# $N_{G}(E)/C_{G}(E)$ is the Weyl group of $G$?

In the algebraic group $$G = \operatorname{PGL}_4(\mathbb{C})$$, let $$E$$ denote the subgroup of elements of order dividing 2 in the diagonal maximal torus; it is generated by the images of the three matrices $$\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix},\quad \begin{bmatrix} 1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & I_2 \end{bmatrix},\quad \begin{bmatrix} -1 & 0 \\ 0 & I_3 \end{bmatrix}.$$ Direct computation shows that $$N_{G}(E)/C_{G}(E) \cong S_4\,$$, which is the Weyl group of $$G$$. Analogous result can be obtained for $$n = 3, 5, 6, 7, 8, 9, 10$$ in $$\operatorname{PGL}_{n}(\mathbb{C})$$ by direct calculations. Is this observation true for any $$n$$? Is there any reference relevant?


Let $$G$$ be a semisimple group of adjoint type over an algebraically closed field $$k$$ of characteristic 0. Let $$T\subset G$$ be a maximal torus. Write $$E=T^{(2)}:= \{t\in T(k)\ |\ t^2=1\}.$$ We wish to compute the centralizer $$\ZZ(E)$$ and the normalizer $$\NN(E)$$.

Observe that $$\NN(E)\supseteq \ZZ(E)\supseteq T$$. We compute $$\zz(E):=\Lie \ZZ(E)$$.

Lemma 1. $$\zz(E)=\tt:=\Lie T$$.

Proof. Since $$\ZZ(E)\supseteq T$$, we have $$\zz(E)\supseteq\tt$$. Write the root decomposition $$\Lie G=\tt\oplus\bigoplus_{\beta\in R}\gg_\beta\,,$$ where $$R=R(G,T)\subset \X^*(T)$$ is the root system. Since $$\zz(E)\supseteq\tt$$, we have $$\zz(E)=\tt\oplus\bigoplus_{\beta\in M}\gg_\beta$$ for some subset $$M\subseteq R$$. Here $$M=\{\beta\in R\ |\ \beta(t)=1\ \forall t\in E\}.$$ Let $$S\subset R$$ be a system of simple roots (a basis of $$R$$). Since $$G$$ is of adjoint type, the set $$S\subset R\subset \X^*(T)$$ is a basis of the character group $$\X^*(T)$$ of $$T$$. It follows that for any simple root $$\alpha\in S$$, there exists $$t\in E=T^{(2)}$$ such that $$\alpha(t)=-1$$. Let $$W=W(G,T)=\NN(T)/T$$ be the Weyl group. The group $$W$$ acts on $$E$$ and on $$R$$, and $$W\cdot S=R$$. Therefore, for any root $$\beta\in R$$, there exists $$t\in E$$ such that $$\beta(t)=-1$$, and therefore $$M=\varnothing$$ and $$\zz(E)=\tt$$, as required.

We compute $$\NN(E)$$. Since $${\rm char}(k)=0$$, it follows from Lemma 1 that the identity component $$\ZZ(E)^0$$ of $$\ZZ(E)$$ is $$T$$. We see that $$\NN(E)$$ normalizes $$\ZZ(E)$$, and hence it normalizes $$\ZZ(E)^0=T$$. It follows that $$\NN(E)\subseteq \NN(T)$$. On the other hand, $$\NN(T)$$ normalizes $$T$$, and therefore, it normalizes $$E=T^{(2)}$$, whence $$\NN(T)\subseteq \NN(E)$$. Thus $$\NN(E)=\NN(T)$$.

We wish to compute $$\ZZ(E)$$. Consider the Weyl group $$W=\NN(T)/T$$. Set $$W_E=\ZZ(E)/T\subseteq \NN(T)/T=W.$$ Then $$W_E$$ is a finite group, the kernel of the natural homomorphism $$W\to{\rm Aut\,} E$$, and $$\ZZ(E)$$ is the preimage of $$W_E$$ in $$\NN(T)$$.

Lemma 2. Let $$W'=S_n$$ (the permutation group on $$n$$ symbols) naturally acting on the set $$E'=\ker\, \Sigma\colon (\Z/2\Z)^n\to \Z/2\Z,$$ where $$\Sigma(x_1,x_2,\dots,x_n)=x_1+x_2+\cdots+x_n\,.$$ If $$n\ge 3$$, then this action is effective (has trivial kernel).

Proof. Left to OP and the reader.

Observe that Lemma 2 is false for $$n=2$$, when $${\rm Aut\,} E'=\{1\}$$ whereas $$S_2\neq \{1\}$$.

Theorem. Let $$G={\rm PGL}_n$$ with $$n\ge 3$$. Then $$\NN(E)/\ZZ(E)=\NN(T)/T=W$$.

Proof. We have seen that $$\NN(E)=\NN(T)$$. By Lemma 2, if $$n\ge 3$$, then $$W_E=\{1\}$$, whence $$\ZZ(E)=T$$, and the theorem follows.

Edit. Observe that the analogues of Lemma 2 and the theorem do not hold for the adjoint group $${\rm SO}(2n+1)$$ of type $${\sf B}_n$$ for $$n\ge 1$$. Indeed, then $$W\simeq(\Z/2\Z)^n\rtimes S_n\,,$$ and the normal subgroup $$(\Z/2\Z)^n\subseteq W$$ acts on $$E=T^{(2)}\cong (\pm1)^n$$ trivially. Therefore, $$\ZZ(E)\neq T$$.

• Thank you for the answer. I got confused by Lemma 2. What is the map $\Sigma$? Jul 31, 2022 at 1:02
• I have added the definition of the map $\Sigma$. Jul 31, 2022 at 5:51
• Is the proof that $C_{G}(E)^{\circ} = T$ a must? Thank you, Jul 31, 2022 at 6:47
• Yes. The argument goes as follows: ${\mathcal N}_G(E)$ normalizes ${\mathcal Z}_G(E)$, and therefore, it normalizes ${\mathcal Z}_G(E)^0=T$. We conclude that ${\mathcal N}_G(E)\subseteq {\mathcal N}_G(T)$. Jul 31, 2022 at 7:44
• @LSpice: In positive characteristic, I cannot write two lines without mistakes. So I suggest that you write, say, in comments, your proof that ${\mathcal Z}_G(E)^0=T$ also in positive characteristic. Aug 1, 2022 at 18:03