$1\leq i,j\leq n$, and define $Vol(v_1,\cdots,v_n)=\sqrt{\det(a_{ij})}$. Note that if $v_1,\cdots,v_n$ is an orthogonal basis, then $Vol(v_1,\cdots,v_n)=|v_1|\cdots|v_n|$. For vectors in $W$ we define the same things. Suppose $A:V\rightarrow W$ is a linear map, define $$\label{E:det}
|\det A|=\frac{Vol(Av_1,\cdots,Av_n)}{Vol(v_1,\cdots,v_n)},$$ where $v_1,\cdots,v_n$ is a basis of $V$. It is easily to check that the definition is independent of the choice of the basis $v_1,\cdots,v_n$. In the special case that $v_1,\cdots,v_n$ is an orthogonal basis of $V$ and $Av_1,\cdots,Av_n$ are mutually orthogonal, then $$\label{E:det2}
|\det A|=\frac{|Av_1|\cdots|Av_n|}{|v_1|\cdots|v_n|}.$$ Note that we can only expect the norm of the determinant $|\det A|$ is well defined, since the sign “$\pm$" depends on the choice of orientations of $V$ and $W$.

\[P:pullback\] Suppose $M, N$ are two $n$-dimensional Riemannian manifolds with the associated Riemannian measures $dx, dy$, repsectively. If $\varphi:M\rightarrow N$ is a smooth map, then $$\label{E:Riemmannpullback}
\varphi^* (dy)=|\det(d\varphi)_x|dx.$$

Suppose that in local coordinate charts the Riemannian metrics on $M$ and $N$ are $ds^2=\Sigma_{ij}g_{ij}(x)dx_idx_j$ and $d\widetilde{s}^2=\Sigma_{ij}\widetilde{g}_{ij}(y)dy_idy_j$, respectively, where $g_{ij}(x)=\langle\frac{\partial}{\partial
x_i},\frac{\partial}{\partial x_j}\rangle$ and $\widetilde{g}_{ij}(y)=\langle\frac{\partial}{\partial
y_i},\frac{\partial}{\partial y_j}\rangle$. Then by definition, the Riemannian measures $dx,dy$ are $dx=\sqrt{\det\big(g_{ij}(x)\big)}dx_1\cdots dx_n$ and $dy=\sqrt{\det\big(\widetilde{g}_{ij}(y)\big)}dy_1\cdots dy_n$. We have $$\begin{aligned}
|\det(d\varphi)_x|^2
=&\frac{Vol\left((d\varphi)_x(\frac{\partial}{\partial
x_1}),\cdots,(d\varphi)_x(\frac{\partial}{\partial
x_n})\right)^2}{Vol\left(\frac{\partial}{\partial
x_1},\cdots,\frac{\partial}{\partial x_n}\right)^2}\\
=&\frac{\det\left(\left\langle\sum_k\frac{\partial y_k}{\partial
x_i}(\frac{\partial}{\partial
y_k})_{\varphi(x)},\sum_l\frac{\partial y_l}{\partial
x_j}(\frac{\partial}{\partial
y_l})_{\varphi(x)}\right\rangle\right)}{\det\left(\left\langle\frac{\partial}{\partial
x_i},\frac{\partial}{\partial x_j}\right\rangle\right)}\\
=&\frac{\det\left(\sum_{kl}\frac{\partial y_k}{\partial
x_i}\frac{\partial y_l}{\partial
