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Definition: Suppose $M\subseteq \mathbb{R}^n$ is a compact smooth k-manifold and $f:M\to \mathbb{R}$ is a continuous function. Since $S=\text{support}(f)$ is compact suppose $\phi:U\to V$ is a chart such that $S\subseteq V$ and $U$ is bounded. We define the integral of $f$ over $M$ to be $$\int_M fdV=\int_{U^{int}} (f\circ \phi)\text{Vol}(D\phi)$$

Suppose $x_1,...x_k\in\mathbb{R}^n$ form the vertices of a k-parallelepiped $P$, and $X$ be the matrix with columns $x_i$ then the volume of $P$ is $\sqrt{\det(X^TX)}$

Prove this definition is independent of coordinate chart $\phi$

Suppose $\psi:\widetilde U\to V$ is coordinate chart, then $\phi^{-1}\circ\psi:\widetilde U\to U$ is a change of variables.

Applying change of variables to the integral $$\int_{U^{int}} (f\circ \phi) \text{Vol}(D\phi)$$

$$\int_{\widetilde U^{\text{int}}} (f\circ\phi\circ(\phi^{-1}\circ\psi))\vert \det D(\phi^{-1}\circ\psi)\vert \text{Vol}(D\phi)$$ $$=\int_{\widetilde U^{\text{int}}} (f\circ\psi)\vert \det D\phi^{-1} D\psi\vert\text{Vol}(D\phi)$$

I'm not sure how to deal with the jacobians here. Since I don't really understand how you can take the determinant of $D\phi$ when $\phi$ is a map from $\mathbb{R}^k$ to $\mathbb{R}^n$ so it isn't a square matrix.

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  • $\begingroup$ Please define $\text{Vol}(D\phi)$. $\endgroup$ – Ted Shifrin Mar 27 at 1:06
  • $\begingroup$ @Ted Shifrin Added definition. $\endgroup$ – AColoredReptile Mar 27 at 1:13
  • $\begingroup$ So now you have some linear algebra to do with what you have written. $\endgroup$ – Ted Shifrin Mar 27 at 1:26
  • $\begingroup$ @Ted Shifrin So I need to show that $\text{Vol}(D\psi)=\vert \det (D\phi^{-1})(D\psi)\vert\text{Vol}(D\phi)$? I guess my confusion is that $D\phi$ is an $n\times k$ matrix, is it not? Otherwise if it was the $k\times k$ independent rows of that matrix then I would have $\text{Vol}(D\phi)=\vert\det (D\phi)\vert$ and $\det (D\phi^{-1}) \det(D\phi)=1$ so $\vert \det (D\phi^{-1})(D\psi)\vert\text{Vol}(D\phi)=\vert\det (D\psi)\vert=\text{Vol}(D\psi)$ $\endgroup$ – AColoredReptile Mar 27 at 1:37
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Applying the change of variables we get $$\int_{U^{int}} (f\circ \phi) \text{Vol}(D\phi)=\int_{\widetilde U^{\text{int}}} (f\circ\psi)\cdot\text{Vol}(D\phi)\circ\phi^{-1}\circ\psi\cdot\vert \det D(\phi^{-1}\circ\psi)\vert$$

So we should show

$$\text{Vol}(D\psi)=\text{Vol}(D\phi)\circ\phi^{-1}\circ\psi\cdot\vert \det D(\phi^{-1}\circ\psi)\vert$$

Writing this out, taking squares and using $|\det A|^2=\det (A\cdot A^T)$ and the multiplicativity of the determinant wee see that this is equivalent to

$$\det (D\psi^T\cdot D\psi) =\det\left(((D\phi)\circ\phi^{-1}\circ\psi)^T\cdot ((D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)^T\right).$$

Since $\psi=\phi\circ(\phi^{-1}\circ\psi)$ by applying the chain rule the right hand side becomes

$$\det\left(((D\phi)\circ\phi^{-1}\circ\psi)^T\cdot D\psi\cdot D(\phi^{-1}\circ\psi)^T\right).$$

Using the commutativity of the determinantant this is equal to $$ \det\left((D(\phi^{-1}\circ\psi)^T\cdot ((D\phi)\circ\phi^{-1}\circ\psi))^T\cdot D\psi\right)$$

which due to $(A\cdot B)^T=B^T\cdot A^T$ is

$$\det\left((((D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi))^T\cdot D\psi\right)$$

and so again by using the chain rule this is just $$ \det\left(D\psi^T\cdot D\psi\right)$$

which is what we wanted to show.

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  • $\begingroup$ What happened to $D(\phi^{-1}\circ\psi)$ in your 2nd step? $\endgroup$ – AColoredReptile Mar 27 at 17:15
  • $\begingroup$ Sorry. You said $\psi=\phi\circ(\phi^{−1}\circ \psi)$ and applied the chain rule. And I assume simplified $(D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)\to D\psi$. But I dont see how.$D(\phi^{-1}\circ\psi)=D(\phi^{-1})\circ\psi\cdot D\psi$ right? $\endgroup$ – AColoredReptile Mar 27 at 21:25
  • $\begingroup$ Maybe you get confused by the notation. $ ((D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)=D\psi$ is an equation of matrix valued functions on $\tilde U$. $D(\phi^{-1}\circ\psi)$ is the function which maps an $x$ to $(D(\phi^{-1}\circ\psi))(x)$ which is the Jacobian of $\phi^{-1}\circ\psi$ at the point $x$ whereas $(D\phi)\circ\phi^{-1}\circ\psi$ maps an $x$ to $(D\phi)(\phi^{-1}(\psi(x))$ which is the Jacobian of $\phi$ at the point $\phi^{-1}(\psi(x))$. $\endgroup$ – Chiara Mar 27 at 23:33
  • $\begingroup$ The chain rule states that for all $x\in \tilde U$ $(D\phi)(\phi^{-1}(\psi(x))\cdot (D(\phi^{-1}\circ\psi))(x)=(D\psi)(x)$ which just means $((D\phi)\circ\phi^{-1}\circ\psi)\cdot D(\phi^{-1}\circ\psi)=D\psi$. I hope this helps! $\endgroup$ – Chiara Mar 27 at 23:42
  • $\begingroup$ Yes I see what you meant now. $\endgroup$ – AColoredReptile Mar 27 at 23:47

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