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Change of variables

Differential Geometry
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Papapetros Vaggelis
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Change of variables

#1

Post by Papapetros Vaggelis » Mon Mar 07, 2016 10:54 pm

Let \(\displaystyle{f:\mathbb{R}^{n}\longrightarrow \mathbb{R}^{n}}\) be a differentiable function.

If \(\displaystyle{\left\{y_1,...,y_n\right\}}\) are the coordinates of the image of \(\displaystyle{f}\) and

\(\displaystyle{\left\{x_1,...,x_n\right\}}\) are the coordinates of the domain of \(\displaystyle{f}\) , then prove

that \(\displaystyle{f^{\star}\,\omega=\left(\det\,\mathrm{d}f\right)\,\mathrm{d}x_1\,\land...\land\,\mathrm{d}x_{n}}\)



where \(\displaystyle{\omega=\mathrm{d}y_1\,\land...\land\,\mathrm{d}y_{n}}\) .
Papapetros Vaggelis
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Re: Change of variables

#2

Post by Papapetros Vaggelis » Mon Mar 07, 2016 10:55 pm

Let \(\displaystyle{p\in\mathbb{R}^{n}}\) and \(\displaystyle{v_1,...,v_n\in(\mathbb{R}^{n})_{p}}\) .

According to the definition,

\(\displaystyle{\begin{aligned}(f^{\star}\omega)_{p}(v_1,...,v_n)&=\omega_{f(p)}((\mathrm{d}f)_{p}(v_1),...,(\mathrm{d}f)_{p}(v_n))\\&=\left(\mathrm{d}y_1\,\land...\land\mathrm{d}y_n \right )_{f(p)}((\mathrm{d}f)_{p}(v_1),...,(\mathrm{d}f)_{p}(v_n))\\&=\begin{vmatrix}
(\mathrm{d}y_1)_{f(p)}((\mathrm{d}f)_{p}(v_1))&... &(\mathrm{d}y_1)_{f(p)}((\mathrm{d}f)_{p}(v_n)) \\
...& ... & ...\\
(\mathrm{d}y_n)_{f(p)}((\mathrm{d}f)_{p}(v_1))&... & (\mathrm{d}y_n)_{f(p)}((\mathrm{d}f)_{p}(v_n))
\end{vmatrix}\,\,(I) \end{aligned}}\)

where, for each \(\displaystyle{i\,,j\in\left\{1,...,n\right\}}\) holds :

\(\displaystyle{\mathrm{d}f)_{p}(v_{j})=\mathrm{d}f\cdot v_{j}=}\)

\(\displaystyle{=\left(\dfrac{\partial{f_1} }{\partial{x_1}}(p)\,(v_{j})_{1}+...+\dfrac{\partial{f_1} }{\partial{x_n}}(p)\,(v_{j})_{n},...,\dfrac{\partial{f_i} }{\partial{x_1}}(p)\,(v_{j})_{1}+...+\dfrac{\partial{f_i} }{\partial{x_n}}(p)\,(v_{j})_{n},...,\dfrac{\partial{f_n} }{\partial{x_1}}(p)\,(v_{j})_{1}+...+\dfrac{\partial{f_n} }{\partial{x_n}}(p)\,(v_{j})_{n} \right )}\)


so:

\(\displaystyle{\begin{aligned} \left(\mathrm{d}y_{i}\right)_{f(p)}(\mathrm{d}f)_{p}(v_{j}))&=<\rm{grad}y_{i}(f(p)),(\mathrm{d}f)_{p}(v_{j}))>\\&=<e_{i},\mathrm{d}f\cdot v_{j}>\\&=\dfrac{\partial{f_i} }{\partial{x_1}}(p)\,(v_{j})_{1}+...+\dfrac{\partial{f_i} }{\partial{x_n}}(p)\,(v_{j})_{n},...,\end{aligned}}\)

which is the \(\displaystyle{i\,j}\) element of the matrix

\(\displaystyle{(\mathrm{d}f)_{p}\cdot \begin{pmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{pmatrix}}\)

Therefore, the relation \(\displaystyle{(I)}\) gives us:

\(\displaystyle{(f^{\star}\omega)_{p}(v_1,...,v_n)=\rm{det}\left((\mathrm{d}f)_{p}\cdot \begin{pmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{pmatrix}\right)=}\)

\(\displaystyle{=\rm{det}(\mathrm{d}f)_{p}\cdot \begin{vmatrix}
v_{1\,1}&... &v_{n\,1} \\
...& ... & ...\\
v_{1\,n}&... & v_{n\,n}
\end{vmatrix}=}\)

\(\displaystyle{=\left(\rm{det}(\mathrm{d}f)_{p}\right)\,\left(\mathrm{d}x_1\,\land...\land \mathrm{d}x_n\right)_{p}(v_1,...,v_n)}\)

since,

\(\displaystyle{\begin{aligned} v_{ij}&=<e_{j},\left(v_{i\,1},...,v_{i\,j},...,v_{i\,n}\right)>\\&=<\rm{grad}x_{j}(p),\left(v_{i\,1},...,v_{i\,j},...,v_{i\,n}\right)>\\&=\left(\mathrm{d}x_j\right)_{p}(v_{i})\end{aligned}}\)

for \(\displaystyle{i\,,j\in\left\{1,...,n\right\}}\)

and so: \(\displaystyle{f^{\star}\omega=\left(\rm{det}\mathrm{d}f\right)\,\mathrm{d}x_1\,\land...\land\mathrm{d}x_n}\) .
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