# Choice of volume forms in the Weyl Integration Formula

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:

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

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 $$|W|$$?