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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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  • $\begingroup$ Do you really mean power series $\mathbb{Q}[[x_1, \ldots, x_n]]$ ? $\endgroup$
    – Ralph
    Dec 20, 2012 at 19:23
  • $\begingroup$ 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. $\endgroup$
    – Paul Siegel
    Dec 20, 2012 at 19:39
  • $\begingroup$ 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 $H^\ast(\mathbb{C}P^n)=\mathbb{Z}[x]/(x^{n+1})$. $\endgroup$
    – Ralph
    Dec 20, 2012 at 19:55
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    $\begingroup$ 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 $H^n(X)$. $\endgroup$
    – Craig Westerland
    Dec 20, 2012 at 20:04
  • $\begingroup$ 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. $\endgroup$
    – Paul Siegel
    Dec 20, 2012 at 20:16

3 Answers 3

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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: Richard Gonzales, Localization in equivariant cohomology and GKM theory (cf. Remark 9, Lemma 5). Other approaches and more information can be found in this answer.

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 doi:10.1016/0040-9383(81)90015-X,

characterizes when the induced map $$\bar{\rho}^\ast: H^\ast(BG;\mathbb{Z})/Tors \to H^\ast(BT;\mathbb{Z})$$ is an isomorphism:

$\bar{\rho}^\ast$ 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, $\mathrm{Spin}(12)$, that shows that $\bar{\rho}^\ast$ is not always an isomorphism.

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  • $\begingroup$ Thank you for all of the great references! In particular, I plan to spend some time with Feshbach's paper. $\endgroup$
    – Paul Siegel
    Dec 20, 2012 at 22:43
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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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  • $\begingroup$ 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... $\endgroup$
    – Paul Siegel
    Dec 20, 2012 at 20:21
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    $\begingroup$ 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. $\endgroup$
    – Craig Westerland
    Dec 20, 2012 at 20:27
  • $\begingroup$ For posterity, here is a proof borrowed from Mimura and Toda. It relies on already knowing $H^*(G/T)$ is concentrated in even degree, something I believe was initially due to Borel. First, show the only fixed point of left multiplication by any $t \in T \setminus \{1\}$ on $G/N$ is the identity coset $1N$. Now $T$ acts by isometries on the normal $\mathfrak n^\perp \cong \mathfrak g /\mathfrak n$, preserving a small sphere. The exponential of this sphere is another sphere, $T$-invariantly dividing $G/N$ into a disk and a complement $C$. $\endgroup$
    – jdc
    Mar 1, 2016 at 1:29
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    $\begingroup$ Because $T$ is path-connected, left multiplication by $t$ has Lefschetz number equal to the Euler characteristic. Since $t \neq 1 \in T$ acts without fixed points on the sphere and $C$, these have Euler characteristic 0. In particular, the sphere is odd-dimensional. By excision, the relative Euler characteristic $\chi(G/N,C) = 1$. The long exact sequence of a pair then gives $\chi(G/N) = 1$. Since $G/T \to G/N$ is a finite covering with fiber $W$ and $H^{\mathrm{odd}}(G/T) = 0$, it follows $H^{\mathrm{odd}}(G/N) = 0$ and so $\dim_{\mathbb Q} H^*(G/N) = 1$. $\endgroup$
    – jdc
    Mar 1, 2016 at 1:31
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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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  • $\begingroup$ 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. $\endgroup$
    – Craig Westerland
    Dec 20, 2012 at 20:16
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    $\begingroup$ 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. $\endgroup$
    – Craig Westerland
    Dec 20, 2012 at 20:23

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