# Naive Question About Volume Forms and Integration

So given a compact $$n$$ dimensional Riemannian manifold $$(M,g)$$ there exists a volume form $$\omega$$ given in coordinates by: $$\omega=\det{g}dx^1\wedge\cdots\wedge dx^n$$ That allows us to define the volume on $$M$$ via: $$\text{vol}(M)=\int_M\omega$$ However, when actually trying to calculate the volume of $$M$$ we need a partition of unity to actually integrate over all of $$M$$...except it appears that sometimes we do not. For example, take $$S^2$$ with the induced metric from $$\mathbb{R}^3$$, then we have that with the angle parameterization of $$S^2$$: $$\omega=\sin\theta d\theta\wedge d\phi$$ in the chart $$\theta\in(0,\pi)$$, $$\phi\in (0,2\pi)$$. This covers $$\textit{most}$$ of the manifold, and naively integrating this volume form over $$M$$ as we would in multivariable calculus: $$\int_M\omega=\int_0^\pi\int_0^{2\pi}\sin\theta d\phi d\theta$$ $$=4\pi$$ which is the volume, or in this case the surface area, of $$S^2$$. I suspect this works because the chart is "good enough" at getting "most" of the manifold, but I'm unsure of how to make it precise. Is it perhaps because the arc that isn't covered by this parameterization has volume zero? My knowledge on measure theory is quite lacking so I'm unsure of what's actually going on here...

• Indeed, the volume form induces a measure on the sphere, and one can always remove subsets of measure zero when computing integrals with respect to a measure. Aug 31, 2022 at 19:07
• Great thank you, that was what I was hoping for. Aug 31, 2022 at 19:59
• btw, you have a typo, it should be $\sqrt{\det g}$. Sep 1, 2022 at 0:09

Suppose $$\omega$$ is any (Lebesgue-measurable) $$n$$-form on an $$n$$-dimensional smooth manifold $$M$$. Then, there is a unique positive measure $$\mu_{\omega}$$ defined on the Lebesgue $$\sigma$$-algebra of $$M$$, such that for any chart $$(U,x)$$ and any Lebesgue-measurable set $$A\subset M$$, we have \begin{align} \mu_{\omega}(A\cap U)&=\int_{x[A\cap U]}|f_{(x)}\circ x^{-1}|\,d\lambda_n,\tag{*} \end{align} where $$\omega=f_{(x)}\,dx^1\wedge\cdots\wedge dx^n$$ on $$U$$, and $$\lambda_n$$ is the $$n$$-dimensional Lebesgue measure on $$\Bbb{R}^n$$. Note that such measures can be defined regardless of whether or not $$\omega$$ is a volume form (i.e regardless of whether or not it is nowhere-vanishing).

A measure $$\mu$$ which arises in this fashion has the property that $$\mu(S)=0$$ for any submanifold $$S\subset M$$ of strictly smaller dimension. The proof of this is pretty simple. For each point $$p\in S$$, choose a chart $$(U_p,x_p)$$ of $$M$$ around $$p$$ with the submanifold property for $$S$$, i.e that $$x_p[S\cap U_p]=x_p[U_p]\cap (\Bbb{R}^k\times \{0\})$$ whereby $$k=\dim S$$. Due to the second-countability assumption of manifolds, we can extract a countable collection of submanifold charts $$(U,x)$$ whose domains cover $$S$$. As a result of having countably many charts, it thus suffices to prove that $$\mu(S\cap U)=0$$ for such charts (because due to countable (sub)-additivity of measures, $$\mu(S)\leq\sum \mu(U\cap S)=0$$, where the sum is over the countable many charts which cover $$S$$). But now, this is very trivial from the defining property $$(*)$$ above because \begin{align} \mu(S\cap U)&=\int_{x[U]\cap (\Bbb{R}^k\times\{0\})}(\text{some non-negative function})\,d\lambda_n=0, \end{align} because we’re integrating a function over a subset of Lebesgue $$n$$-dimensional measure zero. This completes the proof.

Note that this is a very convenient criterion for integration, and it justifies a lot of the calculations often made. In your case, the portion left out by the spherical coordinate parametrization is a certain arc of a great circle, which is obviously contained in a $$1$$-dimensional submanifold. Hence, for the purposes of integration, it is negligible.

You write

However, when actually trying to calculate the volume of $$M$$ we need a partition of unity to actually integrate over all of $$M$$...except it appears that sometimes we do not.

Well yes, we use a partition of unity to define integration of differential forms, but no one ever uses them in concrete calculations. Partitions of unity are only a (very useful) technical tool used for proving theorems (eg Stokes) because it allows us to chop up functions/tensor fields/differential forms defined on the manifold, rather than chopping up the domains. However, no one uses them for concrete calculations. Often, we get by with one (or two at worst, in my experience) coordinate charts which cover ‘most’ of the manifold, in the sense that the complement is contained in a set of measure zero.

Let me now remark that although I defined a measure arising from a differential form above, I didn’t really have to (I only did it to show you some of the details since you mention you’re not too adept with measure theory). I didn’t have to define the measure, because the notion of measure-zero can be introduced in a very elementary way:

A subset $$A\subset M$$ is said to have $$n$$-dimensional Lebesgue measure zero in $$M$$ if for every coordinate chart $$(U,x)$$, we have that $$x[A\cap U]$$ has $$n$$-dimensional Lebesgue measure zero in $$\Bbb{R}^n$$.

For completeness, let me recall that a set $$B\subset\Bbb{R}^n$$ has $$n$$-dimensional Lebesgue measure zero if for every $$\epsilon>0$$, there are countably many rectangles $$\{Q_i\}$$ which cover $$B$$, such that $$\sum_i\text{vol}(Q_i)<\epsilon$$. With this definition, it is also easy to show that lower-dimensional submanifolds have measure-zero.