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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.

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  • $\begingroup$ 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. $\endgroup$ – Jason DeVito Dec 10 '19 at 2:24
  • $\begingroup$ @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 $W$ $\endgroup$ – C. Zhihao Dec 10 '19 at 13:37
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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. $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} T & \ra{i} & G \\ \da{c_n|_T} & & \da{c_n} \\ T & \ras{i} & G \\ \end{array}$$

This diagram induces another commutitive diagram. $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} H^\ast(BG) & \ra{p^\ast} & H^\ast(BT) \\ \da{c_n^\ast} & & \da{(c_n|_T)^\ast} \\ H^\ast(BG) & \ras{p^\ast} & H^\ast(BT) \\ \end{array}$$

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 $p^\ast(H^\ast(BG))\subseteq H^\ast(BT)^W$.

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