Lecture 16: The Volume Measure and Integration on Manifolds (Contd.)
KSM3E04: Analysis on Manifolds — Fall 2025 | September 19, 12:00 PM–1:00 PM
In Lecture 11, we formally defined a $k$-manifold $M$ as a geometric subset of $\mathbf{R}^n$ that is locally Euclidean, described by an atlas of compatible charts. This intrinsic viewpoint frees us from relying on a single, global parametrization and allows us to study the manifold as a geometric object in its own right.
This raises a profound question: how do we measure the 'size'—the volume, area, or length—of such a curved object? Our goal today is to construct a canonical volume measure, denoted by $d\mu$, on $M$. This measure will be our ultimate "ruler" for manifolds. With it, we can define the integral of a scalar function, allowing us to calculate total quantities like mass, charge, or heat distributed over curved domains.
1. Constructing the Volume Measure on a Manifold
The Core Idea: Building a Globe from Flat Maps
Imagine you want to make a globe using flat pieces of paper. A single rectangular sheet cannot be wrapped around a sphere without tearing or stretching it too much. Instead, we cut the surface into smaller pieces, each of which can be flattened without too much distortion. These pieces are like the charts $U_i \subset \mathbf{R}^k$, and the way we place them on the manifold is given by the chart maps $\alpha_i : U_i \to M$.
Each flat piece has to stretch or compress slightly when we fit it onto the manifold. The amount of this local stretching is measured by the volume element $|V(D\alpha_i)|$. To compute the total volume, we add up contributions from each chart, weighted by how much of the manifold is covered there. Since charts overlap, we need a smooth way to divide up the contribution so that we don’t double-count regions. This is achieved by using a partition of unity, a collection of smooth functions $\{\phi_i\}$ that “split” the manifold into weighted regions subordinate to the open sets $\alpha_i(U_i) \subset M$.
The General Definition for Any Manifold
The following definition works for any $k$-dimensional manifold, compact or not.
Definition: The Volume Measure on a Manifold
Let $M$ be a $k$-manifold in $\mathbf{R}^n$. Suppose $\{\alpha_i: U_i \to M\}_{i \in I}$ is a countable atlas, and let $\{\phi_i\}_{i \in I}$ be a smooth partition of unity on $M$ such that $\text{supp}(\phi_i) \subseteq \alpha_i(U_i)$. The volume measure $\mu$ on $M$ is defined for any Borel set $S \subseteq M$ by:
This makes $(M, \mathcal{B}, \mu)$ into a measure space. The $k$-dimensional volume of $M$ is $v(M) = \mu(M)$, which may be infinite if $M$ is not compact.
A Crucial Simplification: The Compact Case
If $M$ is compact, things become simpler. Compactness ensures that we only need a finite atlas and a finite partition of unity. The infinite sum in the definition above then reduces to a finite sum, so the total volume $v(M)$ is guaranteed to be finite and easy to compute.
At first glance, the definition might seem to depend on the choice of atlas or the partition of unity. However, the following theorem shows that the volume measure is intrinsic to $M$ itself.
Theorem: The Volume Measure is Well-Defined
The value of $\mu(S)$ does not depend on which atlas or partition of unity is used. This follows because transition maps between charts are diffeomorphisms, as we proved in Lecture 11.
Proof (why the volume measure is independent of the atlas and partition of unity)
Let $M$ be a $k$-manifold in $\mathbf{R}^n$. Suppose we construct two measures:
- Using an atlas $\{\alpha_i: U_i \to M\}$ with partition of unity $\{\phi_i\}$, producing measure $\mu$.
- Using another atlas $\{\beta_j: V_j \to M\}$ with partition $\{\psi_j\}$, producing measure $\mu'$.
Step 1 — Insert the second partition into the first sum.
Since $\sum_j \psi_j = 1$ on $M$, we can write $\phi_i = \sum_j (\phi_i\psi_j)$. Substituting this into the definition of $\mu$ rewrites the integral as a double sum over $i$ and $j$.
Step 2 — Change variables on overlaps.
On the overlap of $\alpha_i(U_i)$ and $\beta_j(V_j)$, the transition map $\tau_{ij} = \beta_j^{-1} \circ \alpha_i$ is a diffeomorphism. A change of variables transfers the integral from $\alpha_i$-coordinates to $\beta_j$-coordinates, and the Jacobian ensures that the volume element transforms correctly.
Step 3 — Simplify the double sum.
After the change of variables, the double sum becomes \[ \sum_{j}\int_{V_j} (\chi_S \circ \beta_j)(y)\, \Big(\sum_i (\phi_i \circ \beta_j)(y)\Big)\, (\psi_j \circ \beta_j)(y)\, |V(D\beta_j(y))|\,dy. \] Since $\sum_i \phi_i = 1$, the inner sum reduces to $1$, leaving exactly the definition of $\mu'(S)$.
Conclusion
Thus $\mu(S) = \mu'(S)$ for every Borel set $S$, showing that the volume measure is intrinsic to $M$ and does not depend on the chosen atlas or partition of unity.
2. Integration of a Scalar Function
The "What" vs. The "How": Abstract Definition vs. Practical Computation
Now that we have a measure space $(M, \mu)$, we can define what it means to integrate a function. This allows us to calculate total quantities (like mass or charge) that are distributed unevenly over the manifold. We first define the integral abstractly, then provide the tools for its computation.
Definition: The Integral (Applies to all Manifolds)
Let $f: M \to \mathbf{R}$ be a continuous function. The integral of $f$ over $M$ is defined in the standard sense of Lebesgue integration with respect to our volume measure $\mu$:
This abstract definition needs a practical method for computation. The partition of unity framework provides the essential tool.
Theorem: Computation of the Integral
The integral $\int_M f \, d\mu$ is computed via the formula:
The function $f$ is called integrable if this sum converges absolutely (i.e., if $\int_M |f| \, d\mu < \infty$).
Defining the Infinite Sum: The Case of Non-Compact Manifolds
If $M$ is not compact, the sum above is an infinite series. As with improper integrals in calculus, this series needs a formal definition via a limit. The method of exhaustion by compact sets provides the rigorous framework.
Definition: Integration via Exhaustion
Let $\{K_j\}_{j=1}^\infty$ be an exhaustion of $M$ by a sequence of nested compact sets. The value of the infinite sum is defined as the limit of finite sums:
On each compact set $K_j$, the integral is a well-defined finite sum because $K_j$ only intersects a finite number of the supports of the $\{\phi_i\}$. The overall integral exists only if this limit converges.
Worked Example: A Simple Case (Compact Manifold)
Let's apply this theory to the simplest case: a compact manifold where the integral can be computed directly without needing infinite sums or limits. The hemisphere $M$ can be covered by a single chart, so the computational formula simplifies beautifully.
- Step 1: Choose a Chart. The chart $\alpha: U \to M$ maps $U = [0, \pi/2] \times [0, 2\pi]$ to the hemisphere: $$ \alpha(\phi, \theta) = (R\sin\phi \cos\theta, R\sin\phi \sin\theta, R\cos\phi) $$
- Step 2: Compute the Volume Element. For a sphere of radius $R$, $|V(D\alpha)| = R^2 \sin\phi$.
- Step 3: Pull Back the Function. $f(x,y,z)=z$ becomes: $$ (f \circ \alpha)(\phi, \theta) = z(\phi,\theta) = R\cos\phi $$
- Step 4: Solve the Integral. $$ \int_M z \, d\mu = \int_U (f \circ \alpha) \cdot |V(D\alpha)| \, dx = \int_0^{2\pi} \int_0^{\pi/2} (R\cos\phi) \cdot (R^2 \sin\phi) \, d\phi \, d\theta = \pi R^3 $$
3. Problems for Practice
Exercise 1: Center of Mass (Compact Manifold)
This problem applies the direct computational method for a compact manifold. Consider the uniform hemispherical shell $M$ from our example. Its center of mass $(\bar{x}, \bar{y}, \bar{z})$ is given by $\bar{z} = \frac{1}{\text{Area}(M)} \int_M z \, d\mu$.
- Using the integral calculated in the lecture, find the $\bar{z}$ coordinate of the center of mass.
- Explain by symmetry why the coordinates $\bar{x}$ and $\bar{y}$ must be zero.
Exercise 2: Integration on a Non-Compact Manifold
This exercise is a practical application of the exhaustion method required for non-compact manifolds. Consider $M = \mathbf{R}^2$ and $f(x,y) = e^{-(x^2+y^2)}$. Using the exhaustion by disks $K_j = \{(x,y) \mid x^2+y^2 \le j^2\}$, show that $f$ is integrable and compute $\int_{\mathbf{R}^2} f \, d\mu$.
Exercise 3: Integration and Topology (Compact Manifold)
This problem returns to a compact manifold to hint at a deep connection between analysis and topology. The Gaussian curvature $K$ is a function on a 2-manifold. For the torus $T$, $K = \frac{\cos\theta}{r(R+r\cos\theta)}$.
Calculate the total curvature $\int_T K \, d\mu$. The Gauss-Bonnet theorem states that for any compact surface $M$, $\int_M K \, d\mu = 2\pi \chi(M)$, where $\chi(M)$ is the Euler characteristic (for a torus, $\chi(T)=0$). Does your calculation agree?