Change of variables when integrating over a manifold

Question

Suppose $M_1$ is an $n$-dimensional smooth manifold in $\mathbb{R}^m$, $n < m$. Suppose we have a diffeomorphism $f: \mathbb{R}^m\to\mathbb{R}^m$ that maps $M_1$ to the $n$-dimensional manifold $M_2$. I want to find the relation between the integral over $M_1$, $\int_{M_1}g(x)dV(x)$ to the integral over $M_2$. If $n = m$ then this is a familiar case for me, we just multiply by $|det(Df^{-1})|$ after substituting $x = f^{-1}(y)$. When $n < m$ I thought about using parameterization: Suppose $\Omega \subset \mathbb{R}^n,\varphi:\Omega \to M_1$. Then, $f \circ \varphi$ is a parameterization of $M_2$. Using the chain rule, $D(f \circ \varphi) = D(f)D(\varphi)$, so that an infinitesimal volume element is $dV(x) = \sqrt{\det(D\varphi ^TDf^TDfD\varphi)}du$, and: $$ \int_{M_2} g(x)dV(x) = \int_{\Omega}(g\circ f\circ\varphi)(u)\sqrt{\det(D\varphi ^TDf^TDfD\varphi)}du $$ but I cannot seem to relate this expression to the integral over $M_1$: $$ \int_{M_1}g(x)dV(x) = \int_{\Omega}(g\circ\varphi)(u)\sqrt{\det(D\varphi^TD\varphi)}du $$ How can I "pull out" $Df^TDf$ from the determinant and the square root?

Answer 1

First, your uses of $M_1,M_2$ is confusing because you're mixing them up. Also, your function $g$ appears twice: first as a function on $M_1$, but then later on $M_2$. Which is it?

Anyway, I'll assume $g:M_2\to\Bbb{R}$, and $f:\Bbb{R}^m\to\Bbb{R}^m$ is a diffeomorphism which maps $M_1$ onto $M_2$, and $\phi:\Omega\subset\Bbb{R}^m\to \phi(\Omega)\subset M_1$ a parametrization. Then, given your coordinate calculation, you just multiply and divide by $\sqrt{\det D\phi^tD\phi}=\sqrt{\det \left\langle\frac{\partial \phi}{\partial x^i},\frac{\partial\phi}{\partial x^j}\right\rangle}$ to get your answer. More explicitly, \begin{align} \int_{(f\circ \phi)(\Omega)}g(x)\,dV_{M_2}(x)&=\int_{\Omega}(g\circ (f\circ \phi))(u)\sqrt{\det (D\phi_u)^t(Df_{\phi(u)})^t(Df_{\phi(u)})D\phi_u}\,\,\,du\\ &=\int_{\Omega}(g\circ f)(\phi(u))\cdot J_f(\phi(u))\cdot \sqrt{\det (D\phi_u)^t(D\phi_u)}\,du\\ &=\int_{\phi(\Omega)}(g\circ f)(y)\cdot J_f(y)\,dV_{M_1}(y), \end{align} where \begin{align} J_f(\phi(u)):=\frac{\sqrt{\det (D\phi_u)^t(Df_{\phi(u)})^t(Df_{\phi(u)})D\phi_u}}{\sqrt{\det (D\phi_u)^t(D\phi_u)}}, \end{align} i.e the function $J_f:\phi(\Omega)\subset M_1\to\Bbb{R}$ is given as \begin{align} J_f(y)&=\frac{\sqrt{\det \left(D\phi_{\phi^{-1}(y)}\right)^t(Df_y)^t(Df_y)\left(D\phi_{\phi^{-1}(y)}\right)}}{\sqrt{\det \left(D\phi_{\phi^{-1}(y)}\right)^t\left(D\phi_{\phi^{-1}(y)}\right)}}. \end{align} You can now check by hand (an exercise in linear algebra/chain rule) that if you had started with a different $\phi$ in the beginning, then the map $J_f$ you get is actually the same: $J_f$ does not depend on the $\phi$, which is why I didn't include $\phi$ in the notation. Note that, both the numerator and denominator alone will depend on $\phi$, but their quotient does not, and this is why $J_f$ can actually be extended to a well-defined smooth function $M_1\to\Bbb{R}$.

Hence the change of variables formula is \begin{align} \int_{M_2}g\,dV_{M_2}&=\int_{M_1}(g\circ f)\cdot J_f\,dV_{M_1}. \end{align} Looks an awful lot like your usual change of variable formula doesn't it? Yes, the only difference is that the "Jacobian factor" $J_f$ is not as simple as you'd expect. But note that if you specialize to the case where $M_1,M_2$ are open subsets of $\Bbb{R}^m$ (and hence $n=m$), then this $J_f$ reduces exactly to $|\det Df|$, which is the usual thing we learn. More abstractly, $J_f$ is the Radon-Nikodym derivative of two different Riemannian volume measures on $M_1$: \begin{align} J_f&=\frac{dV_{f^*\delta_{\Bbb{R}^m}}}{dV_{M_1}}, \end{align} where $dV_{f^*\delta_{\Bbb{R}^m}}$ is the Riemannian measure induced on $M_1$ by the Riemannian metric $f^*(\delta_{\Bbb{R}^m})$ (with $\delta_{\Bbb{R}^m}$ being the usual Riemannian metric on the ambient $\Bbb{R}^m$). See Change of variables in integration on manifolds: what's wrong? for a similar discussion.