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The following proof is adapted from Bump's Lie Groups. I've tried to rewrite the parts I find unclear. However, I seem to end up with a discrepancy. Please help me complete the proof or point out any mistakes.

Theorem: Let $G$ be a compact connected Lie group, and $T$ be a maximal torus. Let $f$ be a function constant on conjugacy classes of $G$. Give $G$ and $T$ normalized Haar measures, and write $\mathfrak{g}=\mathfrak{p} \oplus \mathfrak{t}$ where $\mathfrak{p}$ is the orthogonal complement of $\mathfrak{t}$ for some choice of invariant inner product. Then $$\int_G fdg= \frac{1}{|W|}\int_{T}f(t)\det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)dt$$ where $W=N(T)/T$, the Weyl group.

Proof: Let $\omega$, $\beta$ be left-invariant top forms on $G$, $T$ respectively which integrate to give $1$. These then give the Haar measures on $G$, $T$. These forms are also right-invariant since both $G$ and $T$ are compact.

Let $d=\dim(G/T)$. We can also construct a left-invariant top form $\alpha$ on $G/T$ as follows:

Choose bases $\lbrace A_i \rbrace$ and $\lbrace B_i \rbrace$ of $\mathfrak{p}$ and $\mathfrak{t}$ respectively such that $\beta_e(\wedge B_i)=1, \omega_e(\wedge A_i \bigwedge \wedge B_i)=1$. Identifying $\bigwedge^{d} T_{eT}(G/T) = \bigwedge^d \mathfrak{p}$, define $\alpha_{eT}(\wedge A_i) = 1$. We then extend $\alpha_{eT}$ to a differentiable section by setting $\alpha_{xT}\left(\wedge V_i \right) = \alpha_{eT}(\bigwedge dl_{x^{-1}}V_i)$. Here $V_i$ are tangent vectors at $xT$ and $dl_x$ is the differential of left translation.

It remains to check $\alpha$ is well defined and smooth. If $x_1T = x_2T$, then we would have $x_2 = x_1t$ for some $t \in T$. so $dl_{x_2^{-1}} = dl_{t^{-1}} \circ dl_{x_1^{-1}}$. Thus it suffices to show $\bigwedge dl_t (A_i)= \wedge A_i$. This follows since we may compute $dl_t = Ad(t)|_\mathfrak{p} \in Aut(\mathfrak{p})$ and since $\det\left( Ad(t)|_{\mathfrak{p}}\right)=1$. $\alpha$ is smooth since the left action of $G$ on $G/T$ is smooth. Thus $\alpha$ is a non-zero, left-invariant, top form and gives us a left invariant measure on $G/T$. Note, I am aware that left-invariant measures can be constructed using the Riesz representation theorem however, since I intend to use the jacobian change of variables formula, it seems necessary to show that these measures are induced by differential forms.

Define $\phi: G/T \times T \to G, \phi(xT,t)=xtx^{-1}$. We now compute the pullback $\phi^*\omega$ in terms of $\alpha \wedge \beta$ as follows: Fix $(xT,t) \in G/T \times T$.

Identify $\mathfrak{p}\oplus \mathfrak{t}$ with $T_{(xT,t)}\left(G/T\times T\right)$ by $A_i \mapsto \frac{d}{ds}(xe^{sA_i}T,t)$ and $B_i \mapsto \frac{d}{ds}(xT,te^{sB_i})$.

Since $\phi(xe^{sA_i}T,t)=xe^{sA_i}te^{-sA_i}x^{-1} = xt(t^{-1}s^{sA_i}t)e^{-sA_i}x^{-1}$ and $\phi(xT,te^{sB_i})= xte^{sB_i}x^{-1}$, we see that the differential of $\phi$ is given as in the following commutative diagram: $\require{AMScd}$ \begin{CD} T_{(eT,e)}(G/T\times T)=\mathfrak{p}\oplus \mathfrak{t} @>{(Ad(t^{-1})-I)\oplus I}>> \mathfrak{p}\oplus \mathfrak{t}=T_e(G)\\ @VVV @V{dl_{xt}\circ dr_x^{-1}}VV \\ T_{(xT,t)}\left(G/T\times T\right) @>{d\phi}>> T_{xtx^{-1}}(G) \end{CD}

Taking exterior powers we get,

\begin{CD} \wedge \mathfrak{p}\otimes \wedge\mathfrak{t} @>{\det\left(Ad(t^{-1})-I\right)}>> \wedge \mathfrak{p}\otimes \wedge\mathfrak{t}=\wedge T_e(G)\\ @VVV @V{\wedge (dl_{xt}\circ dr_x^{-1})}VV \\ \wedge T_{(xT,t)}\left(G/T\times T\right) @>{J\phi}>> \wedge T_{xtx^{-1}}(G) \end{CD}

After dualizing the above diagram, we see that since $\omega$ is bi-invariant, $\omega_{xtx^{-1}}$ pulls back vertically to $\omega_e$. Since $(\alpha\wedge\beta)_{(eT,e)}(\wedge A_i \otimes \wedge B_i)= \omega_e(\wedge A_i \bigwedge \wedge B_i)=1$, we see that $\omega_e$ pulls back along the top arrow to $\det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)(\alpha\wedge\beta)_{(eT,e)}$. By the left invariance of $\alpha$ and $\beta$, we see that $(\alpha \wedge \beta)_{(xT,t)}$ pulls back along the left vertical arrow to $(\alpha\wedge\beta)_{(eT,e)}$. The commutativity of the above diagram thus shows $\phi^*\omega=\det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)(\alpha\wedge\beta)$.

We now use the following fact: There exist dense open sets $U\subset T$ and $V\subset G$ such that $G/T\times U \to V$ is a $|W|$-fold cover and $\det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)$ never vanishes on $U$. For a proof, see (Bump's Lie Groups $2$nd Edition, prop. $17.3$).

We now complete the proof by computing:

\begin{equation} \int_G fdg=\int_G fw=\int_V fw = \frac{1}{|W|}\int_{G/T\times U}\phi^*(fw) = \frac{1}{|W|}\int_{G/T\times U}f(t)\det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)(\alpha\wedge\beta)\\ =\frac{1}{|W|}\left(\int_T f(t)\det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)dt\right) \left(\int_{G/T}\alpha\right). \end{equation}

Question: How do I show the form $\alpha$ constructed above integrates to give $1$?

Edit: I feel like some computation of $d(t) := det\left(Ad(t^{-1})|_{\mathfrak{p}}-I_{\mathfrak{p}}\right)$ is required in the proof. If $\Phi$ denotes the set of weights of $T$ in the adjoint representation on $\mathfrak{p}\otimes_{\mathbb{R}} \mathbb{C}$, we can write $$d(t) = \prod\limits_{\alpha\in\Phi}\left(\alpha(t^{-1})-1\right) = \prod\limits_{\alpha\in\Phi^+}|\alpha(t)-1|^2.$$ Is there any way to see that the integral of this is equal to $|W|$?

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