Lecture 11: The Intrinsic Geometry of Manifolds

KSM3E04: Analysis on Manifolds — Fall 2025   |   September 15, 4:30–6:00 PM

Our journey so far has focused on parametrized-manifolds, where the shape and the map describing it were intertwined. Today, we take a significant step forward by liberating the geometry from its description. We will define a manifold not as a map, but as the underlying geometric set itself—a space that can be viewed through many different local lenses, which we call charts.

This new perspective allows us to ask a deeper question: What properties belong to the manifold itself, independent of how we choose to look at it? To answer this, we will build the essential tools for studying local geometry: the tangent space and the volume element. Our ultimate goal is to use these local pieces to define a global notion of volume and to prove the central theorem of this lecture: that this volume is an intrinsic property of the manifold, a true measure of the set that remains unchanged no matter which charts we use for the calculation.

1. What is a Manifold?

Intuition: From Globes to Geometry

The core idea of a manifold is that it's a space that is locally simple but may be globally complex. The best analogy is the Earth: while we know it's a sphere, our immediate surroundings appear as a flat plane. We can navigate our local area using a simple, flat map.

In mathematics, we call such a local, flat map a chart. A chart takes a piece of our curved space and gives it coordinates from a simple, flat Euclidean space ($\mathbf{R}^k$). Just as no single flat map can accurately represent the entire spherical Earth without distortion or tearing, no single chart can typically cover a whole manifold. Therefore, we need a collection of charts that overlap, covering the entire space. This collection is called an atlas.

The Formal Definition

The following definition formalizes this idea. It states that for any point on our set, we can find a chart that makes the neighborhood of that point look like a piece of flat space.

Definition: $k$-Manifold of Class $C^r$ in $\mathbf{R}^n$

A subset $M \subset \mathbf{R}^n$ is a $k$-dimensional manifold of class $C^r$ if, for each point $p \in M$, there exists a chart consisting of:

  • an open set $U \subset \mathbf{R}^k$ (the flat "coordinate domain"),
  • an open neighborhood $V \subset \mathbf{R}^n$ of $p$ (a "local window" in the ambient space),
  • a map $\alpha: U \to V \cap M$ (the "coordinate system"),

such that the map $\alpha$ satisfies three crucial conditions:

  1. Regularity: $\alpha$ is a $C^r$ map (it's smooth enough for calculus).
  2. Homeomorphism: $\alpha$ is a bijection onto its image with a continuous inverse (it preserves local topology without tearing or gluing).
  3. Full Rank: The derivative $D\alpha(x)$ has rank $k$ everywhere (it's a non-degenerate parametrization).

Dissecting the Conditions: Why Each One Matters

Dropping any of these conditions allows for "pathological" spaces that we don't want to consider manifolds.

  • Regularity ($C^r$) is the bedrock of calculus. Without it, we couldn't meaningfully define derivatives on the manifold. A shape with a sharp corner, like the absolute value graph $|x|$, is continuous but not smooth at the origin.
  • The Homeomorphism condition ensures the chart is a faithful representation of the local space. A failure of this condition occurs in the "figure-eight" curve at its self-intersection. Any neighborhood of the crossing point on the curve doesn't look like a simple line segment; it looks like two lines crossing. No chart can map a simple interval to that crossing point in a way that is one-to-one and has a continuous inverse.
  • The Full Rank condition prevents the parametrization from collapsing dimensions. Consider the curve given by $\alpha(t) = (t^2, t^3)$. At $t=0$, the derivative $D\alpha(0) = (0,0)$ has rank 0, not 1. This corresponds to the "cusp" at the origin, a point where the curve isn't locally flat. A valid chart must be an "immersion," never stopping or turning back on itself in a degenerate way.

Ensuring Consistency: The Role of Transition Maps

To perform calculus on a manifold $M$, we must use charts to translate problems into the familiar setting of $\mathbf{R}^k$. For instance, to decide if a function $f: M \to \mathbf{R}$ is "smooth," we examine its composition with a chart, $f \circ \alpha: U \to \mathbf{R}$. But this raises a critical problem: what if our choice of chart affects the answer? If $f \circ \alpha$ is smooth, can we be sure that for another overlapping chart $\beta$, the function $f \circ \beta$ is also smooth? If not, the very notion of smoothness on $M$ would be ambiguous and ill-defined.

This is where transition maps become essential. A transition map $g = \alpha^{-1} \circ \beta$ is the "conversion formula" between the coordinate system of $\beta$ and that of $\alpha$. We can relate the two compositions by the identity: $$ f \circ \beta = f \circ (\alpha \circ (\alpha^{-1} \circ \beta)) = (f \circ \alpha) \circ g $$ The proof below shows that our definition of a manifold forces $g$ to be a $C^r$ map (in fact, a diffeomorphism). This is the key. Since $g$ is smooth, composing with it preserves smoothness. Therefore, $f \circ \beta$ is smooth if and only if $f \circ \alpha$ is smooth.

The smoothness of transition maps provides the fundamental coherence needed for an atlas. It guarantees that all charts are compatible, allowing us to define intrinsic properties of the manifold without ambiguity. This consistency is what elevates a manifold from being a mere collection of parametrized pieces into a single, unified geometric object on which we can do calculus.

A Proof that Transition Maps are $C^r$

Let $\alpha: U_1 \to M$ and $\beta: U_2 \to M$ be two $C^r$ charts whose images overlap. We want to show the transition map $g = \alpha^{-1} \circ \beta$ is $C^r$ on its domain. The strategy is to show that $\alpha^{-1}$ can be locally extended to a $C^r$ map defined on an open set in the ambient space $\mathbf{R}^n$.

  1. Set up the problem. Let $p$ be a point in the overlap of the chart images, with $p = \alpha(x_0)$ for some $x_0 \in U_1$. Our goal is to find a $C^r$ map $\Phi$ defined on a neighborhood of $p$ in $\mathbf{R}^n$ such that $\Phi$ agrees with $\alpha^{-1}$ on the manifold.
  2. Use the rank condition. Since $D\alpha(x_0)$ is an $n \times k$ matrix of rank $k$, it has $k$ linearly independent rows. By reordering the coordinates in $\mathbf{R}^n$ (which is a smooth operation), we can assume without loss of generality that the first $k$ rows are linearly independent.
  3. Define a projection. Let $\pi: \mathbf{R}^n \to \mathbf{R}^k$ be the projection onto the first $k$ coordinates: $\pi(y_1, \dots, y_n) = (y_1, \dots, y_k)$.
  4. Apply the Inverse Function Theorem. Consider the composite map $h = \pi \circ \alpha: U_1 \to \mathbf{R}^k$. By the Chain Rule, its derivative at $x_0$ is $Dh(x_0) = D\pi(\alpha(x_0)) \circ D\alpha(x_0)$. This is simply the $k \times k$ matrix formed by the first $k$ rows of $D\alpha(x_0)$. By our assumption, this matrix is invertible. The Inverse Function Theorem now applies to $h$. Thus, there is a local $C^r$ inverse $h^{-1}$ defined in a neighborhood of $h(x_0) = \pi(p)$.
  5. Construct the local extension. For any point $q$ on the manifold near $p$, we have $q = \alpha(x)$ and $\pi(q) = \pi(\alpha(x)) = h(x)$. Applying the local inverse gives $x = h^{-1}(\pi(q))$, which is just $\alpha^{-1}(q) = h^{-1}(\pi(q))$. We can now define our extension $\Phi$ for any point $y$ in a neighborhood of $p$ in $\mathbf{R}^n$ by the formula: $$\Phi(y) = h^{-1}(\pi(y))$$ Since $\pi$ and $h^{-1}$ are $C^r$ maps, their composition $\Phi$ is a $C^r$ local extension of $\alpha^{-1}$.
  6. Conclude. The transition map is $g = \alpha^{-1} \circ \beta$. Locally, this is equivalent to $\Phi \circ \beta$. Since both $\Phi$ and $\beta$ are $C^r$, their composition is $C^r$. This proves that all transition maps are smooth.

Key Examples

Example 1: Graphs of Functions

If $f: U \subset \mathbf{R}^k \to \mathbf{R}^{n-k}$ is a $C^r$ function, its graph $M = \{(x, f(x)) \mid x \in U\}$ is a $k$-manifold in $\mathbf{R}^n$. The single chart $\alpha(x) = (x, f(x))$ covers the entire set. Its derivative is $$D\alpha(x) = \begin{pmatrix} I_k \\ Df(x) \end{pmatrix}$$ which always has rank $k$ because the identity matrix $I_k$ sits at the top. This is the simplest way manifolds appear.

Example 2: The Sphere $S^2$

The unit sphere $S^2 = \{ (x,y,z) \in \mathbf{R}^3 \mid x^2+y^2+z^2=1 \}$ is a classic 2-manifold. It can be covered by an atlas of six "graph charts" (for the upper/lower, front/back, left/right hemispheres). Another common and powerful chart is stereographic projection, which maps the entire sphere except for one point to the flat plane $\mathbf{R}^2$. Two such charts, one from the north pole and one from the south, are sufficient to form a complete atlas for the sphere.

3. Manifolds with Boundary

Motivation: Including Objects with Edges

Our current definition of a manifold describes idealized, edgeless spaces like a sphere or a torus. However, many important objects in physics and mathematics have smooth edges—for example, a closed disk, a cylinder with its ends, or a hemisphere. Our goal is to extend our framework to include these objects in a rigorous way. The key idea is to allow some points to be modeled not on all of $\mathbf{R}^k$, but on the edge of a flat space.

The Model Space: The Upper Half-Space

The simplest space with a flat boundary is the upper half-space, defined as: $$ \mathbf{H}^k = \{ (x_1, \dots, x_k) \in \mathbf{R}^k \mid x_k \ge 0 \} $$ This space serves as our new model for local neighborhoods. Points in $\mathbf{H}^k$ with $x_k > 0$ are interior points, as they have full open neighborhoods around them. Points with $x_k = 0$ lie on the boundary of $\mathbf{H}^k$ and have "half-neighborhoods." This structure perfectly models the distinction between the inside of an object and its edge.

A Technical Point on Smoothness

How do we define the derivative of a function at a boundary point? A function $\alpha: U \subset \mathbf{H}^k \to \mathbf{R}^n$ is said to be of class $C^r$ if, for every point in its domain, it can be extended to a $C^r$ function on an open neighborhood in the full space $\mathbf{R}^k$. This ensures that derivatives are well-defined even at the edge.

Formal Definition and Examples

With the upper half-space as our local model, the definition is a natural extension of what we had before.

Definition: $k$-Manifold with Boundary

A subset $M \subset \mathbf{R}^n$ is a $k$-dimensional manifold with boundary if for each point $p \in M$, there is a chart $\alpha: U \to M$ mapping from an open set $U \subset \mathbf{H}^k$ to a neighborhood of $p$ in $M$. This chart must satisfy the same conditions of being a $C^r$ homeomorphism with a full-rank derivative.

  • A point $p = \alpha(x)$ is an interior point of $M$ if $x$ is an interior point of $\mathbf{H}^k$ (i.e., $x_k > 0$).
  • A point $p = \alpha(x)$ is a boundary point of $M$ if $x$ is a boundary point of $\mathbf{H}^k$ (i.e., $x_k = 0$).

The Boundary as a Manifold

One of the most elegant results in this area is that the boundary of a manifold is itself a well-behaved manifold, but of one lower dimension and without a boundary of its own.

Theorem: The Boundary of a manifold is a manifold and the Boundary of a Boundary is Empty

If $M$ is a $k$-dimensional manifold with boundary, its boundary $\partial M$ is a $(k-1)$-dimensional manifold without boundary. That is, $\partial(\partial M) = \emptyset$.

Click to expand the proof of the theorem

The proof relies on constructing a chart for any point on the boundary $\partial M$ and showing that this chart's domain is an open set in $\mathbf{R}^{k-1}$, which is the model space for an interior point of a $(k-1)$-manifold.

  1. Setup. Let $M$ be a $k$-manifold with boundary and let $p$ be an arbitrary point in its boundary, $\partial M$.
  2. Use the Chart for $M$. Since $p$ is a boundary point of $M$, there must exist a chart for $M$, say $\alpha: U \to M$, where $U$ is an open set in $\mathbf{H}^k$ and there is a point $x_0 \in U$ on the boundary of $\mathbf{H}^k$ (meaning its last coordinate is $0$) such that $\alpha(x_0) = p$.
  3. Construct a Chart for $\partial M$. We restrict the chart $\alpha$ to only the boundary part of its domain. Let $\partial\mathbf{H}^k$ be the boundary of the upper half-space, which we identify with $\mathbf{R}^{k-1}$. Define a new domain $U' = U \cap \partial\mathbf{H}^k$. Because $U$ is open in $\mathbf{H}^k$, its intersection $U'$ is an open set in $\mathbf{R}^{k-1}$. We can now define a new chart for $\partial M$ by $\alpha' : U' \to \partial M$.
  4. Conclusion. The new chart $\alpha'$ for the $(k-1)$-manifold $\partial M$ has a domain $U'$ that is open in $\mathbf{R}^{k-1}$. By definition, a point is an interior point of a manifold if it can be charted from an open set in the corresponding Euclidean space. Since our chart $\alpha'$ for $p$ has a domain open in $\mathbf{R}^{k-1}$, the point $p$ is an interior point of $\partial M$. As $p$ was arbitrary, every point in $\partial M$ is an interior point, and thus the boundary of $\partial M$ is empty.

The fact that $\partial M$ is a boundary-less manifold is not just a mathematical curiosity; it is the geometric foundation for the General Stokes' Theorem.

A Powerful Construction: Using Local Defining Functions

Now that we have established that the boundary of a manifold is itself a proper manifold, we can revisit a powerful tool for constructing such objects. Recall the concept of a local defining function from Lecture 3. This is a smooth function $r: U \to \mathbf{R}$ that carves out a set $M$ and its boundary $\partial M$ by the conditions $r(\mathbf{x}) \le 0$ and $r(\mathbf{x}) = 0$, respectively, with the crucial property that its derivative $dr$ does not vanish on the boundary.

This setup provides a practical and powerful method for proving that a set is a manifold with a well-behaved boundary, without needing to construct an entire atlas of charts explicitly. The non-vanishing derivative condition is precisely what allows the Inverse Function Theorem to guarantee the local structure we need.

Proof that a Defining Function creates a Manifold

We need to show that for any point $p$ in the set $M = \{ \mathbf{x} \mid r(\mathbf{x}) \le 0 \}$, there is a valid chart.

  1. Interior Points ($r(p) < 0$): If $r(p) < 0$, then by continuity, there is a neighborhood of $p$ where $r$ remains negative. This neighborhood is an open set in $\mathbf{R}^n$, which is trivially an $n$-manifold.
  2. Boundary Points ($r(p) = 0$): This is the crucial case. We are given that $dr(p) \neq 0$. By reordering coordinates if necessary, we can assume $\frac{\partial r}{\partial x_n}(p) \neq 0$.
  3. Constructing a Diffeomorphism: Define a new map $F: U \to \mathbf{R}^n$ by: $$ F(x_1, \dots, x_n) = (x_1, \dots, x_{n-1}, r(x_1, \dots, x_n)) $$ The Jacobian determinant of this map at $p$ is $\frac{\partial r}{\partial x_n}(p) \neq 0$. By the Inverse Function Theorem, $F$ is a local diffeomorphism—a valid, smooth change of coordinates in a neighborhood of $p$.
  4. Straightening the Boundary: In the new coordinate system $y = F(x)$, the condition $r(x) \le 0$ becomes simply $y_n \le 0$. The map $F$ transforms the neighborhood of $p$ in $M$ into a neighborhood of $F(p)$ in the standard upper half-space $\mathbf{H}^n$.
  5. Conclusion: The inverse map $F^{-1}$ serves as a chart that maps a standard open set of $\mathbf{H}^n$ to a neighborhood of $p$ in $M$. This proves that $M$ is an $n$-manifold with boundary.

3. Problems for Practice

The following problems explore the subtleties of the definitions of manifolds and their boundaries. They require a careful application of the concepts of charts, homeomorphisms, the rank condition, and the consequences of the Inverse and Implicit Function Theorems.

Exercise 1: Parametrized vs. Intrinsic Manifolds (The Cusp)

This problem explores the crucial difference between the image of a smooth map and a true manifold. Consider the curve in $\mathbf{R}^2$ given by the parametrization $\alpha: \mathbf{R} \to \mathbf{R}^2$, $\alpha(t) = (t^2, t^3)$. The image set is $M = \alpha(\mathbf{R})$.

  1. Show that the map $\alpha$ is smooth and injective, but that this specific parametrization fails the full-rank condition at the origin.
  2. The failure of one parametrization does not mean the set is not a manifold. Prove, however, that the set $M$ is not a 1-manifold. (Hint: Assume for contradiction that a valid chart $\gamma: (-\epsilon, \epsilon) \to M$ exists around the origin with $\gamma(0) = (0,0)$. Analyze the behavior of the derivative $\gamma'(0)$ and what it implies about the local geometry.)

Exercise 2: The Orthogonal Group O(n)

The set of all $n \times n$ orthogonal matrices, denoted $O(n)$, is a subset of the space of all $n \times n$ matrices $\mathbf{R}^{n^2}$ defined by the condition $A^T A = I$. Prove that $O(n)$ is a manifold and find its dimension.
(Hint: Define a function $f: M(n, \mathbf{R}) \to S(n, \mathbf{R})$, where $S(n, \mathbf{R})$ is the space of $n \times n$ symmetric matrices, such that $O(n)$ is the level set $f^{-1}(I)$. Show that the identity matrix $I$ is a regular value of $f$ and apply the Preimage Theorem.)

Exercise 3: Transversal Intersections

Let $M^k$ and $N^l$ be manifolds in $\mathbf{R}^n$. They are said to intersect transversally at a point $p \in M \cap N$ if their tangent spaces at that point span the ambient space, i.e., $T_p(M) + T_p(N) = \mathbf{R}^n$.

Prove that if $M$ and $N$ intersect transversally at every point of their intersection, then the set $M \cap N$ is a $(k+l-n)$-dimensional manifold.
(Hint: Near $p$, locally represent $M$ as the zero set of a function $f: U \to \mathbf{R}^{n-k}$ and $N$ as the zero set of $g: U \to \mathbf{R}^{n-l}$. Consider the combined map $h = (f, g)$ and use the Implicit Function Theorem.)

Exercise 4: Non-Transversal Intersections

Explore what happens when the condition in the previous problem is not met.

  1. Consider the intersection of the sphere $S^2 \subset \mathbf{R}^3$ with the plane $P = \{ (x,y,z) \mid z=1 \}$. Describe the intersection set. Show that the tangent spaces are not transversal at the point of intersection. Is the intersection set a manifold?
  2. Construct two smooth surfaces in $\mathbf{R}^3$ whose intersection is precisely the figure-eight curve, which we know is not a manifold.

Exercise 5: Manifolds with Corners

Our definition of a manifold with boundary models objects with smooth edges, locally represented by the half-space $\mathbf{H}^k$. Consider the set $C = \{ (x, y) \in \mathbf{R}^2 \mid x \ge 0, y \ge 0 \}$, the first quadrant of the plane, which has a "corner" at the origin.

Prove that $C$ is not a 2-manifold with boundary according to our definition.
(Hint: Assume for contradiction that a chart $\alpha: U \to C$ exists for a neighborhood of the origin, where $U$ is an open set in $\mathbf{H}^2$. Show that whether the origin's preimage $\alpha^{-1}(0,0)$ lies in the interior of $\mathbf{H}^2$ or on its boundary, the map $\alpha$ cannot be a homeomorphism.)

Exercise 6: The Irrational Winding on a Torus

Let the 2-torus be defined as the quotient space $T^2 = \mathbf{R}^2 / \mathbb{Z}^2$. Consider the map $\alpha: \mathbf{R} \to T^2$ given by $\alpha(t) = (t, \gamma t) \pmod 1$, where $\gamma$ is an irrational number.

  1. Show that $\alpha$ is an injective immersion (i.e., it is one-to-one and its derivative is everywhere non-zero).
  2. The image set $M = \alpha(\mathbf{R})$ is a dense subset of the torus. Explain why the topology that $M$ inherits as a subspace of $T^2$ is different from the standard topology of the real line $\mathbf{R}$.
  3. Conclude that the set $M$ is not a 1-manifold. Which specific condition in the definition of a manifold fails?

Exercise 7: The Space of Lines (The Grassmannian)

Let $G(k, n)$ be the set of all $k$-dimensional linear subspaces of $\mathbf{R}^n$. Consider the specific case of $G(2, 4)$, the set of all 2D planes through the origin in $\mathbf{R}^4$.

  1. A 2D plane $P$ can be represented as the row space of a $2 \times 4$ matrix $A$ of rank 2. Show that two full-rank $2 \times 4$ matrices, $A$ and $B$, represent the same plane if and only if $A = GB$ for some invertible $2 \times 2$ matrix $G$.
  2. Let $P_0$ be the plane spanned by $\{e_1, e_2\}$. Show that any plane $P$ "close" to $P_0$ can be uniquely represented as the graph of a linear map $L: P_0 \to P_0^\perp$, where $P_0^\perp$ is the orthogonal complement of $P_0$. The space of such linear maps can be identified with $\mathbf{R}^{2 \times 2} \cong \mathbf{R}^4$.
  3. Use this idea to construct an atlas for $G(2, 4)$, proving that it is a 4-dimensional manifold. What is the dimension of $G(k, n)$ in general?
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