Question

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?

Answer 1

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 $x_i$.

Answer 2

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.

Answer 3

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.

Answer 4

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.)

Answer 5

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.