To my students in MATH 4230/5230, Fall 2026—
may these notes help you see geometry in motion.

Preface

These notes accompany MATH 4230/5230, Differential Geometry of Curves and Surfaces, at the University of Colorado Boulder in Fall 2026. They are organized lecture by lecture and follow the course’s progression from parametrized curves in Euclidean space to the local and global geometry of surfaces.

The primary text is Kristopher Tapp’s Differential Geometry of Curves and Surfaces [Tap16]. Andrew Browder’s Mathematical Analysis: An Introduction [Bro96] serves as a supplementary reference for analytic background, especially for students enrolled in MATH 5230.

The aim of the notes is not to replace either the textbook or the classroom. Instead, each lecture records the principal definitions, examples, statements, and calculations that organize the day’s discussion. Proofs may occasionally be abbreviated in class and completed here; conversely, some geometric pictures and informal explanations will remain most natural at the board.

Organization of the notes

Each lecture begins with the assigned reading, the related homework, and a short list of goals. Definitions, examples, and results are numbered within that lecture. The index is meant to make the notes useful as a reference near the end of the semester, while the bibliography records the sources used in preparing them.

These notes will be revised throughout the semester. The date on the title page identifies the current working edition.

Course Information

Course

MATH 4230/5230, Differential Geometry of Curves and Surfaces.

Semester

Fall 2026.

Instructor

Sebastian Casalaina-Martin.

Class meetings

Monday, Wednesday, and Friday, 11:15 a.m.–12:05 p.m., in ECCR 151.

Primary text

Kristopher Tapp, Differential Geometry of Curves and Surfaces [Tap16].

Supplementary analysis reference

Andrew Browder, Mathematical Analysis: An Introduction [Bro96].

Scope

The course studies the geometry of plane and space curves and regular surfaces in ℝ3. Central topics include speed and arc length, curvature and torsion, the Frenet frame, the first and second fundamental forms, Gaussian and mean curvature, geodesics, parallel transport, and the Gauss–Bonnet theorem.

How to use these notes

Read the assigned section of Tapp before class. During class, use these notes to organize definitions and calculations rather than as a substitute for working them out. After class, complete the assigned exercises and add your own examples, diagrams, and corrections in the margins.

Conventions and Notation

Unless stated otherwise, all maps are smooth. An interval means a connected subset of ℝ, usually open unless endpoints are explicitly included. A parametrized curve is generally denoted

𝜸: I→ℝn,

with velocity 𝜸′(t) and acceleration 𝜸″(t). The standard Euclidean inner product and norm are written

⟨𝐮,𝐯⟩and ‖𝐮‖ = ⟨𝐮 , 𝐮⟩.

We follow the typography of Tapp’s curve chapters: names of vectors and vector-valued functions are boldface. Thus 𝜸 is a curve, 𝐯 and 𝐚 are its velocity and acceleration, and the lowercase bold Fraktur symbols 𝖙 and 𝖓 denote the unit tangent and unit normal vectors. The curvature function is κ; the vector 𝖓(t) is defined only when κ(t)≠0.

For plane curves, R𝜃 denotes counterclockwise rotation through angle 𝜃, and R90 = Rπ∕2. The signed curvature is κs, an angle function for the unit tangent is 𝜃, and rot ⁡ (𝜸) denotes the rotation index of a closed plane curve.

For the global theory of plane curves, the unit circle is

S1 = {𝐮 ∈ ℝ2 : |𝐮| = 1}.

If 𝐟 : [a,b] →S1 is continuous with 𝐟(a) = 𝐟(b), then deg ⁡ (𝐟) denotes its degree. For a simple closed plane curve with trace C and a point 𝐩∉C,

WC(𝐩) = deg ⁡ (t⟼ 𝜸(t) −𝐩 |𝜸(t) −𝐩| )

is the winding number of C about 𝐩. The bounded and unbounded components of ℝ2 −C are called the interior and exterior. For a piecewise-regular plane curve, αi denotes the signed turning angle at the ith corner.

For space curves in ℝ3, 𝖇 = 𝖙 ×𝖓 is the unit binormal vector, and {𝖙,𝖓,𝖇} is the Frenet frame. The torsion function is denoted by τ, using Tapp’s sign convention

τ = −⟨𝖇′,𝖓⟩ |𝐯| .

The osculating plane is spanned by 𝖙 and 𝖓, with 𝖇 as its unit normal.

For rigid motions, O(n) denotes the orthogonal group. If A ∈O(n), then LA(𝐱) = A𝐱, while T𝐪(𝐱) = 𝐱 + 𝐪 denotes translation by 𝐪. Every rigid motion has the form T𝐪 ∘LA; it is proper when det ⁡ A = 1 and improper when det ⁡ A = −1.

For differential forms on an open set U ⊆ ℝn, Ωk(U) denotes the vector space of smooth k-forms. The coordinate one-forms are dx1,…, dxn, and for an ordered index set I = (i1 < ⋯ < ik),

dxI = dxi1 ∧⋯ ∧ dxik.

The symbol d also denotes the exterior derivative. A vector field 𝐅 = (F1,…,Fn) corresponds, through the Euclidean inner product, to the one-form

ω𝐅 = ∑ i=1nF i dxi.

Its line integral along a curve 𝜸 is written

∫ 𝜸ω𝐅 = ∫ 𝜸𝐅 ⋅ d𝜸.

For a smooth function f : U → ℝ, its gradient is ∇ ⁡f, and

D𝐯f(𝐩) = (𝑑𝑓)𝐩(𝐯) = ⟨∇ ⁡f(𝐩),𝐯⟩

denotes the directional derivative. A conservative field has the form 𝐅 = ∇ ⁡f, equivalently ω𝐅 = 𝑑𝑓. In the energy interpretation, ρ = −f denotes potential energy.

For a positively oriented plane boundary C, the outward unit normal is 𝐧 = R−90(𝐯∕ |𝐯|), where R−90(x1,x2) = (x2,−x1). This 𝐧 is distinct from the Frenet normal 𝖓. Flux is written ∮ ⁡ C𝐅 ⋅𝐧 𝑑𝑠, where 𝑑𝑠 = |𝐯(t)| 𝑑𝑡. The divergence is ∇ ⁡⋅𝐅 = ∑ ⁡ i∂Fi∕∂xi.

The symbol ∗ denotes the Euclidean Hodge star with the standard orientation. In the plane,

∗ω𝐅 = F1 dx2 −F2 dx1, d(∗ω𝐅) = (∇ ⁡⋅𝐅) dx1 ∧ dx2.

For a function f, the notation 𝐷𝑓 denotes its derivative as a linear map. Later, when covariant differentiation is introduced, its notation will be stated explicitly to avoid confusing it with the ordinary Euclidean derivative.

Beginning with Lecture 16, we follow Tapp’s surface chapters in using ordinary, rather than boldface, symbols for vectors and vector-valued maps. For a smooth map f : U ⊆ ℝn → ℝm, the differential at p is (𝐷𝑓)p : ℝn → ℝm, represented by the m ×n Jacobian J(f)(p). Its value on v is the directional derivative (Dvf)(p). Tapp denotes the same differential by dfp. A coordinate chart for a regular d-fold is denoted by φ, with inverse ψ; the domain of φ is open in ℝd, and its image is open in the d-fold.

The end of a proof is marked by □. Definitions, examples, and results are numbered by lecture; for example, Proposition 4.2 is the second numbered item in Lecture 4.

Contents

Preface
Course Information
Conventions and Notation
1 Parametrized Curves: Regularity, Speed, and Arc Length
1.1 Review
1.1.1 Number systems and Euclidean space
1.1.2 Intervals
1.1.3 Smooth functions and maps
1.2 Parametrized curves
1.3 Velocity, regularity, and speed
1.4 Arc length
1.5 Summary
2 Inner Products, Projections, and Orthonormal Bases
2.1 The Euclidean inner product
2.2 Orthogonal projection
2.2.1 Review: Components and projections (not covered in class)
2.2.2 Review: Orthonormal sets and bases
2.3 Differentiating inner products
2.4 A geometric application: straight lines are shortest (not covered in class)
2.5 Summary
3 Acceleration and Reparametrization
3.1 Acceleration
3.1.1 Tangential and perpendicular acceleration
3.2 Trace and reparametrization
3.3 More on reparameterization (not covered in class)
3.4 Reparametrization by arc length (Covered on Friday August 28)
3.5 Closed curves (not covered in class)
3.6 Summary
4 Curvature
4.1 The curvature function
4.2 Unit tangent and unit normal vectors
4.3 Curvature of a graph at a critical point (not covered in class)
4.3.1 Computational formula
4.4 A worked example (not covered in class)
4.5 Further topics
4.6 Summary
5 Plane Curves
5.1 Rotation matrices
5.2 Signed curvature
5.2.1 The unit-speed case
5.3 Angle functions
5.4 Rotation index
5.5 Summary
6 Space Curves: Frenet Frame, Torsion, and Frenet Equations
6.1 The cross product
6.1.1 Rotations and the cross product
6.2 The Frenet frame
6.3 The Frenet equations
6.4 Some further topics
6.4.1 Zero torsion
6.4.2 Product rule
6.4.3 A curvature formula in ℝ3
6.5 Summary
7 Rigid Motions and Fundamental Theorems
7.1 Rigid motions of Euclidean space
7.2 Geometric quantities under rigid motion
7.3 The fundamental theorems of curves
7.3.1 The plane-curve construction
7.4 A short overview of curvature formulas
7.5 Summary
8 Hopf’s Umlaufsatz and the Jordan Curve Theorem
8.1 Degree and the unit tangent map
8.2 Hopf’s Umlaufsatz
8.3 The Jordan curve theorem
8.4 A first look at the piecewise-regular formula
8.5 Summary
9 Differential Forms, Vector Fields, and Line Integrals
9.1 Coordinate forms and the wedge product
9.1.1 Wedge products of forms
9.2 Smooth differential forms
9.3 The exterior derivative
9.4 Vector fields and one-forms
9.5 Line integrals
9.6 Summary
10 Conservative Vector Fields and Green’s Theorem
10.1 Conservative vector fields
10.2 The chain rule and directional derivatives
10.3 Geometry of the gradient
10.4 Integrating a conservative field
10.5 Path independence
10.6 Conservation of energy
10.7 Example: gravity
11 Green’s Theorem III: Circulation, Divergence, and Flux
11.1 Review: the exterior derivative squares to zero
11.2 Green’s theorem
11.2.1 The differential-form expression
11.2.2 Infinitesimal circulation
11.3 The flux version of Green’s theorem
11.3.1 The outward normal and flux
11.3.2 Divergence and the star operation
11.3.3 The planar calculation and the flux theorem
11.4 Summary
16 Derivatives of maps from Euclidean space to Euclidean space
16.1 Smooth maps
16.2 Differentials of smooth maps
16.2.1 Directional derivatives
16.3 The chain rule
16.4 The inverse and implicit function theorems
16.5 Diffeomorphisms
16.5.1 Open subsets and relative neighborhoods
16.5.2 Smooth maps on subsets
16.6 Summary
17 Regular surfaces: patches, level sets, ruled surfaces, and examples
17.1 Review: smooth maps and diffeomorphisms
17.1.1 Open sets and neighborhoods
17.1.2 Diffeomorphisms and local coordinates
17.2 Regular d-folds
17.2.1 Graphs
17.3 A regular d-fold is locally a graph
17.4 Local level-set descriptions
17.5 Summary
19 Tangent planes and derivatives of maps between surfaces
19.1 Review: charts and transition maps
19.1.1 Example: two coordinate systems on a cylinder
19.2 Tangent planes
19.3 Differentials of maps between manifolds
19.4 Submanifolds and curves
19.4.1 Tangent lines of curves fill the tangent plane
19.5 Example: the tangent plane to the cylinder
19.6 Summary
20 Area distortion, orientation, and orientable surfaces
20.1 Determinants, area, and volume
20.1.1 Linear maps and the unit parallelepiped
20.1.2 The two-dimensional wedge calculation
20.1.3 The differential and infinitesimal distortion
20.2 Orientations of vector spaces
20.2.1 A plane in ℝ3: a choice of unit normal
20.3 Orientations of manifolds
20.3.1 Surfaces and unit normal fields
20.4 Examples and regular level sets
20.5 A nonexample: the Möbius strip
20.6 Summary
21 Surface area, isometries, and the first fundamental form
21.1 Inner products
21.1.1 The matrix of an inner product
21.1.2 Subspaces and injective linear maps
21.2 Lengths and angles
21.3 Linear isometries and volume
21.3.1 The Gram-determinant formula
21.3.2 The volume density in coordinates
21.4 Summary
22 The first fundamental form in local coordinates
22.1 Review: inner products and volumes
22.2 The first fundamental form
22.2.1 A chart and its differential
22.2.2 The matrix in local coordinates
22.3 Example: rolling a strip onto the cylinder
22.4 Surface area
22.4.1 Independence of coordinates
22.5 Arc length in local coordinates
22.6 Summary
Bibliography
Index

Lecture 1
Parametrized Curves: Regularity, Speed, and Arc Length

Friday, August 21, 2026

Reading. [Tap16, §1.1, pp. 2–8].

After class. Tapp, Exercises 1.5–1.10.

Goals. Review intervals and smooth maps, describe curves as smooth maps, distinguish regular from singular parametrizations, compute speed, and define arc length.

1.1 Review

1.1.1 Number systems and Euclidean space

Recall the familiar inclusions

ℕ ⊆ ℤ ⊆ ℚ ⊆ ℝ,

where ℕ = {1,2,3,…} are the natural numbers, ℤ = {⋯,−1,0,1,2,⋯} are the integers, ℚ = {a∕b : a,b ∈ ℤ,b≠0} are the rational numbers, and ℝ is the set of real numbers (all the numbers on the number line).

For each positive integer n, Euclidean n-space is

ℝn = {(x 1,…,xn) : xi ∈ ℝ}.

In particular, ℝ1 = ℝ. For 𝐱 = (x1,…,xn) ∈ ℝn, the Euclidean norm is

‖𝐱‖ = x1 2 + ⋯ + xn 2,

and the distance between 𝐱,𝐲 ∈ ℝn is ‖𝐱 −𝐲‖.

1.1.2 Intervals

An interval is a subset I ⊆ ℝ with the property that, if a,b ∈I and a < c < b, then c ∈I. A bounded interval with endpoints a < b has one of the four forms

(a,b),(a,b],[a,b),[a,b].

We also use −∞ and ∞ to describe unbounded intervals, for example (a,∞), [a,∞), and (−∞,b]. The symbols ±∞ are never included as endpoints.

1.1.3 Smooth functions and maps

Let I ⊆ ℝ be an interval. A function f : I → ℝ is smooth if it is infinitely differentiable at every x ∈I; i.e., for all x ∈I and all integers k ≥ 0, we have f(k)(x) exists. For intervals that include one or both of their endpoints, then at these endpoints, we mean that the derivatives exist as one sided limits; for instance, if I = [a,b), then to say that f′(a) exists means that lim ⁡ h→0+f(a+h)−f(a) h exists. We write

C∞(I) = {f : I → ℝ : f is smooth}.

In particular, C∞(ℝ) denotes the smooth real-valued functions on ℝ. A map

𝐟 = (f1,…,fn) : I→ℝn
𝐟(t) = (f1(t),…,fn(t))

is smooth when each coordinate function fi : I → ℝ is smooth.

Remark 1.1 (Smoothness at a boundary point). The one-sided definition of smoothness above agrees with the extension definition of smoothness used in [Tap16, Definition/Exercise 1.2]. That is to say, 𝐟 : I → ℝn is smooth if and only if there exists an open interval Î ⊆ ℝ with I ⊆Î and a smooth map 𝐟^ : Î → ℝn such that for all x ∈I and all integers k ≥ 0, one has 𝐟(k)(x) = 𝐟^(k)(x). (Note that 𝐟(k)(x) is a one-sided limit if x is an endpoint of I, whereas 𝐟^(k)(x) is a two-sided limit at such a point.) For instance, if 𝐟 : [a,b) → ℝn is smooth, then there is a real number 𝜖 > 0 and a smooth map 𝐟^ : (a −𝜖,b) → ℝn such that for all x ∈ [a,b) and all integers k ≥ 0, one has 𝐟(k)(x) = 𝐟^(k)(x). A similar statement holds for intervals (a,b], and [a,b]. We will use this extension fact without proof.

Exercise 1.2 (For students in MATH 5230). Let I be an interval containing 0 in its interior, and consider the linear map

𝒯 : C∞(I)→ℝ[[x]],𝒯(f) = ∑ k=0∞f(k)(0) k! xk,

which sends a smooth function to its formal Taylor series at 0. Borel’s lemma says that 𝒯 is surjective. Look up the statement of Borel’s lemma; no proof is required. Then use that to show that our definition of smoothness agrees with the definition in Tapp; i.e., prove that the remark above is true. For instance, if I = [0,b), then let g(x) ∈C∞(ℝ) be any function such that 𝒯(g) = ∑ ⁡ k=0∞f(k)(0) k! xk; then define f^(x) = f(x) for x ∈I = [0,b) and f^(x) = g(x) for x ∈ (−𝜖,0]. Also describe ker⁡(𝒯): what does it mean for a smooth function to have zero Taylor series at 0? Is the kernel finite dimensional?

Remark 1.3 (Analysis background for MATH 5230). Graduate students should review [Bro96, Chapter 8, especially §§ 8.1–8.2 and 8.4] and learn the definition of the differential of a map f : ℝn → ℝm, its interpretation as a linear approximation, its matrix representation, and the chain rule. They should also work several relevant exercises from § 8.6.

1.2 Parametrized curves

A geometric curve is a path traced through space. A parametrization records both the points of the path and the manner in which the path is traversed. Thus two maps can have the same image while moving along it at different speeds, in different directions.

The basic object in this course is therefore a map rather than merely a subset of Euclidean space.

Definition 1.4 (Parametrized curve). A smooth parametrized curve in ℝn is a smooth map

𝜸 : I→ℝn,

where I ⊆ ℝ is an interval.

Example 1.5 (Lines). Fix 𝐩,𝐯 ∈ ℝn with 𝐯≠0. The map

𝜸(t) = 𝐩 + t𝐯

parametrizes the line through 𝐩 in direction 𝐯. Its velocity is constant:

𝜸′(t) = 𝐯.

Example 1.6 (A circle). For r > 0, the map

𝜸(t) = (rcos⁡t,rsin⁡t),0 ≤t ≤ 2π,

traverses the circle of radius r once counterclockwise.

A straight line through p in direction v, parametrized by gamma(t) = p + tv.

A line through 𝐩 in direction 𝐯.

A radius-r circle parametrized counterclockwise by gamma(t) = (r cos t, r sin t). A circle traversed counterclockwise.
Figure 1.1: Two basic parametrized curves.

Example 1.7 (The graph of a function). Let f : I → ℝ be smooth. The curve

𝜸(t) = (t,f(t))

parametrizes the graph y = f(x). Its derivative is

𝜸′(t) = (1,f′(t)).

The graph y = f(x), with the position and tangent vectors of gamma(t) = (t, f(t)).

Figure 1.2: The graph of a smooth function can be viewed as a parametrized curve.

1.3 Velocity, regularity, and speed

Definition 1.8 (Velocity and regularity). The velocity of 𝜸 at time t is 𝜸′(t). The curve is regular at t if 𝜸′(t)≠0, and it is regular if it is regular at every point of its domain.

Regularity says that the parametrization never comes instantaneously to a complete stop. It is a condition on the parametrization, not only on its image.

Definition 1.9 (Speed). The speed of a parametrized curve 𝜸 is the function

v𝜸(t) = ‖𝜸′(t)‖.

Remark 1.10. A parametrized curve is regular if and only if its speed ‖𝜸′(t)‖ is never zero. This is the definition used in Tapp.

Example 1.11 (Speed of the circle). For 𝜸(t) = (rcos⁡t,rsin⁡t),

𝜸′(t) = (−rsin⁡t,rcos⁡t)and hence ‖𝜸′(t)‖ = r.

Thus the standard angular parametrization has constant speed r.

Example 1.12 (A nonregular parametrization). Consider

𝜸(t) = (t2,t3),t ∈ ℝ.

Then

𝜸′(t) = (2t,3t2),

so 𝜸′(0) = 0. The curve is regular away from t = 0, but it is not regular at 0. Its image is the semicubical cusp y2 = x3, with x ≥ 0.

The semicubical cusp gamma(t) = (t squared, t cubed) and its tangent line at the origin.

Figure 1.3: A semicubical cusp: the image is a curve, but the parametrization is not regular at t = 0.

1.4 Arc length

Definition 1.13 (Arc length). Let 𝜸 : [a,b] → ℝn be a smooth parametrized curve. Its arc length is

L(𝜸) = ∫ ab ‖𝜸′(t)‖ 𝑑𝑡.

The formula expresses a familiar physical principle: distance traveled is the integral of speed over time.

A curve from gamma(a) to gamma(b), with a velocity vector tangent to the path.

Figure 1.4: A curve from 𝜸(a) to 𝜸(b), with instantaneous velocity vector 𝜸′(t).

Example 1.14 (Length of a line segment). If 𝜸(t) = 𝐩 + t(𝐪 −𝐩) for 0 ≤t ≤ 1, then

L(𝜸) = ∫ 01 ‖𝐪 −𝐩‖ 𝑑𝑡 = ‖𝐪 −𝐩‖.

Example 1.15 (Circumference of a circle). For the circle above,

L(𝜸) = ∫ 02πr 𝑑𝑡 = 2𝜋𝑟.

1.5 Summary

Lecture 2
Inner Products, Projections, and Orthonormal Bases

Monday, August 24, 2026

Reading. [Tap16, §1.2, pp. 9–15].

After class. Tapp, Exercises 1.15–1.20.

Goals. Use the Euclidean inner product to measure angles and orthogonality, decompose vectors into parallel and perpendicular parts, compute projections onto subspaces using orthonormal bases, and apply these ideas to vector-valued functions and curves.

2.1 The Euclidean inner product

The norm introduced in Lecture 1 measures the length of a vector. The inner product contains additional information: it measures how much one vector points in the direction of another.

Definition 2.1 (Euclidean inner product). For

𝐱 = (x1,…,xn),𝐲 = (y1,…,yn) ∈ ℝn,

the Euclidean inner product of 𝐱 and 𝐲 is

⟨𝐱,𝐲⟩ = x1y1 + ⋯ + xnyn.

It is also commonly called the dot product.

Proposition 2.2 (Basic properties). For all 𝐱,𝐲,𝐳 ∈ ℝn and a,b ∈ ℝ,

⟨a𝐱 + b𝐲,𝐳⟩ = a ⟨𝐱,𝐳⟩ + b ⟨𝐲,𝐳⟩, ⟨𝐱,a𝐲 + b𝐳⟩ = a ⟨𝐱,𝐲⟩ + b ⟨𝐱,𝐳⟩, ⟨𝐱,𝐲⟩ = ⟨𝐲,𝐱⟩, ⟨𝐱,𝐱⟩ ≥ 0, and  ⟨𝐱,𝐱⟩ = 0 if and only if 𝐱 = 0.

Proof. Each identity follows directly by expanding the coordinate formula. For the final assertion,

⟨𝐱,𝐱⟩ = x12 + ⋯ + x n2,

which is nonnegative and vanishes exactly when every coordinate of 𝐱 vanishes. □

Remark 2.3. The first two conditions say that ⟨−,−⟩ is bilinear, the next that it is symmetric, and the third that it is positive definite.

Theorem 2.4 (Cauchy–Schwarz inequality). For all 𝐱,𝐲 ∈ ℝn,

|⟨𝐱,𝐲⟩| ≤ ‖𝐱‖ ‖𝐲‖.

Equality holds precisely when 𝐱 and 𝐲 are linearly dependent.

Remark 2.5. The proof of the inequality is Tapp, Exercise 1.15. It should be proved algebraically rather than by appealing to the angle formula below, since that formula is justified using the Schwarz inequality.

Proof. If 𝐱 and 𝐲 are linearly dependent, then one can easily show that one has equality above. So assume that 𝐱 and 𝐲 are not linearly dependent. Then for t ∈ ℝ, consider

0 < ⟨𝐱 + t𝐲,𝐱 + t𝐲⟩ = ⟨𝐱,𝐱⟩ + 2 ⟨𝐱,𝐲⟩t + ⟨𝐲,𝐲⟩t2,

where the strict inequality holds since 𝐱 and 𝐲 are not linearly dependent. A real quadratic polynomial az2 + 𝑏𝑧 + c has no zeros if and only if b2 − 4𝑎𝑐 < 0; so therefore, we have

4 ⟨𝐱,𝐲⟩2 − 4 ⟨𝐲,𝐲⟩ ⟨𝐱,𝐱⟩ < 0.

The result then follows from some algebra. □

Note that the theorem above implies that

−1 ≤ |⟨𝐱,𝐲⟩| ‖𝐱‖ ‖𝐲‖ ≤ 1

Definition 2.6 (Angle, orthogonality, and parallelism). If 𝐱,𝐲 ∈ ℝn are nonzero, their angle is the unique 𝜃 ∈ [0,π] satisfying

cos⁡𝜃 = ⟨𝐱,𝐲⟩ ‖𝐱‖ ‖𝐲‖.

Equivalently,

⟨𝐱,𝐲⟩ = ‖𝐱‖ ‖𝐲‖cos⁡𝜃.

The vectors 𝐱 and 𝐲 are orthogonal if ⟨𝐱,𝐲⟩ = 0, and they are parallel if one is a scalar multiple of the other.

Thus nonzero vectors are orthogonal exactly when their angle is π∕2. The sign of ⟨𝐱,𝐲⟩ records whether the angle is acute, right, or obtuse.

Example 2.7. Let 𝐱 = (1,2,−1) and 𝐲 = (2,0,2). Then

⟨𝐱,𝐲⟩ = 2 + 0 − 2 = 0,

so 𝐱 and 𝐲 are orthogonal.

2.2 Orthogonal projection

Our goal is to understand and prove the following lemma:

Lemma 2.8. Let V ⊆ ℝn be a subspace. There exists a unique linear map

ΠV : ℝn→ℝn

with the following properties:

  1. Im ⁡ (ΠV ) = V , i.e., the image of the linear map is V ;
  2. ΠV |V = Id ⁡ V , i.e., for all 𝐯 ∈V , we have ΠV (𝐯) = 𝐯;
  3. ker⁡ΠV = V ⊥, i.e., if ΠV (𝐱) = 0, then 𝐱 ∈V ⊥.

Proof. Exercise. □

Recall that the kernel of a linear map L (denoted ker⁡L) is sometimes also referred to as the null space. We will recall the definition of the orthogonal complement later. For those of you who know what an orthonormal basis is (this definition is recalled below also):

Exercise 2.9. Show that if (𝐮1,…,𝐮n) is an orthonormal basis for V , then for all 𝐱 ∈ ℝn,

ΠV (𝐱) = ∑ i=1n ⟨𝐱,𝐮 1⟩ 𝐮1.

Exercise 2.10 (For students in MATH 5230). Show that there is a unique isomorphism

L : ℝn→∼V ⊕V ⊥

such that L|V = Id ⁡ V and L|V ⊥ = Id ⁡ V ⊥.

In the subsections below, we review the terminology above, as well as some of the notation and terminology in Tapp.

2.2.1 Review: Components and projections (not covered in class)

Fix a nonzero vector 𝐲. We want to split an arbitrary vector 𝐱 into one part that points along 𝐲 and another part perpendicular to 𝐲.

Definition 2.11 (Component and projection in a direction). Let 𝐱,𝐲 ∈ ℝn, with 𝐲≠0. The component of 𝐱 in the direction of 𝐲 is the signed scalar

comp ⁡ 𝐲(𝐱) = ⟨𝐱,𝐲⟩ ‖𝐲‖ .

The projection of 𝐱 in the direction of 𝐲 is the vector

proj ⁡ 𝐲(𝐱) = ⟨𝐱,𝐲⟩ ‖𝐲‖2 𝐲.

If 𝐲 is a unit vector, these formulas simplify to

comp ⁡ 𝐲(𝐱) = ⟨𝐱,𝐲⟩,proj ⁡ 𝐲(𝐱) = ⟨𝐱,𝐲⟩𝐲.

Proposition 2.12 (Parallel–perpendicular decomposition). Let 𝐱,𝐲 ∈ ℝn, with 𝐲≠0. There is a unique decomposition

𝐱 = 𝐱∥ + 𝐱⊥,

where 𝐱∥ is parallel to 𝐲 and 𝐱⊥ is orthogonal to 𝐲. It is given by

𝐱∥ = proj ⁡ 𝐲(𝐱),𝐱⊥ = 𝐱 − proj ⁡ 𝐲(𝐱).

Proof. Since 𝐱∥ is parallel to 𝐲, it must have the form 𝐱∥ = c𝐲. The condition that 𝐱⊥ = 𝐱 −c𝐲 be orthogonal to 𝐲 gives

0 = ⟨𝐱 −c𝐲,𝐲⟩ = ⟨𝐱,𝐲⟩ −c ‖𝐲‖2.

Hence

c = ⟨𝐱,𝐲⟩ ‖𝐲‖2 ,

which proves both existence and uniqueness. □

Example 2.13. Let 𝐱 = (3,1) and 𝐲 = (1,1). Since ⟨𝐱,𝐲⟩ = 4 and ‖𝐲‖2 = 2,

𝐱∥ = proj ⁡ 𝐲(𝐱) = 2(1,1) = (2,2),

and therefore

𝐱⊥ = 𝐱 −𝐱∥ = (1,−1).

Indeed, ⟨(1,−1),(1,1)⟩ = 0.

Orthogonal projection of x onto the line spanned by y, with parallel and perpendicular components.

Figure 2.1: Projection splits a vector into a part parallel to 𝐲 and a part perpendicular to 𝐲.

2.2.2 Review: Orthonormal sets and bases

Definition 2.14 (Orthonormal set). A finite set {𝐮1,…,𝐮k}⊆ ℝn is orthonormal if

⟨𝐮i,𝐮j⟩ = { 1,i = j, 0, i≠ j.

Thus each 𝐮i is a unit vector, and distinct members are mutually orthogonal. An orthonormal set that is a basis of a subspace V is an orthonormal basis of V .

Every orthonormal set is linearly independent; this is Tapp, Exercise 1.19. The standard orthonormal basis of ℝn is

𝐞1 = (1,0,…,0),𝐞2 = (0,1,0,…,0),…,𝐞n = (0,…,0,1).

If 𝐱 = (x1,…,xn), then

𝐱 = ⟨𝐱,𝐞1⟩ 𝐞1 + ⋯ + ⟨𝐱,𝐞n⟩ 𝐞n.

In other words, the usual coordinates of 𝐱 are precisely its components in the directions of the standard basis vectors.

For a subspace V ⊆ ℝn, its orthogonal complement is

V ⊥ = {𝐰 ∈ ℝn : ⟨𝐰,𝐯⟩ = 0 for every 𝐯 ∈V }.

Proposition 2.15 (Projection onto a subspace). Let V ⊆ ℝn be a subspace with orthonormal basis {𝐮1,…,𝐮k}. Every 𝐱 ∈ ℝn has a unique decomposition

𝐱 = 𝐱V + 𝐱V ⊥,

where 𝐱V ∈V and 𝐱 V ⊥ is orthogonal to every vector in V . The projection of 𝐱 onto V is

𝐱V = proj ⁡ V (𝐱) = ∑ i=1k ⟨𝐱,𝐮 i⟩ 𝐮i,

and

𝐱V ⊥ = 𝐱 − proj ⁡ V (𝐱).

Proof. Write a prospective vector in V as

𝐱V = a1𝐮1 + ⋯ + ak𝐮k.

The vector 𝐱 −𝐱V is orthogonal to all of V exactly when it is orthogonal to each basis vector 𝐮j. Orthonormality gives

0 = ⟨𝐱 −𝐱V ,𝐮j⟩ = ⟨𝐱,𝐮j⟩ −aj.

Thus the only possible coefficients are aj = ⟨𝐱,𝐮j⟩, and these coefficients do give the desired decomposition. □

Example 2.16 (Projection onto a plane). In ℝ3, let

𝐮1 = 1 2(1,1,0),𝐮2 = (0,0,1),

and let V = span ⁡ {𝐮1,𝐮2}. The set {𝐮1,𝐮2} is an orthonormal basis of V . For 𝐱 = (2,0,3),

proj ⁡ V (𝐱) = ⟨𝐱,𝐮1⟩ 𝐮1 + ⟨𝐱,𝐮2⟩ 𝐮2 = 2𝐮1 + 3𝐮2 = (1,1,3).

Hence the perpendicular part is

𝐱 − proj ⁡ V (𝐱) = (1,−1,0).

Remark 2.17 (Gram–Schmidt). Every finite-dimensional subspace of ℝn has an orthonormal basis. Starting from a basis 𝐯1,…,𝐯k, normalize 𝐯1, then subtract from each later vector its projections onto the orthonormal vectors already constructed, and normalize the result. For the first two steps,

𝐮1 = 𝐯1 ‖𝐯1‖ ,𝐰2 = 𝐯2 −⟨𝐯2,𝐮1⟩ 𝐮1,𝐮2 = 𝐰2 ‖𝐰2‖ .

Continuing this process gives the Gram–Schmidt process; its proof is Tapp, Exercise 1.20.

2.3 Differentiating inner products

The inner product also interacts well with differentiation.

Lemma 2.18 (Product rules). Let 𝐱,𝐲 : I → ℝn and c : I → ℝ be smooth maps. Then

d 𝑑𝑡 ⟨𝐱(t),𝐲(t)⟩ = ⟨𝐱′(t),𝐲(t)⟩ + ⟨𝐱(t),𝐲′(t)⟩, d 𝑑𝑡(c(t)𝐱(t)) = c′(t)𝐱(t) + c(t)𝐱′(t).

Proof. Differentiate each coordinate and use the ordinary product rule for real-valued functions. □

Proposition 2.19 (Derivatives of orthogonality relations). Let 𝐱,𝐲 : I → ℝn be smooth.

  1. If ‖𝐱(t)‖ is constant, then ⟨𝐱(t),𝐱′(t)⟩ = 0 for every t ∈I.
  2. If ⟨𝐱(t),𝐲(t)⟩ = 0 for every t ∈I, then

    ⟨𝐱′(t),𝐲(t)⟩ = −⟨𝐱(t),𝐲′(t)⟩.

Proof. For the first statement, differentiate the constant function

⟨𝐱(t),𝐱(t)⟩ = ‖𝐱(t)‖2.

The product rule gives

0 = 2 ⟨𝐱(t),𝐱′(t)⟩.

For the second, differentiate the identity ⟨𝐱(t),𝐲(t)⟩ = 0. □

Example 2.20 (Motion on a sphere). If 𝐱 : I → ℝ3 satisfies ‖𝐱(t)‖ = r, then its image lies on the sphere of radius r centered at the origin. The proposition says that 𝐱′(t) is orthogonal to the radius vector 𝐱(t). Thus the velocity is tangent to the sphere.

A curve on a circle with its velocity perpendicular to its constant-length position vector.

Figure 2.2: If ∥𝐱(t)∥ = r is constant, then the velocity is tangent to the sphere and orthogonal to the radius vector.

Corollary 2.21 (Constant-speed curves). If a curve 𝜸 : I → ℝn has constant speed, then

⟨𝜸′(t),𝜸″(t)⟩ = 0

for every t ∈I. Thus its velocity and acceleration are orthogonal.

Proof. Apply the first part of the proposition to the vector-valued function 𝐱 = 𝜸′. □

2.4 A geometric application: straight lines are shortest (not covered in class)

Theorem 2.22 (Length dominates endpoint distance). Let 𝜸 : [a,b] → ℝn be a smooth curve, with 𝐩 = 𝜸(a) and 𝐪 = 𝜸(b). Then

L(𝜸) ≥ ‖𝐪 −𝐩‖.

Thus no curve joining 𝐩 to 𝐪 is shorter than the line segment between them.

Proof. The result is immediate if 𝐩 = 𝐪, so assume 𝐩≠𝐪 and set

𝐮 = 𝐪 −𝐩 ‖𝐪 −𝐩‖.

This is the unit vector pointing from 𝐩 toward 𝐪. By Cauchy–Schwarz,

⟨𝜸′(t),𝐮⟩ ≤ ‖𝜸′(t)‖ ‖𝐮‖ = ‖𝜸′(t)‖.

Therefore

L(𝜸) = ∫ ab ‖𝜸′(t)‖ 𝑑𝑡 ≥∫ ab ⟨𝜸′(t),𝐮⟩ 𝑑𝑡 = ⟨𝜸(b) −𝜸(a),𝐮⟩ = ⟨𝐪 −𝐩, 𝐪 −𝐩 ‖𝐪 −𝐩‖ ⟩ = ‖𝐪 −𝐩‖. □

Remark 2.23. The proof illustrates a recurring strategy: choose a unit vector or an orthonormal basis adapted to the geometry of the problem, and then use inner products to extract the relevant components.

A curved path between p and q compared with the straight line segment.

Figure 2.3: Among smooth curves from 𝐩 to 𝐪, the line segment gives the shortest possible length.

2.5 Summary

Lecture 3
Acceleration and Reparametrization

Wednesday, August 26, 2026

Reading. [Tap16, §§ 1.3–1.4, pp. 16–23].

After class. Tapp, Exercises 1.26–1.29, 1.31, and 1.33.

Goals. Interpret acceleration geometrically, split it into components parallel and perpendicular to velocity, relate tangential acceleration to the rate of change of speed, define reparametrization and orientation, prove invariance of arc length, construct unit-speed parametrizations, and discuss closed curves.

3.1 Acceleration

Let 𝜸 : I → ℝn be a regular curve. Following the notation used in physics, we write

𝐯(t) = 𝜸′(t)and𝐚(t) = 𝜸″(t).

The vector 𝐯(t) is the velocity, while 𝐚(t) measures the instantaneous change in velocity. Thus acceleration can result from a change in speed, a change in direction, or both.

Definition 3.1 (Acceleration). The acceleration of a smooth parametrized curve 𝜸 at time t is

𝐚(t) = 𝜸″(t).

The physical interpretation is supplied by Newton’s equation

𝐅(t) = m𝐚(t).

For a particle of mass m, the acceleration points in the direction of the net force. A force may pull partly along the direction of motion, thereby changing speed, and partly across the direction of motion, thereby bending the path.

Example 3.2 (Circular motion). For

𝜸(t) = (rcos⁡t,rsin⁡t),

we have

𝐯(t) = (−rsin⁡t,rcos⁡t),𝐚(t) = (−rcos⁡t,−rsin⁡t) = −𝜸(t).

Thus the velocity is tangent to the circle, while the acceleration points toward its center. Moreover,

⟨𝐚(t),𝐯(t)⟩ = 0,

so the acceleration changes the direction of motion but not the speed.

Velocity tangent to a circle and acceleration pointing toward its center.

Figure 3.1: For circular motion, velocity is tangent to the path and acceleration is center-pointing.

3.1.1 Tangential and perpendicular acceleration

At each time, decompose the acceleration into the part parallel to velocity and the part perpendicular to velocity.

Definition 3.3 (Tangential and perpendicular components). For a regular curve, define

𝐚∥(t) = proj ⁡ 𝐯(t)(𝐚(t)) = ⟨𝐚(t),𝐯(t)⟩ ‖𝐯(t)‖2 𝐯(t), 𝐚⊥(t) = 𝐚(t) −𝐚∥(t).

Then

𝐚(t) = 𝐚∥(t) + 𝐚⊥(t),

where 𝐚∥(t) is parallel to 𝐯(t) and 𝐚⊥(t) is orthogonal to 𝐯(t).

Acceleration resolved into components parallel and perpendicular to velocity.

Figure 3.2: Acceleration decomposes into a component parallel to velocity and a component perpendicular to velocity.

The parallel component controls the speed.

Proposition 3.4 (Derivative of speed). Let 𝜸 : I → ℝn be regular. Then

d 𝑑𝑡 ‖𝐯(t)‖ = ⟨𝐚(t),𝐯(t)⟩ ‖𝐯(t)‖ = comp ⁡ 𝐯(t)(𝐚(t)).

Thus the signed scalar component of acceleration in the direction of velocity is exactly the rate of change of speed.

Proof. We start with ‖𝐯‖2 = ⟨𝐯,𝐯⟩. We then differentiate both sides. On the left we have

d 𝑑𝑡 ‖𝐯‖2 = 2 ‖𝐯‖ d 𝑑𝑡 ‖𝐯‖.

On the right we have

d 𝑑𝑡 ⟨𝐯,𝐯⟩ = ⟨𝐯′,𝐯⟩ + ⟨𝐯,𝐯′⟩ = 2 ⟨𝐚,𝐯⟩.

The result follows from some algebra. □

Corollary 3.5. For a regular curve:

  1. the speed is increasing when ⟨𝐚(t),𝐯(t)⟩ > 0 and decreasing when ⟨𝐚(t),𝐯(t)⟩ < 0;
  2. the speed is constant if and only if 𝐚∥(t) = 0 for every t, equivalently 𝐚(t) ⊥𝐯(t) for every t.

Example 3.6 (Acceleration along a parabola). Let

𝜸(t) = (t,t2).

Then

𝐯(t) = (1,2t),𝐚(t) = (0,2).

At t = 1,

⟨𝐚(1),𝐯(1)⟩ = 4, ‖𝐯(1)‖2 = 5,

so

𝐚∥(1) = 4 5(1,2) = (4 5, 8 5 ), 𝐚⊥(1) = (0,2) −𝐚∥(1) = (−4 5, 2 5 ).

The positive value

d 𝑑𝑡 ‖𝐯(t)‖|t=1 = 4 5

confirms that the particle is speeding up at this time.

The perpendicular component changes the direction of motion. Its magnitude is influenced both by how sharply the path bends and by how quickly the path is traversed. Reparametrization will allow us to separate these two effects.

3.2 Trace and reparametrization

Definition 3.7 (Trace). The trace of a parametrized curve 𝜸 : I → ℝn is the subset

𝜸(I) = {𝜸(t) : t ∈I}⊆ ℝn.

The trace remembers the path but forgets the clock: it does not record where the particle is at a particular time, how fast it moves, or which direction it travels.

Example 3.8 (Same trace, different clocks). Let

𝜸(s) = (s,s2), − 2 ≤s ≤ 2,

and let

ϕ(t) = t + t3, − 1 ≤t ≤ 1.

Since ϕ′(t) = 1 + 3t2 > 0, the map ϕ : [−1,1] → [−2,2] is a smooth bijection. The curve

𝜸~(t) = 𝜸(ϕ(t)) =(t + t3,(t + t3)2)

has the same trace as 𝜸, but its speed varies according to the new clock t.

A curve with position markers at equal increments of its original parameter.

The same curve with markers at equal increments of the new parameter.
Figure 3.3: A reparametrization keeps the trace but changes the association between parameter values and points on the trace.

Definition 3.9 (Reparametrization). Let 𝜸 : I → ℝn be a regular curve. A reparametrization of 𝜸 is a curve of the form

𝜸~ = 𝜸∘ϕ : I~→ℝn,

where ϕ : I~ →I is a smooth bijection with ϕ′(t)≠0 for every t ∈I~.

In diagram form we say that the following diagram commutes:

A commuting diagram: gamma from I to Euclidean space and gamma tilde = gamma composed with phi.

3.3 More on reparameterization (not covered in class)

The derivative condition ensures that the new curve is still regular.

Proposition 3.10 (Velocity and acceleration under reparametrization). If 𝜸~ = 𝜸∘ϕ, then

𝜸~′(t) = ϕ′(t)𝜸′(ϕ(t)), 𝜸~″(t) =(ϕ′(t))2𝜸″(ϕ(t)) + ϕ″(t)𝜸′(ϕ(t)).

Consequently,

‖𝜸~′(t)‖ = |ϕ′(t)| ‖𝜸′(ϕ(t))‖.

Proof. Apply the ordinary chain rule componentwise to obtain the first formula, then differentiate it using the product rule to obtain the second. Taking norms in the first formula gives the speed transformation law. □

Remark 3.11 (What changes and what does not). The term ϕ″(t)𝜸′(ϕ(t)) in the acceleration formula is parallel to velocity. Therefore the perpendicular component satisfies

𝐚~⊥(t) =(ϕ′(t))2𝐚⊥(ϕ(t)).

Since speed squared transforms by the same factor,

‖𝐚~⊥(t)‖ ‖𝜸~′(t)‖2 = ‖𝐚⊥(ϕ(t))‖ ‖𝜸′(ϕ(t))‖2.

This reparametrization-invariant ratio will become the curvature in the next lecture.

Because I~ is an interval and ϕ′ never vanishes, the intermediate value theorem implies that ϕ′ has a constant sign.

Definition 3.12 (Orientation). A reparametrization 𝜸~ = 𝜸∘ϕ is orientation-preserving if ϕ′(t) > 0 everywhere and orientation-reversing if ϕ′(t) < 0 everywhere.

The reparametrization in the previous example preserves orientation. By contrast,

𝜸^(t) = 𝜸(−t)

reverses orientation.

Proposition 3.13 (Arc length is independent of parametrization). Let

𝜸~ = 𝜸∘ϕ : [c,d] → ℝn

be a reparametrization. Set

u0 = min ⁡ {ϕ(c),ϕ(d)},u1 = max ⁡ {ϕ(c),ϕ(d)}.

Then

∫ cd ‖𝜸~′(t)‖ 𝑑𝑡 = ∫ u0u1 ‖𝜸′(u)‖ 𝑑𝑢.

Thus corresponding portions of the two parametrizations have the same arc length.

Proof. If ϕ′> 0, then the speed transformation formula and the substitution u = ϕ(t) give

∫ cd ‖𝜸~′(t)‖ 𝑑𝑡 = ∫ cdϕ′(t) ‖𝜸′(ϕ(t))‖ 𝑑𝑡 = ∫ ϕ(c)ϕ(d) ‖𝜸′(u)‖ 𝑑𝑢.

If ϕ′< 0, then |ϕ′| = −ϕ′, and the same substitution gives

∫ cd ‖𝜸~′(t)‖ 𝑑𝑡 = ∫ ϕ(d)ϕ(c) ‖𝜸′(u)‖ 𝑑𝑢.

These are precisely the asserted formulas in the two cases. □

3.4 Reparametrization by arc length (Covered on Friday August 28)

Unit-speed curves are especially convenient: the parameter difference equals the distance traveled along the curve. Every regular curve admits such a parametrization.

Theorem 3.14 (Arc-length parametrization). Every regular curve can be reparametrized by arc length. Moreover, for such a parameterization, the speed is constant and equal to 1.

Proof. For I = (a,b) (for [a,b), etc., a similar argument works) define the signed arc-length function

s(t) = ∫ at ‖𝜸′(u)‖ 𝑑𝑢.

By the fundamental theorem of calculus,

s′(t) = ‖𝜸′(t)‖ > 0. (3.1)

Thus s : I →J = s(I) = (0,s(b)) is a smooth increasing bijection with a smooth inverse ϕ : J →I. Define

𝜸¯ = 𝜸∘ϕ : J → ℝn.

Next, since for all v ∈J we have s(ϕ(v)) = v, differentiating with respect to v gives

s′(ϕ(v))ϕ′(v) = 1.

Therefore

‖𝜸¯′(v)‖ = ϕ′(v) ‖𝜸′(ϕ(v))‖ 3.10 = ϕ′(v)s′(ϕ(v)) (3.1) = 1.

From this we can check that 𝜸¯ gives a paramerization by arclength. In other words, for all v ∈J, we have ∫ 0v ‖𝜸¯′(u)‖ 𝑑𝑢 = ∫ 0v1 𝑑𝑢 = v. □

Remark 3.15. There are some details for you to check in the argument above. First, to prove that s is smooth, you must check that if 𝜸 is regular, then ‖𝜸′‖ is smooth. Then you should review the proof from your calculus class that s′> 0 implies that s is strictly increasing (see for instance [Bro96, Cor. 4.23]). Then you should check that s is invertible, and the inverse is differentiable. Then you should check that the inverse of s is smooth.

Example 3.16 (A circle parametrized by arc length). For

𝜸(t) = (rcos⁡t,rsin⁡t),0 ≤t ≤ 2π,

the speed is r. Taking t0 = 0,

s(t) = 𝑟𝑡,ϕ(s) = s r.

Hence

𝜸¯(s) = (rcos⁡ s r,rsin⁡ s r ),0 ≤s ≤ 2𝜋𝑟,

and ‖𝜸¯′(s)‖ = 1.

Remark 3.17. The proof gives an explicit procedure, but the inverse of the arc-length function often cannot be written in elementary closed form. The theorem is consequently more useful in theoretical arguments than as a computational recipe.

3.5 Closed curves (not covered in class)

A loop should meet itself smoothly at its beginning and ending point, not just have the same endpoint twice.

Definition 3.18 (Closed and simple closed curves). A closed curve is a regular curve 𝜸 : [a,b] → ℝn such that

𝜸(a) = 𝜸(b)

and every derivative agrees at the endpoints:

𝜸(k)(a) = 𝜸(k)(b)for every k ≥ 1.

It is simple closed if it is one-to-one on [a,b).

The standard circle is simple closed. The figure-eight curve

𝜼(t) = (sin⁡t,sin⁡2t),0 ≤t ≤ 2π,

closes smoothly but is not simple because it passes through the origin more than once.

A simple closed curve without self-intersection.

A closed curve with a self-intersection.
Figure 3.4: A simple closed curve and a closed curve with a self-intersection.

Proposition 3.19 (Periodic extension). A regular curve 𝜸 : [a,b] → ℝn is closed if and only if there is a periodic regular curve 𝜸^ : ℝ → ℝn of period b −a whose restriction to [a,b] is 𝜸.

Proof. This is Tapp, Exercise 1.33. The matching of every derivative at the two endpoints is precisely what permits the translated copies of 𝜸 to glue together smoothly. □

Remark 3.20 (Reparametrizing a closed curve). For closed curves, changing the clock should also allow a cyclic shift of the starting point. If 𝜸^ is the periodic extension above and 𝜸λ is its restriction to [a + λ,b + λ], then a closed reparametrization has the form

𝜸~ = 𝜸λ ∘ϕ.

In addition to being a smooth bijection with nonzero derivative, the derivatives of ϕ must match at its two endpoints so that the composite also closes smoothly.

Proposition 3.21. Two simple closed curves have the same trace if and only if each is a reparametrization of the other.

For nonsimple closed curves, the trace does not contain enough information: one trip around a circle and two trips around the same circle have identical traces, but they are not reparametrizations of one another.

3.6 Summary

Lecture 4
Curvature

Friday, August 28, 2026

Reading. [Tap16, § 1.5, pp. 24–31].

After class. HW 1 is due. Begin Tapp, Exercises 1.36, 1.37, 1.38, and 1.43.

Goals. Define curvature in a parametrization-independent way, define the unit tangent and unit normal vectors, and relate curvature to the change in the direction of motion.

Remark 4.1 (Scope). We omit the discussion of osculating planes and osculating circles. The notation, order of presentation, and principal definitions below follow Tapp’s treatment of curvature.

4.1 The curvature function

Let

𝜸 : I→ℝn

be a regular curve. Following Tapp, write the velocity and acceleration as

𝐯(t) = 𝜸′(t),𝐚(t) = 𝜸″(t).

Decompose acceleration into parts parallel and perpendicular to velocity:

𝐚(t) = 𝐚∥(t) + 𝐚⊥(t).

The perpendicular part detects turning, but its length also depends on how quickly the curve is traversed. If 𝜸~ = 𝜸∘ϕ is a regular reparametrization, then

𝐯~(t) = ϕ′(t)𝐯(ϕ(t)),𝐚~⊥(t) = ϕ′(t)2𝐚⊥(ϕ(t)).

Thus the quotient below is unchanged by reparametrization.

Definition 4.2: Curvature function

Let 𝜸 : I → ℝn be a regular curve. Its curvature vector 𝜿 : I → ℝn is

𝜿(t) = 𝐚⊥(t) |𝐯(t)|2.

Its curvature function κ : I → [0,∞) is

κ(t) = |𝐚⊥(t)| |𝐯(t)|2 .

The value κ(t) measures how sharply the trace bends near 𝜸(t). Solving the definition for the perpendicular acceleration gives

|𝐚⊥(t)| = κ(t) |𝐯(t)|2.

Thus the perpendicular acceleration increases both with curvature and with the square of the speed.

Proposition 4.3: Unit-speed curves

If 𝜸 is parametrized by arc length, then

κ(t) = |𝐚(t)|.

Proof. For a unit-speed curve, |𝐯| = 1 and 𝐚 ⊥𝐯. Hence 𝐚⊥ = 𝐚, and the result follows directly from the definition. □

Example 4.4 (A circle). For r > 0, consider

𝜸(t) = (rcos⁡t,rsin⁡t).

Then

𝐯(t) = (−rsin⁡t,rcos⁡t),𝐚(t) = (−rcos⁡t,−rsin⁡t).

Since 𝐚(t) ⊥𝐯(t), one has 𝐚⊥(t) = 𝐚(t). Therefore

κ(t) = r r2 = 1 r.

A circle of radius r has constant curvature 1∕r.

4.2 Unit tangent and unit normal vectors

Definition 4.5: Unit tangent and unit normal vectors

Let 𝜸 : I → ℝn be regular. Define

𝖙(t) = 𝐯(t) |𝐯(t)|,𝖓(t) = 𝐚⊥(t) |𝐚⊥(t)|.

The vector 𝖙(t) is the unit tangent vector. The vector 𝖓(t) is the unit normal vector, and it is defined only at times for which κ(t)≠0.

Whenever the input is clear, we suppress it and write simply 𝖙 and 𝖓. Where 𝖓 is defined, the set {𝖙,𝖓} is orthonormal. The path t↦𝖙(t) records the direction in which the curve is moving. Its derivative measures how quickly that direction changes.

Proposition 4.6: Curvature and change of direction

For every regular curve,

𝐚 = |𝐯|′𝖙⏟ 𝐚∥ + |𝐯|𝖙′⏟ 𝐚⊥ = |𝐯|′𝖙 + κ|𝐯|2𝖓.

and

κ(t) = |𝖙′(t)| |𝐯(t)|.

Proof. Since 𝐯 = |𝐯|𝖙, differentiation gives

𝐚 = |𝐯|′𝖙⏟ 𝐚∥ + |𝐯|𝖙′⏟ 𝐚⊥

The identity |𝖙| = 1 implies 𝖙′ ⊥𝖙. The first term above is therefore parallel to 𝐯, while the second is perpendicular to 𝐯. Hence

𝐚⊥ = |𝐯|𝖙′.

Substitution into the definition of curvature proves the formula κ(t) = |𝖙′(t)| |𝐯(t)| . Then, putting this formula for curvature into the right hand side of 𝐚 = |𝐯|′𝖙⏟ 𝐚∥ + |𝐯|𝖙′⏟ 𝐚⊥ = |𝐯|′𝖙 + κ|𝐯|2𝖓 shows that the right and middle terms are equal. □

At every time for which κ(t)≠0, the same calculation gives

𝖓(t) := 𝐚⊥(t) |𝐚⊥(t)| = 𝖙′(t) |𝖙′(t)|

and then, clearing denominators and using the formula for curvature in the proposition gives,

𝖙′(t) = κ(t) |𝐯(t)|𝖓(t).

4.3 Curvature of a graph at a critical point (not covered in class)

Example 4.7. Let f : I → ℝ be smooth, let t0 ∈I satisfy f′(t0) = 0, and consider the natural parametrization of its graph,

𝜸(t) = (t,f(t)).

At t0,

𝐯(t0) = (1,0),𝐚(t0) = (0,f″(t0)).

These vectors are orthogonal, so

κ(t0) = |f″(t0)|.

Thus, at a critical point of a graph, curvature is the absolute value of its concavity.

For the parabola f(t) = t2 at t0 = 0,

𝖙(0) = (1,0),𝖓(0) = (0,1),κ(0) = 2.

The parabola at its vertex with unit tangent and unit normal vectors.

Figure 4.1: For the parabola at its vertex, the unit tangent is horizontal and the unit normal points toward the side on which the curve bends.

4.3.1 Computational formula

The perpendicular component can be computed directly from velocity and acceleration:

Proposition 4.8 (Computational formula). For a regular curve,

κ(t) = ‖𝜸 ′ (t) ‖ 2 ‖𝜸 ″ (t) ‖ 2 − ⟨𝜸 ′ (t), 𝜸 ″ (t) ⟩ 2 ‖𝜸′(t)‖3 .

Proof. Since 𝐚∥ = proj ⁡ 𝐯(𝐚), orthogonality gives

‖𝐚⊥‖2 = ‖𝐚‖2 −‖𝐚∥‖2 = ‖𝐚‖2 −⟨𝐚,𝐯⟩2 ‖𝐯‖2 .

Substitute this into κ = ‖𝐚⊥‖∕ ‖𝐯‖2 and simplify. □

4.4 A worked example (not covered in class)

Example 4.9 (A parabola). Consider

𝜸(t) = (t,t2).

Then

𝜸′(t) = (1,2t),𝜸″(t) = (0,2).

The computational formula gives

κ(t) = (1 + 4t2 )(4) − (4t)2 (1 + 4t2)3∕2 = 2 (1 + 4t2)3∕2.

The curvature is largest at the vertex t = 0, where

𝖙(0) = (1,0),𝐍(0) = (0,1),κ(0) = 2.

The parabola in the worked example, showing tangent and principal normal at the vertex.

Figure 4.2: At the vertex of the parabola, the unit tangent is horizontal and the principal normal points toward the side on which the curve bends.

4.5 Further topics

Here are a few things to think about going forward. We will return to this later.

Recall that we have

𝖙′ = κ|𝐯|𝖓.

What about 𝖓′? We have

𝖓′ = −κ|𝐯|𝖙 + 𝐱

for some 𝐱 ∈ (Span ⁡ 𝖙)⊥. In fact, we know that ⟨𝖓′,𝖓⟩ = 0, so that 𝐱 ∈ (Span ⁡ (𝖙,𝖓))⊥.

Write 𝖇 = 𝐱∕ ‖𝐱‖, assuming 𝐱 is non-zero. Write

τ = |𝐱|∕|𝐯|.

Then we have by definition

𝖓′ = −κ|𝐯|𝖙 + τ|𝐯|𝖇

for some scalar function τ(t).

Exercise 4.10. What happens with the above under orientation preserving reparameterization? In other words, if we take 𝜸~ = 𝜸∘ϕ, then how are 𝐱~, 𝖇~, and τ~ related to 𝐱, 𝖇, and τ change?

Solution. If we write t = ϕ(u), so that 𝜸~(u) = 𝜸(t), then we know that 𝖙~(u) = 𝖙(t), since this is the unit tangent direction at 𝜸~(u) = 𝜸(t). Similarly 𝖓~(u) = 𝖓(t). So on the one hand we have

d 𝑑𝑢(𝖓~(u)) = (𝖓(ϕ(u))′ = 𝖓′(ϕ(u))ϕ′(u)

Recall that we have

𝐯~(u) = d 𝑑𝑢(𝜸~(u)) = d 𝑑𝑢(𝜸(ϕ(u))) = (𝜸′(ϕ(u))ϕ′(u) = 𝐯(ϕ(u))ϕ′(u),

so that

|𝐯~(u)| = |𝐯(ϕ(u))|ϕ′(u).

Since κ~(u) = κ(ϕ(u)), we have

𝐱~(u) := 𝖓~′(u) + κ~(u)|𝐯~(u)|𝖙~(u) = 𝖓′(ϕ(u))ϕ′(u) + κ(ϕ(u))|𝐯(ϕ(u))|ϕ′(u)𝖙(ϕ(u)) = (𝖓′ + κ|𝐯|𝖙)ϕ′(u) = 𝐱(ϕ(u))ϕ′(u).

Thus 𝐱~(u) and 𝐱(ϕ(u)) point in the same direction (with the exact scalar above), and so 𝖇~(u) = 𝖇(ϕ(u)). With this and the formula 𝖓′ = −κ|𝐯|𝖙 + τ|𝐯|𝖇 defining τ, we can conclude that τ~(u) = τ(ϕ(u)). □

Then what about 𝖇′? We have ⟨𝖇′,𝖇⟩ = 0. We have ⟨𝖇,𝖙⟩ = 0, so that ⟨𝖇′,𝖙⟩ = −⟨𝖇,𝖙′⟩ = 0 (from formula above). We have ⟨𝖇′,𝖓⟩ = −⟨𝖇,𝖓′⟩ = −τ|𝐯|. So we have

𝖇′ = −τ|𝐯|𝖓 + 𝐲

where 𝐲 ∈ (Span ⁡ (𝖙,𝖓,𝖇))⊥

4.6 Summary

Lecture 5
Plane Curves

Monday, August 31, 2026

Reading. [Tap16, § 1.6, pp. 32–40].

After class. Begin HW 2(b): Tapp, Exercises 1.44–1.49.

Goals. Use counterclockwise rotations in the plane, define signed curvature, construct an angle function for the velocity of a unit-speed plane curve, and define the rotation index of a closed plane curve.

The plane has an extra feature that is not available in higher-dimensional Euclidean space: once an orientation is chosen, it makes sense to distinguish counterclockwise turning from clockwise turning. This gives curvature a sign.

5.1 Rotation matrices

For 𝜃 ∈ ℝ, define the counterclockwise rotation through angle 𝜃 by

R𝜃 = ( cos⁡𝜃 −sin⁡𝜃 sin⁡ 𝜃 cos⁡ 𝜃 ).

Thus R𝜃 : ℝ2 → ℝ2 is a linear map. The rotation through 90∘ is

R90 = Rπ∕2 = ( 0 −1 1 0 ),R90(x,y) = (−y,x).

For every 𝐰 ∈ ℝ2,

⟨𝐰,R90(𝐰)⟩ = 0, |R90(𝐰)| = |𝐰|,R902(𝐰) = −𝐰.

Consequently, if 𝐰≠0, then R90(𝐰) spans the line perpendicular to 𝐰.

A vector in the plane and its counterclockwise rotation by ninety degrees.

Figure 5.1: The operator R90 rotates vectors counterclockwise by 90∘.

5.2 Signed curvature

Let

𝜸 : I→ℝ2

be a regular plane curve, and write

𝐯 = 𝜸′,𝐚 = 𝜸″.

The vector

R90 ( 𝐯 |𝐯| )

is a unit vector spanning (span ⁡ {𝐯})⊥. Orthogonal projection therefore gives

𝐚⊥ = ⟨𝐚,R 90 ( 𝐯 |𝐯| )⟩R90 ( 𝐯 |𝐯| ).

Dividing by |𝐯|2, the curvature vector from Lecture 4 becomes

𝐚⊥ |𝐯|2 = ⟨ 𝐚 |𝐯|2,R90 ( 𝐯 |𝐯|)⟩⏟κsR90 ( 𝐯 |𝐯| ).

The scalar coefficient records not only the amount of turning, but also its direction.

Definition 5.1: Signed curvature

Let 𝜸 : I → ℝ2 be a regular plane curve. Its signed curvature function is

κs(t) = ⟨ 𝐚(t) |𝐯(t)|2,R90 ( 𝐯(t) |𝐯(t)|)⟩ = ⟨𝐚(t),R90(𝐯(t))⟩ |𝐯(t)|3 .

The curvature vector and the signed curvature are related by

𝐚⊥ |𝐯|2 = κsR90 ( 𝐯 |𝐯| ),κ = |κs| .

Thus κs > 0 means that the curve is turning counterclockwise (to the left), while κs < 0 means that it is turning clockwise (to the right).

5.2.1 The unit-speed case

If 𝜸 has unit speed, then |𝐯| = 1. Since differentiating ⟨𝐯,𝐯⟩ = 1 gives ⟨𝐚,𝐯⟩ = 0, all of the acceleration is perpendicular to the velocity. Hence

𝐚 = κsR90(𝐯) ,κs = ⟨𝐚,R90(𝐯)⟩,κ = |κs| .

Example 5.2 (A circle parametrized by arc length). Let r > 0 and

𝜸(t) = (rcos⁡ t r,rsin⁡ t r ).

Then

𝐯(t) = (−sin⁡ t r,cos⁡ t r ),𝐚(t) = 1 r (−cos⁡ t r,−sin⁡ t r ) = 1 rR90(𝐯(t)).

Therefore κs = 1∕r. Reversing the orientation of the circle changes the signed curvature to −1∕r, while its ordinary curvature remains 1∕r.

5.3 Angle functions

For a unit-speed plane curve, the velocity takes values in the unit circle

S1 = {𝐩 ∈ ℝ2 : |𝐩| = 1}.

Thus we have a smooth map 𝐯 : I →S1. An angle function is a smooth lift of this map through the standard parametrization of the circle:

An angle function theta lifting the unit tangent map to the real line.

Definition 5.3: Angle function

Let 𝜸 : I → ℝ2 be a unit-speed plane curve with velocity 𝐯. An angle function for 𝜸 is a smooth function 𝜃 : I → ℝ such that

𝐯(t) = (cos⁡𝜃(t),sin⁡𝜃(t))

for every t ∈I.

A single inverse-trigonometric formula does not work globally: for example, arccos⁡(vx) loses the sign of vy, and forcing 𝜃 ∈ [0,2π) creates jumps whenever the tangent direction passes through angle 2π. Locally one may choose a suitable branch of arcsin⁡ or arccos⁡; the next result shows that these local choices can be assembled into a smooth function on the whole interval.

Proposition 5.4: Existence of a global angle function

Every unit-speed plane curve has a smooth angle function. It is unique up to addition of a constant integer multiple of 2π. Moreover,

𝜃′(t) = κ s(t).

Proof. Suppose first that 𝜃 is a local angle function. Differentiating 𝐯 = (cos⁡𝜃,sin⁡𝜃) gives

𝐚 = 𝐯′ = 𝜃′(−sin⁡𝜃,cos⁡𝜃) = 𝜃′R 90(𝐯).

Comparing this with 𝐚 = κsR90(𝐯) shows that 𝜃′ = κs.

Choose t0 ∈I and 𝜃0 ∈ ℝ with 𝐯(t0) = (cos⁡𝜃0,sin⁡𝜃0), and define

𝜃(t) = 𝜃0 + ∫ t0tκ s(u) 𝑑𝑢.

Let

A = {t ∈I : 𝐯(t) = (cos⁡𝜃(t),sin⁡𝜃(t))}.

The set A contains t0 and is closed by continuity. If t ∈A, choose a local inverse-trigonometric branch giving an angle function near t. After adding a multiple of 2π, the local angle agrees with 𝜃 at t; since both have derivative κs, they agree on a neighborhood of t. Thus A is also open. Because the interval I is connected, A = I. The same derivative argument proves uniqueness up to a constant multiple of 2π. □

The formula

𝜃(t) = 𝜃(t0) + ∫ t0tκ s(u) 𝑑𝑢

says that signed curvature is the instantaneous counterclockwise turning rate of the unit tangent.

Remark 5.5 (A general parametrization). If the curve is regular but not necessarily unit speed, an angle function for its unit tangent

𝖙(t) = 𝐯(t) |𝐯(t)| = (cos⁡𝜃(t),sin⁡𝜃(t))

satisfies

𝜃′(t) = κ s(t) |𝐯(t)|.

This follows by passing to an orientation-preserving unit-speed reparametrization.

5.4 Rotation index

For a closed plane curve, the unit tangent returns to its initial direction. The angle function itself may nevertheless change by one or more full turns.

Definition 5.6: Rotation index

Let 𝜸 : [a,b] → ℝ2 be a unit-speed closed plane curve, and let 𝜃 be an angle function. The rotation index of 𝜸 is

rot ⁡ (𝜸) = 𝜃(b) −𝜃(a) 2π .

For a regular closed plane curve of variable speed, the rotation index means the rotation index of an orientation-preserving unit-speed reparametrization.

Because 𝐯(a) = 𝐯(b), the difference 𝜃(b) −𝜃(a) is an integer multiple of 2π. Thus the rotation index is an integer. It counts the net number of times the unit tangent winds counterclockwise around the origin. In particular, a counterclockwise circle has rotation index 1, while the same circle with the opposite orientation has rotation index −1.

Corollary 5.7 (Total signed curvature). For a unit-speed closed plane curve,

rot ⁡ (𝜸) = 1 2π∫ abκ s(t) 𝑑𝑡.

For a regular closed plane curve with arbitrary parameter,

rot ⁡ (𝜸) = 1 2π∫ abκ s(t) |𝐯(t)| 𝑑𝑡.

Proof. Integrate 𝜃′ = κs in the unit-speed case. For a general parameter, integrate 𝜃′ = κs |𝐯|. □

5.5 Summary

Lecture 6
Space Curves: Frenet Frame, Torsion, and Frenet Equations

Wednesday, September 2, 2026

Reading. [Tap16, § 1.7, pp. 41–47].

After class. Tapp, Exercises 1.62–1.66 and 1.68.

Goals. Review the cross product in ℝ3, build the Frenet frame of a space curve, define torsion using Tapp’s sign convention, derive the Frenet equations, and explain why vanishing torsion characterizes planar curves.

6.1 The cross product

The special geometry of curves in ℝ3 depends on the cross product. Throughout this lecture, vectors and vector-valued functions are written in boldface, following Tapp.

Definition 6.1: Cross product

Let

𝐚 = (a1,a2,a3),𝐛 = (b1,b2,b3) ∈ ℝ3.

Their cross product is

𝐚×𝐛 = | 𝐢 𝐣 𝐤 a 1 a2 a3 b1 b2 b3 | = (a2b3−a3b2,a3b1−a1b3,a1b2−a2b1).

The determinant in the middle is the standard mnemonic for the coordinate formula on the right.

For 𝐚,𝐛,𝐜 ∈ ℝ3, you can check that

⟨𝐜,𝐚 ×𝐛⟩ = det ⁡ ( c1 c2 c3 a1 a2 a3 b1 b2 b3 ) . (6.1)

Lemma 6.2 (Basic properties of the cross product). For 𝐚,𝐛,𝐜 ∈ ℝ3 and λ,μ ∈ ℝ, the following hold.

  1. 𝐚 ×𝐛 is orthogonal to both 𝐚 and 𝐛. Thus

    𝐚 ×𝐛 ∈ span ⁡ {𝐚,𝐛}⊥.
  2. 𝐚 ×𝐛 = −(𝐛 ×𝐚).
  3. The cross product is linear in each input; for example,

    (λ𝐚 + μ𝐛) ×𝐜 = λ(𝐚 ×𝐜) + μ(𝐛 ×𝐜).
  4. |𝐚 ×𝐛|2 = |𝐚|2 |𝐛|2 −⟨𝐚,𝐛⟩2.

    If 𝐚 and 𝐛 are nonzero and 𝜃 = ∠(𝐚,𝐛), then

    |𝐚 ×𝐛| = |𝐚| |𝐛|sin⁡𝜃.
  5. The direction of 𝐚 ×𝐛 is determined by the right-hand rule.

Proof. The first statement follows from (6.1), because a repeated row makes the determinant zero. The algebraic identities follow by expanding the coordinate formula. Finally,

|𝐚|2 |𝐛|2 −⟨𝐚,𝐛⟩2 = |𝐚|2 |𝐛|2(1 −cos⁡2𝜃),

which gives the length formula. □

Two vectors and their cross product, perpendicular to their spanning plane; its length is the parallelogram area.

Figure 6.1: The cross product is perpendicular to the spanned plane, and its length is the area of the parallelogram.

6.1.1 Rotations and the cross product

A matrix R ∈M3×3(ℝ) is a rotation matrix if

RT R = Iand det ⁡ R = 1.

Proposition 6.3 (Rotations preserve the cross product). If R is a rotation matrix, then

R(𝐚 ×𝐛) = (R𝐚) × (R𝐛)

for all 𝐚,𝐛 ∈ ℝ3.

Proof. For arbitrary 𝐜 ∈ ℝ3, use preservation of the inner product and the scalar triple product:

⟨𝐜,(R𝐚) × (R𝐛)⟩ = det ⁡ (𝐜,R𝐚,R𝐛) = det ⁡ (R(RT 𝐜),R𝐚,R𝐛) = det ⁡ (R)det ⁡ (RT 𝐜,𝐚,𝐛) = ⟨RT 𝐜,𝐚 ×𝐛⟩ = ⟨𝐜,R(𝐚 ×𝐛)⟩.

Since this holds for every 𝐜, the two vectors are equal. □

Remark 6.4. The condition det ⁡ R = 1 matters. More generally, for an orthogonal matrix Q ∈O(3), i.e., QQT = QT Q = I,

(Q𝐚) × (Q𝐛) = det ⁡ (Q)Q(𝐚 ×𝐛).

Thus a reflection reverses the cross-product direction.

6.2 The Frenet frame

Let 𝜸 : I → ℝ3 be a regular space curve. At a time t for which κ(t)≠0, the unit tangent and unit normal are already defined by

𝖙(t) = 𝐯(t) |𝐯(t)|,𝖓(t) = 𝐚⊥(t) |𝐚⊥(t)| = 𝖙′(t) |𝖙′(t)|.

The cross product supplies a third vector.

Definition 6.5: Frenet frame

Let 𝜸 : I → ℝ3 be a regular space curve and suppose κ(t)≠0. The unit binormal vector is

𝖇(t) = 𝖙(t) ×𝖓(t).

The ordered orthonormal basis

{𝖙(t),𝖓(t),𝖇(t)}

is called the Frenet frame at time t.

The tangent, normal, and binormal vectors of a space curve at a point.

Figure 6.2: The Frenet frame: 𝖙 and 𝖓 span a plane, with 𝖇 orthogonal to this plane.

6.3 The Frenet equations

Proposition 6.6: The Frenet equations

Let 𝜸 : I → ℝ3 be a regular space curve. At every time for which κ≠0,

𝖙′ = |𝐯|κ𝖓, 𝖓′ = −|𝐯|κ𝖙 + |𝐯|τ𝖇, 𝖇′ = −|𝐯|τ𝖓.

where τ is defined to make the second equation be true. Equivalently,

( 𝖙 𝖓 𝖇 )′ = |𝐯| ( 0 κ 0 −κ 0 τ 0 −τ 0 ) ( 𝖙 𝖓 𝖇 ).

Proof. The first Frenet equation was proved in Lecture 4, but let us recall quickly. Since 𝐯 = |𝐯|𝖙, differentiation gives

𝐚 = |𝐯|′𝖙⏟ 𝐚∥ + |𝐯|𝖙′⏟ 𝐚⊥

The identity |𝖙| = 1 implies 𝖙′ ⊥𝖙. The first term above is therefore parallel to 𝐯, while the second is perpendicular to 𝐯. Hence

𝐚⊥ = |𝐯|𝖙′.

Thus we have

𝜿 = 𝐚⊥ |𝐯|2 = |𝐯|𝖙′ |𝐯|2 = 𝖙′ |𝐯| = κ𝖓.

Clearing denominators in the last equation gives 𝖙′ = |𝐯|κ𝖓.

For the second Frenet equation, compute the three components of 𝖓′ in the orthonormal basis {𝖙,𝖓,𝖇}:

⟨𝖓′,𝖙⟩ = −⟨𝖓,𝖙′⟩ = −|𝐯|κ, ⟨𝖓′,𝖓⟩ = 0, ⟨𝖓′,𝖇⟩ =: |𝐯|τ.

In other words, we define τ := ⟨𝖓′,𝖇⟩ |𝐯| .

This gives

𝖓′ = −|𝐯|κ𝖙 + |𝐯|τ𝖇.

Finally, 𝖇′ is orthogonal to both 𝖇 and 𝖙, so it is parallel to 𝖓. By the definition of τ, we have

⟨𝖇′,𝖓⟩ = −⟨𝖇,𝖓′⟩ = −|𝐯|τ,

which proves the third equation. □

The definition of τ above agrees with the definition in Tapp:

Definition 6.7: Torsion (Tapp)

Let 𝜸 : I → ℝ3 be a regular space curve, and let t ∈I satisfy κ(t)≠0. The torsion of 𝜸 at t is

τ(t) = −⟨𝖇′(t),𝖓(t)⟩ |𝐯(t)| .

Remark 6.8 (Sign convention). This is Tapp’s convention. Some authors define torsion with the opposite sign. The factor |𝐯| makes torsion independent of the speed of parametrization.

Remark 6.9 (Behavior under reparametrization). Torsion is independent of parametrization. Under an orientation-preserving reparametrization, all three Frenet vectors simply compose with the change of parameter. Under an orientation-reversing reparametrization, 𝖙 and 𝖇 change sign while 𝖓 does not; the signs in the definition of τ therefore cancel.

6.4 Some further topics

6.4.1 Zero torsion

Proposition 6.10. Let 𝜸 : I → ℝ3 be a regular curve with κ(t)≠0 for all t ∈I. The trace of 𝜸 is constrained to a plane if and only if τ(t) = 0 for all t ∈I.

Proof. If 𝜸 is constrained to a plane, then it is easy to see that 𝖇′ = 0, so that using the Frenet equations, we have that τ(t) = 0 for all t ∈I.

Conversely, if τ(t) = 0 for all t ∈I, then the Frenet equations tell us that 𝖇′ = 0, so that 𝖇 is constant. The claim is then that ⟨𝜸(t),𝖇⟩ = C is constant for all t (which, since 𝖇 is constant, implies that 𝜸 is in a translate of the plane orthogonal to 𝖇).

To see that ⟨𝜸(t),𝖇⟩ is constant, we have

⟨𝜸(t),𝖇⟩′ = ⟨𝐯,𝖇⟩ + ⟨𝜸(t),𝖇′⟩ = ⟨𝐯,𝖇⟩ = 0

since 𝐯 is a scalar multiple of 𝖙, and 𝖙 ⊥𝖇. □

6.4.2 Product rule

Lemma 6.11 (Product rule). If 𝜶,𝜷 : I → ℝ3 are smooth, then

d 𝑑𝑡(𝜶(t) ×𝜷(t)) = 𝜶′(t) ×𝜷(t) + 𝜶(t) ×𝜷′(t).

Proof. Differentiate the coordinate formula component by component and apply the ordinary product rule. □

6.4.3 A curvature formula in ℝ3

Proposition 6.12: Curvature from the cross product

Let 𝜸 : I → ℝ3 be a regular space curve, with

𝐯 = 𝜸′,𝐚 = 𝜸″.

Then

κ(t) = |𝐯(t) ×𝐚(t)| |𝐯(t)|3 .

Proof. If 𝐚 = 0, both sides vanish. Otherwise, let 𝜃 be the angle between 𝐯 and 𝐚. Then |𝐚⊥| = |𝐚|sin⁡𝜃. Therefore

κ = |𝐚⊥| |𝐯|2 = |𝐯| |𝐚|sin⁡𝜃 |𝐯|3 = |𝐯 ×𝐚| |𝐯|3 .
□

6.5 Summary

Lecture 7
Rigid Motions and Fundamental Theorems

Friday, September 4, 2026

Reading. [Tap16, §§ 1.8–1.9, pp. 48–60].

After class. HW 2 is due. Begin HW 3(a): Tapp, Exercises 1.70, 1.71, 1.75, 1.76, 1.78, and 1.83.

Goals. Define and classify rigid motions of Euclidean space, determine how curvature, signed curvature, and torsion behave under rigid motions, and state the fundamental theorems of plane and space curves. We prove the plane-curve theorem by reconstructing a curve from its signed curvature.

A rigid motion changes the position of a curve, and may also rotate or reflect it, without changing any distances. Thus rigid motions formalize the idea that two curves have the same shape but are viewed in different coordinate systems. We have been studying curves embedded in Euclidean space ℝn, and we will be interested in this lecture in understanding how our notions such as curvature and torsion change (or rather do not change) under rigid motions of the curve. In other words, we want to see that curvature and torsion are intrinsic to the curve in space, and not to its parameterization, or our coordinates (so long as we change coordinates in a way that preserves distances).

7.1 Rigid motions of Euclidean space

Definition 7.1: Rigid motion

A rigid motion of ℝn is a function

f : ℝn→ℝn

that preserves distances. Equivalently,

|f(𝐩) −f(𝐪)| = |𝐩 −𝐪|

for all 𝐩,𝐪 ∈ ℝn.

The inner product can be recovered from distances by the polarization identity

⟨𝐩,𝐪⟩ = 1 2 (|𝐩|2 + |𝐪|2 −|𝐩 −𝐪|2) . (7.1)

In particular, if a rigid motion fixes the origin, then it preserves norms and inner products.

For a matrix A ∈Mn×n(ℝ), write

LA(𝐱) = A𝐱.

Recall that A is orthogonal if

AT A = AAT = I.

The set of orthogonal n ×n matrices is denoted by O(n); this is called the orthogonal group. The following conditions are equivalent (if they hold for all 𝐩,𝐪 ∈ ℝn):

A ∈O(n), ⟨A𝐩,A𝐪⟩ = ⟨𝐩,𝐪⟩, |A𝐩| = |𝐩|.

Thus the linear rigid motions are exactly the maps LA with A ∈O(n).

For 𝐪 ∈ ℝn, let

T𝐪(𝐱) = 𝐱 + 𝐪

denote translation by 𝐪.

Theorem 7.2: Classification of rigid motions

Every rigid motion f : ℝn → ℝn has a unique expression

f = T𝐪 ∘LA,

where 𝐪 ∈ ℝn and A ∈O(n). Equivalently,

f(𝐱) = A𝐱 + 𝐪.

Proof. Set 𝐪 = f(0) and define

g = T−𝐪 ∘f.

Then g is a rigid motion with g(0) = 0. By (7.1),

⟨g(𝐩),g(𝐫)⟩ = ⟨𝐩,𝐫⟩

for all 𝐩,𝐫 ∈ ℝn.

Let 𝐞1,…,𝐞n be the standard basis, and let A be the matrix whose ith column is g(𝐞i). These columns form an orthonormal basis, so A ∈O(n). Define

h = LA−1 ∘g.

Then h preserves inner products and fixes every 𝐞i. If 𝐩 = ∑ ⁡ ipi𝐞i and h(𝐩) = ∑ ⁡ ibi𝐞i, then

bi = ⟨h(𝐩),𝐞i⟩ = ⟨h(𝐩),h(𝐞i)⟩ = ⟨𝐩,𝐞i⟩ = pi.

Thus h is the identity, so g = LA and f = T𝐪 ∘LA.

For uniqueness, evaluating at 0 determines 𝐪 = f(0), after which the linear part A is determined as well. □

Since AT A = I,

1 = det ⁡ (AT A) = det ⁡ (A)2,

so det ⁡ A = 1 or det ⁡ A = −1.

Recall that GL ⁡ n(ℝ) is the set of n ×n matrices with non-zero determinant, and is called the general linear group. A matrix A ∈ GL ⁡ n(ℝ) is said to be orientation preserving if det ⁡ (A) > 0 (and is orientation reversing if det ⁡ (A) < 0). A rigid motion is called orientation preserving if the matrix A in the theorem above has determinant 1.

Definition 7.3: Proper and improper rigid motions

Suppose f = T𝐪 ∘LA is a rigid motion. Following Tapp, f is called proper if det ⁡ A = 1, and improper if det ⁡ A = −1. Proper rigid motions preserve orientation; improper rigid motions reverse orientation.

A curve and its image under a rigid motion, with corresponding points p and q.

Figure 7.1: A rigid motion changes the position of a curve but preserves the distance between every pair of corresponding points.

7.2 Geometric quantities under rigid motion

Let 𝜸 : I → ℝn be regular, let f = T𝐪 ∘LA be a rigid motion, and define

𝜸^ = f ∘𝜸 = A𝜸 + 𝐪.

Writing 𝐯 = 𝜸′ and 𝐚 = 𝜸″ = 𝐯′, and similarly for 𝜸^, differentiation gives

𝐯^ := 𝜸^′ = (A𝜸 + 𝐪)′ = A′𝜸 + A𝜸′ + 𝐪′ = A𝜸′ = A𝐯, (7.2) 𝐚^ := (𝐯^)′ = (A𝐯)′ = A′𝐯 + A𝐯′ = A𝐚, (7.3)

since A and 𝐪 are constant.

Because A is orthogonal, it preserves inner products, lengths, and orthogonality.

Proposition 7.4: Invariance under rigid motions

For a regular curve, the following statements hold.

  1. Curvature is invariant under every rigid motion:

    κ^ = κ.
  2. For a space curve, torsion is preserved by proper rigid motions and changes sign under improper rigid motions:

    τ^ = det ⁡ (A)τ.
  3. For a plane curve, signed curvature is preserved by proper rigid motions and changes sign under improper rigid motions:

    κ^s = det ⁡ (A)κs.

Proof. Equation (7.3) implies

𝐚^⊥ = A𝐚⊥, |𝐯^| = |𝐯|.

Therefore, since A preserves lengths,

κ^ = |𝐚^⊥| |𝐯^|2 = |A𝐚⊥| |A𝐯|2 = |𝐚⊥| |𝐯|2 = κ.

For a space curve at a point where κ≠0,

𝖙^ = A𝖙,𝖓^ = A𝖓.

The transformation rule for the cross product is

(A𝐱) × (A𝐲) = det ⁡ (A)A(𝐱 ×𝐲),

so

𝖇^ = det ⁡ (A)A𝖇.

Using Tapp’s torsion convention,

τ^ = −⟨𝖇^′,𝖓^⟩ |𝐯^| = −det ⁡ (A) ⟨A𝖇′,A𝖓⟩ |𝐯| = det ⁡ (A)τ.

In the plane,

R90(A𝐰) = det ⁡ (A)AR90(𝐰).

Substituting this identity into the formula for signed curvature gives

κ^s = ⟨A𝐚,R90(A𝐯)⟩ |A𝐯|3 = det ⁡ (A)κs.
□

Thus ordinary curvature records bending without orientation, while signed curvature and torsion also detect handedness. A reflection preserves the amount of bending but reverses the relevant sign.

7.3 The fundamental theorems of curves

The preceding proposition shows that curvature data are unchanged by proper rigid motions. The converse is the fundamental result: the appropriate curvature data determine the curve up to a proper rigid motion.

Theorem 7.5: The fundamental theorems of plane and space curves

Let I ⊆ ℝ be an interval.

  1. If κs : I → ℝ is smooth, then there exists a unit-speed plane curve 𝜸 : I → ℝ2 whose signed curvature is κs. If 𝜸 and 𝜸^ are two such curves, then

    𝜸^ = f ∘𝜸

    for some proper rigid motion f of ℝ2.

  2. If κ,τ : I → ℝ are smooth and κ > 0, then there exists a unit-speed space curve 𝜸 : I → ℝ3 whose curvature and torsion are κ and τ. If 𝜸 and 𝜸^ are two such curves, then

    𝜸^ = f ∘𝜸

    for some proper rigid motion f of ℝ3.

The theorem has two parts. Existence says that allowable curvature data come from a curve. Uniqueness says that these data determine the curve’s shape: the only remaining freedom is its initial position and initial orientation.

7.3.1 The plane-curve construction

Proof of the plane-curve statement. Fix t0 ∈I. We first construct the curve with initial conditions

𝜸(t0) = (0,0),𝜸′(t 0) = (1,0).

Define

𝜃(t) := ∫ t0tκ s(u) 𝑑𝑢, (7.4)

then define the velocity and the curve by

𝐯(t) := (cos⁡𝜃(t),sin⁡𝜃(t)),𝜸(t) := ∫ t0t𝐯(u) 𝑑𝑢. (7.5)

The vector integral is taken componentwise. The construction can be remembered as

κs →integrate𝜃 →(cos⁡𝜃,sin⁡𝜃)𝐯 →integrate𝜸.

By the Fundamental Theorem of Calculus,

𝜸′ = 𝐯, |𝐯| = 1,

so 𝜸 has unit speed. Furthermore,

𝐚 = 𝐯′ = 𝜃′(−sin⁡𝜃,cos⁡𝜃) = κ sR90(𝐯).

For a unit-speed plane curve, this equation says exactly that the signed curvature is κs. This proves existence. Adding a constant to 𝜃 and a constant vector to 𝜸 gives arbitrary initial direction and initial position.

For uniqueness, let 𝜸 and 𝜸^ be unit-speed plane curves with the same signed curvature κs, and choose angle functions 𝜃 and 𝜃^. After applying a rigid motion to each curve, we can assume that 𝜸(t0) = 𝜸^(t0) = 0, and 𝜸′(t0) = 𝜸^′(t0) = [ 1 0 ]. Since

𝜃′ = κ s = 𝜃^′,

there is a constant 𝜃0 such that

𝜃^ = 𝜃 + 𝜃0.

But then, plugging into 𝐯(t) := (cos⁡𝜃(t),sin⁡𝜃(t)) and 𝐯^(t) := (cos⁡𝜃^(t),sin⁡𝜃^(t)), and setting t = t0, we see that 𝜃0 is a multiple of 2π, so we can take 𝜃0 = 0. Therefore 𝐯 = 𝐯^.

Finally, since

𝜸′(t) =: 𝐯(t) = 𝐯^(t) := 𝜸^′(t)

we see that 𝜸^(t) = 𝜸(t) + 𝜸0. Setting t = t0, and using that 𝜸(t) = 𝜸^(t) = 0, we see that 𝜸^ = 𝜸. □

Remark 7.6 (Sketch of proof of the space-curve statement and ODE). Tapp states the space-curve theorem without proof because its proof uses existence and uniqueness for ordinary differential equations. For unit speed, the Frenet equations are

𝖙′ = κ𝖓,𝖓′ = −κ𝖙 + τ𝖇,𝖇′ = −τ𝖓.

In matrix form, in the form you would use the result from ODE, this is (in block form)

( 𝖙 𝖓 𝖇 )′⏟ 𝐅′(t) = ( 0 κ 0 − κ 0 τ 0 −τ 0 )⏟Ω ( 𝖙 𝖓 𝖇 )⏟ 𝐅(t),

where the 9 × 9 (skew symmetric) matrix Ω and the 9 × 1 column matrix 𝐅(t) are defined as indicated.

Given κ > 0, τ, and an initial positively oriented orthonormal frame 𝐅0 = (𝖙0,𝖓0,𝖇0), this linear first-order system has a unique solution 𝐅(t) = (𝖙(t),𝖓(t),𝖇(t)) with 𝐅(t0) = (𝖙(t0),𝖓(t0),𝖇(t0)) = 𝐅0 = (𝖙0,𝖓0,𝖇0).

Note that since Ω = −ΩT , we have that

(𝐅(t)T 𝐅(t))′ = (𝐅T )′𝐅 + 𝐅T 𝐅′ = (𝐅′)T 𝐅 + 𝐅T 𝐅′ = (Ω𝐅)T 𝐅 + 𝐅T (Ω𝐅) = 𝐅T ΩT 𝐅 + 𝐅T Ω𝐅 = 0,

so that 𝐅T (t)𝐅(t) is constant. Since 𝐅(t0)T 𝐅(t0) = I (since 𝖙0, 𝖓0, and 𝖇0 form an orthonormal basis), we see that 𝐅T (t)𝐅(t) = I for all t, so that for all t, we have that 𝖙(t), 𝖓(t), and 𝖇(t) form an orthonormal basis. For the orientation, make a new 3 × 3 matrix G(t) with columns given by 𝖙, 𝖓, and 𝖇. We know that GT G = I for all t, so that G(t) ∈O(3) for all t; therefore det ⁡ (G(t)) = ±1. Since the determinant is is a polynomial in the entries, it is continuous, and therefore since det ⁡ (G(t0)) = 1, we know that 𝖙(t), 𝖓(t), and 𝖇(t) are positively oriented for all t.

Defining

𝜸(t) = 𝐩0 + ∫ t0t𝖙(u) 𝑑𝑢

then produces a curve with 𝜸(t0) = 𝐩0, and, from the Fundamental Theorem of Calculus, we have that 𝜸′(t) = 𝖙, so that the curve has unit speed. Then the differential equations above imply that 𝖙′ = κ𝖓, so that the unit normal for 𝜸 is equal to 𝖓, and the curvature is κ. The differential equations above also give us that 𝖓′ = −κ𝖙 + τ𝖇, so that the torsion is τ.

Uniqueness for the ODE, after using a proper rigid motion to align the initial points and Frenet frames, gives uniqueness of the curve up to a proper rigid motion.

7.4 A short overview of curvature formulas

Tapp’s § 1.9 collects the formulas developed in the preceding lectures. For a regular curve in ℝn, (Theorem 4.6)

κ = |𝐚⊥| |𝐯|2 = |𝖙′| |𝐯|, κ = |𝐚|when the curve has unit speed.

For a plane curve 𝜸(t) = (x(t),y(t)),

κs = ⟨𝐚,R90(𝐯)⟩ |𝐯|3 = x′y″ −x″y′ ((x′)2 + (y′)2)3∕2.

In particular, for the graph parametrization 𝜸(t) = (t,f(t)),

κs(t) = f″(t) (1 + (f′(t))2)3∕2.

For a space curve (see Section 6.4.3),

κ = |𝐯 ×𝐚| |𝐯|3 .

7.5 Summary

Lecture 8
Hopf’s Umlaufsatz and the Jordan Curve Theorem

Wednesday, September 9, 2026

Reading. [Tap16, § 2.1, pp. 61–67].

After class. HW 3(b): Tapp, Exercises 1.52, 1.55, 1.58, 2.1, 2.2, and 2.3.

Goals. Interpret the rotation index as the degree of the unit tangent map, prove Hopf’s Umlaufsatz using a family of secant directions, and use winding number together with a tubular neighborhood to explain the Jordan curve theorem. We conclude with the piecewise-regular formula recorded in class.

The results in the preceding lectures were principally local: they described what a curve is doing near one point. The two theorems in this lecture are global. They constrain the behavior of an entire simple closed plane curve. The German word Umlaufsatz means, roughly, a “going-around theorem.”

Theorem 8.1: Theorems of Hopf and Jordan

Let 𝜸 : [a,b] → ℝ2 be a simple closed plane curve, and let

C = 𝜸([a,b])

denote its trace.

  1. Hopf’s Umlaufsatz. The rotation index of 𝜸 is either 1 or −1.
  2. Jordan curve theorem. The complement ℝ2 −C has exactly two path-connected components. Their common boundary is C. One component is bounded and is called the interior; the other is unbounded and is called the exterior.

8.1 Degree and the unit tangent map

Write

S1 = {(x,y) ∈ ℝ2 : x2 + y2 = 1}.

The rotation index introduced in Lecture 5 is a special case of a topological integer attached to any closed path in the circle.

Definition 8.2: Degree of a circle-valued loop

Let 𝐟 : [a,b] →S1 be continuous and suppose that 𝐟(a) = 𝐟(b). A continuous angle function for 𝐟 is a continuous function 𝜃 : [a,b] → ℝ such that

𝐟(t) = (cos⁡𝜃(t),sin⁡𝜃(t)).

Such an angle function exists and is unique up to adding a constant integer multiple of 2π. The degree of 𝐟 is

deg ⁡ (𝐟) = 𝜃(b) −𝜃(a) 2π ∈ ℤ.

The integer condition follows from 𝐟(a) = 𝐟(b): the two endpoint angles differ by an integer multiple of 2π. Existence of a global angle function can be proved by covering S1 with semicircles, choosing local inverse branches of 𝜃↦(cos⁡𝜃,sin⁡𝜃), and matching the branches on finitely many subintervals.

If 𝜸 is a regular closed plane curve, its unit tangent map is

𝖙(t) = 𝜸′(t) |𝜸′(t)| ∈S1.

The rotation index is exactly

rot ⁡ (𝜸) = deg ⁡ (𝖙).

For a unit-speed curve, the formula 𝜃′ = κs gives

rot ⁡ (𝜸) = 1 2π∫ abκ s(t) 𝑑𝑡.

For an arbitrary regular parametrization, the corresponding formula is

rot ⁡ (𝜸) = 1 2π∫ abκ s(t) |𝜸′(t)| 𝑑𝑡.

A simple closed plane curve and its unit tangent vectors.

The corresponding unit tangent map taking values on the unit circle.
Figure 8.1: The rotation index is the degree of the unit tangent map.

The next lemma is the basic stability principle used in both global proofs.

Lemma 8.3 (Antipodal-value lemma). Let 𝐟0,𝐟1 : [a,b] →S1 be continuous loops. If deg ⁡ (𝐟0)≠deg ⁡ (𝐟1), then for some t0 ∈ [a,b],

𝐟0(t0) = −𝐟1(t0).

Equivalently, if the two vectors are never opposite, then the two maps have the same degree.

Proof. Choose angle functions 𝜃0 and 𝜃1, and set δ = 𝜃1 −𝜃0. If the degrees differ, then

|δ(b) −δ(a)| ≥ 2π.

Consequently the interval between δ(a) and δ(b) contains an odd multiple of π. By the intermediate value theorem, δ(t0) is an odd multiple of π for some t0, which is exactly the assertion that 𝐟0(t0) = −𝐟1(t0). □

Corollary 8.4 (Homotopy invariance of degree). Suppose 𝐇 : [0,1] × [a,b] →S1 is continuous and 𝐇(s,a) = 𝐇(s,b) for every s. Then

deg ⁡ (𝐇(s, ⋅))

is independent of s.

Proof. Uniform continuity implies that for nearby values of s, the two loops are pointwise close and hence never antipodal. Lemma 8.3 therefore shows that their degrees agree. Thus the integer-valued degree is locally constant as a function of s, and hence constant on [0,1]. □

8.2 Hopf’s Umlaufsatz

Proof of Hopf’s Umlaufsatz. Reparametrize 𝜸 by arc length. Choose a point 𝐩 ∈C such that the entire trace lies on one side of the tangent line at 𝐩. Such a point exists: place the curve inside a large circle and shrink the circle until it first touches the trace. After shifting the parameter, assume

𝜸(a) = 𝜸(b) = 𝐩.

Consider the closed triangle

T = {(t1,t2) : a ≤t1 ≤t2 ≤b}.

Define the secant-direction map 𝝍 : T →S1 by

𝝍(t1,t2) = { 𝜸(t2) −𝜸(t1) |𝜸(t2) −𝜸(t1)|,t1 < t2,(t1,t2)≠(a,b), 𝖙(t1), t1 = t2, −𝖙(a), (t1,t2) = (a,b).

The derivative definition of 𝖙 gives continuity along the diagonal, and the special value at (a,b) makes the map continuous at the remaining corner.

Let 𝜶0 traverse the diagonal of T from (a,a) to (b,b). Then

𝝍∘𝜶0 = 𝖙,

so

deg ⁡ (𝝍∘𝜶0) = rot ⁡ (𝜸).

Let 𝜶1 instead travel from (a,a) to (a,b) and then from (a,b) to (b,b). Since T is convex, the paths

𝜶s = (1 −s)𝜶0 + s𝜶1,0 ≤s ≤ 1,

remain in T and give a homotopy through loops in S1 after composing with 𝝍. Corollary 8.4 gives

deg ⁡ (𝝍∘𝜶0) = deg ⁡ (𝝍∘𝜶1).

Along the first edge of 𝜶1, the secant vectors point from 𝐩 to the rest of the curve. Because the curve lies on one side of the supporting tangent line, these vectors remain in one semicircle and turn from 𝖙(a) to −𝖙(a). Along the second edge, their negatives traverse the other semicircle. The total change in angle is therefore 2π or −2π, depending on the orientation. Hence

rot ⁡ (𝜸) = 1or rot ⁡ (𝜸) = −1.
□

A supporting tangent line to a simple closed curve at its starting point.

A supporting tangent line at 𝐩.

The triangular parameter domain and the two boundary paths used in the proof of Hopf’s theorem. The diagonal path is deformed to the two outer edges.
Figure 8.2: The two geometric ingredients in the proof of Hopf’s Umlaufsatz.

8.3 The Jordan curve theorem

The proof uses two ideas from the handwritten notes: a narrow tubular neighborhood of the curve and the winding number of the curve about a point.

Proposition 8.5: Tubular neighborhood of a simple closed curve

After an arc-length reparametrization, let 𝖙 be the unit tangent of the simple closed plane curve 𝜸. For sufficiently small 𝜀 > 0, the map

Φ : (−𝜀,𝜀) × [a,b)→ℝ2,Φ(s,t) = 𝜸(t) + sR 90(𝖙(t)),

is injective.

Its image is a tubular neighborhood of C. Removing C leaves two path-connected strips,

Φ((0,𝜀) × [a,b))andΦ((−𝜀,0) × [a,b)).

Tapp postpones the proof of injectivity until the inverse function theorem is available; locally it follows from regularity, while compactness and simplicity prevent different normal segments from meeting globally.

For 𝐩 ∈ ℝ2 −C, define

𝐟𝐩(t) = 𝜸(t) −𝐩 |𝜸(t) −𝐩| ∈S1,

and define the winding number of C about 𝐩 by

W(𝐩) = deg ⁡ (𝐟𝐩).

Lemma 8.6. The function

W : ℝ2 −C→ℤ

is locally constant. Consequently it is constant on every path-connected component of ℝ2 −C.

Proof. Fix 𝐩∉C. Because C is compact, d = dist ⁡ (𝐩,C) > 0. If 𝐪 is sufficiently close to 𝐩, the segment from 𝐩 to 𝐪 does not meet C. Moving the observation point along that segment gives a homotopy from 𝐟𝐩 to 𝐟𝐪. Their degrees are therefore equal by Corollary 8.4. □

Proof of the Jordan curve theorem. First choose a point 𝐪 very far from C. All vectors 𝜸(t) −𝐪 then point into a single open semicircle, so 𝐟𝐪 has degree zero:

W(𝐪) = 0.

Next choose two nearby points 𝐩+ and 𝐩− on opposite sides of C inside a tubular neighborhood. Zooming in near the intervening point of C, the curve is nearly its tangent line. Over the short local arc, the two direction maps traverse opposite semicircles; away from that arc they are arbitrarily close. The antipodal-value lemma allows these maps to be perturbed without changing degree, and the local semicircle computation gives

|W(𝐩+) −W(𝐩−)| = 1.

Thus 𝐩+ and 𝐩− lie in different path-connected components, so the complement has at least two components.

Now let 𝐩 ∈ ℝ2 −C be arbitrary. Since C is compact, there is a shortest line segment from 𝐩 to C. Before reaching its endpoint, that segment enters the tubular neighborhood and remains on one of its two sides. Within that side strip it can be joined to either 𝐩+ or 𝐩−. Hence every point in the complement lies in one of these two components, and there are exactly two.

A sufficiently large ball contains C. The component containing the complement of that ball is unbounded and is the exterior; the other is bounded and is the interior. Finally, every point of C has tubular-neighborhood points from both components arbitrarily close to it, so C is the common boundary of the two components. □

A tubular neighborhood of a simple closed curve, with interior and exterior test points.

Figure 8.3: The winding number is constant on each complementary component, and it changes by one when one crosses the curve.

The two theorems give equivalent ways to choose an orientation.

Definition 8.7: Positive orientation

A simple closed plane curve is positively oriented if its rotation index is 1. Equivalently, the interior stays on one’s left while the curve is traversed: for every t, the vector R90(𝖙(t)) points into the interior. A curve with the opposite orientation has rotation index −1.

The equivalence follows from the same local picture used above. The vector R90(𝖙) points consistently into one of the two tubular strips; at the supporting point used in the proof of Hopf’s theorem, the sign of the single full turn identifies that strip as the interior or exterior.

8.4 A first look at the piecewise-regular formula

The final pages of the handwritten notes record the version needed for curves with corners. We include that formula here before turning to differential forms and line integrals in Lecture 9.

Definition 8.8: Piecewise-regular curve and signed corner angle

A continuous curve 𝜸 : [a,b] → ℝ2 is piecewise regular if there is a partition

a = t0 < t1 < ⋯ < tm = b

such that each restriction to [ti,ti+1] is regular. At a corner ti, let 𝐯i− and 𝐯i+ be the one-sided velocity vectors. The signed angle αi ∈ [−π,π] is the smallest signed rotation carrying 𝐯i− to 𝐯i+, positive for counterclockwise rotation and negative for clockwise rotation.

Theorem 8.9: Generalized Hopf Umlaufsatz

Let 𝜸 : [a,b] → ℝ2 be a positively oriented, unit-speed, piecewise-regular simple closed plane curve. If κs is its signed curvature on the smooth pieces and αi are its signed corner angles, then

∫ abκ s(t) 𝑑𝑡 + ∑ iαi = 2π.

The integral means the sum of the integrals over the regular pieces.

The idea is to round each corner in a very small neighborhood. The signed curvature accumulated along the rounded portion approaches the signed corner angle. The resulting smooth simple closed curve has total signed curvature 2π by Hopf’s Umlaufsatz.

Example 8.10 (A rectangle). Orient the boundary of a rectangle counterclockwise. The signed curvature is zero on every open side. At each of the four corners the tangent turns counterclockwise through π∕2. Therefore

∫ κs 𝑑𝑡 + ∑ i=14α i = 0 + 4 (π 2 ) = 2π.

A positively oriented square, with zero curvature along its sides and a quarter-turn at each corner.

Figure 8.4: For a positively oriented rectangle, all of the turning occurs at the four corners.

8.5 Summary

Lecture 9
Differential Forms, Vector Fields, and Line Integrals

Friday, September 11, 2026

Reading. Related reading: [Tap16, § 2.4, pp. 81–89].

After class. HW 3 due; review the wedge-product and line-integral computations from class.

Goals. Introduce differential forms in Euclidean coordinates, compute with the wedge product and exterior derivative, identify vector fields with one-forms using the Euclidean inner product, and define line integrals in a parametrization-independent way.

The handwritten lecture begins the language needed for Green’s theorem. Rather than starting immediately with circulation and flux, we first record a coordinate model for differential forms. In this lecture, the symbols dx1,…, dxn are algebraic objects that encode oriented infinitesimal directions. The wedge product keeps track of the order of those directions.

9.1 Coordinate forms and the wedge product

Let x1,…,xn denote the standard coordinate functions on ℝn. Associated to them are the coordinate one-forms (just consider these as symbols) dx1,…,dxn. Let V be the real vector space with basis dx1,…,dxn. Then for any non-negative integer k, consider the real vector space, denoted by ⊗ ⁡ kV , which is defined to be the real vector space with basis the symbols

dxi1 ⊗dxi2 ⊗⋯ ⊗dxik,

where i1,i2,…,ik ∈{1,…,n}. This vector space has dimension nk. By convention, ⊗ ⁡ 0V = ℝ.

Now, inside of ⊗ ⁡ kV , for any i1,…,ik, with i1,…,ik ∈ (1,…,n), then if σ is a permutation of {1,…,k} that interchanges exactly two of the numbers (e.g., (1,2,3,4)↦(2,1,3,4)), consider the vector

dxiσ(1) ⊗⋯ ⊗ dxiσ(k) + dxi1 ⊗⋯ ⊗ dxik.

The span of these vectors is a vector subspace Sk of ⊗ ⁡ kV . This subspace defines an equivalence relation on ⊗ ⁡ kV ; two vectors in ⊗ ⁡ kV are equivalent if their difference lies in Sk. We define ∧ ⁡ kV to be the vector space that has elements that are equivalence classes of elements in ⊗ ⁡ kV . We denote the equivalence class of dxi1 ⊗dxi2 ⊗⋯ ⊗dxik by

dxi1 ∧dxi2 ∧⋯ ∧dxik.

The basic consequence of this construction is that we have the following relation between the symbols above: Given i1,…,ik, with i1,…,ik ∈ (1,…,n), then if σ is a permutation of {1,…,k} that interchanges exactly two of the numbers (e.g., (1,2,3,4)↦(2,1,3,4)), we have

dxiσ(1) ∧⋯ ∧ dxiσ(k) = −dxi1 ∧⋯ ∧ dxik.

In other words, for example, we have

dxi ∧ dxj = −dxj ∧ dxi.

This implies that

dxi ∧ dxi = −dxi ∧ dxi,

so

dxi ∧ dxi = 0.

Thus any wedge product containing a repeated coordinate differential is zero.

For example, in ℝ3,

dx2 ∧ dx1 ∧ dx3 = −dx1 ∧ dx2 ∧ dx3, dx3 ∧ dx1 ∧ dx2 = dx1 ∧ dx2 ∧ dx3, dx3 ∧ dx2 ∧ dx1 = −dx1 ∧ dx2 ∧ dx3.

If

I = (i1,⋯,ik),

we use the abbreviation

dxI = dxi1 ∧⋯ ∧ dxik.

Examples

The coordinate basis forms in low dimensional examples are:

one-forms two-forms three-forms ℝ1 dx1 0 0 ℝ2 dx1, dx2 dx1 ∧ dx2 0 ℝ3 dx1, dx2, dx3 dx1 ∧ dx2, dx1 ∧ dx3, dx2 ∧ dx3 dx1 ∧ dx2 ∧ dx3.

There are no nonzero k-forms on ℝn when k > n, because every wedge of more than n coordinate differentials must repeat an index.

Exercise 9.1. Show that a basis for ∧ ⁡ kV is given by the collection of dxI for the set of (i1,…,ik) with 1 ≤i1 < i2 < ⋯ < ik ≤n. Therefore the dimension of ∧ ⁡ kV is (n k) .

9.1.1 Wedge products of forms

One can take wedges of differential forms. If ω is a k-form and η is an ℓ-form, then one gets a (k + ℓ)-form

ω ∧η.

For instance, if

ω = f1(x1,x2,x3)dx1 + f2(x1,x2,x3)dx2

and

η = g13(x1,x2,x3)dx1 ∧dx3,

then

ω ∧η = (f1(x1,x2,x3)dx1 + f2(x1,x2,x3)dx2) ∧ (g13(x1,x2,x3)dx1 ∧dx3) = f1(x1,x2,x3)dx1 ∧g13(x1,x2,x3)dx1 ∧dx3 + f2(x1,x2,x3)dx2 ∧g13(x1,x2,x3)dx1 ∧dx3 = 0 + f2(x1,x2,x3)g13(x1,x2,x3)dx2 ∧dx1 ∧dx3 = −f2(x1,x2,x3)g13(x1,x2,x3)dx1 ∧dx2 ∧dx3.

You can check, in this example, and in general, that if ω is a k-form and η is an ℓ-form, then

ω ∧η = (−1)𝑘ℓη ∧ω.

This is called graded commutativity.

9.2 Smooth differential forms

Definition 9.2: Differential form in Euclidean coordinates

Let U ⊆ ℝn be open. A smooth k-form on U is an expression

ω = ∑ |I|=kfI(𝐱) dxI,

where the sum runs over ordered index sets I = (i1 < ⋯ < ik), and every coefficient fI : U → ℝ is smooth. The vector space of smooth k-forms on U is denoted by Ωk(U).

A smooth zero-form is simply a smooth function:

Ω0(U) = C∞(U).

The low-dimensional forms in the notes can therefore be written as follows.

Example 9.3 (Forms on ℝ). Every one-form on an interval has the form

ω = f(x) 𝑑𝑥.

Example 9.4 (Forms on ℝ2). Writing the coordinates as (x,y), a one-form and a two-form have the respective forms

ω = P(x,y) 𝑑𝑥 + Q(x,y) 𝑑𝑦

and

η = h(x,y) 𝑑𝑥 ∧ 𝑑𝑦.

Example 9.5 (Forms on ℝ3). Writing the coordinates as (x,y,z), the possible degrees are

Ω0(U) : f, Ω1(U) : P 𝑑𝑥 + Q 𝑑𝑦 + R 𝑑𝑧, Ω2(U) : A 𝑑𝑦 ∧ 𝑑𝑧 + B 𝑑𝑧 ∧ 𝑑𝑥 + C 𝑑𝑥 ∧ 𝑑𝑦, Ω3(U) : h 𝑑𝑥 ∧ 𝑑𝑦 ∧ 𝑑𝑧.

9.3 The exterior derivative

The exterior derivative turns a k-form into a (k + 1)-form.

Definition 9.6: Exterior derivative

For a smooth function f : U → ℝ, define

𝑑𝑓 = ∑ i=1n ∂𝑓 ∂xi dxi.

For a general differential form

ω = ∑ IfI dxI,

define

𝑑𝜔 = ∑ I dfI ∧ dxI.

For example, if

ω = P(x,y) 𝑑𝑥 + Q(x,y) 𝑑𝑦

is a one-form on ℝ2, then

𝑑𝜔 = 𝑑𝑃 ∧ 𝑑𝑥 + 𝑑𝑄 ∧ 𝑑𝑦 = (Px 𝑑𝑥 + Py 𝑑𝑦) ∧ 𝑑𝑥 + (Qx 𝑑𝑥 + Qy 𝑑𝑦) ∧ 𝑑𝑦 = (Qx −Py) 𝑑𝑥 ∧ 𝑑𝑦.

The coefficient Qx −Py is the planar curl that will appear in Green’s theorem.

The exterior derivative obeys the graded product rule recorded in the handwritten notes.

Proposition 9.7 (Graded Leibniz rule). If ω ∈Ωk(U) and η ∈Ωℓ(U), then

d(ω ∧η) = 𝑑𝜔 ∧η + (−1)kω ∧ 𝑑𝜂.

For k = 0, this is the ordinary product rule

d(𝑓𝜂) = 𝑑𝑓 ∧η + f 𝑑𝜂.

The sign in the general rule occurs because d must pass across a k-form before differentiating the second factor.

9.4 Vector fields and one-forms

Tapp formulates line integrals using vector fields. The Euclidean inner product identifies vector fields with one-forms.

Definition 9.8: Vector field and associated one-form

A smooth vector field on an open set U ⊆ ℝn is a smooth map

𝐅 = (F1,…,Fn) : U→ℝn.

Its associated one-form is

ω𝐅 = F1 dx1 + ⋯ + Fn dxn.

At 𝐩 ∈U, this one-form acts on 𝐮 ∈ ℝn by

(ω𝐅)𝐩(𝐮) = ⟨𝐅(𝐩),𝐮⟩.

Thus a vector field and its associated one-form contain the same information once the Euclidean inner product has been fixed. The one-form notation is especially useful because it makes the wedge product and exterior derivative available.

9.5 Line integrals

Let 𝜸 : [a,b] →U be a smooth curve. The line integral of a one-form measures the form on the oriented velocity vector and then integrates over the parameter interval.

Definition 9.9: Line integral of a one-form

If ω ∈Ω1(U), define

∫ 𝜸ω = ∫ abω 𝜸(t)(𝜸′(t)) 𝑑𝑡.

If ω = ω𝐅 is associated to a vector field, then

∫ 𝜸ω𝐅 = ∫ 𝜸𝐅 ⋅ d𝜸 = ∫ ab ⟨𝐅(𝜸(t)),𝜸′(t)⟩ 𝑑𝑡.

The formal notation d𝜸 = 𝜸′(t) 𝑑𝑡 encodes this formula.

A vector field along a curve, with the velocity tangent to the curve.

Figure 9.1: A line integral accumulates the component of the vector field in the oriented tangent direction of the curve.

The definition depends on the orientation of the curve, but not on the particular orientation-preserving parametrization.

Proposition 9.10 (Change of parameter). Let ϕ : [α,β] → [a,b] be a smooth diffeomorphism and let 𝜸~ = 𝜸∘ϕ. If ϕ is increasing, then

∫ 𝜸~ω = ∫ 𝜸ω.

If ϕ is decreasing, then

∫ 𝜸~ω = −∫ 𝜸ω.

Proof. For an increasing reparametrization and ω = ω𝐅, the chain rule gives

∫ 𝜸~ω = ∫ αβ ⟨𝐅(𝜸(ϕ(s))),𝜸′(ϕ(s))ϕ′(s)⟩ 𝑑𝑠 = ∫ ab ⟨𝐅(𝜸(t)),𝜸′(t)⟩ 𝑑𝑡,

where the second equality is the substitution t = ϕ(s). If ϕ is decreasing, the same substitution reverses the limits of integration and introduces a minus sign. □

Example 9.11 (The work–energy identity). Suppose a particle of mass m travels along 𝜸 : [a,b] → ℝn. Let

𝐯 = 𝜸′,𝐚 = 𝜸″,𝐅 = m𝐚.

The work done by the force along the path is

∫ 𝜸𝐅 ⋅ d𝜸 = m∫ ab ⟨𝐚(t),𝐯(t)⟩ 𝑑𝑡 = m∫ ab ⟨𝐯′(t),𝐯(t)⟩ 𝑑𝑡 = m 2 ∫ ab d 𝑑𝑡 |𝐯(t)|2 𝑑𝑡 = m 2 |𝐯(b)|2 −m 2 |𝐯(a)|2.

Thus the work equals the change in kinetic energy.

9.6 Summary

Lecture 10
Conservative Vector Fields and Green’s Theorem

Monday, September 14, 2026

Reading. [Tap16, § 2.4, pp. 86–91].

After class. HW 4(b): Tapp, Exercises 2.18, 2.20(3), and 2.22.

Goals. Relate gradients to directional derivatives and level sets, prove the endpoint formula and the path-independence characterization of conservative fields, and apply these ideas to conservation of energy and gravity.

10.1 Conservative vector fields

Definition 10.1: Gradient and conservative vector field

Let U ⊆ ℝn be open and let f : U → ℝ be smooth. Its gradient is the vector field

∇ ⁡f = ( ∂𝑓 ∂x1,…, ∂𝑓 ∂xn ) .

A smooth vector field 𝐅 : U → ℝn is conservative if 𝐅 = ∇ ⁡f for some smooth f : U → ℝ. Such an f is called a potential function of 𝐅.

In the differential-form notation of Lecture 9, the vector field 𝐅 = (F1,…,Fn) corresponds to

ω𝐅 = ∑ i=1nF i dxi.

Thus 𝐅 = ∇ ⁡f if and only if

ω𝐅 = 𝑑𝑓 = ∑ i=1n ∂𝑓 ∂xi dxi.

A one-form of the form 𝑑𝑓 is called exact. These are the same conservative fields as in [Tap16, Definition 2.22], expressed using one-forms.

10.2 The chain rule and directional derivatives

Lemma 10.2 (Chain rule along a curve). Let f : U → ℝ be smooth and let 𝜸 : I →U be a smooth curve. Then

d 𝑑𝑡f(𝜸(t)) = ⟨∇ ⁡f(𝜸(t)),𝜸′(t)⟩.

Proof. Write 𝜸 = (γ1,…,γn). The multivariable chain rule gives

d 𝑑𝑡f(𝜸(t)) = ∑ i=1n ∂𝑓 ∂xi(𝜸(t))γi′(t) = ⟨∇ ⁡f(𝜸(t)),𝜸′(t)⟩.
□

Fix 𝐩0 ∈U and 𝐯0 ∈ ℝn. Since U is open, the line

ℓ(t) = 𝐩0 + t𝐯0

lies in U for all sufficiently small |t|. The directional derivative of f at 𝐩0 in the direction 𝐯0 is

D𝐯0f(𝐩0) := d 𝑑𝑡 |t=0f(𝐩0 + t𝐯0).

Applying the chain rule to f ∘ℓ yields

D𝐯0f(𝐩0) = ∑ i=1n ∂𝑓 ∂xi(𝐩0)(v0)i = ⟨∇ ⁡f(𝐩0),𝐯0⟩ .

This is also written (𝑑𝑓)𝐩0(𝐯0) in [Tap16, Lemma 2.23 and Eq. (2.3)]. The vector 𝐯0 need not have unit length. More generally, any smooth curve through 𝐩0 with initial velocity 𝐯0 gives the same initial derivative of f.

10.3 Geometry of the gradient

For a real number c, the level set of f at height c is

f−1(c) = {𝐩 ∈U : f(𝐩) = c}.

If the trace of 𝜸 lies in this level set, then f(𝜸(t)) = c, and hence

0 = d 𝑑𝑡f(𝜸(t)) = ⟨∇ ⁡f(𝜸(t)),𝜸′(t)⟩.

Thus the gradient is orthogonal to every velocity vector of a curve in the level set.

Lemma 10.3 (Direction and rate of greatest increase). Let 𝐩 ∈U and assume that ∇ ⁡f(𝐩)≠0.

  1. The vector ∇ ⁡f(𝐩) is orthogonal to the level set through 𝐩, in the sense just described.
  2. Among unit vectors 𝐮, the derivative D𝐮f(𝐩) is largest when

    𝐮 = ∇ ⁡f(𝐩) |∇ ⁡f(𝐩)|.
  3. The maximum directional derivative is |∇ ⁡f(𝐩)|.

Proof. The level-set calculation proves the first assertion. For a unit vector 𝐮, the Schwarz inequality gives

D𝐮f(𝐩) = ⟨∇ ⁡f(𝐩),𝐮⟩ ≤ |∇ ⁡f(𝐩)|.

Equality holds for 𝐮 = ∇ ⁡f(𝐩)∕ |∇ ⁡f(𝐩)|, which proves the other two assertions. □

A curve within a level set, with its gradient perpendicular to its tangent.

Figure 10.1: Along a level curve, the tangent direction and gradient direction are perpendicular.

10.4 Integrating a conservative field

Proposition 10.4: The endpoint formula

Let f : U → ℝ be smooth, and let C be an oriented piecewise-regular curve in U, parametrized by 𝜸 : [a,b] →U. Then

∫ C∇ ⁡f ⋅ d𝜸 = f(𝜸(b)) −f(𝜸(a)).

Equivalently,

∫ C 𝑑𝑓 = f(𝜸(b)) −f(𝜸(a)).

Proof. For a smooth curve, the chain rule and the fundamental theorem of calculus give

∫ C∇ ⁡f ⋅ d𝜸 = ∫ ab ⟨∇ ⁡f(𝜸(t)),𝜸′(t)⟩ 𝑑𝑡 = ∫ ab d 𝑑𝑡f(𝜸(t)) 𝑑𝑡 = f(𝜸(b)) −f(𝜸(a)).

For a piecewise-regular curve, apply this calculation on each smooth piece and add; all intermediate endpoint values cancel. □

Only the initial and final points matter. In particular, if C is a closed path, then

∮ C∇ ⁡f ⋅ d𝜸 = 0.

In coordinates the same calculation reads

∫ C∇ ⁡f ⋅ d𝜸 = ∫ C ∑ i=1n ∂𝑓 ∂xi dxi = ∫ C 𝑑𝑓.

10.5 Path independence

Proposition 10.5: Equivalent characterizations of a conservative field

Let U ⊆ ℝn be open and path-connected, and let 𝐅 be a smooth vector field on U. The following conditions are equivalent.

  1. 𝐅 = ∇ ⁡f for some smooth f : U → ℝ; equivalently, ω𝐅 = 𝑑𝑓.
  2. For every piecewise-regular closed path C in U,

    ∮ C𝐅 ⋅ d𝜸 = 0.
  3. Line integrals of 𝐅 are path-independent: two piecewise-regular paths in U with the same initial and final points have equal line integrals.

Proof. (1) ⇒ (2) follows from the endpoint formula.

(2) ⇒ (3): given two paths C1,C2 with the same endpoints, traverse C1 and then traverse C2 in reverse. The resulting path is closed. Additivity and reversal of orientation give

0 = ∫ C1𝐅 ⋅ d𝜸−∫ C2𝐅 ⋅ d𝜸.

(3) ⇒ (1): fix 𝐩0 ∈U. Define

f(𝐱) = ∫ 𝐩0 𝐱𝐅 ⋅ d𝜸,

where the integral may be taken along any piecewise-regular path in U from 𝐩0 to 𝐱. Such paths exist in an open, path-connected set, and path independence makes f well defined.

To compute its ith partial derivative, keep 𝐱 fixed and let 𝐞i be the ith standard basis vector. For sufficiently small h, the line segment from 𝐱 to 𝐱 + h𝐞i lies in U. By adjoining this segment to a path ending at 𝐱,

f(𝐱 + h𝐞i) −f(𝐱) = ∫ 0h ⟨𝐅(𝐱 + s𝐞 i),𝐞i⟩ 𝑑𝑠 = ∫ 0hF i(𝐱 + s𝐞i) 𝑑𝑠.

Dividing by h and taking the limit gives

∂𝑓 ∂xi(𝐱) = Fi(𝐱).

Since the Fi are smooth, f is smooth, and ∇ ⁡f = 𝐅. □

Remark 10.6 (The Green’s theorem aside). In the plane, write ω = P 𝑑𝑥 + Q 𝑑𝑦. Recall that

𝑑𝜔 = (Qx −Py) 𝑑𝑥 ∧ 𝑑𝑦.

If C = ∂𝐷 is a positively oriented piecewise-regular simple closed curve and ω is smooth on an open set containing D ∪C, Green’s theorem takes the form

∮ ∂𝐷ω = ∫ D 𝑑𝜔 = ∬D(Qx −Py) 𝑑𝑥 𝑑𝑦.

This is the connection between boundary integrals and exterior derivatives noted in class; see [Tap16, Theorem 2.32].

For an exact form ω = 𝑑𝑓, equality of mixed partial derivatives implies 𝑑𝜔 = 0. A form satisfying 𝑑𝜔 = 0 is called closed. One domain qualification is important: closedness alone does not imply global exactness on an arbitrary open set. The preceding proposition requires vanishing integrals around every closed path in the domain, not just boundaries of regions contained in that domain; see [Tap16, Exercises 2.25–2.26].

10.6 Conservation of energy

Suppose that 𝐅 = ∇ ⁡f is conservative. Set

ρ = −f.

The function ρ is the potential energy function; its sign is opposite to that of the potential f used in 𝐅 = ∇ ⁡f. Assume that a particle of mass m > 0 moves under this force alone, so

𝐅(𝜸(t)) = m𝐚(t) = m𝜸″(t),𝐯(t) = 𝜸′(t).

By the work–energy calculation from Lecture 9,

∫ C𝐅 ⋅ d𝜸 = ∫ abm ⟨𝐚(t),𝐯(t)⟩ 𝑑𝑡 = m 2 |𝐯(b)|2 −m 2 |𝐯(a)|2.

On the other hand, the endpoint formula gives

∫ C𝐅 ⋅ d𝜸 = f(𝜸(b)) −f(𝜸(a)) = −ρ(𝜸(b)) + ρ(𝜸(a)).

Equating these expressions and rearranging yields

m 2 |𝐯(b)|2 + ρ(𝜸(b)) = m 2 |𝐯(a)|2 + ρ(𝜸(a)).

Thus the total energy

m 2 |𝐯(t)|2⏟ kinetic energy + ρ(𝜸(t))⏟potential energy

is constant. This is the conservation law associated with a conservative force field.

10.7 Example: gravity

Place a mass M at the origin of ℝ3, and a mass m at 𝐩≠0. Let G denote the gravitational constant and set r = |𝐩|. The gravitational force on the second mass has magnitude 𝑚𝑀𝐺∕r2 and points toward the origin. Therefore

𝐅(𝐩) = 𝑚𝑀𝐺 r2 (− 𝐩 |𝐩| ) = −𝑚𝑀𝐺 |𝐩|3 𝐩.

For 𝐩 = (x,y,z), this reads

𝐅(x,y,z) = −𝑚𝑀𝐺(x2 + y2 + z2)−3∕2(x,y,z).

A radial gravitational force pointing from a mass at p toward a mass at the origin.

Figure 10.2: The gravitational force points from 𝐩 toward the mass at the origin.

On U = ℝ3 ∖{0}, choose the potential

f(𝐩) = 𝑚𝑀𝐺 |𝐩| −K = 𝑚𝑀𝐺(x2 + y2 + z2)−1∕2 −K,

where K ∈ ℝ is an arbitrary constant. For example,

∂𝑓 ∂𝑥 = −𝑚𝑀𝐺x(x2 + y2 + z2)−3∕2,

and the analogous y- and z-derivatives show that ∇ ⁡f = 𝐅. Hence gravity is conservative, with potential energy

ρ(𝐩) = −f(𝐩) = K −𝑚𝑀𝐺 |𝐩| .

For a trajectory 𝐩(t) = 𝜸(t) subject only to this force, the conservation law becomes

1 2m |𝐯(t)|2 + K −𝑚𝑀𝐺 |𝐩(t)| = constant.

This is the gravity example of [Tap16, Example 2.28].

Lecture 11
Green’s Theorem III: Circulation, Divergence, and Flux

Wednesday, September 16, 2026

Reading. [Tap16, § 2.4, pp. 92–96].

After class. HW 4(c): Tapp, Exercises 2.21(3), 2.23, 2.24, 2.25, 2.26, and 2.27.

Goals. Express Green’s theorem in differential-form notation, interpret infinitesimal circulation, and derive the flux version using the outward normal, divergence, and the star operation on forms.

11.1 Review: the exterior derivative squares to zero

Recall from Lecture 9 that if ω is a smooth k-form on an open set U ⊆ ℝn, then 𝑑𝜔 is a (k + 1)-form. Applying the exterior derivative twice always gives zero:

d(𝑑𝜔) = 0.

Example 11.1 (A zero-form on the plane). Let ω = f be a smooth function on U ⊆ ℝ2. Write fxi = ∂𝑓∕∂xi. Then

𝑑𝑓 = fx1 dx1 + fx2 dx2,

and the calculation on the board becomes

d(𝑑𝑓) = d(fx1) ∧ dx1 + d(fx2) ∧ dx2 = ( ∂2f ∂x12 dx1 + ∂2f ∂x2∂x1 dx2) ∧ dx1 + ( ∂2f ∂x1∂x2 dx1 + ∂2f ∂x22 dx2) ∧ dx2 = (− ∂2f ∂x2∂x1 + ∂2f ∂x1∂x2 ) dx1 ∧ dx2 = 0.

The terms with dxi ∧ dxi vanish. The remaining terms cancel because mixed partial derivatives agree and dx2 ∧ dx1 = −dx1 ∧ dx2.

11.2 Green’s theorem

Theorem 11.2: Green’s theorem: circulation form

Let C be a positively oriented piecewise-regular simple closed plane curve with interior D. Let 𝐅 = (F1,F2) be a smooth vector field on an open set containing D ∪C. Then

∮ C𝐅 ⋅ d𝜸 = ∬D (∂F2 ∂x1 −∂F1 ∂x2 ) 𝑑𝐴.

Positive orientation means that D lies on the left as C is traversed. The left side is the circulation of 𝐅 around C. This is [Tap16, Theorem 2.32], written with component names F1,F2 and coordinates x1,x2.

11.2.1 The differential-form expression

Associate to 𝐅 the one-form

ω𝐅 = F1 dx1 + F2 dx2.

Along a parametrization 𝜸 = (γ1,γ2),

𝐅 ⋅ d𝜸 = ⟨𝐅(𝜸(t)),𝜸′(t)⟩ 𝑑𝑡 =(F1(𝜸(t))γ1′(t) + F 2(𝜸(t))γ2′(t)) 𝑑𝑡.

This is exactly how ω𝐅 is integrated along C. Its exterior derivative is

dω𝐅 = dF1 ∧ dx1 + dF2 ∧ dx2 = (∂F2 ∂x1 −∂F1 ∂x2 ) dx1 ∧ dx2.

With the standard orientation of the plane, dx1 ∧ dx2 is the oriented area form. Thus Green’s theorem has the compact form

∮ ∂𝐷ω𝐅 = ∫ D dω𝐅.

The boundary integral and the integral of the exterior derivative are the two sides of the same theorem.

11.2.2 Infinitesimal circulation

The scalar function

∂F2 ∂x1 −∂F1 ∂x2

is called the infinitesimal circulation of 𝐅 in [Tap16, Definition 2.30]. The following limit explains the name.

Corollary 11.3 (Circulation around a small circle). Suppose 𝐅 is smooth near 𝐩 ∈ ℝ2, and let Cr,𝐩 be the counterclockwise circle of radius r centered at 𝐩. Then

(∂F2 ∂x1 −∂F1 ∂x2 ) (𝐩) = lim ⁡ r→0+ 1 πr2 ∮ Cr,𝐩𝐅 ⋅ d𝜸.

Proof. Write g = ∂F2∕∂x1 −∂F1∕∂x2, and let Br(𝐩) be the disk bounded by Cr,𝐩. For sufficiently small r, Green’s theorem gives

1 πr2 ∮ Cr,𝐩𝐅 ⋅ d𝜸 = 1 πr2 ∬Br(𝐩)g 𝑑𝐴.

The right side is the average of g over the disk. By continuity, this average tends to g(𝐩) as r → 0+. □

A planar region D bounded by C and a small circular region centered at p.

Figure 11.1: The small circle lies inside the region and is oriented counterclockwise. Its circulation divided by πr2 approaches the infinitesimal circulation at 𝐩.

Recall from Lecture 10 that a conservative field 𝐅 = ∇ ⁡f has zero circulation around every closed path in its domain. Consequently, its infinitesimal circulation is zero. In forms, this is the review identity once more:

ω∇ ⁡f = 𝑑𝑓, dω∇ ⁡f = d(𝑑𝑓) = 0.

11.3 The flux version of Green’s theorem

11.3.1 The outward normal and flux

Let C = ∂𝐷 have a positive parametrization 𝜸 : [a,b] → ℝ2. At a regular point, write 𝐯(t) = 𝜸′(t). The outward unit normal 𝐧(t) is

𝐧(t) = R−90 ( 𝐯(t) |𝐯(t)| ),R−90 = ( 0 1 −1 0 ),R−90(x1,x2) = (x2,−x1).

The clockwise quarter-turn points outward because the interior is on the left. Here 𝐧 is the outward boundary normal, not the curvature normal 𝖓 used in the Frenet frame.

Definition 11.4: Flux across a closed plane curve

The flux of 𝐅 across C is

∮ C𝐅 ⋅𝐧 𝑑𝑠 := ∫ ab ⟨𝐅(𝜸(t)),𝐧(t)⟩ |𝐯(t)| 𝑑𝑡.

Here 𝑑𝑠 = |𝐯(t)| 𝑑𝑡 is the arc-length element. For a piecewise-regular boundary, the integral is the sum over its regular pieces, on which 𝐧 is defined.

This integrates the outward component of the field with respect to length; compare [Tap16, Definition 2.34].

A positively oriented closed curve, its unit tangent, its outward unit normal, and a vector field.

Figure 11.2: For a positively oriented boundary, 𝐧 = R−90(𝖙). The displayed unit tangent and outward unit normal are perpendicular; the interior is on the left.

Substitute the formula for 𝐧 and cancel the factor |𝐯|. Since R−90T = R90,

∮ C𝐅 ⋅𝐧 𝑑𝑠 = ∫ ab ⟨𝐅(𝜸(t)),R −90(𝐯(t))⟩ 𝑑𝑡 = ∫ ab ⟨R 90(𝐅(𝜸(t))),𝐯(t)⟩ 𝑑𝑡 = ∫ ab( −F 2(𝜸(t))γ1′(t) + F 1(𝜸(t))γ2′(t)) 𝑑𝑡 = ∮ C( −F2 dx1 + F1 dx2).

Thus flux is a line integral of the one-form associated with R90(𝐅) = (−F2,F1).

11.3.2 Divergence and the star operation

For a smooth vector field 𝐅 = (F1,…,Fn) on U ⊆ ℝn, its divergence is the function

∇ ⁡⋅𝐅 = ( ∂ ∂x1,…, ∂ ∂xn ) ⋅ (F1,…,Fn) = ∑ i=1n∂Fi ∂xi .

In the plane, this is ∂F1∕∂x1 + ∂F2∕∂x2, as in [Tap16, Definition 2.35].

The coordinate operation denoted by ∗ on the board is the Hodge star for the standard Euclidean inner product and orientation. The cases needed here are

∗dxi = (−1)i−1 dx 1 ∧⋯ ∧dxi^ ∧⋯ ∧ dxn, ∗ (dx1 ∧⋯ ∧ dxn) = 1,

where a hat means that the indicated factor is omitted. Extend these formulas linearly over smooth coefficient functions. In particular,

𝐅 ⟼ω𝐅 = ∑ i=1nF i dxi, ω𝐅 ⟼ ∗ω𝐅 = ∑ i=1n(−1)i−1F i dx1 ∧⋯ ∧dxi^ ∧⋯ ∧ dxn.

Thus ∗ω𝐅 is an (n − 1)-form.

On differentiating, every term of dFi except its dxi term gives a repeated wedge factor and vanishes. Moving the remaining dxi into its usual position gives another factor of (−1)i−1. Hence

d(∗ω𝐅) = ∑ i=1n(−1)i−1∂Fi ∂xi dxi ∧ dx1 ∧⋯ ∧dxi^ ∧⋯ ∧ dxn = ∑ i=1n∂Fi ∂xi dx1 ∧⋯ ∧ dxn = (∇ ⁡⋅𝐅) dx1 ∧⋯ ∧ dxn.

Equivalently, applying ∗ to this top-degree form gives

∗d(∗ω𝐅) = ∇ ⁡⋅𝐅.

11.3.3 The planar calculation and the flux theorem

For n = 2, the star operation reads

∗dx1 = dx2, ∗dx2 = −dx1.

Therefore

ω𝐅 = F1 dx1 + F2 dx2, ∗ω𝐅 = F1 dx2 −F2 dx1,

which is precisely the one-form obtained from the flux calculation. Direct differentiation gives

d(∗ω𝐅) = dF1 ∧ dx2 − dF2 ∧ dx1 = (∂F1 ∂x1 dx1 + ∂F1 ∂x2 dx2) ∧ dx2 −(∂F2 ∂x1 dx1 + ∂F2 ∂x2 dx2) ∧ dx1 = (∂F1 ∂x1 + ∂F2 ∂x2 ) dx1 ∧ dx2 = (∇ ⁡⋅𝐅) 𝑑𝐴.

Theorem 11.5: Green’s theorem: flux form

Let C be a piecewise-regular simple closed plane curve with interior D, and let 𝐧 be its outward unit normal on each regular piece. If 𝐅 is smooth on an open set containing D ∪C, then

∮ C𝐅 ⋅𝐧 𝑑𝑠 = ∬D∇ ⁡⋅𝐅 𝑑𝐴.

Proof. Choose the positive orientation on C = ∂𝐷. Applying Green’s theorem gives

∬D∇ ⁡⋅𝐅 𝑑𝐴 = ∫ D d(∗ω𝐅) = ∮ C ∗ω𝐅 = ∮ C(F1 dx2 −F2 dx1) = ∮ C𝐅 ⋅𝐧 𝑑𝑠.

This is the final calculation in the notes. It is also the rotation argument in [Tap16, Theorem 2.37], expressed using forms. □

11.4 Summary

Lecture 16
Derivatives of maps from Euclidean space to Euclidean space

Monday, September 28, 2026

Reading. [Tap16, § 3.1, pp. 113–124].

After class. HW 5(a): Tapp, Exercises 3.1, 3.2, 3.3, 3.4, 3.5, and 3.9 (tentative on the course schedule).

Goals. Describe smooth maps componentwise, compute their differentials, express the chain rule as composition of linear maps, and define smooth maps and diffeomorphisms on subsets of Euclidean space.

As we begin the surface chapters, we follow Tapp in no longer using boldface for vector-valued quantities; see the footnote on [Tap16, p. 114]. The dimensions of the domain and codomain will indicate the type of each map.

16.1 Smooth maps

Let U ⊆ ℝn be open, and write a map f : U → ℝm in components as

f(x1,…,xn) =(f1(x1,…,xn),…,fm(x1,…,xn)).

Definition 16.1: Smooth map

The map f is smooth if all partial derivatives of all orders of its component functions exist and are continuous on U. Thus, for every i ∈{1,…,m}, every r ≥ 1, and every choice of j1,…,jr ∈{1,…,n}, the function

∂rfi ∂xj1⋯∂xjr

exists and is continuous on U.

Lemma 16.2. A map f = (f1,…,fm) : U → ℝm is smooth if and only if each component fi : U → ℝ is smooth.

This is the componentwise meaning of smoothness used in [Tap16, § 3.1].

Example 16.3. The map

f : ℝ3→ℝ3,f(x 1,x2,x3) =(x1x2x3,x2 + cos⁡x3,ex1−x3 )

is smooth. For instance,

∂ ∂x2 (∂f1 ∂x1 ) = x3,∂2f3 ∂x22 = 0.

Lemma 16.4 (Composition of smooth maps). Let g : U ⊆ ℝn → ℝm and f : V ⊆ ℝm → ℝℓ be smooth maps, where U,V are open and g(U) ⊆V . Then f ∘g : U → ℝℓ is smooth.

This is [Tap16, Proposition 3.2].

Example 16.5. Every linear map ℝn → ℝm is smooth. More generally, a map whose component functions are polynomials is smooth.

16.2 Differentials of smooth maps

Definition 16.6: Differential and Jacobian matrix

Let f : U ⊆ ℝn → ℝm be smooth, with U open, and let p ∈U. The differential of f at p is the linear map

(𝐷𝑓)p : ℝn→ℝm

represented in the standard bases by the m ×n Jacobian matrix

J(f)(p) = ( ∂f1 ∂x1(p) ⋯ ∂f1 ∂xn(p) ⋮ ⋱ ⋮ ∂fm ∂x1 (p) ⋯ ∂fm ∂xn (p) ).

In particular, (𝐷𝑓)p(v) = J(f)(p)v, with v written as a column.

Tapp writes dfp for this linear map; compare [Tap16, Proposition and Definition 3.5]. Our convention is that rows correspond to output components and columns to input coordinates; no transpose is intended in the identification (𝐷𝑓)p = J(f)(p).

Example 16.7 (The Jacobian of the preceding map). For the map in Example 16.3,

J(f)(x1,x2,x3) = ( x2x3 x1x3 x1x2 0 1 −sin⁡x3 ex1−x3 0 −ex1−x3 ) .

Evaluating this matrix at p gives (𝐷𝑓)p.

16.2.1 Directional derivatives

For v ∈ ℝn, the differential applied to v is the directional derivative:

(Dvf)(p) = (𝐷𝑓)p(v) = lim ⁡ h→0f(p + h𝑣) −f(p) h .

There is no requirement that v have unit length. This is also d 𝑑h | h=0f(p + h𝑣): the derivative of f along the line through p with velocity v. Compare [Tap16, Definition 3.3 and Proposition 3.4].

16.3 The chain rule

The differential respects composition: it is functorial.

Proposition 16.8: Chain rule

Let g : U ⊆ ℝn → ℝm and f : V ⊆ ℝm → ℝℓ be smooth, with U,V open and g(U) ⊆V . For p ∈U,

(D(f ∘g))p = (𝐷𝑓)g(p) ∘ (𝐷𝑔)p.

In terms of Jacobian matrices, composition is matrix multiplication:

J(f ∘g)(p) = J(f)(g(p))J(g)(p).

This is [Tap16, Proposition 3.9]. The two diagrams display the maps and their differentials:

The chain rule: differentiating the composition of two maps gives the composition of their differentials.

Example 16.9 (The one-variable chain rule). When n = m = ℓ = 1, the differentials are multiplication by the ordinary derivatives. Thus

(f ∘g)′(p) = f′(g(p))g′(p),

which is the familiar chain rule from calculus.

16.4 The inverse and implicit function theorems

We record the smooth versions needed below. For the analytic details and proofs, see Browder, Mathematical Analysis: An Introduction, [Bro96, § 8.5, “Inverse and Implicit Functions”].

Theorem 16.10: Inverse function theorem

Let f : U ⊆ ℝn → ℝn be smooth, with U open. If (𝐷𝑓)p is invertible at p ∈U, then there are open neighborhoods Up ⊆U of p and V of f(p) such that

f|Up : Up→V

is bijective and has a smooth inverse. Moreover,

(D((f|Up)−1)) f(p) =((𝐷𝑓)p)−1.

The inverse exists locally, not necessarily on all of U. The final identity follows from the chain rule. Compare also [Tap16, Theorem 3.10].

Theorem 16.11: Implicit function theorem: a level set is locally a graph

Let F : Ω ⊆ ℝn → ℝm be smooth, with Ω open and 1 ≤m ≤n. Suppose p ∈Ω and

rank ⁡ (𝐷𝐹)p = m.

After reordering the input coordinates, write p = (a,b) ∈ ℝn−m × ℝm so that the m ×m matrix of partial derivatives with respect to the last m coordinates is invertible at p. There are open neighborhoods A of a and B of b, with A ×B ⊆Ω, and a smooth map h : A →B, with h(a) = b, such that

F−1(F(p)) ∩ (A ×B) = {(x,h(x)) : x ∈A}.

Here maximal rank means that the differential is onto ℝm. Locally, the m equations F = F(p) solve uniquely for m coordinates as smooth functions of the remaining n −m coordinates.

16.5 Diffeomorphisms

16.5.1 Open subsets and relative neighborhoods

Recall that U ⊆ ℝn is open if, for every p ∈U, there is an 𝜀p > 0 such that

B𝜀p(p) := {x ∈ ℝn : |x −p| < 𝜀 p}⊆U.

If X ⊆ ℝn, a subset V ⊆X is open in X if V = U ∩X for some open U ⊆ ℝn. An open subset of X containing p is an open neighborhood of p in X. These are the relative notions in [Tap16, Definition 3.18].

16.5.2 Smooth maps on subsets

A subset X ⊆ ℝn need not be open, so partial derivatives of a map defined only on X may not make sense in all ambient coordinate directions. We use local extensions instead.

Definition 16.12: Smooth map on a subset

Let X ⊆ ℝn. A map f : X → ℝm is smooth if, for every p ∈X, there are an open neighborhood U of p in ℝn and a smooth map f~ : U → ℝm such that

f~|X∩U = f|X∩U.

This is [Tap16, Definition 3.12]. The extension may depend on p; a single extension on all of ℝn is not required.

Definition 16.13: Diffeomorphism

Let X1 ⊆ ℝn1 and X2 ⊆ ℝn2. A diffeomorphism f : X1 →X2 is a smooth bijection whose inverse g = f−1 : X2 →X1 is also smooth. If such a map exists, X1 and X2 are called diffeomorphic.

Smoothness of each map is understood using local extensions into its ambient Euclidean space; compare [Tap16, Definition 3.13].

Example 16.14 (The sphere and an ellipsoid). For a,b,c > 0, let

X1 = {(x,y,z) ∈ ℝ3 : x2 + y2 + z2 = 1}, X2 = {(x,y,z) ∈ ℝ3 : (x a )2 + (y b )2 + (z c )2 = 1}.

The maps

f : X1→X2,f(x,y,z) = (𝑎𝑥,𝑏𝑦,𝑐𝑧), g : X2→X1,g(x,y,z) = (x∕a,y∕b,z∕c)

are inverse bijections. Both extend to linear, hence smooth, maps on ℝ3. Thus the sphere and ellipsoid are diffeomorphic, as in [Tap16, Example 3.14].

16.6 Summary

Smooth maps are described componentwise, and their differentials are linear maps represented by Jacobian matrices. The chain rule becomes composition of these linear maps. The inverse and implicit function theorems turn rank conditions into local inverse and graph descriptions. Local smooth extensions let us speak of diffeomorphisms even when the domain is not open in its ambient space.

Lecture 17
Regular surfaces: patches, level sets, ruled surfaces, and examples

Wednesday, September 30, 2026

Reading. [Tap16, § 3.2, pp. 125–140].

After class. HW 5(b): Tapp, Exercises 3.13, 3.15, 3.18, 3.20, 3.23, and 3.29 (tentative on the course schedule).

Goals. Review smooth maps on subsets, define regular d-folds and coordinate charts, and relate the local descriptions by charts, graphs, and level sets.

17.1 Review: smooth maps and diffeomorphisms

For X ⊆ ℝn, a map f : X → ℝm is smooth if it has a smooth extension near every point of X. More explicitly, for each p ∈X, there are an open neighborhood U ⊆ ℝn of p and a smooth map f~ : U → ℝm such that

f~|X∩U = f|X∩U,that is,f~(x) = f(x)for all x ∈X ∩U.

The extension is smooth in the ordinary Euclidean sense. Recall Definition 16.12 and [Tap16, Definition 3.12].

17.1.1 Open sets and neighborhoods

An open set U ⊆ ℝn contains a ball around each of its points:

for every p ∈U,there is 𝜀p > 0 with B𝜀p(p) = {x : |x −p| < 𝜀p}⊆U.

For X ⊆ ℝn, a subset V ⊆X is open in X if V = X ∩U for some open U ⊆ ℝn. If also p ∈V , then V is an open neighborhood of p in X. This is the notion of neighborhood used throughout this lecture; see [Tap16, Definition 3.18].

17.1.2 Diffeomorphisms and local coordinates

A diffeomorphism f : X1 →X2, where Xi ⊆ ℝni, is a smooth bijection with smooth inverse. In particular, if Up ⊆X is a neighborhood of p and W ⊆ ℝd is open, a diffeomorphism

φ : W→Up,ψ = φ−1 : U p→W

allows us to describe points of Up by d coordinates in W. The map φ is smooth as a map into ℝn; smoothness of ψ means that it has local smooth extensions off Up.

A coordinate neighborhood U in X and its diffeomorphic open parameter domain W, with phi(a) = p and psi(p) = a.

Figure 17.1: A coordinate neighborhood is diffeomorphic to an open set in ℝd. The corresponding points satisfy φ(a) = p and ψ(p) = a. The picture is schematic.

17.2 Regular d-folds

Definition 17.1: Regular d-fold, coordinate chart, and atlas

A subset X ⊆ ℝn is a regular d-fold if every p ∈X has an open neighborhood Up in X and a diffeomorphism

φp : V p→Up,V p ⊆ ℝd open.

Such a map φp is a coordinate chart, with inverse ψp. A collection of coordinate charts whose images cover X is an atlas.

A regular surface in ℝ3 is the case d = 2, n = 3. In this case Tapp also calls a coordinate chart a surface patch; see [Tap16, Definition 3.19]. The neighborhood Up is open in X, not necessarily in ℝn.

17.2.1 Graphs

Example 17.2 (The graph of a smooth function). Let U ⊆ ℝ2 be open and f : U → ℝ smooth. Its graph is

Γf = {(x,y,f(x,y)) : (x,y) ∈U}⊆ ℝ3.

The map

φ : U→Γf,φ(x,y) = (x,y,f(x,y))

is a smooth bijection, with inverse

ψ : Γf→U,ψ(x,y,z) = (x,y).

The inverse extends to the smooth coordinate projection on ℝ3. Thus φ is a diffeomorphism and Γf is a regular surface covered by one chart. Compare [Tap16, Lemma 3.17].

The same argument applies to a smooth map f : U ⊆ ℝd → ℝn−d: its graph is a regular d-fold.

Remark 17.3 (The square-root example as written). The handwritten example on the unit disk B1(0) = {(x,y) : x2 + y2 < 1} reads f(x,y) = x2 + y2. This formula is not smooth at the origin, since f(t,0) = |t|. With the formula as written, the smooth-graph argument applies on the punctured disk B1(0) ∖{(0,0)}, not on the entire disk. The formula is retained here rather than inferred from the accompanying sketch.

17.3 A regular d-fold is locally a graph

Theorem 17.4: Local graph description

Let X ⊆ ℝn be a regular d-fold and p ∈X. There are an open neighborhood V of p in X, a linear isomorphism

A : ℝn→ℝd × ℝn−d,

an open set W ⊆ ℝd, and a smooth map h : W → ℝn−d such that

A(V ) = {(u,h(u)) : u ∈W}.

Equivalently, u↦A−1(u,h(u)) is a coordinate chart for V . Thus X is locally a graph after a linear change of ambient coordinates.

Proof using the inverse function theorem. Choose a chart φ : U →V 0 ⊆X, with φ(a) = p, and let ψ = φ−1. Extend ψ smoothly to an ambient neighborhood of p, writing the extension as ψ~. After shrinking U, ψ~ ∘φ = Id ⁡ U. The chain rule gives

(Dψ~)p(𝐷𝜑)a = Id.

Hence P := (Dψ~)p has rank d. Complete its rows to a basis of the row vectors on ℝn: choose a linear map L : ℝn → ℝn−d such that

A = ( P L )

is invertible. The augmented map from the handwritten proof is

H(q) =(ψ~(q),𝐿𝑞),(𝐷𝐻)p = A.

By the inverse function theorem, H is a diffeomorphism between sufficiently small ambient neighborhoods. On the chart, H(φ(t)) = (t,𝐿𝜑(t)), which is a graph in these smooth coordinates.

To obtain the stated linear graph coordinates, use the same row completion A and set g = P ∘φ. Since

(𝐷𝑔)a = P(𝐷𝜑)a = Id,

the inverse function theorem makes g : U′→W a diffeomorphism after shrinking U to U′. Define

h(u) = 𝐿𝜑(g−1(u)),u ∈W.

Then

𝐴𝜑(g−1(u)) = (u,h(u)).

Taking V = φ(U′), which is open in X, proves the claim. □

17.4 Local level-set descriptions

Theorem 17.5: Local level-set characterization

A subset X ⊆ ℝn is a regular d-fold if and only if, for every p ∈X, there are an open neighborhood O of p in ℝn and a smooth map F : O → ℝn−d such that

X ∩O = F−1(0),rank ⁡ (𝐷𝐹) q = n −dfor all q ∈O.

For the converse, it suffices that the rank condition hold at p.

Proof. The case d = n is the open-set case, with defining map valued in ℝ0; assume d < n. First suppose X is a regular d-fold. By Theorem 17.4, choose linear coordinates (x,y) ∈ ℝd × ℝn−d in which a neighborhood in X is the graph of a smooth map h = (h1,…,hn−d) : W → ℝn−d. In these coordinates, set

F : W × ℝn−d→ℝn−d,F(x,y) = y −h(x).

Then F−1(0) = Γh, and its Jacobian is the block matrix from the notes:

J(F)(x,y) = ( −∂h1 ∂x1(x) ⋯ −∂h1 ∂xd(x) 1 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ −∂hn−d ∂x1 (x) ⋯ −∂hn−d ∂xd (x) 0 ⋯ 1 ) =(−J(h)(x)∣In−d).

The identity block gives rank n −d everywhere. Composing with the linear coordinate change and restricting the ambient neighborhood gives the desired defining map for X ∩O.

Conversely, suppose X ∩O = F−1(0) and rank ⁡ (𝐷𝐹)p = n −d. The implicit function theorem (Theorem 16.11) says that this level set is a graph near p, after reordering coordinates. A smooth graph is a regular d-fold by the argument of Example 17.2. Apply this at each point of X. □

In the surface case, a smooth scalar function F on an open subset of ℝ3 has a regular level surface near any point where its gradient is nonzero. Compare [Tap16, Definition 3.26 and Theorem 3.27].

17.5 Summary

A regular d-fold is locally diffeomorphic to an open subset of ℝd. The inverse function theorem turns this chart description into a local graph description. A graph is the zero set of F(x,y) = y −h(x), whose differential has full rank; conversely, the implicit function theorem turns a full-rank level set into a graph.

Lecture 19
Tangent planes and derivatives of maps between surfaces

Monday, October 5, 2026

Reading. [Tap16, § 3.3, pp. 141–146].

After class. HW 6(b): Tapp, Exercises 3.35, 3.36, 3.38, 3.40, 3.43, and 3.46 (tentative on the course schedule).

Goals. Review transition maps, define tangent spaces using charts, and use commuting diagrams to understand chart independence, differentials, and tangent spaces of submanifolds.

19.1 Review: charts and transition maps

Let S ⊆ ℝ3 be a regular surface. In the notation of today’s handwritten diagrams, write two coordinate charts and their inverses as

σi : V i→Ui ⊆S,φi = σi−1 : U i→V i,i = 1,2,

where V i ⊆ ℝ2 is open and Ui is open in S. On an overlap, the transition map is

τ𝑗𝑖 = φj ∘σi : σi−1(U i ∩Uj)→σj−1(U i ∩Uj).

It is a diffeomorphism with inverse τ𝑖𝑗. The identities

σj∘τ𝑗𝑖 = σi, τ𝑘𝑖 = τ𝑘𝑗 ∘τ𝑗𝑖

say that transitions change coordinates, not the point on S, and compose on triple overlaps. All maps are restricted to their appropriate overlap domains. The same statements hold for regular d-folds.

Two overlapping surface charts. The transition map sends q1 to q2 while both charts send these coordinates to the same point p on S.

Figure 19.1: Two coordinate descriptions of the same point: σ1(q1) = p = σ2(q2) and τ21(q1) = q2. The transition map is defined only on the shaded preimages of the overlap.

19.1.1 Example: two coordinate systems on a cylinder

Consider the cylinder

S = {(x,y,z) ∈ ℝ3 : x2 + y2 = 1} = F−1(0),F(x,y,z) = x2 + y2 − 1.

Since J(F)(x,y,z) = (2x,2y,0) has rank one on S, this is a regular surface by the level-set description in Lecture 17.

For the portion U = {(x,y,z) ∈S : y > 0, − 1 < z < 1}, take

σ1 : (0,π) × (−1,1) →U, σ1(𝜃,h) = (cos⁡𝜃,sin⁡𝜃,h), σ2 : (−1,1) × (−1,1) →U, σ2(u,v) = (u,1 − u2,v).

The inverses are

φ1(x,y,z) = (arccos⁡x,z),φ2(x,y,z) = (x,z).

The domain 0 < 𝜃 < π makes sin⁡𝜃 > 0 and ensures that arccos⁡(cos⁡𝜃) = 𝜃. The two transition maps are therefore

τ21(𝜃,h) = (cos⁡𝜃,h),τ12(u,v) = (arccos⁡u,v).

Their differentials are the inverse matrices

(Dτ21)(𝜃,h) = ( −sin⁡𝜃 0 0 1 ),(Dτ12)(u,v) = ( −1∕1 − u2 0 0 1 ),u = cos⁡𝜃.

This is the coordinate-change calculation from the handwritten example; compare the cylindrical chart in [Tap16, Example 3.23].

Angle-height and graph coordinates on the same open cylinder patch, related by tau21(theta,h) = (cos(theta),h).

Figure 19.2: Angle–height coordinates and graph coordinates on the same open portion of the cylinder. Rectangle boundaries are excluded.

19.2 Tangent planes

Let σ : V ⊆ ℝ2 →U ⊆S be a chart, with σ(q) = p. Its differential is a linear map

(𝐷𝜎)q : ℝ2→ℝ3.

It is injective: if φ~ is a local smooth extension of φ = σ−1, then the chain rule applied to φ~ ∘σ = Id ⁡ gives (Dφ~)p(𝐷𝜎)q = I2.

Definition 19.1: Tangent plane

The tangent plane to S at p is

TpS = Im ⁡ (𝐷𝜎)q = span ⁡ {σu(q),σv(q)}⊆ ℝ3.

It is a two-dimensional linear subspace. Its translate

p + TpS = {p + w : w ∈TpS}

is the affine plane through p drawn tangent to the surface.

We must check that this definition does not depend on the chart.

Proposition 19.2 (Independence of the chart). If σ1(q1) = σ2(q2) = p, then

Im ⁡ (Dσ1)q1 = Im ⁡ (Dσ2)q2.

Proof by differentiating the chart diagram. Write V 𝑖𝑗 = σi−1(Ui ∩Uj). The transition map τ21 : V 12 →V 21 satisfies τ21(q1) = q2. The left diagram commutes; differentiating it at q1 gives the right diagram:

Chart-independence proof: the commuting diagram for two charts and their transition map, followed by the commuting diagram of their differentials.

Thus

(Dσ1)q1 = (Dσ2)q2 ∘ (Dτ21)q1.

Because τ21 is a diffeomorphism, its differential is an isomorphism of ℝ2. Consequently, as the input runs over all of ℝ2, both downward maps have exactly the same image in ℝ3. □

Compare [Tap16, Lemma 3.31]. Tapp begins with a description by curves; here we begin with chart differentials and recover the curve description below. For a regular d-fold X ⊆ ℝn, the same definition and diagram give a d-dimensional subspace TpX = Im ⁡ (𝐷𝜎)q ⊆ ℝn.

19.3 Differentials of maps between manifolds

Let f : X →Y be smooth, where X ⊆ ℝn and Y ⊆ ℝm are regular d- and e-folds. Choose charts σ : V →U ⊆X and η : W →U′⊆Y , with σ(a) = p and η(b) = f(p). After shrinking the neighborhoods so that f(U) ⊆U′, its coordinate expression is

f^ = η−1 ∘f ∘σ : V →W,f^(a) = b.

The handwritten diagram and its differential take the form

A smooth map between manifolds in local charts, and the corresponding commuting diagram between tangent spaces and coordinate vector spaces.

The vertical maps in the differential diagram are isomorphisms onto the tangent spaces. Therefore the bottom arrow determines the top arrow.

Definition 19.3: Differential between tangent spaces

The differential of f at p is the linear map

(𝐷𝑓)p : TpX→Tf(p)Y

determined by

(𝐷𝑓)p ∘ (𝐷𝜎)a = (𝐷𝜂)b ∘ (Df^)a.

Equivalently,

(𝐷𝑓)p = (𝐷𝜂)b ∘ (Df^)a ∘ ((𝐷𝜎)a)−1,

where the inverse on the right has domain TpX.

Changing either chart introduces a transition map and its inverse into this formula. The chain rule cancels their differentials, so the top arrow is independent of the charts. This is the same diagrammatic principle used for TpS.

There is also an ambient description. If f~ is any local smooth extension of f, then differentiating f~ ∘σ = η ∘f^ gives

(𝐷𝑓)p = (Df~)p|TpX.

Thus the restriction is independent of the extension, and its image lies in Tf(p)Y . For surfaces, compare [Tap16, Definition 3.33 and Proposition 3.34].

19.4 Submanifolds and curves

Proposition 19.4: Tangent spaces of submanifolds

If Y ⊆X ⊆ ℝn is a smooth embedded submanifold and p ∈Y , then

TpY ⊆TpX.

Proof using a subchart. Suppose dim⁡Y = k and dim⁡X = d. Choose a chart σ : V ⊆ ℝd →U ⊆X adapted to the submanifold: with

j : ℝk→ℝd,j(u) = (u,0),W = {u ∈ ℝk : (u,0) ∈V },

the map ρ = σ ∘j : W →Y ∩U is a chart for Y . Write ρ(a) = p, so σ(j(a)) = p. The subchart diagram and its differentiated version are

A submanifold in a compatible subchart. Differentiating the coordinate-subspace inclusion gives inclusion of its tangent space in the ambient tangent space.

The upper arrow on the right is the ambient identity: the inclusion Y ∩U↪U is the restriction of the identity on ℝn. Commutativity gives

(𝐷𝜌)a = (𝐷𝜎)j(a) ∘j.

Taking images proves TpY = Im ⁡ (𝐷𝜌)a ⊆ Im ⁡ (𝐷𝜎)j(a) = TpX. In words, the coordinate subspace downstairs maps into the ambient tangent space upstairs. □

19.4.1 Tangent lines of curves fill the tangent plane

Proposition 19.5. For a regular surface S and p ∈S,

TpS = ⋃ C⊆S a smooth embedded curve p∈C TpC.

Equivalently,

TpS = {γ′(0) : γ : (−𝜀,𝜀) →S smooth,γ(0) = p}.

Proof. Every tangent line TpC is contained in TpS by the submanifold diagram. Conversely, write a nonzero w ∈TpS as w = (𝐷𝜎)q(v). For sufficiently small 𝜀 > 0,

γ(t) = σ(q + 𝑡𝑣),|t|< 𝜀,

is a regular embedded curve through p, with γ′(0) = (𝐷𝜎)q(v) = w. The zero vector lies in every tangent line and is also the velocity of a constant smooth curve. Finally, any smooth curve in a chart is σ ∘β, so its velocity (𝐷𝜎)qβ′(0) lies in TpS. □

Compare [Tap16, Definition 3.30 and Lemma 3.31].

19.5 Example: the tangent plane to the cylinder

Return to p = (cos⁡𝜃,sin⁡𝜃,h) on the cylinder. The matrix requested at the end of the handwritten notes is

(Dσ1)(𝜃,h) = ( −sin⁡𝜃 0 cos⁡𝜃 0 0 1 ).

Its columns span

TpS = span ⁡ {(−sin⁡ ⁡ 𝜃,cos⁡ ⁡ 𝜃,0),(0,0,1)}.

For p = (x,y,z), this is equivalently

TpS = {(a,b,c) ∈ ℝ3 : 𝑥𝑎 + 𝑦𝑏 = 0}.

The affine tangent plane therefore has equation

x(X −x) + y(Y −y) = 0,or, since x2 + y2 = 1,𝑥𝑋 + 𝑦𝑌 = 1.

Here (X,Y,Z) denotes a variable point of the affine plane.

On the overlap with the graph chart, its differential is

(Dσ2)(u,v) = ( 1 0 −u∕1 − u2 0 0 1 ).

The differentiated chart diagram is visible directly in the product

( 1 0 − cot⁡ ⁡ 𝜃 0 0 1 )⏟ (Dσ2)(cos⁡ ⁡𝜃,h) ( −sin⁡ ⁡𝜃 0 0 1 )⏟ (Dτ21)(𝜃,h) = ( −sin⁡ ⁡𝜃 0 cos⁡ ⁡𝜃 0 0 1 )⏟ (Dσ1)(𝜃,h).

The middle factor changes a basis of the coordinate plane, not the resulting tangent plane in ℝ3.

19.6 Summary

Transition maps change coordinates, and their differentials explain why tangent spaces and differentials are independent of those coordinates. A subchart gives a tangent subspace. In particular, the tangent lines of curves through a point fill the tangent plane of a surface at that point.

Lecture 20
Area distortion, orientation, and orientable surfaces

Wednesday, October 7, 2026

Reading. [Tap16, §§ 3.4–3.5, pp. 147–159].

After class. HW 6(c): Tapp, Exercises 3.51, 3.54, 3.57, 3.59, 3.63, and 3.72 (tentative on the course schedule).

Goals. Interpret the absolute determinant geometrically, distinguish the two orientations of a vector space, and describe orientations of manifolds by compatible charts and, for surfaces, by unit normal fields.

20.1 Determinants, area, and volume

Let A = (v1⋯vn) be the square matrix with columns v1,…,vn ∈ ℝn. The geometric quantity we use is |det ⁡ A|, interpreted using familiar length, area, and volume, without developing a general higher-dimensional volume theory.

Proposition 20.1: The absolute determinant

For n ∈{1,2,3}, the parallelepiped spanned by the columns v1,…,vn of A has length, area, or volume |det ⁡ A|, respectively.

Sketch by elementary column reduction. For n = 1, the interval from 0 to a has length |a|. For n = 2,3, the essential operation is

Cj←Cj + cCi(i≠j).

Adding a multiple of one column to another preserves the determinant. Geometrically, the corresponding edge slides in a direction parallel to the unchanged base, without changing the height. Thus the area in ℝ2, or the volume in ℝ3, also stays the same.

For an invertible matrix, elementary column additions, with column interchanges when needed, reduce the matrix to a diagonal matrix. Interchanging columns leaves the parallelepiped unchanged and only changes the sign of the determinant. For the resulting diagonal matrix diag ⁡ (a1,… ⁡ ,an), the parallelepiped is a rectangle or rectangular box, so its size is

|a1| ⋯ |an| = |a1⋯an| = |det ⁡ A|.

If the columns are dependent, both the determinant and the corresponding area or volume are zero. □

Here Cj ←Cj + cCi leaves Ci unchanged. Scaling a column by itself by c would instead multiply the determinant by c and the geometric size by |c|; it is not a size-preserving operation.

A parallelogram and its shear to a rectangle. Adding a multiple of one spanning column to another preserves the area.

Figure 20.1: Adding a multiple of the base edge to the other edge changes neither base nor height. This is the geometric step in the column-reduction argument; the three-dimensional step uses the same base–height idea.

20.1.1 Linear maps and the unit parallelepiped

If L : ℝn → ℝn is represented by A, then L(ei) = vi. Thus L sends the standard unit parallelepiped to the one spanned by the columns of A. More generally, if B contains the edge vectors of another parallelepiped, its image has edge matrix 𝐴𝐵, and

|det ⁡ (𝐴𝐵)| = |det ⁡ A| |det ⁡ B|.

In the dimensions just discussed, |det ⁡ A| is therefore the length, area, or volume scaling factor of L.

20.1.2 The two-dimensional wedge calculation

For the linear map in the handwritten notes,

L : ℝ2→ℝ2,(x 1,x2)⟼(y1,y2) = (x1+2x2,3x1+4x2),A = ( 1 2 3 4 ),

we have

dy1 = dx1 + 2 dx2, dy2 = 3 dx1 + 4 dx2.

Using bilinearity and the alternating rules dxi ∧ dxi = 0 and dx2 ∧ dx1 = −dx1 ∧ dx2, we obtain

dy1 ∧ dy2 = (dx1 + 2 dx2) ∧ (3 dx1 + 4 dx2) = 4 dx1 ∧ dx2 + 6 dx2 ∧ dx1 = (4 − 6) dx1 ∧ dx2 = −2 dx1 ∧ dx2.

The coefficient is det ⁡ A. The negative sign records an orientation reversal, while |det ⁡ A| = 2 is the ordinary area scaling factor. The same expansion for a general 2 × 2 matrix gives

dy1 ∧ dy2 = det ⁡ (A) ⁡ dx1 ∧ ⁡ dx2.

Here the yi are viewed as functions of x; equivalently, the left side is the pullback of the target area form by L.

20.1.3 The differential and infinitesimal distortion

For a smooth map f : Ω ⊆ ℝn → ℝn, its differential is the first-order linear approximation:

f(p + w) = f(p) + (𝐷𝑓)p(w) + o(|w|).

Consequently, in dimensions 1,2,3, |det ⁡ J(f)(p)| is the length, area, or volume scaling factor of this linear approximation. This is the infinitesimal distortion viewpoint; no change-of-variables theorem is needed here. For a map between surfaces, represent (𝐷𝑓)p : TpS →Tf(p)S~ in orthonormal bases of the two tangent planes. The absolute determinant of that 2 × 2 matrix gives its area distortion; compare [Tap16, Proposition 3.38]. Arbitrary chart bases need not be orthonormal.

20.2 Orientations of vector spaces

An ordered basis B = (v1,…,vd) of a d-dimensional vector space V ⊆ ℝn gives an isomorphism

ℝd→V,(a 1,…,ad)⟼a1v1 + ⋯ + advd.

Any other ordered basis B′ is obtained by an invertible d ×d change-of-basis matrix A: in column notation, B′ = 𝐵𝐴.

Definition 20.2: Orientation of a vector space

Two ordered bases B,B′ of a positive-dimensional real vector space have the same orientation if the change-of-basis matrix from one to the other has positive determinant. An orientation is a choice of one of the resulting two classes of ordered bases.

This is an equivalence relation because determinants multiply under composition and det ⁡ (A−1) = 1∕det ⁡ A. Relative to a fixed basis, every ordered basis belongs to one of the classes det ⁡ A > 0 or det ⁡ A < 0. Once a class is chosen, its bases are called positively oriented; the other bases are negatively oriented. No orthonormality is required for this sign distinction.

20.2.1 A plane in ℝ3: a choice of unit normal

For a two-dimensional subspace V ⊆ ℝ3, an ordered basis (v1,v2) determines

ν = v1 ×v2 |v1 ×v2| .

If (w1,w2) = (v1,v2)A, bilinearity and alternation give

w1 ×w2 = det ⁡ (A)(v1 ×v2).

Thus positively related bases determine the same unit normal, while negatively related bases determine opposite normals. Choosing an orientation of V is therefore equivalent to choosing one of its two unit normals, ν or −ν; compare [Tap16, Definition 3.40].

An ordered basis v1, v2 of a plane in three-dimensional space determines a unit normal N.

Figure 20.2: The two orientations of a plane correspond to its two unit normals. Reversing the order of a basis reverses the normal.

20.3 Orientations of manifolds

Let X ⊆ ℝn be a regular d-fold, with coordinate charts σi : V i →Ui ⊆X. As in Lecture 19, let τ𝑗𝑖 = σj−1 ∘σi on the overlap.

Definition 20.3: Compatible orientations and an oriented atlas

The charts σi,σj are compatibly oriented if

det ⁡ ((Dτ𝑗𝑖)q) > 0for every q ∈σi−1(U i ∩Uj).

An oriented atlas is a collection of pairwise compatibly oriented charts whose images cover X. The manifold X is orientable if it admits such an atlas; choosing such an atlas determines an orientation of X.

Two oriented atlases describe the same orientation when their union is still compatibly oriented. The positivity condition is imposed at every point of each overlap, including every component if an overlap is disconnected.

A chart gives the ordered tangent basis ((Dσi)q(e1),…,(Dσi)q(ed)). The differentiated transition diagram gives

(Dσi)q = (Dσj)τ𝑗𝑖(q)(Dτ𝑗𝑖)q.

Thus the two bases determine the same orientation of TpX precisely when the transition differential has positive determinant. An oriented atlas makes these local choices agree over all of X.

20.3.1 Surfaces and unit normal fields

Proposition 20.4: Orientation by a unit normal field

For a regular surface S ⊆ ℝ3, giving an orientation is equivalent to giving a smooth map

ν : S→S2 ⊆ ℝ3,ν(p) ⊥T pSfor every p ∈S.

Such a map is a unit normal field.

Proof. A chart σi(u,v) gives the smooth local unit normal

νi(σi(q)) = (σi)u(q) × (σi)v(q) |(σi)u(q) × (σi)v(q)|.

On an overlap, the transition identity above and the cross-product calculation give

(σi)u(q) × (σi)v(q) = det ⁡ ((Dτ𝑗𝑖)q)((σj)u(τ𝑗𝑖(q)) × (σj)v(τ𝑗𝑖(q))).

For an oriented atlas the determinant is positive, so the local normals agree and define a smooth global field.

Conversely, given ν, choose small connected chart neighborhoods. The chart normal is either ν or −ν, with a constant sign on each such neighborhood. Reverse one coordinate when necessary so that all chart normals agree with ν. Their transition determinants are then positive. □

Tapp uses unit normal fields as the definition of an orientation; see [Tap16, Definitions 3.42–3.44].

20.4 Examples and regular level sets

Example 20.5 (Euclidean space and graphs). The identity chart gives the usual orientation of ℝd. A smooth graph Γh ⊆ ℝd × ℝn−d is also orientable: the single chart u↦(u,h(u)) already covers it, so there are no conflicting coordinate changes. For surfaces, compare [Tap16, Example 3.48].

Example 20.6 (Every manifold is locally orientable). Every point of a regular d-fold has a neighborhood covered by a single chart. Such a neighborhood is orientable. The possible obstruction is therefore global: orientations chosen in different neighborhoods may fail to fit together.

Theorem 20.7: Regular level sets are orientable

Let F : Ω ⊆ ℝn → ℝm be smooth, with Ω open and 1 ≤m < n. If 0 is a regular value, that is,

rank ⁡ (𝐷𝐹)p = mfor every p ∈X := F−1(0),

then X is an orientable regular (n −m)-fold.

Proof in the surface case. For F : Ω ⊆ ℝ3 → ℝ, the rank assumption says ∇ ⁡F(p)≠0 on X. Since F ∘σ = 0 for a chart of X, differentiation gives

TpX ⊆ker⁡(𝐷𝐹)p.

Both spaces have dimension two, so equality holds. Using (𝐷𝐹)p(w) = ⟨∇ ⁡F(p),w⟩, we see that

ν(p) = ∇ ⁡F(p) |∇ ⁡F(p)|

is a smooth unit normal field on X. Hence X is orientable. This is [Tap16, Example 3.46]. □

Remark 20.8 (The higher-codimension sketch from the lecture). The same kernel argument gives TpX = ker⁡(𝐷𝐹)p in the general case. The bundle shorthand in the handwritten notes records the exact sequence

0→𝑇𝑋→X × ℝn →𝐷𝐹X × ℝm→0,(p,w)⟼(p,(𝐷𝐹) p(w)).

Here 𝑇𝑋 collects the tangent spaces of X. Taking determinant lines (the highest exterior powers) gives

det ⁡ (𝑇𝑋)≅ ⁡ det ⁡ (X × ℝn) ⊗ det ⁡ (X × ℝm)∗≅ ⁡ X × ℝ.

The fixed standard orientations on ℝn and ℝm therefore induce a smooth orientation of 𝑇𝑋, proving the general statement. This is the abbreviated argument from the lecture, not a development of the theory of vector bundles. In linear-algebra terms, orient the kernel so that a positive kernel basis, followed by lifts of a positive target basis, forms a positive basis of ℝn.

20.5 A nonexample: the Möbius strip

The Möbius strip is obtained by identifying the two ends of a rectangle with a half-twist, as indicated by the opposite arrows in Figure 20.3. To obtain a regular surface in the sense used here, omit the boundary edge.

The Möbius strip: opposite ends of a rectangle are identified with reversed directions; a locally chosen normal reverses after one circuit.

Figure 20.3: The end identifications reverse direction. After one trip around the central circle of the resulting strip, a continuously transported normal points opposite to its starting direction. The two arrows at the marked point represent the start and end of this trip.

Choose a unit normal at one point on the central circle and continue it continuously as we travel once around the strip. The half-twist brings us back to the same point with the opposite normal. A globally defined continuous unit normal would have to assign the same vector at the start and finish, which is impossible. Thus the Möbius strip is not orientable, even though every sufficiently small chart neighborhood is orientable. Compare [Tap16, § 3.5, pp. 156–157].

20.6 Summary

The magnitude of a determinant measures linear area or volume scaling; its sign distinguishes orientation. Positive transition determinants make local tangent-space orientations agree. On a surface in ℝ3, this is equivalent to a smooth global unit normal field. Regular level sets are orientable, while the Möbius strip demonstrates that local orientability does not imply global orientability.

Lecture 21
Surface area, isometries, and the first fundamental form

Friday, October 9, 2026

Reading. [Tap16, §§ 3.6–3.7, pp. 160–169].

After class. HW 6 due. HW 7(a): Tapp, Exercises 3.74, 3.76, 3.79, 3.80, 3.83, and 3.85 (tentative on the course schedule).

Goals. Describe inner products by matrices, recover lengths and angles, and derive the volume formula needed for the first fundamental form.

This lecture develops the linear algebra of inner products and volumes. The next lecture applies it to tangent planes and surface coordinates.

21.1 Inner products

Definition 21.1: An inner product

An inner product on a real vector space V is a map

⟨⋅, ⋅⟩ : V ×V →ℝ

that is bilinear, symmetric, and positive definite. Thus it is linear in each variable, ⟨u,v⟩ = ⟨v,u⟩, and

⟨v,v⟩ ≥ 0, ⟨v,v⟩ = 0⟺v = 0.

The standard example on ℝn is the dot product

u ⋅v = uT v = ∑ i=1nu ivi.

In this lecture, a dot denotes this Euclidean product, whereas angle brackets may denote a different inner product. Vectors in matrix products are regarded as columns.

21.1.1 The matrix of an inner product

A real n ×n matrix A is symmetric positive definite if AT = A and vT 𝐴𝑣 > 0 for every nonzero v ∈ ℝn. Such a matrix determines an inner product by

⟨u,v⟩A = uT 𝐴𝑣.

Bilinearity follows from matrix multiplication; symmetry follows from AT = A; and positive definiteness is exactly the condition just stated. Also, for the standard basis vectors,

⟨ei,ej⟩ A = eiT Ae j = A𝑖𝑗.

Example 21.2 (The matrix from the lecture). For

A = ( 2 1 1 1 ),

the corresponding inner product is

⟨(u1,u2),(v1,v2)⟩A = ( u1 u2 ) ( 2 1 1 1 ) ( v1 v2 ) = (2u1 + u2)v1 + (u1 + u2)v2.

In particular, ⟨e1,e2⟩ A = 1. To check positive definiteness,

⟨u,u⟩A = 2u12 + 2u 1u2 + u22 = u 12 + (u 1 + u2)2,

which is positive unless u = 0.

Lemma 21.3. Every inner product on ℝn has the form ⟨u,v⟩ = uT 𝐴𝑣 for a unique symmetric positive-definite matrix A, namely A𝑖𝑗 = ⟨ei,ej⟩.

Indeed, bilinearity gives

⟨u,v⟩ = ∑ i,juivj ⟨ei,ej⟩ = uT 𝐴𝑣,

and symmetry and positive definiteness of the inner product give the corresponding properties of A.

21.1.2 Subspaces and injective linear maps

If V ⊆ ℝN is a vector subspace, then restricting the Euclidean dot product to V gives an inner product on V :

⟨u,v⟩V = u ⋅v(u,v ∈V ).

This is the construction we will use on tangent planes.

More generally, let L : ℝn↪ℝN be an injective linear map, and use L also for its N ×n matrix. Define

⟨u,v⟩L = (𝐿𝑢) ⋅ (𝐿𝑣).

This is positive definite because 𝐿𝑢 = 0 implies u = 0. In matrix notation,

⟨u,v⟩L = (𝐿𝑢)T (𝐿𝑣) = uT LT 𝐿𝑣 = uT 𝐵𝑣, B = LT L.

The entry B𝑖𝑗 is (Lei) ⋅ (Lej): it records the inner product of the ith and jth columns of L.

21.2 Lengths and angles

An inner product determines a length function by

‖u‖ = ⟨u, u⟩, ‖u‖2 = ⟨u,u⟩.

The following inequality makes it possible to define angles as well.

Theorem 21.4: Cauchy–Schwarz

For vectors u,v in a real inner-product space,

|⟨u,v⟩|2 ≤ ‖u‖2 ‖v‖2.

In particular, if u,v≠0, then

−1 ≤ ⟨u,v⟩ ‖u‖ ‖v‖ ≤ 1.

For nonzero vectors, the angle 𝜃 ∈ [0,π] is therefore specified by

cos⁡𝜃 = ⟨u,v⟩ ‖u‖ ‖v‖.

Both length and angle are determined by the chosen inner product; compare the Euclidean formulas in [Tap16, § 3.7, p. 165].

21.3 Linear isometries and volume

Definition 21.5: A linear isometry

Let V,W be real inner-product spaces. A linear map L : V →W is inner-product-preserving, or a linear isometry, if

⟨𝐿𝑢,𝐿𝑣⟩W = ⟨u,v⟩V for all u,v ∈V.

Such a map preserves lengths, since

‖𝐿𝑢‖W 2 = ⟨𝐿𝑢,𝐿𝑢⟩ W = ⟨u,u⟩V = ‖u‖V 2.

It also preserves angles, by the preceding cosine formula. In particular, it is injective, although it need not be onto W.

Theorem 21.6: Euclidean realization

If V is an n-dimensional real inner-product space, there is a linear isomorphism T : V → ℝn such that

⟨u,v⟩ = (𝑇𝑢) ⋅ (𝑇𝑣)for all u,v ∈V.

Equivalently, V admits an orthonormal basis for its inner product.

21.3.1 The Gram-determinant formula

Let v1,…,vn ∈V . Use an isomorphism T as above to measure the volume of their parallelepiped in Euclidean coordinates:

Vol ⁡ ⟨⋅,⋅⟩(v1,… ⁡ ,vn) := |det ⁡ ( Tv1 ⋯ Tvn ) | .

Here the parallelepiped consists of the vectors ∑ ⁡ i=1ntivi with 0 ≤ti ≤ 1. The next calculation shows that this definition does not depend on which isometry T we choose.

Proposition 21.7: Volume from inner products

The volume of the parallelepiped spanned by v1,…,vn is

Vol ⁡ ⟨⋅,⋅⟩(v1,… ⁡ ,vn) = det ⁡ G,G =( ⟨vi,vj⟩ )i,j=1n.

The matrix G is called the Gram matrix of these vectors.

The isometry calculation. Put M = ( Tv1 ⋯ Tvn ) . Its columns are vectors in ℝn, so

Vol ⁡ ⟨⋅,⋅⟩(v1,… ⁡ ,vn) = |det ⁡ M | = (det ⁡ M)2 = det ⁡ (MT M).

The (i,j) entry of the matrix inside the determinant is

(MT M) 𝑖𝑗 = (Tvi) ⋅ (Tvj) = ⟨vi,vj⟩ ,

because T preserves inner products. Thus MT M = G, proving the formula. Its right side uses only the original inner product and the vectors vi, not the auxiliary choice of T. □

21.3.2 The volume density in coordinates

Return to an injective linear map L : ℝn↪ℝN and the inner product ⟨u,v⟩L = uT 𝐵𝑣, where B = LT L. The standard basis (e1,…,en) has Gram matrix B, so

Vol ⁡ ⟨⋅,⋅⟩L(e1,… ⁡ ,en) = det ⁡ B = det ⁡ (LT L).

The coordinate unit parallelepiped has Euclidean volume 1, but its volume for this inner product is det ⁡ B.

Consequently, for a region R ⊆ ℝn for which the integral is defined, the corresponding volume is

Vol ⁡ ⟨⋅,⋅⟩L(R) = ∫ Rdet ⁡ B dx1⋯ dxn.

Here L is linear, so B is constant. In the next lecture, L will be the differential of a surface chart at a point, and B will vary from point to point.

21.4 Summary

An inner product is represented in coordinates by a symmetric positive-definite matrix. It determines lengths, angles, and volumes. For an injective linear map L, the induced inner product has matrix LT L, and its coordinate volume density is det ⁡ (LT L). This is the linear algebra behind the surface area formula.

Lecture 22
The first fundamental form in local coordinates

Monday, October 12, 2026

Reading. [Tap16, § 3.9, pp. 182–187].

After class. HW 7(b): Tapp, Exercises 3.100, 3.101, 3.103, 3.105, 3.107, and 3.108 (tentative on the course schedule).

Goals. Express the first fundamental form as B = J(σ)TJ(σ), compute it for the cylinder, and use it to measure surface area and the lengths of curves.

22.1 Review: inner products and volumes

An injective linear map L : ℝn↪ℝN induces the inner product

⟨u,v⟩L = (𝐿𝑢) ⋅ (𝐿𝑣) = uT LT L⏟ Bv.

The n ×n matrix B is symmetric positive definite. By the calculation in Lecture 21,

Vol ⁡ ⟨⋅,⋅⟩L(v1,… ⁡ ,vn) = det ⁡ ( ⟨vi , vj ⟩ L ) ,Vol ⁡ ⟨⋅,⋅⟩L(e1,… ⁡ ,en) = det ⁡ B.

We now apply this construction with L equal to the differential of a surface chart.

22.2 The first fundamental form

Let S ⊆ ℝ3 be a regular surface. At each p ∈S, the tangent plane TpS is a two-dimensional vector subspace of ℝ3. It therefore inherits the Euclidean inner product.

Definition 22.1: The first fundamental form

The first fundamental form of S is the family of inner products

Ip : TpS ×TpS→ℝ,Ip(v1,v2) = v1 ⋅v2,p ∈S.

Thus it is the restriction of the Euclidean dot product to each tangent plane, taken over all points of S.

The restriction can be displayed as the commuting diagram

The first fundamental form Ip is the restriction of the Euclidean dot product from R3 times R3 to TpS times TpS.

We use the bilinear convention of the lecture. Tapp instead names the associated quadratic form v↦Ip(v,v) = |v|2 the first fundamental form; these contain the same information [Tap16, Definition 3.61 and § 3.7].

22.2.1 A chart and its differential

Take a chart

σ : V ⊆ ℝ2→U ⊆S,q = (u,v),σ(q) = p,

where V is open in ℝ2 and U is open in S. Its inverse is φ = σ−1 : U →V . The differential

(𝐷𝜎)q : ℝ2→T pS ⊆ ℝ3

is an isomorphism onto TpS. Writing σ = (σ1,σ2,σ3), its matrix in ambient coordinates is the 3 × 2 Jacobian

J(σ)(q) = ( ∂σ1∕∂𝑢 ∂σ1∕∂𝑣 ∂σ2∕∂𝑢 ∂σ2∕∂𝑣 ∂σ3∕∂𝑢 ∂σ3∕∂𝑣 )q = ( σu(q) σv(q) ).

Its columns are (𝐷𝜎)q(e1) and (𝐷𝜎)q(e2). They form a basis for the tangent plane, but are not generally orthonormal.

A surface chart sigma and its inverse phi. The chart differential maps the coordinate basis at q to sigma_u(q) and sigma_v(q) in the tangent plane at p.

Figure 22.1: A chart identifies the coordinate domain with a surface neighborhood. Its differential sends the coordinate basis to the two tangent vectors shown. The tangent plane is drawn translated to p.

22.2.2 The matrix in local coordinates

Pull the inner product on TpS back to ℝ2 using (𝐷𝜎)q. For a,b ∈ ℝ2,

Ip((𝐷𝜎)q(a),(𝐷𝜎)q(b)) = (J(σ)(q)a)T (J(σ)(q)b) = aT B(q)b,

where

B(q) = J(σ)(q)T J(σ)(q).

Thus

B11(q) = σu(q) ⋅σu(q), B12(q) = B21(q) = σu(q) ⋅σv(q), B22(q) = σv(q) ⋅σv(q).

This is a 2 × 2 symmetric positive-definite matrix, since (𝐷𝜎)q is injective. In particular, det ⁡ B(q) > 0. The matrix varies with q; it describes the first fundamental form in the chosen chart, not a new inner product on the surface.

For comparison with Tapp’s notation,

E = B11,F = B12,G = B22.

Tapp writes the associated quadratic expression as E du2 + 2F 𝑑𝑢 𝑑𝑣 + G dv2 [Tap16, Definition 3.78]. We keep the matrix notation B used in the lecture.

22.3 Example: rolling a strip onto the cylinder

Consider

S = {(x,y,z) ∈ ℝ3 : x2 + y2 = 1},σ(u,v) = (cos⁡u,sin⁡u,v),

with V = (0,2π) × ℝ. This chart covers the cylinder except for the vertical line {(1,0,z) : z ∈ ℝ}. The variable u is an angle measured in radians, and v is height.

The Jacobian is

J(σ)(u,v) = ( −sin⁡u 0 cos⁡ u 0 0 1 ).

Multiplying the matrices gives

B(u,v) = ( −sin⁡u cos⁡u 0 0 0 1 ) ( −sin⁡u 0 cos⁡ u 0 0 1 ) = ( sin⁡2u + cos⁡2u 0 0 1 ) = I2 .

Consequently, for a,b ∈ ℝ2,

Iσ(q)((𝐷𝜎)q(a),(𝐷𝜎)q(b)) = aT I 2b = a ⋅b.

At every point, the chart differential is a linear isometry from the coordinate plane to the tangent plane. This explains the picture of rolling up a flat strip without changing its tangent-vector lengths or angles; compare [Tap16, Example 3.63].

The cylinder chart rolls the open strip (0,2pi) times R onto a cylinder with one vertical seam omitted. A square Q becomes the curved patch sigma(Q).

Figure 22.2: The cylinder chart rolls the strip onto the cylinder with one vertical seam omitted. A square in the strip becomes a curved patch; its area is computed below. The diagram is schematic.

22.4 Surface area

Let R ⊆U be a region covered by the chart σ : V →U. As in the lecture, we work with polygonal regions: the inverse image σ−1(R) is the interior together with the boundary of a piecewise-regular simple closed plane curve. The name does not require the boundary curves to be straight line segments; compare [Tap16, Definition 3.55].

Definition 22.2: Area in local coordinates

For such a region R ⊆S, its surface area is

Area ⁡ (R) = ∬σ−1(R)det ⁡ B(u, v) 𝑑𝑢 𝑑𝑣.

Equivalently, the area density in the chart is det ⁡ B = B11 B22 − B12 2.

Indeed, B is the Gram matrix of the two tangent vectors σu,σv. The volume calculation from Lecture 21 therefore gives det ⁡ B for the area of their tangent parallelogram. Integrating this density gives the area of the region. In Tapp’s notation the integrand is 𝐸𝐺 − F2 [Tap16, Lemma 3.77(4)].

22.4.1 Independence of coordinates

The handwritten notes point out that this definition is independent of coordinates by change of variables. To spell out that observation, write σ~ = σ ∘τ, where τ is the transition diffeomorphism between the coordinate domains. The chain rule gives

B~(w) = J(τ)(w)T B(τ(w))J(τ)(w),

and hence

det ⁡ B~(w) = det ⁡ B(τ(w)) |det ⁡ J(τ)(w)|.

This is exactly the factor in the ordinary change-of-variables theorem, so the two area integrals agree. The absolute value also shows that no orientation choice is needed for area.

Example 22.3 (A square rolled onto the cylinder). For the cylinder chart, B = I2, so det ⁡ B = 1. If Q = [a,a + s] × [b,b + s] ⊆V is a square of side s > 0, then

Area ⁡ (σ(Q)) = ∬Q1 𝑑𝑢 𝑑𝑣 = s2.

Thus the square has the same area after it is rolled onto the cylinder.

22.5 Arc length in local coordinates

Let γ : [a,b] →V be a regular plane curve with γ(t) = (u(t),v(t)), and let

β = σ ∘γ : [a,b]→S

be the corresponding curve on the surface. The composition diagram and its differentiated version are

The curve beta equals sigma composed with gamma. The differential diagram commutes: D beta equals D sigma composed with D gamma, with the stated evaluation points.

The chain rule makes the second diagram commute. Applying it to 1 ∈ ℝ gives

β′(t) = (𝐷𝜎) γ(t)(γ′(t)) = u′(t)σ u(γ(t)) + v′(t)σ v(γ(t)).

In matrix form, suppressing the evaluation point γ(t) in the partial derivatives, this is

( ∂σ1∕∂𝑢 ∂σ1∕∂𝑣 ∂σ2∕∂𝑢 ∂σ2∕∂𝑣 ∂σ3∕∂𝑢 ∂σ3∕∂𝑣 ) ( u′(t) v′(t) ) = ( d 𝑑𝑡σ1(u(t),v(t)) d 𝑑𝑡σ2(u(t),v(t)) d 𝑑𝑡σ3(u(t),v(t)) ).

Squaring the speed and using B = J(σ)T J(σ), we obtain

|β′(t)|2 = γ′(t)T B(γ(t))γ′(t) = B11(γ(t))u′(t)2 + 2B 12(γ(t))u′(t)v′(t) + B 22(γ(t))v′(t)2.

The two mixed terms combine because B12 = B21.

Proposition 22.4: Arc length from the first fundamental form

With the notation above,

L(σ ∘γ) = ∫ abB 11(u′)2 + 2B12u′v′ + B22(v′)2 𝑑𝑡,

where u′,v′ are evaluated at t, and each B𝑖𝑗 is evaluated at γ(t) = (u(t),v(t)).

This follows by integrating the speed just computed; compare [Tap16, Lemma 3.77(3)]. In the cylinder chart, B = I2, so the expression reduces to

L(σ ∘γ) = ∫ abu′ (t)2 + v′ (t)2 𝑑𝑡 = L(γ).

Thus rolling the strip onto the cylinder preserves curve lengths as well as the area calculated above.

22.6 Summary

The first fundamental form is the Euclidean inner product restricted to each tangent plane. In coordinates it is represented by B = J(σ)T J(σ). Its determinant supplies the surface area density, and evaluating its quadratic expression on the coordinate velocity supplies the squared speed of a curve.

Bibliography

[Bro96]   Andrew Browder. Mathematical Analysis: An Introduction. Undergraduate Texts in Mathematics. Springer, New York, first edition, 1996.

[Tap16]   Kristopher Tapp. Differential Geometry of Curves and Surfaces. Undergraduate Texts in Mathematics. Springer, Cham, first edition, 2016.

Index

Index links lead to lecture sections.

C∞(I), §1.1.3
O(n), §7.1
R90, §5.1
Ωk(U), §9.2

acceleration, §3.1
    orthogonal to velocity, §2.3
    perpendicular component, §3.1.1
    tangential component, §3.1.1
affine tangent plane, §19.2
angle
    between vectors, §2.1
    from an inner product, §21.2
angle function, §5.3, §8.1
antisymmetry, §9.1
arc length, §1.4
    first fundamental form, §22.5
    invariance under reparametrization, §3.3
    lower bound by endpoint distance, §2.4
arc-length element, §11.3.1
arc-length parametrization, §3.4
area density, §22.4
area distortion
    of a differential, §20.1.3
area form, §11.2.1
atlas, §17.2
    oriented, §20.3

binormal vector, §6.2
Borel’s lemma, §1.1.3

Cauchy–Schwarz inequality, §2.1, §21.2
chain rule, §16.3
    along a curve, §10.2
    for reparametrization, §3.3
circle
    curvature of, §4.1
circulation, §11.2
closed curve, §3.5
    periodic extension, §3.5
closed one-form, §10.5
cocycle identity, §19.1
component
    of a vector, §2.2.1
conservation of energy, §10.6
conservative vector field, §10.1
    equivalent characterizations, §10.5
    zero infinitesimal circulation, §11.2.2
constant-speed curve, §2.3
coordinate chart, §17.2
    compatibly oriented, §20.3
    transition maps, §19.1
corner
    of a curve, §8.4
cross product, §6.1
curvature
    cross-product formula, §6.4.3
    formulas for, §7.4
    invariance under rigid motion, §7.2
    of a curve, §4.1
    signed, §5.2
curve
    acceleration, §3.1
    arc length, §1.4
    closed, §3.5
    curvature, §4.1
    nonregular, §1.3
    parametrized, §1.2
    regular, §1.3
    reparametrization, §3.2
    speed, §1.3
    trace, §3.2
    unit normal vector, §4.2
    unit tangent vector, §4.2
cusp
    semicubical, §1.3
cylinder
    coordinate charts, §19.1.1
    first fundamental form, §22.3
    tangent plane, §19.5

degree
    of a map to the circle, §8.1
    winding number, §8.3
determinant
    length, area, and volume, §20.1
determinant line
    regular level-set sketch, §20.4
diffeomorphism, §16.5.2
differential
    between manifolds, §19.3
    of a map, §1.1.3
    of a smooth map, §16.2
differential form, §9.2
    coordinate form, §9.1
    exact, §10.1
directional derivative, §10.2, §16.2.1
distance
    Euclidean, §1.1.1
    preserved by a rigid motion, §7.1
divergence, §11.3.2
dot product, §2.1

ellipsoid, §16.5.2
Euclidean norm, §1.1.1
exact one-form, §10.1
extension
    of a smooth function, §1.1.3
exterior
    of a Jordan curve, Lecture 8
exterior derivative, §9.3
    squares to zero, §11.1

first fundamental form, §22.2
    matrix in local coordinates, §22.2.2
flat function, §1.1.3
flux, §11.3.1
    and divergence, §11.3.3
flux form, §11.3.2
Frenet equations, §6.3
Frenet frame, §6.2
fundamental theorem
    of line integrals, §10.4
    of plane curves, §7.3
    of space curves, §7.3

graded commutativity, §9.1.1
gradient, §10.1
    direction of greatest increase, §10.3
    orthogonal to level sets, §10.3
Gram determinant
    volume formula, §21.3.1
Gram matrix, §21.1.2
Gram–Schmidt process, §2.2.2
graph
    as a parametrized curve, §1.2
    of a smooth function, §17.2.1
    orientability, §20.4
gravitational potential energy, §10.7
gravity, §10.7
Green’s theorem
    circulation form, §11.2
    differential-form notation, §10.5
    flux form, §11.3.3

Hodge star
    on coordinate forms, §11.3.2
Hopf’s Umlaufsatz, Lecture 8
    generalized, §8.4

implicit function theorem, §16.4
infinitesimal circulation, §11.2.2
inner product
    Euclidean, §2.1
    general definition, §21.1
    induced by a linear map, §21.1.2
    matrix of, §21.1.1
interior
    of a Jordan curve, Lecture 8
interval, §1.1.2
inverse function theorem, §16.4
isometry
    cylinder chart, §22.3

Jacobian matrix, §16.2
Jordan curve theorem, Lecture 8

kinetic energy, §9.5, §10.6

Leibniz rule
    graded, §9.3
level set, §10.3
    locally a graph, §16.4
    orientability, §20.4
    regular, §17.4
line integral, §9.5
    of a gradient, §10.4
    path independence, §10.5
    reparametrization invariance, §9.5
line segment
    shortest path, §2.4
linear isometry, §21.3
linear map
    area and volume scaling, §20.1.1

Möbius strip, §20.5

negative orientation, §8.3
neighborhood
    relative, §16.5.1
nonorientable surface, §20.5
norm
    from an inner product, §21.2

one-form
    associated to a vector field, §9.4
open set
    in Euclidean space, §16.5.1
    relative, §16.5.1
ordered basis
    orientation, §20.2
orientability
    local, §20.4
orientable manifold, §20.3
orientation
    of a curve, §3.3
    of a manifold, §20.3
    of a simple closed plane curve, §8.3
    of a surface, §20.3.1
    of a vector space, §20.2
    preserved by a rigid motion, §7.1
    reversal of a line integral, §9.5
oriented area, §20.1.2
orthogonal complement, §2.2.2
orthogonal decomposition, §2.2.1
    of acceleration, §3.1.1
orthogonal matrix, §7.1
orthogonal vectors, §2.1
orthonormal basis, §2.2.2
orthonormal set, §2.2.2
outward unit normal, §11.3.1

parallel vectors, §2.1
parallelepiped, §20.1
parametrized curve, §1.2
partial derivative, §16.1
path independence, §10.5
periodic curve, §3.5
piecewise-regular curve, §8.4
polygonal region
    on a surface, §22.4
positive orientation, §8.3
potential energy, §10.6
potential function, §10.1
product rule
    for the inner product, §2.3
projection
    in a direction, §2.2.1
    onto a subspace, §2.2.2

regular curve, §1.3
regular d-fold, §17.2
    as a level set, §17.4
    locally a graph, §17.3
regular surface, §17.2
regular value, §20.4
reparametrization, §3.2
    orientation-preserving, §3.3
    orientation-reversing, §3.3
rigid motion, §7.1
    classification, §7.1
    improper, §7.1
    proper, §7.1
rotation
    clockwise quarter-turn, §11.3.1
rotation index, §5.4
    as a degree, §8.1
rotation matrix, §5.1
    cross product, §6.1.1

signed angle
    at a corner, §8.4
signed curvature, §5.2
    invariance under rigid motion, §7.2
simple closed curve, §3.5, Lecture 8
smooth extension
    local, §16.5.2
smooth function, §1.1.3
smooth map, §1.1.3
    composition, §16.1
    on a subset, §16.5.2
    on an open set, §16.1
smoothness
    at a boundary point, §1.1.3
speed, §1.3
    derivative of, §3.1.1
sphere
    diffeomorphic to an ellipsoid, §16.5.2
    velocity tangent to, §2.3
submanifold
    tangent-space inclusion, §19.4
surface area
    in local coordinates, §22.4
    independence of coordinates, §22.4.1
surface patch, §17.2
symmetric positive-definite matrix, §21.1.1

tangent plane, §19.2
    characterization by curves, §19.4.1
tangent space, §19.2
Taylor series
    formal, §1.1.3
torsion, §6.3
    invariance under rigid motion, §7.2
total signed curvature, §5.4, §8.4
trace of a curve, §3.2
transition map, §19.1
translation, §7.1
tubular neighborhood
    of a curve, §8.3
turning number, see rotation index

unit binormal vector, §6.2
unit normal
    orientation of a plane, §20.2.1
unit normal field, §20.3.1
unit normal vector
    of a curve, §4.2
unit tangent
    angle function, §5.3
unit tangent map, §8.1
unit tangent vector, §4.2
unit-speed curve
    existence of reparametrization, §3.4

vector field, §9.4
velocity of a curve, §1.3
volume
    from an inner product, §21.3.1
volume density, §21.3.2

wedge product, §9.1
    determinant calculation, §20.1.2
winding number, §8.3
    of the unit tangent, §5.4
work integral, §9.5
work–energy identity, §9.5