\xi\in {\mathfrak{l}}.$$ If $y\in Y'$, then $\dim{\mathfrak{l}}=\dim T_y O_y$. We choose an inner product on ${\mathfrak{l}}$, and endow the inner product on $T_y O_y$ induced from the Riamannian structure on $X$. Then the “determinants" $|\det\Psi_y|$ and $|\det((d\pi)_e|_{\mathfrak{l}})|$ make sense.

\[T:det\] Under the above assumptions, we have $$\label{F:J=det}
J(y)=C|\det\Psi_y|,$$ for $y\in Y'$, where $C=|\det((d\pi)_e|_{\mathfrak{l}})|^{-1}$ is a constant.

By the transversality condition (b), the tangent map $$(d\varphi)_{([e],y)}:T_{([e],y)}(G/K\times Y)\rightarrow T_yX$$ of $\varphi$ at the point $([e], y)$ ($y\in Y'$) can be regarded as $$(d\varphi)_{([e],y)}:T_{[e]}(G/K) \oplus T_yY \rightarrow T_yO_y
\oplus T_yY .$$ Denote $\widetilde{\Psi}_y=(d\varphi)_{([e],y)}|_{T_{[e]}(G/K)}:T_{[e]}(G/K)\rightarrow
T_yO_y$, then it is obvious that $\Psi_y=\widetilde{\Psi}_y\circ(d\pi)_e|_{\mathfrak{l}}$, and one can easily show that in the matrix form, $$(d\varphi)_{([e],y)}=\left(\begin{array}{cc}\widetilde{\Psi}_y&0\\0&id
\end{array}\right).$$

0.3cm Since $d\mu$ is the associated Riemannian measure on $G/K$, the product measure $d\mu dy$ is the the associated Riemannian measure on the product Riemannian manifold $G/K\times
Y'$. By Proposition \[P:pullback\] and the orthogonality condition (d), $$\begin{aligned}
J(y)=&\left|\det(d\varphi)_{([e],y)}\right|\\
=&\left|\det\left(\begin{array}{cc}\widetilde{\Psi}_y&0\\0&id
\end{array}\right)\right|\\
=&|\det\widetilde{\Psi}_y|\\
=&|\det(\Psi_y\circ((d\pi)_e|_{\mathfrak{l}})^{-1})|\\
=&C|\det\Psi_y|,\end{aligned}$$ where $C=|\det((d\pi)_e|_{\mathfrak{l}})|^{-1}$. This proves the theorem.

Although formula only hold on $Y'$, since $Y'$ is dense in $Y$ and $J\in C^\infty(Y)$, we can get $J(y)$ for all $y\in Y$ by smooth continuation.

0.5cm

Integrations over $G$-spaces
============================

0.5cm Occasionally we will be interested in some kinds of integration 