Let $T$ be a maximal torus of a compact Lie group $G$. We define the Weyl group of $G$ as the quotient space $N(T)/T$ where $N(T)$ is the normalizer of $T$ in $G$. I have to prove that if $G$ is $U(n)$ we can identify $W$ with the permuation group $S_{n}$. Could you give me a reference or a proof of this claim?

  • 1
    $\begingroup$ Well, just do what the theorem says: 1. find (any) torus, 2. find its normalizer, 3. make the quotient, 4. compare it with $S_n$. Which step are you having trouble with? $\endgroup$
    – Marek
    Jan 16, 2013 at 19:13
  • $\begingroup$ @Marek: Number 4: compare with $S_{n}$. $\endgroup$
    – user58536
    Jan 16, 2013 at 23:20
  • $\begingroup$ I'd like to know 2,3 and 4 step. $\endgroup$ Jan 17, 2013 at 0:15

1 Answer 1


First let's fix some torus, e.g. one consisting of the matrices of the form $$T(\phi_1, \dots, \phi_n) = \pmatrix { e^{i\phi_1} & 0 & \cdots & 0 \cr 0& e^{i\phi_2} & \cdots & 0 \cr \vdots & \vdots & \ddots & 0 \cr 0 & 0 & \cdots & e^{i\phi_n} \cr }$$ where $\phi_i \in [0, 2\pi)$ for all $1 \leq i \leq n$.

Now, you should be familiar with the fact that for any $A, B \in GL(n, \mathbb C)$ we have that $B$ and $ABA^{-1}$ have the same eigenvalues. In particular, this holds when $B \in T$ and so every $C \in N_{U_n}(T)$ must therefore preserve the eigenvalues of $B$. Since eigenvalues of $CBC^{-1}$ and $B$ are the same, we must have $CBC^{-1} = T(\phi_{j_1}, \dots \phi_{j_n})$ for some permutation $j \in S_n$. Therefore $N_{U_n}(T) / T \leq S_n$. Moreover, it is easy to see that each permutation of entries of $T(\phi_1, \dots, \phi_n)$ gives us a different element of $N_{U_n}(T) / T$ and therefore the Weyl group is indeed equal to $S_n$.


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.