# Weyl group action on maximal torus

This is a classical result in the theory of Lie groups, but I am stuck on trying to understand the action of the Weyl group on the classifying space of the torus. Namely,

Let $$G$$ be a compact connected Lie Group and $$T$$ be a maximal torus in $$G$$. Let $$W = N(T)/T$$ be the Weyl group of $$T$$ in $$G$$. This groups acts on $$T$$ by conjugation (the action is well defined as $$T$$ is abelian) and so it induces an action on the classifying space $$BT$$.

This is the argument that I still do not completely get

Mimura M, Toda H. Topology of Lie groups, I and II. American Mathematical Soc.; 1991.

Let $$p: BT \rightarrow BG$$ be the map induced by the inclusion, and $$f_n : BT \rightarrow BT$$ the map induced by the conjugation of $$nT \in W$$. Then $$f_n(eT) = enT$$ for $$e \in ET$$ and so $$(p\circ f_n)(eT) = p(enT) = p(enG) = eG = p(eT)$$ Therefore, $$p^*(H^*(BG)) \subseteq H^*(BT)^W$$

It is not clear for me why the conjugation action induces the map $$f_n$$ defined above. I'd appreciate if someone has an insight of how to compute the map out of the action.

• What do you already know about classifying spaces? The existence of $f_n$ is coming from the more general fact that a homomorphism a between Lie groups induces a map between their classifying spaces. – Jason DeVito Dec 10 '19 at 2:24
• @JasonDeVito I know that. My question is why specifically the map $f_n$ is given by that when it comes from conjugation; this is crucial to show that (the image of) $H^*(BG)$ is invariant under the action of $W$ – C. Zhihao Dec 10 '19 at 13:37

I'm not sure how Toda computed the map $$f_n$$, but here is an alternate proof that $$p^\ast(H^\ast(BG))\subseteq H^\ast(BT)^W$$. I found the proof in the notes of Dwyer and Wilkerson
Let $$n\in N(T)$$ and consider the function $$c_n:G\rightarrow G$$ which is conjugation by $$n$$. Because there is a path in $$G$$ from $$n$$ to the identity, we see that the induced map $$c_n:BG\rightarrow BG$$ is homotopic to the identity, and hence, that $$c_n^\ast: H^\ast(BG)\rightarrow H^\ast(BG)$$ is the identity map.
Now, consider the following commutative diagram. $$\newcommand{\ra}{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T & \ra{i} & G \\ \da{c_n|_T} & & \da{c_n} \\ T & \ras{i} & G \\ \end{array}$$
This diagram induces another commutitive diagram. $$\newcommand{\ra}{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} H^\ast(BG) & \ra{p^\ast} & H^\ast(BT) \\ \da{c_n^\ast} & & \da{(c_n|_T)^\ast} \\ H^\ast(BG) & \ras{p^\ast} & H^\ast(BT) \\ \end{array}$$
We have already argued that $$c_n^\ast$$ is the identity. So, $$(c_n|_T)^\ast p^\ast = p^\ast$$.
In other words, each $$n\in N(T)$$ acts trivially on the image of $$p^\ast$$. Thus, all of $$W = N(T)/T$$ acts trivially on the image of $$p^\ast$$ so $$p^\ast(H^\ast(BG))\subseteq H^\ast(BT)^W$$.