## The Cohomology of Quotients

We’ve organized a mostly informal Topics in Noncommutative Algebra seminar this semester, and I’m talking first in it.  I’m eventually going to be talking about a paper of Ginzburg’s connecting Hochschild and cyclic cohomology to the equivariant cohomology of representation schemes.  Unfortunately, the trouble about talking about fun results like that is that you need to cover alot of background material; as such, I’m doing what is turning out to be a two lecture series on equivariant cohomology and its deRham version.  I figured I’d mirror these talks with a couple of posts, and maybe even talking about Ginzburg’s paper if I get enough prereqs covered.

Today I’m just going to be talking about topological equivariant cohomology.  Let’s start with a nice space (say, a CW complex$M$ and a Lie group $G$ which acts on $M$.  Unless this action is free and proper, the quotient space $M/G$ might be a poorly behaved space.  Take, for example, $\mathbb{Z}$ acting on $S^1$ by some irrational rotation; the quotient isn’t even Hausdorff.

The motivating question of equivariant cohomology is: “Is there a good cohomology theory for the pair $(M,G)$, which is $H^\bullet_{CW}(M/G)$ if $G$ acts freely and properly?” The hope is that this will shine some more light on the hidden internal structure of the bad quotients.

The idea is to replace the space $M$ with a related space on which the action of $G$ is free and proper.  First, lets assume the existance of a space $E$ with the following two crucial properties:

1) $E$ is a contractible space.

2) $G$ has a prescribed action on $E$ which is free and proper.

Readers might recognize the quotient space $E/G$; it is called the classifying space or the (first) Eilenberg-MacLane space of $G$, depending on whether $G$ is a connected Lie group or a discrete group.  The space $E$ exists in most reasonable cases (discrete groups, compact finite-dimensional Lie groups), but its almost always unavoidably infinite-dimensional.

Next, let us consider the space $M\times E$ equipped with the diagonal action of $G$ (so $g\cdot(m,e)=(gm,ge)$).  This action is free and proper, and so the quotient $(M\times E)/G$ is a CW complex, though sadly still infinite dimensional.  However, in the case that $G$ acted freely and properly on $M$, this space is weak homotopy equivalent to $M/G$!  To see this, note that the projection $M\times E\rightarrow M$ induces a map $(M\times E)/G\rightarrow M/G$ which is a fiber bundle with fiber $E$$E$ is contractible, so the projection map is a weak homotopy equivalence.

This inspires the definition of the $G$-equivariant cohomology of $M$ $H^\bullet_G(M):=H^\bullet_{CW}((M\times E)/G)$.  I haven’t justified that this is well-defined; it would seem to depend on the choice of $E$.  It is independant of this choice, since it turns out any two $E$ are weakly homotopic.  Also, note that this definition is functorial with respect to $G$-equivariant maps between spaces, as any good cohomology theory should be.

Let’s see some examples.

Example 1.  $M=S^1$, $G=\mathbb{Z}$ and $\mathbb{Z}$ acts by rotation by $\gamma$ (note this is not free and proper for any $\gamma$).  This is the first example because it was one of the few cases with a finite dimensional $E=\mathbb{R}$, where $\mathbb{Z}$ acts by translation by 1.

Then $M\times E=S^1\times \mathbb{R}$, and $n\cdot (\theta , r)=(e^{2\pi i \gamma n}\theta, r+n)$.  We can make this simpler by conjugating the action by the homeomorphism $S^1\times \mathbb{R}$: $(\theta, r)\rightarrow (e^{2\pi i \gamma r}\theta,r)$.  This turns the action into $n\cdot(\theta,r)\rightarrow (\theta,r+n)$, and didn’t change the topology of the quotient.  Therefore, $S^1\times \mathbb{R}/\mathbb{Z}=S^1\times S^1$, and so $H^\bullet_{\mathbb{Z}}(S^1)=H^\bullet_{CW}(S^1\times S^1)$.

Note that $S^1$ was only one dimensional, and yet the equivariant cohomology had components in degree 2.  This is a reoccuring theme, that the dimension of $M$ is no longer a useful bound on degree of the cohomology.

Example 2.  $M=\{pt\}$, and $G$ is any group with a trivial action on $\{pt\}$.  Then $H^\bullet_G(\{pt\})=H^\bullet(E/G)$.  Note that this cohomology ring is most likely not trivial, despite $M=\{pt\}$.  Since this cohomology ring only depends on the choice of group $G$, it is called the group cohomology of $G$ (and this coincides with most other notions of group cohomology you might know).  Therefore, equivariant cohomology is a generalization of the theory of group cohomologies.

The fact that points have interesting cohomology has some enjoyable consequences.  Any $M$ with a $G$ action has a trivial map to $\{pt\}$ which is equivariant.  Therefore, there is a canonical map $H^\bullet_G(\{pt\})\rightarrow H^\bullet_G(M)$ called the characteristic homomorphism

Now lets consider the case of $G$ acting freely and properly on $M$.  Since $H^\bullet_G(M)=H^\bullet(M/G)$, we get a map from the group cohomology of $G$ to the CW cohomology of $M/G$.

However, when $G$ acts freely and properly, the quotient map $M\rightarrow M/G$ often goes by a different name: principal $G$-bundle.  This is because it is a fiber bundle with each fiber a $G$-torsor.  Principal bundles come up in all sorts of places, in particular because they contain all the important information of a vector bundle with structure group $G$.

Thus, we can do the following.  Starting with a vector bundle over some space $X$ with structure group $G$, we can reduce it to a principal $G$ bundle $M$ over $X$ (note that the quotient $M/G=X$).  We then forget about the bundle map and only think about the total space $M$ and its $G$ action (which is free and proper).  The characteristic homomorphism in this case is a map:

$H^\bullet_G(\{pt\})\rightarrow H^\bullet_{CW}(M/G)=H^\bullet_{CW}(X)$

Hence, the vector bundle determined a map from the group cohomology of $G$ to the topological cohomology of $X$.  If we go a step further, and agree upon some canonical generators for the group cohomology, we can just talk about the images of these generators as cohomology classes on $X$; these are called characteristic classes.  A brief list of some well-known groups and their characteristic classes:

• $O(n)$: Pontraygin classes
• $SO(2n)$: Pontraygin classes, plus the Euler class (one extra generator is needed)
• $U(n)$: Chern classes

(Does anyone know the group for Steifel-Whitney classes?  My guess is that its something like $(\mathbb{Z}/2\mathbb{Z})^n$, but I don’t know.)

Anyway, so equivariant cohomology is clearly very neat.  Although, if you are like me, you’d prefer all of your topological cohomologies to be deRham cohomology.  It would seem like this is a doomed undertaking, because $E$ is infinite-dimensional. Yet, as we will see, it is not only possible, but in some ways more natural, to phrase this theory in terms of a kind of deRham theory.

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### 8 Responses to “The Cohomology of Quotients”

1. toomuchcoffeeman Says:

minor quibble: principal bundle’ not principle bundle’

2. Ben Webster Says:

Aren’t the Steifel-Whitney classes the characteristic classes for O(n) with $\mathbb{Z}/2$ coefficients?

3. TW Says:

Clearly $\mathbb{C}/S^1 = [0,\infty)$. Hence it is Hausdorff. For a non-Hausdorff quotient consider $\mathbb{C}^2/S^1$ where $z.(x,y) := (zx,y/z)$ for $z \in S^1$ and $(x,y) \in \mathbb{C}^2$.

4. TW Says:

Sorry. You have to consider the “same” action but for C* instead of S¹.

5. Greg Muller Says:

Oh, yeah, thats a good point. I was moving a little too quickly and trying to strip down the example $\mathbb{C}^n/\mathbb{C}^*$.

Great, I was going mad yesterday because I couldn’t figure out why $\mathbb{C} / S^1$ wasn’t Hausdorff! No I can stop worrying about it!
The Steifel-Whitney classes are the $\mathbb{Z}/2$ characteristic classes of O(n). The Pontryagin classes might by the rational characteristic classes, but I’m not remembering for sure. The construction I remember is that you take your real vector bundle, tensor with $\mathbb{C}$ and take the (even) Chern classes of that. I also remember that there is another type of Pontryagin class for quaternionic bundles and also that the integral cohomology of BO(n) is a mess (hence that integral characteristic classes are a mess).