this i!*OY ? For that matter, explaining either the * or ! extension instead would be a helpful start if its easier).

My question is this: Since X is affine, D(X) is just an associative algebra, generated by O(X) and Vect(X) by the usual construction. My question is how can I understand i!*OY as a module the associative algebra D(X), supposing I understand D(X)?

In other words, what is the vector space underlying i!*OY, how do functions in O(X) and vector fields act?

I probably need to invest some serious time with a textbook to answer this question myself, but any help getting started would be most appreciated!

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Just for completeness: D(X) is only generated by O(X) and derivations, if X is defined over a field of characteristic 0. –  Lars Nov 1 '09 at 18:57
And if X is smooth. –  Greg Muller Nov 1 '09 at 22:04
Yes, thanks on both counts. I'm really interested in the situation on the link, so both of these are true. –  David Jordan Nov 1 '09 at 23:24

3 Answers 3

up vote 4 down vote accepted

Have you looked at Bernstein's lectures on D-modules? He proves a result relevant to your questoin in Lec. 3, Sec. 14: for an affine embedding Y --> X with Y irreducible, if E is an OY-coherent DY module (i.e. a vector bundle with a flat connection) then the !* direct image from E can be characterized as the unique irreducible subquotient of either the * or ! direct image which has nonzero restriction to Y. Since it can be easy in such situations to compute the * direct image, this may be a good way to get a handle on the other functors too. For example, I think one can see from this that if we take the embedding of the origin into the affine line, then both the * and !* direct images of OY=k (the ground field) are A(1)/A(1)t, where t is the coordinate on the affine line and A(1)=DX is the 1-dimensional Weyl algebra k[t,d/dt]. This quotient is rightly considered the "delta-function" A(1)-module, since its generator δ satisfies tδ=0. I'm not sure how similar to this case your general situation will be. But certainly by Kashiwara's theorem one knows that the * direct image (and hence the !* direct image) will be supported on the subvariety Y.

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Thanks! I had looked over these a few years ago in an intro course, but not using it hadn't known exactly where to look. –  David Jordan Nov 1 '09 at 19:52

Since Y is closed, the !, *, and !* extensions coincide, and there should be a straightforward (but confusing for me) way of doing what you want. If Y is smooth, then I wonder if you can identify the vector space you're talking about with something like sections of the conormal bundle to Y in X, but this is not quite right (e.g. when Y = X). When Y is singular, there is a more complicated story, having to do with the fact that the obvious definition of D-module on a singular variety is no good.

When Y is only locally closed, describing the !* extension is a difficult problem. Vilonen's thesis, the only reference I know about this, is available here:


There is even something nontrivial to say about the ! and * extensions--the characteristic cycles of such D-modules were computed by Schmid and Vilonen. Maybe there is a more elementary answer to your question about these extensions, though.

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Thanks, this is helpful! As you saw in the other post, the ones I'm interested in eventually aren't closed. –  David Jordan Nov 1 '09 at 23:26
OK not "sections of the conormal bundle" but "functions on the total space of the conormal bundle." A vector field on X determines a function on the conormal bundle to Y by contraction, and operates on the vector space of such functions by multiplication. –  David Treumann Nov 2 '09 at 4:28

I think I can answer my subquestion about * extension after thinking about it, but it's really the ! and !* extensions I'd like to understand. I think I remember the * extension is just the pushforward.

So i*O(Y) = O(X) ⊗i O(Y), and the D-module action is always happening on the O(X) factor.

Sorry to answer (a tiny part of) my own question, but I couldn't do it in a comment because of the formatting, and maybe this will save some kind respondent some time.

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