This question should have a good answer somewhere on here, but as of yet I've been unable to find one. Any links to existing writings would be very welcome.

My question relates to how exactly one should interpret the Borel cohomology $H^*_{G}X$ of a $G$-space $X$.

Coming from non-equivariant homotopy theory, I have come to recognise ordinary cohomology as a powerful tool for probing the topological structure of spaces. Using its equivalence with cellular cohomology on the category of CW spaces, its graded structure tells me in which dimensions cells live, and the torsion, cup products and Steenrod algebra tell me how they are attached to one another. The information it provides often makes explicitly constructing the space a fairly straightforward task.

But how should I interpret the Borel cohomology $H^*_{G}X$ of a $G$-space $X$? Exactly what information does it tell us about the $G$-space $X$?

This is to be compared with the Bredon cohomology $\mathcal{H}^*_G(X)$, which is constructed with a given equivariant cellular structure on $X$ and does give explicit information relating to what cells are glued together, and how so, to form $X$. Although extracting concrete statements from the information it provides is more delicate than the non-equivariant case, the interpretation of Bredon cohomology is fairly clear.

One most often defines Borel cohomology by moving to the homotopy orbit space, that is, setting $H^*_{G}X=H^*(EG\times_GX)$, so that the Borel cohomology of a $G$-space $X$ is the ordinary cohomology of the homotopy orbit $EG\times_GX=EG\times X/[(eg,x)\sim (e,g^{-1}x)]$. If the $G$-action is free then $EG\times_GX\simeq X/G$, so $H^*_{G}X\cong H^*(X/G)$, and in this case the Borel cohomology does say something direct about the topological structure of $X$.

On the other hand, if the $G$-action is trivial, then $EG\times_GX\simeq BG\times X$, so $H^*_G(X)\cong H^*(BG\times X)$, and the Borel cohomology says nothing useful about the topological structure of $X$.

Obviously most cases of interest lie somewhere inbetween these two extremes. Is it therefore the case that the Borel cohomology should be interpreted as a deviation of the $G$-action itself from being free or trivial, rather than directly saying something about the topological structure of $X$?

So what about, say, semi-free actions? These enjoy both the best and worst properties of free and trivial actions (depending on how you look at things). What can we say about $X$ - or the $G$-action upon it - in this case?


1 Answer 1


Borel cohomology $H^*_G(X)$ is the cohomology of the homotopy quotient $X//G := EG \times_G X$.

The usual quotient, $X/G$, is bad from a homotopy theoretical point of view: it does not preserve equivariant homotopy equivalences. In other words, given two $G$-spaces $X$ and $Y$ and a homotopy equivalence $f : X \to Y$ which is $G$-equivariant, it is not necessarily true that the induced map $X/G \to Y/G$ is a homotopy equivalence. For an easy example, consider $X = \mathbb{R}$ with the $\mathbb{Z}$-action given by translation, and $Y = *$ the point with the trivial $\mathbb{Z}$-action.

The homotopy quotient fixes this problem. It is homotopy invariant: the map $f$ above will always induce a homotopy equivalence $X//G \to Y//G$.

Moreover there is a canonical map $X//G \to X/G$. If the $G$-action is free, then this map is a homotopy equivalence. So the way to think about this is the following: the homotopy quotient $(-)//G$ is an approximation of the usual quotient which has the good taste of being homotopy invariant. Spaces with a free $G$-action are the nice $G$-spaces (think about quotient of manifolds, for example), and for these nice spaces it's exactly the usual quotient; but for non-nice spaces, if you want something functorial and preserving homotopy, then you need to consider the homotopy quotient.

So in this sense, Borel cohomology is a bit like relative cohomology. Given a pair $(X,A)$, you want to say what's the cohomology of $X$ forgetting what's happening in $A$. For good pairs (CW pairs for example), this is just the cohomology $H^*(X/A)$. But for nasty pairs, you need to look at relative cohomology, $H^*(X,A)$. The same idea is at work for Borel cohomology.

This is the first motivation for $X//G$. But there's more.

Moreover the homotopy quotient is not just a space. It is a space equipped with a map to $BG = EG/G$. This map to $BG$ is simply the map $EG \times_G X \to EG \times_G * = EG/G = BG$ induced by the terminal map $X \to *$.

Conversely, given a space $Y$ equipped with a map to $BG$ you can recover a $G$-space, simply given by the pullback $EG \times_{BG} Y$. (Attention, the notation is awful! $(- \times_G -)$ is a quotient by the $G$-action, while $(- \times_{BG} -)$ is a pullback. But it's the standard notation.)

And this correspondence is an equivalence on homotopy categories. Given a $G$-space $X$, there is a zigzag of $G$-equivalences $X \gets EG \times X \to EG \times_{BG} (X//G)$. Conversely, given a space $Y \to BG$, the canonical map $(EG \times_{BG} Y)//G \to Y$ is a homotopy equivalence of spaces, compatible with the two given maps to $BG$.

[There may be better references, but a possible one is found in Section 4 of the paper Real models for the framed little $n$-disks operads by Khoroshkin–Willwacher. But this isn't the main point of that paper, and I'm sure textbooks about equivariant cohomology mention this, but I haven't checked.]

So to summarize, given a $G$-space $X$, the homotopy quotient $X//G$ equipped with its map to $BG$ tells you everything there is to know about $X$ and its $G$-action (up to homotopy).

So when you take its cohomology, you have the cohomology of a space that tells you everything about $X$ and its $G$-action... If you do not forget the canonical map $H^*(BG) \to H^*_G(X)$! The cohomology $H^*(BG)$ is simply the equivariant cohomology of a point $H^*_G(*)$, and this is often even abbreviated $H^*_G$.

In other words, you should not think about $H^*_G(X)$ as just an algebra. You should think about it as an $H^*_G$-algebra. And the data $H^*_G \to H^*_G(X)$ tells you everything cohomology knows about $X$ with its $G$-action.

For example in your question, you said that for a space with trivial $G$-action, $H^*_G(X) = H^*_G \otimes H^*(X)$ doesn't tell you much about the topology of $X$. That's not true. It's an $H^*_G$-algebra, and if you consider the tensor product $$\Bbbk \otimes_{H^*_G} (H^*_G \otimes H^*(X)) = \Bbbk \otimes H^*(X) = H^*(X),$$ you recover the cohomology of $X$. So if you agree that cohomology tells you something useful about a space, then you agree that $H^*_G(X)$ tells you something useful about $X$ too.

Also, you see that above the group cohomology of $G$ with trivial coefficients $H^*_G$ appears. This is also very nice, because equivariant cohomology can go both ways. It can tell you stuff about spaces with $G$-action, but it can also tell you stuff about group cohomology of $G$ if you choose your spaces well! For example there is a theorem of Symonds that says (among other things!) that the Castelnuovo–Mumford regularity of $H^*_G$ for a finite group $G$ is nonpositive. This completely algebraic statement, rather amazingly, is proved using equivariant cohomology of positive-dimensional manifolds.

  • $\begingroup$ I'd like to note I'm not an expert on equivariant cohomology. For example, regarding the theorem of Symonds, the only reason I know it is because I participated in a reading seminar in Lille on the topic. But as a non-expert, that's how I view it. $\endgroup$ Apr 28, 2018 at 10:23
  • $\begingroup$ Thank you for the detailed answer! Maybe I have misinterpreted the notation, but are you sure that the map $X\rightarrow EG\times_{BG}(EG\times_GX)$ is a $G$-equivalence? The right-hand space is $G$-equivalent to $EG\times X$ with the diagonal action, and since $EG$ has no $G$-fixed basepoint, the inclusion $X\hookrightarrow EG\times X$ isn't even a $G$-map. $\endgroup$
    – Tyrone
    Apr 28, 2018 at 15:40
  • $\begingroup$ I suppose what I'm struggling with is not that the homotopy quotient $X//G$ isn't a natural object to study homologically, rather it's the inherent difference between the spaces $X$ and $X//G$. I'm in a situation where I have knowledge of $H^*_G(X)$, and very little knowledge of the $G$-space $X$. $X$ is a finite, semi-free $G$-CW complex and the $H^*_G$-module $H^*_G(X)$ is certainly an (interesting) invariant of it, but what more does it really tell me about $X$? $\endgroup$
    – Tyrone
    Apr 28, 2018 at 15:48
  • $\begingroup$ @Tyrone I was a bit too hasty. There is no direct map $X \to EG \times_{BG} (X//G)$ actually. It's a zigzag $X \gets EG \times X \to EG \times_{BG} (X//G)$. Now both of these are $G$-equivalences and everything should be fine... $\endgroup$ Apr 28, 2018 at 17:18
  • $\begingroup$ @Tyrone My point is basically that there is no difference between the data "$G$-space $X$" and "space $X//G$ equipped with a map to $BG$". It's all a question of point of view. But the point of view "$X//G$ allows you to define an invariant, equivariant cohomology, which is interesting. $\endgroup$ Apr 28, 2018 at 17:19

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