MATH 4230/5230 — Summary Review, Lectures 1

01Parametrized curves: regularity, speed, and arc length

1.1 Review

1.1.1 Number systems and Euclidean space

Definition — Number systems

Recall the familiar inclusions

N \subseteq Z \subseteq Q \subseteq R,

where $N = {1, 2, 3, \dots}$ are the natural numbers, $Z = {\dots, - 1, 0, 1, 2, \dots}$ are the integers, $Q = {a / b : a, b \in Z, b \neq 0}$ are the rational numbers, and $R$ is the set of real numbers (all the numbers on the number line).

Lecture notes, §1.1.1 · unnumbered.

Definition — Euclidean space

For each positive integer $n$ , Euclidean $n$ -space is

R^{n} = {(x_{1}, \dots, x_{n}) : x_{i} \in R} .

Lecture notes, §1.1.1 · unnumbered.

Definition — Euclidean norm and distance

In particular, $R^{1} = R$ . For $x = (x_{1}, \dots, x_{n}) \in R^{n}$ , the Euclidean norm is

‖ x ‖ = \sqrt{x_{1}^{2} + \dots + x_{n}^{2}},

and the distance between $x, y \in R^{n}$ is $‖ x - y ‖$ .

Lecture notes, §1.1.1 · unnumbered.

1.1.2 Intervals

Definition — Intervals

An interval is a subset $I \subseteq R$ with the property that, if $a, b \in I$ and $a < c < b$ , then $c \in I$ .

Lecture notes, §1.1.2 · unnumbered.

1.1.3 Smooth functions and maps

Definition — Smooth functions and one-sided derivatives

Let $I \subseteq R$ be an interval. A function $f : I \to R$ is smooth if it is infinitely differentiable at every $x \in I$ ; i.e., for all $x \in I$ and all integers $k \geq 0$ , 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 $\lim_{h \to 0^{+}} \frac{f (a + h) - f (a)}{h}$ exists. We write

C^{\infty} (I) = {f : I \to R : f is smooth} .

Lecture notes, §1.1.3 · unnumbered.

Definition — Smooth maps into Euclidean space

A map

f = (f_{1}, \dots, f_{n}) : I \to R^{n}

f (t) = (f_{1} (t), \dots, f_{n} (t))

is smooth when each coordinate function $f_{i} : I \to R$ is smooth.

Lecture notes, §1.1.3 · unnumbered.

1.2 Parametrized curves

Definition 1.4 — Parametrized curve

A smooth parametrized curve in $R^{n}$ is a smooth map

γ : I \to R^{n},

where $I \subseteq R$ is an interval.

Lecture notes, §1.2 · Definition 1.4.

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 $γ^{'} (t) \neq 0$ , and it is regular if it is regular at every point of its domain.

Lecture notes, §1.3 · Definition 1.8.

Definition 1.9 — Speed

The speed of a parametrized curve $γ$ is the function

v_{γ} (t) = ‖ γ^{'} (t) ‖ .

Lecture notes, §1.3 · Definition 1.9.

1.4 Arc length

Definition 1.13 — Arc length

Let $γ : [a, b] \to R^{n}$ be a smooth parametrized curve. Its arc length is

L (γ) = \int_{a}^{b} ‖ γ^{'} (t) ‖ d 𝑡.

Lecture notes, §1.4 · Definition 1.13.

1.5 Summary

02Inner products, projections, and orthonormal bases

2.1 The Euclidean inner product

Definition 2.1 — Euclidean inner product

For

x = (x_{1}, \dots, x_{n}), y = (y_{1}, \dots, y_{n}) \in R^{n},

the Euclidean inner product of $x$ and $y$ is

⟨ x, y ⟩ = x_{1} y_{1} + \dots + x_{n} y_{n} .

It is also commonly called the dot product.

Lecture notes, §2.1 · Definition 2.1.

Proposition 2.2 — Basic properties

For all $x, y, z \in R^{n}$ and $a, b \in R$ ,

\begin{array}{l} ⟨ 𝑎𝑥 + 𝑏𝑦, z ⟩ & = a ⟨ x, z ⟩ + b ⟨ y, z ⟩, \\ ⟨ x, 𝑎𝑦 + 𝑏𝑧 ⟩ & = a ⟨ x, y ⟩ + b ⟨ x, z ⟩, \\ ⟨ x, y ⟩ & = ⟨ y, x ⟩, \\ ⟨ x, x ⟩ & \geq 0, and ⟨ x, x ⟩ = 0 if and only if x = 0 . \end{array}

Lecture notes, §2.1 · Proposition 2.2.

Proof

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

⟨ x, x ⟩ = x_{1}^{2} + \dots + x_{n}^{2},

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

Theorem 2.4 — Cauchy–Schwarz inequality

For all $x, y \in R^{n}$ ,

| ⟨ x, y ⟩ | \leq ‖ x ‖ ‖ y ‖ .

Equality holds precisely when $x$ and $y$ are linearly dependent.

Lecture notes, §2.1 · Theorem 2.4.

Proof

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 $t \in R$ , consider

0 < ⟨ x + 𝑡𝑦, x + 𝑡𝑦 ⟩ = ⟨ x, x ⟩ + 2 ⟨ x, y ⟩ t + ⟨ y, y ⟩ t^{2},

where the strict inequality holds since $x$ and $y$ are not linearly dependent. A real quadratic polynomial $a z^{2} + 𝑏𝑧 + c$ has no zeros if and only if $b^{2} - 4 𝑎𝑐 < 0$ ; so therefore, we have

4 {⟨ x, y ⟩}^{2} - 4 ⟨ y, y ⟩ ⟨ x, x ⟩ < 0 .

The result then follows from some algebra. □

Definition 2.6 — Angle, orthogonality, and parallelism

If $x, y \in R^{n}$ are nonzero, their angle is the unique $𝜃 \in [0, π]$ satisfying

\cos 𝜃 = \frac{⟨ 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.

Lecture notes, §2.1 · Definition 2.6.

2.2 Orthogonal projection

Lemma 2.8 — Orthogonal projection

Let $V \subseteq R^{n}$ be a subspace. There exists a unique linear map

Π_{V} : R^{n} \to R^{n}

with the following properties:

$Im (Π_{V}) = V$ , i.e., the image of the linear map is $V$ ;
$Π_{V} |_{V} = {Id}_{V}$ , i.e., for all $v \in V$ , we have $Π_{V} (v) = v$ ;
$\ker Π_{V} = V^{⊥}$ , i.e., if $Π_{V} (x) = 0$ , then $x \in V^{⊥}$ .

Lecture notes, §2.2 · Lemma 2.8.

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

Definition 2.11 — Component and projection in a direction

Let $x, y \in R^{n}$ , with $y \neq 0$ . The component of $x$ in the direction of $y$ is the signed scalar

{comp}_{y} (x) = \frac{⟨ x, y ⟩}{‖ y ‖} .

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

{proj}_{y} (x) = \frac{⟨ x, y ⟩}{{‖ y ‖}^{2}} 𝑦.

Lecture notes, §2.2.1 · Definition 2.11.

Proposition 2.12 — Parallel–perpendicular decomposition

Let $x, y \in R^{n}$ , with $y \neq 0$ . 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) .

Lecture notes, §2.2.1 · Proposition 2.12.

Proof

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 = \frac{⟨ x, y ⟩}{{‖ y ‖}^{2}},

which proves both existence and uniqueness. □

2.2.2 Review: Orthonormal sets and bases

Definition 2.14 — Orthonormal set

A finite set ${u_{1}, \dots, u_{k}} \subseteq R^{n}$ is orthonormal if

⟨ u_{i}, u_{j} ⟩ = {\begin{matrix} 1, & i = j, \\ 0, & i \neq 𝑗. \end{matrix}

Thus each $u_{i}$ 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 $V$ .

Lecture notes, §2.2.2 · Definition 2.14.

Definition — Standard orthonormal basis

The standard orthonormal basis of $R^{n}$ is

e_{1} = (1, 0, \dots, 0), e_{2} = (0, 1, 0, \dots, 0), \dots, e_{n} = (0, \dots, 0, 1) .

Lecture notes, §2.2.2 · unnumbered.

Definition — Orthogonal complement

For a subspace $V \subseteq R^{n}$ , its orthogonal complement is

V^{⊥} = {w \in R^{n} : ⟨ w, v ⟩ = 0 for every v \in V} .

Lecture notes, §2.2.2 · unnumbered.

Proposition 2.15 — Projection onto a subspace

Let $V \subseteq R^{n}$ be a subspace with orthonormal basis ${u_{1}, \dots, u_{k}}$ . Every $x \in R^{n}$ has a unique decomposition

x = x_{V} + x_{V^{⊥}},

where $x_{V} \in V$ and $x_{V^{⊥}}$ is orthogonal to every vector in $V$ . The projection of $x$ onto $V$ is

x_{V} = {proj}_{V} (x) = ∑_{i = 1}^{k} ⟨ x, u_{i} ⟩ u_{i},

and

x_{V^{⊥}} = x - {proj}_{V} (x) .

Lecture notes, §2.2.2 · Proposition 2.15.

Proof

Proof. Write a prospective vector in $V$ as

x_{V} = a_{1} u_{1} + \dots + a_{k} u_{k} .

The vector $x - x_{V}$ is orthogonal to all of $V$ exactly when it is orthogonal to each basis vector $u_{j}$ . Orthonormality gives

0 = ⟨ x - x_{V}, u_{j} ⟩ = ⟨ x, u_{j} ⟩ - a_{j} .

Thus the only possible coefficients are $a_{j} = ⟨ x, u_{j} ⟩$ , and these coefficients do give the desired decomposition. □

2.3 Differentiating inner products

Lemma 2.18 — Product rules

Let $x, y : I \to R^{n}$ and $c : I \to R$ be smooth maps. Then

\begin{array}{l} \frac{d}{d t} ⟨ x (t), y (t) ⟩ & = ⟨ x^{'} (t), y (t) ⟩ + ⟨ x (t), y^{'} (t) ⟩, \\ \frac{d}{d t} (c (t) x (t)) & = c^{'} (t) x (t) + c (t) x^{'} (t) . \end{array}

Lecture notes, §2.3 · Lemma 2.18.

Proof

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

Proposition 2.19 — Derivatives of orthogonality relations

Let $x, y : I \to R^{n}$ be smooth.

If $‖ x (t) ‖$ is constant, then $⟨ x (t), x^{'} (t) ⟩ = 0$ for every $t \in I$ .
If $⟨ x (t), y (t) ⟩ = 0$ for every $t \in I$ , then
$⟨ x^{'} (t), y (t) ⟩ = - ⟨ x (t), y^{'} (t) ⟩ .$

Lecture notes, §2.3 · Proposition 2.19.

Proof

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 $⟨ x (t), y (t) ⟩ = 0$ . □

Corollary 2.21 — Constant-speed curves

If a curve $γ : I \to R^{n}$ has constant speed, then

⟨ γ^{'} (t), γ^{″} (t) ⟩ = 0

for every $t \in I$ . Thus its velocity and acceleration are orthogonal.

Lecture notes, §2.3 · Corollary 2.21.

Proof

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 $γ : [a, b] \to R^{n}$ be a smooth curve, with $p = γ (a)$ and $q = γ (b)$ . Then

L (γ) \geq ‖ q - p ‖ .

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

Lecture notes, §2.4 · Theorem 2.22.

Proof

Proof. The result is immediate if $p = q$ , so assume $p \neq q$ and set

u = \frac{q - p}{‖ q - p ‖} .

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

⟨ γ^{'} (t), u ⟩ \leq ‖ γ^{'} (t) ‖ ‖ u ‖ = ‖ γ^{'} (t) ‖ .

Therefore

\begin{array}{l} L (γ) & = \int_{a}^{b} ‖ γ^{'} (t) ‖ d t \\ \geq \int_{a}^{b} ⟨ γ^{'} (t), u ⟩ d t \\ = ⟨ γ (b) - γ (a), u ⟩ \\ = ⟨ q - p, \frac{q - p}{‖ q - p ‖} ⟩ \\ = ‖ q - p ‖ . \end{array}

□

2.5 Summary

03Acceleration and reparametrization

3.1 Acceleration

Definition 3.1 — Acceleration

The acceleration of a smooth parametrized curve $γ$ at time $t$ is

a (t) = γ^{″} (t) .

Lecture notes, §3.1 · Definition 3.1.

3.1.1 Tangential and perpendicular acceleration

Definition 3.3 — Tangential and perpendicular components

For a regular curve, define

\begin{array}{l} a^{∥} (t) & = {proj}_{v (t)} (a (t)) = \frac{⟨ a (t), v (t) ⟩}{{‖ v (t) ‖}^{2}} v (t), \\ a^{⊥} (t) & = a (t) - a^{∥} (t) . \end{array}

Then

a (t) = a^{∥} (t) + a^{⊥} (t),

where $a^{∥} (t)$ is parallel to $v (t)$ and $a^{⊥} (t)$ is orthogonal to $v (t)$ .

Lecture notes, §3.1.1 · Definition 3.3.

Proposition 3.4 — Derivative of speed

Let $γ : I \to R^{n}$ be regular. Then

\frac{d}{d t} ‖ v (t) ‖ = \frac{⟨ 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.

Lecture notes, §3.1.1 · Proposition 3.4.

Proof

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

\frac{d}{d t} {‖ v ‖}^{2} = 2 ‖ v ‖ \frac{d}{d t} ‖ v ‖ .

On the right we have

\frac{d}{d t} ⟨ v, v ⟩ = ⟨ v^{'}, v ⟩ + ⟨ v, v^{'} ⟩ = 2 ⟨ a, v ⟩ .

The result follows from some algebra. □

Corollary 3.5 — Acceleration and change of speed

For a regular curve:

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

Lecture notes, §3.1.1 · Corollary 3.5.

3.2 Trace and reparametrization

Definition 3.7 — Trace

The trace of a parametrized curve $γ : I \to R^{n}$ is the subset

γ (I) = {γ (t) : t \in I} \subseteq R^{n} .

Lecture notes, §3.2 · Definition 3.7.

Definition 3.9 — Reparametrization

Let $γ : I \to R^{n}$ be a regular curve. A reparametrization of $γ$ is a curve of the form

\tilde{γ} = γ \circ ϕ : \tilde{I} \to R^{n},

where $ϕ : \tilde{I} \to I$ is a smooth bijection with $ϕ^{'} (t) \neq 0$ for every $t \in \tilde{I}$ .

Lecture notes, §3.2 · Definition 3.9.

3.3 More on reparameterization (not covered in class)

Proposition 3.10 — Velocity and acceleration under reparametrization

If $\tilde{γ} = γ \circ ϕ$ , then

\begin{array}{l} {\tilde{γ}}^{'} (t) & = ϕ^{'} (t) γ^{'} (ϕ (t)), \\ {\tilde{γ}}^{″} (t) & = {(ϕ^{'} (t))}^{2} γ^{″} (ϕ (t)) + ϕ^{″} (t) γ^{'} (ϕ (t)) . \end{array}

Consequently,

‖ {\tilde{γ}}^{'} (t) ‖ = | ϕ^{'} (t) | ‖ γ^{'} (ϕ (t)) ‖ .

Lecture notes, §3.3 · Proposition 3.10.

Proof

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. □

Definition 3.12 — Orientation

A reparametrization $\tilde{γ} = γ \circ ϕ$ is orientation-preserving if $ϕ^{'} (t) > 0$ everywhere and orientation-reversing if $ϕ^{'} (t) < 0$ everywhere.

Lecture notes, §3.3 · Definition 3.12.

Proposition 3.13 — Arc length is independent of parametrization

Let

\tilde{γ} = γ \circ ϕ : [c, d] \to R^{n}

be a reparametrization. Set

u_{0} = \min {ϕ (c), ϕ (d)}, u_{1} = \max {ϕ (c), ϕ (d)} .

Then

\int_{c}^{d} ‖ {\tilde{γ}}^{'} (t) ‖ d t = \int_{u_{0}}^{u_{1}} ‖ γ^{'} (u) ‖ d 𝑢.

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

Lecture notes, §3.3 · Proposition 3.13.

Proof

Proof. If $ϕ^{'} > 0$ , then the speed transformation formula and the substitution $u = ϕ (t)$ give

\int_{c}^{d} ‖ {\tilde{γ}}^{'} (t) ‖ d t = \int_{c}^{d} ϕ^{'} (t) ‖ γ^{'} (ϕ (t)) ‖ d t = \int_{ϕ (c)}^{ϕ (d)} ‖ γ^{'} (u) ‖ d 𝑢.

If $ϕ^{'} < 0$ , then $| ϕ^{'} | = - ϕ^{'}$ , and the same substitution gives

\int_{c}^{d} ‖ {\tilde{γ}}^{'} (t) ‖ d t = \int_{ϕ (d)}^{ϕ (c)} ‖ γ^{'} (u) ‖ d 𝑢.

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

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

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 $1$ .

Lecture notes, §3.4 · Theorem 3.14.

Proof

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

s (t) = \int_{a}^{t} ‖ γ^{'} (u) ‖ d 𝑢.

By the fundamental theorem of calculus,

s^{'} (t) = ‖ γ^{'} (t) ‖ > 0 .

(3.1)

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

\bar{γ} = γ \circ ϕ : J \to R^{n} .

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

s^{'} (ϕ (v)) ϕ^{'} (v) = 1 .

Therefore

\begin{array}{l} ‖ {\bar{γ}}^{'} (v) ‖ & = ϕ^{'} (v) ‖ γ^{'} (ϕ (v)) ‖ & 3.10 \\ = ϕ^{'} (v) s^{'} (ϕ (v)) & (3.1) \\ = 1 . \end{array}

From this we can check that $\bar{γ}$ gives a paramerization by arclength. In other words, for all $v \in J$ , we have $\int_{0}^{v} ‖ {\bar{γ}}^{'} (u) ‖ d u = \int_{0}^{v} 1 d u = v$ . □

3.5 Closed curves (not covered in class)

Definition 3.18 — Closed and simple closed curves

A closed curve is a regular curve $γ : [a, b] \to R^{n}$ such that

γ (a) = γ (b)

and every derivative agrees at the endpoints:

γ^{(k)} (a) = γ^{(k)} (b) for every k \geq 1 .

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

Lecture notes, §3.5 · Definition 3.18.

Proposition 3.19 — Periodic extension

A regular curve $γ : [a, b] \to R^{n}$ is closed if and only if there is a periodic regular curve $\hat{γ} : R \to R^{n}$ of period $b - a$ whose restriction to $[a, b]$ is $γ$ .

Lecture notes, §3.5 · Proposition 3.19.

Proof

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. □

Proposition 3.21 — Simple closed curves with the same trace

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

Lecture notes, §3.5 · Proposition 3.21.

3.6 Summary

04Curvature, unit tangent and normal vectors, and osculating circles

4.1 The curvature function

Definition 4.2 — Curvature function

Let $𝜸 : I \to R^{n}$ be a regular curve. Its curvature vector $κ : I \to R^{n}$ is

\vec{κ} (t) = \frac{𝐚^{⊥} (t)}{{| 𝐯 (t) |}^{2}} .

Its curvature function $κ : I \to [0, \infty)$ is

κ (t) = \frac{| 𝐚^{⊥} (t) |}{{| 𝐯 (t) |}^{2}} .

Lecture notes, §4.1 · Definition 4.2.

Proposition 4.3 — Unit-speed curves

If $𝜸$ is parametrized by arc length, then

κ (t) = | 𝐚 (t) | .

Lecture notes, §4.1 · Proposition 4.3.

Proof

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

4.2 Unit tangent and unit normal vectors

Definition 4.5 — Unit tangent and unit normal vectors

Let $𝜸 : I \to R^{n}$ be regular. Define

𝐭 (t) = \frac{𝐯 (t)}{| 𝐯 (t) |}, 𝐧 (t) = \frac{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 $κ (t) \neq 0$ .

Lecture notes, §4.2 · Definition 4.5.

Proposition 4.6 — Curvature and change of direction

For every regular curve,

𝐚 = \underset{𝐚^{∥}}{\underset{⏟}{{| 𝐯 |}^{'} 𝐭}} + \underset{𝐚^{⊥}}{\underset{⏟}{| 𝐯 | 𝐭^{'}}} = | 𝐯 |^{'} 𝐭 + κ | 𝐯 |^{2} 𝐧 .

and

κ (t) = \frac{| 𝐭^{'} (t) |}{| 𝐯 (t) |} .

Lecture notes, §4.2 · Proposition 4.6.

Proof

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

𝐚 = \underset{𝐚^{∥}}{\underset{⏟}{{| 𝐯 |}^{'} 𝐭}} + \underset{𝐚^{⊥}}{\underset{⏟}{| 𝐯 | 𝐭^{'}}}

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 $κ (t) = \frac{| 𝐭^{'} (t) |}{| 𝐯 (t) |}$ . Then, putting this formula for curvature into the right hand side of $𝐚 = \underset{𝐚^{∥}}{\underset{⏟}{{| 𝐯 |}^{'} 𝐭}} + \underset{𝐚^{⊥}}{\underset{⏟}{| 𝐯 | 𝐭^{'}}} = | 𝐯 |^{'} 𝐭 + κ | 𝐯 |^{2} 𝐧$ shows that the right and middle terms are equal. □

Unit normal and derivative of the unit tangent

At every time for which $κ (t) \neq 0$ , the same calculation gives

𝐧 (t) : = \frac{𝐚^{⊥} (t)}{| 𝐚^{⊥} (t) |} = \frac{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) .

Lecture notes, §4.2 · unnumbered.

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

4.3.1 Computational formula

Proposition 4.8 — Computational formula

For a regular curve,

κ (t) = \frac{\sqrt{{‖ γ^{'} (t) ‖}^{2} {‖ γ^{″} (t) ‖}^{2} - {⟨ γ^{'} (t), γ^{″} (t) ⟩}^{2}}}{{‖ γ^{'} (t) ‖}^{3}} .

Lecture notes, §4.3.1 · Proposition 4.8.

Proof

Proof. Since $a^{∥} = {proj}_{v} (a)$ , orthogonality gives

\begin{array}{l} {‖ a^{⊥} ‖}^{2} & = {‖ a ‖}^{2} - {‖ a^{∥} ‖}^{2} \\ = {‖ a ‖}^{2} - \frac{{⟨ a, v ⟩}^{2}}{{‖ v ‖}^{2}} . \end{array}

Substitute this into $κ = ‖ a^{⊥} ‖ / {‖ v ‖}^{2}$ and simplify. □

4.4 A worked example (not covered in class)

4.5 Further topics

Derivative of the normal and the remaining component

𝐧^{'} = - κ | 𝐯 | 𝐭 + 𝐱

for some $𝐱 \in {(Span 𝐭)}^{⊥}$ . In fact, we know that $⟨ 𝐧^{'}, 𝐧 ⟩ = 0$ , so that $𝐱 \in {(Span (𝐭, 𝐧))}^{⊥}$ .

Lecture notes, §4.5 · unnumbered.

Definition — Auxiliary binormal and scalar function

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 $τ (t)$ .

Lecture notes, §4.5 · unnumbered.

Derivative of the auxiliary binormal

𝐛^{'} = - τ | 𝐯 | 𝐧 + 𝐲

where $𝐲 \in {(Span (𝐭, 𝐧, 𝐛))}^{⊥}$

Lecture notes, §4.5 · unnumbered.

Calculation

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 $𝐲 \in {(Span (𝐭, 𝐧, 𝐛))}^{⊥}$

4.6 Summary

05Plane curves: signed curvature, angle functions, and rotation index

5.1 Rotation matrices

Definition — Rotation matrices in the plane

For $𝜃 \in R$ , define the counterclockwise rotation through angle $𝜃$ by

R_{𝜃} = (\begin{matrix} \cos 𝜃 & - \sin 𝜃 \\ \sin 𝜃 & \cos 𝜃 \end{matrix}) .

Thus $R_{𝜃} : R^{2} \to R^{2}$ is a linear map. The rotation through $9 0^{\circ}$ is

R_{90} = R_{π / 2} = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}), R_{90} (x, y) = (- y, x) .

Lecture notes, §5.1 · unnumbered.

Properties of the counterclockwise quarter-turn

For every $𝐰 \in R^{2}$ ,

⟨ 𝐰, R_{90} (𝐰) ⟩ = 0, | R_{90} (𝐰) | = | 𝐰 |, R_{90}^{2} (𝐰) = - 𝐰 .

Consequently, if $𝐰 \neq 0$ , then $R_{90} (𝐰)$ spans the line perpendicular to $𝐰$ .

Lecture notes, §5.1 · unnumbered.

5.2 Signed curvature

Definition 5.1 — Signed curvature

Let $𝜸 : I \to R^{2}$ be a regular plane curve. Its signed curvature function is

κ_{s} (t) = ⟨ \frac{𝐚 (t)}{{| 𝐯 (t) |}^{2}}, R_{90} (\frac{𝐯 (t)}{| 𝐯 (t) |}) ⟩ = \frac{⟨ 𝐚 (t), R_{90} (𝐯 (t)) ⟩}{{| 𝐯 (t) |}^{3}} .

Lecture notes, §5.2 · Definition 5.1.

Signed curvature and the curvature vector

The curvature vector and the signed curvature are related by

\frac{𝐚^{⊥}}{{| 𝐯 |}^{2}} = κ_{s} R_{90} (\frac{𝐯}{| 𝐯 |}), κ = | κ_{s} | .

Lecture notes, §5.2 · unnumbered.

5.2.1 The unit-speed case

Signed curvature for unit-speed curves

If $𝜸$ has unit speed, then $| 𝐯 | = 1$ .

𝐚 = κ_{s} R_{90} (𝐯), κ_{s} = ⟨ 𝐚, R_{90} (𝐯) ⟩, κ = | κ_{s} | .

Lecture notes, §5.2.1 · unnumbered.

Calculation

Since differentiating $⟨ 𝐯, 𝐯 ⟩ = 1$ gives $⟨ 𝐚, 𝐯 ⟩ = 0$ , all of the acceleration is perpendicular to the velocity. Hence

𝐚 = κ_{s} R_{90} (𝐯), κ_{s} = ⟨ 𝐚, R_{90} (𝐯) ⟩, κ = | κ_{s} | .

5.3 Angle functions

Definition 5.3 — Angle function

Let $𝜸 : I \to R^{2}$ be a unit-speed plane curve with velocity $𝐯$ . An angle function for $𝜸$ is a smooth function $𝜃 : I \to R$ such that

𝐯 (t) = (\cos 𝜃 (t), \sin 𝜃 (t))

for every $t \in I$ .

Lecture notes, §5.3 · Definition 5.3.

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 $2 π$ . Moreover,

𝜃^{'} (t) = κ_{s} (t) .

Lecture notes, §5.3 · Proposition 5.4.

Proof

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 $𝐚 = κ_{s} R_{90} (𝐯)$ shows that $𝜃^{'} = κ_{s}$ .

Choose $t_{0} \in I$ and $𝜃_{0} \in R$ with $𝐯 (t_{0}) = (\cos 𝜃_{0}, \sin 𝜃_{0})$ , and define

𝜃 (t) = 𝜃_{0} + \int_{t_{0}}^{t} κ_{s} (u) d 𝑢.

Let

A = {t \in I : 𝐯 (t) = (\cos 𝜃 (t), \sin 𝜃 (t))} .

The set $A$ contains $t_{0}$ and is closed by continuity. If $t \in A$ , choose a local inverse-trigonometric branch giving an angle function near $t$ . After adding a multiple of $2 π$ , the local angle agrees with $𝜃$ at $t$ ; since both have derivative $κ_{s}$ , 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 $2 π$ . □

5.4 Rotation index

Definition 5.6 — Rotation index

Let $𝜸 : [a, b] \to R^{2}$ be a unit-speed closed plane curve, and let $𝜃$ be an angle function. The rotation index of $𝜸$ is

rot (𝜸) = \frac{𝜃 (b) - 𝜃 (a)}{2 π} .

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

Lecture notes, §5.4 · Definition 5.6.

Corollary 5.7 — Total signed curvature

For a unit-speed closed plane curve,

rot (𝜸) = \frac{1}{2 π} \int_{a}^{b} κ_{s} (t) d 𝑡.

For a regular closed plane curve with arbitrary parameter,

rot (𝜸) = \frac{1}{2 π} \int_{a}^{b} κ_{s} (t) | 𝐯 (t) | d 𝑡.

Lecture notes, §5.4 · Corollary 5.7.

Proof

Proof. Integrate $𝜃^{'} = κ_{s}$ in the unit-speed case. For a general parameter, integrate $𝜃^{'} = κ_{s} | 𝐯 |$ . □

5.5 Summary

06Space curves: Frenet frame, torsion, and Frenet equations

6.1 The cross product

Definition 6.1 — Cross product

Let

𝐚 = (a_{1}, a_{2}, a_{3}), 𝐛 = (b_{1}, b_{2}, b_{3}) \in R^{3} .

Their cross product is

𝐚 \times 𝐛 = | \begin{matrix} 𝐢 & 𝐣 & 𝐤 \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix} | = (a_{2} b_{3} - a_{3} b_{2}, a_{3} b_{1} - a_{1} b_{3}, a_{1} b_{2} - a_{2} b_{1}) .

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

Lecture notes, §6.1 · Definition 6.1.

Scalar triple product

For $𝐚, 𝐛, 𝐜 \in R^{3}$ , you can check that

⟨ 𝐜, 𝐚 \times 𝐛 ⟩ = \det (\begin{matrix} c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}) .

(6.1)

Lecture notes, §6.1 · unnumbered.

Lemma 6.2 — Basic properties of the cross product

For $𝐚, 𝐛, 𝐜 \in R^{3}$ and $λ, μ \in R$ , the following hold.

$𝐚 \times 𝐛$ is orthogonal to both $𝐚$ and $𝐛$ . Thus
$𝐚 \times 𝐛 \in span {𝐚, 𝐛}^{⊥} .$
$𝐚 \times 𝐛 = - (𝐛 \times 𝐚)$ .
The cross product is linear in each input; for example,
$(λ 𝐚 + μ 𝐛) \times 𝐜 = λ (𝐚 \times 𝐜) + μ (𝐛 \times 𝐜) .$
${| 𝐚 \times 𝐛 |}^{2} = {| 𝐚 |}^{2} {| 𝐛 |}^{2} - {⟨ 𝐚, 𝐛 ⟩}^{2} .$

If $𝐚$ and $𝐛$ are nonzero and $𝜃 = ∠ (𝐚, 𝐛)$ , then
$| 𝐚 \times 𝐛 | = | 𝐚 | | 𝐛 | \sin 𝜃.$
The direction of $𝐚 \times 𝐛$ is determined by the right-hand rule.

Lecture notes, §6.1 · Lemma 6.2.

Proof

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^{2} 𝜃),

which gives the length formula. □

6.1.1 Rotations and the cross product

Definition — Rotation matrices in three dimensions

A matrix $R \in M_{3 \times 3} (R)$ is a rotation matrix if

R^{T} R = I and \det R = 1 .

Lecture notes, §6.1.1 · unnumbered.

Proposition 6.3 — Rotations preserve the cross product

If $R$ is a rotation matrix, then

R (𝐚 \times 𝐛) = (R 𝐚) \times (R 𝐛)

for all $𝐚, 𝐛 \in R^{3}$ .

Lecture notes, §6.1.1 · Proposition 6.3.

Proof

Proof. For arbitrary $𝐜 \in R^{3}$ , use preservation of the inner product and the scalar triple product:

\begin{array}{l} ⟨ 𝐜, (R 𝐚) \times (R 𝐛) ⟩ & = \det (𝐜, R 𝐚, R 𝐛) \\ = \det (R (R^{T} 𝐜), R 𝐚, R 𝐛) \\ = \det (R) \det (R^{T} 𝐜, 𝐚, 𝐛) \\ = ⟨ R^{T} 𝐜, 𝐚 \times 𝐛 ⟩ \\ = ⟨ 𝐜, R (𝐚 \times 𝐛) ⟩ . \end{array}

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

6.2 The Frenet frame

Definition 6.5 — Frenet frame

Let $𝜸 : I \to R^{3}$ be a regular space curve and suppose $κ (t) \neq 0$ . The unit binormal vector is

𝐛 (t) = 𝐭 (t) \times 𝐧 (t) .

The ordered orthonormal basis

{𝐭 (t), 𝐧 (t), 𝐛 (t)}

is called the Frenet frame at time $t$ .

Lecture notes, §6.2 · Definition 6.5.

6.3 The Frenet equations

Proposition 6.6 — The Frenet equations

Let $𝜸 : I \to R^{3}$ be a regular space curve. At every time for which $κ \neq 0$ ,

\begin{aligned} 𝐭^{'} & = | 𝐯 | κ 𝐧, \\ 𝐧^{'} & = - | 𝐯 | κ 𝐭 + | 𝐯 | τ 𝐛, \\ 𝐛^{'} & = - | 𝐯 | τ 𝐧 . \end{aligned}

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

{(\begin{matrix} 𝐭 \\ 𝐧 \\ 𝐛 \end{matrix})}^{'} = | 𝐯 | (\begin{matrix} 0 & κ & 0 \\ - κ & 0 & τ \\ 0 & - τ & 0 \end{matrix}) (\begin{matrix} 𝐭 \\ 𝐧 \\ 𝐛 \end{matrix}) .

Lecture notes, §6.3 · Proposition 6.6.

Proof

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

𝐚 = \underset{𝐚^{∥}}{\underset{⏟}{{| 𝐯 |}^{'} 𝐭}} + \underset{𝐚^{⊥}}{\underset{⏟}{| 𝐯 | 𝐭^{'}}}

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

\vec{κ} = \frac{𝐚^{⊥}}{| 𝐯 |^{2}} = \frac{| 𝐯 | 𝐭^{'}}{| 𝐯 |^{2}} = \frac{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 ${𝐭, 𝐧, 𝐛}$ :

\begin{array}{l} ⟨ 𝐧^{'}, 𝐭 ⟩ & = - ⟨ 𝐧, 𝐭^{'} ⟩ = - | 𝐯 | κ, \\ ⟨ 𝐧^{'}, 𝐧 ⟩ & = 0, \\ ⟨ 𝐧^{'}, 𝐛 ⟩ & = : | 𝐯 | 𝜏. \end{array}

In other words, we define $τ : = \frac{⟨ 𝐧^{'}, 𝐛 ⟩}{| 𝐯 |}$ .

This gives

𝐧^{'} = - | 𝐯 | κ 𝐭 + | 𝐯 | τ 𝐛 .

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. □

Definition 6.7 — Torsion (Tapp)

Let $𝜸 : I \to R^{3}$ be a regular space curve, and let $t \in I$ satisfy $κ (t) \neq 0$ . The torsion of $𝜸$ at $t$ is

τ (t) = - \frac{⟨ 𝐛^{'} (t), 𝐧 (t) ⟩}{| 𝐯 (t) |} .

Lecture notes, §6.3 · Definition 6.7.

6.4 Some further topics

6.4.1 Zero torsion

Proposition 6.10 — Zero torsion and planar curves

Let $γ : I \to R^{3}$ be a regular curve with $κ (t) \neq 0$ for all $t \in I$ . The trace of $γ$ is constrained to a plane if and only if $τ (t) = 0$ for all $t \in I$ .

Lecture notes, §6.4.1 · Proposition 6.10.

Proof

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 $t \in I$ .

Conversely, if $τ (t) = 0$ for all $t \in I$ , 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 $𝜶, 𝜷 : I \to R^{3}$ are smooth, then

\frac{d}{d t} (𝜶 (t) \times 𝜷 (t)) = 𝜶^{'} (t) \times 𝜷 (t) + 𝜶 (t) \times 𝜷^{'} (t) .

Lecture notes, §6.4.2 · Lemma 6.11.

Proof

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

6.4.3 A curvature formula in

R^{3}

Proposition 6.12 — Curvature from the cross product

Let $𝜸 : I \to R^{3}$ be a regular space curve, with

𝐯 = 𝜸^{'}, 𝐚 = 𝜸^{″} .

Then

κ (t) = \frac{| 𝐯 (t) \times 𝐚 (t) |}{{| 𝐯 (t) |}^{3}} .

Lecture notes, §6.4.3 · Proposition 6.12.

Proof

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

κ = \frac{| 𝐚^{⊥} |}{{| 𝐯 |}^{2}} = \frac{| 𝐯 | | 𝐚 | \sin 𝜃}{{| 𝐯 |}^{3}} = \frac{| 𝐯 \times 𝐚 |}{{| 𝐯 |}^{3}} .

□

6.5 Summary

07Rigid motions, fundamental theorems, and curvature formulas

7.1 Rigid motions of Euclidean space

Definition 7.1 — Rigid motion

A rigid motion of $R^{n}$ is a function

f : R^{n} \to R^{n}

that preserves distances. Equivalently,

| f (𝐩) - f (𝐪) | = | 𝐩 - 𝐪 |

for all $𝐩, 𝐪 \in R^{n}$ .

Lecture notes, §7.1 · Definition 7.1.

Polarization identity

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

⟨ 𝐩, 𝐪 ⟩ = \frac{1}{2} ({| 𝐩 |}^{2} + {| 𝐪 |}^{2} - {| 𝐩 - 𝐪 |}^{2}) .

(7.1)

Lecture notes, §7.1 · unnumbered.

Definition — Linear map associated with a matrix

For a matrix $A \in M_{n \times n} (R)$ , write

L_{A} (𝐱) = A 𝐱 .

Lecture notes, §7.1 · unnumbered.

Definition — Orthogonal matrices and the orthogonal group

Recall that $A$ is orthogonal if

A^{T} A = A A^{T} = 𝐼.

The set of orthogonal $n \times n$ matrices is denoted by $O (n)$ ; this is called the orthogonal group. The following conditions are equivalent (if they hold for all $𝐩, 𝐪 \in R^{n}$ ):

A \in O (n), ⟨ A 𝐩, A 𝐪 ⟩ = ⟨ 𝐩, 𝐪 ⟩, | A 𝐩 | = | 𝐩 | .

Thus the linear rigid motions are exactly the maps $L_{A}$ with $A \in O (n)$ .

Lecture notes, §7.1 · unnumbered.

Definition — Translation

For $𝐪 \in R^{n}$ , let

T_{𝐪} (𝐱) = 𝐱 + 𝐪

denote translation by $𝐪$ .

Lecture notes, §7.1 · unnumbered.

Theorem 7.2 — Classification of rigid motions

Every rigid motion $f : R^{n} \to R^{n}$ has a unique expression

f = T_{𝐪} \circ L_{A},

where $𝐪 \in R^{n}$ and $A \in O (n)$ . Equivalently,

f (𝐱) = A 𝐱 + 𝐪 .

Lecture notes, §7.1 · Theorem 7.2.

Proof

Proof. Set $𝐪 = f (0)$ and define

g = T_{- 𝐪} \circ 𝑓.

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

⟨ g (𝐩), g (𝐫) ⟩ = ⟨ 𝐩, 𝐫 ⟩

for all $𝐩, 𝐫 \in R^{n}$ .

Let $𝐞_{1}, \dots, 𝐞_{n}$ be the standard basis, and let $A$ be the matrix whose $i$ th column is $g (𝐞_{i})$ . These columns form an orthonormal basis, so $A \in O (n)$ . Define

h = L_{A}^{- 1} \circ 𝑔.

Then $h$ preserves inner products and fixes every $𝐞_{i}$ . If $𝐩 = \sum_{i} p_{i} 𝐞_{i}$ and $h (𝐩) = \sum_{i} b_{i} 𝐞_{i}$ , then

b_{i} = ⟨ h (𝐩), 𝐞_{i} ⟩ = ⟨ h (𝐩), h (𝐞_{i}) ⟩ = ⟨ 𝐩, 𝐞_{i} ⟩ = p_{i} .

Thus $h$ is the identity, so $g = L_{A}$ and $f = T_{𝐪} \circ L_{A}$ .

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

Determinant of an orthogonal matrix

Since $A^{T} A = I$ ,

1 = \det (A^{T} A) = \det {(A)}^{2},

so $\det A = 1$ or $\det A = - 1$ .

Lecture notes, §7.1 · unnumbered.

Definition — General linear group and orientation

Recall that ${GL}_{n} (R)$ is the set of $n \times n$ matrices with non-zero determinant, and is called the general linear group. A matrix $A \in {GL}_{n} (R)$ 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 $1$ .

Lecture notes, §7.1 · unnumbered.

Definition 7.3 — Proper and improper rigid motions

Suppose $f = T_{𝐪} \circ L_{A}$ 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.

Lecture notes, §7.1 · Definition 7.3.

7.2 Geometric quantities under rigid motion

Velocity and acceleration under a rigid motion

Let $𝜸 : I \to R^{n}$ be regular, let $f = T_{𝐪} \circ L_{A}$ be a rigid motion, and define

\hat{𝜸} = f \circ 𝜸 = 𝐴𝛾 + 𝐪 .

Writing $𝐯 = 𝜸^{'}$ and $𝐚 = 𝜸^{″} = 𝐯^{'}$ , and similarly for $\hat{γ}$ , differentiation gives

\begin{align} \hat{𝐯} & : = {\hat{γ}}^{'} = {(𝐴𝛾 + 𝐪)}^{'} = A^{'} γ + A γ^{'} + 𝐪^{'} = A γ^{'} = A 𝐯, & (7.2) \\ \hat{𝐚} & : = {(\hat{𝐯})}^{'} = {(A 𝐯)}^{'} = A^{'} 𝐯 + A 𝐯^{'} = A 𝐚, & (7.3) \end{align}

since $A$ and $𝐪$ are constant.

Lecture notes, §7.2 · unnumbered.

Proposition 7.4 — Invariance under rigid motions

For a regular curve, the following statements hold.

Curvature is invariant under every rigid motion:
$\hat{κ} = 𝜅.$
For a space curve, torsion is preserved by proper rigid motions and changes sign under improper rigid motions:
$\hat{τ} = \det (A) 𝜏.$
For a plane curve, signed curvature is preserved by proper rigid motions and changes sign under improper rigid motions:
${\hat{κ}}_{s} = \det (A) κ_{s} .$

Lecture notes, §7.2 · Proposition 7.4.

Proof

Proof. Equation (7.3) implies

{\hat{𝐚}}^{⊥} = A 𝐚^{⊥}, | \hat{𝐯} | = | 𝐯 | .

Therefore, since $A$ preserves lengths,

\hat{κ} = \frac{| {\hat{𝐚}}^{⊥} |}{{| \hat{𝐯} |}^{2}} = \frac{| A 𝐚^{⊥} |}{{| A 𝐯 |}^{2}} = \frac{| 𝐚^{⊥} |}{{| 𝐯 |}^{2}} = 𝜅.

For a space curve at a point where $κ \neq 0$ ,

\hat{𝐭} = A 𝐭, \hat{𝐧} = A 𝐧 .

The transformation rule for the cross product is

(A 𝐱) \times (A 𝐲) = \det (A) A (𝐱 \times 𝐲),

so

\hat{𝐛} = \det (A) A 𝐛 .

Using Tapp’s torsion convention,

\hat{τ} = - \frac{⟨ {\hat{𝐛}}^{'}, \hat{𝐧} ⟩}{| \hat{𝐯} |} = - \det (A) \frac{⟨ A 𝐛^{'}, A 𝐧 ⟩}{| 𝐯 |} = \det (A) 𝜏.

In the plane,

R_{90} (A 𝐰) = \det (A) A R_{90} (𝐰) .

Substituting this identity into the formula for signed curvature gives

{\hat{κ}}_{s} = \frac{⟨ A 𝐚, R_{90} (A 𝐯) ⟩}{{| A 𝐯 |}^{3}} = \det (A) κ_{s} .

□

7.3 The fundamental theorems of curves

Theorem 7.5 — The fundamental theorems of plane and space curves

Let $I \subseteq R$ be an interval.

If $κ_{s} : I \to R$ is smooth, then there exists a unit-speed plane curve $𝜸 : I \to R^{2}$ whose signed curvature is $κ_{s}$ . If $𝜸$ and $\hat{𝜸}$ are two such curves, then
$\hat{𝜸} = f \circ 𝜸$

for some proper rigid motion $f$ of $R^{2}$ .
If $κ, τ : I \to R$ are smooth and $κ > 0$ , then there exists a unit-speed space curve $𝜸 : I \to R^{3}$ whose curvature and torsion are $κ$ and $τ$ . If $𝜸$ and $\hat{𝜸}$ are two such curves, then
$\hat{𝜸} = f \circ 𝜸$

for some proper rigid motion $f$ of $R^{3}$ .

Lecture notes, §7.3 · Theorem 7.5.

Proof — plane-curve statement

Proof of the plane-curve statement. Fix $t_{0} \in I$ . We first construct the curve with initial conditions

𝜸 (t_{0}) = (0, 0), 𝜸^{'} (t_{0}) = (1, 0) .

Define

𝜃 (t) : = \int_{t_{0}}^{t} κ_{s} (u) d u,

(7.4)

then define the velocity and the curve by

𝐯 (t) : = (\cos 𝜃 (t), \sin 𝜃 (t)), 𝜸 (t) : = \int_{t_{0}}^{t} 𝐯 (u) d 𝑢.

(7.5)

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

κ_{s} \to_{}^{integrate} 𝜃 \to_{}^{(\cos 𝜃, \sin 𝜃)} 𝐯 \to_{}^{integrate} 𝜸 .

By the Fundamental Theorem of Calculus,

𝜸^{'} = 𝐯, | 𝐯 | = 1,

so $𝜸$ has unit speed. Furthermore,

𝐚 = 𝐯^{'} = 𝜃^{'} (- \sin 𝜃, \cos 𝜃) = κ_{s} R_{90} (𝐯) .

For a unit-speed plane curve, this equation says exactly that the signed curvature is $κ_{s}$ . 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 $\hat{𝜸}$ be unit-speed plane curves with the same signed curvature $κ_{s}$ , and choose angle functions $𝜃$ and $\hat{𝜃}$ . After applying a rigid motion to each curve, we can assume that $γ (t_{0}) = \hat{γ} (t_{0}) = 0$ , and $γ^{'} (t_{0}) = {\hat{γ}}^{'} (t_{0}) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ . Since

𝜃^{'} = κ_{s} = {\hat{𝜃}}^{'},

there is a constant $𝜃_{0}$ such that

\hat{𝜃} = 𝜃 + 𝜃_{0} .

But then, plugging into $𝐯 (t) : = (\cos 𝜃 (t), \sin 𝜃 (t))$ and $\hat{𝐯} (t) : = (\cos \hat{𝜃} (t), \sin \hat{𝜃} (t))$ , and setting $t = t_{0}$ , we see that $𝜃_{0}$ is a multiple of $2 π$ , so we can take $𝜃_{0} = 0$ . Therefore $𝐯 = \hat{𝐯}$ .

Finally, since

γ^{'} (t) = : 𝐯 (t) = \hat{v} (t) : = {\hat{γ}}^{'} (t)

we see that $\hat{γ} (t) = γ (t) + γ_{0}$ . Setting $t = t_{0}$ , and using that $γ (t) = \hat{γ} (t) = 0$ , we see that $\hat{γ} = γ$ . □

Proof sketch — space-curve statement

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^{'} = κ 𝐧, 𝐧^{'} = - κ 𝐭 + τ 𝐛, 𝐛^{'} = - τ 𝐧 .

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

\underset{𝐅^{'} (t)}{\underset{⏟}{{(\begin{matrix} 𝐭 \\ 𝐧 \\ 𝐛 \end{matrix})}^{'}}} = \underset{Ω}{\underset{⏟}{(\begin{matrix} 0 & κ & 0 \\ - κ & 0 & τ \\ 0 & - τ & 0 \end{matrix})}} \underset{𝐅 (t)}{\underset{⏟}{(\begin{matrix} 𝐭 \\ 𝐧 \\ 𝐛 \end{matrix})}},

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

Given $κ > 0$ , $τ$ , and an initial positively oriented orthonormal frame $𝐅_{0} = (𝔱_{0}, 𝔫_{0}, 𝔟_{0})$ , this linear first-order system has a unique solution $𝐅 (t) = (𝔱 (t), 𝔫 (t), 𝔟 (t))$ with $𝐅 (t_{0}) = (𝔱 (t_{0}), 𝔫 (t_{0}), 𝔟 (t_{0})) = 𝐅_{0} = (𝔱_{0}, 𝔫_{0}, 𝔟_{0})$ .

Note that since $Ω = - Ω^{T}$ , we have that

{(𝐅 {(t)}^{T} 𝐅 (t))}^{'} = {(𝐅^{T})}^{'} 𝐅 + 𝐅^{T} 𝐅^{'} = {(𝐅^{'})}^{T} 𝐅 + 𝐅^{T} 𝐅^{'} = {(Ω 𝐅)}^{T} 𝐅 + 𝐅^{T} (Ω 𝐅) = 𝐅^{T} Ω^{T} 𝐅 + 𝐅^{T} Ω 𝐅 = 0,

so that $𝐅^{T} (t) 𝐅 (t)$ is constant. Since $𝐅 {(t_{0})}^{T} 𝐅 (t_{0}) = I$ (since $𝔱_{0}$ , $𝔫_{0}$ , and $𝔟_{0}$ form an orthonormal basis), we see that $𝐅^{T} (t) 𝐅 (t) = I$ 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 $3 \times 3$ matrix $𝐆 (t)$ with columns given by $𝔱$ , $𝔫$ , and $𝔟$ . We know that $𝐆^{T} 𝐆 = I$ for all $t$ , so that $𝐆 (t) \in O (3)$ for all $t$ ; therefore $\det (𝐆 (t)) = \pm 1$ . Since the determinant is is a polynomial in the entries, it is continuous, and therefore since $\det (𝐆 (t_{0})) = 1$ , we know that $𝔱 (t)$ , $𝔫 (t)$ , and $𝔟 (t)$ are positively oriented for all $t$ .

Defining

𝜸 (t) = 𝐩_{0} + \int_{t_{0}}^{t} 𝐭 (u) d u

then produces a curve with $γ (t_{0}) = 𝐩_{0}$ , 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.

From Remark 7.6 in the uploaded notes.

7.3.1 The plane-curve construction

Theorem 7.5 — The fundamental theorems of plane and space curves

Let $I \subseteq R$ be an interval.

If $κ_{s} : I \to R$ is smooth, then there exists a unit-speed plane curve $𝜸 : I \to R^{2}$ whose signed curvature is $κ_{s}$ . If $𝜸$ and $\hat{𝜸}$ are two such curves, then
$\hat{𝜸} = f \circ 𝜸$

for some proper rigid motion $f$ of $R^{2}$ .
If $κ, τ : I \to R$ are smooth and $κ > 0$ , then there exists a unit-speed space curve $𝜸 : I \to R^{3}$ whose curvature and torsion are $κ$ and $τ$ . If $𝜸$ and $\hat{𝜸}$ are two such curves, then
$\hat{𝜸} = f \circ 𝜸$

for some proper rigid motion $f$ of $R^{3}$ .

Lecture notes, §7.3.1 · Theorem 7.5.

Proof — plane-curve statement

Proof of the plane-curve statement. Fix $t_{0} \in I$ . We first construct the curve with initial conditions

𝜸 (t_{0}) = (0, 0), 𝜸^{'} (t_{0}) = (1, 0) .

Define

𝜃 (t) : = \int_{t_{0}}^{t} κ_{s} (u) d u,

(7.4)

then define the velocity and the curve by

𝐯 (t) : = (\cos 𝜃 (t), \sin 𝜃 (t)), 𝜸 (t) : = \int_{t_{0}}^{t} 𝐯 (u) d 𝑢.

(7.5)

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

κ_{s} \to_{}^{integrate} 𝜃 \to_{}^{(\cos 𝜃, \sin 𝜃)} 𝐯 \to_{}^{integrate} 𝜸 .

By the Fundamental Theorem of Calculus,

𝜸^{'} = 𝐯, | 𝐯 | = 1,

so $𝜸$ has unit speed. Furthermore,

𝐚 = 𝐯^{'} = 𝜃^{'} (- \sin 𝜃, \cos 𝜃) = κ_{s} R_{90} (𝐯) .

For a unit-speed plane curve, this equation says exactly that the signed curvature is $κ_{s}$ . 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 $\hat{𝜸}$ be unit-speed plane curves with the same signed curvature $κ_{s}$ , and choose angle functions $𝜃$ and $\hat{𝜃}$ . After applying a rigid motion to each curve, we can assume that $γ (t_{0}) = \hat{γ} (t_{0}) = 0$ , and $γ^{'} (t_{0}) = {\hat{γ}}^{'} (t_{0}) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ . Since

𝜃^{'} = κ_{s} = {\hat{𝜃}}^{'},

there is a constant $𝜃_{0}$ such that

\hat{𝜃} = 𝜃 + 𝜃_{0} .

But then, plugging into $𝐯 (t) : = (\cos 𝜃 (t), \sin 𝜃 (t))$ and $\hat{𝐯} (t) : = (\cos \hat{𝜃} (t), \sin \hat{𝜃} (t))$ , and setting $t = t_{0}$ , we see that $𝜃_{0}$ is a multiple of $2 π$ , so we can take $𝜃_{0} = 0$ . Therefore $𝐯 = \hat{𝐯}$ .

Finally, since

γ^{'} (t) = : 𝐯 (t) = \hat{v} (t) : = {\hat{γ}}^{'} (t)

we see that $\hat{γ} (t) = γ (t) + γ_{0}$ . Setting $t = t_{0}$ , and using that $γ (t) = \hat{γ} (t) = 0$ , we see that $\hat{γ} = γ$ . □

7.4 A short overview of curvature formulas

Curvature of a regular curve

Tapp’s § 1.9 collects the formulas developed in the preceding lectures. For a regular curve in $R^{n}$ , (Theorem 4.6)

κ = \frac{| 𝐚^{⊥} |}{{| 𝐯 |}^{2}} = \frac{| 𝐭^{'} |}{| 𝐯 |}, κ = | 𝐚 | when the curve has unit speed .

Lecture notes, §7.4 · unnumbered.

Signed curvature in coordinates

For a plane curve $𝜸 (t) = (x (t), y (t))$ ,

κ_{s} = \frac{⟨ 𝐚, R_{90} (𝐯) ⟩}{{| 𝐯 |}^{3}} = \frac{x^{'} y^{″} - x^{″} y^{'}}{{({(x^{'})}^{2} + {(y^{'})}^{2})}^{3 / 2}} .

Lecture notes, §7.4 · unnumbered.

Signed curvature of a graph

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

κ_{s} (t) = \frac{f^{″} (t)}{{(1 + {(f^{'} (t))}^{2})}^{3 / 2}} .

Lecture notes, §7.4 · unnumbered.

Curvature from velocity and acceleration

For a space curve (see Theorem 6.12),

κ = \frac{| 𝐯 \times 𝐚 |}{{| 𝐯 |}^{3}} .

Lecture notes, §7.4 · unnumbered.

7.5 Summary

08Hopf’s Umlaufsatz, the Jordan curve theorem, and the generalized Umlaufsatz

Theorem 8.1 — Theorems of Hopf and Jordan

Let $𝜸 : [a, b] \to R^{2}$ be a simple closed plane curve, and let

C = 𝜸 ([a, b])

denote its trace.

Hopf’s Umlaufsatz. The rotation index of $𝜸$ is either $1$ or $- 1$ .
Jordan curve theorem. The complement $R^{2} - C$ 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.

Lecture notes, Lecture 8 · Theorem 8.1.

Proof — Hopf’s Umlaufsatz

Proof of Hopf’s Umlaufsatz. Reparametrize $𝜸$ by arc length. Choose a point $𝐩 \in C$ 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 = {(t_{1}, t_{2}) : a \leq t_{1} \leq t_{2} \leq b} .

Define the secant-direction map $𝝍 : T \to S^{1}$ by

𝝍 (t_{1}, t_{2}) = {\begin{matrix} \frac{𝜸 (t_{2}) - 𝜸 (t_{1})}{| 𝜸 (t_{2}) - 𝜸 (t_{1}) |}, & t_{1} < t_{2}, (t_{1}, t_{2}) \neq (a, b), \\ 𝐭 (t_{1}), & t_{1} = t_{2}, \\ - 𝐭 (a), & (t_{1}, t_{2}) = (a, b) . \end{matrix}

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 $𝜶_{0}$ traverse the diagonal of $T$ from $(a, a)$ to $(b, b)$ . Then

𝝍 \circ 𝜶_{0} = 𝐭,

so

\deg (𝝍 \circ 𝜶_{0}) = rot (𝜸) .

Let $𝜶_{1}$ instead travel from $(a, a)$ to $(a, b)$ and then from $(a, b)$ to $(b, b)$ . Since $T$ is convex, the paths

𝜶_{s} = (1 - s) 𝜶_{0} + s 𝜶_{1}, 0 \leq s \leq 1,

remain in $T$ and give a homotopy through loops in $S^{1}$ after composing with $𝝍$ . Corollary 8.4 gives

\deg (𝝍 \circ 𝜶_{0}) = \deg (𝝍 \circ 𝜶_{1}) .

Along the first edge of $𝜶_{1}$ , 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 $2 π$ or $- 2 π$ , depending on the orientation. Hence

rot (𝜸) = 1 or rot (𝜸) = - 1 .

□

Proof — Jordan curve theorem

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 $f_{𝐪}$ 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 $𝐩 \in R^{2} - C$ 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. □

8.1 Degree and the unit tangent map

Definition 8.2 — Degree of a circle-valued loop

Let $f : [a, b] \to S^{1}$ be continuous and suppose that $f (a) = f (b)$ . A continuous angle function for $f$ is a continuous function $𝜃 : [a, b] \to R$ 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 $2 π$ . The degree of $f$ is

\deg (f) = \frac{𝜃 (b) - 𝜃 (a)}{2 π} \in 𝑍.

Lecture notes, §8.1 · Definition 8.2.

Proof sketch — angle-function existence and the integer condition

The integer condition follows from $f (a) = f (b)$ : the two endpoint angles differ by an integer multiple of $2 π$ . Existence of a global angle function can be proved by covering $S^{1}$ with semicircles, choosing local inverse branches of $𝜃 \mapsto (\cos 𝜃, \sin 𝜃)$ , and matching the branches on finitely many subintervals.

Rotation index as the degree of the unit tangent

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

𝐭 (t) = \frac{𝜸^{'} (t)}{| 𝜸^{'} (t) |} \in S^{1} .

The rotation index is exactly

rot (𝜸) = \deg (𝐭) .

Lecture notes, §8.1 · unnumbered.

Lemma 8.3 — Antipodal-value lemma

Let $f_{0}, f_{1} : [a, b] \to S^{1}$ be continuous loops. If $\deg (f_{0}) \neq \deg (f_{1})$ , then for some $t_{0} \in [a, b]$ ,

f_{0} (t_{0}) = - f_{1} (t_{0}) .

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

Lecture notes, §8.1 · Lemma 8.3.

Proof

Proof. Choose angle functions $𝜃_{0}$ and $𝜃_{1}$ , and set $δ = 𝜃_{1} - 𝜃_{0}$ . If the degrees differ, then

| δ (b) - δ (a) | \geq 2 𝜋.

Consequently the interval between $δ (a)$ and $δ (b)$ contains an odd multiple of $π$ . By the intermediate value theorem, $δ (t_{0})$ is an odd multiple of $π$ for some $t_{0}$ , which is exactly the assertion that $f_{0} (t_{0}) = - f_{1} (t_{0})$ . □

Corollary 8.4 — Homotopy invariance of degree

Suppose $H : [0, 1] \times [a, b] \to S^{1}$ is continuous and $H (s, a) = H (s, b)$ for every $s$ . Then

\deg (H (s, \cdot))

is independent of $s$ .

Lecture notes, §8.1 · Corollary 8.4.

Proof

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

Theorem 8.1 — Theorems of Hopf and Jordan

Let $𝜸 : [a, b] \to R^{2}$ be a simple closed plane curve, and let

C = 𝜸 ([a, b])

denote its trace.

Hopf’s Umlaufsatz. The rotation index of $𝜸$ is either $1$ or $- 1$ .
Jordan curve theorem. The complement $R^{2} - C$ 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.

Lecture notes, §8.2 · Theorem 8.1.

Proof — Hopf’s Umlaufsatz

Proof of Hopf’s Umlaufsatz. Reparametrize $𝜸$ by arc length. Choose a point $𝐩 \in C$ 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 = {(t_{1}, t_{2}) : a \leq t_{1} \leq t_{2} \leq b} .

Define the secant-direction map $𝝍 : T \to S^{1}$ by

𝝍 (t_{1}, t_{2}) = {\begin{matrix} \frac{𝜸 (t_{2}) - 𝜸 (t_{1})}{| 𝜸 (t_{2}) - 𝜸 (t_{1}) |}, & t_{1} < t_{2}, (t_{1}, t_{2}) \neq (a, b), \\ 𝐭 (t_{1}), & t_{1} = t_{2}, \\ - 𝐭 (a), & (t_{1}, t_{2}) = (a, b) . \end{matrix}

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 $𝜶_{0}$ traverse the diagonal of $T$ from $(a, a)$ to $(b, b)$ . Then

𝝍 \circ 𝜶_{0} = 𝐭,

so

\deg (𝝍 \circ 𝜶_{0}) = rot (𝜸) .

Let $𝜶_{1}$ instead travel from $(a, a)$ to $(a, b)$ and then from $(a, b)$ to $(b, b)$ . Since $T$ is convex, the paths

𝜶_{s} = (1 - s) 𝜶_{0} + s 𝜶_{1}, 0 \leq s \leq 1,

remain in $T$ and give a homotopy through loops in $S^{1}$ after composing with $𝝍$ . Corollary 8.4 gives

\deg (𝝍 \circ 𝜶_{0}) = \deg (𝝍 \circ 𝜶_{1}) .

Along the first edge of $𝜶_{1}$ , 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 $2 π$ or $- 2 π$ , depending on the orientation. Hence

rot (𝜸) = 1 or rot (𝜸) = - 1 .

□

8.3 The Jordan curve theorem

Theorem 8.1 — Theorems of Hopf and Jordan

Let $𝜸 : [a, b] \to R^{2}$ be a simple closed plane curve, and let

C = 𝜸 ([a, b])

denote its trace.

Hopf’s Umlaufsatz. The rotation index of $𝜸$ is either $1$ or $- 1$ .
Jordan curve theorem. The complement $R^{2} - C$ 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.

Lecture notes, §8.3 · Theorem 8.1.

Proof — Jordan curve theorem

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 $f_{𝐪}$ 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 $𝐩 \in R^{2} - C$ 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. □

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) \times [a, b) \to R^{2}, Φ (s, t) = 𝜸 (t) + s R_{90} (𝐭 (t)),

is injective.

Lecture notes, §8.3 · Proposition 8.5.

Definition — Winding number

For $𝐩 \in R^{2} - C$ , define

f_{𝐩} (t) = \frac{𝜸 (t) - 𝐩}{| 𝜸 (t) - 𝐩 |} \in S^{1},

and define the winding number of $C$ about $𝐩$ by

W (𝐩) = \deg (f_{𝐩}) .

Lecture notes, §8.3 · unnumbered.

Lemma 8.6 — Local constancy of the winding number

The function

W : R^{2} - C \to Z

is locally constant. Consequently it is constant on every path-connected component of $R^{2} - C$ .

Lecture notes, §8.3 · Lemma 8.6.

Proof

Proof. Fix $𝐩 \notin C$ . 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 $f_{𝐩}$ to $f_{𝐪}$ . Their degrees are therefore equal by Corollary 8.4. □

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 $R_{90} (𝐭 (t))$ points into the interior. A curve with the opposite orientation has rotation index $- 1$ .

Lecture notes, §8.3 · Definition 8.7.

Proof sketch — equivalence of the two orientations

The equivalence follows from the same local picture used above. The vector $R_{90} (𝐭)$ 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

Definition 8.8 — Piecewise-regular curve and signed corner angle

A continuous curve $𝜸 : [a, b] \to R^{2}$ is piecewise regular if there is a partition

a = t_{0} < t_{1} < \dots < t_{m} = b

such that each restriction to $[t_{i}, t_{i + 1}]$ is regular. At a corner $t_{i}$ , let $𝐯_{i}^{-}$ and $𝐯_{i}^{+}$ be the one-sided velocity vectors. The signed angle $α_{i} \in [- π, π]$ is the smallest signed rotation carrying $𝐯_{i}^{-}$ to $𝐯_{i}^{+}$ , positive for counterclockwise rotation and negative for clockwise rotation.

Lecture notes, §8.4 · Definition 8.8.

Theorem 8.9 — Generalized Hopf Umlaufsatz

Let $𝜸 : [a, b] \to R^{2}$ be a positively oriented, unit-speed, piecewise-regular simple closed plane curve. If $κ_{s}$ is its signed curvature on the smooth pieces and $α_{i}$ are its signed corner angles, then

\int_{a}^{b} κ_{s} (t) d t + ∑_{i} α_{i} = 2 𝜋.

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

Lecture notes, §8.4 · Theorem 8.9.

Proof idea

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 $2 π$ by Hopf’s Umlaufsatz.

8.5 Summary

09Green’s theorem I: vector fields and line integrals

9.1 Coordinate forms and the wedge product

Definition — Coordinate one-forms and tensor powers

Let $x_{1}, \dots, x_{n}$ denote the standard coordinate functions on $R^{n}$ . Associated to them are the coordinate one-forms (just consider these as symbols) $d x_{1}, \dots, d x_{n}$ . Let $V$ be the real vector space with basis $d x_{1}, \dots, d x_{n}$ . Then for any non-negative integer $k$ , consider the real vector space, denoted by $\otimes^{k} V$ , which is defined to be the real vector space with basis the symbols

d x_{i_{1}} \otimes d x_{i_{2}} \otimes \dots \otimes d x_{i_{k}},

where $i_{1}, i_{2}, \dots, i_{k} \in {1, \dots, n}$ . This vector space has dimension $n^{k}$ . By convention, $\otimes^{0} V = R$ .

Lecture notes, §9.1 · unnumbered.

Definition — Exterior powers as a quotient

Now, inside of $\otimes^{k} V$ , for any $i_{1}, \dots, i_{k}$ , with $i_{1}, \dots, i_{k} \in (1, \dots, n)$ , then if $σ$ is a permutation of ${1, \dots, k}$ that interchanges exactly two of the numbers (e.g., $(1, 2, 3, 4) \mapsto (2, 1, 3, 4)$ ), consider the vector

d x_{i_{σ (1)}} \otimes \dots \otimes d x_{i_{σ (k)}} + d x_{i_{1}} \otimes \dots \otimes d x_{i_{k}} .

The span of these vectors is a vector subspace $S_{k}$ of $\otimes^{k} V$ . This subspace defines an equivalence relation on $\otimes^{k} V$ ; two vectors in $\otimes^{k} V$ are equivalent if their difference lies in $S_{k}$ . We define $\land^{k} V$ to be the vector space that has elements that are equivalence classes of elements in $\otimes^{k} V$ . We denote the equivalence class of $d x_{i_{1}} \otimes d x_{i_{2}} \otimes \dots \otimes d x_{i_{k}}$ by

d x_{i_{1}} \land d x_{i_{2}} \land \dots \land d x_{i_{k}} .

Lecture notes, §9.1 · unnumbered.

Alternating relation and repeated coordinate factors

The basic consequence of this construction is that we have the following relation between the symbols above: Given $i_{1}, \dots, i_{k}$ , with $i_{1}, \dots, i_{k} \in (1, \dots, n)$ , then if $σ$ is a permutation of ${1, \dots, k}$ that interchanges exactly two of the numbers (e.g., $(1, 2, 3, 4) \mapsto (2, 1, 3, 4)$ ), we have

d x_{i_{σ (1)}} \land \dots \land d x_{i_{σ (k)}} = - d x_{i_{1}} \land \dots \land d x_{i_{k}} .

In other words, for example, we have

d x_{i} \land d x_{j} = - d x_{j} \land d x_{i} .

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

Lecture notes, §9.1 · unnumbered.

Calculation — repeated coordinate factors

This implies that

d x_{i} \land d x_{i} = - d x_{i} \land d x_{i},

so

d x_{i} \land d x_{i} = 0 .

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

Definition — Multi-index notation for coordinate forms

If

I = (i_{1}, \dots, i_{k}),

we use the abbreviation

d x_{I} = d x_{i_{1}} \land \dots \land d x_{i_{k}} .

Lecture notes, §9.1 · unnumbered.

Examples

Vanishing above the ambient dimension

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

Lecture notes, Lecture 9 · unnumbered.

9.1.1 Wedge products of forms

Degree of a wedge product

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 \land 𝜂.

Lecture notes, §9.1.1 · unnumbered.

Graded commutativity

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

ω \land η = {(- 1)}^{𝑘ℓ} η \land 𝜔.

This is called graded commutativity.

Lecture notes, §9.1.1 · unnumbered.

9.2 Smooth differential forms

Definition 9.2 — Differential form in Euclidean coordinates

Let $U \subseteq R^{n}$ be open. A smooth $k$ -form on $U$ is an expression

ω = ∑_{| I | = k} f_{I} (𝐱) d x_{I},

where the sum runs over ordered index sets $I = (i_{1} < \dots < i_{k})$ , and every coefficient $f_{I} : U \to R$ is smooth. The vector space of smooth $k$ -forms on $U$ is denoted by $Ω^{k} (U)$ .

Lecture notes, §9.2 · Definition 9.2.

Definition — Smooth zero-forms

A smooth zero-form is simply a smooth function:

Ω^{0} (U) = C^{\infty} (U) .

Lecture notes, §9.2 · unnumbered.

9.3 The exterior derivative

Definition 9.6 — Exterior derivative

For a smooth function $f : U \to R$ , define

d f = ∑_{i = 1}^{n} \frac{∂𝑓}{\partial x_{i}} d x_{i} .

For a general differential form

ω = ∑_{I} f_{I} d x_{I},

define

d ω = ∑_{I} d f_{I} \land d x_{I} .

Lecture notes, §9.3 · Definition 9.6.

Proposition 9.7 — Graded Leibniz rule

If $ω \in Ω^{k} (U)$ and $η \in Ω^{ℓ} (U)$ , then

d (ω \land η) = d ω \land η + {(- 1)}^{k} ω \land d 𝜂.

Lecture notes, §9.3 · Proposition 9.7.

9.4 Vector fields and one-forms

Definition 9.8 — Vector field and associated one-form

A smooth vector field on an open set $U \subseteq R^{n}$ is a smooth map

𝐅 = (F_{1}, \dots, F_{n}) : U \to R^{n} .

Its associated one-form is

ω_{𝐅} = F_{1} d x_{1} + \dots + F_{n} d x_{n} .

At $𝐩 \in U$ , this one-form acts on $𝐮 \in R^{n}$ by

{(ω_{𝐅})}_{𝐩} (𝐮) = ⟨ 𝐅 (𝐩), 𝐮 ⟩ .

Lecture notes, §9.4 · Definition 9.8.

9.5 Line integrals

Definition 9.9 — Line integral of a one-form

If $ω \in Ω^{1} (U)$ , define

\int_{𝜸} ω = \int_{a}^{b} ω_{𝜸 (t)} (𝜸^{'} (t)) d 𝑡.

If $ω = ω_{𝐅}$ is associated to a vector field, then

\int_{𝜸} ω_{𝐅} = \int_{𝜸} 𝐅 \cdot d 𝜸 = \int_{a}^{b} ⟨ 𝐅 (𝜸 (t)), 𝜸^{'} (t) ⟩ d 𝑡.

The formal notation $d 𝜸 = 𝜸^{'} (t) d t$ encodes this formula.

Lecture notes, §9.5 · Definition 9.9.

Proposition 9.10 — Change of parameter

Let $ϕ : [α, β] \to [a, b]$ be a smooth diffeomorphism and let $\tilde{𝜸} = 𝜸 \circ 𝜙.$ If $ϕ$ is increasing, then

\int_{\tilde{𝜸}} ω = \int_{𝜸} 𝜔.

If $ϕ$ is decreasing, then

\int_{\tilde{𝜸}} ω = - \int_{𝜸} 𝜔.

Lecture notes, §9.5 · Proposition 9.10.

Proof

Proof. For an increasing reparametrization and $ω = ω_{𝐅}$ , the chain rule gives

\begin{array}{l} \int_{\tilde{𝜸}} ω & = \int_{α}^{β} ⟨ 𝐅 (𝜸 (ϕ (s))), 𝜸^{'} (ϕ (s)) ϕ^{'} (s) ⟩ d s \\ = \int_{a}^{b} ⟨ 𝐅 (𝜸 (t)), 𝜸^{'} (t) ⟩ d t, \end{array}

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. □

9.6 Summary

10Green’s theorem II: conservative fields and Green’s theorem

10.1 Conservative vector fields

Definition — Gradient and conservative vector field

Let $U\subseteq\mathbb{R}^n$ be open and let $f:U\to\mathbb{R}$ be smooth. Its gradient is the vector field $\nabla f=\left(\frac{\partial f}{\partial x_1},\ldots, \frac{\partial f}{\partial x_n}\right).$ A smooth vector field $\mathbf F:U\to\mathbb{R}^n$ is conservative if $\mathbf F=\nabla f$ for some smooth $f:U\to\mathbb{R}$ . Such an $f$ is called a potential function of $\mathbf F$ .

Lecture notes, §10.1 · unnumbered.

Definition — Exact one-forms

In the differential-form notation of Lecture 9, the vector field $\mathbf F=(F_1,\ldots,F_n)$ corresponds to $\omega_{\mathbf F}=\sum_{i=1}^n F_i\,\mathrm{d} x_i.$ Thus $\mathbf F=\nabla f$ if and only if $\omega_{\mathbf F}=\mathrm{d} f =\sum_{i=1}^n\frac{\partial f}{\partial x_i}\,\mathrm{d} x_i.$ A one-form of the form $\mathrm{d} f$ is called exact.

Lecture notes, §10.1 · unnumbered.

10.2 The chain rule and directional derivatives

Lemma — Chain rule along a curve

Let $f:U\to\mathbb{R}$ be smooth and let $\boldsymbol\gamma:I\to U$ be a smooth curve. Then $\frac{\mathrm{d}}{\mathrm{d} t}f(\boldsymbol\gamma(t)) =\langle \nabla f(\boldsymbol\gamma(t)), \boldsymbol\gamma'(t)\rangle.$

Lecture notes, §10.2 · unnumbered.

Proof

Proof. Write $\boldsymbol\gamma=(\gamma_1,\ldots,\gamma_n)$ . The multivariable chain rule gives $\frac{\mathrm{d}}{\mathrm{d} t}f(\boldsymbol\gamma(t)) =\sum_{i=1}^n \frac{\partial f}{\partial x_i}(\boldsymbol\gamma(t))\, \gamma_i'(t) =\langle \nabla f(\boldsymbol\gamma(t)), \boldsymbol\gamma'(t)\rangle.$ ◻

Definition — Directional derivative

Fix $\mathbf p_0\in U$ and $\mathbf v_0\in\mathbb{R}^n$ . Since $U$ is open, the line $\boldsymbol\ell(t)=\mathbf p_0+t\mathbf v_0$ lies in $U$ for all sufficiently small $\abs t$. The directional derivative of $f$ at $\mathbf p_0$ in the direction $\mathbf v_0$ is $D_{\mathbf v_0}f(\mathbf p_0) :=\left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} f(\mathbf p_0+t\mathbf v_0).$ Applying the chain rule to $f\circ\boldsymbol\ell$ yields $\boxed{ D_{\mathbf v_0}f(\mathbf p_0) =\sum_{i=1}^n\frac{\partial f}{\partial x_i}(\mathbf p_0)(v_0)_i =\langle \nabla f(\mathbf p_0), \mathbf v_0\rangle.}$ This is also written $(\mathrm{d} f)_{\mathbf p_0}(\mathbf v_0)$ in Tapp, Lemma 2.23 and Eq. (2.3). The vector $\mathbf v_0$ need not have unit length. More generally, any smooth curve through $\mathbf p_0$ with initial velocity $\mathbf v_0$ gives the same initial derivative of $f$ .

Lecture notes, §10.2 · unnumbered.

10.3 Geometry of the gradient

Definition — Level sets

For a real number $c$ , the level set of $f$ at height $c$ is $f^{-1}(c)=\{\mathbf p\in U:f(\mathbf p)=c\}.$

Lecture notes, §10.3 · unnumbered.

Lemma — Direction and rate of greatest increase

Let $\mathbf p\in U$ and assume that $\nabla f(\mathbf p)\neq0$ .

The vector $\nabla f(\mathbf p)$ is orthogonal to the level set through $\mathbf p$ , in the sense just described.
Among unit vectors $\mathbf u$ , the derivative $D_{\mathbf u}f(\mathbf p)$ is largest when $\mathbf u=\frac{\nabla f(\mathbf p)}{\lvert \nabla f(\mathbf p)\rvert}.$
The maximum directional derivative is $\lvert \nabla f(\mathbf p)\rvert$ .

Lecture notes, §10.3 · unnumbered.

Level-set calculation used in the proof

If the trace of $\boldsymbol\gamma$ lies in this level set, then $f(\boldsymbol\gamma(t))=c$ , and hence $0=\frac{\mathrm{d}}{\mathrm{d} t}f(\boldsymbol\gamma(t)) =\langle \nabla f(\boldsymbol\gamma(t)), \boldsymbol\gamma'(t)\rangle.$ Thus the gradient is orthogonal to every velocity vector of a curve in the level set.

Proof

Proof. The level-set calculation proves the first assertion. For a unit vector $\mathbf u$ , the Schwarz inequality gives $D_{\mathbf u}f(\mathbf p) =\langle \nabla f(\mathbf p), \mathbf u\rangle \leq\lvert \nabla f(\mathbf p)\rvert.$ Equality holds for $\mathbf u=\nabla f(\mathbf p)/\lvert \nabla f(\mathbf p)\rvert$ , which proves the other two assertions. ◻

10.4 Integrating a conservative field

Proposition — The endpoint formula

Let $f:U\to\mathbb{R}$ be smooth, and let $C$ be an oriented piecewise-regular curve in $U$ , parametrized by $\boldsymbol\gamma:[a,b]\to U$ . Then $\int_C\nabla f\cdot\mathrm{d}\boldsymbol\gamma =f(\boldsymbol\gamma(b))-f(\boldsymbol\gamma(a)).$ Equivalently, $\int_C\mathrm{d} f=f(\boldsymbol\gamma(b))-f(\boldsymbol\gamma(a)).$

Lecture notes, §10.4 · unnumbered.

Proof

Proof. For a smooth curve, the chain rule and the fundamental theorem of calculus give $\begin{aligned} \int_C\nabla f\cdot\mathrm{d}\boldsymbol\gamma &=\int_a^b \langle \nabla f(\boldsymbol\gamma(t)), \boldsymbol\gamma'(t)\rangle\,\mathrm{d} t\\ &=\int_a^b\frac{\mathrm{d}}{\mathrm{d} t}f(\boldsymbol\gamma(t))\,\mathrm{d} t\\ &=f(\boldsymbol\gamma(b))-f(\boldsymbol\gamma(a)). \end{aligned}$ For a piecewise-regular curve, apply this calculation on each smooth piece and add; all intermediate endpoint values cancel. ◻

10.5 Path independence

Proposition — Equivalent characterizations of a conservative field

Let $U\subseteq\mathbb{R}^n$ be open and path-connected, and let $\mathbf F$ be a smooth vector field on $U$ . The following conditions are equivalent.

$\mathbf F=\nabla f$ for some smooth $f:U\to\mathbb{R}$ ; equivalently, $\omega_{\mathbf F}=\mathrm{d} f$ .
For every piecewise-regular closed path $C$ in $U$ , $\oint_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma=0.$
Line integrals of $\mathbf F$ are path-independent: two piecewise-regular paths in $U$ with the same initial and final points have equal line integrals.

Lecture notes, §10.5 · unnumbered.

Proof

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

(2) ⇒ (3): given two paths $C_1,C_2$ with the same endpoints, traverse $C_1$ and then traverse $C_2$ in reverse. The resulting path is closed. Additivity and reversal of orientation give $0=\int_{C_1}\mathbf F\cdot\mathrm{d}\boldsymbol\gamma -\int_{C_2}\mathbf F\cdot\mathrm{d}\boldsymbol\gamma.$

(3) ⇒ (1): fix $\mathbf p_0\in U$ . Define $f(\mathbf x)=\int_{\mathbf p_0}^{\mathbf x} \mathbf F\cdot\mathrm{d}\boldsymbol\gamma,$ where the integral may be taken along any piecewise-regular path in $U$ from $\mathbf p_0$ to $\mathbf x$ . Such paths exist in an open, path-connected set, and path independence makes $f$ well defined.

To compute its $i$ th partial derivative, keep $\mathbf x$ fixed and let $\mathbf e_i$ be the $i$ th standard basis vector. For sufficiently small $h$ , the line segment from $\mathbf x$ to $\mathbf x+h\mathbf e_i$ lies in $U$ . By adjoining this segment to a path ending at $\mathbf x$ , $f(\mathbf x+h\mathbf e_i)-f(\mathbf x) =\int_0^h\langle \mathbf F(\mathbf x+s\mathbf e_i), \mathbf e_i\rangle\,\mathrm{d} s =\int_0^h F_i(\mathbf x+s\mathbf e_i)\,\mathrm{d} s.$ Dividing by $h$ and taking the limit gives $\frac{\partial f}{\partial x_i}(\mathbf x)=F_i(\mathbf x).$ Since the $F_i$ are smooth, $f$ is smooth, and $\nabla f=\mathbf F$ . ◻

10.6 Conservation of energy

Conservation of energy

Suppose that $\mathbf F=\nabla f$ is conservative. Set $\rho=-f.$ The function $\rho$ is the potential energy function; its sign is opposite to that of the potential $f$ used in $\mathbf F=\nabla f$ . Assume that a particle of mass $m>0$ moves under this force alone, so $\mathbf F(\boldsymbol\gamma(t)) =m\mathbf a(t)=m\boldsymbol\gamma''(t), \qquad \mathbf v(t)=\boldsymbol\gamma'(t).$

Thus the total energy $\underbrace{\frac m2\lvert \mathbf v(t)\rvert^2}_{\text{kinetic energy}} +\underbrace{\rho(\boldsymbol\gamma(t))}_{\text{potential energy}}$ is constant. This is the conservation law associated with a conservative force field.

Lecture notes, §10.6 · unnumbered.

Derivation

By the work–energy calculation from Lecture 9, $\begin{aligned} \int_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma &=\int_a^b m\langle \mathbf a(t), \mathbf v(t)\rangle\,\mathrm{d} t\\ &=\frac m2\lvert \mathbf v(b)\rvert^2-\frac m2\lvert \mathbf v(a)\rvert^2. \end{aligned}$ On the other hand, the endpoint formula gives $\int_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma =f(\boldsymbol\gamma(b))-f(\boldsymbol\gamma(a)) =-\rho(\boldsymbol\gamma(b))+\rho(\boldsymbol\gamma(a)).$ Equating these expressions and rearranging yields $\boxed{ \frac m2\lvert \mathbf v(b)\rvert^2+\rho(\boldsymbol\gamma(b)) =\frac m2\lvert \mathbf v(a)\rvert^2+\rho(\boldsymbol\gamma(a)).}$

10.7 Example: gravity

11.2 Green’s theorem

Theorem — Green’s theorem: circulation form

Let $C$ be a positively oriented piecewise-regular simple closed plane curve with interior $D$ . Let $\mathbf F=(F_1,F_2)$ be a smooth vector field on an open set containing $D\cup C$ . Then $\boxed{ \oint_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma =\iint_D\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right)\mathrm{d} A.}$

Lecture notes, §11.2 · unnumbered.

11.2.1 The differential-form expression

Green’s theorem in differential-form notation

Let $C$ be a positively oriented piecewise-regular simple closed plane curve with interior $D$ . Let $\mathbf F=(F_1,F_2)$ be a smooth vector field on an open set containing $D\cup C$ . Then $\boxed{ \oint_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma =\iint_D\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right)\mathrm{d} A.}$

With the standard orientation of the plane, $\mathrm{d} x_1\wedge\mathrm{d} x_2$ is the oriented area form. Thus Green’s theorem has the compact form $\boxed{\oint_{\partial D}\omega_{\mathbf F} =\int_D\mathrm{d}\omega_{\mathbf F}.}$

Lecture notes, §11.2.1 · unnumbered.

Differential-form calculation

Associate to $\mathbf F$ the one-form $\omega_{\mathbf F}=F_1\,\mathrm{d} x_1+F_2\,\mathrm{d} x_2.$ Along a parametrization $\boldsymbol\gamma=(\gamma_1,\gamma_2)$ , $\begin{aligned} \mathbf F\cdot\mathrm{d}\boldsymbol\gamma &=\langle \mathbf F(\boldsymbol\gamma(t)), \boldsymbol\gamma'(t)\rangle\,\mathrm{d} t\\ &=\bigl(F_1(\boldsymbol\gamma(t))\gamma_1'(t) +F_2(\boldsymbol\gamma(t))\gamma_2'(t)\bigr)\mathrm{d} t. \end{aligned}$ This is exactly how $\omega_{\mathbf F}$ is integrated along $C$ . Its exterior derivative is $\begin{aligned} \mathrm{d}\omega_{\mathbf F} &=\mathrm{d} F_1\wedge\mathrm{d} x_1+\mathrm{d} F_2\wedge\mathrm{d} x_2\\ &=\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right) \mathrm{d} x_1\wedge\mathrm{d} x_2. \end{aligned}$

11.2.2 Infinitesimal circulation

Definition — Infinitesimal circulation

The scalar function $\frac{\partial F_2}{\partial x_1}-\frac{\partial F_1}{\partial x_2}$ is called the infinitesimal circulation of $\mathbf F$ in Tapp, Definition 2.30.

Lecture notes, §11.2.2 · unnumbered.

Corollary — Circulation around a small circle

Suppose $\mathbf F$ is smooth near $\mathbf p\in\mathbb{R}^2$ , and let $C_{r,\mathbf p}$ be the counterclockwise circle of radius $r$ centered at $\mathbf p$ . Then $\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right)(\mathbf p) =\lim_{r\to0^+}\frac{1}{\pi r^2} \oint_{C_{r,\mathbf p}}\mathbf F\cdot\mathrm{d}\boldsymbol\gamma.$

Lecture notes, §11.2.2 · unnumbered.

Proof

Proof. Write $g=\partial F_2/\partial x_1-\partial F_1/\partial x_2$ , and let $B_r(\mathbf p)$ be the disk bounded by $C_{r,\mathbf p}$ . For sufficiently small $r$ , Green’s theorem gives $\frac{1}{\pi r^2}\oint_{C_{r,\mathbf p}} \mathbf F\cdot\mathrm{d}\boldsymbol\gamma =\frac{1}{\pi r^2}\iint_{B_r(\mathbf p)}g\,\mathrm{d} A.$ The right side is the average of $g$ over the disk. By continuity, this average tends to $g(\mathbf p)$ as $r\to0^+$ . ◻

11Green’s theorem III: circulation, divergence, and flux

11.1 Review: the exterior derivative squares to zero

The exterior derivative squares to zero

Recall from Lecture 9 that if $\omega$ is a smooth $k$ -form on an open set $U\subseteq\mathbb{R}^n$ , then $\mathrm{d}\omega$ is a $(k+1)$ -form. Applying the exterior derivative twice always gives zero: $\boxed{\mathrm{d}(\mathrm{d}\omega)=0.}$

Lecture notes, §11.1 · unnumbered.

11.2 Green’s theorem

Theorem — Green’s theorem: circulation form

Let $C$ be a positively oriented piecewise-regular simple closed plane curve with interior $D$ . Let $\mathbf F=(F_1,F_2)$ be a smooth vector field on an open set containing $D\cup C$ . Then $\boxed{ \oint_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma =\iint_D\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right)\mathrm{d} A.}$

Lecture notes, §11.2 · unnumbered.

11.2.1 The differential-form expression

Green’s theorem in differential-form notation

Let $C$ be a positively oriented piecewise-regular simple closed plane curve with interior $D$ . Let $\mathbf F=(F_1,F_2)$ be a smooth vector field on an open set containing $D\cup C$ . Then $\boxed{ \oint_C\mathbf F\cdot\mathrm{d}\boldsymbol\gamma =\iint_D\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right)\mathrm{d} A.}$

With the standard orientation of the plane, $\mathrm{d} x_1\wedge\mathrm{d} x_2$ is the oriented area form. Thus Green’s theorem has the compact form $\boxed{\oint_{\partial D}\omega_{\mathbf F} =\int_D\mathrm{d}\omega_{\mathbf F}.}$

Lecture notes, §11.2.1 · unnumbered.

Differential-form calculation

Associate to $\mathbf F$ the one-form $\omega_{\mathbf F}=F_1\,\mathrm{d} x_1+F_2\,\mathrm{d} x_2.$ Along a parametrization $\boldsymbol\gamma=(\gamma_1,\gamma_2)$ , $\begin{aligned} \mathbf F\cdot\mathrm{d}\boldsymbol\gamma &=\langle \mathbf F(\boldsymbol\gamma(t)), \boldsymbol\gamma'(t)\rangle\,\mathrm{d} t\\ &=\bigl(F_1(\boldsymbol\gamma(t))\gamma_1'(t) +F_2(\boldsymbol\gamma(t))\gamma_2'(t)\bigr)\mathrm{d} t. \end{aligned}$ This is exactly how $\omega_{\mathbf F}$ is integrated along $C$ . Its exterior derivative is $\begin{aligned} \mathrm{d}\omega_{\mathbf F} &=\mathrm{d} F_1\wedge\mathrm{d} x_1+\mathrm{d} F_2\wedge\mathrm{d} x_2\\ &=\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right) \mathrm{d} x_1\wedge\mathrm{d} x_2. \end{aligned}$

11.2.2 Infinitesimal circulation

Definition — Infinitesimal circulation

The scalar function $\frac{\partial F_2}{\partial x_1}-\frac{\partial F_1}{\partial x_2}$ is called the infinitesimal circulation of $\mathbf F$ in Tapp, Definition 2.30.

Lecture notes, §11.2.2 · unnumbered.

Corollary — Circulation around a small circle

Suppose $\mathbf F$ is smooth near $\mathbf p\in\mathbb{R}^2$ , and let $C_{r,\mathbf p}$ be the counterclockwise circle of radius $r$ centered at $\mathbf p$ . Then $\left(\frac{\partial F_2}{\partial x_1} -\frac{\partial F_1}{\partial x_2}\right)(\mathbf p) =\lim_{r\to0^+}\frac{1}{\pi r^2} \oint_{C_{r,\mathbf p}}\mathbf F\cdot\mathrm{d}\boldsymbol\gamma.$

Lecture notes, §11.2.2 · unnumbered.

Proof

Proof. Write $g=\partial F_2/\partial x_1-\partial F_1/\partial x_2$ , and let $B_r(\mathbf p)$ be the disk bounded by $C_{r,\mathbf p}$ . For sufficiently small $r$ , Green’s theorem gives $\frac{1}{\pi r^2}\oint_{C_{r,\mathbf p}} \mathbf F\cdot\mathrm{d}\boldsymbol\gamma =\frac{1}{\pi r^2}\iint_{B_r(\mathbf p)}g\,\mathrm{d} A.$ The right side is the average of $g$ over the disk. By continuity, this average tends to $g(\mathbf p)$ as $r\to0^+$ . ◻

11.3 The flux version of Green’s theorem

11.3.1 The outward normal and flux

Definition — Outward unit normal

Let $C=\partial D$ have a positive parametrization $\boldsymbol\gamma:[a,b]\to\mathbb{R}^2$ . At a regular point, write $\mathbf v(t)=\boldsymbol\gamma'(t)$ . The outward unit normal $\mathbf n(t)$ is $\mathbf n(t) =R_{-90}\left(\frac{\mathbf v(t)}{\lvert \mathbf v(t)\rvert}\right), \qquad R_{-90}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad R_{-90}(x_1,x_2)=(x_2,-x_1).$ The clockwise quarter-turn points outward because the interior is on the left. Here $\mathbf n$ is the outward boundary normal, not the curvature normal $\boldsymbol{\mathfrak n}$ used in the Frenet frame.

Lecture notes, §11.3.1 · unnumbered.

Definition — Flux across a closed plane curve

The flux of $\mathbf F$ across $C$ is $\boxed{ \oint_C\mathbf F\cdot\mathbf n\,\mathrm{d} s :=\int_a^b \langle \mathbf F(\boldsymbol\gamma(t)), \mathbf n(t)\rangle \lvert \mathbf v(t)\rvert\,\mathrm{d} t.}$ Here $\mathrm{d} s=\lvert \mathbf v(t)\rvert\,\mathrm{d} t$ is the arc-length element. For a piecewise-regular boundary, the integral is the sum over its regular pieces, on which $\mathbf n$ is defined.

Lecture notes, §11.3.1 · unnumbered.

Flux as a line integral

Thus flux is a line integral of the one-form associated with $R_{90}(\mathbf F)=(-F_2,F_1)$ .

Lecture notes, §11.3.1 · unnumbered.

Calculation

Substitute the formula for $\mathbf n$ and cancel the factor $\lvert \mathbf v\rvert$ . Since $R_{-90}^{T}=R_{90}$ , $\begin{aligned} \oint_C\mathbf F\cdot\mathbf n\,\mathrm{d} s &=\int_a^b \langle \mathbf F(\boldsymbol\gamma(t)), R_{-90}(\mathbf v(t))\rangle \,\mathrm{d} t\\ &=\int_a^b \langle R_{90}(\mathbf F(\boldsymbol\gamma(t))), \mathbf v(t)\rangle \,\mathrm{d} t\\ &=\int_a^b \bigl(-F_2(\boldsymbol\gamma(t))\gamma_1'(t) +F_1(\boldsymbol\gamma(t))\gamma_2'(t)\bigr)\mathrm{d} t\\ &=\oint_C\bigl(-F_2\,\mathrm{d} x_1+F_1\,\mathrm{d} x_2\bigr). \end{aligned}$

11.3.2 Divergence and the star operation

Definition — Divergence

For a smooth vector field $\mathbf F=(F_1,\ldots,F_n)$ on $U\subseteq\mathbb{R}^n$ , its divergence is the function $\boxed{ \nabla\cdot\mathbf F =\left(\frac{\partial}{\partial x_1},\ldots, \frac{\partial}{\partial x_n}\right)\cdot(F_1,\ldots,F_n) =\sum_{i=1}^n\frac{\partial F_i}{\partial x_i}.}$ In the plane, this is $\partial F_1/\partial x_1+\partial F_2/\partial x_2$ , as in Tapp, Definition 2.35.

Lecture notes, §11.3.2 · unnumbered.

Definition — Hodge star in Euclidean coordinates

The coordinate operation denoted by $\ast$ on the board is the Hodge star for the standard Euclidean inner product and orientation. The cases needed here are $\ast\,\mathrm{d} x_i =(-1)^{i-1}\mathrm{d} x_1\wedge\cdots\wedge \widehat{\mathrm{d} x_i}\wedge\cdots\wedge\mathrm{d} x_n, \qquad \ast(\mathrm{d} x_1\wedge\cdots\wedge\mathrm{d} x_n)=1,$ where a hat means that the indicated factor is omitted. Extend these formulas linearly over smooth coefficient functions. In particular, $\begin{aligned} \mathbf F&\longmapsto \omega_{\mathbf F}=\sum_{i=1}^n F_i\,\mathrm{d} x_i,\\ \omega_{\mathbf F}&\longmapsto \ast\omega_{\mathbf F} =\sum_{i=1}^n(-1)^{i-1}F_i\, \mathrm{d} x_1\wedge\cdots\wedge\widehat{\mathrm{d} x_i} \wedge\cdots\wedge\mathrm{d} x_n. \end{aligned}$ Thus $\ast\omega_{\mathbf F}$ is an $(n-1)$ -form.

Lecture notes, §11.3.2 · unnumbered.

Divergence and the exterior derivative

Equivalently, applying $\ast$ to this top-degree form gives $\boxed{\ast\mathrm{d}(\ast\omega_{\mathbf F})=\nabla\cdot\mathbf F.}$

Lecture notes, §11.3.2 · unnumbered.

Calculation

On differentiating, every term of $\mathrm{d} F_i$ except its $\mathrm{d} x_i$ term gives a repeated wedge factor and vanishes. Moving the remaining $\mathrm{d} x_i$ into its usual position gives another factor of $(-1)^{i-1}$ . Hence $\begin{aligned} \mathrm{d}(\ast\omega_{\mathbf F}) &=\sum_{i=1}^n(-1)^{i-1}\frac{\partial F_i}{\partial x_i}\, \mathrm{d} x_i\wedge\mathrm{d} x_1\wedge\cdots\wedge \widehat{\mathrm{d} x_i}\wedge\cdots\wedge\mathrm{d} x_n\\ &=\sum_{i=1}^n\frac{\partial F_i}{\partial x_i}\, \mathrm{d} x_1\wedge\cdots\wedge\mathrm{d} x_n\\ &=(\nabla\cdot\mathbf F)\,\mathrm{d} x_1\wedge\cdots\wedge\mathrm{d} x_n. \end{aligned}$

11.3.3 The planar calculation and the flux theorem

The star operation and the planar flux form

For $n=2$ , the star operation reads $\ast\,\mathrm{d} x_1=\mathrm{d} x_2, \qquad \ast\,\mathrm{d} x_2=-\mathrm{d} x_1.$ Therefore $\omega_{\mathbf F}=F_1\,\mathrm{d} x_1+F_2\,\mathrm{d} x_2, \qquad \ast\omega_{\mathbf F}=F_1\,\mathrm{d} x_2-F_2\,\mathrm{d} x_1,$ which is precisely the one-form obtained from the flux calculation.

Lecture notes, §11.3.3 · unnumbered.

Planar calculation

Direct differentiation gives $\begin{aligned} \mathrm{d}(\ast\omega_{\mathbf F}) &=\mathrm{d} F_1\wedge\mathrm{d} x_2-\mathrm{d} F_2\wedge\mathrm{d} x_1\\ &=\left(\frac{\partial F_1}{\partial x_1}\,\mathrm{d} x_1 +\frac{\partial F_1}{\partial x_2}\,\mathrm{d} x_2\right) \wedge\mathrm{d} x_2\\ &\quad-\left(\frac{\partial F_2}{\partial x_1}\,\mathrm{d} x_1 +\frac{\partial F_2}{\partial x_2}\,\mathrm{d} x_2\right) \wedge\mathrm{d} x_1\\ &=\left(\frac{\partial F_1}{\partial x_1} +\frac{\partial F_2}{\partial x_2}\right) \mathrm{d} x_1\wedge\mathrm{d} x_2 =(\nabla\cdot\mathbf F)\,\mathrm{d} A. \end{aligned}$

Theorem — Green’s theorem: flux form

Let $C$ be a piecewise-regular simple closed plane curve with interior $D$ , and let $\mathbf n$ be its outward unit normal on each regular piece. If $\mathbf F$ is smooth on an open set containing $D\cup C$ , then $\boxed{\oint_C\mathbf F\cdot\mathbf n\,\mathrm{d} s =\iint_D\nabla\cdot\mathbf F\,\mathrm{d} A.}$

Lecture notes, §11.3.3 · unnumbered.

Proof

Proof. Choose the positive orientation on $C=\partial D$ . Applying Green’s theorem gives $\begin{aligned} \iint_D\nabla\cdot\mathbf F\,\mathrm{d} A &=\int_D\mathrm{d}(\ast\omega_{\mathbf F}) =\oint_C\ast\omega_{\mathbf F}\\ &=\oint_C\bigl(F_1\,\mathrm{d} x_2-F_2\,\mathrm{d} x_1\bigr) =\oint_C\mathbf F\cdot\mathbf n\,\mathrm{d} s. \end{aligned}$ This is the final calculation in the notes. It is also the rotation argument in Tapp, Theorem 2.37, expressed using forms. ◻

11.4 Summary