[proofplan] We show first that the identity component of the normalizer $N_G(T)$ is exactly $T$. The key point is that conjugation by elements of $N_G(T)^\circ$ gives a continuous family of automorphisms of the torus $T$, but the automorphism group of a compact torus is discrete, so this family is constant. Hence $N_G(T)^\circ$ centralizes $T$, and the same-run centralizer theorem for maximal tori gives $N_G(T)^\circ=T$. Since $N_G(T)$ is a compact Lie group, its component group is finite, and this component group is precisely $N_G(T)/T$. [/proofplan] [step:Show that the normalizer is a compact Lie subgroup containing $T$] Define \begin{align*} N:=N_G(T)=\{g\in G:gTg^{-1}=T\}. \end{align*} For each $t\in T$, the map \begin{align*} c_t: G \to G, \quad g \mapsto gtg^{-1} \end{align*} is continuous. If $(g_i)$ is a net in $N$ converging to $g\in G$, then for each $t\in T$ the elements $g_itg_i^{-1}$ lie in the closed subgroup $T$, so continuity gives $gtg^{-1}\in T$. Thus $gTg^{-1}\subset T$. Applying the same argument to the convergent net $(g_i^{-1})$ gives $g^{-1}Tg\subset T$, equivalently $T\subset gTg^{-1}$. Hence $gTg^{-1}=T$, so $g\in N$. Therefore $N$ is closed in $G$. Since $G$ is compact, the closed subgroup $N$ is compact. By the closed subgroup theorem, $N$ is an embedded Lie subgroup of $G$. Moreover $T\le N$, because $T$ is abelian and hence every $t\in T$ satisfies $tTt^{-1}=T$. [/step] [step:Identify the automorphism group of the torus as a discrete group] Let $\mathfrak t:=T_eT$ be the [Lie algebra](/page/Lie%20Algebra) of $T$, and let \begin{align*} \Lambda_T:=\ker(\exp_T:\mathfrak t\to T) \end{align*} be the kernel of the exponential map of the torus $T$. Since $T$ is a compact torus, $\Lambda_T$ is a lattice in the finite-dimensional real [vector space](/page/Vector%20Space) $\mathfrak t$, and every continuous Lie group automorphism $a:T\to T$ has derivative \begin{align*} da_e:\mathfrak t\to\mathfrak t \end{align*} satisfying $da_e(\Lambda_T)=\Lambda_T$. The assignment \begin{align*} \operatorname{Aut}(T) \to \operatorname{Aut}_{\mathbb Z}(\Lambda_T), \quad a \mapsto da_e|_{\Lambda_T} \end{align*} is injective. Indeed, if $da_e|_{\Lambda_T}$ is the identity on $\Lambda_T$, then $da_e$ is the identity on $\mathfrak t$ because $\Lambda_T$ spans $\mathfrak t$ over $\mathbb R$. For every $X\in\mathfrak t$, \begin{align*} a(\exp_T X)=\exp_T(da_eX)=\exp_T X. \end{align*} The exponential map $\exp_T:\mathfrak t\to T$ is surjective for a compact connected abelian Lie group, so $a=\operatorname{id}_T$. Thus the derivative-lattice representation embeds $\operatorname{Aut}(T)$ into the lattice automorphism group $\operatorname{Aut}_{\mathbb Z}(\Lambda_T)$. We will use this only through the following consequence: if a [connected space](/page/Connected%20Space) maps continuously into $GL(\mathfrak t)$ with image contained in the discrete subset $\operatorname{Aut}_{\mathbb Z}(\Lambda_T)$, then the induced lattice action is constant. [guided] The only point we need about torus automorphisms is that they cannot vary continuously unless they are constant. We now justify this from the exponential description of a torus. Let $\mathfrak t:=T_eT$ be the Lie algebra of $T$, and define the lattice \begin{align*} \Lambda_T:=\ker(\exp_T:\mathfrak t\to T). \end{align*} For a compact torus, $\exp_T:\mathfrak t\to T$ is surjective and $\Lambda_T$ is a full lattice in $\mathfrak t$. If $a:T\to T$ is a Lie group automorphism, its derivative at the identity is a real-linear isomorphism \begin{align*} da_e:\mathfrak t\to\mathfrak t. \end{align*} The exponential naturality identity says that, for every $X\in\mathfrak t$, \begin{align*} a(\exp_T X)=\exp_T(da_eX). \end{align*} If $X\in\Lambda_T$, then $\exp_T X=e$, so \begin{align*} e=a(e)=a(\exp_T X)=\exp_T(da_eX). \end{align*} Hence $da_eX\in\Lambda_T$. Applying the same reasoning to $a^{-1}$ gives $da_e(\Lambda_T)=\Lambda_T$. Therefore every automorphism of $T$ determines an automorphism of the lattice $\Lambda_T$ via the map $\operatorname{Aut}(T)\to \operatorname{Aut}_{\mathbb Z}(\Lambda_T)$, $a\mapsto da_e|_{\Lambda_T}$. This map is injective. If $da_e$ is the identity on $\Lambda_T$, then $da_e$ is the identity on all of $\mathfrak t$, because a full lattice spans the ambient real vector space. The exponential naturality identity then gives \begin{align*} a(\exp_T X)=\exp_T X \end{align*} for every $X\in\mathfrak t$. Since $\exp_T$ is surjective, this forces $a=\operatorname{id}_T$. Thus $\operatorname{Aut}(T)$ injects into the lattice automorphism group $\operatorname{Aut}_{\mathbb Z}(\Lambda_T)$. After choosing a $\mathbb Z$-basis of $\Lambda_T$, this lattice automorphism group is $GL(r,\mathbb Z)$, where $r=\dim T$. Its image inside $GL(\mathfrak t)$ is discrete. Consequently, if a connected [topological space](/page/Topological%20Space) maps continuously into $GL(\mathfrak t)$ and the image lies in $\operatorname{Aut}_{\mathbb Z}(\Lambda_T)$, then that map is constant. This is the topology statement we need for the conjugation family below. [/guided] [/step] [step:Prove that the identity component of the normalizer centralizes $T$] Let $N^\circ$ denote the identity component of the compact Lie group $N$. For each $n\in N^\circ$, define the conjugation automorphism \begin{align*} \Phi_n:T\to T, \quad t\mapsto ntn^{-1}. \end{align*} This is well-defined because $n\in N$ implies $nTn^{-1}=T$. Define the derivative-action map \begin{align*} \Psi:N^\circ\to GL(\mathfrak t), \quad n\mapsto d(\Phi_n)_e. \end{align*} The conjugation map $N^\circ\times T\to T$, $(n,t)\mapsto ntn^{-1}$, is smooth as the restriction of the smooth multiplication and inversion maps on $G$; therefore its derivative in the $T$-variable at $e$ depends smoothly, hence continuously, on $n$. Thus $\Psi$ is continuous as a map into $GL(\mathfrak t)$. For each $n\in N^\circ$, the previous step applied to the torus automorphism $\Phi_n$ gives \begin{align*} \Psi(n)(\Lambda_T)=\Lambda_T. \end{align*} Hence $\Psi(N^\circ)$ is contained in the discrete subgroup $\operatorname{Aut}_{\mathbb Z}(\Lambda_T)\subset GL(\mathfrak t)$. Since $N^\circ$ is connected, the continuous map $\Psi$ is constant. Because $e\in N^\circ$ and $\Psi(e)=\operatorname{id}_{\mathfrak t}$, we have \begin{align*} \Psi(n)=\operatorname{id}_{\mathfrak t} \end{align*} for every $n\in N^\circ$. Now define the centralizer $C_G(T):=\{g\in G:gt=tg\text{ for every }t\in T\}$ of $T$ in $G$. Let $n\in N^\circ$ and $t\in T$. Since $T$ is a compact connected abelian Lie group, $\exp_T:\mathfrak t\to T$ is surjective, so choose $X\in\mathfrak t$ with $t=\exp_T X$. Naturality of the exponential map for the Lie group automorphism $\Phi_n$ gives $ntn^{-1}=\Phi_n(\exp_T X)=\exp_T(\Psi(n)X)=\exp_T X=t$. Thus every $n\in N^\circ$ commutes with every $t\in T$, so $N^\circ\le C_G(T)$. [guided] The subtle point is the topology. We do not claim that the map $n\mapsto \Phi_n$ is continuous after putting a [discrete topology](/page/Discrete%20Topology) on $\operatorname{Aut}(T)$. Instead, we pass to the derivative action on the Lie algebra, where continuity is a standard smooth-dependence statement. For each $n\in N^\circ$, define \begin{align*} \Phi_n:T\to T, \quad t\mapsto ntn^{-1}. \end{align*} This is a Lie group automorphism because $n\in N$ means $nTn^{-1}=T$. Now define \begin{align*} \Psi:N^\circ\to GL(\mathfrak t), \quad n\mapsto d(\Phi_n)_e. \end{align*} The map $(n,t)\mapsto ntn^{-1}$ from $N^\circ\times T$ to $T$ is smooth, since it is obtained by restricting the smooth multiplication and inversion maps of the Lie group $G$. Taking the derivative in the $T$-variable at the identity element $e\in T$ is a smooth operation in local coordinates, so $n\mapsto d(\Phi_n)_e$ is continuous as a map into $GL(\mathfrak t)$. Why does this continuous map have discrete image? The previous step showed that every torus automorphism preserves the exponential kernel $\Lambda_T$. Applying that result to $\Phi_n$ gives \begin{align*} \Psi(n)(\Lambda_T)=\Lambda_T. \end{align*} Thus $\Psi(N^\circ)$ lies in $\operatorname{Aut}_{\mathbb Z}(\Lambda_T)$. After choosing a $\mathbb Z$-basis of the full lattice $\Lambda_T$, this group is identified with $GL(r,\mathbb Z)$, where $r=\dim T$, sitting as a discrete subset of $GL(\mathfrak t)$. Since $N^\circ$ is connected, a continuous map from $N^\circ$ into this discrete subset must be constant. The constant is determined at the identity element. For $e\in N^\circ$, conjugation by $e$ is the identity automorphism of $T$, so \begin{align*} \Psi(e)=\operatorname{id}_{\mathfrak t}. \end{align*} Therefore \begin{align*} \Psi(n)=\operatorname{id}_{\mathfrak t} \end{align*} for every $n\in N^\circ$. It remains to lift this infinitesimal statement back to the torus. Let $n\in N^\circ$ and $t\in T$. Since $T$ is compact, connected, and abelian, the exponential map $\exp_T:\mathfrak t\to T$ is surjective; choose $X\in\mathfrak t$ with $t=\exp_T X$. Naturality of the exponential map for the Lie group automorphism $\Phi_n$ gives \begin{align*} ntn^{-1}=\Phi_n(\exp_T X)=\exp_T(\Psi(n)X)=\exp_T X=t. \end{align*} Hence $n$ commutes with $t$. Since $n\in N^\circ$ and $t\in T$ were arbitrary, $N^\circ\le C_G(T)$. [/guided] [/step] [step:Use maximality of $T$ to identify the identity component of the normalizer] The hypotheses of [citetheorem:9722] are exactly the hypotheses in force here: $G$ is a compact connected Lie group and $T\le G$ is a maximal torus. Therefore the centralizer of $T$ in $G$ is the torus itself: \begin{align*} C_G(T)=T. \end{align*} The preceding step gives $N^\circ\le C_G(T)$, hence $N^\circ\le T$. On the other hand $T\le N$ by the first step, and $T$ is connected, so $T$ is contained in the identity component $N^\circ$ of $N$. Therefore \begin{align*} N^\circ=T. \end{align*} [/step] [step:Identify the Weyl group with a finite component group] Since $N$ is a compact Lie group, it has finitely many connected components. Indeed, the identity component $N^\circ$ is an open subgroup of the Lie group $N$, so the connected components of $N$ form an [open cover](/page/Open%20Cover) of the [compact space](/page/Compact%20Space) $N$. A [finite subcover](/page/Finite%20Subcover) exists, and distinct connected components are disjoint, so there are only finitely many components. From the previous step, $N^\circ=T$. Therefore \begin{align*} W(G,T)=N_G(T)/T=N/N^\circ. \end{align*} The quotient $N/N^\circ$ is the set of connected components of $N$, with its natural group structure. Since $N$ has finitely many connected components, $W(G,T)$ is finite. [/step]