[proofplan] We first show that the regular subset $T_{\mathrm{reg}}$ is open in $T$ by using the root-space description of the adjoint action of $T$. Then we compute the differential of the conjugation map and show that the condition $C_G(t)=T$ makes the differential invertible along $G/T\times T_{\mathrm{reg}}$. This proves that the image $G_{\mathrm{reg}}$ is open and that $\Phi_{\mathrm{reg}}$ is a local diffeomorphism. Finally, we identify each fibre with the Weyl group $W=N_G(T)/T$ and apply the standard finite-fibre local-diffeomorphism covering criterion. [/proofplan] [step:Show that the regular part of the maximal torus is open] Let $\mathfrak g$ denote the [Lie algebra](/page/Lie%20Algebra) of $G$, and let $\mathfrak t$ denote the Lie algebra of $T$. Choose an $\operatorname{Ad}(G)$-invariant real [inner product](/page/Inner%20Product) on $\mathfrak g$, which exists by averaging any inner product over the compact group $G$ as in [citetheorem:9712]. Let $\mathfrak m$ denote the orthogonal complement of $\mathfrak t$ in $\mathfrak g$, so \begin{align*} \mathfrak g=\mathfrak t\oplus \mathfrak m. \end{align*} Since $T$ is a compact torus and the restricted adjoint representation $T\to GL(\mathfrak g_{\mathbb C})$ is a continuous finite-dimensional complex representation, the weight-space decomposition [citetheorem:9726] applies. With $R\subset X^*(T)$ the non-zero weights declared in the theorem statement, this gives the root-space decomposition \begin{align*} \mathfrak g_{\mathbb C}=\mathfrak t_{\mathbb C}\oplus \bigoplus_{\alpha\in R}\mathfrak g_\alpha, \end{align*} where $X^*(T)=\operatorname{Hom}_{\mathrm{cts}}(T,S^1)$ and $\mathfrak g_\alpha:=\{Z\in\mathfrak g_{\mathbb C}:\operatorname{Ad}_s Z=\alpha(s)Z\text{ for every }s\in T\}$. For $t\in T$, differentiating the conjugation equation $gtg^{-1}=t$ at the identity gives the standard fixed-point formula \begin{align*} \operatorname{Lie}(C_G(t))=\{X\in\mathfrak g:\operatorname{Ad}_t X=X\}. \end{align*} After complexification, this fixed-point space is \begin{align*} \mathfrak t_{\mathbb C}\oplus \bigoplus_{\substack{\alpha\in R, \alpha(t)=1}}\mathfrak g_\alpha. \end{align*} Thus $C_G(t)$ has Lie algebra $\mathfrak t$ exactly when $\alpha(t)\neq 1$ for every $\alpha\in R$. Let $S_t:=\overline{\{t^m:m\in\mathbb Z\}}\le T$ be the closed torus generated by $t$, and use the subset-centralizer notation \begin{align*} C_G(S_t):=\{g\in G:gs=sg\text{ for every }s\in S_t\}. \end{align*} Since commutation is preserved under taking closures, $C_G(t)=C_G(S_t)$. Because $G$ is compact and connected and $S_t\le G$ is a closed torus, [citetheorem:9721] implies that $C_G(t)$ is connected. If its Lie algebra is $\mathfrak t$, then $T\subset C_G(t)$ because $T$ is abelian, and the inclusion $T\hookrightarrow C_G(t)$ has derivative an isomorphism $\mathfrak t\to\mathfrak t$. Hence $T$ is an open Lie subgroup of $C_G(t)$. Since $C_G(t)$ is connected and $T$ is non-empty, this open subgroup must be all of $C_G(t)$, so $C_G(t)=T$. Conversely, if $C_G(t)=T$, then its Lie algebra is $\mathfrak t$. Hence $C_G(t)=T$ is equivalent to $\alpha(t)\neq 1$ for every $\alpha\in R$. Hence \begin{align*} T_{\mathrm{reg}}=T\setminus \bigcup_{\alpha\in R}\{t\in T:\alpha(t)=1\}. \end{align*} For each root $\alpha\in R$, the character $\alpha:T\to S^1$ is continuous, so $\{t\in T:\alpha(t)=1\}$ is closed in $T$. Since $R$ is finite, $T_{\mathrm{reg}}$ is the complement of a finite union of closed subsets of $T$. Therefore $T_{\mathrm{reg}}$ is open in $T$. [guided] The regularity condition $C_G(t)=T$ says that $t$ has no centralizer directions beyond the torus directions. To detect those extra directions, we look infinitesimally. Let $\mathfrak g$ be the Lie algebra of $G$ and $\mathfrak t$ the Lie algebra of $T$. Because $G$ is compact, we may average an arbitrary inner product over $G$ and obtain an $\operatorname{Ad}(G)$-invariant inner product by [citetheorem:9712]. Let $\mathfrak m$ be the orthogonal complement of $\mathfrak t$ in $\mathfrak g$; then \begin{align*} \mathfrak g=\mathfrak t\oplus \mathfrak m. \end{align*} Now consider the complexified adjoint action of $T$ on \begin{align*} \mathfrak g_{\mathbb C}:=\mathfrak g\otimes_{\mathbb R}\mathbb C. \end{align*} Because $T$ is a compact torus and $T\to GL(\mathfrak g_{\mathbb C})$ is a continuous finite-dimensional complex representation, the weight-space decomposition [citetheorem:9726] applies to the complexified adjoint action. With $R\subset X^*(T)$ the non-zero weights declared in the theorem statement, it gives \begin{align*} \mathfrak g_{\mathbb C}=\mathfrak t_{\mathbb C}\oplus \bigoplus_{\alpha\in R}\mathfrak g_\alpha, \end{align*} where $X^*(T)=\operatorname{Hom}_{\mathrm{cts}}(T,S^1)$ and $\mathfrak g_\alpha:=\{Z\in\mathfrak g_{\mathbb C}:\operatorname{Ad}_s Z=\alpha(s)Z\text{ for every }s\in T\}$. The point of this decomposition is that each $t\in T$ acts by scalar multiplication on each root space: \begin{align*} \operatorname{Ad}_t X=\alpha(t)X \end{align*} for $X\in \mathfrak g_\alpha$. The Lie algebra of the centralizer $C_G(t)$ is the fixed-point space of $\operatorname{Ad}_t$. This follows by differentiating the equation $gtg^{-1}=t$ along curves through the identity in $G$, and gives \begin{align*} \operatorname{Lie}(C_G(t))=\{X\in\mathfrak g:\operatorname{Ad}_tX=X\}. \end{align*} After complexifying, the fixed-point space is therefore \begin{align*} \mathfrak t_{\mathbb C}\oplus \bigoplus_{\substack{\alpha\in R, \alpha(t)=1}}\mathfrak g_\alpha. \end{align*} Thus extra infinitesimal centralizer directions occur exactly when some root satisfies $\alpha(t)=1$. To pass from the Lie algebra statement to the group centralizer, let $S_t:=\overline{\{t^m:m\in\mathbb Z\}}\le T$ be the closed torus generated by $t$, and define \begin{align*} C_G(S_t):=\{g\in G:gs=sg\text{ for every }s\in S_t\}. \end{align*} An element of $G$ commutes with $t$ if and only if it commutes with every element of $S_t$, so $C_G(t)=C_G(S_t)$. The theorem on connectedness of torus centralizers [citetheorem:9721] applies because $G$ is compact and connected and $S_t\le G$ is a closed torus. Hence $C_G(S_t)$ is connected. Therefore, once the fixed-point Lie algebra is $\mathfrak t$, the whole centralizer is forced to be $T$: we have $T\subset C_G(t)$ because $T$ is abelian, and the inclusion $T\hookrightarrow C_G(t)$ has derivative an isomorphism on the common Lie algebra $\mathfrak t$. Thus $T$ is an open Lie subgroup of the connected group $C_G(t)=C_G(S_t)$, and a non-empty open subgroup of a connected topological group is the whole group. Conversely, if $C_G(t)=T$, then the fixed-point Lie algebra is $\mathfrak t$. Hence \begin{align*} T_{\mathrm{reg}}=T\setminus \bigcup_{\alpha\in R}\{t\in T:\alpha(t)=1\}. \end{align*} Each root $\alpha:T\to S^1$ is continuous, and $\{1\}\subset S^1$ is closed. Therefore each set $\{t\in T:\alpha(t)=1\}$ is closed in $T$. Because the root system $R$ is finite, the union \begin{align*} \bigcup_{\alpha\in R}\{t\in T:\alpha(t)=1\} \end{align*} is closed. Its complement $T_{\mathrm{reg}}$ is open in $T$. [/guided] [/step] [step:Compute the differential of the conjugation map on the regular locus] Define the smooth map \begin{align*} \Phi:G/T\times T\to G,\qquad ([g],t)\mapsto gtg^{-1}. \end{align*} It restricts to $\Phi_{\mathrm{reg}}$ on $G/T\times T_{\mathrm{reg}}$. We compute the differential at $(eT,t)$, where $e\in G$ is the identity element. Using the splitting $T_{eT}(G/T)\cong \mathfrak g/\mathfrak t\cong \mathfrak m$, a tangent vector to $G/T\times T$ at $(eT,t)$ may be written as $(X,A)$ with $X\in\mathfrak m$ and $A\in\mathfrak t$. Here $A$ represents the tangent vector at $t$ given by the curve $s\mapsto t\exp(sA)$. Let \begin{align*} \gamma:\mathbb R\to G,\qquad s\mapsto \exp(sX)t\exp(sA)\exp(-sX). \end{align*} Then $\gamma(0)=t$. Left-translating $d\Phi_{(eT,t)}(X,A)$ by $t^{-1}$ identifies it with \begin{align*} \operatorname{Ad}_{t^{-1}}X-X+A. \end{align*} Since $\operatorname{Ad}_t$ fixes $\mathfrak t$ and preserves $\mathfrak m$, this expression lies in the [direct sum](/page/Direct%20Sum) $\mathfrak m\oplus\mathfrak t$, with $\operatorname{Ad}_{t^{-1}}X-X\in\mathfrak m$ and $A\in\mathfrak t$. If $t\in T_{\mathrm{reg}}$ and $d\Phi_{(eT,t)}(X,A)=0$, then \begin{align*} \operatorname{Ad}_{t^{-1}}X-X=0 \end{align*} and \begin{align*} A=0. \end{align*} The first equality says $X\in \operatorname{Lie}(C_G(t))$. Since $C_G(t)=T$, we have $\operatorname{Lie}(C_G(t))=\mathfrak t$. But $X\in\mathfrak m$ and $\mathfrak m\cap\mathfrak t=\{0\}$, so $X=0$. Therefore $d\Phi_{(eT,t)}$ is injective. The dimensions agree: \begin{align*} \dim(G/T\times T_{\mathrm{reg}})=\dim G-\dim T+\dim T=\dim G. \end{align*} Hence $d\Phi_{(eT,t)}$ is an isomorphism. By $G$-equivariance of $\Phi$ under left multiplication on $G/T$ and conjugation on $G$, the same conclusion holds at every point $([g],t)\in G/T\times T_{\mathrm{reg}}$. Thus $\Phi_{\mathrm{reg}}$ is a local diffeomorphism onto its image. [/step] [step:Deduce that the regular conjugation locus is an open submanifold] Since $\Phi_{\mathrm{reg}}$ is a local diffeomorphism, every point $([g],t)\in G/T\times T_{\mathrm{reg}}$ has an open neighbourhood $U_{([g],t)}\subset G/T\times T_{\mathrm{reg}}$ such that \begin{align*} \Phi_{\mathrm{reg}}\big|_{U_{([g],t)}}:U_{([g],t)}\to \Phi_{\mathrm{reg}}(U_{([g],t)}) \end{align*} is a diffeomorphism onto an open subset of $G$. Since \begin{align*} G_{\mathrm{reg}}=\Phi_{\mathrm{reg}}(G/T\times T_{\mathrm{reg}}), \end{align*} the set $G_{\mathrm{reg}}$ is a union of open subsets of $G$. Therefore $G_{\mathrm{reg}}$ is open in $G$. With its open-submanifold structure, $\Phi_{\mathrm{reg}}$ is a smooth local diffeomorphism \begin{align*} \Phi_{\mathrm{reg}}:G/T\times T_{\mathrm{reg}}\to G_{\mathrm{reg}}. \end{align*} [/step] [step:Identify each fibre with the Weyl group] Fix $y\in G_{\mathrm{reg}}$. By definition of $G_{\mathrm{reg}}$, choose $g_0\in G$ and $t_0\in T_{\mathrm{reg}}$ such that \begin{align*} y=g_0t_0g_0^{-1}. \end{align*} Define a map \begin{align*} \Theta_y:W\to \Phi_{\mathrm{reg}}^{-1}(y),\qquad nT\mapsto ([g_0n^{-1}],nt_0n^{-1}). \end{align*} This is well-defined because $n\in N_G(T)$ implies $nt_0n^{-1}\in T$, and \begin{align*} C_G(nt_0n^{-1})=nC_G(t_0)n^{-1}=nTn^{-1}=T, \end{align*} so $nt_0n^{-1}\in T_{\mathrm{reg}}$. Also, \begin{align*} \Phi_{\mathrm{reg}}([g_0n^{-1}],nt_0n^{-1})=g_0n^{-1}nt_0n^{-1}ng_0^{-1}=g_0t_0g_0^{-1}=y. \end{align*} If $nT=n'T$, then $n'=na$ for some $a\in T$. Since $T$ is abelian, $n't_0(n')^{-1}=nt_0n^{-1}$, and $[g_0(n')^{-1}]=[g_0n^{-1}]$. Thus $\Theta_y$ is well-defined on the quotient $N_G(T)/T$. We prove that $\Theta_y$ is injective. Suppose \begin{align*} \Theta_y(nT)=\Theta_y(n'T). \end{align*} Then $[g_0n^{-1}]=[g_0(n')^{-1}]$, so there exists $a\in T$ such that \begin{align*} g_0(n')^{-1}=g_0n^{-1}a. \end{align*} Cancelling $g_0$ gives $n'=a^{-1}n$. Since $n\in N_G(T)$, we have $n^{-1}a^{-1}n\in T$, and therefore $a^{-1}n=n(n^{-1}a^{-1}n)\in nT$. Hence $n'T=nT$. Therefore $\Theta_y$ is injective. We prove that $\Theta_y$ is surjective. Let $([g],t)\in \Phi_{\mathrm{reg}}^{-1}(y)$. Then \begin{align*} gtg^{-1}=g_0t_0g_0^{-1}. \end{align*} Set \begin{align*} x:=g_0^{-1}g. \end{align*} Then \begin{align*} xtx^{-1}=t_0. \end{align*} Taking centralizers gives \begin{align*} xC_G(t)x^{-1}=C_G(t_0). \end{align*} Since $t,t_0\in T_{\mathrm{reg}}$, both centralizers equal $T$. Hence \begin{align*} xTx^{-1}=T, \end{align*} so $x\in N_G(T)$. Put $n:=x^{-1}\in N_G(T)$. Then $g=g_0x=g_0n^{-1}$ and \begin{align*} t=nt_0n^{-1}. \end{align*} Thus $([g],t)=\Theta_y(nT)$. Therefore $\Theta_y$ is bijective. By [citetheorem:9723], the Weyl group $W=N_G(T)/T$ is finite. Hence every fibre of $\Phi_{\mathrm{reg}}$ has cardinality $|W|$. [/step] [step:Apply properness to obtain evenly covered neighbourhoods] We prove the covering property directly using properness near the chosen fibre. Fix $y\in G_{\mathrm{reg}}$ and write \begin{align*} \Phi_{\mathrm{reg}}^{-1}(y)=\{x_1,\dots,x_d\}, \end{align*} where $d=|W|$ by the fibre computation above. For each $i\in\{1,\dots,d\}$, choose an open neighbourhood $U_i\subset G/T\times T_{\mathrm{reg}}$ of $x_i$ such that the sets $U_i$ are pairwise disjoint and \begin{align*} \Phi_{\mathrm{reg}}\big|_{U_i}:U_i\to V_i \end{align*} is a diffeomorphism onto an open neighbourhood $V_i\subset G_{\mathrm{reg}}$ of $y$. Since $G$ is a compact smooth manifold, it is locally compact and regular. Because $G_{\mathrm{reg}}\subset G$ is open and $y\in G_{\mathrm{reg}}$, choose an open neighbourhood $V_0\subset G_{\mathrm{reg}}$ of $y$ whose closure $K:=\overline{V_0}$ is compact and contained in $G_{\mathrm{reg}}$. Consider \begin{align*} E:=\Phi_{\mathrm{reg}}^{-1}(K)\subset G/T\times T_{\mathrm{reg}}. \end{align*} If $([g],t)\in E$, then $gtg^{-1}\in K\subset G_{\mathrm{reg}}$. By definition of $G_{\mathrm{reg}}$, choose $h\in G$ and $s\in T_{\mathrm{reg}}$ such that $gtg^{-1}=hsh^{-1}$. Then \begin{align*} C_G(gtg^{-1})=hC_G(s)h^{-1}=hTh^{-1}. \end{align*} Since $t\in T$, every element of $gTg^{-1}$ commutes with $gtg^{-1}$, so $gTg^{-1}\subset hTh^{-1}$. Both groups are compact connected abelian Lie subgroups of dimension $\dim T$; the inclusion has derivative an isomorphism of Lie algebras, hence has open image, and the image is a non-empty open subgroup of the connected group $hTh^{-1}$. Therefore equality holds. Conjugating by $g^{-1}$ gives $C_G(t)=T$, so $t\in T_{\mathrm{reg}}$. Thus the full inverse image of $K$ under the continuous map $G/T\times T\to G$ lies in $G/T\times T_{\mathrm{reg}}$ and equals $E$. Since $K$ is closed and $G/T\times T$ is compact, $E$ is compact. We claim that, after shrinking $V_0$ if necessary, no point over $V_0$ lies outside $\bigcup_{i=1}^d U_i$. If not, there exist points $z_j\in E\setminus \bigcup_{i=1}^d U_i$ with $\Phi_{\mathrm{reg}}(z_j)\to y$. Compactness of $E$ gives a convergent subsequence $z_{j_k}\to z\in E$. Continuity gives $\Phi_{\mathrm{reg}}(z)=y$, so $z=x_i$ for some $i$. This contradicts $z_{j_k}\notin U_i$ for all $k$ sufficiently large, because $U_i$ is an open neighbourhood of $x_i$. Hence there is an open neighbourhood \begin{align*} V\subset V_0\cap\bigcap_{i=1}^d V_i \end{align*} of $y$ such that \begin{align*} \Phi_{\mathrm{reg}}^{-1}(V)\subset \bigcup_{i=1}^d U_i. \end{align*} Then \begin{align*} \Phi_{\mathrm{reg}}^{-1}(V)=\bigcup_{i=1}^d \bigl(U_i\cap \Phi_{\mathrm{reg}}^{-1}(V)\bigr), \end{align*} and each restriction \begin{align*} \Phi_{\mathrm{reg}}:U_i\cap \Phi_{\mathrm{reg}}^{-1}(V)\to V \end{align*} is a diffeomorphism. Thus $V$ is evenly covered. Since $y\in G_{\mathrm{reg}}$ was arbitrary, $\Phi_{\mathrm{reg}}$ is a smooth covering map, and the fibre computation gives degree $|W|$. [guided] The only delicate point in turning a local diffeomorphism into a covering map is preventing extra sheets from appearing arbitrarily close to the chosen fibre. Fix $y\in G_{\mathrm{reg}}$ and write its fibre as \begin{align*} \Phi_{\mathrm{reg}}^{-1}(y)=\{x_1,\dots,x_d\}, \end{align*} with $d=|W|$. Because $\Phi_{\mathrm{reg}}$ is a local diffeomorphism, for each $x_i$ we can choose an open neighbourhood $U_i\subset G/T\times T_{\mathrm{reg}}$ such that the $U_i$ are pairwise disjoint and \begin{align*} \Phi_{\mathrm{reg}}\big|_{U_i}:U_i\to V_i \end{align*} is a diffeomorphism onto an open neighbourhood $V_i\subset G_{\mathrm{reg}}$ of $y$. We now supply the compactness that justifies the shrinking. Since $G$ is a compact smooth manifold, it is locally compact and regular. Because $G_{\mathrm{reg}}$ is open in $G$ and $y\in G_{\mathrm{reg}}$, choose an open neighbourhood $V_0\subset G_{\mathrm{reg}}$ of $y$ whose closure $K:=\overline{V_0}$ is compact and still contained in $G_{\mathrm{reg}}$. Define \begin{align*} E:=\Phi_{\mathrm{reg}}^{-1}(K)\subset G/T\times T_{\mathrm{reg}}. \end{align*} This set is compact. Indeed, if $([g],t)\in E$, then $gtg^{-1}\in K\subset G_{\mathrm{reg}}$. By definition of $G_{\mathrm{reg}}$, there are $h\in G$ and $s\in T_{\mathrm{reg}}$ such that $gtg^{-1}=hsh^{-1}$. Because $s\in T_{\mathrm{reg}}$, we have $C_G(s)=T$, and therefore \begin{align*} C_G(gtg^{-1})=hC_G(s)h^{-1}=hTh^{-1}. \end{align*} On the other hand, $t\in T$, so every element of $gTg^{-1}$ commutes with $gtg^{-1}$. Hence $gTg^{-1}\subset hTh^{-1}$. These are compact connected abelian Lie subgroups of the same dimension. The inclusion has derivative an isomorphism of Lie algebras, so its image is open; because $hTh^{-1}$ is connected and the image is a non-empty subgroup, the inclusion is equality. Conjugating back gives $C_G(t)=T$, hence $t\in T_{\mathrm{reg}}$. Therefore the full inverse image of $K$ under the continuous conjugation map $G/T\times T\to G$ is contained in $G/T\times T_{\mathrm{reg}}$ and is exactly $E$. Since $K$ is closed and $G/T\times T$ is compact, $E$ is compact. Now suppose no shrinking removed the unwanted points outside the chosen sheets. Then there would be a sequence $z_j\in E\setminus\bigcup_{i=1}^d U_i$ such that $\Phi_{\mathrm{reg}}(z_j)\to y$. Compactness of $E$ gives a convergent subsequence $z_{j_k}\to z\in E$. By continuity, $\Phi_{\mathrm{reg}}(z)=y$, so $z$ must be one of the listed fibre points $x_i$. But $U_i$ is an open neighbourhood of $x_i$, so $z_{j_k}\in U_i$ for all sufficiently large $k$, contradicting the choice of the sequence. Hence we may choose an open neighbourhood \begin{align*} V\subset V_0\cap\bigcap_{i=1}^d V_i \end{align*} of $y$ such that \begin{align*} \Phi_{\mathrm{reg}}^{-1}(V)\subset \bigcup_{i=1}^d U_i. \end{align*} It follows that \begin{align*} \Phi_{\mathrm{reg}}^{-1}(V)=\bigcup_{i=1}^d \bigl(U_i\cap \Phi_{\mathrm{reg}}^{-1}(V)\bigr), \end{align*} and each restriction \begin{align*} \Phi_{\mathrm{reg}}:U_i\cap \Phi_{\mathrm{reg}}^{-1}(V)\to V \end{align*} is a diffeomorphism. This is exactly the definition of an evenly covered neighbourhood. Since every $y\in G_{\mathrm{reg}}$ has such a neighbourhood and every fibre has $|W|$ points, $\Phi_{\mathrm{reg}}$ is a smooth covering map of degree $|W|$. [/guided] [/step]