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

This is a follow-up to this post on the Decomposition Theorem. Hopefully, this will also invite some discussion about the theorem and perverse sheaves in general.

My question is how does one use the Decomposition Theorem in practice? Is there any way to pin down the subvarieties and local systems that appear in the decomposition. For example, how do you compute intesection homology complexes using this theorem? Does anyone have a link to a source with worked out examples?

Another related question: What is the deep part of the theorem? Is it the fact that the pushforward of a perverse sheaf is isomorphic to its perverse hypercohomology? Is it the fact that these pieces are semisimple? Or are these both hard statements? And what is so special about algebraic varieties?

share|improve this question

4 Answers 4

up vote 8 down vote accepted

To supplement Ben's answer, basically every aspect of the decomposition theorem is hard.

To give you a simple example of something which is implied by the decomposition theorem but is far from trivial is the following statement: given a proper smooth map of smooth varieties f : X -> Y the direct image of the constant sheaf splits as a direct sum of local systems. Note that this implies (but is stronger than) the degeneration of the Leray-Serre spectral sequence for the fibration. This answers to some extent your question "what is so special about algebraic varieties" because Leray-Serre just doesn't degenerate in general.

I think the situation has been cleared up considerably by the work of de Cataldo and Migliorini which (IMHO) is the first genuinely geometric proof of the decomposition theorem.

One might think of the "smooth map" case above as the "easiest case" (and indeed it does have an easier proof). However de Cataldo and Migliorini point out that in fact the "easiest case" is the case of a semi-small map, for which the decomposition theorem can be deduced from the non-degeneracy of certain bilinear forms. In a difficult work, they deduce the general case by reducing to this case by induction on the "defect of semi-smallness" (how far away a map is from being semi-small) and by taking hyperplane sections to reduce this defect.

An excellent informal survey about the decomposition theorem, with lots of wonderful examples can be found in The decomposition theorem, perverse sheaves and the topology of algebraic maps by de Cataldo and Migliorini.

Note that there are really three statements in the decomposition theorem, all of which are hard:

  1. the direct image is the sum of its perverse cohomology groups;
  2. each perverse cohomology is a direct sum of IC extensions of a local system;
  3. each local system is semi-simple.

As is often the case in mathematics, a nice way to learn why the decomposition theorem is hard is to go to situations when it fails. This occurs when one takes perverse sheaves with coefficients in positive characteristic (or even Z). Daniel Juteau, Carl Mautner and I have written a survey called "Perverse sheaves and modular representation theory" which contains lots of examples of the failure of the decomposition theorem. (Note that all of 1), 2) and 3) above can fail!)

share|improve this answer
It's one my favorite articles -- I couldn't resist linkifying it. –  Ilya Nikokoshev Nov 1 '09 at 20:42
And I basically recommend reading it from the first page until the last. –  Ilya Nikokoshev Nov 1 '09 at 20:43
I too like that article very much. –  Thomas Riepe Nov 14 '09 at 13:18

The short answer is that in general its very hard. For special classes of maps like semi-small ones, it's not so bad (see the book of Chriss and Ginzburg), but for an arbitrary projective map, I don't know any reliable way of dealing with it. If you know the IC sheaves downstairs well, you can use point counting (there's an example of a computation like this in my paper with Geordie Williamson; it's in Section 4, I think).

To answer your last point:

The hard part of the theorem is developing the theory of weights, and proving that IC sheaves are pure. Since you can't have any extensions between pure sheaves of the same weight (this is roughly because it would require the Frobenius to act trivially on Ext^1), this proves that the pushforward (which is pure because proper pushforward preserves purity) is a direct sum of shifts of perverse sheaves, so really one proves both of those pieces simultaneously.

share|improve this answer
"The hard part of the theorem is developing the theory of weights" -- I think you're referring to Saito's proof using the Mixed Hodge Structures (correct me if I'm wrong). Perhaps the hardest part for me in that proof is that it works at all. –  Ilya Nikokoshev Nov 1 '09 at 20:46
Nope, I'm refering to Delige's theory of weights, which while not easy, makes a lot more sense to me than mixed hodge modules. –  Ben Webster Nov 1 '09 at 22:15
Ok! Does that prove Decomposition Theorem? –  Ilya Nikokoshev Nov 1 '09 at 23:50
If you develop it properly; I gave the very short explanation in my answer. That's certainly what Beilinson, Bernstein and Deligne used in their original proof. Mixed Hodge modules hadn't been invented yet. –  Ben Webster Nov 2 '09 at 0:15

There's a recent paper by Mark Andrea de Cataldo and Luca Migliorini (http://arxiv.org/abs/0712.0349) which gives an excellent introduction to the decomposition theorem. In particular, they discuss semi-small maps in the context of Springer theory and Hilbert schemes.

share|improve this answer

There is a winter school on the decomposition theorem in Freiburg, Germany, 22-26 Feb 2010. Link.

share|improve this answer
And there will soon be lecture notes, as well as videos online! –  Jan Weidner Feb 24 '10 at 18:41
Are there any note available online now? –  Mikhail Bondarko Jul 18 '10 at 21:31

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.