## cohomology of BG, G compact Lie group

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It has been stated in several papers that $H^{odd}(BG,\mathbb{R})=0$ for compact Lie group $G$. However, I've still not found a proof of this. I believe that the proof is as follows:

--> $G$ compact $\Rightarrow$ it has a maximal toral subgroup, say $T$

--> the inclusion $T\hookrightarrow G$ induces inclusion $H^k(BG,\mathbb{R})\hookrightarrow H^k(BT,\mathbb{R})$

--> $H^*(BT,\mathbb{R})\cong \mathbb{R}[c_1,...,c_n]$ where the $c_i$'s are Chern classes of degree $\deg(c_k)=2k$

--> Thus, any polys in $\mathbb{R}[c_1,...,c_n]$ are necessarily of even degree. Hence, $H^{odd}(BG,\mathbb{R})=0$

Is this the correct reasoning? Could someone fill in the gaps; i.e., give a formal proof of this statement?

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Why is this tagged mathematical-physics? – Andy Putman Apr 15 2011 at 5:48
@Andy - perhaps it is the direction the OP is coming from... which doesn't make it right however. – David Roberts Apr 15 2011 at 6:34
Yes, sorry about tagging it mp. I was thinking about where this surfaces - which happened to be some papers on Chern-Simons theory. – Kevin Wray Apr 15 2011 at 6:43
For the groups $U(n), SU(n) and Sp(n)$ the result follows from calculating the integral cohomology rings and seeing that they are polynomial rings on even degree generators. This is 4D.4 in Hatcher. But I believe the OP is thinking of the approach outlined in e.g. the first paragraph of arxiv.org/abs/0903.4865 - this is due to Borel I believe. – David Roberts Apr 15 2011 at 6:45
Retagged . – David Roberts Apr 15 2011 at 6:46
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## 5 Answers

For $G$ compact (and connected), $H(BG,\mathbb{R})$ is the $G$-equivariant cohomology of a point. It can be computed via infinitesimal methods and it is isomorphic to the cohomology of the Weil algebra of the Lie algebra of $G$. However there is an equivalent model computing the same cohomology, known as the Cartan model. The grading in the Cartan model is such that all cochains have even degree.

Since you tagged this "mathematical physics", I should add that the Weyl algebra is the dga generated by the connection one-form (i.e., gauge field) on the universal $G$-bundle and the point behind the Cartan model is that the only appearance of the gauge field in an equivariant cocycle is via "minimal coupling" or via the curvature. There's nothing in a point for the gauge field to couple minimally to, hence all you get are curvatures. Being 2-forms all polynomial expressions in the curvatures have even degree.

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If you know a bit of Algebraic Topology (in particular the dreaded spectral sequences), the following is a nice way to see this.

The well known Hopf Theorem states that for $G$ a compact connected Lie group the real cohomology $H^*(G;\mathbb{R})\cong\wedge(y_1,\ldots , y_r)$ is an external algebra on odd dimensional generators $y_i$. This is proved using the Hopf algebra structure on $H^*(G;\mathbb{R})$. A good reference is Chapter 1 of the book Algebraic Models in Geometry by Félix, Oprea and Tanré (which I believe also discusses the approach mentioned in José's answer).

Since $BG$ is simply-connected, it is a nice exercise using the Serre spectral sequence of the universal $G$-bundle $G\to EG\to BG$ to see that $H^*(BG;\mathbb{R})\cong\mathbb{R}[x_1,\ldots x_r]$ is a polynomial algebra on even dimensional generators $x_i$.

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See also McCleary's User's guide to spectral sequences, Theorem 7.30 for a result attributed to Borel: If $G$ is a topological group with $H_\ast(G;k) = \Lambda(x_1,...,x_r)$ where $deg(x_i)$ is odd for all $i$ and $k$ is a field, then $H^\ast(BG;k) = k[y_1,...,y_r]$ with $deg(y_i) = deg(x_i) + 1.$ – Ralph Apr 15 2011 at 10:33

For a reference see "Hsiang: Cohomology theory of topological transformation groups" (chapter III, §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).

The idea is roughly as follows: Let $G$ be a compact Lie group and $G_0$ be the connecting component of the identity element. Then $BG_0 \to BG$ is a covering and $$H^\ast(BG;\mathbb{R}) = H^*(BG_0;\mathbb{R})^{\Gamma}$$ where $\Gamma = G/G_0$ is a finite group. Let $T$ be a maximal torus of $G_0$. Then one shows that $$H^\ast(BG_0;\mathbb{R}) = H^\ast(BT;\mathbb{R})^W$$ where $W$ is the Weyl group of $G_0$. Thus $H^\ast(BG;\mathbb{R})$ can be identified with a subring of $H^\ast(BT;\mathbb{R})$ that is a polynomial ring with generators of degree two. So $H^{odd}(BG;\mathbb{R}) = 0$ follows.

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I added the missing asterisks: it is safer to use `\ast` than `*`. – José Figueroa-O'Farrill Apr 15 2011 at 9:38
Good to know. José, thanks. – Ralph Apr 15 2011 at 9:49
The reference I know for the general fact $H^\ast(X/\Gamma;{\Bbb Q}) = H^\ast(X;{\Bbb Q})^\Gamma$ (for $\Gamma$ a finite group acting freely) is Grothendieck's Tohoku paper, which of course uses a lot of spectral sequences. Grothendieck remarks that this can be proved much more easily, presumably along the lines Allen K. indicates. Does anyone know an earlier/easier reference? (Cf. the answer here: mathoverflow.net/questions/18898/… ) – Dave Anderson Apr 17 2011 at 5:47

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@ Dave Anderson: The isomorphism $H^\ast(X/\Gamma;{\Bbb Q}) = H^\ast(X;{\Bbb Q})^\Gamma$ is a simple application of the transfer homomorphism. For a textbook proof see my algebraic topology book, Proposition 3G.1, page 321. – Allen Hatcher Apr 17 2011 at 14:40
@Allen, thanks! – Dave Anderson Apr 17 2011 at 15:45

Recall the Chern-Weil homomorphism $Sym^{\ast} \mathfrak{g}^{\vee} \to H^{\ast}(BG; \mathbb{R})$ for each Lie group $G$. If $G$ is compact, it is an isomorphism. See Dupont, Curvature and Characteristic classes (a great book). This includes the desired statement, because the CW-homomorphism doubles the degree. Here is a sketch, taken from that book.

One step is fairly easy, namely that $H^{\ast}(BG) \to H^{\ast} (BT)$ is injective: There is a fibre bundle $f:BT \to BG$ with fibre $G/T$. Let $\chi$ be the Euler class of the vertical tangent bundle of $BT$. Then the transfer $\tau_{f}: H^{\ast}(BT) \to H^{\ast}(BG)$ is defined as $x \mapsto f_{!} (x \chi)$. It is not hard to see (exercise, use $f_{!} (f^{\ast} x \cdot y) =x f_{!} (y)$ that $\tau_f \circ f^{\ast}: H^{\ast} (BG) \to H^{\ast} (BG)$ is the multiplication by Euler number of $G/T$. It is a classical result that the Euler number of $G/T$ is the order of the Weyl group, in particular positive, in particular nonzero. Hence $f^{\ast}: H^{\ast}(BG) \to H^{\ast}(BT)$ is injective.

Moreover, the Chern-Weil homomorphism is an iso for a torus by a direct computation (which ultimately boils down to the computation $\frac{1}{2 \pi i} \int_{S^1} \frac{dz}{z}=1$).

To prove that $f^{\ast}$ is surjective on the $W$-invariants is harder. It remains to show (write down a diagram) that the restriction $Sym^{\ast} (\mathfrak{g}^{\vee}) \to Sym^{\ast} \mathfrak{t}^{W}$ is an isomorphism. This is a theorem by Chevalley and uses quite a bit of structure theory of Lie groups.

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Yes, I like your argument for injectivity and this is all that I need to show $H^{odd}(BG,\mathbb{R})=0$. Thanks. – Kevin Wray Apr 17 2011 at 6:33

Here's the argument I know that avoids spectral sequences, based on the little-known space $G/N(T)$.

In between $T$ and $G$ is $N(T)$. Note that $EG$ "is an" $ET$ and $EN(T)$, since it's contractible and $T,N(T)$ act freely on it, so we can identify $BT, BN(T)$ with $EG/T, EG/N(T)$.

Now consider the two maps $EG/T \to EG/N(T) \to EG/G$, with fibers $W = N(T)/T$ and $G/N(T)$ respectively. The first case divides by a free action of $W$, so we can identify $H^\ast(BN(T);{\mathbb Q}) = H^\ast(BT; {\mathbb Q})^W$ by pushing and pulling. (Actually we only need to invert $|W|$, and generally less; for $G=U(n)$ it's true over $\mathbb Z$.) In particular, there is only even cohomology.

So let's look at the space $G/N(T) = (G/T)/W$. The space $G/T$ has a Bruhat decomposition, hence only even-degree cohomology (even over $\mathbb Z$), which you can prove via Morse theory on a generic adjoint orbit if you don't want to bring in algebraic geometry, and its Euler characteristic is $|W|$.

Hence the space $(G/T)/W$ has (rationally) only even-degree cohomology, and Euler characteristic $1$. So it has the rational cohomology of a point! For $G=SU(2)$ this space is ${\mathbb RP}^2$.

By a particularly trivial application of Leray-Hirsch (which I think is the only remainder of the spectral sequence argument Mark Grant gave), $H^\ast(EG/G; {\mathbb Q}) \cong H^\ast(EG/N(T); {\mathbb Q})$.

(Oops: I guess this answer isn't so different from Ralph's.)

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Allen, I can see why push-pull gives an inclusion of $H^\ast(BN(T);{\Bbb Q})$ in $H^\ast(BT;{\Bbb Q})^W$, but why is the map surjective?  Am I missing something obvious?  This almost proves the general fact about finite quotients from Grothendieck's Tohoku paper, and I'd like to know the "easy" proof! – Dave Anderson Apr 17 2011 at 5:54
Isn't pull-push just multiplication by $|W|$? – Allen Knutson Apr 17 2011 at 12:40
I like this argument. If $p:Y \to X$ is a finite Galois covering of manifolds with group $W$, then $H^{\ast} (X) \to H^{\ast}(Y)^W$ is an isomorphism (with real coeff.). Injectivity is ''pull-push''. Surjectivity: if a form $\omega$ represents and element in $H^{\ast}(Y)^G$, then $\frac{1}{|W|} \sum_g g^{\ast} \omega$ represents the same class and is invariant. But invariant forms on $Y$ descend to forms on $X$. $BG$ and the likes are not manifolds, but one can find Hilbert manifold models, where de Rham theory works well.  – Johannes Ebert Apr 17 2011 at 14:06
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