[proofplan] We first set $C=C_G(T)$ and observe that $T\le C$ because $T$ is abelian. The centralizer connectedness theorem implies that $C$ is connected. We then compute the [Lie algebra](/page/Lie%20Algebra) of $C$ as the set of elements of $\mathfrak g$ commuting with $\mathfrak t=\operatorname{Lie}(T)$, and use maximality of $\mathfrak t$ as an abelian subalgebra to force this Lie algebra to equal $\mathfrak t$. Since $T\le C$ and both groups are connected Lie subgroups with the same Lie algebra, they coincide. [/proofplan]

[step:Place the maximal torus inside its centralizer] Let \begin{align*} C:=C_G(T)=\{g\in G:gt=tg\text{ for every }t\in T\}. \end{align*} Since $T$ is a torus, it is abelian. Hence every element of $T$ commutes with every element of $T$, so $T\le C$. By [citetheorem:9721] applied to the torus $T\le G$, the subgroup $C=C_G(T)$ is connected. Since $C$ is a closed subgroup of the compact Lie group $G$, it is also a compact Lie subgroup of $G$. [/step]

[step:Identify the Lie algebra of the centralizer]Let $\mathfrak g:=\operatorname{Lie}(G)$, let $\mathfrak t:=\operatorname{Lie}(T)$, and let $\mathfrak c:=\operatorname{Lie}(C)$. Write $\exp_G:\mathfrak g\to G$ and $\exp_T:\mathfrak t\to T$ for the Lie exponential maps; under the inclusion $T\le G$, $\exp_T(Y)=\exp_G(Y)$ for $Y\in\mathfrak t$. We claim that \begin{align*} \mathfrak c=\{X\in\mathfrak g:[X,Y]=0\text{ for every }Y\in\mathfrak t\}. \end{align*} Indeed, let $X\in\mathfrak c$. For every $s\in\mathbb R$, the element $\exp_G(sX)$ lies in $C$, so for every $t\in T$, \begin{align*} t\exp_G(sX)t^{-1}=\exp_G(sX). \end{align*} Equivalently, \begin{align*} \exp_G(s\operatorname{Ad}_tX)=\exp_G(sX). \end{align*} Differentiating at $s=0$ gives $\operatorname{Ad}_tX=X$ for every $t\in T$. Now let $Y\in\mathfrak t$. Applying this to $t=\exp_G(rY)$ for $r\in\mathbb R$ gives \begin{align*} \operatorname{Ad}_{\exp_G(rY)}X=X. \end{align*} Differentiating at $r=0$ gives $[Y,X]=0$, hence $[X,Y]=0$. Conversely, suppose $X\in\mathfrak g$ satisfies $[X,Y]=0$ for every $Y\in\mathfrak t$. Since $T$ is a torus, the exponential map $\exp_T:\mathfrak t\to T$ is surjective. For any $t\in T$, choose $Y\in\mathfrak t$ such that $t=\exp_G(Y)$. Since $[X,Y]=0$, the one-parameter subgroups $\exp_G(sX)$ and $\exp_G(rY)$ commute for all $r,s\in\mathbb R$. In particular, \begin{align*} \exp_G(sX)t=t\exp_G(sX) \end{align*} for every $s\in\mathbb R$ and every $t\in T$. Thus $\exp_G(sX)\in C$ for every $s\in\mathbb R$, so $X\in\operatorname{Lie}(C)=\mathfrak c$.[/step]

[guided]The goal of this step is to translate the group-theoretic condition “commutes with every element of $T$” into the infinitesimal condition “commutes with every element of $\mathfrak t$.” Keep the notation $C=C_G(T)$, $\mathfrak g=\operatorname{Lie}(G)$, $\mathfrak t=\operatorname{Lie}(T)$, and $\mathfrak c=\operatorname{Lie}(C)$. Write $\exp_G:\mathfrak g\to G$ and $\exp_T:\mathfrak t\to T$ for the Lie exponential maps. We prove both inclusions. First take $X\in\mathfrak c$. By definition of the Lie algebra of a Lie subgroup, the one-parameter subgroup \begin{align*} s\mapsto \exp_G(sX) \end{align*} has image in $C$. Since $C$ centralizes $T$, for every $t\in T$ and every $s\in\mathbb R$ we have \begin{align*} t\exp_G(sX)t^{-1}=\exp_G(sX). \end{align*} Conjugation by $t$ differentiates to the [linear map](/page/Linear%20Map) $\operatorname{Ad}_t:\mathfrak g\to\mathfrak g$, and the exponential map is equivariant under conjugation: \begin{align*} t\exp_G(sX)t^{-1}=\exp_G(s\operatorname{Ad}_tX). \end{align*} Therefore \begin{align*} \exp_G(s\operatorname{Ad}_tX)=\exp_G(sX) \end{align*} for all $s$. Differentiating this identity at $s=0$ gives \begin{align*} \operatorname{Ad}_tX=X. \end{align*} Now choose $Y\in\mathfrak t$ and put $t=\exp_G(rY)$, where $r\in\mathbb R$. The identity above becomes \begin{align*} \operatorname{Ad}_{\exp_G(rY)}X=X. \end{align*} Differentiating at $r=0$ gives \begin{align*} [Y,X]=0. \end{align*} By skew-symmetry of the Lie bracket, $[X,Y]=0$. Since $Y\in\mathfrak t$ was arbitrary, $X$ lies in the Lie algebra centralizer of $\mathfrak t$. For the reverse inclusion, suppose $X\in\mathfrak g$ satisfies \begin{align*} [X,Y]=0 \end{align*} for every $Y\in\mathfrak t$. We need to prove that the one-parameter subgroup generated by $X$ lies in $C$. Let $t\in T$. Because $T$ is a torus, its exponential map $\exp_T:\mathfrak t\to T$ is surjective, so there exists $Y\in\mathfrak t$ such that \begin{align*} t=\exp_G(Y). \end{align*} The hypothesis $[X,Y]=0$ implies that the one-parameter subgroups generated by $X$ and $Y$ commute: \begin{align*} \exp_G(sX)\exp_G(rY)=\exp_G(rY)\exp_G(sX) \end{align*} for all $r,s\in\mathbb R$. Taking $r=1$ gives \begin{align*} \exp_G(sX)t=t\exp_G(sX) \end{align*} for every $s\in\mathbb R$. Since $t\in T$ was arbitrary, $\exp_G(sX)$ centralizes all of $T$, so $\exp_G(sX)\in C$ for every $s$. Hence $X\in\operatorname{Lie}(C)=\mathfrak c$. Combining the two inclusions proves \begin{align*} \mathfrak c=\{X\in\mathfrak g:[X,Y]=0\text{ for every }Y\in\mathfrak t\}. \end{align*}[/guided]

[step:Use maximality of $\mathfrak t$ to force $\operatorname{Lie}(C)=\mathfrak t$] By [citetheorem:9719], the Lie algebra $\mathfrak t$ of the maximal torus $T$ is a maximal abelian subalgebra of $\mathfrak g$. Let $X\in\mathfrak c$. By the preceding step, \begin{align*} [X,Y]=0 \end{align*} for every $Y\in\mathfrak t$. Define the real vector subspace \begin{align*} \mathfrak a:=\mathfrak t+\mathbb RX\subseteq\mathfrak g. \end{align*} The subspace $\mathfrak a$ is an abelian Lie subalgebra: brackets inside $\mathfrak t$ vanish because $\mathfrak t$ is abelian, brackets between $X$ and elements of $\mathfrak t$ vanish by the choice of $X$, and $[X,X]=0$ by skew-symmetry. Since $\mathfrak t$ is maximal abelian and $\mathfrak t\subseteq\mathfrak a$, we must have $\mathfrak a=\mathfrak t$. Hence $X\in\mathfrak t$. Thus $\mathfrak c\subseteq\mathfrak t$. Since $T\le C$, differentiating the inclusion gives $\mathfrak t\subseteq\mathfrak c$. Therefore \begin{align*} \mathfrak c=\mathfrak t. \end{align*} [/step]

[step:Conclude equality from connectedness and equality of Lie algebras] We have shown that $T\le C$, that $C$ is connected, and that \begin{align*} \operatorname{Lie}(C)=\mathfrak c=\mathfrak t=\operatorname{Lie}(T). \end{align*} For connected Lie subgroups of a Lie group, the Lie algebra determines the subgroup: if $H_1$ and $H_2$ are connected Lie subgroups with $\operatorname{Lie}(H_1)=\operatorname{Lie}(H_2)$, then $H_1=H_2$. Applying this to $T$ and $C$ gives \begin{align*} C=T. \end{align*} Since $C=C_G(T)$ by definition, this proves \begin{align*} C_G(T)=T. \end{align*} [/step]