Htykuut

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 $Phi colon Omega rightarrow Phi(Omega)subseteqmathbb{R}^d$ a diffeomorphism. The function $f$ is integrable on $Phi(Omega)$ if and only if $xmapsto f(Phi(x))|det(DPhi(x))|$ is on $Omega$. It then holds that

$displaystyle int_{Phi(Omega)}f(y),mathrm{d}y=int_{Omega} f(Phi(x))|det(DPhi(x))|,mathrm{d}x$

where $DPhi(x)$ 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 $Psicolon Trightarrow Vsubseteq M, Tsubseteqmathbb{R}^k$ 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?

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

$$ int_{M} f , dS := int_V f(Psi(y)) , sqrt{det (DPsi^T(y)cdot DPsi(y))} , dy. $$

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

$$ int_V f(Psi(y)) , sqrt{det left( DPsi^T(y) cdot DPsi(y) right)} , dy = int_{V'} f(Psi(g(x))) sqrt{det left( DPsi^T(g(x)) cdot DPsi(g(x)) right)} left| det Dg(x) right| dx $$

Note that

$$ Psi(g(x)) = Psi(Psi^{-1}(Phi(x))) = Phi(x), \
f(Psi(g(x)) = f(Phi(x)), \
DPhi(x) = DPsi(g(x)) cdot Dg(x), \
sqrt{det left( DPsi^T(g(x)) cdot DPsi(g(x)) right)} left| det Dg(x) right| = \
sqrt{ det(Dg^T(x)) det left( DPsi^T(g(x)) cdot DPsi(g(x)) right) det(Dg(x)) } = \
sqrt{det left( Dg^T(x) cdot DPsi^T(g(x)) cdot DPsi(g(x)) cdot Dg(x) right)} = \
sqrt{ det left( DPhi^T(x) cdot DPhi(x) right) } $$

and so we also get

$$ int_V f(Psi(y)) , sqrt{det left( DPsi^T(y) cdot DPsi(y) right)} , dy = int_{V'} f(Phi(x)) , sqrt{ det left( DPhi^T(x) cdot DPhi(x) right) } , dx. $$

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.