# Show that integral of $f$ over a manifold $M$ is independent of coordinate chart

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.

• Please define $\text{Vol}(D\phi)$. – Ted Shifrin Mar 27 at 1:06
• @Ted Shifrin Added definition. – AColoredReptile Mar 27 at 1:13
• So now you have some linear algebra to do with what you have written. – Ted Shifrin Mar 27 at 1:26
• @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 $\vert \det (D\phi^{-1})(D\psi)\vert\text{Vol}(D\phi)=\vert\det (D\psi)\vert=\text{Vol}(D\psi)$ – AColoredReptile Mar 27 at 1:37

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.

• What happened to $D(\phi^{-1}\circ\psi)$ in your 2nd step? – AColoredReptile Mar 27 at 17:15
• 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? – AColoredReptile Mar 27 at 21:25
• 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 $\phi^{-1}(\psi(x))$. – Chiara Mar 27 at 23:33
• 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! – Chiara Mar 27 at 23:42
• Yes I see what you meant now. – AColoredReptile Mar 27 at 23:47