Let $M$ be an $n$ dimensional smooth manifold in $\mathbb{R}^N$, $n In a simple case, suppose $M$ is a line in $\mathbb{R}^2$ connecting $(0,0)$ and $(1,1)$. Let $f(x_1,x_2)=(\alpha x_1, \beta x_2)^\prime$, i.e. $f$ rescales the x,y coordinate by $\alpha,\beta>0$ respectively. Then $J_f$ should be $\alpha\beta$. Since $f(M)$ is a line connecting $(0,0)$ and $(\alpha,\beta)$, we should have $\int_{f(M)} dV(z) = \sqrt{\alpha^2+\beta^2}$, while $ \int_{M} |J_f(x)| dV(x) = \sqrt{2}\alpha\beta$. What should be the correct form of the above change of variables formula, using the ''Jacobian'' $J_f$?
Change of variables for integration on manifolds
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differential-geometry
riemannian-geometry
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0What do you mean by the Jacobian? The partial derivatives matrix is not square. – 2017-02-04
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0I have edited the question, hope it is clearer. – 2017-02-05
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1OK, what you wrote now makes sense, but it gives a wrong answer. You can see this immediately from the fact that $vol(f(M))$ should depend only on the restriction $f|_M$, while your formula does not. However, you can use square root of the determinant of a suitable Gramm matrix. That will give you the right answer. – 2017-02-05
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0Thank you. If I do not have any parameterization for the manifold, is a formula possible? I agree that $Vol(f(M))$ should only depend on $f|M$. In the example, if we change $M$ to be a line between $(0,0)$ and $(1,0)$ (or $(0,1)$), then the length of $f(M)$ only depends on $\alpha$ (or $\beta$). – 2017-02-06
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1Suppose we have a parameterization $\phi$ for $M$ such that $\phi(\Omega) = M$ for $\Omega\subset\mathbb{R}^n$. Then I think we should have $\int_{f(M)} g(z) dV(z) = \int_{\Omega} g(f(\phi(y)) \sqrt{|D\phi^TDf^TDfD\phi|}dV(y)$....Here we assumed a parameterization. – 2017-02-06