x_j}\widetilde{g}_{kl}(\varphi(x))\right)}{\det\big(g_{ij}(x)\big)}\\
=&\frac{\det\left(\left(\frac{\partial y_k}{\partial x_i}\right)^t
\Big(\widetilde{g}_{kl}(\varphi(x))\Big)\left(\frac{\partial
y_l}{\partial
x_j}\right)\right)}{\det\big(g_{ij}(x)\big)}\\
=&\frac{\left(\det\left(\frac{\partial y}{\partial
x}\right)\right)^2\det\big(\widetilde{g}_{ij}(\varphi(x))\big)}{\det\big(g_{ij}(x)\big)}.\end{aligned}$$ Hence $$\begin{aligned}
\varphi^*(dy)=&\sqrt{\det\big(\widetilde{g}_{ij}(\varphi(x))\big)}\left|\det\left(\frac{\partial
y}{\partial x}\right)\right|dx_1\cdots dx_n\\
=&|\det(d\varphi)_x|\sqrt{\det\big(g_{ij}(x)\big)}dx_1\cdots dx_n\\
=&|\det(d\varphi)_x|dx.\end{aligned}$$

Now we come to the main geometric problems which will be concerned in the following sections. Let $G$ be a Lie group, which acts on an $n$-dimensional smooth manifold $X$. The action is denoted by $\sigma: G\times X\rightarrow X$, and we denote $\sigma_g(x)=\sigma(g,x)$. Our first goal is, roughly speaking, to choose a representation point in each $G$-orbit $O_x = \{
\sigma_g(x): g\in G \}$, and the representation points should depend smoothly on the orbits. But in general this aim can only be achieved partially. So suppose we have a closed submanifold $Y$ of $X$, which consists of the representation points of the orbits in one’s mind, such that $Y$ intersects “almost all" orbits transversally. More precisely, we suppose there are closed zero measure subsets $X_{\mathrm{z}}\subset X$, $Y_{\mathrm{z}}\subset Y$. Let $X' = X
\setminus X_{\mathrm{z}}$, $Y' = Y \setminus Y_{\mathrm{z}}$, and suppose that

0.3cm (*invariance condition*) $X' = {\displaystyle}\bigcup_{y\in Y'} O_y$.

0.3cm (*transversality condition*) $T_y X=T_y O_y\oplus T_y Y$, $\forall y\in
Y'$.

0.3cm [It]{} is clear that (a) implies $Y'=Y\cap
X'$, and then $Y_{\mathrm{z}}=Y\cap X_{\mathrm{z}}$. Notice that $X'$ and $Y'$ are open 