# Cohomology ring of BG

Let $G$ be a compact Lie group, let $T$ be a maximal torus, and let $W$ be the Weyl group. My main question is as follows:

• How does one prove that $H^\ast(BG,\mathbb{Q})$ is isomorphic to the $W$-invariant part of $H^\ast(BT,\mathbb{Q}) \cong \mathbb{Q}[[x_1, \ldots, x_n]]$? This is apparently basic knowledge in algebraic topology, because I keep reading "recall that..." followed by some version of this statement and no references. But I can't find a proof in any of my textbooks.

I would ideally like a reference which also addresses the following secondary question:

• When is the natural map $H^\ast(BG,\mathbb{Z}) \to H^\ast(BT,\mathbb{Z})^W$ an isomorphism, and what can one say about the integral cohomology ring of $BG$ when it is not? Note the fact that the map above is an isomorphism for $G = U(n)$ is equivalent to the statement that the Chern classes are integral.

Thanks!

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Do you really mean power series $\mathbb{Q}[[x_1, \ldots, x_n]]$ ? –  Ralph Dec 20 '12 at 19:23
I think so... for example, the classifying space of $S^1$ is $\mathbb{C}P^\infty$, and the cohomology of $\mathbb{C}P^\infty$ is well known to be the ring of formal power series in one variable. –  Paul Siegel Dec 20 '12 at 19:39
Paul, sorry, but I doubt that this is well-known. The cohomology of $BS^1$ is a polynomial ring in one variable. Also note that $H^\ast(\mathbb{C}P^n)=\mathbb{Z}[x]/(x^{n+1})$. –  Ralph Dec 20 '12 at 19:55
The differences between the power series and polynomial rings in this case depend upon your choice to define $H^\ast(X)$ as either the product or sum over all $n$ of $H^n(X)$. –  Craig Westerland Dec 20 '12 at 20:04
Ah, quite right - I should have specified that I was defining $H^*(X)$ to be the product of $H^n(X)$ over all $n$. This question came up as I was working through some computations with characteristic classes, and this is apparently a common convention in that context. –  Paul Siegel Dec 20 '12 at 20:16

## 3 Answers

Notice that there is a sequence of homomorphisms $T \to N \to G$, where $N$ is the maximal torus normaliser (so $W = N/T$). $W$ acts on $BT$ (because it acts on $T$ by conjugation through group homomorphisms), and there is an equivalence from the classifying space of $N$ to the Borel construction for this action:

$$BN \simeq EW \times_W BT.$$

Consequently, we can compute the cohomology of $BN$ from the Leray-Serre spectral sequence

$$H^\ast(W; H^\ast(BT)) \implies H^\ast(BN).$$

Taking rational cohomology, this spectral sequence is concentrated in group-cohomological degree $0$, since $W$ is a finite group. Therefore the spectral sequence collapses at $E_2$, which is $H^0(W, H^\ast BT) = H^\ast(BT)^W$.

It therefore suffices to show that the map $BN \to BG$ is an isomorphism in rational cohomology. If we write $BN$ as $EG / N$, this map is a fibre bundle with fibre $G / N$, so it's enough to show that $G/N$ has the rational homology of a point.

For instance, if $G = SU(2)$, $N = \mathbb{Z} / 2 \ltimes T$, and $T = S^1$. Then $G/T = \mathbb{C} P^1$, and the action of $\mathbb{Z} / 2$ is antipodal, giving $G / N = \mathbb{R} P^2$, which is indeed rationally a point. I don't remember the argument in general, but I think this is always true.

Hopefully this indicates how the corresponding integral statement can fail - there can be torsion contributions from the higher group cohomology of $W$, which needs to be exactly cancelled (via a differential in the second spectral sequence above) with a torsion cohomology class from $G/N$.

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This looks very nice, thanks! My only apprehension - apart from the fact that I'm not sure either how to calculate the rational homology of $G/N$ in general - is that I would like to avoid spectral sequences when I type this up. That might be awkward in this case... –  Paul Siegel Dec 20 '12 at 20:21
You can avoid the first spectral sequence if you have another way of talking yourself into believing that the cohomology of $BN$ is the same as the $W$-invariants of $H^*(BT)$, e.g., using the transfer. For the latter spectral sequence, you can use the fact that the Euler class of $G/T$ is nonzero, as Chris Gerig indicates in his answer. –  Craig Westerland Dec 20 '12 at 20:27

I don't remember where I heard the following proof/sketch:

Using the fibering $G/T\to BT\to BG$ and the fact that the Euler class of $G/T$ is nonzero, we have that $H^\ast(BG)$ embeds into $H^\ast(BT)$ (it composes with the transfer map to be multiplication by the Euler class); and the desired isomorphism comes from the fact that $W$ acts on $H^*(G/T)$ as the regular representation.

This is actually a special case of equivariant cohomology, where we instead use the Borel construction and the fibering $G/T\to M_T\to M_G$.

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Chris, this was my memory, too, but it's hard to imagine that it's exactly true: for the case $G=SU(2)$, $N=\mathbb{Z}/2$ and $G/T=P^1$, whose cohomology is indeed free of rank two. However there's no way that $W$ can act on it by the regular representation, since one generator is in dimension 0, and the other in dimension 2. –  Craig Westerland Dec 20 '12 at 20:16
Oh, maybe this is nonsense -- the 0 dimensional class is the trivial factor in the regular representation, and the 2 dimensional factor is the reduced regular representation. What is subtle here is that how the regular representation knows which summands correspond to which cohomological degrees. I suppose that this should be related to the Bruhat order on the Weyl group. –  Craig Westerland Dec 20 '12 at 20:23

Q1: Let me first note, that the statement $$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BT,\mathbb{Q})^W\tag{\ast}$$ made in the question requires in addition that $G$ is connected. However, the general case can be reduced to the connected case since $$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BG_0,\mathbb{Q})^{G/G_0}$$ where $G_0$ is the identity component of $G$.

A text book reference for $(\ast)$ can be found in [Hsiang: Cohomology theory of topological transformation groups, Chapter III, §1, Lemma 1.1]. The results of the book that are relevant for your question can also be found in the following paper: http://www.math.uwo.ca/~rgonzal3/qfy.pdf (cf. Remark 9, Lemma 5). Other approaches and more information can be found in

Q2: First note that the kernel of the restriction map $$\rho^\ast: H^\ast(BG;\mathbb{Z}) \to H^\ast(BT;\mathbb{Z})$$ is the tosion subgroup $Tors$ of $H^\ast(BG;\mathbb{Z})$. So, $\rho^\ast$ is injective iff $H^\ast(BG;\mathbb{Z})$ is torsion-free (there is a lot of literature about torsion of $H^\ast(BG;\mathbb{Z})$ so I won't discuss it here).

A short paper of Feshbach [The image of $H^\ast(BG;\mathbb{Z})$ in $H^\ast(BT;\mathbb{Z})$ for $G$ a compact Lie group with maximal torus $T$. Topology 20(1981),93-95] characterizes when the induced map $$\bar{\rho}^\ast: H^\ast(BG;\mathbb{Z})/Tors \to H^\ast(BT;\mathbb{Z})$$ is an isomorphism:

$\quad\bar{\rho}^\ast$ is an isomorphism iff $H^\ast(BG;\mathbb{Z})/Tors \otimes \mathbb{F}_p$ is an integral domain for each prime $p$.

There is also a counter-example that shows that $\bar{\rho}^\ast$ is not always an isomorphism.

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Thank you for all of the great references! In particular, I plan to spend some time with Feshbach's paper. –  Paul Siegel Dec 20 '12 at 22:43