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.

  • $\begingroup$ 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. $\endgroup$ Dec 10, 2019 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, 2019 at 13:37

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.