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Change of variables

Differential Geometry
Papapetros Vaggelis
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Change of variables

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

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