Lecture 10: From Euclidean Space to Manifolds

KSM3E04: Analysis on Manifolds — Fall 2025   |   September 12, 3:00–4:30 PM

In calculus, one studies the notion of length not only for subsets of $\mathbf{R}^1$, but also for smooth curves in $\mathbf{R}^2$ and $\mathbf{R}^3$. And one studies the notion of area not only for subsets of $\mathbf{R}^2$, but also for smooth surfaces in $\mathbf{R}^3$.

In this lecture, we introduce the $k$-dimensional analogues of curves and surfaces; they are called parametrized $k$-manifolds in $\mathbf{R}^n$. Our first step is to develop a consistent way to measure the "volume" of flat, $k$-dimensional objects in a higher-dimensional space. With this foundational tool, we can then rigorously define the volume of these more general curved objects and the integral of a function over them.

1. The $k$-Dimensional Volume of a Parallelopiped

Our goal is to define what it means for a $k$-dimensional object to have “volume” inside a higher-dimensional space $\mathbf{R}^n$. For example, a line segment ($k=1$) in $\mathbf{R}^3$ clearly has a length, and a parallelogram ($k=2$) floating in $\mathbf{R}^4$ should also have a natural notion of area—even though in the full $n$-dimensional sense these objects have measure zero. The challenge is to capture their intrinsic $k$-dimensional size.

To make progress, let us restrict attention to the simplest $k$-dimensional shapes: parallelopipeds spanned by $k$ linearly independent vectors in $\mathbf{R}^n$. We seek a function $V$ that assigns to each such parallelopiped a non-negative real number $V(P)$, its $k$-dimensional volume.

What conditions must $V$ satisfy? Two natural requirements are enough:

Theorem: Existence and Uniqueness of $k$-Volume

There exists a unique function $V$ assigning to every $k$-parallelopiped $P \subset \mathbf{R}^n$ a number $V(P) \geq 0$ such that:

  1. Invariance under isometries: $V(P)$ is unchanged if $P$ is rotated or reflected by an orthogonal transformation of $\mathbf{R}^n$.
  2. Agreement with standard volume: If $P$ lies entirely in the coordinate subspace $\mathbf{R}^k \times \{0\}$, then $V(P)$ is exactly the usual $k$-dimensional Euclidean volume of $P$ in $\mathbf{R}^k$, i.e. the absolute value of the determinant of its spanning vectors.
Click to expand the proof of the theorem

Let $P$ be spanned by vectors $\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ and form the matrix $X$ with these columns. Define $V(P) = \sqrt{\det(X^T X)}$. We check the two properties:

  1. Consistency: If $P \subset \mathbf{R}^k \times \{0\}$, then $X$ is block-structured with a $k \times k$ square matrix on top and zeros below. In this case $V(P) = |\det(X_{\text{square}})|$, the standard $k$-dimensional volume.
  2. Invariance: If we apply an orthogonal matrix $A$, then $P$ is spanned by $AX$. The formula gives $V(AP) = \sqrt{\det((AX)^T(AX))} = \sqrt{\det(X^T A^T A X)} = V(P)$.
  3. Uniqueness: Any $k$-dimensional subspace can be rotated into $\mathbf{R}^k \times \{0\}$. By invariance, $V(P)$ must equal $V(AP)$, and by consistency, $V(AP)$ is fixed by the standard determinant formula. Thus no other definition is possible.

Definition: $k$-Dimensional Volume (Theoretical Formula)

The unique function $V$ described above is called the $k$-dimensional volume. For a parallelopiped $P = \{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ we have

$$V(P) = \sqrt{\det(X^T X)},$$

where $X$ is the $n \times k$ matrix whose columns are the spanning vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$.

A useful perspective is to study the effect of a linear map. Let $T$ be an invertible $n \times n$ matrix. If we apply $T$ to the unit $k$-parallelopiped in $\mathbf{R}^k$, the result is another parallelopiped whose $k$-volume we denote by $V(T)$. Thus $V(T)$ measures exactly how $T$ distorts $k$-dimensional volume. It can be seen that $V(T)$ connects directly to determinants and to the change of variables formula.

2. Parametrized Manifolds, Volume, and Integration

We now generalize the notions of arc length, surface area, and higher-dimensional volume. The guiding idea is that curved objects can be described by a smooth parametrization from a flat domain, and the volume can be recovered by transferring the familiar Lebesgue measure on flat space to the manifold.

Definition: Parametrized $k$-Manifold

Let $A \subset \mathbf{R}^k$ be open, and let $\alpha: A \to \mathbf{R}^n$ be a $C^1$ map. The image $Y = \alpha(A)$, together with the parametrization $\alpha$, is called a parametrized $k$-manifold and is denoted by $Y_{\alpha}$.

From Lebesgue Measure to Volume Measure

The domain $A$ carries the standard Lebesgue measure $m$ on $\mathbf{R}^k$. To transfer this structure to $Y$, we define a measure $\mu$ on subsets of $Y$ as follows. For $S \subseteq Y$ measurable, set

$$ \mu(S) = \int_{\alpha^{-1}(S)} |V(D\alpha(x))| \, dm(x). $$

The function $x \mapsto |V(D\alpha(x))|$ serves as the Radon–Nikodym derivative that describes how measure transforms under the map $\alpha$. Intuitively, it is the local “stretching factor” that converts flat volume on $A$ into curved volume on $Y$.

Definition: Volume of a Parametrized Manifold

The $k$-dimensional volume of $Y_{\alpha}$ is the total measure of $Y$ under $\mu$. By definition:

$$ v(Y_{\alpha}) = \mu(Y) = \int_{A} |V(D\alpha(x))| \, dm(x). $$

This formula tells us that the volume of a parametrized manifold is obtained by integrating the Jacobian determinant of the parametrization over the flat domain $A$.

Integrating Functions on the Manifold

The construction of $\mu$ provides more than just a way to measure total volume; it equips $Y$ with a full-fledged volume measure. This makes it possible to integrate scalar functions defined on $Y$. Let $f: Y \to \mathbf{R}$ be continuous. The integral of $f$ over $Y_{\alpha}$ is defined in the abstract sense as

$$ \int_{Y_{\alpha}} f \, d\mu. $$

By construction of $\mu$, this integral can be computed concretely by pulling everything back to $A$:

$$ \int_{Y_{\alpha}} f \, d\mu = \int_{A} (f \circ \alpha)(x) \cdot |V(D\alpha(x))| \, dm(x). $$

Thus integration on $Y$ reduces to an ordinary Lebesgue integral on the flat domain, with the Jacobian determinant providing the density correction.

Invariance Under Reparametrization

A natural concern is whether this construction depends on the choice of parametrization. Suppose $Y$ is described by two different parametrizations $\alpha$ and $\beta$. Do they yield the same measure and hence the same integrals? The following theorem shows that the answer is yes: the construction depends only on the geometry of $Y$, not on the chosen parametrization.

Theorem: Invariance of the Integral Under Reparametrization

Let $g: A \to B$ be a diffeomorphism between open sets in $\mathbf{R}^k$, and let $\beta: B \to \mathbf{R}^n$ be $C^1$. Define $\alpha = \beta \circ g$. Then both maps trace out the same image $Y = \alpha(A) = \beta(B)$.

For any continuous function $f: Y \to \mathbf{R}$, the integrals with respect to the two parametrizations coincide:

$$ \int_{Y_{\alpha}} f \, d\mu = \int_{Y_{\beta}} f \, d\mu. $$

In particular, the volume itself is independent of parametrization: $$ v(Y_{\alpha}) = v(Y_{\beta}). $$

Click to expand the proof of the theorem

The proof combines the chain rule for derivatives with the Change of Variables theorem for diffeomorphisms.

  1. Relating the densities: Since $\alpha = \beta \circ g$, the Jacobians satisfy
    $$ |V(D\alpha(x))| = |V(D\beta(g(x)))| \cdot |\det(Dg(x))|. $$
  2. Set up the integral:
    $$ \int_{Y_{\alpha}} f \, d\mu = \int_{A} (f \circ \alpha)(x) \, |V(D\alpha(x))| \, dm(x). $$
  3. Substitute the density relation:
    $$ = \int_{A} (f \circ \beta \circ g)(x) \, \big( |V(D\beta(g(x)))| \cdot |\det(Dg(x))| \big) \, dm(x). $$
  4. Apply Change of Variables: With $y = g(x)$, the integral becomes
    $$ \int_{B} (f \circ \beta)(y) \, |V(D\beta(y))| \, dm(y). $$
  5. Conclude: This is exactly the integral of $f$ over $Y_{\beta}$. Thus the construction is invariant under reparametrization.

4. Problems for Practice

Exercise 1: Alternate Formula for $k$-Volume

Use the Cauchy–Binet formula to show that for any $n \times k$ matrix $X$, $$\det(X^T X) = \sum_{[I]} (\det X_I)^2,$$ and verify this identity by computing the area of the parallelogram spanned by $\mathbf{v}_1 = (1,1,0)$ and $\mathbf{v}_2 = (0,1,1)$ in $\mathbf{R}^3$.

Exercise 2: Volume of the Torus

A torus $T$ in $\mathbf{R}^3$ is a classic parametrized 2-manifold, given by $\alpha(\theta, \phi) = ((R + r\cos\theta)\cos\phi, (R + r\cos\theta)\sin\phi, r\sin\theta)$ for parameters $\theta, \phi \in [0, 2\pi]$ and constants $0 < r < R$. Its volume (surface area) is given by the integral of its volume element.

Calculate the surface area of the torus by setting up and solving the integral $v(T_{\alpha}) = \int_0^{2\pi} \int_0^{2\pi} V(D\alpha) \,d\theta d\phi$. Your final answer should be $4\pi^2 Rr$.

Exercise 3: Homogeneity of the Volume Function

This problem verifies that our definition of $k$-volume behaves as expected under scaling. Let $P$ be a $k$-parallelopiped spanned by vectors $\{\mathbf{v}_1, \dots, \mathbf{v}_k\}$. For a scalar $c > 0$, let $cP$ be the parallelopiped spanned by the scaled vectors $\{c\mathbf{v}_1, \dots, c\mathbf{v}_k\}$.

Prove directly from the theoretical definition of $k$-volume, $V(P) = \sqrt{\det(X^T X)}$, that $V(cP) = c^k V(P)$. This confirms that scaling a $k$-dimensional object by a factor of $c$ scales its $k$-volume by $c^k$.

Problem: The Schwarz Lantern Paradox

Let \(A = (0,1) \times (0,1)\) be the open unit square in \(\mathbb{R}^2\). For each integer \(n \ge 1\), consider the sequence of smooth parametrizations \(\alpha_n: A \to \mathbb{R}^3\) given by:

\[ \alpha_n(u,v) = \left( u, v, \frac{1}{2^n} \sin(2\pi \cdot 4^n u) \right) \]

Let \(Y_n = \alpha_n(A)\) denote the parametrized 2-manifold (surface) for each \(n\). The 2-dimensional volume (surface area) of \(Y_n\) is denoted by \(v(Y_n)\).

  1. (a) Derive the explicit integral expression for the surface area, \(v(Y_n)\), of the manifold \(Y_n\). Simplify the resulting integrand as much as possible. [3]
  2. (b) Determine the pointwise limit of the sequence of parametrizations, defined as \(\alpha_\infty(u,v) = \lim_{n\to\infty} \alpha_n(u,v)\). Describe the resulting geometric set \(Y_\infty = \alpha_\infty(A)\) and state its elementary surface area. [3]
  3. (c) Prove that the limit of the surface areas of the manifolds \(Y_n\) diverges to infinity. That is, prove:
    \[ \lim_{n\to\infty} v(Y_n) = \infty \]
    [3]
  4. (d) Reconcile the results of parts (b) and (c). Explain this paradox by analyzing the behavior of the volume element, \(\sqrt{\det(D\alpha_n^T D\alpha_n)}\), as \(n \to \infty\). Your explanation should clarify why the derivatives of the parametrizations prevent the area of the limiting surface from being the limit of the sequence of areas. [4]

Conclusion: This problem demonstrates the Schwarz Lantern paradox, which reveals that surface area is not a continuous functional. It proves that the convergence of a sequence of shapes does not guarantee the convergence of their areas, because the volume integral is fundamentally sensitive to the limiting behavior of the parametrization's derivatives, not just the limiting points of the set itself.

← Lecture 9 Lecture 11 →