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Suppose $M=\varphi(A)$ a parametrized $k$-manifold in $\mathbb R^n$, given a diffeomorphism $g$ in $\mathbb R^n$, $N=g(M)$ is still a parametrized $k$-manifold. Now for a scalar function $f:N\to\mathbb R$, is it true that $$\int_N fdV=\int_{M}(f\circ g)|\det Dg|dV$$?

What I have concluded so far: I am able to show that $$\int_N fdV=\int_A (f\circ g\circ\varphi)V(D(g\circ\varphi)=\int_A(f\circ g\circ\varphi)\frac{V(D(g\circ\varphi))}{V(D\varphi)}V(D\varphi)=\int_A(f\circ g\circ\varphi)(\tilde J\circ\varphi)V(D\varphi)=\int_M(f\circ g)\tilde JdV$$ where $\tilde J$ is defined to be $$\tilde J=\frac{V(D(g\circ\varphi))}{V(D\varphi)}\circ\varphi^{-1}=\frac{\det\left[(D\varphi_{\varphi^{-1}})^T(Dg)^T(Dg)(D\varphi_{\varphi^{-1}})\right]^{1/2}}{\det\left[(D\varphi_{\varphi^{-1}})^T(D\varphi_{\varphi^{-1}})\right]^{1/2}}$$ It leaves to show $\tilde J=|\det Dg|$. If $D\varphi$ are square matrices, then the determinant can be factored which gives us the result but that is just a regular change of variables.

I know from computation experiences that, for example, one can change a 2-surface in $\mathbb R^3$ from Cartesian coordinate to spherical coordinate by simply multiplying $|\det Dg|=\rho^2\sin\theta$ just like treating them as open sets in $\mathbb R^3$. I have also tried a few computations for spherical coordinates, it seems that regardless of the choice of $\varphi$, $\tilde J$ defined above indeed equals to $|\det Dg|$. I wonder if this generally holds for all parametrized manifolds? My intuition tells me an $n$-diffeomorphism does not necessarily work for $k$-volume. Thank you in advance for answering.

sudeep5221
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rpoc1231
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  • I take it $dV$ is actually $i_N^\ast(dV)$ or $i_M^\ast(dV)$ where $i_N:N \to \mathbb{R}^n$ and $i_M:M \to \mathbb{R}^n$ are the inclusion functions? – aschepler Jun 02 '23 at 17:35
  • @aschepler I suppose? Sorry I am using a quite elementary textbook where they just defined $\int_{\varphi(A)}fdV=\int_{A}(f\circ\varphi)V(D\varphi)$ as a notation instead of treating it as a form. – rpoc1231 Jun 02 '23 at 17:45
  • @aschepler no. pulling back a top form of an ambient manifold (in particular a volume form of the ambient manifold) to a submanifold of strictly smaller dimension always gives zero for obvious reasons. The $dV$’s should be the appropriate induced volume elements/ surface measures. – peek-a-boo Jun 02 '23 at 17:49
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    OP: I think Change of variables in integration on manifolds: what’s wrong should answer your question. Granted I talk a little more abstractly but you should be able to piece together the general idea. Bottom line is yes there are some other weird terms in the change of volume factor. – peek-a-boo Jun 02 '23 at 17:53
  • @peek-a-boo Thank you so much for the link! Do you mind provide me some insight to why this Radon-Nikodym derivative does correspond to |det Dg| for spherical or cylindrical coordinates? Computationally I get the correct result but I wonder if there is a general criteria for that. – rpoc1231 Jun 02 '23 at 18:10
  • @peek-a-boo Right. I thought I recalled some straightforward way to get a "natural" volume form on an embedded manifold. Or that's only simple when dimensions are the same? – aschepler Jun 02 '23 at 18:39
  • I believe you’re wrong about the spherical coordinates example. What happens when the surface is the cylinder $r=a$ and you use spherical coordinates $\theta,\phi$? I don’t have the patience to decipher your notation. – Ted Shifrin Jun 02 '23 at 23:33
  • @aschepler Yes, you contract (interior product) with an orthonormal frame for the normal bundle of the submanifold. – Ted Shifrin Jun 02 '23 at 23:34
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    Your experience is misleading, since you may be only thinking of very simple examples where things may seem to simplify naively. See the example above by Ted Shifrin. Oh and btw, I found another answer of mine, that is more at the level of this question, to which (modulo notation… I use $f$ instead of $g$) I think this can be considered a dupe: Change of variables when integrating over a manifold. As I mentioned in a comment to one of my answers, $\det Dg$ does not make sense between general manifolds. – peek-a-boo Jun 03 '23 at 03:21
  • @peek-a-boo Thank you, I think I understand now. After a few experiments I think simple cases work because the parametrization I used preserves coordinates. Thanks again for your help! – rpoc1231 Jun 03 '23 at 04:15

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