Understanding Integration on an Orientable Manifold

[Silviu]

Hello! I am reading how to integrate on an orientable manifold. So we have $f:M \to R$ and an m-form (m is the dimension of M): $\omega = h(p)dx^1 \wedge ... \wedge dx^m$, where $h(p)$ is another function on the manifold which is always positive as the manifold is orientable. The way integral is defined is like this: $\int_{U_i} f\omega = \int_{\phi(U_i)}f(\phi^{-1}(x))h(\phi^{-1}(x))dx^1...dx^m$, where $U_i$ are the coordinate neighborhoods and $\phi$ is the mapping from M to $R^m$. The definition makes sense, intuitively, by making an analogy with the surface or volume integrals. However I am not sure formally, how did the wedge product of $\omega$, transformed into simple multiplication. Can someone explain this to me? Thank you!

 

[Orodruin]

In general, the integration of an $r$-form $\omega$ over an $r$-dimensional sub-manifold parametrised by coordinates $t_1, \ldots t_r$ on the sub-manifold is defined as
$$
\int \omega = \int \omega(\dot \gamma_1, \ldots, \dot \gamma_r) dt_1 \ldots dt_r,
$$
where $\dot\gamma_i$ are the tangent vectors of the corresponding coordinate lines. In your case, you use the coordinates $x^i$ for the integral as well as for the basis and therefore by the definition
$$
\int f\omega = \int f(\phi^{-1}(x))h(\phi^{-1}(x)) [dx^1\wedge\ldots \wedge dx^m](\partial_1,\ldots, \partial_m) dx^1 \ldots dx^m
$$
where we have used that the tangent vector of the coordinate line of $x^i$ is just $\partial_i$. Now, it holds that $[dx^1\wedge\ldots \wedge dx^m](\partial_1,\ldots, \partial_m) = 1$ and from there you obtain your sought result.