The problem is this: Let $f:M^m\to\mathbb{R}$ a continuous function with compact support and $h:N\to M$ a diffeomorphism between Riemannian compact manifolds. Then $$\int_{M}fdV^M=\int_N f\circ h|\det dh|dV^N$$ The proof using differential forms is easy, but I'm trying to prove this without using forms, just the definition.


Let $\Omega$ be the support of $f$, $\Theta=h^{-1}(\Omega)$ the support of $f\circ h$, $\textbf{x}_\alpha:U_\alpha\to M$ and $\textbf{y}_\alpha:V_\alpha\to N$ parametrizations and $\{\phi_\alpha\}_{\alpha\in A}$ a partition of unity subordinate to $\{U_\alpha\}_{\alpha\in A}$. Let $g^M$ the metric of $M$ and $g^N$ the metric of $N$.Then $$\int_M fdV^M=\sum_{\alpha}\int_{\textbf{x}_\alpha^{-1}(\Omega)}\phi_\alpha f\circ\textbf{x}_\alpha\sqrt{\det g^M}dx^M$$ Denoting $\Omega_\alpha=\textbf{x}_\alpha^{-1}(\Omega)\subset U_\alpha$,$\Theta_\alpha=\textbf{y}^{-1}_\alpha(\Theta)\subset V_\alpha$ and $T_\alpha=\textbf{x}_\alpha^{-1}\circ h\circ\textbf{y}_\alpha:V_\alpha\to U_\alpha$ we will have $$\int_M fdV^M=\sum_{\alpha}\int_{\Omega_\alpha}\phi_\alpha f\circ\textbf{x}_\alpha\sqrt{\det g^M}dx^M=\sum_{\alpha}\int_{T_\alpha(\Theta_\alpha)}\phi_\alpha f\circ\textbf{x}_\alpha\sqrt{\det g^M}dx^M$$ by the variable change theorem, $$\int_M fdV^M=\sum_{\alpha}\int_{\Theta_\alpha}(\phi_\alpha\circ T_\alpha)(f\circ\textbf{x}_\alpha\circ T_\alpha)\sqrt{\det g^M\circ T_\alpha}|\det dT_\alpha|dy^N$$ How \begin{eqnarray*} \sqrt{\det g^M\circ T_\alpha}&=&\mathrm{vol}\ P\left[\frac{\partial}{\partial x_1},\cdots,\frac{\partial}{\partial x_m}\right] \\ &=&\mathrm{vol}\ P\left[dh\left(\frac{\partial}{\partial y_1}\right),\cdots,dh\left(\frac{\partial}{\partial y_m}\right)\right] \\ &=&\mathrm{vol}\ P\left[\frac{\partial}{\partial y_1},\cdots,\frac{\partial}{\partial y_m}\right]|\det dh| \\ &=&\sqrt{g^N}|\det dh| \end{eqnarray*} and $$\det dT_\alpha=\det dh$$ We have \begin{eqnarray*} \int_M fdV^M&=&\sum_{\alpha}\int_{\Theta_\alpha}(\phi_\alpha\circ T_\alpha)(f\circ\textbf{x}_\alpha\circ T_\alpha)\sqrt{\det g^N}|\det dh|^2dy^N \\ &=&\sum_{\alpha}\int_{\textbf{y}_\alpha^{-1}(\Theta)}(\phi_\alpha\circ T_\alpha)(f\circ h\circ \textbf{y}_\alpha)\sqrt{\det g^N}|\det dh|^2dy^N \\ &=&\int_N f\circ h|\det dh|^2dV^N \end{eqnarray*} Question: where is the error? It must probably be something stupid but I can't see it.

  • $\begingroup$ my suggestion is to assume first of all that $M,N$ can be covered by a single chart, and then see if your calculation lines up, because clearly, for the general case you just have to put a few extra $\sum$ signs, so that can't be where the issue is. $\endgroup$
    – peek-a-boo
    Mar 12, 2021 at 20:55
  • $\begingroup$ @peek-a-boo In fact this was the first path I followed, and gave the same problem. $\endgroup$ Mar 13, 2021 at 18:21
  • 1
    $\begingroup$ By the way, writing $\det dh$ abstractly like this is incorrect (at the very least extremely misleading) because $dh$ is a linear transformation (pointwise) between different vector spaces and as such, the determinant is not well-defined. It is only once you specify a basis on the domain and target space that you can consider the determinant of the matrix representation. Strictly speaking the extra factor which should appear is a Radon-Nikodym derivative. Or alternatively, you can write the rule as $\int_Mf\, dV_g = \int_Nf\circ h \,dV_{h^*g}$, so that the "effect of $h$" is already encoded... $\endgroup$
    – peek-a-boo
    Mar 13, 2021 at 18:49
  • $\begingroup$ [cont.] inside volume measure induced by the metric $h^*g$. In any case, I highly suspect your calculation in local coordinates is going wrong because you're mixing up the roles of $h$ and $h^{-1}$. I did the calculation in my own notation and got the right answer. $\endgroup$
    – peek-a-boo
    Mar 13, 2021 at 18:53
  • $\begingroup$ I understand what you mean. Would it be possible to post your calculations? (although I never studied measure theory as this discipline was never offered at the university, but I would like to see the reasoning used.) $\endgroup$ Mar 14, 2021 at 19:58

1 Answer 1


Suppose $(M,g)$ is a Riemannian manifold, and $h:N\to M$ a diffeomorphism. Then, we can consider the Riemannian manifold $(N,h^*g)$. Then, for any smooth function $f:M\to\Bbb{R}$, we can write the change of variables formula as \begin{align} \int_Mf\,dV_g &=\int_{N}f\circ h\, dV_{h^*g}\tag{i} \end{align} This should remind you very much of the differential form version where $\int_M\omega = \int_{h(N)}\omega = \int_Nh^*\omega$. Here, you can think of $dV_g$ as the Riemannian volume form on $M$ induced by the metric $g$ (provided $M$ is orientable, in which case the choice of volume form provides the orientation). Or you can also think of this as a measure (which you can do even when $M$ is not orientable).

As I mentioned in the comments, writing $\det dh$ abstractly as you've done is incorrect (at the very least very bad practice) because (pointwise) $dh$ is a linear transformations between different vector spaces. Now, you seem to be starting with two Riemannian manifolds $(M,g^M)$ and $(N,g^N)$, and then considering a diffeomorphism $h:N\to M$. In this case, one can also write down a change of variables formula: \begin{align} \int_Mf\,dV_{g^M} &=\int_{N}f\circ h\cdot \left(\dfrac{dV_{h^*(g^M)}}{dV_{g^N}}\right)\, dV_{g^N}\tag{ii} \end{align} Here, the bracketed term should be interpreted as a Radon-Nikodym derivative of the two measures $dV_{h^*(g^M)}$ and $dV_{g^N}$ (which one can show are positive $\sigma$-finite measures which are mutually absolutely continuous, hence the Radon-Nikodyn theorem can be applied). This is the correct factor which should appear (very roughly speaking, it's like writing $dy=\dfrac{dy}{dx}dx$ inside the integral sign, or writing $d^ny=\left|\det\frac{\partial y}{\partial x}\right|\,d^nx$ in the multivariable setting). As you can see from the notation itself, this Radon-Nikodym derivative involves all three things: $h,g^M$ and $g^N$, and this takes into account the "change of measure" factor (which you incorrectly just wrote as $\det dh$). Of course, if one works things out in local coordinates, one will get some sort of determinant of a coordinate representation of derivatives of $h$, but in addition, one will also get extra factors involving the determinants of $g^M, g^N$ in appropriate coordinates.

As you can see (ii) is not the nicest way of writing things down; (i) is the nice way of doing things. Now, let us forget about (ii) and try to prove (i), atleast locally. So, let's forget the excess baggage of a partition of unity and work with a single chart (because like I said, once we do this, we just put in a few extra summations and we're done).

To this end, let $(U,x)$ be a chart on $M$, and $(V,y)$ a chart on $N$ such that $h(V)\subset U$ (for my own sanity, I'm working with charts... i.e the function $x:U\to x[U]\subset\Bbb{R}^n$ goes from the manifold to the Euclidean space, rather than the other way around). Also, I shall denote the metric on $M$ simply by $g$, and I consider the pull-back metric $h^*g$ on $N$. We now want to show that \begin{align} \int_{h(V)}f\,dV_g =\int_{V}f\circ h\,dV_{h^*g}. \end{align}

Ok so let's start on the right. We have \begin{align} \int_{V}f\circ h\,dV_{h^*g} &:= \int_{y(V)}(f\circ h)\circ y^{-1}\cdot \sqrt{\left| \det [(h^*g)_{ij}]\right|\circ y^{-1}}\, dy\tag{$*$} \end{align} Let's calculate what the entries of the matrix are at a point $b\in y(V)$: \begin{align} (h^*g)_{ij}(y^{-1}(b))&:=(h^*g)\left( \frac{\partial}{\partial y^i}\bigg|_{y^{-1}(b)}, \frac{\partial}{\partial y^j}\bigg|_{y^{-1}(b)}\right)\\ &=g\left[h_*\left(\frac{\partial}{\partial y^i}\bigg|_{y^{-1}(b)}\right), h_*\left(\frac{\partial}{\partial y^j}\bigg|_{y^{-1}(b)}\right) \right]\\ &= \frac{\partial (x^{\mu}\circ h)}{\partial y^i}\bigg|_{(x\circ h\circ y^{-1})(b)}\cdot \frac{\partial (x^{\nu}\circ h)}{\partial y^j}\bigg|_{(x\circ h\circ y^{-1})(b)}\cdot g_{\mu\nu}(x\circ h\circ y^{-1}(b)) \end{align} The efficient way of writing this down is that the matrices are related via a congruence (I'm omitting the points of evaluation) \begin{align} [h^*g] &= [D(x\circ h\circ y^{-1})]^t\cdot [g_{\mu\nu}]\cdot [D(x\circ h\circ y^{-1})] \end{align} (This should hopefully be familiar from linear algebra: the matrix representations of a bilinear map are related by congruence). Now, since a matrix and its transpose have the same determinant, it follows that the determinants are related as: \begin{align} \sqrt{\left|\det (h^*g)_{ij}(y^{-1}(b))\right|}&= \sqrt{\left|\det g_{\mu\nu}((x\circ h\circ y^{-1})(b))\right|}\cdot \left|\det D(x\circ h\circ y^{-1})_b\right| \end{align} We now have everything we need to use the standard change of variables in the RHS of $(*)$: \begin{align} \text{RHS of $(*)$} &= \int_{y(V)} (f\circ x^{-1})\circ (x\circ h\circ y^{-1})\cdot \sqrt{\left|\det g_{\mu\nu}\right|}\circ (x\circ h\circ y^{-1})\cdot \left|\det D(x\circ h\circ y^{-1})\right|\,dy\\ &=\int_{x(h(V))}f\circ x^{-1}\cdot \sqrt{\left|\det g_{\mu\nu}\right|}\,dx\\ &:=\int_{h(V)}f\, dV_g. \end{align} This completes the proof.

  • $\begingroup$ @peak-a-boo, you are the man! $\endgroup$ Mar 15, 2021 at 12:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .