suppose $M$ is a manifold , $G$ is a lie group (may be finite) ,then let $G$ act on $M$ freely , $N=M/G$ is then a manifold ,so my question is what relations may be between the homology and cohomology of $M$ and $N$ ? In the surface case ,if $G$ is a finite group ,then we can get no differences between the cohomolgy and homology of $M$ and $N$ then the euler number will be the same what will mean that the order of the group must be one ,and so there are no actions of finite groups freely acting on a surface and it is a very beautiful claim.

Does there any effective way to compute the cohomology of a quotient space when the action is not free?

So let us consider more about the question raised above , references about this question are also welcomed!!

  • $\begingroup$ 1) why is this community wiki? Don't you expect a definite answer? 2) Are you sure about your claim concerning surfaces? $H_1(\mathbb{RP}^2) \neq H_1(S^2)$. $\endgroup$ – José Figueroa-O'Farrill Dec 9 '10 at 10:26
  • $\begingroup$ sorry, i didn't know what the wiki is ,then i think it as a good choice . So , if adding a condition that if $N $ is oriented ,so the situation works as i concern works? $\endgroup$ – HKSHLZW Dec 9 '10 at 10:31
  • $\begingroup$ Or, if essentially my assertion about surface is wrong in any way , so what i concern is just the question: can we analys the cohomology of $N$ from the one of the $M$? $\endgroup$ – HKSHLZW Dec 9 '10 at 10:33
  • $\begingroup$ For a finite group acting on a reasonable space, the rational cohomology of the quotient is the fixed points in the rational cohomology of the total space (this works with any field coefficients). See my answer here: mathoverflow.net/questions/18898/… $\endgroup$ – Dan Ramras Dec 9 '10 at 15:34
  • $\begingroup$ Let us try to use a sequence of cutting and you get that you want. $\endgroup$ – user2464 Dec 10 '10 at 1:00

To elaborate a little on Kostya's answer, for the case where $G$ acts freely you may want to look up the Cartan-Leray spectral sequence (eg Chapter 7 of the book by Brown). Also, transfer homomorphisms are a useful elementary tool (see section 3.G of Hatcher's book).

When $G$ doesn't act freely you're into the realm of orbifolds, about which I have much to learn!

  • 2
    $\begingroup$ To elaborate a bit on Mark's answer, when G is finite cyclic, the Smith (Smith-Richardson) sequences are an elementary but quite powerful tool ($G=\Bbb Z/2$ and integer coefficients are in Hatcher's Example 3H.3; $G=\Bbb Z/p$ and mod $p$ coefficients are e.g. in V. Prasolov's Elements of homology theory). For an $n$-fold ramified cover $X\to Y$, Dmitry Gugnin gave a cool inequality for rational (or mod $p$, where $p>n$) cohomology cup-lengths: $l(Y)+1\ge (l(X)+1)/n$. He essentially uses Buchstaber's $n$-valued group theory. This is his PhD thesis, I guess not translated into English yet. $\endgroup$ – Sergey Melikhov Dec 9 '10 at 21:49

I think that you have to investigate in the direction of fibrations. And then, maybe, spectral sequences.

edit: Concernig your comment -- I am interested in that as well.

Right now my investigation got me there -- check the book "Modern Geometry - Methods and Applications: Part 3: Introduction to Homology Theory" by: B.A. Dubrovin, A.T. Fomenko, S.P. Novikov.

Chapter 20.

It is quite difficult for me to read the book (Yeah, you got me, I'm a physicist.) so I cannot give you anything but the direction.

  • $\begingroup$ yeah,i have got it ! But it may doesn't anything to the case when the action is not free which is what i want to consider! Any ideas? $\endgroup$ – HKSHLZW Dec 9 '10 at 12:13

Although of course Kostya's answer is a good definitive one, I thought best to mention that there are tools available other than spectral sequences depending on exactly what you want.

For example, if $M$ is an orientable surface, $G$ is finite, the action is not neccessarly free and the quotient is still smooth, then you can use the Riemann-Hurwitz formula.

There are also higher dimensional analogues of this which give you information about the canonical bundle of $M$ if $M$ is a complex manifold.


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