Spacetime Geometry

Franz Lemmermeyer September 13, 2003

Preface

These are the lecture notes to my MATH552 class ‘Spacetime Geometry’ (CSU San Marcos, Spring Semester 2003) on differential geometry with the aim of applications to general relativity. Since I intended to emphasize ideas rather than rigour, there are various places in which I argue using a lot of handwaving; anyone with a decent command of real analysis should be able to fill these gaps without problems.

Contents

1 Jan 21, 2003 6 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Jan 23, 2003 10 2.1 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Jan 28, 2003 14 3.1 Reparametrizations . . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Jan 30, 2003 17 4.1 Parametrization by Arc Length . . . . . . . . . . . . . . . . . . . 17 4.2 Frenet’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Feb 04, 2003 20 5.1 Frenet’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

6 Feb 06, 2003 22 6.1 Application of Frenet’s formulas . . . . . . . . . . . . . . . . . . . 22 6.2 Frenet’s Formulas for curves with arbitrary speed . . . . . . . . . 23 6.3 Euclidean Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 23

7 Feb 11, 2003 24 7.1 The hyperbolic plane . . . . . . . . . . . . . . . . . . . . . . . . . 24

8 Feb 13, 2003 26 8.1 PSL_{2}(ℝ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 8.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 26

9 Feb 18, 2003 28 9.1 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 9.2 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

10 Feb 20, 2003 31 10.1 Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 11 March 11, 2003 34

11.1 Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 11.2 Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . . . 34

12 March 13, 2003 38 12.1 Kepler’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . 38

13 March 18, 2003 41 13.1 Galilean Spacetime . . . . . . . . . . . . . . . . . . . . . . . . . . 41 13.2 Minkowski Spacetime . . . . . . . . . . . . . . . . . . . . . . . . 42

14 March 25, 2003 44 14.1 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 14.2 Minkowski Spacetime . . . . . . . . . . . . . . . . . . . . . . . . 45 14.3 Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . . 46

15 March 27, 2003 47 15.1 Moving Clocks Run Slow . . . . . . . . . . . . . . . . . . . . . . 47 15.2 Moving Rulers Contract . . . . . . . . . . . . . . . . . . . . . . . 47 15.3 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 15.4 Inertial Worldlines have Maximal Length . . . . . . . . . . . . . 48 15.5 Newton versus Einstein . . . . . . . . . . . . . . . . . . . . . . . 49

16 April 10, 2003 50 16.1 Curvature of Surfaces: Euler’s Formula . . . . . . . . . . . . . . . 50 16.2 Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 16.3 The metric tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

17 April 15, 2003 53 17.1 Computing Arc Lengths . . . . . . . . . . . . . . . . . . . . . . . 53 17.2 Computing Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

18 April 21, 2003 56 18.1 Curvature of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 56 18.2 A surface with negative curvature . . . . . . . . . . . . . . . . . . 58

19 April 29, 2003 59 19.1 Some Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . 59 19.2 Theorema Egregium . . . . . . . . . . . . . . . . . . . . . . . . . 59 19.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

20 May 1, 2003 64 20.1 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 20.2 General Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 65

21 May 8, 2003 67 21.1 Weak Gravity Approximation . . . . . . . . . . . . . . . . . . . . 67 22 Homework 69

22.1 Homework 1, Jan. 21, 2003 . . . . . . . . . . . . . . . . . . . . . 69 22.2 Homework 2, Jan. 28, 2003 . . . . . . . . . . . . . . . . . . . . . 71 22.3 Homework 3, Feb 13, 2003 . . . . . . . . . . . . . . . . . . . . . . 72 22.4 Homework 4, Feb 13, 2003 . . . . . . . . . . . . . . . . . . . . . . 73 22.5 Homework 5, Feb 27, 2003 . . . . . . . . . . . . . . . . . . . . . . 75 22.6 Homework 6, March 12, 2003 . . . . . . . . . . . . . . . . . . . . 76 22.7 Homework 7, March 26, 2003 . . . . . . . . . . . . . . . . . . . . 78 22.8 Homework 8, April , 2003 . . . . . . . . . . . . . . . . . . . . . . 79

23 Exams 80 23.1 Midterm 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 23.2 Midterm 2 (May 2, 2003) . . . . . . . . . . . . . . . . . . . . . . 82 23.3 Final . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Chapter 1

Jan 21, 2003

1.1 Introduction

This semester we will learn about the geometric background needed for understanding the basic idea behind Einstein’s general relativity. We start by carefully introducing some of the main players: curves, lengths, angles, and curvature.

Linear Algebra

The space in which everything happens will be the ℝ^{n}, at least at first. Its elements are vectors x = (x_{1}, …, x_{n}) with coordinates x_{i} ∈ ℝ. On ℝ^{n} we have the standard dot product: dot (or scalar) products are maps ℝ^{n} × ℝ^{n} ⟶ ℝ (that is, machines that eat two vectors and spit out a real number) with the following properties:

• Linearity: (λu + μv) · w = λ(u · w) + μ(v · w) for λ, μ ∈ ℝ and u, v, w ∈ V = ℝ^{n};

• Symmetry: u · v = v · u for u, v ∈ V.

The dot product is called nondegenerate if u · v = 0 for all v ∈ V implies that u = 0.

The standard dot product is given by the formulas (x_{1}, …, x_{n})·(y_{1}, …, y_{n}) = x_{1}y_{1} + … + x_{n}y_{n}.

Given a dot product, we can define the length of a vector x ∈ ℝ^{n} by ||x|| = \sqrt{x · x} as long as the dot product is positive semidefinite, that is, if x · x ≥ 0 for all x ∈ V.

Exercise. Show that the standard dot product on ℝ^{n} is positive definite, that is, it is positive semidefinite and x · x = 0 implies x = 0.

Exercise. Show that the dot product on ℝ^{4} given by

(t_{1}, x_{1}, y_{1}, z_{1}) · (t_{2}, x_{2}, y_{2}, z_{2}) = t_{1}t_{2} − x_{1}x_{2} − y_{1}y_{2} − z_{1}z_{2}

is not positive definite.

Proposition 1.1 (Cauchy-Schwarz Inequality). Let V be a real vector space with a positive definite dot product. Then |u · v| ≤ ||u|| · ||v||.

Proof. There are two cases:

1. u and v are linearly dependent, i.e., there is a λ ∈ ℝ such that v = λu (if u = 0, you have to switch u and v). Then |u · v| = |u · λu| = |λ| · ||u||^{2} and ||u|| · ||v|| = ||u|| · |λ|||u|| = |λ| · ||u||^{2}, hence we have equality.

2. u and v are linearly independent. Consider the quadratic polynomial f(x) = ||x · u + v||^{2} = (xu + v) · (xu + v) = x^{2}||u||^{2} + 2xu · v + ||v||^{2}. Since u and v are linearly independent, xu + v ≠ 0 and therefore f(x) > 0 for all x ∈ ℝ. This implies that f has negative discriminant, that is, 0 > disc f = 4(u · v)^{2} − 4||u||^{2}||v||^{2} or (u · v)^{2} < ||u||^{2}||v||^{2}, which implies the strict Cauchy-Schwarz inequality |u · v| < ||u|| · ||v||.

We now can define the angle α between two vectors u and v by

cos α = \frac{u · v}{||u|| · ||v||}.

A vector space V is called normed if there is a map || · || : V ⟶ ℝ with the following properties:

• ||x|| ≥ 0; ||x|| = 0 if and only if x = 0;

• ||λx|| = |λ| · ||x|| for λ ∈ ℝ;

• ||x + y|| ≤ ||x|| + ||y||.

Exercise. Any positive definite dot product on V gives rise to a norm via ||x|| = \sqrt{x · x}.

In particular, the norm of x in the usual (Euclidean) vector space ℝ^{n} is given by ||x|| = \sqrt{x_{1}^{2} + … + x_{n}^{2}}.

Exercise. Define ||x||_{0} = max\{|r|, |s|\} and ||x||_{1} = |x| + |y| for x = ( \begin{matrix}r \\ s \end{matrix}) ∈ ℝ^{2}.

Show that || · ||_{0} and || · ||_{1} define norms on ℝ^{2}.

Metric Spaces

Let X be a set of points. A function d : X ⟶ ℝ is called a metric on X if, for x, y, z ∈ X,

1. d(x, y) ≥ 0; d(x, y) = 0 if and only if x = y;

2. d(x, y) = d(y, x);

3. d(x, y) + d(y, z) ≥ d(x, z) (triangle inequality).

Exercise. Let (V, || · ||) be a normed vector space; show that d(x, y) = ||x − y|| defines a metric on V.

Curves

A curve c in ℝ^{n} is (the image of) a continuous map c : I ⟶ ℝ^{n} from some interval I in ℝ to ℝ^{n}. In other words: a curve consists of n continuous functions c^{i} : I ⟶ ℝ^{n} with c(t) = (c_{1}(t), …, c_{n}(t)). The curve c is called smooth if the coordinate functions c^{i} are differentiable (at least once, sometimes twice, sometimes infinitely often).

Given a smooth curve c, we define its velocity vector in t by

c′(t) = (c′_{1}(t), …, c′_{n}(t)).

A line is (part of) a curve c(t) = u + tv for fixed vectors u, v ∈ ℝ^{n}. With u = (u_{1}, …, u_{n}) and v = (v_{1}, …, v_{n}) we have c(t) = (c_{1}(t), …, c_{n}(t) with c_{i}(t) = u_{i} + tv_{i}. Note that c′(t) = v and c′′(t) = 0 (the acceleration vector vanishes).

Exercise. The curve c : [0, 2π) ⟶ ℝ^{2} given by c(t) = (r cos t, r sin t) is the circle with radius r and center 0. Compute c′(t), and draw a diagram including the tangent vectors c′(kπ) for k = 0, \frac{1}{2}, 1, \frac{3}{2}.

Preliminary Definition of Curvature

Consider a smooth function y = f(x). From calculus you know how to define (and compute) the slope of the function f: one starts with the slope of lines (which are easy to define), then approximates an (almost) arbtirary function in the vicinity of a point (x, f(x)) by a line, and if this can be done in a unique way (i.e., if f is differentiable) one defines the slope of f to be the slope of this unique line (called the tangent to f at P).

We now want to play the same game with curvature: just as the slope is easily defined for lines, we need curves to which we can easily attach a curvature. The simplest such curves seem to be circles; since the curvature of circles seems to diminish as their radius grows, we define the curvature κ (this is a Greek ‘k’) of a circle with radius r to be κ = \frac{1}{r}.

As above, we would like to measure curvature of f in P = (x, y) by comparing f with a circle that approximates f at P as good as possible. The idea is simple: the circle should pass through P, and it’s tangent in P should be the same as the tangent to f in P. A simple diagram shows that there should be infinitely many such circles. If, however, we demand that circle and function also have the same second derivative (“passing through P ” is a condition on the values of f and the circle, and “having the same tangent” is equivalent to demanding that the first derivatives agree).

The calculations are straightforward: write the equation of a circle with radius r and center (a, b) as

(x − a)^{2} + (y − b)^{2} = r^{2}.  (1.1)

Taking the derivative with respect to x yields 2(x − a) + 2(y − b)y′ = 0, that is,

x − a + y′(y − b) = 0.  (1.2)

Taking the derivative of (1.2) then yields

1 + y′′(y − b) + y′^{2} = 0.  (1.3)

Now (1.2) gives x − a = −y′(y − b), hence

r = \sqrt{(x − a)^{2} + (y − b)^{2}} = |y − b|\sqrt{1 + y′^{2}}.  (1.4)

Similarly, from (1.3) we get y − b = −\frac{1+y′^{2}}{y′′}. Thus

r = |y − b|\sqrt{1 + y′^{2}} = \frac{(1 + y′^{2})^{3/2}}{|y′′|} .

This allows us to define the curvature of f in (x, y) as the inverse of r, or by introducing a sign, as

κ = \frac{y′′}{(1 + y′^{2})^{3/2}} .

Exercise. Compute the curvature κ(x) of the parabola y = x^{2} in P = (x, y). Where is the curvature maximal?

Exercise. Compute the curvature κ(x) of the graph of the sine function y = sin x in P = (x, y). Where is the curvature maximal (minimal)?

The problem with this definition is that it is not clear how to generalize curvature to curves in higher dimension, and that we can’t apply the formulas to parametrized curves.

Jan 23, 2003

2.1 Arc Length

From calculus you know that the arc length of the graph of a smooth function y = f(x) between a and b is given by

L = ∫_{a}^{b} \sqrt{1 + f′(x)^{2}}dx.

As an example, consider the semicircle y = f(x) = \sqrt{1 − x^{2}}. We find 1 + f′(x)^{2} = \frac{1}{\sqrt{1−x^{2}}} , hence

L = ∫_{−1}^{1} \frac{dx}{\sqrt{1 − x^{2}}} = ∫_{−π/2}^{π/2} dt = π,

where we have substituted x = sin t. In particular, the circumference of the unit circle equals 2π, where π is the smallest positive zero of the sine function.

The need to cut the circle into two parts comes from our inability to consider parametrized curves (we need functions!). Thus our first job is to generalize the definition of arc length to parametrized curves, and then to check that the formulas agree in the case of curves that have a parametrization of the form (t, f(t)).

There’s another problem we have to deal with: the parametrizations

c : [0, 2π] ⟶ ℝ^{2} :  t ⟼ (cos t, sin t)

and

C : [0, π] ⟶ ℝ^{2} :  t ⟼ (cos 2t, sin 2t)

both parametrize the circle; the only difference is the speed with which the circle is traversed: the velocity of c in the point c(π) = (−1, 0) is c′(π) = (0, −1), whereas the velocity of C at the same point C(\frac{π}{2} ) = (−1, 0) is C′(\frac{π}{2} ) = (0, −2). Thus C runs through the circle with twice the speed of c.

Of course we expect that the arc length is independent of the chosen parametrization: but this needs to be proved.

The Riemann Integral

Recall how we defined the area below the graph of a curve: we partitioned the interval [a, b] into n parts (x_{0} = a < x_{1} < … < x_{n} = b) and approximated the area in the subinterval [x_{i}, x_{i+1}] by f(ξ_{i})(x_{i+1} − x_{i}) for some ξ_{i} ∈ (x_{i}, x_{i+1}). The approximation for the total area then was A ≈ ∑_{i=0}^{n−1} f(ξ_{i})(x_{i+1} − x_{i}), and for continuous functions this expression has a limit for n → ∞, namely A = ∫_{a}^{b} f(x)dx.

Arc Length of Graphs of Functions

Now let us similarly compute the length of the graph between a and b. The approximate length in the subinterval [x_{i}, x_{i+1}] turns out to be

\sqrt{(x_{i+1} − x_{i})^{2} + (f(x_{i+1}) − f(x_{i}))^{2}} = \sqrt{1 + (\frac{f(x_{i+1}) − f(x_{i})}{x_{i+1} − x_{i}})^{2}} · (x_{i+1} − x_{i});

from analysis we know that if f has a continuous derivative, then there is a ξ_{i} ∈ (x_{i}, x_{i+1}) such that \frac{f(x_{i+1})−f(x_{i})}{x_{i+1}−x_{i}} = f′(ξ_{i}). Thus the arc length is approximated by

L ≈ ∑_{i=0}^{n} g(ξ_{i})(x_{i+1} − x_{i}),

where g(ξ) = \sqrt{1 + f′(ξ)^{2}}, and by the theory of the Riemann integral this expression has the limit

L = ∫_{a}^{b} \sqrt{1 + f′(x)^{2}} dx

as we refine the partitions by letting n go to ∞. This expression is defined to be the arc lenght of the graph of a function with a continuous derivative.

The Arc Length of Curves

Now consider a smooth curve c : I = [a, b] ⟶ ℝ^{m}. In order to simplify the notation, let us assume that we are working in only three dimensions. In this case, we can put c(t) = (x(t), y(t), z(t)); we think of the curve as the path of an object whose position at the time t is given by (x(t), y(t), z(t)). In order to have a well defined tangent vector everywhere we will assume that the derivatives x′(t), y′(t), z′(t) do not vanish simultaneously.

We start with a partition a = t_{0} < t_{1} < … < t_{n} = b of the interval I and approximate the curve by the polygonal curve S_{n} defined by the points c(t_{0}), c(t_{1}), …, c(t_{n}). The Euclidean length of the segement between c(t_{i}) and c(t_{i+1}) is

L_{i} = ||c(t_{i+1}) − c(t_{i})||
= \sqrt{(x(t_{i+1}) − x(t_{i}))^{2} + (y(t_{i+1}) − y(t_{i}))^{2} + (z(t_{i+1}) − z(t_{i}))^{2}},

hence the length of the polygonal curve S_{n} is L(S_{n}) = ∑_{i=0}^{n−1} L_{i}. Since the coordinate functions x(t), y(t), z(t) have a continuous derivative, there exist ξ_{i}, η_{i}, ζ_{i} ∈ (t_{i}, t_{i+1}) (depending on the partition) such that

x(t_{i+1}) − x(t_{i})) = x′(ξ_{i})(t_{i+1} − t_{i}),
y(t_{i+1}) − y(t_{i})) = x′(η_{i})(t_{i+1} − t_{i}),
z(t_{i+1}) − z(t_{i})) = x′(ζ_{i})(t_{i+1} − t_{i}).

Thus we find

L(S_{n}) = ∑_{i=0}^{n−1} \sqrt{x′(ξ_{i})^{2} + y′(η_{i})^{2} + z′(ζ_{i})^{2}}(t_{i+1} − t_{i}).

But this is a Riemann sum, so if the functions occurring in these expressions are continuous (and they will be continuous if we assume that c(t) has a continuous first derivative), there exists a limit as we let the partitions become finer and finer, and we see that the length of the polygonal curve tends to

L = lim_{n→∞} L(S_{n}) = ∫_{a}^{b} \sqrt{x′(t)^{2} + y′(t)^{2} + z′(t)^{2}} dt,

which is called the arc length of the curve c. Since the expression we integrate over is nothing but \sqrt{c′(t) · c′(t)} = ||c′(t)||, we can write

L = ∫_{a}^{b} ||c′(t)||dt.

This agrees with the physical point of view that the distance an object travels is the integral over its speed (the absolute value of the velocity).

We can also define a function that measures the arc length as we move through the curve from t = a to b by putting

s(t) = ∫_{a}^{t} ||c′(u))|| du.

What happens if the curve is the graph of a function y = f(x)? Then we can choose the simple parametrization t ⟼ (t, f(t)); the coordinate functions are c_{1}(t) = t and c_{2}(t) = f(t), hence we find

L = ∫_{a}^{b} ||c′(t)|| dt = ∫_{a}^{b} \sqrt{1 + f′(t)^{2}}dt

2.1. ARC LENGTH 13 in agreement with our results above.

Consider the circle with radius r, as given by the parametrization

c : [0, 2π] ⟶ ℝ^{2} : t ⟼ (r cos t, r sin t).

We find that the circumference of the circle is L = ∫_{0}^{2π} \sqrt{r^{2} sin^{2} t + r^{2} cos^{2} t} dt = ∫_{0}^{2π} r dt = 2πr.

Jan 28, 2003

3.1 Reparametrizations

Every curve has lots of different parametrizations. For example,

c : [0, 2π] ⟶ ℝ^{2} :t ⟼ (cos t, sin t),
C : [0, 1] ⟶ ℝ^{2} :t ⟼ (cos 2πt, sin 2πt)

both parametrize the unit circle. We find

L(c) = ∫_{0}^{2π} ||c′(t)||dt = ∫_{0}^{2π} dt = 2π,
L(C) = ∫_{0}^{1} ||C′(t)||dt = ∫_{0}^{1} 2πdt = 2π.

Thus the arc length does not seem to depend on the parametrization (it wouldn’t be much of a notion of lenght if it would). We will now prove that the formula for arc length is independent of the parametrization of the curve.

Suppose C(T) is a smooth curve defined on J = [c, d]. We can introduce a new parameter τ by taking any smooth bijective function ϕ : I ⟶ J from another interval I = [a, b] to J such that ϕ′(t) > 0 on I. Then we define a parametrization c = c(t) by putting T = ϕ(t) and c(t) = C(ϕ(t)). Here’s a diagram with the maps we are dealing with:

I ⟶^{c} ℝ^{n}
↓^{ϕ}  ↓^{=}
J ⟶_{C} ℝ^{n}

By the chain rule we have c′(t) = \frac{d}{dt} c(t) = \frac{d}{dt} C(ϕ(t)) = \frac{d}{dT} C(T)\frac{d}{dt} ϕ(t), hence ||c′(t)|| = ||C′(T)ϕ′(t)|| = ||C′(T)||ϕ′(t) since ϕ′(t) > 0. Now recall that

L(c) = ∫_{a}^{b} ||\frac{d}{dt}c(t)|| dt, L(C) = ∫_{c}^{d} ||\frac{d}{dT} C(T)|| dT.

The substitution T = ϕ(t) gives

L(C) = ∫_{c}^{d} ||C′(T)|| dT = ∫_{a}^{b} ||C′(ϕ(t))|| ϕ′(t) dt = ∫_{a}^{b} ||c′(t)|| dt = L(c).

3.2 Geodesics

Now that we have defined the arc length of smooth curves, we can ask which are the curves of minimal length connecting any two given points. For the usual Euclidean space, this is the line:

Theorem 3.1. The shortest path between the points p, q ∈ ℝ^{n} is the line given by ℓ(t) = p + t(q − p).

Proof. Let us start with a very simple proof that lines are shortest paths: let c : [a, b] ⟶ ℝ^{n} be any smooth curve connecting p = c(a) and q = c(b); then we have, by the ‘triangle inequality’ for integrals:

∫_{a}^{b} ||c′(t)|| dt ≥ || ∫_{a}^{b} c′(t)dt|| = ||c(b) − c(a)|| = ||q − p|| = d(p, q),

so lines minimize the length.

We want to show more, however: namely that any shortest path is a line. To this end, let us first compute the length of the the line connecting p and q; it is given by ℓ : [0, 1] ⟶ ℝ^{n} : t ⟼ p + t(q − p). We find

L(ℓ) = ∫_{0}^{1} ||ℓ′(t)|| dt = ||q − p|| ∫_{0}^{1} dt = ||q − p||

as expected.

Now let c : [a, b] ⟶ ℝ^{n} be an arbitrary smooth curve with c(a) = p, c(b) = q and c′(t) ≠ 0. We want to show that L(c) ≥ L(ℓ). How can we measure the difference between c and ℓ? The vector c′(t) points in the direction of c, which will differ from the unit vector u = \frac{q−p}{||q−p||}; thus the dot product c′(t) · u will vanish if and only if c = ℓ.

So let us compute the ‘sum’ of all these dot products, that is, its integral. We find

∫_{a}^{b} c′(t) · u dt = ∫_{a}^{b}(c(t) · u)′ dt = c(b) · u − c(a) · u
= q · u − p · u = (q − p) · u = ||q − p|| = L(ℓ).

On the other hand we have, by ∫ f(t)dt ≤ ∫ |f(t)|dt and the inequality of Cauchy-Schwarz,

∫_{a}^{b} c′(t) · u dt ≤ ∫_{a}^{b} |c′(t) · u| dt ≤ ∫_{a}^{b} ||c′(t)|| · ||u|| dt
= ∫_{a}^{b} ||c′(t)|| dt = L(c),

hence L(ℓ) ≤ L(α). Moreover, we have equality in Cauchy-Schwarz if and only if u and c′(t) are always parallel, that is, if and only if c′(t) = λ(t)u for some realvalued function λ.

The condition c′(t) ≠ 0 implies that λ(t) ≠ 0, hence λ never changes sign. Now q − p = c(b) − c(a) = ∫_{a}^{b} c′(t)dt = ∫_{a}^{b} λ(t)udt = u ∫_{a}^{b} λ(t)dt. From q − p = u · ||q − p|| we deduce that ∫_{a}^{b} λ(t)dt = ||q − p|| > 0, and since λ never changes its sign, we must have λ(t) > 0 for t ∈ [a, b].

Thus if we have equality in Cauchy-Schwarz, then c′(t) = λ(t) · u for some positive function λ (which is continuous since c′ is). But then integration gives c(t) = r + Λ(t) · u for some integration constant r (plugging in t = a shows that p = r+Λ(a)·u) and a function Λ such that Λ′ = λ. But this is just the equation of a line segment; observe that Λ(t) = ∫_{a}^{t} λ(u)du, hence Λ(b)−Λ(a) = ∫_{a}^{b} λ(u)du = ||q − p||, and thus c(b) = r + Λ(b)u = p + Λ(b)u − Λ(a)u = p + ||q − p|| · u = q.

Finally, if c′(t) = λ(t) · u for some positive function λ, then c′(t) · u = λ(t)u · u = λ(t)||u||^{2} > 0, which implies we also have equality in the first inequality.

Jan 30, 2003

4.1 Parametrization by Arc Length

We now show that ‘regular’ curves may be reparametrized by arc length. A parametrization c(t) of a curve is called regular (or nonsingular) if c′(t) ≠ 0. The curve given by c(t) = (t^{2}, t^{3}) has c′(t) = (2t, 3t^{2}), which vanishes at t = 0 and therefore is not regular, although the pieces above and below the x-axis are.

We say that a curve c is parametrized by arc length if ∫_{a}^{b} ||c′(t)|| dt = |b − a| for all a, b in the domain of definition of c, that is, if the arc length of the curve between c(a) and c(t) is as long as the corresponding interval [a, t] for all t. For example, the unit circle c : [0, 2π] ⟶ ℝ^{2} : t ⟼ (cos t, sin t) is parametrized by arc length.

Lemma 4.1. A smooth curve c is parametrized by arc length if and only if we have ||c′(t)|| = 1 for all t.

Proof. Let c be parametrized by arc length, and let L(t) denote the length of the curve from c(a) to c(t); then L(t) = ∫_{a}^{t} ||c′(u)|| du = t − a for all t ≥ a. Thus L′(t) = \frac{d}{dt} (t − a) = 1, as well as L′(t) = \frac{d}{dt} ∫_{a}^{t} ||c′(u)|| du = ||c′(t)||, and comparing both formulas gives the result ||c′(t)|| = 1.

Conversely, if ||c′(t)|| = 1 for all u, then ∫_{a}^{s} ||c′(t)|| dt = ∫_{a}^{s} dt = s − a.

Now we claim

Proposition 4.2. Every regular curve can be parametrized by arc length.

Proof. Consider the smooth curve c : I ⟶ ℝ^{n}, pick some a ∈ I, and put s(t) = ∫_{a}^{t} ||c′(u)|| du. Now s′(t) = ||c′(t)|| > 0, and from calculus we know that the function s(t) has an inverse t(s) whose derivative with respect to s is given by \frac{d}{ds}t(s) = \frac{1}{\frac{d}{dt} s(t)} > 0. We now define a parametrization C : s ⟼ c(t(s)) and

find ||C′(s)|| = ||\frac{d}{ds}c(t(s))|| = ||c′(t)|| · ||\frac{d}{ds}t(s)|| = ||s′(t)||/||s′(t)|| = 1.

Example. Consider the helix c(t) = (a cos t, a sin t, bt). We find

s(t) = ∫_{0}^{t} ||c′(t)|| dt = ct

for c = \sqrt{a^{2} + b^{2}}. Thus t(s) = \frac{1}{c}s, and

C(s) = c(s/c) = (a cos \frac{s}{c}, a sin \frac{s}{c}, \frac{bs}{c})

is an arc length parametrization of the helix.

4.2 Frenet’s Formulas

Let c be a curve parametrized by arc length, that is, with unit speed: ||c′(s)|| = 1. Since c has unit speed, the change of c′(s) is completely due to a change of direction, and therefore should be a good measure of the curvature.

Let t = c′(s) denote the tangent vector at c(s). Taking the derivative of ||t||^{2} = t · t = 1, we find 0 = 2t · t′, so the vector t′ is perpendicular to t, and we define κ(s) = ||t′||.

For circles, this coincides with our old definition: the parametrization c(s) = (r cos \frac{s}{r}, r sin \frac{s}{r}) has unit speed, that is t = c′(s) = (− sin s, cos s) has length 1; moreover, t′(s) = (−\frac{1}{r} cos s, −\frac{1}{r} sin s) has length κ(s) = \frac{1}{r}.

For lines parametrized by ℓ(s) = u + sx with a direction vector x of length 1, we find t(s) = ℓ′(s) = x and therefore t′(s) = 0, so lines have vanishing curvature (the converse is also true).

If the curvature ||t′|| of a curve is ≠ 0, we can define n = \frac{t′}{||t′||}; this is a unit vector perpendicular to the tangent (think of it as an acceleration vector). Finally, if the curve lives in ℝ^{3}, we can form the vector product b = t × n. We find t · n = n · b = b · t = 0, as well as t · t = n · n = b · b = 1. Thus t, n and b form an orthonormal basis of ℝ^{3}. This implies that we should be able to express the derivatives of t, n and b as linear combinations of these vectors. We already know that t′ = κn. Taking the derivative of b · b = 1 shows that b′ · b = 0; taking the derivative of b · t = 0 we find 0 = b′ · t + b · t′ = b′ · t + b · n = b′ · t. Thus b′ is perpendicular to both b and t, and this implies that b′ is a multiple of n. We put b′ = −τn and call τ the torsion of the curve in c(s).

We will later see that c is a plane curve if and only if τ = 0: the torsion is a measure for how fast a curve winds out of the plane formed by the tangent and the normal vector.

Finally we have to take care of n′; since t, n and b form a basis, there are real numbers x, y, z such that n = xt + yn + zb; dotting with t gives x = t · t, and similarly we find y = n · n and z = b · b. Differentiating n · t = 0 we get n′ · t = −n · t′ = −n · κn = −κ. Moreover, n′ · n = 0 since n · n = 1. Finally, differentiating n · b = 0 gives n′ · b = −n · b′ = −n · (−τn) = τ.

Theorem 4.3. If c is a smooth curve in ℝ^{3} parametrized by arc length, then we have Frenet’s formulas

t′ =  κn
n′ = −κt  +τb
b′ =  −τn.

Let us apply these formulas to the circle c(s) = (r cos \frac{s}{r}, r sin \frac{s}{r}, 0). We have t = c′(s) = (− sin \frac{s}{r}, cos \frac{s}{r}, 0) and t′ = (−\frac{1}{r} cos \frac{s}{r}, −\frac{1}{r} sin \frac{s}{r}, 0) = −\frac{1}{r}n for the unit vector n = (cos \frac{s}{r}, sin \frac{s}{r}, 0). Next n′ = (−\frac{1}{r} sin \frac{s}{r}, \frac{1}{r} cos \frac{s}{r}, 0) = −\frac{1}{r}t; this implies τ = 0.

Feb 04, 2003

Cross Products

Let us recall the cross product in ℝ^{3}. Given the standard basis e_{1} = (1, 0, 0), e_{2} = (0, 1, 0) and e_{3} = (0, 0, 1), as well as two vectors u = (u_{1}, u_{2}, u_{3}) and v = (v_{1}, v_{2}, v_{3}), we define the vector

u × v = |\begin{matrix}e_{1} & e_{2} & e_{3} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix}| .

Then u × v = −v × u, and (u × v) · u = (u × v) · v = 0, that is, u × v is perpendicular to u and v. Moreover, ||u × v|| = ||u|| · ||v|| · sin θ, where θ is the angle spanned by u and v; equivalently, the length of u × v is the area of the parallelogram defined by u and v.

5.1 Frenet’s Formulas

Let us summarize the derivation of Frenet’s formulas. First, for a nonzero vector x, we call the vector x/||x|| the normalization of x: it has the same direction as x and length 1.

For a curve in ℝ^{3} parametrized by arc lenght s, let t(s) denote the tangent vector, which has length 1. The normalization n(s) of its derivative t′(s) is a unit vector perpendicular to t. Thus b(s) = t(s)×n(s) is a unit vector perpendicular to both t and n, which makes {t, n, b} into an orthonormal basis of ℝ^{3}.

Example 1. Consider the plane curve c : [0, 5] ⟶ ℝ^{2} : t ⟼ (3t − 2, 4t + 3). Note that c(0) = (−2, 3) = P and that, as the x-coordinate grows by 3, the y-coordinate grows by 4. Thus c is a line through P with slope \frac{4}{3}.

We find s(t) = ∫_{0}^{t} ||c′(u)|| du = 5t, thus t(s) = \frac{1}{5}s. In particular, the parametrization by arc length is given by C(s) = (\frac{3}{5}s − 2, \frac{4}{5}s + 3).

We have t(s) = c′(s) = (\frac{3}{5}, \frac{4}{5}) and n(s) = (0, 0). Thus b(s) = (0, 0), and κ = τ = 0.

Example 2. Consider the curve c(s) = (\frac{4}{5} cos s, 1 − sin s, −\frac{3}{5} cos s). This curve has unit speed: t(s) = c′(s) = (−\frac{4}{5} sin s, − cos s, \frac{3}{5} sin s) and ||t(s)|| = sin^{2}s+cos^{2} s = 1. Next n(s) = t′(s) = (−\frac{4}{5} cos s, sin s, \frac{3}{5} cos s), and ||t′(s)|| = 1, so κ(s) = 1 is constant. Finally n′(s) = (\frac{4}{5} sin s, cos s, −\frac{3}{5} sin s) = −t(s); on the other hand, Frenet’s formulas give n′ = −κt+τb, hence κ = 1 (we already knew that) and τ = 0.

Fact 5.1. For a a unit-speed curve c, we have κ = 0 if and only if c is a line segment.

Proof. Suppose c is a line segment; then it can be parametrized by c : (s) = p+su for some unit vector u. We find t(s) = u, hence 0 = t′ = κn, and therefore κ = 0.

Conversely, assume that κ(s) = 0. Then t′(s) = 0, hence t(s) = u for some constant vector 1. Integrating once more gives c(s) = p+su for some integration constant p.

Fact 5.2. If c is a unit-speed curve with κ > 0, then c is a plane curve if and only if τ = 0.

Proof. Suppose that τ = 0; then b′(s) = 0, so b(s) is a constant. This suggests that the curve lives in a plane perpendicular to b (in the plane spanned by tangent and normal). Let us prove this.

Consider the plane containing c(0) and perpendicular to b. A point p lies in this plane if and only if p − c(0) is perpendicular to b, that is, if and only if (p − c(0)) · b = 0. Thus we have to show that [c(s) − c(0)] · b(s) = 0.

First we show it is constant by forming the derivative; we get {[c(s) − c(0)] · b(s)}′ = [c(s) − c(0)]′ · b(s) + [c(s) − c(0)] · b′(s) = c′(s) · b(s) since b′ = 0. But c′(s) = t(s) and t(s) · b(s) = 0, hence the derivative of [c(s) − c(0)] · b(s) vanishes, and hence this product is a (real) constant. Plugging in s = 0 we find this constant is 0, hence [c(s) − c(0)] · b(s) = 0 as claimed.

Conversely, assume that c lies in a plane; then there is a unit vector u such that [c(s) − c(0)] · u = 0. Differentiating we get c′(s) · u = 0 and c′′(s) · u = 0, that is, t · u = n · u = 0. This implies that b and u are parallel, and since they are unit vectors, we conclude that b = ±u. Thus b is constant and b′(s) = 0, thus τ = 0.

Fact 5.3. If c is a unit-speed curve with κ > 0, and torsion τ = 0, then c is part of a circle with radius r = 1/κ.

Feb 06, 2003

6.1 Application of Frenet’s formulas

Fact 6.1. For a a unit-speed curve c, we have κ = 0 if and only if c is a line segment.

Proof. Suppose c is a line segment; then it can be parametrized by c : (s) = p+su for some unit vector u. We find t(s) = u, hence 0 = t′ = κn, and therefore κ = 0.

Conversely, assume that κ(s) = 0. Then t′(s) = 0, hence t(s) = u for some constant vector 1. Integrating once more gives c(s) = p+su for some integration constant p.

Fact 6.2. If c is a unit-speed curve with κ > 0, then c is a plane curve if and only if τ = 0.

Proof. Suppose that τ = 0; then b′(s) = 0, so b(s) is a constant. This suggests that the curve lives in a plane perpendicular to b (in the plane spanned by tangent and normal). Let us prove this.

Consider the plane containing c(0) and perpendicular to b. A point p lies in this plane if and only if p − c(0) is perpendicular to b, that is, if and only if (p − c(0)) · b = 0. Thus we have to show that [c(s) − c(0)] · b(s) = 0.

First we show it is constant by forming the derivative; we get {[c(s) − c(0)] · b(s)}′ = [c(s) − c(0)]′ · b(s) + [c(s) − c(0)] · b′(s) = c′(s) · b(s) since b′ = 0. But c′(s) = t(s) and t(s) · b(s) = 0, hence the derivative of [c(s) − c(0)] · b(s) vanishes, and hence this product is a (real) constant. Plugging in s = 0 we find this constant is 0, hence [c(s) − c(0)] · b(s) = 0 as claimed.

Conversely, assume that c lies in a plane; then there is a unit vector u such that [c(s) − c(0)] · u = 0. Differentiating we get c′(s) · u = 0 and c′′(s) · u = 0, that is, t · u = n · u = 0. This implies that b and u are parallel, and since they are unit vectors, we conclude that b = ±u. Thus b is constant and b′(s) = 0, thus τ = 0.

Fact 6.3. If c is a unit-speed curve with κ > 0, and torsion τ = 0, then c is part of a circle with radius r = 1/κ.

6.2. FRENET’S FORMULAS FOR CURVES WITH ARBITRARY SPEED23

Proof. If we believe that curvature and torsion are invariant under isometries, this follows from what we did for the plane circle. Now assume that τ = 0 and κ > 0 is constant

6.2 Frenet’s Formulas for curves with arbitrary speed

Theorem 6.4. Let c be a regular curve in ℝ^{3}. Then

T = \frac{c′}{||c′||}, B = \frac{c′ × c′′}{||c′ × c′′||}, N = B × T,

and

κ = \frac{||c′ × c′′||}{||c′||^{3}}, τ = \frac{(c′ × c′′) · c′′′}{||c′ × c′′||^{2}} .

6.3 Euclidean Geometry

Consider the Euclidean plane ℝ^{2}; the distance of two points P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}) is defined to be

d(P_{1}, P_{2}) = \sqrt{(x_{1} − x_{2})^{2} + (y_{1} − y_{2})^{2}}.

Let ϕ : ℝ^{2} ⟶ ℝ^{2} be an isometry, that is, a continuous map that preserves distances:

d(P_{1}, P_{2}) = d(ϕ(P_{1}), ϕ(P_{2})).

Clearly translations (maps defined by T_{a,b} : ((x_{1}, x_{2})) = (x_{1} + a, x_{2} + b) for constants a, b ∈ ℝ) preserve distances. An arbitrary isogeny ϕ will map the origin O = (0, 0) to some point ϕ(O) = (a, b); thus T_{−a,−b} ∘ϕ will be an isometry fixing the origin.

Proposition 6.5. The rotation

R_{α} = (\begin{matrix}cos α & sin α \\ − sin α & cos α \end{matrix})

about an angle α is an isometry fixing O. Moreover, we have R_{α}R_{β} = R_{β}R_{α} = R_{αβ} and R_{α}^{−1} = R_{−α}.

Proof. A simple geometric argument shows that rotations are given by matrices such as those above. Since linear maps fix the origin, we only have to show that R_{α} is an isometry, i.e., that ∥ R_{α}( \begin{matrix}x \\ y \end{matrix}) ∥=∥ ( \begin{matrix}x \\ y \end{matrix}) ∥. But this is straightforward using cos^{2} α + sin^{2} α = 1.

Feb 11, 2003

7.1 The hyperbolic plane

Let ℂ be the complex plane and ℍ = {s ∈ ℂ : s = x + iy, y > 0} be the upper half plane. We will make ℍ into a metric space by defining the ‘metric’

ds = \frac{1}{y} \sqrt{dx^{2} + dy^{2}}.

This means that we define the length of a smooth curve c : [a, b] ⟶ ℝ^{2} ≃ ℂ : c(t) = x(t) + iy(t) by

L(c) = ∫_{a}^{b} \frac{\sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2}}}{y(t)} dt.  (7.1)

Let us see what this means for the line c : [ε, 1] ⟶ ℂ : c(t) = r + it connecting (r, ε) and (r, 1). Observing that x(t) = r and y(t) = t we get

L(c) = ∫_{ε}^{1} \frac{dt}{t} = log 1 − log ε = − log ε.

This means that the length of these lines tends to infinity as ε goes to 0. We now make ℍ into a metric space by defining

d(z, w) = inf L(c),

where the infimum is over all piecewise smooth curves connecting z and w. Of course we have to check a few things.

• d(z, w) ≥ 0: looking at (7.1), observe that the integral is over a nonnegative function since y(t) > 0 for curves in ℍ. Also, equality implies that x(t) and y(t) are constant, and since c(t) = x(t) + iy(t) connects z and w, we must have z = w.

• d(z, w) = d(w, z): any curve connecting z to w also connects w to z.

• The triangle inequality d(z, w) ≤ d(z, v)+d(v, w): Let c and c′ be piecewise smooth curves connecting z to v and v to w. ‘Composing’ these paths gives a piecewise smooth curve C connecting z to w with length L(C) = L(c) + L(c′). This implies that inf{L(C)} = inf L(c) + inf L(c′). we have inf L(c) = d(z, v) and inf L(c′) = d(v, w); moreover, inf L(C) ≥ d(z, w) since the infimum is not over all piecewise smooth curves connecting z and w but only over those that pass through v. This implies that d(z, w) ≤ d(z, v) + d(v, w).

The upper half plane equipped with this metric is called the hyperbolic plane.

Note that the infimum need not be a minimum. In order to see this, consider the space X = ℝ^{2}\{(0, 0)} with the usual Euclidean distance. Then the infimum over the lengths of all piecewise smooth curves connecting (−1, 0) and (0, 1) is 2, but there is no piecewise smooth curve of length 2 in X connecting the two points.

Complex Numbers

For z = x + iy ∈ ℂ and x, y ∈ ℝ, we call x the real part of z (and write x = Re z) and y the imaginary part of z (and write y = Im z). Moreover, \overline{z} = x−iy is called the complex conjugate of z. It is easily seen that Re z = \frac{z+\overline{z}}{2} and Im z = \frac{z−\overline{z}}{2i}. Moreover, the absolute value of a complex number is defined by |z| = \sqrt{z\overline{z}} = \sqrt{x^{2} + y^{2}}; if we identify the complex plane with ℝ^{2} via a+bi ↔ (a, b), then the absolute value of z = a + bi is the Euclidean distance of (a, b) to the origin.

Moebius Transformations

A map z ⟼ \frac{az+b}{cz+d} defined by a, b, c, d ∈ ℝ with ad − bc = 1 is called a Moebius transformation. This is a well defined map on the upper half plane: in fact, if cz + d = 0, then either c = 0 and d = 0, contradicting the fact that ad − bc = 1, or c ≠ 0, and in this case z = −\frac{d}{c} is a real number, that is, not an element of the upper half plane.

Fact 7.1. Moebius transformations map ℍ onto itself.

Proof. Assume that z ∈ ℍ. Then

w = \frac{az + b}{cz + d} = \frac{(az + b)(c\overline{z} + d)}{|cz + d|^{2}} = \frac{ac|z|^{2} + adz + bc\overline{z} + bd}{|cz + d|^{2}},

hence

Im(w) = \frac{w − w}{2i} = \frac{z − \overline{z}}{2i|cz + d|^{2}} = \frac{Im(z)}{|cz + d|^{2}} .

This shows that Im(w) > 0 and thus w ∈ ℍ.

Feb 13, 2003

8.1 PSL_{2}(ℝ)

If we attach the matrix ( \begin{matrix}a & b \\ c & d \end{matrix} ) to the Moebius transformation z ⟼ \frac{az+b}{cz+d}, then each Moebius transformation corresponds to two matrices, namely ±( \begin{matrix}a & b \\ c & d \end{matrix} ). Algebraically this means that the group of Moebius transformations is isomorphic to the group PSL_{2}(ℝ) = SL_{2}(ℝ)/{± id}, where

SL_{2}(ℝ) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) : a, b, c, d ∈ ℝ, ad − bc = 1\}.

This explains why we are going to denote the group of Moebius transformations by PSL_{2}(ℝ).

There are two Moebius transformations with a very simple description: the matrix ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) describes the transformation z ⟼ z + b, hence is a simple horizontal translation. The other is given by the matrix ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) corresponding to the map z ⟼ −\frac{1}{z}.

The two matrices above have integral entries, hence are elements of the subgroup SL_{2}(ℤ). As a matter of fact, it can be shown that they generate the whole group, that is, every element in SL_{2}(ℤ) can be written as a product of these two matrices. Modular functions are functions defined on ℍ that have a power series expansion and are invariant under the action of SL_{2}(ℤ) (this means that f(T (z)) is ‘almost’ the same as f(z) for all T ∈ SL_{2}(ℤ)). The essential step in Wiles’ proof of Fermat’s Last Theorem was to show that every curve y^{2} = x^{3} + ax + b with a, b ∈ ℚ can be ‘parametrized’ by a modular function.

8.2 Complex Analysis

A function f(z) of a complex variable z = x + iy defined on an open subset U ⊆ ℂ is said to be holomorphic at z ∈ U is the limit

f′(z) := lim_{h→0} \frac{f(z + h) − f(z)}{h}

exists.

Of course we can write f(z) = u(z) + iv(z), where u and v are the real and the imaginary part of f; we also can interpret u and v as functions of two real variables by identifying z = x + iy ∈ ℂ with (x, y) ∈ ℝ^{2}. Then f(z) = u(x, y) + iv(x, y). Now there are many ways in which we can h let go to 0 in the definition of the derivative f′(z); one possibility is to let pick h = a real: then

lim_{h→0} \frac{f(z + h) − f(z)}{h} = lim_{h→0} \frac{u(z + h) + iv(z + h) − u(z) − iv(z)}{h}

= lim_{a→0} \frac{u(x + a, y) − u(x, y)}{a} + i lim_{a→0} \frac{v(x + a, y) − v(x, y)}{a}

= \frac{∂}{∂x}u(x, y) + i\frac{∂}{∂x}v(x, y).

On the other hand, we might as well pick h = ib on the imaginary axis; then

lim_{h→0} \frac{f(z + h) − f(z)}{h} = lim_{h→0} \frac{u(z + h) + iv(z + h) − u(z) − iv(z)}{h}

= lim_{b→0} \frac{u(x, y + b) − u(x, y)}{ib} + i lim_{b→0} \frac{v(x, y + b) − v(x, y)}{ib}

= −i\frac{∂}{∂y} u(x, y) + \frac{∂}{∂y} v(x, y).

Comparing real and imaginary parts we see that

\frac{∂}{∂x}u(x, y) = \frac{∂}{∂y} v(x, y), \frac{∂}{∂y} u(x, y) = −\frac{∂}{∂x}v(x, y).

These equations are called the Cauchy-Riemann equations. We have proved that real and imaginary parts of holomorphic functions satisfy the Cauchy-Riemann equations; conversely, a function f(z) = u(z)+iv(z) is holomorphic if and only if u and v are differentiable in the real sense and if the Cauchy-Riemann equations are satisfied.

Example. The function f(z) = \overline{z} is not holomorphic. In fact, f(z) = u(x, y) + iv(x, y) for u(x, y) = x and v(x, y) = −y. Thus

\frac{∂}{∂x}u(x, y) = +1, \frac{∂}{∂y} v(x, y) = −1, \frac{∂}{∂y} u(x, y) = 0, \frac{∂}{∂x}v(x, y) = 0.

Proposition 8.1. Let f_{1} and f_{2} be holomorphic (and defined on a common open set U). Then g = f_{1}f_{2} is holomorphic, and g′ = f′_{1}f_{2} + f_{1}f′_{2}. Moreover, we have \frac{d}{dz} f(w(z)) = \frac{df}{dw} \frac{dz}{dw}. whenever f and w are holomorphic and defined on suitable open sets.

Feb 18, 2003

9.1 Isometries

Fact 9.1. PSL_{2}(ℝ) ⊆ Isom(ℍ): Moebius transformations are isometries.

Proof. Let z : [r, s] ⟶ ℂ be a piecewise smooth curve and T ∈ PSL_{2}(ℝ). With z(t) = x(t) + iy(t), the curve w(t) = T (z(t)) = u(t) + iv(t) is piecewise smooth by the chain rule. We have

L(z) = ∫_{r}^{s} \frac{\sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2}}}{y(t)} dt, L(w) = ∫_{r}^{s} \frac{\sqrt{(\frac{du}{dt})^{2} + (\frac{dv}{dt})^{2}}}{v(t)} dt

Now (\frac{du}{dt})^{2} + (\frac{dv}{dt})^{2} = |\frac{dw}{dt} |^{2} and \frac{dw}{dt} = \frac{dw}{dz} \frac{dz}{dt}, where

\frac{d}{dz} w(t) = \frac{d}{dz} \frac{az + b}{cz + d} = \frac{a(cz + d) − c(az + b)}{(cz + d)^{2}} = \frac{1}{(cz + d)^{2}} .

Moreover v = \frac{y}{(cz+d)^{2}} , so |\frac{dw}{dt} |^{2} = |\frac{dz}{dt} |^{2} \frac{1}{(cz+d)^{2}} = |\frac{dz}{dt} |^{2} \frac{v}{y} or \frac{1}{v} |\frac{dw}{dt} |^{2} = \frac{1}{y} |\frac{dz}{dt} |^{2}.

Exercise: Let z : I ⟶ ℍ : t ⟼ x(t) + iy(t) be a smooth curve in ℍ. Show that w(t) = −\frac{1}{z(t)} is a smooth curve in ℍ, and check directly that it has the same length as z.

9.2 Geodesics

Let us now consider the problem of existence and classification of geodesics in the hyperbolic plane. Observe that there are metric spaces in which not all points are connected by geodesics: consider e.g. X = ℝ^{2} \ {(0, 0)} equipped with the Euclidean metric. Then there is no shortest path connecting (−1, 0) and (1, 0). In fact, defining the distance of two points as above by the infimum over the length of all piecewise smooth curves connecting these points, we can show that (−1, 0) and (1, 0) have distance 2; yet there is no curve with distance 2 connecting them.

Fact 9.2. The geodesics of the hyperbolic plane are lines or semicircles perpendicular to the real axis.

Proof. Assume first that z_{1} = ia and z_{2} = ib for 0 < a < b. Let c : [0, 1] ⟶ ℍ is any piecewise smooth path connecting z_{1} to z_{2}, and if we write c(t) = x(t)+iy(t), then

L(c) = ∫_{0}^{1} \frac{\sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2}}}{y(t)} dt ≥ ∫_{0}^{1} \frac{|\frac{dy}{dt} |}{y(t)} dt ≥ ∫_{0}^{1} \frac{\frac{dy}{dt}}{y(t)} dt = ∫_{0}^{1} \frac{dt}{y(t)} = ln \frac{b}{a}.

On the other hand, ln \frac{b}{a} is the length of the ‘line’ connecting z_{1} and z_{2}.

Actually it is sufficient to prove the last fact only for points on the imaginary axis:

Fact 9.3. Let L be a straight line or a circle (in the usual Euclidean sense), meeting the real line orthogonally at a point α (pick one of the two in the case of a circle). Show that T (z) = −\frac{1}{z−α} + β is a Moebius transformation, and that for a suitable β it maps L to the imaginary axis.

This was proved in the homework.

Corollary 9.4. Any two points z, w ∈ ℍ in the upper half plane can be joined by a unique geodesic, and d(z, w) is equal to the length of the corresponding segment of this geodesic.

Euclid’s axioms:

1. It is possible to draw one and only one straight line from any point to any point.

2. From each end of a finite straight line it is possible to produce it continuously in a straight line by an amount greater than any assigned length.

3. It is possible to describe one and only one circle with any centre and radius.

4. All right angles are equal to one another.

5. If two straight lines lying in a plane are met by another line, and if the sum of the internal angles on one side is less than two right angles, then the straight lines will meet if extended on the side on which the sum of the angles is less than two right angles.

Euclid’s version was occasionally replaced by other versions, like “For each line and a point not on the line, there is exactly one parallel to the line through that point.”

Contrary to what some websites are saying, these two versions are not equivalent!

If you define lines in the hyperbolic plane ℍ as geodesics, then the first four axioms clearly hold in ℍ; the fifth axiom, however, is not true, since (in the second version) there are infinitely many parallels (two hyperbolic lines are parallel if they do not intersect) through any point not on a given line. In particular, this implies that the fifth axiom cannot be proved from the first four axioms (make sure you understand why!).

Feb 20, 2003

10.1 Areas

Let us now briefly discuss the notion of areas in hyperbolic geometry. Take a small ‘rectangle’ whose vertices have coordinates x+iy, x+Δx+iy, x+i(y+Δy), and x + Δx + i(y + Δy). The length of its sides are Δx and Δy in Euclidean, but approximately \frac{Δx}{y} and \frac{Δy}{y} in the hyperbolic plane. Thus the area of small rectangles should be approximately \frac{Δx·Δy}{y^{2}} . This leads us to define the area μ(A) of a figure A to be

A = ∫_{A} \frac{dx dy}{y^{2}}

if this integral exists.

Fact 10.1. The area is invariant under Moebius transformations.

Proof. Write z = x + iy and T (z) = \frac{az+b}{cz+d} with ad − bc = 1, and put w = T (z) = u + iv. The Moebius transformation can be viewed as a differentiable map T : ℝ^{2} ⟶ ℝ^{2} (restricted to the ‘upper half planes’).

From multivariable calculus we know that

∫ f(y)dy = ∫ f(T (x))|J_{T} | dx

, where |J_{T} | denotes the Jacobian of the substitution T.

The Jacobian of the Moebius transformation T is

|J_{T} | = det T′ = |\begin{matrix}\frac{∂u}{∂x} & \frac{∂u}{∂y} \\ \frac{∂v}{∂x} & \frac{∂v}{∂y} \end{matrix} | = \frac{∂u}{∂x} \frac{∂v}{∂y} − \frac{∂u}{∂y} \frac{∂v}{∂x}.

Using the Cauchy-Riemann equations, we find

|J_{T} | = (\frac{∂u}{∂x})^{2} + (\frac{∂v}{∂y} )^{2} = |\frac{dT}{dz} |^{2} = \frac{1}{|cz + d|^{4}} .

32 CHAPTER 10. FEB 20, 2003

Thus μ(T (A)) = ∫_{T (A)} \frac{du dv}{v^{2}} = ∫_{A} |J_{T} |\frac{|cz + d|^{4}}{y^{2}} dx dy = ∫_{A} \frac{dx dy}{y^{2}} = μ(A).

As an application, let us compute the area of the following ‘triangle’ (it is bounded by three hyperbolic lines, namely two lines and a semicircle orthogonal to the real axis):

α α A β β

Assume that the vertical lines are given by x = a and x = b. Then

μ(A) = ∫_{A} \frac{dx dy}{y^{2}} = ∫_{a}^{b} dx ∫_{√\overline{1−x^{2}}}^{∞} \frac{dy}{y^{2}} = ∫_{a}^{b} \frac{dx}{\sqrt{1 − x^{2}}} .

Using the substitution x = cos t we find

μ(A) = − ∫_{π−α}^{β} dt = π − α − β.

More generally, it can be proved

Theorem 10.2. A hyperbolic triangle with angles α, β and γ has area

μ(A) = π − α − β − γ.

Thus the angles of a triangle determine its area (this is horribly false in Euclidean geometry), and the sum of the angles in hyperbolic triangles is always < 180^{o}. There is, however, an analog in spherical geometry:

Theorem 10.3. A triangle on the unit sphere with angles α, β and γ has area

μ(A) = α + β + γ − π.

Note that spherical triangles are formed using spherical lines, that is, great circles.

Both of the theorems above are very special cases of a general result called the Gauss-Bonnet theorem.

March 11, 2003

11.1 Kepler’s Laws

For the ancient Greeks, the orbits on which planets moved had to be circles for philosophical reasons: the heavens were perfect, and the most perfect curve is the circle. It was Johannes Kepler (1571–1630) who realized that observations were better explained by elliptical orbits, and after carefully analyzing the (for their time) extraordinarily exact observations of Tycho Brahe (1546–1601), he formulated the following three laws:

1. Planets orbit the sun in ellipses, with one focus at the center of the sun.

2. The line connecting this focus and the planet sweeps out equal areas in equal intervals of time.

3. The square of the period of the orbit (the time a planet needs for going around once) is proportional to the cube of the length of the orbits major axis.

11.2 Newton’s Laws of Motion

Newton’s laws say that

• objects on which no force acts travel in straight lines and at constant speed;

• if a force F acts on an object with mass m, then its acceleration a is given by F = ma;

• if two objects exert forces on each other, the forces are equal in magnitude and opposite in direction (actio = reactio).

Isaac Newton (1643–1727) realized that Kepler’s laws could be explained by assuming the Law of Gravitation: the force F between two masses m and M with distance r is given by

F = G\frac{mM}{r^{2}} ,

where G is some constant (the gravitational constant). Looking back from our times to Newton’s, it is hard to underestimate the importance attached to Newtons law of gravitation by his contemporaries: essentially it was viewed as a formula that governed the known universe. Later, the electro-magnetic world was discovered, the equivalent of Newton’s law being Maxwell’s equations.

Let us now show that Newton’s law of gravitation explains all three of Kepler’s laws (with a slight correction: it is not the center of the sun that is at one of the foci but the center of mass of sun and planet). To this end, we consider two objects with masses m and M; we assume that they are isolated and affected by no forces except each other’s gravity. Put

\overrightarrow{r} = the vector pointing from 0 to m,
\overrightarrow{f} = the force on m,
\overrightarrow{R} = the vector pointing from 0 to M,
\overrightarrow{F} = the force on M,

and let r = |\overrightarrow{r}|, R = |\overrightarrow{R}|, f = |\overrightarrow{f}|, F = |\overrightarrow{F}|. Note that we regard all these vectors as functions of the time t. The center of mass is the point

\frac{m\overrightarrow{r} + M\overrightarrow{R}}{m + M}.

Check that, for any nonnegative values of m and M, this point lies on the straight line between the points defined by \overrightarrow{r} and \overrightarrow{R}.

Fact 11.1. There is no gravitational force acting on the center of mass.

Proof. By Newton’s second law of motion we have \overrightarrow{f} = m\overrightarrow{r}′′ and \overrightarrow{F} = M\overrightarrow{R}′′. The third law then tells us that \overrightarrow{F} = −\overrightarrow{f}, that is, m\overrightarrow{r}′′ + M\overrightarrow{R}′′ = 0.

In order to simplify our calculations we will now choose a coordinate system with the center of mass at the origin. Thus we have

m\overrightarrow{r} + M\overrightarrow{R} = 0.

According to Newton’s law of gravitation we have

f = G\frac{mM}{(r + R)^{2}} .

Since the vector \overrightarrow{f}, the gravitational force on the object m, points from m towards M, we see that the direction of f is the opposite of the direction of \overrightarrow{r}.

Kepler’s Second Law

Consider the area swept out by \overrightarrow{r} as t varies over a small time interval Δt. The area ΔA swept out by \overrightarrow{r} is approximately equal to the area of the triangle with sides \overrightarrow{r} and Δ\overrightarrow{r} = \overrightarrow{r}(t + Δt) − \overrightarrow{r}(t), hence

ΔA ≈ \frac{1}{2}|\overrightarrow{r} × Δ\overrightarrow{r}|.

Since Δt is small, we have

\frac{Δr}{Δt} ≈ \overrightarrow{r}′,

so

ΔA ≈ \frac{1}{2}|\overrightarrow{r} × \overrightarrow{r}′|Δt.

Dividing through by Δt and letting Δt ⟶ 0, we get

\frac{dA}{dt} = \frac{1}{2}|\overrightarrow{r} × \overrightarrow{r}′|.

Kepler’s claim was that \frac{dA}{dt} is constant; to see this, differentiate the right hand side. We get

\frac{d}{dt}\overrightarrow{r} × \overrightarrow{r}′ = \overrightarrow{r}′ × \overrightarrow{r}′ + \overrightarrow{r} × \overrightarrow{r}′′ = \overrightarrow{r} × \overrightarrow{r}′′ = \overrightarrow{r} × \frac{1}{m}\overrightarrow{f}

by Newton’s second law. Since \overrightarrow{f} is parallel to \overrightarrow{r}, the last cross product vanishes, and we are done.

The orbit lies in a plane.

Now put

\overrightarrow{N} = \overrightarrow{r} × \overrightarrow{r}′;  (11.1)

then \overrightarrow{r} points from 0 to m, and \overrightarrow{r} is perpendicular to \overrightarrow{N}. Thus m lies in the plane perpendiular to \overrightarrow{N}, and since \overrightarrow{N} is constant, this plane contains the whole orbit of m.

Remark. The quantity \overrightarrow{r}×\overrightarrow{r}′ is called the angular momentum in physics. It is, like energy, a preserved quantity: it is therefore no surprise that we can prove it is constant.

Kepler’s First Law

Let us now choose coordinates in such a way that the orbit lies in the x, y-plane. Let \overrightarrow{i}, \overrightarrow{j} and \overrightarrow{k} = \overrightarrow{i} × \overrightarrow{j} be unit vectors pointing in the direction of the x-, y- and z-axis. Write

\overrightarrow{r} = (r cos θ)\overrightarrow{i} + (r sin θ)\overrightarrow{j} = r(cos θ\overrightarrow{i} + sin θ\overrightarrow{j}),  (11.2)

where θ is the angle between \overrightarrow{r} and the x-axis. We find

\overrightarrow{r}′ = r′(cos θ\overrightarrow{i} + sin θ\overrightarrow{j}) + rθ′(− sin θ\overrightarrow{i} + cos θ\overrightarrow{j}),  (11.3)
\overrightarrow{r}′′ = (r′′ − r(θ′)^{2})(cos θ\overrightarrow{i} + sin θ\overrightarrow{j}) + (2r′θ′ + rθ′′)(− sin θ\overrightarrow{i} + cos θ\overrightarrow{j}). (11.4)

Combining (11.1), (11.2), (11.3) we get after a simple calculation

(r^{2}θ′)\overrightarrow{k} = \overrightarrow{N}.

In particular, r^{2}θ′ is constant. Setting a = ||\overrightarrow{N}|| we get θ′ = ar^{−2}, where a is a constant equal to twice the rate of change of the area swept out by \overrightarrow{r}.

Differentiating a = r^{2}θ′ and dividing through by r gives 0 = 2r′θ′ + rθ′′. Plugging this into (11.4) gives

\overrightarrow{r}′′ = (r′′ − r(θ′)^{2})(cos θ\overrightarrow{i} + sin θ\overrightarrow{j}).

Plugging this into Newton’s second law and observing that

\frac{GMm}{(r + R)^{2}} = \frac{1}{r^{2}} \frac{GMm}{(1 + \frac{m}{M})^{2}}

we get

m(r′′ − r(θ′)^{2}) = −\frac{1}{r^{2}} \frac{GMm}{(1 + \frac{m}{M})^{2}} .

It remains to solve this differential equation.

March 13, 2003

12.1 Kepler’s Second Law

First we use the chain rule to replace derivatives with respect to t by derivatives with respect to θ.

\frac{d^{2}r}{dt^{2}} = \frac{d}{dt}(\frac{dr}{dθ} \frac{dθ}{dt} ) = \frac{d}{dt}(\frac{dr}{dθ} \frac{a}{r^{2}} )

= \frac{d}{dt}(\frac{dr}{dθ} )\frac{a}{r^{2}} + \frac{dr}{dθ} \frac{d}{dt}(\frac{a}{r^{2}} )

= \frac{d}{dθ} (\frac{dr}{dθ} )\frac{dθ}{dt} \frac{a}{r^{2}} + \frac{dr}{dθ} (\frac{−2a}{r^{3}} \frac{dr}{dt} )

= \frac{d^{2}r}{dθ^{2}} (\frac{a}{r^{2}} )^{2} + \frac{dr}{dθ} (\frac{−2a}{r^{3}} \frac{dr}{dθ} \frac{dθ}{dt} )

= \frac{d^{2}r}{dθ^{2}} \frac{a^{2}}{r^{4}} − 2(\frac{dr}{dθ} )^{2} \frac{a^{2}}{r^{5}} .

Thus

\frac{d^{2}r}{dθ^{2}} \frac{a^{2}}{r^{4}} − 2(\frac{dr}{dθ} )^{2} \frac{a^{2}}{r^{5}} − \frac{a^{2}}{r^{3}} = −\frac{1}{r^{2}} \frac{GMm}{(1 + \frac{m}{M})^{2}} .

Multiplying through by −r^{2}/a^{2} we get a second order equation governing the orbit of m:

−\frac{1}{r^{2}} \frac{d^{2}r}{dθ^{2}} + \frac{2}{r^{3}} (\frac{dr}{dθ} )^{2} + \frac{1}{r} = \frac{GM}{a^{2}(1 + \frac{m}{M})^{2}} .

This equation can be rewritten as

\frac{d^{2}}{dθ^{2}} (\frac{1}{r}) + \frac{1}{r} = \frac{GM}{a^{2}(1 + \frac{m}{M})^{2}} .

This is a linear second order differential equation with constant coefficients, and its general solution is

\frac{1}{r} = \frac{GM}{a^{2}(1 + \frac{m}{M})^{2}} − B cos(θ − θ_{0}) = \frac{1 − e cos(θ − θ_{0})}{p},  (12.1)

where B and θ_{0} are constants, and

p = \frac{a^{2}(1 + \frac{m}{M})^{2}}{GM} , e = pB.

This is the equation of a conic with a focus at the origin, parameter p, and eccentricity e.

Conics

Geometrically, a conic C may be defined as follows: given a point F (called the focus of C), a line d not containing F (called the directrix of C), and a nonnegative number e ≥ 0 (called the excentricity of C), a conic is the set of all points P such that the distance from P to F is e times the distance from P to d: \overline{P F} = e · dist(d, F).

Using a cartesian coordinate system, it is easy to derive the equations \frac{x^{2}}{a^{2}} + \frac{x^{2}}{b^{2}} = 1 for conics with 0 ≤ e < 1, y^{2} = 2px for e = 1, and \frac{x^{2}}{a^{2}} − \frac{x^{2}}{b^{2}} = 1 for e > 1. Using polar coordinates, on the other hand, we find the equation (12.1) as follows.

• F  d • P• R  • Q • S

We have \overline{F P} = r, \overline{F R} = r cos θ, and by definition of the conic we have r = e\overline{P Q} = e\overline{RS} = e\overline{F S} − e\overline{F R}. Now \overline{F R} = r cos θ, hence r(1 − e cos θ) = e\overline{F S} = p for some constant p (the distance of the focus F from the directrix is constant). This shows that r = \frac{p}{1−e cos θ}.

Kepler’s Third Law

Write t as a function of θ; as θ runs from a to b, the time changes by

Δt = ∫_{w}^{z} \frac{dt}{dθ} dθ.

Since \frac{dt}{dθ} = \frac{r^{2}}{a}, the period of any closed orbit (this only applies to ellipses) is

T = ∫_{0}^{2π} \frac{r^{2}}{a} dθ.

On the other hand, the formula for area in polar coordinates is

A = \frac{1}{2} ∫_{0}^{2π} r^{2}dθ,

hence T = \frac{2}{a}A. The area of an ellipse turns out to be A = \frac{π}{4} LL′, where L and L′ denote the length of the major and the minor axis. Since

L = \frac{2p}{1 − e^{2}} , L′ = \frac{2p}{\sqrt{1 − e^{2}}} ,

we find

A = π \frac{p^{2}}{(1 − e^{2})^{3/2}} = πp^{2}(\frac{L}{2p})^{3/2} = 2^{−3/2}π\sqrt{pL^{3}}.

Thus

T = \frac{2}{a}A = 2^{−1/2}π(\frac{p}{a^{2}} )^{1/2}L^{3/2} = π \frac{1 + \frac{m}{M}}{\sqrt{2GM}} L^{3/2}.

It remains to explain the formulas for L and L′. The major axis of an ellipse is its maximal diameter, the minor axis is the axis through the center of the ellipse orthogonal to the major axis. Given the equation

r = \frac{p}{1 − e cos θ} ,

we see that the maximal distance of a point P on the ellipse corresponds to a minimal denominator, i.e. to θ = 0; the minimal distance corresponds to θ = π, and the length L of the major axis is the sum of these two terms:

L = \frac{p}{1 − e} + \frac{p}{1 + e} = \frac{2p}{1 − e^{2}} .

For computing the maximal diameter orthogonal to the major axis, we have to maximize r sin θ. Differentiating we get r′ sin θ + r cos θ = 0; plugging in r and simplifying we find cos θ = e (note that e < 1 since the conic in question is an ellipse). This shows that

L′ = 2r sin θ = \frac{2p}{1 − e^{2}} \sqrt{1 − e^{2}} = \frac{2p}{\sqrt{1 − e^{2}}} .

A different approach uses the fact that for any point P on an ellipse with foci F and G, the sum \overline{F P} + \overline{P G} is constant. Evaluating this for a point P on the major axis yields \overline{F P} + \overline{P G} = L. If P is a point on the ellipse and the minor axis, then \overline{F P} = \overline{P G} = \frac{1}{2}L. Using the theorem of Pythagoras you can now compute L′.

March 18, 2003

13.1 Galilean Spacetime

Let us now try to model physical reality. Every observer has his own system O of coordinates: at a given time t, observer O measures the coordinates of an event and assignes it the coordinates (x, y, z, t) ∈ ℝ^{4}. The coordinates of this event will, of course, be different for a different observer Õ: assuming they’re using the same clock, Õ will attach coordinates (x̃, ỹ, z̃, t) to the same event. If Õ uses a different clock, his time-coordinate t̃ will differ from that of O by a constant t_{0}, that is, we have t̃= t + t_{0}.

Although O and Õ use different coordinates, they will still agree on some facts: if O measures the distance |x_{2} − x_{1}| between two points x_{1} and x_{2} at the same time t, Õ will produce the same result: observers agree on the distance between simultaneous events.

In order to find the connection between the coordinates of O and Õ, let us assume they’re living in a 2-dimensional spacetime, so coordinates have the form (t, x). We will also assume that O and Õ are inertial observers, that is, they both move at constant speed (no acceleration, no forces); in addition, we assume that the observers meet at t = 0, that is, their coordinates at t = 0 are (0, 0) in both coordinate systems. Since Õ moves with constant velocity v away from O, his x-coordinate at time t (as measured by O) is x(t) = vt.

Now let e be an event with coordinates (t, x) and (t, x̃), respectively. Since x̃ is the x-coordinate as measured by Õ, and since Õ has x-coordinate vt at time t, the x-coordinate of e as measured by O is x = x̃ + vt.

• O  vt  • Õ x̃ • e

Now both observers agree on the distance from e to the origin, that is, the distance x observed by O must coincide with the distance x̃ from e to Õ (origin of the coordinate system of Õ) plus the distance vt between O and Õ: thus x = x̃ + vt.

Thus the coordinate transformations are given by

t̃= t, x̃ = x − vt.

Such a transformation is called a Galilei transformation. Using matrices, we can write them as

(t̃ ; x̃) = (\begin{matrix}1 & 0 \\ v & 1 \end{matrix})(\begin{matrix}t \\ x \end{matrix}).

What is the distance between two simultaneous events in the coordinate system of O? Given the coordinates (t, x) and (t, y), it is d = |y − x|. In the coordinate system of Õ, the coordinates are (x̃, t) and (ỹ, t), and the distance is

d̃= |ỹ − x̃| = |(y − vt) − (x − vt)| = |y − x| = d.

It can be shown that the Newtonian laws are invariant under Galilei transformations. A Galilei spacetime is a spacetime (the set of all possible coordinates (t, x)) in which the coordinates of inertial observers are related by Galilei transformations. A Galilean oberver is an inertial observer in a Galilean spacetime.

Let us now see how addition of velocities works in a Galilean spacetime.

Proposition 13.1. Let O and Õ be Galilean observers, and A an object in a Galilean spacetime. If Õ has velocity v with respect to O, and A is travelling in the same direction with velocity w with respect to Õ, then A is traveling with velocity v + w relative to O.

Proof. The worldline of A satisfies the equation x̃ = wt̃+ x̃_{0} in Õ’s coordinates, where x̃_{0} is a constant (the distance between A and Õ at t = 0. Applying the Galilei transformation from Õ to O this yields x−vt = wt+x̃_{0} for the worldline of A in the coordinates of O. Now x = (v + w)t + x̃_{0}, where x denotes the position of A.

The Galilean spacetime offered a pretty accurate model of the world we live in, until scientists started dealing with velocities close to c, the speed of light. The famous experiment by Michelson and Morley could be interpreted as saying that v + c = c, which contradicts the addition of velocities in a Galilean spacetime. Another problem was the fact that the Maxwell equations, which describe the effects of electromagnetism (including the propagation of light waves) to very high precision, are not invariant under Galilei transformations!

13.2 Minkowski Spacetime

Before we try to motivate why we should look at Minkowski spacetime, let us first describe it.

In Euclidean geometry, the isometries fixing the origin, namely rotations and reflections, preserved circles around the origin.

The space underlying Minkowski geometry is (we only look at 2 dimensions for now) the same as for the Euclidean geometry, namely ℝ^{2}. The difference is the way distances are computed: in Euclidean geometry, we used ||x|| = \sqrt{x · x} as our norm, where x · x was the standard dot product. Now let us see what happens if we replace the standard dot product (x_{1}, y_{1}) · (x_{2}, y_{2}) = x_{1}x_{2} + y_{1}y_{2} with (x_{1}, y_{1}) · (x_{2}, y_{2}) = x_{1}x_{2} − y_{1}y_{2}. Since this dot product is not positive definite, the formula ||x|| = \sqrt{x · x} does not define a norm since the radicand might be negative. Let us agree to call a vector x

• timelike if x · x > 0;

• lightlike if x · x = 0;

• spacelike if x · x < 0.

We still could use this dot product to compute the norm of timelike vectors by putting ||x|| = \sqrt{x · x} in this case. This function || · || is well defined on the set

T^{+} = {(t, z) ∈ ℝ^{2} : t > |z| ≥ 0}.

What do the isometries of this space look like? Are there any isometries at all (except the identity map)?

March 25, 2003

14.1 Isometries

Consider the map

H_{u} = (\begin{matrix}cosh u & sinh u \\ sinh u & cosh u \end{matrix})

defined on ℝ^{2}, where u is any real number. We have ||x||^{2} = t^{2} − z^{2} > 0, as well as

H_{u}(x) = (\begin{matrix}cosh u & sinh u \\ sinh u & cosh u \end{matrix})(\begin{matrix}t \\ z \end{matrix}) = (\begin{matrix}t cosh u + z sinh u \\ t sinh u + z cosh u \end{matrix}),

hence

||H_{u}(x)||^{2} = (t cosh u + z sinh u)^{2} − (t sinh u + z cosh u)^{2}
= t^{2}(cosh^{2} u − sinh^{2} u) − z^{2}(cosh^{2} u − sinh^{2} u)
= t^{2} − z^{2} = ||x||^{2}.

We also have to show that H_{u} preserves T^{+}. Given (t, x) ∈ T^{+}, the t-coordinate of its image is t cosh u + z sinh u > t cosh u − t sinh u = te^{−u} > 0, where we have used that z > −t.

Note that since the hyperbolic rotation H_{u} preserves ||x||, the image of a lightlike vector is again a lightlike vector. Finally, observe that H_{u}^{−1} = H_{−u}, and the isometries H_{u} actually form a group.

Proposition 14.1. For v_{1}, v_{2} ∈ T^{+} we have ||v_{1} + v_{2}|| ≥ ||v_{1}|| + ||v_{2}||.

Proof. We write v_{1} = (x_{1}, y_{1}) and v_{2} = (x_{2}, y_{2}); then v_{1}+v_{2} = (x_{1}+x_{2}, y_{1}+y_{2}), hence ||v_{1} + v_{2}||^{2} = v_{1} · v_{1} + 2v_{1} · v_{2} + v_{2} · v_{2} = ||v_{1}||^{2} + ||v_{2}||^{2} + 2(x_{1}x_{2} − y_{1}y_{2}).

Note that the difference in the bracket is positive since v_{1}, v_{2} ∈ T^{+}. But now

x_{1}x_{2} − y_{1}y_{2} ≥ \sqrt{x_{1}^{2} − y_{1}^{2}}\sqrt{x_{2}^{2} − y_{2}^{2}}  ⟺
(x_{1}x_{2} − y_{1}y_{2})^{2} ≥ (x_{1}^{2} − y_{1}^{2})(x_{2}^{2} − y_{2}^{2})  ⟺
2x_{1}x_{2}y_{1}y_{2} ≤ x_{1}^{2}y_{2}^{2} + x_{2}^{2}y_{1}^{2}  ⟺
0 ≤ (x_{1}y_{2} − x_{2}y_{1})^{2}

14.2 Minkowski Spacetime

When we model physical reality, we will actually use the space ℝ^{4} = ℝ×ℝ^{3} with coordinates (t, x, y, z) and the Minkowski metric defined by the dot product

(t_{1}, x_{1}, y_{1}, z_{1}) · (t_{2}, x_{2}, y_{2}, z_{2}) = c^{2}t_{1}t_{2} − x_{1}x_{2} − y_{1}y_{2} − z_{1}z_{2}.

Now we define a Minkowski spacetime as a spacetime in which the coordinate systems of inertial observers are related by a hyperbolic rotation (also called a Lorentz transformation).

As in Galilean spacetimes, this transformation will depend on the relative speed of Õ with respect to O. In fact, if Õ moves with speed v < c in the direction of the positive x-axis of O, then there is a real number u such that \frac{v}{c} = tanh u, and we have

(ct̃ ; x̃ ) = (\begin{matrix}cosh u & − sinh u \\ − sinh u & cosh u \end{matrix})(\begin{matrix}ct \\ x \end{matrix}).

What happens if v ≪ c? Then cosh u = 1+ \frac{u^{2}}{2!} +… ≈ 1, sinh u = u+ \frac{u^{3}}{3!} +… ≈ u, tanh u = \frac{sinh u}{cosh u} ≈ u, hence

(ct̃ ; x̃ ) = (\begin{matrix}1 & −u \\ −u & 1 \end{matrix})(\begin{matrix}ct \\ x \end{matrix}),

which in turn implies t̃= t + \frac{u}{c} ≈ t and x̃ = x + ctu = x − vt. Thus in the limit v ⟶ 0 the Lorentz transformations become Galilei transformations!

In order to get the actual Lorentz transformations in terms of the velocity v rather than the hyperbolic angle u, we have to express cosh u and sinh u in terms of v. Recall that tanh u = \frac{v}{c}; from the homework you know that sinh u = \frac{v/c}{\sqrt{1−v^{2}/c^{2}}} and cosh u = \frac{1}{\sqrt{1−v^{2}/c^{2}}} . The first equation from the Lorentz

transformations thus reads

ct̃= ct cosh u + x sinh u,

which becomes

t̃= \frac{t − \frac{v}{c^{2}} x}{\sqrt{1 − \frac{v^{2}}{c^{2}}}} .

Similarly, the second equation becomes, after a similar calculation,

x̃ = \frac{x − vt}{\sqrt{1 − \frac{v^{2}}{c^{2}}}} .

From these formulas it is even more obvious that Lorentz transformations turn into Galilei transformations when v ⟶ 0.

14.3 Addition of Velocities

Now let us see how velocities add in Minkowski spacetimes.

Proposition 14.2. Let O and Õ be Galilean (that is, inertial) observers, and A an object in a Minkowski spacetime. If Õ has velocity v with respect to O, and A is travelling in the same direction with velocity w with respect to Õ, then A is traveling with velocity \frac{v+w}{1+ \frac{vw}{c^{2}}} relative to O.

Proof. Let (t, x) and (t̃, x̃) be the coordinates used by O and Õ. The worldline of A satisfies the equation x̃ = wt̃+x̃_{0} in Õ’s coordinates, where x̃_{0} is a constant (the distance between A and Õ at t = 0. Applying the Lorentz transformation from Õ to O this yields

\frac{x − vt}{\sqrt{1 − \frac{v^{2}}{c^{2}}}} + x_{0} = w \frac{t − \frac{v}{c^{2}} x}{\sqrt{1 − \frac{v^{2}}{c^{2}}}} + x_{0}.

Simplifying, we find that the equation for the worldline of A in O’s coordinate system has the form

x = \frac{v + w}{1 + \frac{vw}{c^{2}}} t + constant,

hence its velocity equals

\frac{dx}{dt} = \frac{v + w}{1 + \frac{vw}{c^{2}}}

as claimed.

March 27, 2003

15.1 Moving Clocks Run Slow

Assume Õ is moving with velocity v with respect to Õ. Let E = (t̃, 0) be an event that happens at time t̃ on Õ’s world line. Let tanh u = \frac{v}{c}; then the coordinates in O’s world line are computed by applying a rotation by the hyperbolic angle −u, and we get

(\begin{matrix}ct \\ x \end{matrix}) = (\begin{matrix}cosh u & sinh u \\ sinh u & cosh u \end{matrix})(\begin{matrix}ct̃ \\ 0 \end{matrix} ),

hence t = (cosh u)t = _{\frac{t̃}{\sqrt{1− \frac{v^{2}}{c^{2}}}}} . Since the denominator is less than 1, observer O

will say that Õ’s clock runs slow by a factor of cosh u.

Conversely, of course, the same thing happens: the Lorentz transformation from O to Õ differs only in the sign of u, and this does not affect the final equation. Thus both observers will predict that the other’s clock is running slow.

15.2 Moving Rulers Contract

Assume that Õ has a ruler whose ends have coordinates (ct̃, x̃) = (0, 0) and (0, l). Now Õ moves with velocity v with respect to O, and (0, 0) still has coordinates (0, 0) in O’s system. But when O measures the length of the ruler, he will measure the distance between its ends at t = 0, not at t̃ = 0. The

Lorentz transformation says that, with γ = \sqrt{1 − \frac{v^{2}}{c^{2}}} , we have x̃ = \frac{x−vt}{γ}, so for t = 0 we find x̃ = \frac{x}{γ}, which means that O will observe a length γl. Since γ < 1, this means that moving rulers contract.

15.3 Simultaneity

As we have seen in the last section, there seems to be a problem with simultaneity: Õ′ measures the length of the ruler when t̃ = 0, that is, he computes the spatial distance between (t̃, z̃) = (0, 0) and (0, l). Observer O, however, computes the spatial distance between two other events, namely (t, z) = (0, 0) and (0, γl).

In fact, events that are seen as simultaneous by observer O will in general not be simultaneous for an observer Õ moving with respect to O

15.4 Inertial Worldlines have Maximal Length

Let α(s) = (t(s), x(s)) parametrize the worldline of an inertial observer Õ, where s is the time coordinate of an inertial observer O. Then x(s) = vt(s)+x_{0} in O’s coordinate system. What is the Minkowski length of the vector α(1) − α(0)? It is

||t(1) − t(0), v[t(1) − t(0)])|| = c|t(1) − t(0)|\sqrt{1 − v^{2}}.

Thus as s units pass on O’s clock, we have \frac{1}{c}||α(1)−α(0)|| = cγ, which is exactly the time passing on Õ’s clock. Thus the length \frac{1}{c}||α(s)|| measures the proper time of Õ.

Now assume that Õ is not necessarily inertial. Then for small intervals of time, we have Δα(s) ≈ α′(s)Δ(s), and the proper time of Õ during Δs is \frac{1}{c}||α′(s)||Δs. Thus the proper time elapsed on Õ’s clock as O’s clock goes from 0 to s is

t̃= \frac{1}{c} ∫_{0}^{s} ||α′(s)||ds,

or in other words, \frac{1}{c} of the length of its world line.

Theorem 15.1. An inertial observer takes longer to get from an event A to another event B than any other observer Õ.

Proof. We have to show that ∫_{0}^{s} ||α′(s)||ds ≤ s with equality if α′(s) is constant. Now α′(s) = (1, x′(s)); since Õ travels with speed less than c, we have

c^{2} > (\frac{dx}{ds})^{2}.

Thus

||α′(s)|| = \sqrt{c^{2} − x′(s)^{2}} ≤ c

for every s > 0. It follows that

∫_{0}^{s} ||α′(s)||ds ≤ ∫_{0}^{s} ds = s.

15.5 Newton versus Einstein

In Newton’s law of gravitation, changing the location of the mass m has an instantaneous effect on the gravitational force acting on M: by special relativity, however, exchange of information can only occur with speed less than c. This led Einstein to the idea that the law of gravitation has to be modified.

April 10, 2003

16.1 Curvature of Surfaces: Euler’s Formula

Suppose we are given a smooth surface S in ℝ^{3} and a point P on S. We would like to investigate the curvature of S at P. To this end, we first choose a coordinate system in which P has coordinates (0, 0, 0). Next we turn the coordinate system in such a way that the tangent plane to S at P becomes the x-y-plane. Then we can write the surface, in a small neighborhood of P, as a function z = f(x, y), and develop this into a Taylor series. Since P is at the origin, we have f(0, 0) = 0; moreover, there are no linear terms since the tangent plane in P has equation z = 0. Thus z = f(x, y) = \frac{1}{2}(ax^{2}+2bxy+cy^{2})+O(x, y)^{3} for some a, b, c ∈ ℝ; here O(x, y)^{3} denotes terms of degree ≥ 3.

Now we claim that by rotating the coordinate system around the z-axis, we can make the mixed term bxy disappear. In fact, a rotation around the z-axis will fix the z-coordinates and change (x, y) into (x′, y′). The equation is

(\begin{matrix}x \\ y \end{matrix}) = (\begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix})(\begin{matrix}x′ \\ y′ \end{matrix}).

Plugging this into the equation z = f(x, y) and concentrating on the mixed term b′x′y′, we find that b′ = (a − c) cos θ sin θ + b(cos^{2} θ − sin^{2} θ). We want b′ = 0. If cos θ ≠ 0, divide through by it; we find that we want (a−c) tan θ +b(1−tan^{2} θ). Putting s = tan θ, we end up with the quadratic equation bs^{2} − (a − c)s − b = 0; its discriminant is (a − c)^{2} + 4b^{2} ≥ 0 (actually > 0 except possible when b = 0, but in this case we’re already done). Thus the quadratic has a real solution s, and using the θ with tan θ = s we get b′ = 0.

Now the equation of the surface around P = (0, 0, 0) has the form

z = f(x, y) = \frac{1}{2}(λx^{2} + μy^{2}) + O(x, y)^{3}.

We now intersect the surface with planes orthogonal to z = 0. These are described by x = 0 or y = xt. We introduce a new variable w by putting x = w cos θ, y = w sin θ (note that t = \frac{y}{x} = tan t) and get

z = g(w) = \frac{1}{2}(λ sin^{2} θ + μ cos^{2} θ)w^{2} + O(w^{3}).

This curve has curvature

κ(θ) = λ sin^{2} θ + μ cos^{2} θ.

We find that κ′(θ) = 0 implies sin θ cos θ = 0 (unless a = c, in which case the curvature is constant in every direction), that is, θ = 0 or θ = \frac{π}{2} . Thus if a ≠ c the curvature is extremal for these values of θ (in particular the direction of maximal curvature is perpendicular to the direction of minimal curvature). The values a and c are called the principal curvatures (they are determined up to sign: flipping the coordinate system about the x-y-plane will change (a, c) to (−a, −c)).

This is all due to Euler. Gauss proved that a surface can be mapped to a plane in such a way that distances and angles are preserved if and only if ac = 0. The product κ = ac is called the Gaussian curvature of S at P.

The sphere of radius r has principal curvature a = c = \frac{1}{r} and Gaussian curvature \frac{1}{r^{2}} , the cylinder of radius r has principal curvatures a = \frac{1}{r} and c = 0.

16.2 Surfaces

Informally speaking, a surface is a smooth 2-dimensional object. Just as for curves, there are many ways of describing surfaces (or higher dimensional spaces). In order to get a basic idea of what surfaces are, we start with the description of surfaces as subsets of ℝ^{3}.

The simplest 2-dimensional subsets of ℝ^{3} are planes. They can be described by equations aX + bY + cZ = d, where not all of the real numbers a, b, c are 0. For exampe, the plane given by Z = 0 is the X − Y -plane. Planes also can be described using parametrizations: consider the map p : ℝ^{2} ⟶ ℝ^{3} given by p(x^{1}, x^{2}) = (x^{1}, x^{2}, ax^{1} + bx^{2} + c); its image consists of all (x, y, z) ∈ ℝ^{3} such that ax + by − z = −c and therefore is a plane.

A nontrivial surface is the sphere with radius r: it is described as the set of all (x, y, z) ∈ ℝ^{3} satisfying the equation x^{2} + y^{2} + z^{2} = r^{2}. It also can be parametrized by p(x^{1}, x^{2}) = (r cos x^{1} cos x^{2}, r sin x^{1} cos x^{2}, r sin x^{2}), where −π ≤ x^{1} ≤ π, −\frac{π}{2} ≤ x^{2} ≤ \frac{π}{2} .

In the following, we will use parametrizations

p(q^{1}, q^{2}) = (x(q^{1}, q^{2}), y(q^{1}, q^{2}), z(q^{1}, q^{2}))

to describe surfaces: it has the advantage of giving us tangent vectors almost for free: suppose you put q^{2} = c^{2} for some constant c^{2}; then the map p(q^{1}, c^{2}) = (x(q^{1}, c^{2}), y(q^{1}, c^{2}), z(q^{1}, c^{2})) is a curve in ℝ^{3}, and its tangent vector at q^{1} = c^{1} is given by x_{1} = \frac{∂p}{∂q^{1}} (c^{1}, c^{2}). Similarly, x_{2} = \frac{∂p}{∂q^{2}} (c^{1}, c^{2}) is the tangent vector to the curve p(c^{1}, q^{2}) at q^{2} = c^{2}.

The tangent plane at the point (q^{1}, q^{2}) is the set of all tangent vectors to the surface at this point; we say that a map p as above defines a smooth surface if, first, the partial derivatives exist and are continuous, and, second, if the tangent vectors x_{1} and x_{2} are linearly independent. A different way of expressing the last condition is by demanding that x_{1} × x_{2} ≠ 0. Geometrically it means that the tangent plane at any point of the surface actually is a plane, that is, it has dimension 2 as a vector space. In general, the dimension of a manifold is by definition the definition of its tangent space (which is assumed to be constant for each manifold).

If S is a smooth surface given by p, then for any point P = p(q^{1}, q^{2}) on S we have the tangent plane T_{P} S as the set of all tangent vectors to S at P, or, alternatively, the set of all linear combinations of x^{1} and x^{2}. As before with curves, we think of T_{P} S as a vector space attached to the point P. The abstract collection of data (P, T_{S}P) (this is some 4-dimensional space consisting of points of the manifold plus the corresponding tangent plane) is called the tangent bundle and is an important object in advanced courses in differential geometry.

16.3 The metric tensor

Let us now consider two tangent vectors v = v^{1}x_{1} +v^{2}x_{2} and w = w^{1}x_{1} +w^{2}x_{2} (for the time being, think of objects with lower indices as vectors, and of objects with upper indices as coordinates). We can use the standard dot product to compute the lengths of these vectors. More generally, we can compute their dot product in terms of the coordinates v^{i} and w^{i}: we find

v · w = (v^{1}x_{1} + v^{2}x_{2})(w^{1}x_{1} + w^{2}x_{2})

= v^{1}w^{1}x_{1} · x_{1} + v^{1}w^{2}x_{1} · x_{2} + v^{2}w^{1}x_{2} · x_{1} + v^{2}w^{2}x_{2} · x_{2}

= (v^{1}, v^{2})( \begin{matrix}x_{1}·x_{1} & x_{1}·x_{2} \\ x_{2}·x_{1} & x_{2}·x_{2} \end{matrix} )( \begin{matrix}w^{1} \\ w^{2} \end{matrix} )

= v^{T} Gw.

Thus the standard dot product in ℝ^{3} induces a dot product v·w = v^{T} Gw in the tangent plane T_{P} S of S at P. The matrix G is symmetric since x_{1} · x_{2} = x_{2} · x_{1}, and is called the metric tensor of S at P (tensors are generalizations of matrices). Another name for the metric tensor of surfaces is the first fundamental form (this is Gauss’s terminology).

April 15, 2003

17.1 Computing Arc Lengths

We can consider a parametrized surface S as a smooth map p : D ⟶ ℝ^{3}, where D is an open subset of ℝ^{2}. Suppose that c : I ⟶ D is a curve in the subset D ⊆ ℝ^{2}; then the composition z = p ∘ c : I ⟶ ℝ^{3} will describe a curve on the surface S. If I = [a, b], this curve will have length ∫_{a}^{b} ||z′(t)|| dt. We want to compute this length from c′ and the metric tensor g. To this end, note that z(t) = p(c(t)) = (p(c^{1}(t), c^{2}(t)); now we apply the chain rule:

z′(t) = \frac{dz}{dt} = \frac{dz}{dc^{1}} \frac{dc^{1}}{dt} + \frac{dz}{dc^{2}} \frac{dc^{2}}{dt} = \frac{dc^{1}}{dt} x_{1} + \frac{dc^{2}}{dt} x_{2}.

Thus we have written the tangent vector z′(t) of the curve as a linear combination of x_{1} and x_{2}; now the length of z′(t) is

||z′(t)||^{2} = z′(t) · z′(t) = z′(t)^{T} gz′(t)

= ∑_{i,j} g_{ij}(c^{1}(t), c^{2}(t))\frac{dc^{1}}{dt} (t)\frac{dc^{2}}{dt} (t),

or more briefly as

||z′(t)|| = \sqrt{g_{ij} \frac{dc^{1}}{dt} \frac{dc^{2}}{dt}} ,

where we have used Einstein’s summation convention.

Recap: The Chain Rule

Consider z(t) = p(c^{1}(t), c^{2}(t)) = (x(c^{1}, c^{2}), y(c^{1}, c^{2}), z(c^{1}, c^{2})). What happens if we replace t by t + h for small h? Then c^{i}(t + h) ≈ c^{i}(t) + hċ^{i}(t). Similarly, if we fix c^{2} and consider x as a function of c^{1}, we have

x(c^{1} + h^{1}, c^{2}) = x(c^{1}, c^{2}) + h^{1} \frac{∂x}{∂c^{1}} (c^{1}, c^{2}).

This then implies

x(c^{1} + h^{1}, c^{2} + h^{2}) = x(c^{1} + h^{1}, c^{2}) + h^{2} \frac{∂x}{∂c^{2}} (c^{1} + h^{1}, c^{2})

≈ x(c^{1}, c^{2}) + h^{1} \frac{∂x}{∂c^{1}} (c^{1}, c^{2}) + h^{2} \frac{∂x}{∂c^{2}} (c^{1}, c^{2}),

where we have neglected h^{1}h^{2} ≈ 0. Thus

(x ∘ c)(t + h) ≈ x(c^{1}(t) + h\frac{dc^{1}}{dt} , c^{2}(t) + h\frac{dc^{2}}{dt})

≈ x(c^{1}(t), c^{2}(t)) + h\frac{∂x}{∂c^{1}} \frac{dc^{1}}{dt} + h\frac{∂x}{∂c^{2}} \frac{dc^{2}}{dt}

This suggests that

\frac{dx ∘ c}{dt} = lim_{h⟶0} \frac{x ∘ c(t + h) − x ∘ c(t)}{h} = \frac{∂x}{∂c^{1}} \frac{dc^{1}}{dt} + \frac{∂x}{∂c^{2}} \frac{dc^{2}}{dt} .

As usual, this handwaving argument can be made precise using the intermediate value theorem.

Example: The Unit Sphere

Consider the unit sphere S parametrized by

p(q^{1}, q^{2}) = (cos q^{1} cos q^{2}, sin q^{1} cos q^{2}, sin q^{2})

for −π ≤ q^{1} ≤ π and −\frac{π}{2} ≤ q^{2} ≤ \frac{π}{2} . The metric tensor g is given by

g = (\begin{matrix}cos^{2} q^{2} & 0 \\ 0 & 1 \end{matrix}) .

The set of all points on the unit sphere with q^{2} = ϕ for some constant ϕ form a circle C on the sphere. How long is C? For ϕ = 0, the circle C is given by

C = (cos q^{1}, sin q^{1}, 0),

so C is the equator, and its length is 2π. For ϕ = \frac{π}{2} , cos ϕ = 0, and the ‘circle’ C collapses to a point, namely the north pole, and its ‘length’ accordingly is 0. For values of ϕ between these two values, the length of the circle should be between 0 and 2π.

The actual computation requires us to parametrize the curve q^{2} = ϕ in D; we take c(t) = (t, ϕ) and find

g_{ij} \frac{dc^{1}}{dt} \frac{dc^{2}}{dt} = g_{11}(\frac{dc^{1}}{dt} )^{2} + g_{22}(\frac{dc^{2}}{dt} )^{2} = cos^{2}(ϕ).

Thus the length of the curve is

∫_{−π}^{π} \sqrt{cos^{2} ϕ} dt = ∫_{−π}^{π} cos ϕ dt = 2π cos ϕ.

17.2 Computing Areas

Consider a surface in ℝ^{3} parametrized by the function x(q^{1}, q^{2}). A little rectangle [q^{1}, q^{1} + Δq^{1}] × [q^{2}, q^{2} + Δq^{2}] will get mapped to a little area on the surface. How big is this area? The point (q^{1}, q^{2}) ∈ ℝ^{2} will get mapped to x(q^{1}, q^{2}), and similarly the point (q^{1} + Δq^{1}, q^{2} + Δq^{2}) will go to x(q^{1} + Δq^{1}, q^{2} + Δq^{2}). This last point is close to the point x(q^{1}, q^{2}) + Δq^{1}x_{1}(q^{1}, q^{2}) + Δq^{2}x_{2}(q^{1}, q^{2}) in the tangent plane, and we can approximate the area we are interested in by the area of the parallelogram spanned by the tangent vectors Δq^{1}x_{1}(q^{1}, q^{2}) and Δq^{2}x_{2}(q^{1}, q^{2}):

area ≈ Δq^{1} · Δq^{2} · |x_{1} × x_{2}| = \sqrt{det g}Δq^{1} · Δq^{2},

where det g is the determinant of the metric tensor of the surface at the point x(q^{1}, q^{2}).

Informally: the area Δq^{1} · Δq^{2} of the rectangle in ℝ^{2} gets magnified by the factor \sqrt{det g} under the map x.

One question we have to address is whether det g > 0. This is actually true:

Proposition 17.1. The determinant and the trace of the metric tensor are positive. In other words: the symmetric matrix g has positive eigenvalues.

Proof. The claim is trivial for the trace since ||x_{1}|| > 0 and ||x_{2}|| > 0 by assumption. Next

det g = ||x_{1}||^{2}||x_{2}||^{2} − (x_{1} · x_{2})^{2} ≥ 0

by Cauchy-Schwarz, and the inequality is strict since x_{1} and x_{2} are linearly independent.

Note that

||x_{1}||^{2}||x_{2}||^{2} − (x_{1} · x_{2})^{2} = ||x_{1}||^{2}||x_{2}||^{2} − ||x_{1}||^{2}||x_{2}||^{2} cos^{2} θ = ||x_{1}||^{2}||x_{2}||^{2} sin^{2} θ

is equal to the area of the parallelogram spanned by x_{1} and x_{2}, i.e., equals ||x_{1} × x_{2}||.

Now that we know how little rectangles transform, we can compute areas on the surface S: think of S as the image of some region D in ℝ^{2}; the area of D is simply ∫ ∫_{D} dq^{1}dq^{2}. It is approximated by sums of little rectangles of area Δq^{1} · Δq^{2}; the image of D is approximated by sums of rectangles in the tangent plane with area \sqrt{det g} · Δq^{1} · Δq^{2}, and letting Δq^{1} and Δq^{2} go to 0 gives us the area of the image of D as

area(x(D)) = ∫ ∫_{D} \sqrt{det g(q^{1}, q^{2})} dq^{1}dq^{2}.

As an example, the area of the unit sphere is given by

∫ ∫ \sqrt{det g} dq^{1}dq^{2} = ∫_{−π}^{π} ( ∫_{−π}^{π} cos q^{2} dq^{2}) dq^{1} = ∫_{−π}^{π} 2 dq^{1} = 4π,

since det g = (cos q^{2})^{2}, and since the sphere is covered by the rectangle [−π, π]× [−\frac{π}{2} , \frac{π}{2} ].

April 21, 2003

18.1 Curvature of Surfaces

Curves again

We start by reconsidering one of our definitions of curvature of curves. Map each point c(t) of our smooth curve c to its unit normal vector n(t) at that point (this is called the Gauss map). As t varies over a little interval I = [a, b], the unit normal will trace out a small part L of a circle. Put

κ(c, I) = \frac{L}{ℓ},

where ℓ is the arc length of the curve as t goes from a to b. We might want to call this the average curvature of the curve over this interval. If we form the limit

lim_{b→a} κ(c, [b, a]),

we expect to get the curvature at the point c(a), and we do: first, the length cut out by the unit normal is equal to the length cut out by the unit tangent, which in turn equals ∫_{a}^{b} ||T′(t)|| dt. If we use arc length parametrization, then the length of the curve is just b − a, and we get

lim_{b→a} κ(c, [b, a]) = lim_{b→a} \frac{1}{b − a} ∫_{a}^{b} ||T′(t)|| dt = ||T′(a)|| = κ(a).

Surfaces

Now we do the same thing for surfaces: consider a little region x(D) around the point x(q^{1}, q^{2}); as q^{1} and q^{2} vary over D, the unit normal vector of the surface will trace out a certain part of the unit sphere; we can call

κ(S, D) = \frac{area traced out by unit normal vector}{area of x(D)} the average curvature of S in D, and then

κ(S, P) = lim_{D⟶P} κ(S, D)

can be used to define the curvature of S in some point P ∈ D.

Now let us do the calculation. The basis of the tangent plane in x(P) = x(q^{1}, q^{2}) consists of x_{1} and x_{2}; therefore

n(q^{1}, q^{2}) = \frac{x_{1} × x_{2}}{||x_{1} × x_{2}||}

is the unit normal vector in x(P).

Consider a square D_{ε} with side lengths 2ε and center P = (q^{1}, q^{2}) in D; the area of x(D) is

∫ ∫_{D_{ε}} ||x_{1} × x_{2}|| dq^{1}dq^{2}

(remember that \sqrt{det g} = ||x_{1} × x_{2}||). The area traced out by the unit normal vector is computed in an analogous way as

∫ ∫_{D_{ε}} ||n_{1} × n_{2}|| dq^{1}dq^{2}.

The limit of the quotient as ε → 0 is clearly

κ(S, P) = \frac{||n_{1} × n_{2}||}{||x_{1} × x_{2}||},

and we have proved

Theorem 18.1. The Gaussian curvature of the surface parametrized by x in the point x(q^{1}, q^{2}) equals

κ(S, P) = \frac{||n_{1} × n_{2}||}{||x_{1} × x_{2}||}.

Now we need to compute this fraction. Since n_{1} and n_{2} are tangent vectors on the unit sphere, they are orthogonal to n and therefore live in the tangent plane of x(q^{1}, q^{2}). Thus we can write n_{1} and n_{2} as linear combinations of x_{1} and x_{2}:

n_{1} = b_{1}^{1}x^{1} − b_{1}^{2}x_{2}
n_{2} = b_{2}^{1}x^{1} − b_{2}^{2}x_{2},

or, in matrix form, (n_{1}, n_{2})^{T} = B(x_{1}, x_{2})^{T} for the matrix B = (b_{ij}). A simple calculation shows that

||n_{1} × n_{2}|| = det B · ||x_{1} × x_{2}||,

which implies that κ(S, P) = det B. Thus we have reduced the computation of the Gaussian curvature of a surface to the calculation of a determinant.

18.2 A surface with negative curvature

Consider the surface described by the equation z = xy. We can parametrize it by x(q^{1}, q^{2}) = (q^{1}, q^{2}, q^{1}q^{2}). We find x_{1} = (1, 0, q^{2}) and x_{2} = (0, 1, q^{1}), so the metric tensor is given by

g = ( \begin{matrix}1+(q^{2})^{2} & q^{1}q^{2} \\ q^{1}q^{2} & 1+(q^{1})^{2} \end{matrix} ).

In particular, det g = (1 + (q^{2})^{2})(1 + (q^{1})^{2}) − (q^{1}q^{2})^{2} = 1 + (q^{1})^{2} + (q^{2})^{2} > 0. Next we have to compute the unit normal; we get

x_{1} × x_{2} = | \begin{matrix}i & j & k \\ 1 & 0 & q^{2} \\ 0 & 1 & q^{1} \end{matrix} | = (−q^{2}, −q^{1}, 1).

This is a vector of length Δ := \sqrt{1 + (q^{1})^{2} + (q^{2})^{2}}, hence n = \frac{1}{Δ}(−q^{2}, −q^{1}, 1) and

n_{1} = \frac{1}{Δ^{3}} (q^{1}q^{2}, −1 − (q^{2})^{2}, −q^{1}), n_{2} = \frac{1}{Δ^{3}} (−1 − (q^{1})^{2}, q^{1}q^{2}, −q^{2}).

Now we have to express n_{1} and n_{2} in terms of the basis vectors x_{1} and x_{2} of the tangent plane; we get

n_{1} = \frac{1}{Δ^{3}} (q^{1}q^{2}x_{1} − (1 + (q^{2})^{2})x_{2}),
n_{2} = \frac{1}{Δ^{3}} (−(1 + (q^{1})^{2})x_{1} + q^{1}q_{2}x_{2}).

Thus

B = \frac{1}{Δ^{3}} ( \begin{matrix}q^{1}q^{2} & −1−(q^{2})^{2} \\ −(1+(q^{1})^{2}) & q^{1}q^{2} \end{matrix} )

and det B = \frac{1}{Δ^{6}} (q^{1}q^{2})^{2} −(1+(q^{1})^{2})(1+(q^{2})^{2}) = −\frac{1}{Δ^{6}} (1+(q^{1})^{2} +(q^{2})^{2}) = −\frac{1}{Δ^{4}} . Putting r^{2} = (q^{1})^{2} + (q^{2})^{2} we see that det B = −\frac{1}{(1+r^{2})^{2}} .

April 29, 2003

19.1 Some Linear Algebra

In the proof of Gauss’s theorema egregium below we will come across equations of type A_{jk} = g_{ij}B_{k}^{i}, where the A_{jk}, g_{ij} and B_{k}^{i} are real numbers that we think of being entries of a 2 × 2-matrix. These equations can be written in the form

( \begin{matrix}A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix} ) = ( \begin{matrix}g_{11} & g_{12} \\ g_{21} & g_{22} \end{matrix} )( \begin{matrix}B_{1}^{1} & B_{2}^{1} \\ B_{1}^{2} & B_{2}^{2} \end{matrix} ),

or A = gB; in particular we get det A = det g · det B.

Let g^{−1} = (g^{jk}) denote the inverse of the matrix g = (g_{jk}); then g_{ih}g^{hj} = δ_{i}^{j}, where

δ_{i}^{j} = \begin{cases}1 & if i = j \\ 0 & if i ≠ j \end{cases}

is the Kronecker delta. Note that the matrix (δ_{i}^{j}) is the unit matrix.

Einstein’s summation convention allows us to write the multiplication of matrices in a very suggestive form: if A = (A_{jk}), g = (g_{ij}) and B = (B_{k}^{i}), then A = gB becomes A_{jk} = g_{ij}B_{k}^{i}. Note, however, that in such an equation you are, of course, not allowed to divide through by g_{ij}, because this equation is actually a sum: A_{jk} = ∑_{i} g_{ij}B_{k}^{i}. Multiplication by g_{ij} is called ‘lowering an index’, and multiplication by g^{ij} is called ‘raising an index’.

The summation convention also allows us to write the trace of a matrix A = (a_{j}^{i}) simply as Tr A = a_{i}^{i}. This process is called contraction.

19.2 Theorema Egregium

Gauss gave his main theorem of his 1827 article on curved surfaces the name ‘theorema egregium’, or outstanding theorem (from ‘ex’, which means ‘out of’, and ‘regio’, which means region). Today we interpret his result as saying that ‘Gaussian curvature is intrinsic’: this means that Gaussian curvature can be expressed solely in terms of the metric tensor; note that the metric tensor is something that can be observed by little ants living on the surface: measuring dot products of tangent vectors at a point P means measuring angles and lengths of vectors in sufficiently small neighborhoods of P.

We will now prove Gauss’s theorem by expressing the Gaussian curvature K = κ(S, P) in terms of the entries of the metric tensor. This proof will be a lengthy calculation involving all kinds of functions and coefficients.

Let us start with setting

x_{jk} = \frac{∂^{2}x}{∂q^{k}∂q^{j}} .

The four derivatives x_{jk} are functions of q^{1} and q^{2}, and we assume that the parametrization x of the surface is so smooth that x_{jk} = x_{kj} (we are allowed to interchange the order of partial differentiation if x has a second derivative).

Since x_{1}, x_{2} and n = x_{1} × x_{2} are linearly independent, they form a basis of ℝ^{3}, and this means that there are coefficients Γ_{jk}^{i} and b_{jk} such that

x_{jk} = ∑_{i=1}^{2} Γ_{jk}^{i}x_{i} + b_{jk}n,

or, with Einstein’s summation convention,

x_{jk} = Γ_{jk}^{i}x_{i} + b_{jk}n.

Since x_{jk} = x_{kj}, we have

0 = x_{jk} − x_{kj} = (Γ_{jk}^{i} − Γ_{kj}^{i})x_{i} + (b_{jk} − b_{kj})n.

Since x_{1}, x_{2} and n form a basis of ℝ^{3}, this holds if and only if the components vanish, and we find

Γ_{jk}^{i} = Γ_{kj}^{i}, b_{jk} = b_{kj}.

The Γ_{jk}^{i} are called Christoffel symbols of the second kind; the b_{jk} are related to the entries b_{j}^{i} of the matrix B that occurred in the calculation of the Gaussian curvature.

In fact, we have

x_{jk} · n = (Γ_{jk}^{i}x_{i} + b_{jk}n) · n = b_{jk}

since x_{i} · n = 0 and n · n = 1. Differentiating x_{j} · n = 0 (this looks a lot like the proof of the Frenet formulas) shows

0 = \frac{∂}{∂q^{k}} (x_{j} · n) = x_{jk} · n + x_{j} · n_{k} = b_{jk} + x_{j} · (−b_{k}^{i}x_{i}) = b_{jk} − b_{k}^{i}g_{ji}.

Thus b_{jk} = g_{ji}b_{k}^{i}, hence det B = det g · det B̃, or K = det B̃ = \frac{det B}{det g}. We therefore have to express det B in terms of the metric tensor.

To this end, we introduce the derivatives of the x_{jk}:

x_{jkl} = \frac{∂x_{jk}}{∂q^{l}} = \frac{∂Γ_{jk}^{i}}{∂q^{l}} x_{i} + Γ_{jk}^{i}x_{il} + \frac{∂b_{jk}}{∂q^{l}} n + b_{jk}n_{l}.

Now we play the same game as above: these vectors can be expressed as linear combinations of x_{1}, x_{2} and n. We find

x_{jkl} = \frac{∂Γ_{jk}^{i}}{∂q^{l}} x_{i} + Γ_{jk}^{p}x_{pl} + \frac{∂b_{jk}}{∂q^{l}} n + b_{jk}n_{l}

= \frac{∂Γ_{jk}^{i}}{∂q^{l}} x_{i} + Γ_{jk}^{p}(Γ_{pl}^{i}x_{i} + b_{pl}n) + \frac{∂b_{jk}}{∂q^{l}} n − b_{jk}b_{l}^{i}x_{i}

= (\frac{∂Γ_{jk}^{i}}{∂q^{l}} + Γ_{jk}^{p}Γ_{pl}^{i} − b_{jk}b_{l}^{i})x_{i} + (\frac{∂b_{jk}}{∂q^{l}} + Γ_{jk}^{p}b_{pl})n.

Interchanging k and l we find

x_{jlk} = (\frac{∂Γ_{jl}^{i}}{∂q^{k}} + Γ_{jl}^{p}Γ_{pk}^{i} − b_{jl}b_{k}^{i})x_{i} + (\frac{∂b_{jl}}{∂q^{k}} + Γ_{jl}^{p}b_{pk})n.

Assuming that x has a third derivative, we must have x_{jkl} = x_{jlk}, and as above we find 0 = x_{jlk} − x_{jkl}. The coefficient of x_{i} in this expression must vanish, hence

(\frac{∂Γ_{jl}^{i}}{∂q^{k}} − \frac{∂Γ_{jk}^{i}}{∂q^{l}} + Γ_{jl}^{p}Γ_{pk}^{i} − Γ_{jk}^{p}Γ_{pl}^{i}) − (b_{jl}b_{k}^{i} − b_{jk}b_{l}^{i}) = 0.

If we introduce the abbreviation

R_{jkl}^{i} = \frac{∂Γ_{jl}^{i}}{∂q^{k}} − \frac{∂Γ_{jk}^{i}}{∂q^{l}} + Γ_{jl}^{p}Γ_{pk}^{i} − Γ_{jk}^{p}Γ_{pl}^{i},  (19.1)

then the last equation says

R_{jkl}^{i} = b_{jl}b_{k}^{i} − b_{jk}b_{l}^{i}.

For surfaces in ℝ^{3}, this is a set of 2^{4} = 16 equations; in the 4-dimensional spacetime of general relativity, these will be 4^{4} = 256 equations!

Now define

R_{hjkl} = g_{ih}R_{jkl}^{i};

then

R_{hjkl} = b_{jl}b_{k}^{i}g_{ih} − b_{jk}b_{l}^{i}g_{ih} = b_{jl}b_{kh} − b_{jk}b_{lh}.

In particular,

R_{1212} = b_{22}b_{11} − b_{21}b_{21} = b_{22}b_{11} − b_{21}b_{12} = det B.

Thus we can write

K = \frac{R_{1212}}{det g} .

We have to show that R_{1212} only depends on the metric tensor and its derivatives. But

R_{1212} = g_{11}R_{212}^{1} + g_{12}R_{212}^{2}

together with (19.1) shows that it is sufficient to show that the Γ_{jk}^{i} have this property.

Finally, put

Γ_{jk,h} = Γ_{jk}^{i}g_{ih}.  (19.2)

Then

x_{jk} · x_{h} = (Γ_{jk}^{i}x_{i} + b_{jk}n) · x_{h} = Γ_{jk}^{i}x_{i} · x_{h} + b_{jk}n · x_{h} = Γ_{jk}^{i}g_{ih} = Γ_{jk,h}.

The Γ_{jk,h} are called Christoffel symbols of the first kind.

   This is how far I got today; next time I'll pick it up from here:

Multiplying (19.2) through by g^{hm}, where (g^{hm}) = g^{−1} is the inverse matrix of the metric tensor (recall that det g ≠ 0 since we assumed that x_{1} × x_{2} ≠ 0) we get

Γ_{jk,h}g^{hm} = Γ_{jk}^{i}g_{ih}g^{hm} = Γ_{jk}^{i}δ_{i}^{m} = Γ_{jk}^{m}.

Thus it remains to show that the Γ_{jk,h} and the g^{hm} can be computed from the metric tensor. But this is clear for g^{hm}, since these are the coefficients of g^{−1}, and the claim for the Christoffel symbols will follow from the equations

2Γ_{ij,k} = \frac{∂g_{ik}}{∂q^{j}} + \frac{∂g_{jk}}{∂q^{i}} − \frac{∂g_{ij}}{∂q^{k}} .

In fact, from g_{ik} = x_{i} · x_{k} we get

\frac{∂g_{ik}}{∂q^{j}} = x_{ij} · x_{k} + x_{i} · x_{kj} = Γ_{ij,k} + Γ_{jk,i}.

Similarly,

\frac{∂g_{jk}}{∂q^{i}} = Γ_{ij,k} + Γ_{ik,j}, \frac{∂g_{ij}}{∂q^{k}} = Γ_{ik,j} + Γ_{jk,i}.

This implies the claim.

19.3 Summary

Let us review our proof of the theorema egregium. We came across a variety of functions:

1. The metric tensor (first fundamental form) g_{ij} = x_{i} · x_{j};

2. The second fundamental form b_{ij} = x_{ij} · n.

Raising indices with g^{−1} we get b_{j}^{i} = b_{jk}g^{ki}, and we have n_{j} = −b_{j}^{i}x_{i}.

3. Christoffel symbols of the first kind Γ_{jk,l} = x_{jk} · x_{l}.

4. Christoffel symbols of the second kind Γ_{jk}^{m} = Γ_{jk,h}g^{hm}. We also have x_{jk} = Γ_{jk}^{i}x_{i} + b_{jk}n.

5. The Riemann curvature tensor R_{hjkl} = R_{jkl}^{i}g_{ih}, where

R_{jkl}^{i} = \frac{∂Γ_{jl}^{i}}{∂q^{k}} − \frac{∂Γ_{jk}^{i}}{∂q^{l}} + Γ_{jl}^{p}Γ_{pk}^{i} − Γ_{jk}^{p}Γ_{pl}^{i}.

6. The Ricci tensor will play a role below; it is a contraction of the Riemann tensor, namely R_{ik} = R_{ihk}^{h}.

Given the metric tensor g, compute Γ_{jk,h} from

2Γ_{jk,h} = \frac{∂g_{kh}}{∂q^{j}} + \frac{∂g_{jh}}{∂q^{k}} − \frac{∂g_{jk}}{∂q^{h}} .

Then raise the indices using g^{hm}, and compute the Riemann curvature tensor as in (5). The point is that everything can be computed from g, and that the actual vectors x_{i} (coming from the way the surface is embedded into ℝ^{3}) are not involved. This will allow us to do geometry in any space on which we are given a symmetric matrix g. In particular, we now can generalize hyperbolic geometry considerably.

May 1, 2003

20.1 Geodesics

Now let us think about geodesics on surfaces. One possible way of defining them would be as shortest paths between two points. This is not a good idea for two reasons: first, finding these shortest paths is usually quite difficult, and second, there are metrics (such as the metric in Minkowski spacetime) where there are nonzero vectors of zero length (lightlike vectors) or even vectors of imaginary length (spacelike vectors).

In order to find the correct definition of a geodesic, recall that in Galilean spacetime objects upon which no force acts move on lines. In other words, their world lines z(t) have the property that z̈(t) = 0: these particles do not accelerate.

Now consider a surface S parametrized by x : R ⟶ ℝ^{3}, and let q(t) = (q^{1}(t), q^{2}(t)) : [a, b] ⟶ R denote a curve in the parameter plane ℝ^{2}. Then z = x ∘ q : [a, b] ⟶ S defines a curve on the surface with velocity

ż(t) = \frac{∂x}{∂q^{1}} \frac{dq^{1}}{dt} + \frac{∂x}{∂q^{2}} \frac{dq^{2}}{dt} = x_{1}q̇^{1} + x_{2}q̇^{2} = x_{i}q̇^{i}.

Here we have used the chain rule again. Similarly, the acceleration of the particle moving on the world line z is

z̈(t) = x_{ij}q̇^{i}q̇^{j} + x_{i}q̈^{i}
= (Γ_{ij}^{k}x_{k} + b_{ij}n)q̇^{i}q̇^{j} + x_{k}q̈^{k}
= (q̈^{k} + Γ_{ij}^{k}q̇^{i}q̇^{j})x_{k} + b_{ij}q̇^{i}q̇^{j}n

Now if we want the particle to remain on the surface S, we must allow acceleration in the direction of the normal. But we can demand that the acceleration in the tangent plane be zero: we say that a curve z = x ∘ q on a surface S is a geodesic if the differential equations

\frac{d^{2}q^{k}}{dt^{2}} + Γ_{ij}^{k} \frac{dq^{i}}{dt} · \frac{dq^{j}}{dt} = 0

for k = 1, 2. These are second order nonlinear ordinary differential equations. For spaces with a positive definite metric it can be shown that this definition agrees with our old one; the proof, however, uses the calculus of variations.

20.2 General Geometry

Now that we can compute the Gaussian curvature of a surface from its metric tensor, we can generalize what we have been doing so far: instead of thinking of a curved surface as sitting inside ℝ^{3}, we think of it as an open piece R of ℝ^{2} to which we have attached a metric tensor in the following sense: for each (q^{1}, q^{2}) ∈ R we have a 2 × 2-matrix g = (g_{ij}) with the following properties:

1. each g_{ij} is a smooth function of q^{1} and q^{2};

2. at each point (q^{1}, q^{2}) ∈ R, the matrix g is symmetric;

3. at each point (q^{1}, q^{2}) ∈ R, the two real eigenvalues of g are positive.

As an example, consider the upper half plane ℍ = {(x, y) ∈ ℝ^{2} : y > 0} and define a metric tensor g by

g = (\begin{matrix}\frac{1}{y^{2}} & 0 \\ 0 & \frac{1}{y^{2}} \end{matrix} ).

Here det g = y^{−4} and \sqrt{det g} = y^{−2}. From the general formulas we deduce that the length of a curve c(t) = (x(t), y(t)) is given by

∫_{a}^{b} \frac{\sqrt{ẋ^{2} + ẏ^{2}}}{y} dt.

This agrees with what we had before.

Using the parameters q^{1} = x and q^{2} = y instead of x and y we have

g_{11} = g_{22} = 1/(q^{2})^{2}, g_{12} = g_{21} = 0.

The inverse matrix of a diagonal matrix is easy to compute, and we find

g^{11} = g^{22} = (q^{2})^{2}, g^{12} = g^{21} = 0.

The Christoffel symbols of the first kind are computed from the derivatives of the g_{ij}, and we have

\frac{∂g_{11}}{∂q^{2}} = \frac{∂g_{11}}{∂q^{2}} = \frac{−2}{(q^{2})^{3}} ,

and all other partials vanish. This implies that the only nonzero Christoffel symbols of the first kind are

Γ_{12,1} = Γ_{21,1} = Γ_{22,2} = −\frac{1}{(q^{2})^{3}} , Γ_{11,2} = \frac{1}{(q^{2})^{3}} .

The Christoffel symbols of the second kind are computed by raising indices with the inverse of g; we find (note that we know Γ_{jk}^{i} = Γ_{kj}^{i})

Γ_{11}^{1} = g^{11}Γ_{11,1} + g^{12}Γ_{11,2} = 0,
Γ_{12}^{1} = g^{11}Γ_{12,1} + g^{12}Γ_{12,2} = −\frac{1}{q^{2}} ,
Γ_{22}^{1} = g^{11}Γ_{22,1} + g^{12}Γ_{22,2} = 0,
Γ_{11}^{2} = g^{21}Γ_{11,1} + g^{22}Γ_{11,2} = \frac{1}{q^{2}} ,
Γ_{12}^{2} = g^{21}Γ_{12,1} + g^{22}Γ_{12,2} = 0,
Γ_{22}^{2} = g^{21}Γ_{22,1} + g^{22}Γ_{22,2} = −\frac{1}{q^{2}} .

The geodesic equations now read

0 = q̈^{1} − \frac{2}{q^{2}} q̇^{1}q̇^{2}, 0 = q̈^{2} + \frac{1}{q^{2}} (q̇^{1})^{2} − \frac{1}{q^{2}} (q̇^{2})^{2}.

Are vertical lines geodesics? Consider the lines q^{1} = c, q^{2} = t for t > 0; then q̇^{1} = q̈^{1} = 0 and q̇^{2} = 1, q̈^{2} = 0. The first geodesic equation is obviously satisfied, the second is not: the first two terms on the right hand side vanish, and the third equals −\frac{1}{t} < 0. This does not contradict our old results because the new definition of geodesics depends on the parametrization.

In fact, if we choose an arclength parametrization of the vertical lines, they become geodesics: we know that the length of the line with parametrization q^{1} = c, q^{2} = t, as t runs from a to t, equals s(t) = ln \frac{t}{a}. Picking a = 1, we find s(t) = ln t, hence t = e^{s}, and the arc length parametrization is q^{1} = c, q^{2} = e^{t} (we switched back to t).

Now q̇^{1} = q̈^{1} = 0, q̇^{2} = q̈^{2} = e^{t}, hence the first geodesic equation is trivially satisfied, and the second equation reads 0 = e^{t} − e^{−t}e^{2t}.

May 8, 2003

21.1 Weak Gravity Approximation

Consider a particle of restmass m; its world line in the coordinates of some observer are given by (x^{0}, x^{1}, x^{2}, x^{3}) = (ct, x, y, z) (at a certain time t, the particle has coordinates (x, y, z)). We also need the proper time τ of the particle; remember that we may take t = τ for small velocities.

Newton’s equations of a particle in a constant gravitational field described by the potential Φ(r) (for example, the potential for a spherically symmetric central mass M is Φ(r) = −\frac{GM}{r}) are given by

\frac{d^{2}x^{i}}{dt^{2}} = −\frac{∂Φ}{∂x^{i}} .

Now consider spacetime with coordinates chosen in such a way that c = 1. In flat space, we have the Minkowski metric defined by

η = ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) .

We now change this metric slightly by adding a ‘perturbation’: g = η + ϕ with ϕ = (ϕ_{ij}), |ϕ_{ij}| ≪ 1. Let us figure out how the geodesic equations look like.

General relativity says that particles move on geodesics parametrized by proper time. This means that the world lines have to satisfy the geodesic equations

\frac{d^{2}x^{k}}{dτ^{2}} + Γ_{ij}^{k} \frac{dx^{i}}{dτ} · \frac{dx^{j}}{dτ} = 0

for k = 0, 1, 2, 3. Now we make the assumption that the velocities of the objects are small when compared to c. This means that the geodesic equations simplify to (here we replace τ by t)

\frac{d^{2}x^{k}}{dt^{2}} = −Γ_{00}^{k}(\frac{dx^{0}}{dt} )^{2} = −c^{2}Γ_{00}^{k}.

Assuming that the gravity field does not change with time (this means that the partials of g_{ij} with respect to t vanish), we compute the Γ_{00}^{k}. First observe that

Γ_{00,h} = \frac{1}{2}(\frac{∂g_{0h}}{∂x^{0}} + \frac{∂g_{0h}}{∂x^{0}} − \frac{∂g_{00}}{∂x^{h}} ).

Assuming that the gravity field does not change with time, the derivatives with respect to x^{0} = ct vanish, and we have

Γ_{00,h} = −\frac{1}{2} \frac{∂g_{00}}{∂x^{h}} = −\frac{1}{2} \frac{∂ϕ_{00}}{∂x^{h}} .

Raising the index (with the inverse metric tensor approximated by the Minkowski metric) implies

Γ_{00}^{k} = g^{hk}Γ_{00,h} ≈ η^{hk}Γ_{00,h} = \frac{1}{2} \frac{∂ϕ_{00}}{∂x^{k}} .

Plugging this into the geodesic equations gives

\frac{d^{2}x^{k}}{dt^{2}} = −\frac{c^{2}}{2} \frac{∂ϕ_{00}}{∂x^{k}} .

This agrees with the Newtonian equations of motion if we put

ϕ_{00} = \frac{2Φ(r)}{c^{2}}.

We do not get conditions for the other ϕ_{ij}; what this means is that just by changing the g_{00}-component of the Minkowski metric we get, in first approximation, the Newtonian theory of gravity!

Homework

22.1 Homework 1, Jan. 21, 2003

1. Let V be a real vector space with a positive definite dot product. Show that ||x|| := \sqrt{x · x} for x ∈ V defines a norm on V.

2. Define ||x||_{0} = max\{|r|, |s|\} and ||x||_{1} = |r| + |s| for x = ( \begin{matrix}r \\ s \end{matrix}) ∈ ℝ^{2}. Show

that || · ||_{0} and || · ||_{1} define norms on ℝ^{2}.

3. Let (V, || · ||) be a normed vector space; show that d(x, y) := ||x − y|| defines a metric on V.

4. Show that

d(P_{1}, P_{2}) = \sqrt{(x_{1} − x_{2})^{2} + (y_{1} − y_{2})^{2}}

for two points P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}) in ℝ^{2} defines a metric.

5. Check whether

d(P, Q) = \begin{cases}0 & if P = Q \\ 1 & if P ≠ Q \end{cases}

defines a metric on ℝ_{2}.

6. Compute the circumference of a circle.

7. The curve c : [0, 2π) ⟶ ℝ^{2} given by c(t) = (r cos t, r sin t) is the circle with radius r and center 0. Compute c′(t), and draw a diagram including the tangent vectors c′(kπ) for k = 0, \frac{1}{2}, 1, \frac{3}{2}.

8. Compute the curvature κ(x) of the parabola y = x^{2} in P = (x, y). Where is the curvature maximal?

9. Compute the curvature κ(x) of the graph of the sine function y = sin x in P = (x, y). Where is the curvature maximal (minimal)?

10. Show that the set of isometries of a metric space form a group.

11. Convince yourself geometrically that for each line ℓ : ax + by + c = 0 there exist points R = (r, s) and T = (t, u) such that ℓ is the set of all points P ∈ ℝ^{2} such that \overline{P R} = \overline{P T}. Why does this imply that isometries of the Euclidean plane map lines into lines? What about circles?

12. Read Section 1.1 and solve the problems.

22.2 Homework 2, Jan. 28, 2003

1. Compute the arc length of the unit circles c : [0, 2π] ⟶ ℝ^{2} : t ⟼ (cos t, sin t) and C : [0, 1] ⟶ ℝ^{2} : t ⟼ (cos 2πt, sin 2πt).

2. Compute the arc length of a segment of the circular helix

c(t) = (a cos t, a sin t, bt).

Find its parametrization by arc length and compute its curvature and its torsion.

3. Use the following method to compute the curvature of the parabola y = \frac{1}{2p}x^{2}:

(a) Show that the lower semicircle with radius R and center (0, R) has equation y = R − \sqrt{R^{2} − x^{2}} = R(1 − \sqrt{1 − (x/R)^{2}}).

(b) Develop y = y(x) into a power series and compare it to the equation of the parabola.

(c) Determine the radius R for which all terms up to the second order agree.

4. Let c(t) = (c_{1}(t), …, c_{n}(t)) be a smooth curve. Show that f(t) = c(t)·c(t) is a real-valued function with derivative f′(t) = 2c(t) · c′(t).

5. Let c(t) = (c_{1}(t), …, c_{n}(t)) be a smooth curve. If t = t(s), show that \frac{d}{ds}c(t(s)) = c′(t)t′(s) using the chain rule for real-valued functions.

6. Exercise 1, p. 121.

22.3 Homework 3, Feb 13, 2003

1. Show that the set M(ℍ) of Moebius transformations m(z) = \frac{az+b}{cz+d}, ad − bc = 1, form a group with respect to composition.

2. Let G be a group and X a set. We say that G acts on X if

• g(x) ∈ X for every x ∈ X and all g ∈ G;

• 1x = x for the identity element 1 ∈ G;

• g(h(x)) = (gh)(x) for all x ∈ X and g, h ∈ G.

Show that M(ℍ) acts on ℍ.

3. Let SL_{2}(ℝ) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) : a, b, c, d ∈ ℝ, ad − bc = 1\}. Define a map ϕ :

SL_{2}(ℝ) ⟶ M(ℍ) by sending S = ( \begin{matrix}a & b \\ c & d \end{matrix} ) to the Moebius transformation m(z) = \frac{az+b}{cz+d}. Show that ϕ is a homomorphism, and determine its kernel. Recall that a homomorphism is a map between two groups that respects composition (i.e., ϕ(gh) = ϕ(g)ϕ(h)), and that the kernel is the set of all elements that get mapped to the identity.

4. Find a curve c(t) : (−1, 1) ⟶ ℍ that traces the semicircle with radius 1 and center (0, 0). Compute the length of the arc from t = −1 + ε to 1 − ε for small ε > 0. What happens as ε ⟶ 0?

5. Define the derivative of a complex valued function z(t) = x(t) + iy(t), where x and y are real-valued, by z′(t) = x′(t) + iy′(t). State and prove the Leibniz (product) rule.

6. Define a map J(z) = \frac{iz+1}{z+1} and show that it gives a bijection between the upper half plane ℍ and the unit disc 𝔻.

22.4 Homework 4, Feb 13, 2003

1. Show that the set M(ℍ) of Moebius transformations m(z) = \frac{az+b}{cz+d}, where ad − bc = 1, forms a group with respect to composition.

2. Let G be a group and X a set. We say that G acts on X if

• g(x) ∈ X for every x ∈ X and all g ∈ G;

• 1x = x for the identity element 1 ∈ G;

• g(h(x)) = (gh)(x) for all x ∈ X and g, h ∈ G.

Show that M(ℍ) acts on ℍ.

3. Let SL_{2}(ℝ) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) : a, b, c, d ∈ ℝ, ad − bc = 1\}. Define a map ϕ :

SL_{2}(ℝ) ⟶ M(ℍ) by sending S = ( \begin{matrix}a & b \\ c & d \end{matrix} ) to the Moebius transformation m(z) = \frac{az+b}{cz+d}. Show that ϕ is a homomorphism, and determine its kernel. Recall that a homomorphism is a map between two groups that respects composition (i.e., ϕ(gh) = ϕ(g)ϕ(h)), and that the kernel is the set of all elements that get mapped to the identity.

Let A = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) and A′ = \{( \begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix} ) be two matrices in SL_{2}(ℝ). Then

AA′ = \{( \begin{matrix}aa′+bc′ & ab′+bd′ \\ ca′+dc′ & cb′+dd′ \end{matrix} ). On the other hand, ϕ(A) ∘ ϕ(A′) is the Moebius

transformation sending z to

\frac{a\frac{a′z+b′}{c′z+d′} + b}{c\frac{a′z+b′}{c′z+d′} + d} = \frac{a(a′z + b′) + b(c′z + d′)}{c(a′z + b′) + d(c′z + d)} = \frac{(ab′ + bd′)z + ab′ + bd′}{(ca′ + dc′)z + cb′ + dd′} ,

and this shows that ϕ is indeed a homomorphism.

In order to compute its kernel, we hav to determine which matrices get mapped to the identity of the group of Moebius transformations; this identity is given by m(z) = z = \frac{1z+0}{0z+1}. If ϕ(A) = m(z), then ϕ(A) = \frac{az+b}{cz+d} = z for all x ∈ ℍ; this gives az +b = cz^{2} +dz or cz^{2} +(d−a)z −b = 0. This equation must hold for all z ∈ ℍ, hence b = c = 0 and a = d. From ad − bc = 1 we deduce that ad = 1, hence either a = d = 1 or a = d = −1, and so the kernel consists of only two elements, namely I == \{( \begin{matrix}1 & 0 \\ −1 & 0 \end{matrix} )

and −I = \{( \begin{matrix}−1 & 0 \\ 1 & 0 \end{matrix} ).

4. Find a curve c(t) : (−1, 1) ⟶ ℍ that traces the semicircle with radius 1 and center (0, 0). Compute the length of the arc from t = −1 + ε to 1 − ε for small ε > 0. What happens as ε ⟶ 0?

5. Define the derivative of a complex valued function z(t) = x(t) + iy(t), where x and y are real-valued, by z′(t) = x′(t) + iy′(t). State and prove the Leibniz (product) rule.

6. Let L be a straight line or a circle (in the usual Euclidean sense), meeting the real line orthogonally at a point α (pick one of the two in the case of a circle). Show that T (z) = −\frac{1}{z−α} + β is a Moebius transformation, and that for a suitable β it maps L to the imaginary axis.

7. Define a map J(z) = \frac{iz+1}{z+1} and show that it gives a bijection between the upper half plane ℍ and the unit disc 𝔻.

22.5 Homework 5, Feb 27, 2003

1. Define the length of a piecewise smooth curve c : [a, b] ⟶ 𝔻 : t ⟼ z(t) = x(t) + iy(t) in the Poincaré disc by

L(c) = ∫_{a}^{b} \frac{2\sqrt{ẋ^{2} + ẏ^{2}}}{1 − x^{2} − y^{2}} dt.

Compute the length of the line segment from the origin to (r, 0).

2. Show that the map J : ℍ ⟶ 𝔻 : z ⟼ \frac{iz+1}{z+i} preserves the length of curves and thus induces an isometry from ℍ to 𝔻.

3. Show that the Poincaré transformations M(z) = \frac{αz+\overline{β}}{βz+\overline{α}}, where α, β ∈ ℂ with |α|^{2} − |β|^{2} = 1, act on the unit disc 𝔻.

4. Let z, w ∈ 𝔻 be any two points in the unit disc. Show that there is a Poincaré transformation M such that M(z) = w.

5. Let α be a complex number with absolute value 1. Show that the map z ⟼ αz is a rotation of the complex plane as well as a Poincaré transformation.

6. Show that the set of matrices

\{M = (\begin{matrix}α & \overline{β} \\ β & \overline{α} \end{matrix}) : det M = 1\}

form a group called the special unitary group SU(1, 1). Show that the map sending M to the transformation z ⟼ \frac{αz+\overline{β}}{βz+\overline{α}} is a homomorphism, and determine its kernel.

22.6 Homework 6, March 12, 2003

1. There are three types of Moebius transformations M(z) = \frac{az+b}{cz+d}: we call

M

elliptic  if a + d < 2,
parabolic if a + d = 2,
hyperbolic if a + d > 2.

We know that Moebius transformations act on the upper half plane. Actually, they also act on the ‘boundary’ \overline{ℝ} = ℝ ∪ \{∞\}.

(a) Explain what the last assertion means, and prove it.

(b) Determine the conditions under which ∞ is a fixed point of M.

(c) Hyperbolic transformations have two fixed points in \overline{ℝ}.

(d) Parabolic transformations have one fixed point in \overline{ℝ}.

(e) Elliptic transformations have a pair of complex conjugate fixed points, and therefore a single fixed point in ℍ.

2. Let M be a hyperbolic Moebius transformation. The unique geodesic connecting its two fixed points is called the axis of M. Prove that M maps its axis onto itself.

3. Prove that the distance d(z, w) of z, w ∈ ℍ is given by

sinh \frac{1}{2}d(z, w) = \frac{|z − w|}{2\sqrt{Im(z) Im(w)}}.

Show that this implies that

d(z, w) = ln \frac{|z − \overline{w}| + |z − w|}{|z − \overline{w}| − |z − w|}.

Hints:

(a) Show that sinh \frac{1}{2}d(z, w) is invariant under Moebius transformations.

(b) Show that \frac{|z−w|}{2\sqrt{Im(z) Im(w)}} is invariant under Moebius transformations.

(c) Show that it is sufficient to prove the theorem for points z = ia and w = ib on the imaginary axis.

(d) Complete the proof.

4. The final will be a take home exam; here comes the first problem you’re supposed to work on. Define the length of a piecewise smooth curve c(t) = x(t) + iy(t), where t ∈ [a, b], in the complex plane ℂ by

L(c) = ∫_{a}^{b} \frac{\sqrt{ẋ^{2} + ẏ^{2}}}{1 + \sqrt{x^{2} + y^{2}}} dt.

22.6. HOMEWORK 6, MARCH 12, 2003 77

(a) Compute the arclength of a (Euclidean) circle with center 0 and radius r with respect to this definition of length.

(b) Determine the shortest path between 1 − i and 1 + i.

22.7 Homework 7, March 26, 2003

1. Define the length of a piecewise smooth curve c(t) = x(t) + iy(t), where t ∈ [a, b], in the complex plane ℂ by

L(c) = ∫_{a}^{b} \frac{\sqrt{ẋ^{2} + ẏ^{2}}}{1 + \sqrt{x^{2} + y^{2}}} dt.

(a) Express the formula for arc length in polar coordinates.

(b) Determine the shortest path between z = 2 and z = 3.

2. Define the length of a piecewise smooth curve c(t) = x(t) + iy(t), where t ∈ [a, b], in the complex plane ℂ by

L(c) = ∫_{a}^{b}(2 + x^{2} + y^{2})\sqrt{ẋ^{2} + ẏ^{2}} dt.

Show that the shortest path between z = 0 and z = 1 is a line, and compute its length.

3. Show that if an object A has velocity v < c with respect to Õ, and if Õ has velocity w < c with respect to O, then the velocity of A with respect to O is also < c.

4. Page 47, problems 4, 5; Page 71, problems 4, 6, 7.

5. Read Chapter 2.

22.8 Homework 8, April , 2003

1. Let S be the plane parametrized by by p(q^{1}, q^{2}) = (q^{1}, q^{2}, aq^{1} + bq^{2} + c).

(a) Compute the Euclidean distance between the points O = p(0, 0) and P = p(c^{1}, c^{2}).

(b) Compute the metric tensor g of S at P.

(c) Show that v = p(c^{1}, c^{2})−p(0, 0) is a tangent vector in T_{P} S by writing it as a linear combination of x_{1} and x_{2}.

(d) Compute v · v = v^{t}gv and compare the result with (a).

2. Consider the cylinder S with radius a > 0 parametrized by

p(q^{1}, q^{2}) = (a cos q^{1}, a sin q^{1}, q^{2}).

Find a curve c : ℝ ⟶ ℝ^{2} such that z = q ∘ c is the helix parametrized by z(t) = (a cos t, a sin t, bt), and compute the arc length of the helix using the metric tensor g of S.

3. § 5.1, problems 1, 2, 5.

4. Here comes problem # 2 for the take home final.

(a) Prove that the distance between two points in the hyperbolic plane is given by tanh \frac{1}{2}d(z, w) = \frac{|z−w|}{|z−\overline{w}|}.

(b) Let C be a Euclidean circle with Euclidean radius r and Euclidean center (0, b) in the upper half plane. The claim is that C is also a hyperbolic circle with hyperbolic radius s and huperbolic center (0, c). Explain what this means.

(c) Assuming the claim is true, compute the hyperbolic center of C.

(d) Prove the claim.

Exams

23.1 Midterm 1

1. Complete the definition: A function d mapping pairs of elements x, y in a set X to the real numbers is called a metric if d has the following properties:

•

•

•

Give examples of two different metrics on the Cartesian plane X = ℝ^{2} (you do not have to justify your answer).

2. Compute the arc length of the semicircle with radius R in the Euclidean plane.

3. Derive the formula κ = |y′′|(1+y′^{2})^{−3/2} for curvature of functions y = f(x) by approximating f with a circle.

4. Compute the curvature κ(x) of the parabola y = ax^{2} (a > 0) at (x, y) and determine where κ becomes maximal.

5. Consider the curve c(t) = (cos t, sin t, 2t) in ℝ^{3}.

(a) We have shown that a curve C(t) = (x(t), y(t), z(t)) lives on a sphere with radius r > 0 if and only if x(t)^{2} + y(t)^{2} + z(t)^{2} = r^{2}. Write down a similar equation that has to be satisfied for curves C living on a cylinder.

(b) Show that c ‘lives’ on a cylinder.

(c) Compute the arc length of c as t runs through [0, t].

(d) Write down a unit speed parametrization of c.

(e) Compute T and check that ||T || = 1.

(f) Compute N and the curvature of c.

6. Complete:

(a) A smooth curve with curvature κ = 0 is

(b) A smooth plane curve with constant curvature κ is

(c) A smooth curve with nonzero curvature and torsion τ = 0 lives in a (d) The geodesics in the Euclidean plane are

(e) The isometries of the Euclidean plane are

(f) The geodesics in the hyperbolic plane are

(g) The geodesics on a sphere are

7. For a positive real number r, consider the points P = −r+ri and Q = r+ri in the upper half plane.

(a) Write down an equation for the geodesic connecting P and Q; (you don’t have to justify the answer).

(b) Compute the length of P Q (the shortest distance between P and Q).

(c) How does the result depend on r? How can this be explained?

8. Show that the Moebius transformations M(z) = \frac{az+b}{cz+d} preserve the upper half plane.

9. Find a Moebius transformation different from the identity that fixes 1 + i.

10. Consider the map J : ℍ ⟶ 𝔻 defined by J(z) = \frac{z−i}{z+i}.

(a) Show that J is a bijection between ℍ and 𝔻.

Now let I denote the inverse map of J.

(b) Let w(t) = t (−1 < t < 1) denote the diameter of the unit disc on the real axis. Show I maps this diameter bijectively to the positive imaginary axis.

(c) Now consider the image of the diameter w(t) = it (−1 < t < 1) and compute its image under I. What is the geometric shape of this image? (It may help to compute the images of the ‘endpoints’ corresponding to t = −1 and t = +1).

82 CHAPTER 23. EXAMS

23.2 Midterm 2 (May 2, 2003)

Consider the surface parametrized by x(q^{1}, q^{2}) = (q^{1}, q^{2}, (q^{1})^{2} + (q^{2})^{2}).

Part I.

1. Compute the tangent vectors x_{1}, x_{2}, and the unit normal vector n (you may want to use the abbreviation Δ = \sqrt{1 + 4(q^{1})^{2} + 4(q^{2})^{2}}).

2. Compute the metric tensor g = (g_{ij}).

3. Explain why the partial derivatives n_{1} and n_{2} lie in the plane spanned by x_{1} and x_{2}.

4. Find the coordinates b_{j}^{i} by writing n_{j} = b_{j}^{i}x_{i}. Define B̃ = (b_{j}^{i}), and compute the Gaussian curvature K = det B̃.

5. Where is the curvature maximal?

6. The curvature of the parabola y = x^{2} at the origin is 2. What does this imply for the Gaussian curvature of x at the origin (0, 0, 0)? Does this agree with your calculation?

Part II

Now compute the Gaussian curvature from g.

1. Compute the inverse matrix g^{−1} = (g^{ij}) of g.

2. Compute the Christoffel symbols of the first kind Γ_{jk,h} from

2Γ_{jk,h} = \frac{∂g_{kh}}{∂q^{j}} + \frac{∂g_{jh}}{∂q^{k}} − \frac{∂g_{jk}}{∂q^{h}} .

(Γ_{11,1} = 4q^{1})

3. Compute the Christoffel symbols of the second kind by raising indices:

Γ_{jk}^{m} = g^{hm}Γ_{jk,h}.

(Γ_{11}^{1} = \frac{4q^{1}}{Δ^{2}} ).

4. Compute the entries R_{212}^{1} and R_{212}^{2} of the mixed Riemann tensor

R_{jkl}^{i} = \frac{∂Γ_{jl}^{i}}{∂q^{k}} − \frac{∂Γ_{jk}^{i}}{∂q^{l}} + Γ_{jl}^{p}Γ_{pk}^{i} − Γ_{jk}^{p}Γ_{pl}^{i}.

(R_{212}^{1} = 4\frac{1+4(q^{2})^{2}}{Δ^{4}} ).

23.2. MIDTERM 2 (MAY 2, 2003) 83

5. Compute the component R_{1212} of the Riemann tensor R_{hjkl} = g_{ih}R_{jkl}^{i}.

6. Compute K = \frac{R_{1212}}{det g} .

23.3 Final

1. From homework # 6

Define the length of a piecewise smooth curve c(t) = x(t) + iy(t), where t ∈ [a, b], in the complex plane ℂ by

L(c) = ∫_{a}^{b} \frac{\sqrt{ẋ^{2} + ẏ^{2}}}{1 + \sqrt{x^{2} + y^{2}}} dt.

(a) Compute the arclength of a (Euclidean) circle with center 0 and radius r with respect to this definition of length.

(b) Can you determine the shortest path between 1−i and 1+i? (Identify ℂ ≃ ℝ^{2}, and find a metric tensor g that induces this metric. Compute the geodesic equations; you don’t have to solve these differential equations.)

2. From homework # 8

(a) Prove that the distance between two points in the hyperbolic plane is given by tanh \frac{1}{2}d(z, w) = \frac{|z−w|}{|z−\overline{w}|}.

(b) Let C be a Euclidean circle with Euclidean radius r and Euclidean center (0, b) in the upper half plane. The claim is that C is also a hyperbolic circle with hyperbolic radius s and hyperbolic center (0, c). Explain what this means.

(c) Assuming the claim is true, compute the hyperbolic center of C.

(d) Prove the claim.

3. Describe in your own words (without formulas except when absolutely necessary; no teamwork on this one) the different definitions of curvature we have used, and explain their advantages and disadvantages.

4. Consider the torus T_{a,r} (a > r > 0) parametrized by

x(q^{1}, q^{2}) = ((a + r cos q^{2}) cos q^{1}, (a + r cos q^{2}) sin q^{1}, r sin q^{2}).

(a) Compute the Gauss map n(q^{1}, q^{2}) and the Gaussian curvature.

(b) Compute the metric tensor (g_{ij}) and its inverse (g^{ij}).

(c) Compute the b_{jk} and b_{j}^{i}.

(d) Compute the Christoffel symbols of the second kind using i) Γ_{jk,l} = x_{jk} · x_{l}; ii) 2Γ_{jk,l} = \frac{∂g_{jl}}{∂q^{k}} + \frac{∂g_{kl}}{∂q^{j}} − \frac{∂g_{jk}}{∂q^{l}} .

(e) Compute Γ_{jk}^{i} = g^{il}Γ_{jk,l} and verify that x_{jk} = Γ_{jk}^{i}x_{k} + b_{ij}n.

(f) Compute the mixed Riemann curvature tensor R_{jkl}^{i}.

(g) Compute R_{hjkl} = g_{ih}R_{jkl}^{i} and K = R_{1212}/ det g.

5. (p. 291, problem 4): Use the geodesic equations to show that if q(t) = (q^{1}(t), q^{2}(t)) is a geodesic, then so is q_{b}(t) = (b + q^{1}(t), q^{2}(t)).

This is a special case of a much more general result concerning geodesics in the hyperbolic plane: state it (extra credit: prove it).

6. (p. 291, problem 6) Compute the Riemann curvature tensor R_{jkl}^{i} for the hyperbolic plane ℍ, and compute R_{hjkl} = g_{ih}R_{jkl}^{i}. Show that the Gaussian curvature K = R_{1212}/ det g = −1 everywhere on ℍ.

Solutions to the Final Exam

7. From homework # 8

(a) Prove that the distance between two points in the hyperbolic plane is given by tanh \frac{1}{2}d(z, w) = \frac{|z−w|}{|z−\overline{w}|}.

(b) Let C be a Euclidean circle with Euclidean radius r and Euclidean center (0, b) in the upper half plane. The claim is that C is also a hyperbolic circle with hyperbolic radius s and hyperbolic center (0, c). Explain what this means.

(c) Assuming the claim is true, compute the hyperbolic center of C.

(d) Prove the claim.

Bibliography

[1] James J. Callahan, Spacetime Geometry, Springer-Verlag, 2000

The main reference

[2] George A. Jennings, Modern Geometry with Applications, Springer-Verlag, 1994

Spherical Geometry, Kepler's Laws, Special Relativity

[3] Svetlana Katok, Fuchsian Groups, Chicago Lectures in Mathematics, 1992

A book on number theory from which I lifted most of the
hyperbolic geometry.

[4] Barrett O’Neil, Elementary Differential Geometry, Academic Press, 1966

Frenet's Formulas