# rational cohomology of classifying spaces of complex reductive Lie groups

I am looking for a reference or an ad-hoc proof of the following fact, which seems to be known to experts: Let $\mathbf{G}$ be a complex algebraic group with maximal (algebraic) torus $\mathbf{T}$ and Weyl group $W$. Let $G=\mathbf{G}(\mathbb{C})$ and let $\mathrm{B}G$ denote the classifying space of $G$; define $T$ and $\mathrm{B}T$ similarly. Then the canonical map between the rational cohomology rings $\mathrm{H}^*(\mathrm{B}G,\mathbb{Q})\to\mathrm{H}^*(\mathrm{B}T,\mathbb{Q})$ is injective and its image is the $W$-fixed elements in $\mathrm{H}^*(\mathrm{B}T)$.

There are numerous proofs of the analogous result for compact (real) Lie groups, i.e. when $G$ is a compact Lie group and $T$ is maximal compact torus in $G$.

It should be possible to deduce the complex case from the real one because every complex reductive Lie group $G$ contains a maximal compact real Lie group $G_0$ which is a retract, but this fact alone is not sufficient as one has to take into consideration the corresponding maximal tori, and make sure that the corresponding Weyl groups of $G$ and $G_0$ relative to these tori are canonically isomorphic.

• Let $T_0\subset T$ be maximal tori in $G_0$ and $G$, resp. Let $N_0\subset G_0$ and $N\subset G$ be the stabilizers of the corresponding maximal tori. Set $W_0=N_0/T_0$ and $W=N/T$. The embedding $N_0\hookrightarrow N$ induces a homomorphism $W_0\to W$, which is injective because $N_0\cap T=T_0$. To prove the surjectivity, it suffices to show that any reflection corresponding to a root is contained in $W_0$; for this end it suffices to consider $G={\rm SL}_2$ and $G={\rm PGL}_2$. This gives a canonical isomorphism $W_0\to W$. – Mikhail Borovoi Jun 23 '17 at 14:04
• Thanks! This should be enough for an ad hoc proof, provided that the (real) rank of $T_0$ is the same the (complex) rank of $T$, which should be in the literature. – Uriya First Jul 1 '17 at 20:57
• What do you mean by the real rank of $T_0$? The real algebraic torus $T_0$ is compact, hence anisotropic, hence its real rank is zero! – Mikhail Borovoi Jul 2 '17 at 1:58
• The real algebraic torus $T_0$ is a real form of $T$, that is, there is a canonical isomorphism $$T_0\times_{\mathbb R} {\mathbb C} = T.$$ Therefore, the (real) dimension of the real algebraic torus $T_0$ is equal to the (complex) dimension of the complex algebraic torus $T$. Maybe this is what you need. – Mikhail Borovoi Jul 2 '17 at 2:04
• A reference to the existence of a compact real form: Onishchik and Vinberg, Lie Groups and Algebraic Groups (Springer, Berlin, 1990), Section 5.2.3, Theorem 8, p. 244. Note that this book contains very useful tables! – Mikhail Borovoi Jul 7 '17 at 19:48