Ytukyg

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 $$int_G fdg= frac{1}{|W|}int_{T}f(t)detleft(Ad(t^{-1})|_{mathfrak{p}}-I_{mathfrak{p}}right)dt$$ 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 $detleft( Ad(t)|_{mathfrak{p}}right)=1$. $alpha$ is smooth since the left action of $G$ on $G/T$ is smooth. Thus $alpha$ is a 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 $T_{(xT,t)}left(G/Ttimes Tright)$ 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/Ttimes 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/Ttimes Tright) @>{dphi}>> T_{xtx^{-1}}(G)
end{CD}

Taking exterior powers we get,

begin{CD}
wedge mathfrak{p}otimes wedgemathfrak{t} @>{detleft(Ad(t^{-1})-Iright)}>> wedge mathfrak{p}otimes wedgemathfrak{t}=wedge T_e(G)\
@VVV @V{wedge (dl_{xt}circ dr_x^{-1})}VV \
wedge T_{(xT,t)}left(G/Ttimes Tright) @>{Jphi}>> 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 $(alphawedgebeta)_{(eT,e)}(wedge A_i otimes wedge B_i)= omega_e(wedge A_i bigwedge wedge B_i)=1$, we see that $omega_e$ pulls back along the top arrow to $detleft(Ad(t^{-1})|_{mathfrak{p}}-I_{mathfrak{p}}right)(alphawedgebeta)_{(eT,e)}$. By the left invariance of $alpha$ and $beta$, we see that $(alpha wedge beta)_{(xT,t)}$ pulls back along the left vertical arrow to $(alphawedgebeta)_{(eT,e)}$. The commutativity of the above diagram thus shows $phi^*omega=detleft(Ad(t^{-1})|_{mathfrak{p}}-I_{mathfrak{p}}right)(alphawedgebeta)$.

We now use the following fact: There exist dense open sets $Usubset T$ and $Vsubset G$ such that $G/Ttimes U to V$ is a $|W|$-fold cover and $detleft(Ad(t^{-1})|_{mathfrak{p}}-I_{mathfrak{p}}right)$ 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/Ttimes U}phi^*(fw) = int_{G/Ttimes U}f(t)detleft(Ad(t^{-1})|_{mathfrak{p}}-I_{mathfrak{p}}right)(alphawedgebeta)\
=left(int_T f(t)detleft(Ad(t^{-1})|_{mathfrak{p}}-I_{mathfrak{p}}right)dtright) left(int_{G/T}alpharight).
end{equation}

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