Calculating the volume element

Question

Let $M$ be a smooth oriented $k$-manifold on $\mathbb{R}^n$, with a parametrization $g\colon U \to \mathbb{R}^n$ where $U \subseteq \mathbb{R}^k$. Suppose we have a continuous function $f\colon \mathbb{R}^n \to \mathbb{R}$ with compact support, and we wish to integrate it over $M$. We want then to calculate:

$$\int_M f dV$$

Where $dV$ is a volume element, a differential $k$-form particular to this manifold $M$. I've read on wikipedia that the pullback of the volume element is given by:

$$g^*(dV) = \left|g'^T \cdot g'\right|dx_1 \wedge\cdots \wedge dx_k$$

Where $dx_i$ are the usual projection forms in $\mathbb{R}^k$ and $g'$ is the Jacobian matrix of $g$. I would like to understand why though. I'm not even sure what a formal definition of the volume element of a $k$-manifold on $\mathbb{R}^n$ is.

Sure, if $k = n$, then this will simply be $dV = dx_1\wedge \cdots \wedge dx_n$ with the projections happening in $\mathbb{R}^n$. The pullback is easy to calculate then, as it commutes with exterior products and exterior derivatives. Outside this realm, I'm quite lost on what's going on. What is the length element of a curve in $\mathbb{R}^2$? Or the area element of a surface in $\mathbb{R}^3$? How does all of this generalizes to $k$-manifolds over the $n$-space?

Answer 1

The volume form on $M^k$ is a $k$-form ${\rm vol}_M$ such that $({\rm vol}_M)_p(v_1,\cdots,v_k) = 1$ for every orthonormal positive basis $(v_1,\ldots,v_k)$ of $T_pM$. The volume form always exists locally, but $M$ being oriented ensures that we have this globally.

If $C\subseteq \Bbb R^n$ is a curve (read "$1$-dimensional orientable submanifold), then for each parametrization $\alpha:I \to C$ we have $(\alpha^\ast({\rm vol}_C))_t = \|\alpha'(t)\|\,{\rm d}t$. This $\alpha^\ast({\rm vol}_C)$ is what people usually call ${\rm d}s$, as in ${\rm d}s = \|\alpha'(t)\|\,{\rm d}t$. You have to be careful with the identification, most often than not we'll have some pull-back to be understood here. I can give you two more interpretations:

If $S \subseteq \Bbb R^2$ is a surface oriented by an unit normal field $n = (n_x,n_y,n_z)$, then you can write $${\rm vol}_S = n_x\,{\rm d}y \wedge{\rm d}z + n_y \,{\rm d}z\wedge {\rm d}x + n_z\,{\rm d}x \wedge {\rm d}y,$$and this actually corresponds to $({\rm vol}_S)_p(v,w) = \langle v \times w, n(p)\rangle$.

If $M^{n-1}\subseteq \Bbb R^n$ is a hypersurface oriented by an unit normal $(n_1,\ldots, n_n)$, you can do the same as above and write $${\rm vol}_M = \sum_{k=1}^n (-1)^{k-1} n_k \,{\rm d}x_1 \wedge \cdots\wedge \widehat{{\rm d}x_k}\wedge \cdots {\rm d}x_n,$$and the analogous formula $({\rm vol}_M)_p(v_1,\ldots,v_{n-1}) = \langle v_1 \times\cdots\times v_{n-1},n(p)\rangle$.

I think that Spivak's Calculus on Manifolds explains that well. To give formulas for the volume form in terms of normal fields you need some more sofisticated stuff such as Hodge Duality, but I guess this'll be enough for now.