Q:

cohomology of BG, G compact Lie group

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?

A:

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 $x_i$.

A:

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.)

A:

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.

