[proofplan] The Jacobian is invariant under translating the $G/T$ variable, so it is enough to compute the differential at $(eT,t)$. The quotient density identifies $T_{eT}(G/T)$ with $\mathfrak t^\perp$, and left translation identifies $T_tG$ with $\mathfrak g$, so the differential becomes a block triangular [linear map](/page/Linear%20Map) with identity on the $\mathfrak t$ direction and transverse block $\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp}$. Complexifying this transverse block diagonalizes it on the root spaces, and pairing the conjugate root spaces $\alpha$ and $-\alpha$ converts the complex eigenvalues into the real determinant $\prod_{\alpha\in R^+}|1-e^\alpha(t)|^2$. [/proofplan]

[step:Reduce the Jacobian computation to the point $(eT,t)$] Fix $g_0\in G$ and $t\in T_{\mathrm{reg}}$. Define the left translation map \begin{align*} L_{g_0}:G/T\to G/T \end{align*} by \begin{align*} L_{g_0}(hT):=g_0hT. \end{align*} Define the conjugation isometry \begin{align*} C_{g_0}:G\to G \end{align*} by \begin{align*} C_{g_0}(x):=g_0xg_0^{-1}. \end{align*} Because $(\cdot,\cdot)_{\mathfrak g}$ is $\operatorname{Ad}(G)$-invariant, left translations and conjugations preserve the Riemannian volume density on $G$, and $L_{g_0}$ preserves the quotient Riemannian density on $G/T$. Moreover, \begin{align*} \Phi(L_{g_0}(hT),t)=C_{g_0}(\Phi(hT,t)). \end{align*} Taking differentials at $(eT,t)$ gives \begin{align*} d\Phi_{(g_0T,t)}\circ (dL_{g_0})_{eT}=d(C_{g_0})_t\circ d\Phi_{(eT,t)} \end{align*} on the $G/T$ factor and the identity on the $T$ factor. Since both outside maps are isometries for the relevant densities, the absolute Jacobian satisfies \begin{align*} J_\Phi(g_0T,t)=J_\Phi(eT,t). \end{align*} Thus it remains to compute $J_\Phi(eT,t)$. [/step]

[step:Identify the differential with a block triangular map on $\mathfrak t^\perp\oplus\mathfrak t$] Let \begin{align*} q:G\to G/T \end{align*} be the quotient map. By definition of the quotient Riemannian density, the linear map \begin{align*} dq_e|_{\mathfrak t^\perp}:\mathfrak t^\perp\to T_{eT}(G/T) \end{align*} is an isometry. We identify $T_{eT}(G/T)$ with $\mathfrak t^\perp$ through this map. We identify $T_tT$ with $\mathfrak t$ by left translation by $t^{-1}$ in $T$, and identify $T_tG$ with $\mathfrak g$ by left translation by $t^{-1}$ in $G$. Let $X\in\mathfrak t^\perp$ and $Y\in\mathfrak t$. Consider the smooth curve \begin{align*} \gamma_{X,Y}:(-\varepsilon,\varepsilon)\to G \end{align*} defined for sufficiently small $\varepsilon>0$ by \begin{align*} \gamma_{X,Y}(s):=\exp(sX)\exp(sY)t\exp(-sX). \end{align*} This curve represents the tangent vector $(X,Y)\in\mathfrak t^\perp\oplus\mathfrak t$ in the domain of $\Phi$ at $(eT,t)$. Left translating its derivative at $s=0$ by $t^{-1}$ gives \begin{align*} t^{-1}\gamma_{X,Y}'(0)=(\operatorname{Ad}_{t^{-1}}X)-X+Y. \end{align*} Since $T$ is abelian, $\operatorname{Ad}_{t^{-1}}Y=Y$. Since $\operatorname{Ad}_{t^{-1}}$ preserves $\mathfrak t$ and preserves the [inner product](/page/Inner%20Product), it also preserves $\mathfrak t^\perp$. Hence, under the above identifications, \begin{align*} d\Phi_{(eT,t)}(X,Y)=(\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp})X+Y. \end{align*} With respect to the orthogonal decompositions $\mathfrak t^\perp\oplus\mathfrak t$ in the domain and codomain, this map is block triangular, with transverse block $\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp}$ and torus block $\operatorname{id}_{\mathfrak t}$. [/step]

[step:Reduce the absolute Jacobian to the transverse determinant]The preceding block triangular form implies \begin{align*} J_\Phi(eT,t)=\left|\det_{\mathbb R}(\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp})\right|\cdot \left|\det_{\mathbb R}(\operatorname{id}_{\mathfrak t})\right|. \end{align*} Since $\det_{\mathbb R}(\operatorname{id}_{\mathfrak t})=1$, we obtain \begin{align*} J_\Phi(eT,t)=\left|\det_{\mathbb R}(\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp})\right|. \end{align*}[/step]

[guided]The point of the quotient-measure convention is used exactly here. The domain tangent space at $(eT,t)$ is identified isometrically with $\mathfrak t^\perp\oplus\mathfrak t$: the first summand comes from the quotient map $q:G\to G/T$, and the second summand comes from left translation on $T$. The target tangent space at $t\in G$ is identified isometrically with $\mathfrak g=\mathfrak t^\perp\oplus\mathfrak t$ by left translation by $t^{-1}$. In these coordinates the differential is \begin{align*} (X,Y)\mapsto (\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp})X+Y. \end{align*} The first term belongs to $\mathfrak t^\perp$ because $\operatorname{Ad}_{t^{-1}}$ preserves both $\mathfrak t$ and the inner product, hence preserves the orthogonal complement $\mathfrak t^\perp$. The second term belongs to $\mathfrak t$. Therefore the matrix of the differential relative to the [orthogonal decomposition](/theorems/436) $\mathfrak t^\perp\oplus\mathfrak t$ is block triangular. Its diagonal blocks are \begin{align*} \operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp}:\mathfrak t^\perp\to\mathfrak t^\perp \end{align*} and \begin{align*} \operatorname{id}_{\mathfrak t}:\mathfrak t\to\mathfrak t. \end{align*} The determinant of a block triangular linear map is the product of the determinants of its diagonal blocks. Thus \begin{align*} J_\Phi(eT,t)=\left|\det_{\mathbb R}(\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp})\right|\cdot \left|\det_{\mathbb R}(\operatorname{id}_{\mathfrak t})\right|. \end{align*} The identity map on $\mathfrak t$ has determinant $1$, so the entire Jacobian comes from the directions transverse to $T$: \begin{align*} J_\Phi(eT,t)=\left|\det_{\mathbb R}(\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp})\right|. \end{align*} This also explains why no Weyl group factor appears: we are computing a pointwise differential determinant, not counting how many points of $G/T\times T$ map to the same [conjugacy class](/page/Conjugacy%20Class).[/guided]

[step:Diagonalize the transverse operator on the complex root spaces] Complexify the real [vector space](/page/Vector%20Space) $\mathfrak t^\perp$ and the real linear map \begin{align*} A_t:=\operatorname{Ad}_{t^{-1}}-\operatorname{id}_{\mathfrak t^\perp}. \end{align*} The standard root-space decomposition for a compact connected Lie group gives \begin{align*} (\mathfrak t^\perp)_{\mathbb C}=\bigoplus_{\alpha\in R}\mathfrak g_{\mathbb C,\alpha}. \end{align*} For $Z\in\mathfrak g_{\mathbb C,\alpha}$, the defining convention for $e^\alpha$ gives \begin{align*} \operatorname{Ad}_{t^{-1}}Z=e^{-\alpha}(t)Z. \end{align*} Therefore the complexified map $(A_t)_{\mathbb C}$ acts on $\mathfrak g_{\mathbb C,\alpha}$ by multiplication by \begin{align*} e^{-\alpha}(t)-1. \end{align*} Since each root space $\mathfrak g_{\mathbb C,\alpha}$ is one-dimensional, the complex determinant is \begin{align*} \det_{\mathbb C}((A_t)_{\mathbb C})=\prod_{\alpha\in R}(e^{-\alpha}(t)-1). \end{align*} [/step]

[step:Pair opposite roots to compute the real determinant] Because $A_t$ is a real linear map, its real determinant equals the complex determinant of its complexification: \begin{align*} \det_{\mathbb R}(A_t)=\det_{\mathbb C}((A_t)_{\mathbb C}). \end{align*} Pairing the two factors corresponding to $\alpha$ and $-\alpha$ gives \begin{align*} (e^{-\alpha}(t)-1)(e^\alpha(t)-1). \end{align*} Since $e^\alpha(t)\in U(1)$, we have $e^{-\alpha}(t)=\overline{e^\alpha(t)}$. Hence \begin{align*} |(e^{-\alpha}(t)-1)(e^\alpha(t)-1)|=|1-e^\alpha(t)|^2. \end{align*} Taking the product over all positive roots gives \begin{align*} \left|\det_{\mathbb R}(A_t)\right|=\prod_{\alpha\in R^+}|1-e^\alpha(t)|^2. \end{align*} The assumption $t\in T_{\mathrm{reg}}$ says exactly that $e^\alpha(t)\neq 1$ for every $\alpha\in R$, so none of these factors vanishes and the differential is nonsingular at $(eT,t)$. Combining this determinant computation with the reduction above yields, for every $gT\in G/T$, \begin{align*} J_\Phi(gT,t)=\prod_{\alpha\in R^+}|1-e^\alpha(t)|^2. \end{align*} This is the claimed formula. [/step]