Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a differentiable function. If $\{y_1, \dots, y_n \}$ are the coordinates of the image of $f$ and $\{x_1, \dots, x_n\}$ are the coordinates of the domain , prove that:
$$f^{\star}\,\omega=\left(\det\,\mathrm{d}f\right)\,\mathrm{d}x_1\,\land...\land\,\mathrm{d}x_{n}$$
where $\omega= \mathrm{d}y_1\,\land...\land\,\mathrm{d}y_{n}$.
Solution
Let $p\in\mathbb{R}^{n}$ and $v_1,...,v_n\in(\mathbb{R}^{n})_{p}$ .
According to the definition,
$$\begin{aligned}(f^{\star}\omega)_{p}(v_1,...,v_n)&=\omega_{f(p)}((\mathrm{d}f)_{p}(v_1),...,(\mathrm{d}f)_{p}(v_n))\\&=\left(\mathrm{d}y_1\,\land...\land\mathrm{d}y_n \right )_{f(p)}((\mathrm{d}f)_{p}(v_1),...,(\mathrm{d}f)_{p}(v_n))\\&=\begin{vmatrix}
(\mathrm{d}y_1)_{f(p)}((\mathrm{d}f)_{p}(v_1))&... &(\mathrm{d}y_1)_{f(p)}((\mathrm{d}f)_{p}(v_n)) \\
...& ... & ...\\
(\mathrm{d}y_n)_{f(p)}((\mathrm{d}f)_{p}(v_1))&... & (\mathrm{d}y_n)_{f(p)}((\mathrm{d}f)_{p}(v_n))
\end{vmatrix} \end{aligned} \;\;\; (1)$$
where, for each $i\,,j\in\left\{1,...,n\right\}$ holds :
$$\mathrm{d}f)_{p}(v_{j})=\mathrm{d}f\cdot v_{j}= \\
=\left(\dfrac{\partial{f_1} }{\partial{x_1}}(p)\,(v_{j})_{1}+\cdots+\dfrac{\partial{f_1} }{\partial{x_n}}(p)\,(v_{j})_{n},\cdots, \\
\dfrac{\partial{f_i} }{\partial{x_1}}(p)\,(v_{j})_{1}+\cdots+\dfrac{\partial{f_i} }{\partial{x_n}}(p)\,(v_{j})_{n},\cdots,\dfrac{\partial{f_n} }{\partial{x_1}}(p)\,(v_{j})_{1}+\cdots+\dfrac{\partial{f_n} }{\partial{x_n}}(p)\,(v_{j})_{n} \right )$$
so:
$$\begin{aligned} \left(\mathrm{d}y_{i}\right)_{f(p)}(\mathrm{d}f)_{p}(v_{j}))&=<\rm{grad}y_{i}(f(p)),(\mathrm{d}f)_{p}(v_{j}))>\\&=<e_{i},\mathrm{d}f\cdot v_{j}>\\&=\dfrac{\partial{f_i} }{\partial{x_1}}(p)\,(v_{j})_{1}+...+\dfrac{\partial{f_i} }{\partial{x_n}}(p)\,(v_{j})_{n},...,\end{aligned}$$
which is the $i\,j$ element of the matrix
$$(\mathrm{d}f)_{p}\cdot \begin{pmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{pmatrix}$$
Therefore, the relation $(1)$ gives us:
$$(f^{\star}\omega)_{p}(v_1,...,v_n)=\rm{det}\left((\mathrm{d}f)_{p}\cdot \begin{pmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{pmatrix}\right)=\\
=\rm{det}(\mathrm{d}f)_{p}\cdot \begin{vmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{vmatrix}=\\
=\left(\rm{det}(\mathrm{d}f)_{p}\right)\,\left(\mathrm{d}x_1\,\land...\land \mathrm{d}x_n\right)_{p}(v_1,...,v_n)$$
since,
$$\begin{aligned} v_{ij}&=<e_{j},\left(v_{i\,1},...,v_{i\,j},...,v_{i\,n}\right)>\\&=<\rm{grad}x_{j}(p),\left(v_{i\,1},...,v_{i\,j},...,v_{i\,n}\right)>\\&=\left(\mathrm{d}x_j\right)_{p}(v_{i})\end{aligned}$$
for $i\,,j\in\left\{1,...,n\right\}$
and so: $f^{\star}\omega=\left(\rm{det}\mathrm{d}f\right)\,\mathrm{d}x_1\,\land...\land\mathrm{d}x_n$.
$$f^{\star}\,\omega=\left(\det\,\mathrm{d}f\right)\,\mathrm{d}x_1\,\land...\land\,\mathrm{d}x_{n}$$
where $\omega= \mathrm{d}y_1\,\land...\land\,\mathrm{d}y_{n}$.
Solution
Let $p\in\mathbb{R}^{n}$ and $v_1,...,v_n\in(\mathbb{R}^{n})_{p}$ .
According to the definition,
$$\begin{aligned}(f^{\star}\omega)_{p}(v_1,...,v_n)&=\omega_{f(p)}((\mathrm{d}f)_{p}(v_1),...,(\mathrm{d}f)_{p}(v_n))\\&=\left(\mathrm{d}y_1\,\land...\land\mathrm{d}y_n \right )_{f(p)}((\mathrm{d}f)_{p}(v_1),...,(\mathrm{d}f)_{p}(v_n))\\&=\begin{vmatrix}
(\mathrm{d}y_1)_{f(p)}((\mathrm{d}f)_{p}(v_1))&... &(\mathrm{d}y_1)_{f(p)}((\mathrm{d}f)_{p}(v_n)) \\
...& ... & ...\\
(\mathrm{d}y_n)_{f(p)}((\mathrm{d}f)_{p}(v_1))&... & (\mathrm{d}y_n)_{f(p)}((\mathrm{d}f)_{p}(v_n))
\end{vmatrix} \end{aligned} \;\;\; (1)$$
where, for each $i\,,j\in\left\{1,...,n\right\}$ holds :
$$\mathrm{d}f)_{p}(v_{j})=\mathrm{d}f\cdot v_{j}= \\
=\left(\dfrac{\partial{f_1} }{\partial{x_1}}(p)\,(v_{j})_{1}+\cdots+\dfrac{\partial{f_1} }{\partial{x_n}}(p)\,(v_{j})_{n},\cdots, \\
\dfrac{\partial{f_i} }{\partial{x_1}}(p)\,(v_{j})_{1}+\cdots+\dfrac{\partial{f_i} }{\partial{x_n}}(p)\,(v_{j})_{n},\cdots,\dfrac{\partial{f_n} }{\partial{x_1}}(p)\,(v_{j})_{1}+\cdots+\dfrac{\partial{f_n} }{\partial{x_n}}(p)\,(v_{j})_{n} \right )$$
so:
$$\begin{aligned} \left(\mathrm{d}y_{i}\right)_{f(p)}(\mathrm{d}f)_{p}(v_{j}))&=<\rm{grad}y_{i}(f(p)),(\mathrm{d}f)_{p}(v_{j}))>\\&=<e_{i},\mathrm{d}f\cdot v_{j}>\\&=\dfrac{\partial{f_i} }{\partial{x_1}}(p)\,(v_{j})_{1}+...+\dfrac{\partial{f_i} }{\partial{x_n}}(p)\,(v_{j})_{n},...,\end{aligned}$$
which is the $i\,j$ element of the matrix
$$(\mathrm{d}f)_{p}\cdot \begin{pmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{pmatrix}$$
Therefore, the relation $(1)$ gives us:
$$(f^{\star}\omega)_{p}(v_1,...,v_n)=\rm{det}\left((\mathrm{d}f)_{p}\cdot \begin{pmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{pmatrix}\right)=\\
=\rm{det}(\mathrm{d}f)_{p}\cdot \begin{vmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{vmatrix}=\\
=\left(\rm{det}(\mathrm{d}f)_{p}\right)\,\left(\mathrm{d}x_1\,\land...\land \mathrm{d}x_n\right)_{p}(v_1,...,v_n)$$
since,
$$\begin{aligned} v_{ij}&=<e_{j},\left(v_{i\,1},...,v_{i\,j},...,v_{i\,n}\right)>\\&=<\rm{grad}x_{j}(p),\left(v_{i\,1},...,v_{i\,j},...,v_{i\,n}\right)>\\&=\left(\mathrm{d}x_j\right)_{p}(v_{i})\end{aligned}$$
for $i\,,j\in\left\{1,...,n\right\}$
and so: $f^{\star}\omega=\left(\rm{det}\mathrm{d}f\right)\,\mathrm{d}x_1\,\land...\land\mathrm{d}x_n$.
Applications :
ReplyDelete1. Polar coordinates
Let
\(\displaystyle{f:\mathbb{R}^2\longrightarrow \mathbb{R}^2\,,\left(r,\theta\right)\mapsto f(r,\theta)=\left(x(r,\theta),y(r,\theta)\right)=\left(r\,\cos\,\theta,r\,\sin\,\theta\right)}\)
Obviously, the function \(\displaystyle{f}\) is differentiable. We have that :
\(\displaystyle{\dfrac{\partial x}{\partial r}(r,\theta)=\cos\,\theta\,\,,\dfrac{\partial x}{\partial \theta}(r,\theta)=-r\,\sin\,\theta\,\,,\dfrac{\partial y}{\partial r}(r,\theta)=\sin\,\theta\,\,,\dfrac{\partial y}{\partial \theta}(r,\theta)=r\,\cos\,\theta\,\,,\left(r,\theta\right)\in\mathbb{R}^2}\)
So, \(\displaystyle{\left(\mathrm{d}f\right)_{(r,\theta)}:\mathbb{R}^2\longrightarrow \mathbb{R}^2}\)
given by
\(\displaystyle{\left(\mathrm{d}f\right)_{(r,\theta)}(a,b)=\begin{pmatrix}
\cos\,\theta& -r\,\sin\,\theta \\
\sin\,\theta& r\,\cos\,\theta
\end{pmatrix}\cdot \begin{pmatrix}
a\\
b
\end{pmatrix}}\)
with
\(\displaystyle{\rm{det}(\mathrm{d}f)_{(r,\theta)}=\rm{det}\,\begin{pmatrix}
\cos\,\theta& -r\,\sin\,\theta \\
\sin\,\theta& r\,\cos\,\theta
\end{pmatrix}=r}\)
Therefore,
\(\displaystyle{f^{\star}\omega=r\,\mathrm{d}r\,\land \mathrm{d}\theta}\)
2. If \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}\,,x\mapsto y=f(x)}\) is a
differentiable function, then
\(\displaystyle{\left(\mathrm{d}f\right)_{x}(h)=f^\prime(x)\cdot h\,\,,\forall\,x\in\mathbb{R}\,,\forall\,h\in\mathbb{R}}\)
Therefore, if \(\displaystyle{\omega=\mathrm{d}y}\), then
\(\displaystyle{f^{\star}\omega=f^\prime\,\mathrm{d}x}\) .
Note: The differential form \(\displaystyle{v=\mathrm{d}x_1\,\land...\land\,\mathrm{d}x_n}\)
is called volume element.