Q:

Transformation for Integrals over Manifolds

Most of modern books on integration theory, when constructing the Lebesgue integral, do not introduce manifolds prior. The transformation for Lebesgue integrals can then be stated as follows:
Let $\Omega \subseteq \mathbb{R}^d$ be open and $\Phi \colon \Omega \rightarrow \Phi(\Omega)\subseteq\mathbb{R}^d$ a diffeomorphism. The function $f$ is integrable on $\Phi(\Omega)$ if and only if $x\mapsto f(\Phi(x))|\det(D\Phi(x))|$ is on $\Omega$. It then holds that
$\displaystyle \int_{\Phi(\Omega)}f(y)\,\mathrm{d}y=\int_{\Omega} f(\Phi(x))|\det(D\Phi(x))|\,\mathrm{d}x$
where $D\Phi(x)$ is the functional matrix.
When considering integrals over manifolds one could proceed like this (where some details are omitted for the purpose of transparency and only global charts are considered):
Let $\Psi\colon T\rightarrow V\subseteq M, T\subseteq\mathbb{R}^k$ be a global chart.
The integral of $f$ over $M$ is then defined as
$\displaystyle \int_M f(x)\,\mathrm{d}S(x)=\int_T f(\Psi(t))\sqrt{g(t)}\,\mathrm{d}t$
where $g$ is the gramian determinant of the chart $\Psi$.
I am looking for an analogue of the first integral transformation for manifolds. Has this anything to do with a change of charts? Also, if I remember correctly, in the theory of differentialforms this analogue is the pullback of a form, is this correct? Do you have to invoke parametrisations for a pullback, too?

A:

Given a parametric $k$-dimensional manifold $\Psi \colon V \rightarrow \mathbb{R}^n$ where $V \subseteq \mathbb{R}^k$ is an open subset and $\Psi$ satisfies the usual conditions (one-to-one, smooth, full-rank), the integral over $M = \Psi(V)$ of a function $f \colon M \rightarrow \mathbb{R}$ is defined by
$$ \int_{M} f \, dS := \int_V f(\Psi(y)) \, \sqrt{\det (D\Psi^T(y)\cdot  D\Psi(y))} \, dy. $$
To make sure this definition makes sense, we need to verify that it is actually independent of the specific parametrization (chart) of $M$. If $\Phi \colon V' \rightarrow \mathbb{R}^n$ is another parametrization of $M$ then $g := \Psi^{-1} \circ \Phi \colon V' \rightarrow V$ is a diffeomorphism between open sets $V,V' \subseteq \mathbb{R}^k$ and the regular change of variables formula implies that
$$ \int_V f(\Psi(y)) \, \sqrt{\det \left( D\Psi^T(y) \cdot D\Psi(y) \right)} \, dy = \int_{V'} f(\Psi(g(x))) \sqrt{\det \left( D\Psi^T(g(x)) \cdot D\Psi(g(x)) \right)} \left| \det Dg(x) \right| dx $$
Note that
$$ \Psi(g(x)) = \Psi(\Psi^{-1}(\Phi(x))) = \Phi(x), \\
f(\Psi(g(x)) = f(\Phi(x)), \\
D\Phi(x) = D\Psi(g(x)) \cdot Dg(x), \\
\sqrt{\det \left( D\Psi^T(g(x)) \cdot D\Psi(g(x)) \right)} \left| \det Dg(x) \right| = \\
\sqrt{ \det(Dg^T(x)) \det \left( D\Psi^T(g(x)) \cdot D\Psi(g(x)) \right) \det(Dg(x)) } = \\
\sqrt{\det \left( Dg^T(x) \cdot D\Psi^T(g(x)) \cdot D\Psi(g(x)) \cdot Dg(x)  \right)} = \\
\sqrt{ \det \left( D\Phi^T(x) \cdot D\Phi(x) \right) } $$
and so we also get
$$ \int_V f(\Psi(y)) \, \sqrt{\det \left( D\Psi^T(y) \cdot D\Psi(y) \right)} \, dy = \int_{V'} f(\Phi(x)) \, \sqrt{ \det \left( D\Phi^T(x) \cdot D\Phi(x) \right) } \, dx. $$
In other words, the invariance of the integral where changing charts follows from the regular change of variables rule. Note that if $k = n = d$, the independence of the definition on the chart is precisely the statement of the change of variables rule.

