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

Let $G$ be a compact Lie group, $X$ be a (compact, oriented) smooth manifold, with $G$ acts on $X$ smoothly. Then we can talk about the $G$-equivariant homology and cohomology.

My question: In what sense can we have a duality between the equivariant homology and cohomology, in analogue with the Poincare duality between the ordinary homology and cohomology of $X$? In particular, the degree of a nontrivial equivariant (co)homology class could exceed the dimension of the manifold $X$. Then in such case, what does the dual mean, geometrically?

share|improve this question
    
The expressions "equivariant (co)homology" is used to mean more than one thing. The (co)homology of $M\times _GEG$ is not very well suited to duality statements, I believe. –  Tom Goodwillie Oct 18 '10 at 2:37
    
@ Tom Goodwillie What do you mean more than one thing? Yes, for $M\times_G EG$ is not good in that sense. –  Guangbo Xu Oct 18 '10 at 13:02
    
The method I've seen is to use Borel-Moore homology instead of regular homology so that some form of duality is thrust upon you. –  stankewicz Oct 18 '10 at 19:21
    
@ stankewicz Thank you. It seems that the Borel-Moore homology is kind of homology with compact support. But what kind of duality is it? Can you provide some reference? –  Guangbo Xu Oct 19 '10 at 0:52
1  
@stankewicz: I wish I could say "revised and completely error-free notes", but that should be happening soon! –  Dave Anderson Oct 21 '10 at 1:24
show 3 more comments

4 Answers

up vote 13 down vote accepted

There is a paper by Brion: "Poincaré duality and equivariant (co)homology," Michigan Math. J. 48 (2000). As mentioned in one of the comments, he uses Borel-Moore as the homology theory.

That said, one should always be careful about the word "duality" in this context -- often people just mean there's a canonical isomorphism $H_G^k X \rightarrow H^G_{n-k}X$. (Which does exist using equivariant Borel-Moore homology, when $X$ is smooth.) The geometric meaning is the same as in the ordinary case: if $V$ is a $G$-invariant sub(manifold/variety/cycle/etc) of codimension $k$, it defines a fundamental class in $H^G_{n-k} X$, which is identified with $H_G^kX$ by means of the isomorphism.

Duality could also refer to a pairing $H_G^k X \otimes H_G^\ell X \rightarrow H_G^{k+\ell-n}(point)$, given by the equivariant pushforward (integral): $$\alpha\otimes\beta \mapsto \int \alpha\cdot\beta .$$ Using this pairing, and assuming that $H_G^*X$ is free as a module over $H_G^*(point)$, any $H_G^*(point)$-module basis for $H_G^*X$ has a dual basis, in the usual sense of linear algebra. (Arguably, this duality pairing is what should really be called "Poincaré duality".) Again, the geometric meaning is similar to the ordinary case: if you're lucky enough to have invariant subvarieties whose classes form a basis for $H_G^*X$, the Poincaré dual basis is given by classes of subvarieties intersecting the original ones transversally (if they exist). However, the existence of geometrically defined dual bases is a stronger statement, because the equivariant integral is generally nonzero on classes of degree greater than $\dim X$.

(I should say all this is prejudiced toward the equivariant cohomology rings that usually show up in algebraic geometry, e.g., $H_G^*X$ is a free module over $H_G^*(point)$, so that it makes sense to talk about dual bases.)

PS: Defining equivariant Borel-Moore homology requires a little more care, since the spaces $EG\times^G X \rightarrow BG$ are infinite-dimensional fiber bundles. But they have finite-dimensional approximations $EG_m \times^G X \to BG_m$, so it makes sense to define $$H^G_k X = H_{k+\dim BG_m}(EG_m\times^G X)$$ for $m\gg0$. The equivariant Poincaré isomorphism is just the ordinary one for these approximation spaces.

share|improve this answer
add comment

There is a version of equivariant Poincaré duality when $G$ is finite, due to Costenoble and Waner:

Costenoble, S. R.; Waner, S. Equivariant Poincaré duality. Michigan Math. J. 39 (1992), no. 2, 325–351.

I'm unsure as to the geometric interpretation.

share|improve this answer
add comment

Hi, You can have a look at my Phd thesis. A main part deals with equivariant Poincare duality for compact Lie group from a geomertic point of view. It claims that the obstruction for Poincare duality lies in Tate cohomology so generally we have a "duality map" which is may not an isomorphism. The dimension of the group gives a shift in the grading of the duality map. http://hss.ulb.uni-bonn.de/2011/2391/2391.htm

share|improve this answer
add comment

See the paper Equivariant Ordinary Homology and Cohomology by Steve Costenoble and myself for a comprehensive theory of Poincare duality in ordinary equivariant homology and cohomology (as opposed to Borel homology/cohomology) for compact Lie group actions.

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.