Differential Geometry

Andrew Kobin Spring 2015 Contents Contents

Contents

0 Introduction 1

1 Curves 3 1.1 Parametrized Differentiable Curves ...... 3 1.2 Finding a Parametrization ...... 6 1.3 Parametrization by Arc Length ...... 10 1.4 Curvature ...... 11 1.5 Curvature in R3 ...... 14 2 Surfaces 21 2.1 The Unit Sphere ...... 21 2.2 Regular Surfaces ...... 23 2.3 Differentiable Functions on Surfaces ...... 30 2.4 Tangent Spaces ...... 33 2.5 First Fundamental Form ...... 37

3 Curvature of Surfaces 40 3.1 The Gauss Map ...... 40 3.2 Types of Curvature ...... 44 3.3 Second Fundamental Form ...... 46 3.4 Gaussian Curvature and Area ...... 50

4 Intrinsic Geometry 51 4.1 Isometries ...... 51 4.2 Christoffel Symbols and Gauss’ Theorema Egregium ...... 58 4.3 Geodesics and the Covariant Derivative ...... 62 4.4 The Gauss-Bonnet Theorem ...... 71

i 0 Introduction

0 Introduction

The notes contained in this document survey a course on differential geometry taught by Dr. Stephen Robinson in the Spring of 2015 at Wake Forest University. The main text used in the course is Differential Geometry of Curves and Surfaces by Manfredo P. Do Carmo. We begin with two motivating questions:

1 What constitutes a “straight line” on the surface of a sphere?

2 If one draws a triangle on a sphere, what is the sum of the interior angles?

The answer to 1 is a great circle, which is an example of a geodesic, or shortest path between two points on a surface.

The intuition is that any local “snapshot” of the great circle appears as a line segment. How does this relate to 2 ? Consider the following triangle with angles labelled θ1, θ2 and θ3. What is θ1 + θ2 + θ3?

θ3 θ2 θ1

If the triangle were in the plane, the angles would sum to π = 180◦. To get a better handle on the spherical situation, let’s look at a special case. Mark the north pole N and draw two great circles passing through N, which must intersect the equator in right angles. This gives us a triangle:

N

θ1

1 0 Introduction

The sum of the angles here is 180 + θ1 and θ1 can vary from 0 (a degenerate triangle) to 2π (another degenerate triangle), so we see that π < θ1 + θ2 + θ3 < 3π. One thing to notice is that the size of the angles is proportional to the area of the triangle. Another way to view the problem is in terms of area. Look at the wedge shape formed by extending the lines on either side of θ1.

θ3

θ1 θ2

The proportion of the total surface area of the sphere covered by this wedge region is A1 = θ1 2 2 π · 4πR = 4R θ1 where R is the radius of the sphere. Likewise, we can form two more wedges with areas 2 2 A2 = 4R θ2 and A3 = 4R θ3. Adding up all the wedges gives the area of the sphere, but we have counted the area A of 2 2 the triangle θ1θ2θ3 three extra times. Therefore 4πR + 4A = 4R (θ1 + θ2 + θ3). Solving for the sum of the angles gives us a nice formula in terms of the area of the triangle and radius of the sphere: 1 θ + θ + θ = π + A. 1 2 3 R2 This will work for any triangle on the surface of a sphere. Notice that as A gets small (a “local snapshot”) the sum of the angles approaches π as one would see in the plane. On the other hand, as the sphere gets bigger in relation to the size of the triangle, the sum also approaches π. This expresses the property of the sphere having ‘local flatness’. The above result is a special case of the Gauss-Bonnet Formula:

n n X Z si+1 ZZ X kg(s) ds + K dσ + θi = 2π. i=0 si R i=0 It will be our goal to study and prove this result towards the end of the notes. Another special case of Gauss-Bonnet is the formula ZZ K dσ = 2πχ(S) S where S is a surface, K is curvature and χ(S) is the Euler characteristic of the surface. In order to reach a point of understanding these formulas, we will study the following topics: curves, surfaces and their curvature; tangent spaces; the first and second fundamental form; the Gauss map; and finally isometries and the intrinsic geometry of surfaces.

2 1 Curves

1 Curves

1.1 Parametrized Differentiable Curves

In this chapter we will consider three-dimensional Euclidean space, R3, and in particular certain one-dimensional subsets of R3. Definition. A parametrized differentiable curve is a map α : I → R3, where I = (a, b) is an interval of R, such that α is differentiable. We are treating the word differentiable rather loosely — in practice one may assume these differentiable functions to be infinitely differentiable, or at the very least the curves should have enough derivatives to perform the desired computations. A parametrized differentiable curve in R3 is given by three differentiable component functions, α(t) = (x(t), y(t), z(t)), and an interval (a, b) over which t ranges. At various times we may want to compute: ˆ Velocity: α0(t) = (x0(t), y0(t), z0(t)).

ˆ Speed: |α0(t)| = px0(t)2 + y0(t)2 + z0(t)2.

ˆ Acceleration: α00(t) = (x00(t), y00(t), z00(t)). It may also be helpful at times to express acceleration as the linear combination of position α(t) and velocity α0(t).

Z b ˆ Arc length: |α0(t)| dt. It is typical to use ds for arc length integrals. a Definition. If a curve α(t) has constant speed |α0(t)| = 1, we say α is parametrized by arc length. Examples.

1 Consider α(t) = (cos t, sin t, t), −∞ < t < ∞, the standard helix in R3. Velocity and speed are calculated to be p √ α0(t) = (− sin t, cos t, 1) and |α0(t)| = sin2 t + cos2 t + 1 = 2.

The helix is thus our first example of a curve with constant speed. This is especially nice for computing arc length: Z 2π Z 2π√ √ |α0(t)| dt = 2 dt = 2 2π. 0 0 √ If we normalize the curve (e.g. by dividing the components by 2), the helix would be parametrized by arc length. The reason for this is that if |α0(t)| = 1 for all t, then arc length is found by simply subtracting the starting and ending points of the parametric interval: Z b Z b |α0(t)| dt = 1 dt = b − a. a a

3 1.1 Parametrized Differentiable Curves 1 Curves

Next, acceleration is computed to be

α00(t) = (− cos t, − sin t, 0).

Notice that α0(t) and α00(t) are always orthogonal, so that acceleration is only ever measuring the turn of the helix. This will be the case for any curve with constant speed.

2 Let α(t) = (t3, t2), −1 < t < 1. This traces out the curve y = x2/3 in the xy-plane (in the future, trace will refer to the shape in the plane created by running through the values of t for a curve).

Notice that this has a cusp at the origin, but α(t) is differentiable in its given parametric form. How can this be? Intuitively, α0(t) measures the velocity of a particle travelling along the curve and as the particle approaches the origin, it does so at a slower and slower rate, so (x0(t), y0(t)) approaches (0, 0). This illustrates the difference between parametrized differential curves and functions from regular calculus, which here are realized as the trace of pdc’s.

3 Consider α(t) = (cos t, sin t) for 0 < t < π, which traces out the upper half of the unit circle in a counterclockwise direction.

Another possible parametrization for the upper semicircle is √ α(t) = (t, 1 − t2) − 1 < t < 1

but this parametrized differential curve has clockwise orientation, so these are formally not the same curve. However, their traces are the same.

4 1.1 Parametrized Differentiable Curves 1 Curves

4 Let α(t) = (t3 − 4t, t2 − 4), −∞ < t < ∞.

This curve has two self-intersections, at t = ±2. We typically avoid these types of curves and stick to working with simple curves (defined below).

5 Let α(t) be a parametrized curve which does not pass through the origin. If α(t0) 0 is the point of the trace of α closest to the origin and α (t0) 6= 0, then the position 0 2 vector α(t0) is orthogonal to the tangent vector α (t0). To prove this for R (the proof n generalizes to R ), let t0 be the value such that |α(t0)| is minimized. From calculus, we know that the derivative of this term is 0, which may be expanded in the following manner d d 0 = |α(t)| = px(t)2 + y(t)2 dt dt 1 = (x(t)2 + y(t)2)−1/2 · [2x(t)x0(t) + 2y(t)y0(t)] 2 x(t)x0(t) + y(t)y0(t) = . px(t)2 + y(t)2 Since the curve does not pass through the origin, we know the denominator is nonzero. 0 0 Therefore the above shows that x(t)x (t) + y(t)y (t) = 0 (when t = t0), but this is 0 precisely the same as the dot product α(t0) · α (t0) = 0. Hence the position and tangent vectors are orthogonal at t = t0.

6 Let α : I → R3 be a parametrized curve with α0(t) 6= 0 for all t ∈ I. Then |α(t)| is a nonzero constant if and only if α(t) is orthogonal to α0(t) for all t ∈ I: d |α(t)|= 6 0 is constant ⇐⇒ |α(t)| = 0 dt d ⇐⇒ px(t)2 + y(t)2 + z(t)2 = 0 dt ⇐⇒ x(t)x0(t) + y(t)y0(t) + z(t)z0(t) = 0 as above ⇐⇒ α(t) · α0(t) = 0 ⇐⇒ α(t) and α0(t) are orthogonal for all t.

5 1.2 Finding a Parametrization 1 Curves

Two important characteristics of parametrized differential curves are defined below. Definition. A curve α(t) is regular if α0(t) 6= 0 for all t ∈ I. A regular curve will never have corners or cusps like the curve in 2 . Definition. A curve is simple if it has no self-intersections. The curve in 4 is not simple. In general we prefer to work with simple, regular curves. The fundamental theorems described in Sections 1.4 and 1.5, for example, will fully charac- terize regular curves in R2 and R3.

1.2 Finding a Parametrization

In general, it is hard to find a parametrization of an arbitrary curve in Rn. There are some basic curves that are good starting places for finding the parametrization of more complicated curves. These include:

ˆ Lines: `(t) = p0 + (p1 − p0)t, −∞ < t < ∞. For example, `(t) = (3, 4, 6) + (−1, −2, 4)t is a line in R3.

ˆ Circles: α(t) = p0 + (r cos t, r sin t) for a circle of radius r, for 0 < t < 2π. The most important example is the unit circle S1: α(t) = (cos t, sin t), 0 < t < 2π. ˆ Ellipses: a ‘stretched’ circle, obtained by adjusting the coefficients of each component: α(t) = (a cos t, b sin t), 0 < t < 2π. For example, the ellipse below may be parametrized by α(t) = (3 cos t, 2 sin t).

1

1 2

ˆ Helix: α(t) = (cos t, sin t, t) and other variants. z

y x

6 1.2 Finding a Parametrization 1 Curves

Examples.

1 A cycloid is formed by tracing a point on a circle as the circle ‘rolls’ along the x-axis. For example, consider the cycloid formed by the unit circle.

t = 0

To obtain a parametrization, consider adding the following triangle to the diagram of the circle at any given time t.

t

(x, y)

t

We are looking for a formula for (x, y) at any time t; to determine this, notice that the angle formed by the two radii above is equal to the arc length which the circle has rolled so far, that is, t. Since the radius of the circle is 1, trigonometry tells us that the legs of the triangle are sin t and cos t, for the horizontal and vertical legs, respectively. Thus the parametric form of this curve is

α(t) = (t − sin t, 1 − cos t), −∞ < t < ∞.

If we want to compute the arc length of the cycloid through, say, one full rotation of

7 1.2 Finding a Parametrization 1 Curves

the circle, consider Z 2π Z 2πq |α0(t)| dt = (1 − cos t)2 + sin2 t dt 0 0 Z 2πp = 1 − 2 cos t + cos2 t + sin2 t dt 0 Z 2π√ = 2 − 2 cos t dt 0 Z 2π  t  = 2 sin dt by a half-angle identity 0 2   t 2π = −4 cos = −4 cos(π) + 4 cos(0) = 8. 2 0 Notice that the cycloid α(t) = (t − sin t, 1 − cos t) is differentiable, but the trace it produces has corners:

In general, cycloids are examples of differentiable curves that are not regular. 2 The cissoid of Diocles is formed by the following procedure. Let 0A = 2a be the diameter of a circle S1 and 0Y and AV be the tangents to S1 at 0 and A, respectively. A half-line r is drawn from 0 which meets the circle S1 at C and the line AV at B. On 0B mark off the segment 0p = CB. If we rotate r about 0, the point p will describe the cissoid of Diocles. This is pictured below.

B C K

p

0 A J M

8 1.2 Finding a Parametrization 1 Curves

Fill in the triangles 0CM and CBK; the whole figure is enlarged below. If θ is the angle made by 0B and 0A, then the segment AB has length 2a tan θ.

B C K

p

0 θ A J M

By construction the triangles 0pJ and CBK are equal. Set C = (p, q) and t = tan θ; then by the previous statement we have

x = 2a − p and y = 2at − q.

Therefore we will obtain a parametrization for p = (x, y) if we can solve for p and q in terms of t and a. Notice that (p, q) lies on the circle, so

a2 = (p − a)2 + q2.

Expanding this, we have

a2 = p2 − 2ap + a2 + q2 =⇒ 0 = p2 − 2ap + q2.

Also observe that by our choice of t, the slope of the line 0B is t. This means that q = tp, and we can plug this into the circle equation above to solve for p:

0 = p2 − 2ap + (tp)2 = (1 + t2)p2 − 2ap = p[(1 + t2)p − 2a].

2a 2a 2at2 This gives us p = 0 or p = , and consequently x = 2a − = . Using 1 + t2 1 + t2 1 + t2 2at3 the fact that y = tx, we also deduce that y = . Hence the parametrization for 1 + t2 the cissoid of Diocles is  2at2 2at3  α(t) = , , t ∈ . 1 + t2 1 + t2 R

9 1.3 Parametrization by Arc Length 1 Curves

1.3 Parametrization by Arc Length

The main question in this section is: Is it possible to parametrize every regular curve by arc length? Recall that this means |α0(t)| = 1 for all t. In practice, this is difficult but we will show that it is theoretically possible. Parametrizing by arc length is useful because the geometry of the trace of a curve does not depend on the speed of a particle travelling along the curve, so calculations will be made easier by the assumption that |α0(t)| ≡ 1. 3 Assume α :(a, b) → R is a parametrized differentiable curve and choose t0 ∈ (a, b).

α(t0)

t = a α(t) s(t) t = b

To measure arc length s(t) from the point α(t0), we have Z t s(t) = |α0(τ)| dτ. t0

We know that s(t) is nonnegative (when moving forward in time) and s(t0) = 0. Also notice ds 0 ds that dt = |α (t)| by the Fundamental Theorem of Calculus, and since α is regular, dt 6= 0 for any t. Thus s(t) is always increasing. This allows us to think of t as a function of s, and this function t(s) is differentiable with dt 1 = . ds |α0(t)| (This is a single variable version of the Inverse Function Theorem.) Considerα ¯(s), a potential parametrization of α by arc length. This should beα ¯(s) = α(t(s)) on the interval (s(a), s(b)). This really does reproduce the entire curve, and by the Chain Rule, dα¯ dα dt 1 α0(t) = = α0(t) = . ds dt ds |α0(t)| |α0(t)| Thus eachα ¯0 vector is unit length; we have successfully parametrized α by arc length. Example 1.3.1. In practice it may be difficult to come up with such a parametrization, since the integral defining s(t) may not have a closed form. However, the standard helix is a nice curve in the sense that it is easy to parametrize by arc length using the above procedure. Let √α(t) = (cos t, sin t, t), −π < t < π. Speed, one will recall, is constant for the 0 helix: |α (t)| = 2. Choose t0 = 0. Then Z t√ √ s(t) = 2 dt = 2t. 0 Inverting the function gives us t(s) = √1 s so the new parametrization is 2   1   1  1  √ √ α¯(s) = α(t(s)) = cos √ s , sin √ s , √ s − 2π < s < 2π. 2 2 2

10 1.4 Curvature 1 Curves

To see that this parametrizes α by arc length, consider 1   s   s   α¯0(s) = √ − sin √ , cos √ , 1 2 2 2 s 1  s   s  =⇒ |α¯0(s)| = √ sin2 √ + cos2 √ + 1 2 2 2 √ 2 = √ = 1. 2

1.4 Curvature

Given a regular parametrized differential curve, parametrized by arc length, how do we quantify the curves and slope of its trace? Our goal will be to prove

Theorem 1.4.1. In R2, a regular curve is completely characterized by its curvature. Let α(s) = (x(s), y(s)) be a regular curve in the plane.

N(s) = (−y0(x), x0(s))

T (s) = α0(s) = (x0(s), y0(s))

We will denote the tangent vectors by T (s). We also have a normal vector N(s) = (−y0(s), x0(s)) which is unit length and orthogonal to T (s) for every s. Notice that (T · T )0 = 0 so by the chain rule, T 0 · T + T · T 0 = 0 =⇒ T 0 · T = 0. Hence T 0 is parallel to N, i.e. T 0 = kN for some constant k. Definition. The curvature of a regular parametrized differential curve at s is the constant k(s) for which T 0(s) = k(s)N(s). The reason for the construction using T and N is that at any point (x(s), y(s)), {T,N} forms a local, orthonormal basis for the vectors near α(s). Conventionally, a basis is called positively-oriented if the determinant of the matrix whose columns are the basis vectors (alternatively, the change of basis matrix from the standard Euclidean basis) is positive, and negatively-oriented if the determinant is negative. By this convention, the standard basis is always assumed to have positive orientation. Returning to the above work, notice that k = T 0 · N by taking dot products of both sides of the equation in the curvature definition. Also note that since N and T are orthonormal,

11 1.4 Curvature 1 Curves

the rate of change of N is equal to the rate of change of T , but with a sign change: N 0 = −kT . Rigorously, N · T = 0 =⇒ N 0 · T + N · T 0 = 0 =⇒ N 0 · T = −k and by normality of N,

N · N = 1 =⇒ N 0 · N = 0 by the same logic as above so N 0 = aT for some constant a. Taking dot products on both sides again gives us N 0 ·T = a, so it must be that a = −k and we conclude N 0 = −kT . To summarize, we have two important equations T 0 = kN and N 0 = −kT. This is a system of first order differential equations called the Fren´etequations (for R2). Given the orientation chosen (positive via the right-hand rule), k > 0 corresponds to a left turn and k < 0 corresponds to a right turn.

N(s)

θ

T (s)

Let θ(s) denote the angle between T (s) and the fixed vector (1, 0). Then x0 = cos θ and y0 = sin θ. We can differentiate these:

x00 = − sin θ · θ0 and y00 = cos θ · θ0.

Notice that (x00, y00) = T 0 = kN = k(−y0, x0) = (−ky0, kx0). Thus

x00 = − sin θ · θ0 = −y0θ0 and y00 = cos θ · θ0 = x0θ0.

This implies −y0θ0 = −ky0 and x0θ0 = kx0, so we have discovered that curvature describes the rate of change of the angle θ: θ0 = k. This gives a physical interpretation of our intuition about the idea of curvature. This process can be reversed: given a function k(s), a starting point and initial angle θ, there is one and only one function α(s) with k as its curvature and these initial conditions. Thus k(s) can be used to reconstruct α. This discussion leads to the so-called fundamental theorem of curves for R2: Theorem 1.4.2 (Fundamental Theorem of Curves in the Plane). Given a differentiable function k : I → [0, ∞), a point p0 and an angle θ0, there is a unique, regular, parametrized 2 0 differential curve α : I → R with k(s) as its curvature, α(0) = p0 and α (0) = (cos θ0, sin θ0).

12 1.4 Curvature 1 Curves

Example 1.4.3. What if we want to compute some curvature functions for actual parametrized differential curves? Consider 1 The circle α(t) = (R cos t, R sin t), −∞ < t < ∞ is not parametrized by arc length, so our procedure above will not work. Fortunately it is rather simple to parametrize α(t) by arc length, but we will deduce a nicer formula for curvature in a moment. 2 The cycloid α(t) = (t − sin t, 1 − cos t), −∞ < t < ∞ is much harder to parametrize by arc length, so k(s) is difficult to compute until we develop the next formula. We want a practical formula for curvature that does not rely on a complicated reparametriza- tion of α. Given α(t) = (x(t), y(t)) a regular parametrized differential curve, we have

dα dx dy ! dt dt dt T = dα = dα , dα | dt | dt dt dy dx ! − dt dt and N = dα , dα . dt dt Recall from Section 1.3 that in theory any regular curve can be parametrized by arc length, 0 1 0 and t (s) = dα . Since T · N = k, consider dt dT dT dt dT 1 = = · ds dt ds ds dα dt dα d2x dx d dα dα d2y dy d dα ! 2 − · 2 − · 1 = dt dt dt dt dt , dt dt dt dt dt · . dα 2 dα 2 dα dt dt dt Taking the dot product with N looks like

 2 2  0 1 dy dα d x dy dx d dα dx dα d y dx dy d dα T · N = − + + − dα 4 dt dt dt2 dt dt dt dt dt dt dt2 dt dt dt dt dt 1  dy d2x dx d2y  = − + . dα 3 dt dt dt2 dt2 dt In condensed notation, this is x0y00 − y0x00 k = . |α0|3 Example 1.4.4. With this formula in hand we can compute curvature for the earlier exam- ples. 1 For the circle, we have

α(t) = (R cos t, R sin t) α0(t) = (−R sin t, R cos t) α00(t) = (−R cos t, −R sin t).

13 1.5 Curvature in R3 1 Curves

Then curvature is calculated via the formula: (−R sin t)(−R sin t) − (R cos T )(−R cos t) R2(sin2 t + cos2 t) 1 k = = = . R3 R3 R It turns out that circles are the only regular curves in the plane with constant (nonzero) curvature. Lines are the only curves with zero curvature everywhere; in geometric terms, a line can be thought of as a special case of a circle with infinite radius, so together this extended notion of circles describe all constant-curvature curves in R2. 2 For the cycloid,

α(t) = (t − sin t, 1 − cos t) α0(t) = (1 − cos t, sin t) α00(t) = (sin t, cos t)

which allows us to calculate curvature: (1 − cos t) cos t − (sin t)(sin t) cos t − cos2 t − sin2 t =  3 p 3 p(1 − cos t)2 + sin2 t 1 − 2 cos t + cos2 t + sin2 t cos t − 1 = (2 − 2 cos t)3/2 −1 = √ 23/2 1 − cos t −1 = q by half-angle formula 3/2 2 t
2 2 sin 2 −1 = t
. 4 sin 2 In particular, the curvature of α is always negative.

1.5 Curvature in R3

How do we extend the notion of curvature to R3? Assume α(s) is a regular curve parametrized by arc length that satisfies α00(s) 6= 0 for all s. Again consider the tangent vector T = α0. In three dimensions, there are an infinite number of vectors orthogonal to T , so there’s no unique choice of N as before. However, we still can exploit the fact that T · T = 1, which as in Section 1.4 implies T 0 ⊥ T . This gives us a natural choice for N an orthonormal vector to T : T 0 N = . |T 0| If we define k = |T 0| we have the same Fren´etequation: T 0 = kN. Notice that in R3, curvature is always positive, i.e. there is no notion of “clockwise” or “counterclockwise”. Define B = T × N, which is a unit vector orthogonal to both T and N.

14 1.5 Curvature in R3 1 Curves

Definition. For any s, the triple {T (s),N(s),B(s)} is called the Fren´ettrihedron at s. As in the plane, the system {T,N,B} gives us a local description of α at any point α(s) on the curve. We also know that N · N = 1 =⇒ N 0 ⊥ N =⇒ N 0 lies in the T,B plane. In other words, N 0 = aT + bB for some a, b ∈ R. By orthogonality, N 0 · T = a(T · T ) + b(B · T ) = a · 1 + b · 0 = a and N 0 · B = a(T · B) + b(B · B) = a · 0 + b · 1 = b.

We know N · T = 0, so

0 = N 0 · T + N · T 0 = a + N · kN = a + k.

Thus a = −k and we obtain another Fren´etequation:

N 0 = −kT + bB.

We are still left with the mysterious b coefficient, which we turn to now. Proceeding in the same fashion,

B0 = T 0 × N + T × N 0 = kN × N + T × (−kT + bB) = k(N × N) − k(T × T ) + b(T × B) = k · 0 − k · 0 + bN = N.

Hence b is determined by B0 = −bN. To standardize this, we define Definition. The torsion of α at s is the constant τ satisfying B0 = τN, that is, τ = −b from above.

We summarize the Fren´etequations for R3 here. Theorem 1.5.1 (Fren´etEquations). For a regular curve α : I → R3 parametrized by arc T 0 length, let T denote the tangent vector, N = |T 0| the normal vector and B = T × N the binormal vector. Then

T 0 = kN N 0 = −kT − τB B0 = τN.

The tangent-normal plane to α(s) (the plane tangent to the curve formed by T and N) is sometimes called the osculating plane at s. Then B is the normal vector to the osculating plane and torsion describes how this plane changes in relation to s. Theorem 1.5.2 (Fundamental Theorem of Curves). Given differentiable functions k(s) > 0 and τ(s), s ∈ I, there exists a regular curve α : I → R3, parametrized by arc length, such that s is the arc length, k(s) is the curvature and τ(s) is the torsion of α. Moreover, any other α¯ satisfying the same conditions differs from α by a rigid motion; that is, α is unique up to initial point, tangent vector and normal vector.

15 1.5 Curvature in R3 1 Curves

Like many theorems in mathematics, the Fundamental Theorem of Curves has both an existence and uniqueness portion. We won’t prove existence, but it suffices to show that the system of differential equations T 0 = kN N 0 = −kT − τB B0 = τN has a solution. (This is provided by the so-called existence and uniqueness theorem in the study of linear differential equations.) Rigid motions are formally defined as orthogonal linear transformations ρ : R3 → R3 with positive determinant. They are isometries (distance-preserving) and each ρ can be decomposed into a rotation and translation of R3. For a curve α : I → R3, denote its tangent, normal and binormal vectors at t = 0 by T0,N0 and B0. Likewise, given another curveα ¯, denote these vectors by T 0, N 0 and B0. We would like to find a rigid motion ρ such that ρ(T 0) = T0, ρ(N 0) = N0 and ρ(B0) = B0. Moreover, we want ρ to preserve the local structure of α, that is

ρ(aT 0 + bN 0 + cB0) = aT0 + bN0 + cB0

for any a, b, c ∈ R. We will show that such a rigid motion can be found that preserves dot and cross products: ρ(u) · ρ(v) = u · v and ρ(u) × ρ(v) = ρ(u × v), which will imply that angles, distances and orientation are all preserved under ρ.

Proof of Uniqueness. Start with α andα ¯, two curves parametrized by arc length with the same curvature and torsion. We will construct a rigid motion with the desired properties in two steps: first we reorientα ¯ so that it matches the orientation of α, and then we translate it to α. For any a, b, c ∈ R, set ρ(aT 0 + bN 0 + cB0) = aT0 + bN0 + cB0. Claim. For any u, v ∈ R3, (i) ρ(u) · ρ(v) = u · v. (ii) ρ(u) × ρ(v) = ρ(u × v). (iii) (ρ ◦ α¯)0 = ρ ◦ α¯0 and (ρ ◦ α¯)00 = ρ ◦ α¯00.

Proof. (i) Consider the vectors u = a1T 0 + b1N 0 + c1B0 and v = a2T 0 + b2N 0 + c2B0. Using orthonormality, we can expand the dot product:

u · v = (a1T 0 + b1N 0 + c1B0) · (a2T 0 + b2N 0 + c2B0)

= a1a2 + b1b2 + c1c2. On the other hand,

ρ(u) · ρ(v) = (a1T0 + b1N0 + c1B0) · (a2T0 + b2N0 + c2B0)

= a1a2 + b1b2 + c1c2.

16 1.5 Curvature in R3 1 Curves

Hence ρ(u) · ρ(v) = u · v. (ii) In a similar fashion,

ρ(u) × ρ(v) = (a1T0 + b1N0 + c1B0) × (a2T0 + b2N0 + c2B0)

= a1b2(T0 × N0) + a1c2(T0 × B0) + b1a2(N0 × T0) + b1c2(N0 × B0)

+ c1a2(B0 × T0) + c1b2(B0 × N0)

= a1b2B0 − a1c2N0 − b1a2B0 + b1c2T0 + c1a2N0 − c1b2T0

= (b1c2 − c1b2)T0 + (c1a2 − a1c2)N0 + (a1b2 − b1a2)B0,

and

ρ(u × v) = ρ[(a1T 0 + b1N 0 + c1B0) × (a2T 0 + b2N 0 + c2B0)]

= ρ[a1b2B0 − a1c2N 0 − b1a2B0 + b1c2T 0 + c1a2N 0 − c1b2T 0]

= ρ[(b1c2)T 0 + (c1a2 − a1c2)N 0 + (a1b2 − b1a2)B0]

= (b1c2 − c1b2)T0 + (c1a2 − a1c2)N0 + (a1b2 − b1a2)B0.

Hence ρ(u) × ρ(v) = ρ(u × v). (iii) For any s ∈ I we can writeα ¯(s) = a(s)T 0 + b(s)N 0 + c(s)B0 where a, b, c are real-valued functions of s. Then

0 0 0 0
ρ(¯α ) = ρ a T 0 + b N 0 + c B0 0 0 0 = a T0 + b N0 + c B0 0 = (aT0 + bN0 + cB0)

0 = ρ aT 0 + bN 0 + cB0 = (ρ(¯α))0.

The proof with second derivatives follows exactly the same course. The immediate consequence of this claim is that, as mentioned above, angles, distances and orientation are preserved by ρ. To perform the translation ofα ¯, we set

α¯ = ρ ◦ α¯ + α(0) − ρ ◦ α¯(0).

We will show that α ≡ α¯, that is, they are the same curve. From the way we have defined α¯, we have the following implications.

ˆ α¯(0) = ρ ◦ α¯(0) + α(0) − ρ ◦ α¯(0) = α(0) so α¯ starts at the same point as α.

ˆ |α¯0| = 1 so α¯ is parametrized by arc length.

ˆ T 0 = T0, N 0 = N0 and B0 = B0. ˆ α¯ and α have the same curvature and torsion.

17 1.5 Curvature in R3 1 Curves

To show that α¯ and α are the same curve, we differentiate:

d  2 2 2 T − T + N − N + B − B = ds d             T − T · T − T + N − N · N − N + B − B · B − B ds  0    0    0   = 2 T − T · T − T + N − N · N − N + B − B · B − B         = 2 kN − kN · T − T + −kT − τB + (−kT − τB) · N − N     + τN − τN · B − B  = 2 kN · (−T ) + (−kN) · T + (−kT ) · (−N) + (−τB) · (−N) + (−kT ) · N  +(−τB) · N + (τN) · (−B) + (−τN) · B = 2(0) = 0 after cancellation.

2 2 2

So T − T + N − N + B − B is equal to some constant, and since this sum of squared differences is 0 at t = 0, the difference is 0 at every time t. Finally, this gives us

α(0) = α¯(0) and α0(s) = α¯0(s) for all s ∈ I,

so it must be that α(s) = α¯(s) for all s as well. This completes the proof of the uniqueness portion of the Fundamental Theorem of Curves. Examples.

1 Suppose a regular parametrized curve α : I → R3 has the property that all its tangent lines pass through a fixed point. We may assume α is parametrized by arc length s. Then given any point α(s), the first statement says the line determined by T = α0(s) passes through a fixed point c for all s; that is,

α(s) + f(s)T (s) = c

for some constant f(s) that depends on s. Differentiating yields

0 = α0 + f 0T + fT 0 = T + f 0T + fkN = (1 + f 0)T + (fk)N,

but since T and N are orthogonal, we must have 1 + f 0 = 0 and fk = 0. From these we can conclude that k = 0 (so α00 is identically 0). By the characterization of curves in R3, the only curves with k = 0 are straight lines so α must in fact be a straight line.

18 1.5 Curvature in R3 1 Curves

2 Suppose α is a regular parametrized curve with the property that all normal lines pass through a fixed point c. This means α(s) + f(s)N(s) = c for a constant f(s) which depends on s. Differentiating and using the Fren´etequations, we have 0 = α0 + f 0N + fN 0 = T + f 0N + f(−kT − τB) = (1 − fk)T + f 0N − fτB. Since T,N and B are linearly independent, we must have 1 − fk = 0 f 0 = 0 and − fτ = 0 which tell us that fk = 1 and f is a constant. This implies that f is a nonzero constant 1 equal to k , so curvature is constant for α. Moreover, fτ = 0 implies that τ = 0 so by the fundamental theorem for curves (1.5.2), α must be a circle.

c

3 Let α : I → R2 be a regular parametrized plane curve. Assume that k(t) 6= 0 for all t ∈ I. The curve 1 β(t) = α(t) + N(t) k(t) is called the evolute of α. Since α is regular, we may rewrite this formula with arc length s as the parameter. Claim. The tangent at s of the evolute of α is (parallel to) the normal to α at s.

Proof. Since α is parametrized by arc length s we can take advantage of the Fren´et equations. In particular, dβ 10 1 = α0(s) + N + N 0 ds k k 10 1 = T + N + (−kT ) k k 10 10 = T + N − T = N. k k

dβ 1 Therefore ds is parallel to N since the derivative of k is a constant for any s.

19 1.5 Curvature in R3 1 Curves

Consider the normal lines of α at two neighboring points s1 and s, s1 6= s.

α(s1) α(s)

L1 L2

p

Claim. As s approaches s1, the intersection points of the normals converge to a point on the trace of the evolute of α.

Proof. Fix s1 ∈ I and let L1 and L2 be the normal lines at α(s1) and any other α(s), respectively. These lines are parametrized by

L1 = α(s1) + b(s)N(s1) and L2 = α(s) + c(s)N(s) for values b and c that depend on s. The point of intersection of these lines, say p, satisfies α(s1) + b(s)N(s1) = α(s) + c(s)N(s). Differentiating with respect to s, we have

0 0 0 0 b (s)N(s1) = α (s) + c (s)N(s) + c(s)N (s) = T (s) + c0(s)N(s) + c(s)(−kT (s)) 0 0 =⇒ b (s)N(s1) − c (s)N(s) = (1 − c(s)k(s))T (s). Now using the fact that T and N are orthonormal, we can take the dot product of both sides of the equation above with T to yield

0 T (s) · (b (s)N(s1)) − 0 = (1 − c(s)k(s))(T (s) · T (s)) 0 (T (s) · N(s1))b (s) = 1 − c(s)k(s).

Now as s → s1, T (s) · N(s1) → 0 so 1 lim 1 − c(s)k(s) = 0 =⇒ c(s) → . s→s1 k(s)

1 Thus as s approaches s1, L2 becomes α(s1) + N(s1), precisely the formula for a k(s1) point on the evolute of α.

We have brushed over one important detail: as s → s1, we might be worried that b0(s) → ∞. However, the condition that k(s) 6= 0 for all s ∈ I guarantees that this coefficient remains bounded and therefore we are allowed to take the limit above and conclude the left side approaches 0.

20 2 Surfaces

2 Surfaces

2.1 The Unit Sphere

The unit sphere S2 gives us a nice example of a regular surface that is easy to work with.

S2

The simplest mathematical description of S2 is as a level surface of a function in three variables: x2 + y2 + z2 = 1. Although S2 is not a graph (e.g. it fails the vertical line test in R3), the sphere can be locally described as a graph. This will be one of the defining properties of surfaces in general. For example, the upper hemisphere of S2 can be parametrized by z = p1 − x2 − y2; similarly, the lower hemisphere is represented by z = −p1 − x2 − y2 and other hemispheres can be described by functions of other variables. A third way of creating the sphere is by rotating a function of x and z about the z-axis: z

π π (cos u, sin u), − 2 < u < 2

x

Rotation in the x, y coordinates gives us a circle of radius cos u, which can be parametrized by ~x(u, v) = (cos u cos v, cos u sin v, sin u) for 0 < v < 2π. Notice that we have defined our parameters with open intervals, so the parametrization ~x captures almost the entire sphere. It is easy to imagine modifying the formula so that the two parametrizations cover S2. This spherical coordinate system was originally conceived as a way of keeping track of one’s location on the surface of the Earth: longitude and latitude! Notice that in this construction, S2 is the image of a rectangle in the uv-plane: v 2π

u π π − 2 2

21 2.1 The Unit Sphere 2 Surfaces

This will be our prototype for a surface: they will be the image of some rectangular region in Euclidean coordinates. We can use any of the three methods described above in computations. For example, to √ √ 2  6 6 1  find a tangent plane to S at a point, say ~x = 4 , 4 , 2 , we need a normal vector. Recall that the gradient of a function F (x, y, z) is always normal to the level surfaces of F . Since the sphere is a level surface of F = x2 + y2 + z2 (for F = 1), we calculate

∇F = h2x, 2y, 2zi

√ √ 2  6 6  so a normal vector to S at ~x is ~n = 2 , 2 , 1 . Thus the tangent plane at this point is given by the equation

√ √ √ √  6 6   6 6 1  2 , 2 , 1 · x − 4 , y − 4 , z − 2 = 0

√ √ 6 6 which can be written 2 x + 2 y + z − 2 = 0. Alternatively, since S2 can be locally described in terms of a function of x and y, the partial derivatives in the x and y directions will span the tangent plane at ~x. We have ∂z 1 −x = · (−2x) = ∂x 2p1 − x2 − y2 p1 − x2 − y2 ∂z 1 −y = · (−2y) = . ∂y 2p1 − x2 − y2 p1 − x2 − y2

√ √  6 6 1  At ~x = 4 , 4 , 2 , these partials are √ √ ∂z 6 ∂z 6 = − and = − . ∂x 2 ∂y 2

Then a local approximation of S2 by the tangent plane at ~x is

√ √ √ √ 1 6  6  6  6  z = 2 − 2 x − 4 − 2 y − 4 .

It is clear that this is the same plane as above. Given the parametrization ~x(u, v) = (cos u cos v, cos u sin v, sin u), we can compute a tan- gent plane at ~x using the tangent vectors in the u and v directions. Fixing a value for u, the derivative in the v direction is

~xv = (− cos u sin v, cos u cos v, 0).

Similarly, holding v fixed and differentiating with respect to u gives us

~xu = (− sin u cos v, − sin u sin v, cos u).

22 2.2 Regular Surfaces 2 Surfaces

Then taking the cross product of these two vectors will produce a vector normal to the π π tangent plane at ~x. In our example above, u = 6 and v = 4 , so we have

π π π π π
~xu = − sin 6 cos 4 , − sin 6 sin 4 , cos 6 √ √ √  2 2 3  = − 4 , − 4 , 4 π π π π
and ~xv = − cos 6 sin 4 , cos 6 cos 4 , 0 √ √  6 6  = − 4 , 4 , 0 .

Therefore

~n = ~xu × ~xv √ √ √ √ √  2 2 3   6 6  = − 4 , − 4 , 4 × − 4 , 4 , 0 √ √ √ √  18 18 12 12  = 0 − 16 , − 16 − 0, − 16 − 16 √ √ √  3 2 3 2 3  = − 16 , − 16 , − 4 .

As this is parallel to the normal vectors calculated with the other two methods, we see that once again we have described the tangent plane (or at least the normal vector – putting this together with the point ~x gives the full plane). Notice that any of these methods can go wrong if ~n = 0, which can happen if ∇F = 0 or, in the latter method, if ~xu and ~xv are parallel. We will define a surface with this in mind.

2.2 Regular Surfaces

Definition. A subset S ⊂ R3 is a regular surface if for each p ∈ S, there is a neighborhood V ⊂ R3 of p and a surjection ~x : U → V ∩ S on an open set U ⊂ R2 satisfying (1) ~x is differentiable.

(2) ~x is a homeomorphism, that is, ~x is continuous and invertible, and its inverse is continuous.

2 3 2 (3) The differential d~xq : R → R is one-to-one for each q ∈ R .

v

S ~x(u, v) U p

u

23 2.2 Regular Surfaces 2 Surfaces

Remark. A differentiable homeomorphism is called a diffeomorphism, so (1) and (2) say that every neighborhood V ∩ S is diffeomorphic to some open subset of the plane. Furthermore, (3) is equivalent to ~xu × ~xv 6= 0.

Proof. The only statement requiring proof is the second one: d~xq is one-to-one for every 2 q ∈ R ⇐⇒ ~xu × ~xv 6= 0. First, at a point q the differential is a matrix of derivatives:   xu xv d~xq = yu yv  evaluated at q. zu zv

This matrix is one-to-one exactly when

x x  x x  u v a u v c a c y y = y y =⇒ = .  u v  b  u v  d b d zu zv zu zv

This matrix multiplication can be written as

a~xu + b~xv = c~xu + d~xv

⇐⇒ (a − c)~xu + (b − d)~xv = 0.

a c So if the above must imply = for all choices of a, b, c, d then this is equivalent to b d saying ~xu and ~xv are linearly independent, that is, ~xu × ~xv 6= 0. Example 2.2.1. Consider the sphere S2. p ~x(u, v) U

If p is a point in the upper hemisphere and V = {(x, y, z) | z > 0}, then a good choice of 2 parametrization√ for S is to let U be the open unit disk in the uv-plane and set ~x(u, v) = u, v, 1 − u2 − v2
.

Our first question is, are there any easy ways to check if S is a regular surface? The next few propositions introduce some shortcuts for proving something is a regular surface.

Proposition 2.2.2. Graphs are regular surfaces. That is, if f : U ⊂ R2 → R is a differentiable function then the graph of f is a regular surface parametrized by ~x(u, v) = (u, v, f(u, v)).

24 2.2 Regular Surfaces 2 Surfaces

Proof. Assume f(x, y) is differentiable over R2 – the proof is easily adapted for any U ⊂ R2. We want to show S = {(x, y, z) | z = f(x, y)} is a regular surface. Let ~x(u, v) = (u, v, f(u, v)) be defined on U = R2. First, (1) is satisfied because ~x is differentiable. Next, by the remark, (3) is equivalent to showing ~xu × ~xv 6= 0, so consider

h1, 0, fui × h0, 1, fvi = h−fu, −fv, 1i. Since the third component is nonzero, the cross product can never be the zero vector, so (3) holds. To prove (2), we must show ~x is one-to-one on its image and has a continuous inverse. First, suppose (u1, v1, f(u1, v2)) = (u2, v2, f(u2, v2)) for (u1, v1), (u2, v2) ∈ U. Then clearly −1 u1 = u2 and v1 = v2, so ~x is one-to-one. Instead of defining ~x directly, we define a function ψ(x, y, z) = (x, y) on R3, the entire codomain of ~x and check that it restricts to an inverse on the image of ~x. On one hand,

ψ(~x(u, v)) = ψ(u, v, f(u, v)) = (u, v).

On the other hand, ~x(ψ(x, y, z)) = ~x(x, y) = (x, y, f(x, y)) holds on the set S = {(x, y, z) | z = f(x, y)} so ψ restricts to the inverse of ~x on the set S. Clearly ψ is continuous (even better, it’s differentiable), so we have proven S is a regular surface. √ 2 2 Example 2.2.3. Consider f(u, v) = 1 −√u − v on U the open unit disk. As we saw above, the parametrization ~x(u, v) = u, v, 1 − u2 − v2
describes the unit sphere. What’s more, we can easily compute the directional derivatives  −u   −v  ~xu = 1, 0, √ and ~xv = 0, 1, √ . 1 − u2 − v2 1 − u2 − v2 The second method used in Section 2.1 to describe S2 took advantage of level curves of a function in three variables. The same holds for regular surfaces in general, provided the level curve is computed for a regular value of the function:

Definition. A number a ∈ R is a regular value of F : U ⊂ R3 → R2 if ∇F (x, y, z) 6= 0 for all (x, y, z) ∈ F −1(a).

Proposition 2.2.4. If F : U ⊂ R3 → R is a differentiable function and a ∈ R is a regular value of F , then F −1(a) is a regular surface. Remark. Proposition 2.2.4 implies Proposition 2.2.2. Proof. Suppose f(x, y) is differentiable. We want to show that the graph T = {(x, y, z): z = f(x, y)} is a regular surface. We turn T into a level surface of a real-valued function by setting h(x, y, z) = f(x, y) − z, and by Proposition 2.2.4 is suffices to show a = 0 is a regular value of h. Now h is clearly differentiable, and we have ∇h = (fx, fy, −1) which is possible to compute in every case since f is differentiable. Notice that ∇h 6= (0, 0, 0) since the 3rd component is always −1. Hence every value is a regular value for h, including a = 0, so we conclude that {(x, y, z) | f(x, y) − z = 0} is a regular surface.

25 2.2 Regular Surfaces 2 Surfaces

The next proposition says that every regular surface is locally a graph, which means that S can be broken up into ‘patches’ which are parametrized as the graphs of some functions.

Proposition 2.2.5. Every regular surface is locally a graph.

To prove Proposition 2.2.5, we need the inverse function theorem.

Theorem 2.2.6 (Inverse Function Theorem). Let f : U ⊂ Rn → Rn be a differentiable func- tion and suppose p ∈ U such that df(p) is an invertible matrix. Then there is a neighborhood V of p and a neighborhood W of f(p) such that f : V → W has a differentiable inverse.

Examples.

1 The simplest case is for a real-valued function f :(a, b) ⊆ R → R. Let p ∈ (a, b) such that f 0(p) 6= 0. There is an interval (c, d) ⊆ (a, b) with p ∈ (c, d) and an interval (e, g) containing f(p) such that f :(c, d) → (e, g) has a differentiable inverse. Consider the following linear approximation at p:

f(x) ≈ f(p) + f 0(p)(x − p).

Solving for x produces the inverse function 1 x = p + (f(x) − f(p)). f 0(p)

2 For the two-dimensional case, suppose f : R2 → R2 is a vector valued function given 2 by f(x, y) = (f1(x, y), f2(x, y)). Take p = (x0, y0) ∈ R . The linear approximation at p is ∂f ∂f f (x, y) = f (x , y ) + i (x , y )(x − x ) + i (x , y )(y − y ) + error for i = 1, 2. i i 0 0 ∂x 0 0 0 ∂y 0 0 0 This can be written in terms of matrices and vectors, dropping the error term:

" ∂f1 ∂f1 #   (x0, y0) (x0, y0) x − x f(x, y) = f(x , y ) + ∂x ∂y 0 0 0 ∂f2 ∂f2 y − y ∂x (x0, y0) ∂y (x0, y0) 0 f(~x) = f(p) + Df(p)(~x − p).

As in the one-dimensional case, we can rearrange terms,

f(~x) − f(p) = Df(p)(~x − p)

and if Df(p) is invertible, we have an inverse function

~x = p + Df(p)−1(f(~x) − f(p)).

26 2.2 Regular Surfaces 2 Surfaces

Proof of Prop. 2.2.5. We will apply the inverse function theorem to the parametrization   ~x(u, v) = (x(u, v), y(u, v), z(u, v)) of a regular surface S. The differential d~xq = ~xu ~xv is a 3 × 2 matrix with independent columns, that is ~xu × ~xv 6= 0. This means one of the 2 × 2 minors of d~xq is nonzero, so without loss of generality, assume

xu xv = xuyv − xvyu 6= 0. yu yv

The proof is the same for the other 2 × 2 minors. Let π : R3 → R2 be the projection (x, y, z) 7→ (x, y). In order to prove Proposition 2.2.5, we will show that, restricted to a neighborhood of S parametrized by ~x, π is one-to-one. Consider f(u, v) = π ◦ ~x(u, v) = (x(u, v), y(u, v)). Notice that the linear operator

x x  Df = u v yu yv

−1 is invertible by hypothesis. Let ~x (p) = (u0, v0). By the IFT, there are neighborhoods about (u0, v0) and π(p) such that f is invertible with differentiable inverse on these two neighborhoods.

S

p ~x (u, v π )

f

f −1

U V

Note that ~x ◦ f −1(x, y) = (x, y, z(x, y)) is well-defined, with z some function of x and y. Set π−1(x, y) = (x, y, z(x, y)). Then we have successfully inverted the projection π on a local patch of S. The formula for π−1 shows that S is a graph in the neighborhood around p.

Proposition 2.2.7. Let p ∈ S for a regular surface S and let ~x : U ⊂ R2 → R3 such that (1) and (3) are satisfied. If ~x is one-to-one the ~x−1 is continuous.

Proposition 2.2.7 will make it easier to prove condition (2) for a regular surface.

27 2.2 Regular Surfaces 2 Surfaces

Examples. 1 Consider the function F (x, y, z) = x2 + y2 + z2. As we saw before, the level curve of F at a = 1 describes the regular surface S2. The level curve is given by

F −1(1) = {(x, y, z): F (x, y, z) = 1} = {(x, y, z): x2 + y2 + z2 = 1} = S2.

Notice that ∇F = (2x, 2y, 2z) = 0 if and only if x, y and z all equal 0, so in particular a = 1 is a regular value of F .

2 The infinite cylinder is defined as the set of points C = {(x, y, z) ∈ R3 : x2+y2 = 1}. To prove C is a regular surface, it suffices by Proposition 2.2.4 to show F (x, y, z) = x2 +y2 is differentiable (obvious) and a = 1 is a regular value of F . We have ∇F = (2x, 2y, 0). The critical points of ∇F satisfy x = 0, y = 0 so the only critical value is 0; hence 1 is regular. This proves the cylinder is a regular surface. One parametrization of the cylinder is ~x(u, v) = (cos u, sin u, v) for 0 < u < 2π and −∞ < v < ∞. Similarly, we can take ~y(u, v) = (cos u, sin u, v) to be the same function, π π but defined on a different domain such as − 2 < u < 2 , −∞ < v < ∞ so that together ~x and ~y cover the cylinder.

3 The two-sheeted cone is a figure in R3 defined by {(x, y, z) ∈ R3 : x2 + y2 − z2 = 0}.

Let’s solve the cone’s defining equation for z: z = ±px2 + y2. By Proposition 2.2.5, if the cone is a surface it must locally be a graph. In other words, any neighborhood (in the subspace topology) of the origin must be the image in R3 of some open disk in R2. However, one can see that any potential mapping f : U ⊂ R2 → R3 is not well-defined – in particular, there will be points (x, y, z) ∈ f(U) such that (x, y, −z) ∈ f(U) as well. Therefore the two-sheeted cone fails to be a regular surface.

4 Let f(x, y, z) = z2. Differentiating f gives us ∇f = (0, 0, 2z) and the critical points all lie on the plane z = 0. This shows that 0 is not a regular value of f. However, the level surface of f at 0 is {(x, y, z) ∈ R3 : z2 = 0} = {(x, y, z) ∈ R3 : z = 0} which is the xy-plane and clearly this is a regular surface since it is identically an open set of R2. This shows that the converse to Proposition 2.2.4 is false in general.

28 2.2 Regular Surfaces 2 Surfaces

5 Suppose S is a regular surface with the property that the normal vector to S at every point on S passes through a fixed point c ∈ R3. This is analagous to example 2 in Section 1.5, which described a regular curve with the same property. At a point p ∈ S, there is a parametrization ~x(u, v) which induces a unit normal vector ~x × ~x ~n(u, v) = u v . Then the line passing through p and c is of the form |~xu × ~xv| ` : ~x(u, v) + r~n(u, v) = c

for some parameter r. We differentiate to obtain

~xu + ru~n + r~nu = 0

and ~xv + rv~n + r~nv = 0.

Taking the inner product of each equation with ~n, and reducing accordingly, gives

0 + ru(1) + r~n · ~nu = 0

and 0 + rv(1) + r~n · ~nv = 0,

but as in the earlier example for curves, it’s easy to check that ~n is orthogonal to both ~nu and ~nv, so these equations become

ru = 0 and rv = 0.

Hence r must be a constant, so S is the set of points with constant distance r from c, that is, S is a sphere of radius r.

We take a moment to connect regular curves from Chapter 1 with the regular surfaces we have defined here.

Definition. Let C ⊂ R3 be a subset. We say C is a regular curve if for any point p ∈ C there is a parametrization α :(a, b) ⊂ R → C such that α(t0) = p for some t0 ∈ (a, b) and (1) α is differentiable.

(2) α is a homeomorphism onto its image in C.

(3) α0(t) 6= 0 for any t ∈ (a, b).

Notice that this coincides with the definition of a simple regular curve in Chapter 1. Condition (3) can be cast in terms of the differential dα, so in this language, we can realize a regular curve as the one-dimensional analog of a regular surface.

29 2.3 Differentiable Functions on Surfaces 2 Surfaces

2.3 Differentiable Functions on Surfaces

For a regular surface S, we want to be able to describe differentiable functions f : S → R that might correspond to some real world attributes of the surface.

Example 2.3.1. On the sphere, we may be interested in measuring temperature, wind speed, average sunlight, gravitational force, etc.

f S R

The idea is to use a parametrization ~x(u, v) of the surface to describe differentiable ∂f ∂f functions with partials ∂u and ∂v . We will express f in terms of u and v, giving a composite f ◦ ~x which we want to be differentiable. This raises a technical issue: there are infinitely many parametrizations of the sphere (and any surface S), so we must develop derivatives that have the same properties regardless of the choice of parametrization.

t p v ) ~y( ~x(u, v s, t ) s u

For the sphere, consider the parametrization π ~x(u, v) = (cos u cos v, cos u sin v, sin u), 0 < u < , 0 < v < 2π, 2 which maps onto√ the upper half of S minus the x = 0 seam along the front. Also consider ~y(s, t) = s, t, 1 − s2 − t2
mapping from the open unit disk minus the segment from (0, 0) to (1, 0). Given p ∈ S, we first find the u and v such that ~x(u, v) = p. To get from the left parametrization to the right one, we compose: ~y−1(~x(u, v)). Since ~y is a diffeomorphism, it has a continuous inverse which allows us to define such a composition, called a change of variables. The next result says that a change of variables for a regular surface is always differentiable. We will prove this towards the end of the section.

Proposition 2.3.2. If ~x and ~y are any two parametrizations of a regular surface S, the change of variables ~y−1 ◦ ~x is differentiable.

30 2.3 Differentiable Functions on Surfaces 2 Surfaces

S U

) p V ~y (s, t (u, v ) ~x

~y−1 ◦ ~x

~x−1(U ∩ V ) ~y−1(U ∩ V )

The natural way to describe differentiability on a surface S is to choose a point p ∈ S and a neighborhood U of p equipped with a parametrization ~x(u, v) and consider the composite f ◦ ~x(u, v).

Definition. A function f : S → R is differentiable at a point p ∈ S if f ◦~x is differentiable on ~x−1(U ∩ V ), where U, V are any neighborhoods of p and ~x is a parametrization of U.

As suggested above, what happens if ~y(s, t) is another parametrization of a region about p? Note that f ◦ ~y = (f ◦ ~x) ◦ (~x−1 ◦ ~y) and by Proposition 2.3.2, ~x−1 ◦ ~y is differentiable. Moreover, the composite of differentiable functions is differentiable as well, so this shows that it suffices to check differentiability on a single parametrization of a neighborhood of p. Next, we would like to extend the notion of differentiability to functions between surfaces.

Definition. Let S1 and S2 be two regular surfaces. We say a continuous function f : S1 → S2 −1 −1 −1 is differentiable at p ∈ S1 if ~y ◦ f ◦~x is differentiable on ~x (f (V ) ∩ U) for any choices of parametrizations ~x for U 3 p and ~y for V 3 f(p).

Recall that for a function f on Euclidean space, differentiability implies continuity. We are interested in generalizing this to differentiable functions on surfaces. First, we have

Proposition 2.3.3. If f : S → R is a differentiable function on a surface S then f is continuous.

Proof. The assumption that f is differentiable implies there exists a parametrization ~x of S such that f ◦ ~x is differentiable. Since f ◦ ~x is a real-valued function, it is also continuous. Also, ~x is differentiable and therefore continuous, so the composition f = (f ◦ ~x) ◦ ~x−1 of continuous, real-valued functions is continuous. The proof is almost the same for a differentiable function between any two surfaces but as one might expect, we need a parametrization for each surface.

31 2.3 Differentiable Functions on Surfaces 2 Surfaces

S1 f −1(V ) ∩ U S2 f

U p f(p) V

~x ~y

v t

u s ~y−1 ◦ f ◦ ~x

Proposition 2.3.4. For a function f : S1 → S2, where S1 and S2 are regular surfaces, differentiability implies continuity.

2 Proof. As above, f differentiable implies there are parametrizations ~x1 : U ⊂ R → S1 and 2 −1 ~x2 : V ⊂ R → S2 such that the composite ~x2 ◦ f ◦ ~x1 is differentiable. Consider

−1 −1 f = ~x2 ◦ (~x2 ◦ f ◦ ~x1) ◦ ~x1 .

This is the composition of three continuous functions in Euclidean space (the middle via differentiability =⇒ continuity), so the entire function is continuous.

Proposition 2.3.5. Compositions of differentiable functions are differentiable.

Proof. Suppose f : S1 → S2 and g : S2 → S3 are differentiable. Then there are parametriza- 2 2 2 −1 tions ~x1 : U ⊂ R → S1, ~x2 : V ⊂ R → S2 and ~x3 : W ⊂ R → S3 such that ~x2 ◦ f ◦ ~x1 −1 and ~x3 ◦ g ◦ ~x2 are both differentiable. Consider

−1 −1 −1 ~x3 ◦ (g ◦ f) ◦ ~x1 = (~x3 ◦ g ◦ ~x2) ◦ (~x2 ◦ f ◦ ~x1).

This is the composition of two differentiable functions in Euclidean space and hence the function on the left is differentiable. By definition this proves g ◦ f is differentiable.

32 2.4 Tangent Spaces 2 Surfaces

Examples. 1 Navigation on the surface of the Earth

ϕ S

To make useful maps for navigation, it is common to project the sphere onto a different surface such as the unit cylinder. This is done by fixing the height of a point on the sphere and projecting the point to the closest point on the cylinder with the same height – the points on the equator are fixed by this expansion. Later we will want to see if this function is differentiable on the sphere.

2 Stereographic projection is a way to bijectively map the punctured sphere S2 r {N} (where N = (0, 0, 1) is the north pole) to the xy-plane. Given a point p on the sphere, there is a line containing N and p which intersects the plane at a point xp. The 2 2 stereographic projection f : S → R is then defined by f(p) = xp. One of the nicest results of this construction is that f preserves angles, lines and circles.

N

x

f(x)

3 Some of the most interesting functions on surfaces are those that take S onto itself (e.g. rotations of the sphere). We will investigate when these functions are differentiable.

2.4 Tangent Spaces

Given a regular surface S, we want to find an equation for the tangent plane at any point p ∈ S. If we have a nice parametrization ~x(u, v) of a neighborhood U of p, things are simple: the tangent vectors in each coordinate direction describe a plane that is tangent to S at p. Formally, we define the tangent space to S at p is the set of all linear combinations of the tangent vectors along the coordinate curves:

{s~xu + t~xv | s, t ∈ R}.

33 2.4 Tangent Spaces 2 Surfaces

Notice that this linear system does not exactly describe the actual plane at p, but in reality this is the only information that’s important. Of course, to describe the actual plane tangent to S at p, we can simply translate by p: {p + s~xu + ~xv | s, t ∈ R}. Knowing ~xu and ~xv also makes it easy to give a normal form for the tangent plane, e.g. by computing ~xu × ~xv. An important question is: If ~x(u, v) and ~y(q, r) are parametrizations of neighborhoods of p, then are the tangent spaces spanned by {~xu, ~xv} and {~yq, ~yr} the same? Again we want to be sure that choice of parametrization does not affect any of our computations. In order to get away from specific choices of parametrization, we define the tangent space more generally below. Afterwards, we will show that our two notions of tangent space coincide.

Definition. Let S be a regular surface and let p ∈ S. We say a vector ~w ∈ R3 is a tangent vector to S at p if there is some curve α :(−ε, ε) → S for some ε > 0 such that α(0) = p and α0(0) = ~w.

Definition. The tangent space of a regular surface S at a point p ∈ S is the collection of all tangent vectors at p, that is,

3 Tp(S) = {~w ∈ R | ~w is a tangent vector to S at p}. The next proposition unites our two separate ideas of tangent space.

Proposition 2.4.1. For any parametrization ~x(u, v) of S about p,

Tp(S) = {s~xu + t~xv | s, t ∈ R}.

∗ Proof. For a regular surface S, let p ∈ S and let Tp(S) denote the vector space defined by ∗ a parametrization ~x of S at p. Suppose ~w ∈ Tp(S) . Then ~w = s~xu + t~xv for some s, t ∈ R. Let (a, b) = ~x−1(p) ∈ R2. Define a curve α :(−ε, ε) → R3 by α(τ) = ~x(a + sτ, b + tτ). Notice that at τ = 0, α(0) = ~x(a, b) = p and since α is defined in terms of ~x, its image lies in S.   0 du dv Finally, by the chain rule α (0) = ~xu + ~xv = ~xus + ~xvt = ~w. Hence ~w ∈ Tp(S). dτ dτ τ=0 Conversely, assume ~w is a tangent vector to S at p. Then there is a curve α :(−ε, ε) → S such that α(0) = p and α0(0) = ~w. Letα ¯(t) = ~x−1(α(t)). Thenα ¯ is a regular differentiable curve since ~x is one-to-one and the composition of differentiable functions is differentiable. We can writeα ¯(t) = (u(t), v(t)) since the image ofα ¯ lies in the uv-plane. Thus α(t) = ~x(u(t), v(t)) which means   0 du dv ~w = α (0) = ~xu + ~xv = ~xuc1 + ~xvc2 dt dt t=0

∗ ∗ for some constants c1, c2. This shows ~w ∈ Tp(S) and hence Tp(S) and Tp(S) are equal. Corollary 2.4.2. If ~x(u, v) and ~y(s, t) are parametrizations of neighborhoods of p ∈ S, then

{a~xu + b~xv | a, b ∈ R} = {c~ys + d~yt | c, d ∈ R}

and hence the definition of the tangent space Tp(S) is independent of parametrization.

34 2.4 Tangent Spaces 2 Surfaces

The differential d~x is a matrix of partial derivatives:   xu xv   d~x = yu yv  = ~xu ~xv . zu zv

a a If q = in the uv-plane, then d~x = ~x ~x  = a~x + b~x so the differential is a b q u v b u v map from vectors in the plane to vectors in the tangent space Tp(S). Suppose we have a differentiable function ϕ : S1 → S2.

S1 S2 ~w ϕ dϕ(~w) p ϕ(p)

If Tp(S1) is the tangent space to S1 at p, it would be ideal if dϕ acted on vectors in Tp(S1) in a ‘nice way’, e.g. by sending them to vectors in the tangent space to S2 at ϕ(p). This turns out to be precisely what happens. We show this below. Take ~w ∈ Tp(S1). Then there is a curve α whose trace lies in S1 satisfying the given conditions for a tangent vector. Set β(t) = ϕ ◦ α(t). Then dϕ(~w) = β0(0) which lies in Tϕ(p)(S2) so dϕ preserves tangent vectors.

~w dϕ dϕ(~w) p f(p)

d~x d~y

v t

u s dϕ¯

To see this via the underlying parametrizations, suppose ~x is a parametrization about p ∈ S1 and ~y is a parametrization about ϕ(p) ∈ S2. Denote the map in Euclidean coordinates

35 2.4 Tangent Spaces 2 Surfaces

−1 −1 byϕ ¯ = ~y ◦ ϕ ◦ ~x. Given ~w ∈ Tp(S1), we can write ~w = a~xu + b~xv, so d~x (~w) = (a, b). Denote the image of (a, b) underϕ ¯ by (c, d). Then d~y(c, d) = c~ys +d~yt which lies in Tϕ(p)(S2). We set dϕ(~w) equal to this tangent vector, so we have in general dϕ = d~y ◦ dϕ¯ ◦ d~x−1.

Proposition 2.4.3. If f : S1 → S2 is a function between regular surfaces and there is 3 3 a function F : U ⊂ R → R such that F is differentiable on a neighborhood U of S1, F (U) ⊃ S and F | = f, then f is differentiable. 2 S1 Examples.

2 2 2 x2 y2 z2 1 Consider the unit sphere S1 : x + y + z = 1 and any ellipsoid S2 : a2 + b2 + c2 = 1. A diffeomorphism ϕ : S1 → S2 may be defined as ϕ(x, y, z) = (ax, by, cz).

3 Since ϕ is differentiable on all of R and ϕ(S1) = S2, Proposition 2.4.3 shows that −1 x y z
ϕ : S1 → S2 is differentiable. Moreover, an inverse for ϕ is ϕ (x, y, z) = a , b , c . −1 Then ϕ is also differentiable, so S1 and S2 are diffeomorphic. 2 Recall the map-making projection of the sphere S onto the unit cylinder C.

ϕ C S

Consider the parametrizations of S and C about some points p ∈ S and ϕ(p) ∈ C:

π π S : ~x(u, v) = (cos u cos v, cos u sin v, sin u), − 2 < u < 2 , 0 < v < 2π C : ~y(s, t) = (cos t, sin t, s), −1 < s < 1, 0 < t < 2π. We define ϕ : S → C by ! x y ϕ(x, y, z) = , , z . px2 + y2 px2 + y2

 2  2 Notice that √ x + √ y = 1 so ϕ(U) ⊂ C, where U is the neighborhood x2+y2 x2+y2 of S parametrized by ~x. Hence ϕ is well-defined. Also, the x and y components of ϕ(x, y, z) are just multiples of the original x and y components on the sphere, so geometrically, ϕ carries out the prescribed transformation. To write downϕ ¯ : R2 → R2 between the coordinate systems of S and C, we first compute ϕ ◦ ~x(u, v) = (cos v, sin v, sin u). Then applying ~y−1 gives ϕ¯(u, v) = ~y−1 ◦ ϕ ◦ ~x(u, v) = ~y−1(cos v, sin v, sin u) = (sin u, v).

36 2.5 First Fundamental Form 2 Surfaces

cos u 0 Therefore the differential ofϕ ¯ is the 2 × 2 matrix dϕ¯ = . Accordingly, we 0 1 can describe the action of dϕ on tangent vectors to S in terms of the action of dϕ¯ on the coordinate vectors. Let ~w = a~xu + b~xv ∈ Tp(S). Then

dϕ(~w) = d~y ◦ dϕ¯ ◦ d~x−1(~w) = d~y ◦ dϕ¯(a, b) cos u 0 a a cos u = d~y ◦ = d~y 0 1 b b

= a cos u~ys + b~yt,

which lies in Tϕ(p)(C). One can verify that our mapping ϕ(x, y, z) is indeed a diffeo- morphism between the sphere (minus poles) and the cylinder (minus the centers of its top and bottom disks).

2.5 First Fundamental Form

Using a fixed parametrization, we would like to compute angles, vector lengths, arc lengths, surface area and similar quantities on a surface S. This boils down to being able to compute dot products between the tangent vectors to the surface.

Definition. Let S be a regular surface and suppose ~x : U ⊂ R2 → S is a parametrization about p ∈ S. The first fundamental form associated with ~x is the triple (E,F,G):

E = ~xu · ~xu,F = ~xu · ~xv and G = ~xv · ~xv.

The first fundamental form sometimes (as in Do Carmo) refers to the inner product Ip(w) = hw, wi evaluated at tangent vectors w ∈ Tp(S). Then the triple (E,F,G) is an evaluation of Ip for a particular parametrization of S at p. Example 2.5.1. Consider the unit sphere with ~x(u, v) = (cos u cos v, cos u sin v, sin u) for π π − 2 < u < 2 , 0 < v < 2π. The first fundamental form for S is computed as E = (− sin u cos v, − sin u sin v, cos u) · (− sin u cos v, − sin u sin v, cos u) = sin2 u cos2 v + sin2 u sin2 v + cos2 u = sin2 u + cos2 u = 1; F = (− sin u cos v, − sin u sin v, cos u) · (− cos u sin v, cos u cos v, 0) = cos u sin u cos v sin v − cos u sin u cos v sin v + 0 = 0; G = (− cos u sin v, cos u cos v, 0) · (− cos u sin v, cos u cos v, 0) = cos2 u sin2 v + cos2 u cos2 v + 0 = cos2 u.

The condition that F = ~xu · ~xv = 0 tells us that ~xu and ~xv are orthogonal. In general, this is characteristic of a nice choice of parametrization in that the linear algebra is easier to compute with an orthogonal basis.

37 2.5 First Fundamental Form 2 Surfaces

Using the definition of the ordinary dot product of vectors, we can express the angle θ between ~xu and ~xv via the equation√ ~xu · ~xv = |~xu| |~xv| cos θ. Written in terms of the first fundamental form, this is F = EG cos θ, so in general the angle may be computed as  F  θ = cos−1 √ . EG

Notice that, as the definition of F suggests, ~xu and ~xv are orthogonal ⇐⇒ F = 0. To compute arc length, suppose α(t) is a regular curve lying on the surface of S. From multivariable calculus we know arc length may be expressed as Z b |α0(t)| dt a but notice that α(t) = (u(t), v(t)), i.e. α has an expression in terms of the coordinate curves. Then we see that

0 p 0 0 0 0 p 0 2 0 0 0 2 |α | = hu ~xu + v ~xv, u ~xu + v ~xvi = (u ) E + 2u v F + (v ) G. Therefore the arc length of a regular curve lying on a surface S is given by the formula Z b pu0(t)2E + 2u0(t)v0(t)F + v0(t)2G dt. a Example 2.5.2. Let α(t) = (u(t), v(t)) be a curve on the unit sphere S defined by u = t and v = c, a constant. This corresponds to a constant-height circle drawn on the surface of S.

α

Z b √ In this case u0(t) = 1 and v0(t) = 0, so the arc length formula reduces to E dt. a Next we want to compute surface area over S. Take an arbitrarily small patch of S, which is locally homeomorphic to a parallelogram bounded by the coordinate curves:

~xv dv

~xu du

38 2.5 First Fundamental Form 2 Surfaces

The area of such a parallelogram is calculated by a cross product |~xu × ~xv| du dv, so it makes sense to approximate the surface area over S with an integral, ZZ |~xu × ~xv| du dv, U where U is an open set in the uv-plane which parametrizes the region of integration. To show that this integral does not depend on the parametrization chosen, let ~y(s, t) be another ∂(u,v) parametrization of the patch of S. Let ∂(s,t) be the Jacobian of the change of variables ~x−1 ◦ ~y. Then ZZ ZZ ZZ ∂(u, v) |~ys × ~yt| ds dt = |~xu × ~xv| ds dt = |~xu × ~xv| du dv. U Q ∂(s, t) U Thus the surface area formula is independent of parametrization, so we can define

Definition. For a bounded region D of a regular surface S, where D is parametrized by ~x : U ⊂ R2 → S, the surface area of D is defined by ZZ A(D) = |~xu × ~xv| du dv. U

2 2 2 2 2 Notice that |~xu × ~xv| = |~xu| |~xv| − h~xu, ~xvi = EG − F , so the formula for surface area can be written in terms of the first fundamental form: ZZ √ A(D) = EG − F 2 du dv. U As noted above, a parametrization with F = 0 is particularly nice for computation. It turns out that E = G and F = 0 if and only if ~x is a conformal map. Surfaces (or rather, parametrizations of surfaces) where E = G = 1 and F = 0 are rare since this means the parametrization is an isometry between S and the plane.

Example 2.5.3. The unit cylinder, under the usual parametrization ~x(u, v) = (cos u, sin u, v), 0 < u < 2π, −1 < v < 1, has partials ~xu = (− sin u, cos u, 0) and ~xv = (0, 0, 1). The first fundamental form is

E = sin2 u + cos2 u = 1; F = 0; G = 1.

Thus the cylinder is one of those particularly rare surfaces possessing an isometry with R2.

39 3 Curvature of Surfaces

3 Curvature of Surfaces

3.1 The Gauss Map

In this section we introduce the central object in the study of surface curvature. We have already seen a version of the Gauss map in Chapter 1. Suppose α(s) is a regular plane curve parametrized by arc length. We saw that the rate of change of the angle θ(s) between the unit tangent vector T (s) and (1, 0) is given by curvature k(s). Formally, this corresponds to a map from α to the unit circle S1, which takes unit tangent vectors to points on the circle at the head of T (s).

T (s) T (s)

The Gauss map gives us a way of keeping track of the normal vectors to a surface S. We will then use this to describe the curvature of S. Definition. Let S be a regular surface. The Gauss map N : S → S2 from the surface to the unit sphere is the function p 7→ N(p) where N(p) is the unit normal vector to S at p. The Gauss map is well-defined, since every unit normal vector corresponds to a point on the unit sphere S2 at the head of the vector. Even though N is not defined via coordinates, we can obtain a useful formula for what we will denote N(u, v), which is shorthand for N ◦ ~x(u, v) for a parametrization ~x of a neighborhood of p ∈ S. The formula, as one might expect from examples in Chapter 2, is ~x × ~x N(u, v) = u v . |~xu × ~xv| Example 3.1.1. The M¨obiusstrip is a non-orientable surface. This is actually an example of a regular surface for which the Gauss map is not well-defined on the entire object. However, N may still be defined locally. Consider the differential of the Gauss map:

T (S) T (S2) p dN(p) N(p) p f(p)

Conveniently, the normal vector to N(p) on the sphere is always just N(p). This shows 2 that the tangent planes to S at p and to S at N(p) are parallel – in other words, Tp(S) = 2 TN(p)(S ). Even nicer, it turns out that dN(p) is a symmetric linear operator.

40 3.1 The Gauss Map 3 Curvature of Surfaces

Example 3.1.2. Consider the graph of f(x, y) = y2 − x2.

N(p)

p N

By Proposition 2.2.2, this represents a regular surface S parametrized by

2 2 ~x(u, v) = (u, v, f(u, v)) = (u, v, v − u ), u, v ∈ R. The partial derivatives for this parametrization are

~xu = (1, 0, −2u) and ~xv = (0, 1, 2v), so the formula for the unit normal vector is ~x × ~x (2u, −2v, 1) N(u, v) = u v = √ . |~xu × ~xv| 4u2 + 4v2 + 1 For p = (0, 0, 0) we see that N(p) = N(0, 0) = (0, 0, 1) as shown in the figure above. In addition, one can see from the figure or the formula that the coordinate curve along y = 0 (in red) gets mapped to a curve on S2 with orientation reversed (also in red). On the other hand, the coordinate curve x = 0 maps to the arc on S2, shown below, with orientation preserved. We will use these principal curves to describe the curvature of the surface.

N

At p = (0, 0, 0), ~xu = (1, 0, 0), ~xv = (0, 1, 0) and N(p) = (0, 0, 1). The tangent space 2 Tp(S) is spanned by (1, 0, 0) and (0, 1, 0). Recall that Tp(S) and TN(p)(S ) are the same 2 vector space (the tangent planes are parallel), so ~xu and ~xv also span TN(p)(S ). We want to describe the image of these coordinate tangent vectors under dN.

41 3.1 The Gauss Map 3 Curvature of Surfaces

T (S) T (S2) p dN(p) N(p) ~xv ~xv ~xu f(p) p ~xu

0 Consider the coordinate curve α(t) = ~x(t, 0). Then α (0) = ~xu and N ◦ α(t) = N(t, 0) so the differential of the Gauss map acts on ~xu in the following way √   4t2 + 1(2, 0, 0) − (2t, 0, 1) √ 8t d 2 4t2+1 dN · ~xu = N(t, 0)| = dt t=0 4t2 + 1 t=0 (2, 0, 0) − (0, 0, 0) = = (2, 0, 0). 1 + 0

From this we can see that dN scales ~xu by the positive value λ = 2. Similarly, d dN · ~x = N(0, r)| = (0, −2, 0), v dt r=0 so dN scales ~xv by λ = −2. Now we know what dN does to a basis of Tp(S): Nu(0, 0) = 2~xu and Nv(0, 0) = −2~xv. So for any a~xu + b~xv ∈ Tp(S),

dN(a~xu + b~xv) = a dN · ~xu + b dN · ~xv = 2a~xu − 2b~xv.

In matrix-vector form, this can be written

2 0  a  2a  = 0 −2 b −2b

1 so dN is a symmetric matrix with eigenvalues λ = ±2 and eigenvectors ~x = and u 0 0 ~x = . It turns out that these eigenvalues are the curvatures of the principal (in this v 1 case coordinate) curves through p = (0, 0, 0) on the surface of S. In Section 3.2, these eigenvalues will be used to define the principal curvatures of S and the determinant of dN will correspond to the notion of Gaussian curvature for S.

Note that dNp is not always represented by a symmetric matrix – this is highly sensitive to the chosen basis of Tp(S), which of course depends on the parametrization chosen. However, dNp is always symmetric with respect to the inner product, i.e. dNp is self-adjoint. Theorem 3.1.3. Suppose S is a regular surface and let N be the Gauss map for S. Then for any p ∈ S, dNp is a self-adjoint linear operator.

Proof. We must show that hdNp · w1, w2i = hw1, dNp · w2i for any w1, w2 ∈ Tp(S). For a parametrization ~x(u, v) about p ∈ S, {~xu, ~xv} is a basis for Tp(S). Then w1 = a1~xu + b1~xv

42 3.1 The Gauss Map 3 Curvature of Surfaces

and w2 = a2~xu + b2~xv so the two sides of the desired equality are

hdNp · w1, w2i = hdNp(a1~xu + b1~xv), a2~xu + b2~xvi

= ha1Nu + b1Nv, a2~xu + b2~xvi

= a1a2hNu, ~xui + a1b2hNu, ~xvi + b1a2hNv, ~xui + b1b2hNv, ~xvi

hw1, dNp · w2i = ha1~xu + b1~xv, dNp(a2~xu + b2~xv)i

= ha1~xu + b1~xv, a2Nu + b2Nvi

= a1a2h~xu,Nui + a1b2h~xu,Nvi + b1a2h~xv,Nui + b1b2h~xv,Nvi

= a1a2hNu, ~xui + a1b2hNv, ~xui + b1a2hNu, ~xvi + b1b2hNv, ~xvi by symmetry of the inner product. Clearly if we can show hNu, ~xvi = hNv, ~xui then the middle terms will be equal and so both expressions will be equal. By definition of the Gauss map, hN, ~xui = 0 and hN, ~xvi = 0 so differentiating each with respect to the opposite parameter gives us

0 = hNv, ~xui + hN, ~xuvi

and 0 = hNu, ~xvi + hN, ~xvui.

By equality of mixed partials, ~xuv = ~xvu so setting the two equations equal produces

hNu, ~xvi = hNv, ~xui.

Therefore dNp is self-adjoint.

Corollary 3.1.4. dNp has two real eigenvalues and there exist two eigenvectors of dNp that form an orthonormal basis of Tp(S).

1 Example 3.1.5. Suppose the eigenvalues of dNp are λ1 = 2 and λ2 = 2 .

T (S) T (S2) p dN(p) N(p) ~xv 1 2~xv ~xu 2~xu

Eigenvalues and eigenvectors give us a nice, geometric interpretation of how dN stretches tangent vectors around a given point p ∈ S.

43 3.2 Types of Curvature 3 Curvature of Surfaces

3.2 Types of Curvature

Let p be a point on a regular surface S and let v be a unit tangent vector at p; that is v ∈ 0 Tp(S). Then there is a regular differentiable curve α(s) with α(0) = p and t(0) = α (0) = v — to distinguish between tangent/normal vectors for curves and surfaces, we will use lower case t(s) and n(s) for α(s) and upper case T (s) and N(s) for the surface. By the Fr´enet equations for α, α00 = kn. Note that the unit normal vector n may not be normal to S; however, n has a component normal to the surface and a component tangent to the surface. Write n = kN N + kgw for w ∈ Tp(S). The coefficient kN in this expression can be written 00 as an inner product: kN = hα ,Ni.

00 Definition. The constant kN = hα ,Ni, where N is the normal vector to S at the point p, is called the normal curvature at p. Notice that we have defined normal curvature without appealing to any specific parametriza- tion of S. This is useful for theoretical results, but in practice we may want more explicit 0 formulas for kN . For example, differentiating hN ◦ α(s), α (s)i = 0 with respect to s gives hdN ◦ α0(s), α0(s)i + hN ◦ α(s), α00(s)i = 0, so

00 0 0 kN = hα ,Ni = −hdN ◦ α , α i. The right side is a more useful formula for normal curvature in actual computations. It 0 represents the curvature in a specific direction, α = v ∈ Tp(S). From Theorem 3.1.3 we know that for each p ∈ s, dNp is a self-adjoint operator and therefore it has real eigenvalues λ1 and λ2 and corresponding unit eigenvectors v1 and v2 which are orthogonal. The formulas above give us

kN (v1) = −hdN · v1, v1i = −λ1 and kN (v2) = −hdN · v2, v2i = −λ2. From this, we define

Definition. For a point p on a regular surface S, let λ1, λ2 and v1, v2 be the eigenvalues and corresponding orthonormal eigenvectors for dNp. We call k1 = kN (v1) = −λ1 and k2 = kN (v2) = −λ2 the principal curvatures of S at p. Euler provided the following description for normal curvature in any (nonprincipal) di- rection in terms of the principal curvatures of a surface. Let (λ1, v1) and (λ2, v2) be the eigenpairs for dNp. The eigenvectors v1 and v2 form an orthonormal basis for Tp(S) so if w is a tangent vector at p, w = cos θv1 + sin θv2, where θ is the angle between w and v1.

v2

w θ p

v1

44 3.2 Types of Curvature 3 Curvature of Surfaces

The normal curvature in the direction of w is

kN (w) = −hdNp · w, wi

= −hdNp(cos θv1 + sin θv2), cos θv1 + sin θv2i

= −hλ1 cos θv1 + λ2 sin θv2, cos θv1 + sin θv2i 2 2 = −(λ1 cos θ + λ2 sin θ) by orthonormality 2 2 = k1 cos θ + k2 sin θ. This equation is called Euler’s formula, one of many formulas named for the famous Swiss mathematician. Since the formula only depends on the angle between the tangent vector and one of the principal directions, we usually write kN as a function of θ:

Theorem 3.2.1 (Euler’s Formula). For a surface S with principal curvatures k1 and k2 at a point p, the normal curvature in any direction θ from v1 is

2 2 kN (θ) = k1 cos θ + k2 sin θ. We can see that the normal curvature reaches its maximum and minimum values in the principal directions: Suppose without loss of generality that k1 ≤ k2. It is easy to plot kN (θ) and determine its intervals of increase and decrease after computing the first derivative:

0 kN (θ) = −2k1 cos θ sin θ + 2k2 sin θ cos θ = 2 sin θ cos θ(k2 − k1).

kN

k2

k1 θ 0 π 2π

π 3π The maxima of kN are at θ = 2 , 2 ,..., which correspond to the second principal direction with curvature k2, and the minima are at θ = 0, π, 2π, . . ., which correspond to the first principal direction with curvature k1.

Definition. Let S be a surface with principal curvatures k1 and k2 at a point p ∈ S. The determinant det dNp is called the Gaussian curvature of S at p, denoted K. One-half of the negative of the trace Tr dNp is called the mean curvature of S at p, denoted H.

Notice that since dNp is self-adjoint, the determinant of the matrix for dNp in the eigen- basis {v1, v2} is just the product of the eigenvalues, so K = (−λ1)(−λ2) = k1k2. Similarly, k1+k2 mean curvature can be written H = 2 . The next definition classifies some types of surfaces in terms of their principal curvatures. Many of these can be stated in terms of Gaussian curvature as well.

45 3.3 Second Fundamental Form 3 Curvature of Surfaces

Definition. Let p ∈ S have principal curvatures k1 and k2.

ˆ p is elliptic if K > 0, that is, if k1, k2 > 0 or k1, k2 < 0.

ˆ p is hyperbolic if K < 0, that is, if k1 < 0 < k2.

ˆ p is parabolic if k1 = 0 and k2 6= 0.

ˆ p is planar if k1 = k2 = 0.

ˆ p is an umbilical point if k1 = k2 > 0 or k1 = k2 < 0. Example 3.2.2. The unit sphere S2 has Gaussian curvature K = 1 everywhere. Are there other surfaces with K = 1 everywhere? What about K = −1?

3.3 Second Fundamental Form

In Section 3.1, we saw that there is a nice choice of basis {v1, v2} in eigenvectors that make dNp a diagonal matrix. These types of matrices are especially nice – we have already seen some important linear-algebraic properties that make theoretical constructions easy – however, not every basis of Tp gives us such a nice matrix representation. Take a parametrization ~x(u, v) about a point p ∈ S. Then {~xu, ~xv} is a basis for Tp(S) but dNp is usually not diagonal. In particular, we must write dNp · ~xu = Nu = a~xu + b~xv and dNp · ~xv = Nv = c~xu + d~xv for some a, b, c, d. To find these constants, apply the inner product with ~xu and ~xv to the first equation:

h~xu,Nui = ah~xu, ~xui + bh~xu, ~xvi = aE + bF

and h~xv,Nui = ah~xv, ~xui + bh~xv, ~xvi = aF + bG, where (E,F,G) is the first fundamental form for ~x(u, v) (see Section 2.5). Likewise, taking inner products of the expression for Nv gives

h~xu,Nvi = ch~xu, ~xui + dh~xu, ~xvi = cE + dF

and h~xv,Nvi = ch~xv, ~xui + dh~xv, ~xvi = cF + dG.

2 2 Recall that EG − F = |~xu × ~xv| 6= 0 so the above systems of equations are linearly independent and therefore they have unique solutions (a, b) and (c, d).

Definition. The second fundamental form for a parametrization ~x(u, v) of S is the triple (e, f, g) defined by

e = −h~xu,Nui, f = −h~xv,Nui = −h~xu,Nvi and g = −h~xv,Nvi.

46 3.3 Second Fundamental Form 3 Curvature of Surfaces

One can solve the systems of equations above to determine the coefficients a, b, c, d in terms of the first and second fundamental forms for ~x(u, v), yielding solutions collectively known as the Weingarten equations: fF − eG eF − fE a = b = EG − F 2 EG − F 2 gF − fG fF − gE c = d = . EG − F 2 EG − F 2 Notice that normal curvature has a nice expression in terms of the second fundamental form as well: 0 2 0 0 0 2 kN (w) = −hdNp(w), wi = e(u ) + 2fu v + g(v ) , where ~x(u, v) is a parametrization about p such that (~xu(0), ~xv(0)) = w. We also have Proposition 3.3.1. For a surface S and a parametrization ~x(u, v) about a point p ∈ S,

eg − f 2 eG − 2fF + gE K = and H = . EG − F 2 2(EG − F 2)

a b Proof. By definition K = det dN. Write dN = . Then the Weingarten equations give c d us

a b K = c d  1 2 = ((f 2F 2 − fgEF − efF G + egEG) EG − F 2 − (egF 2 − efF G − fgEG + f 2EG)) 1 = ((f 2 − eg)F 2 − (f 2 − eg)EG) (EG − F 2)2 (eg − f 2)(EG − F 2) = (EG − F 2)2 eg − f 2 = . EG − F 2

1 Moreover, by definition H is − 2 times the trace of this matrix, so we have 1 a b H = − T r 2 c d −1 = (fF − eG + fF − gE) 2(EG − F 2) eG − 2fF + gE = . 2(EG − F 2)

47 3.3 Second Fundamental Form 3 Curvature of Surfaces

Next, notice that since K = ad − bc for the matrix above, we can write

|Nu × Nv| = |(a~xu + b~xv) × (c~xu + d~xv)|

= |ad(~xu × ~xv) + bc(~xv × ~xu)|

= |(ad − bc)(~xu × ~xv|

= |ad − bc| |~xu × ~xv|

= |K| |~xu × ~xv|. The first (as we’ve seen before) and second fundamental forms are especially nice in some cases. We will see that this is true for surfaces of revolution as we endeavor to compute the Gaussian and mean curvatures for such a surface. Consider a curve α in the xz-plane, parametrized by arc length.

z

α = (x, z)

x

A surface of revolution may be parametrized as a regular surface by ~x(u, v) = (x(u) cos v, x(u) sin v, z(u)).

0 0 0 This has the partial derivatives ~xu = (x (u) cos v, x (u) sin v, z (u)) and ~xv = (−x sin v, x cos v, 0). The Gauss map is then computed (suppressing the functions of u) as ~x × ~x (−xz0 cos v, −xz0 sin v, xx0 cos2 v + xx0 sin2 v) N = u v = |~xu × ~xv| |~xu × ~xv| (−xz0 cos v, −xz0 sin v, xx0) = p(xz0 cos v)2 + (−xz0 sin v)2 + (xx0)2 (−xz0 cos v, −xz0 sin v, xx0) = xp(x0)2 + (z0)2 (−xz0 cos v, −xz0 sin v, xx0) = since |α0| = 1 x = (−z0 cos v, −z0 sin v, x0). Using the Fr´enetformulas, we compute

0 0 0 0 0 Nu = (z sin v, −z cos v, 0) = (−kx cos v, −kx sin v, −kz ) 0 0 0 = −k(x cos v, x sin v, z ) = −k~xu

48 3.3 Second Fundamental Form 3 Curvature of Surfaces

Therefore k is a principal curvature of the surface and ~xu is a principal direction (eigenvector). Likewise, −z0 −z0 N = (z sin v, −z0 cos v, 0) = (−x sin v, x cos v, 0) = ~x , v x x v −z0 so x is also a principal curvature and ~xv is a principal direction. Thus by the alternate formulations of Gaussian and mean curvature, we have z0 x00 1  z0  K = k = − and H = k + . x x 2 x In a moment we will verify these using the fundamental forms. It is evident that the first fundamental form is E = 1,F = 0 and G = x2. The second fundamental form is computed to be

e = −h~xu,Nui = kE = k

f = −~xu,Nui = 0 z0 g = −h~x ,N i = G = z0x. v v x eg − f 2 kz0x kz0 Therefore K = = = . Mean curvature is computed in a similar fashion. EG − F 2 x2 x In some texts (e.g. Do Carmo), the second fundamental form refers to the function IIp(v) = −hdNp · v, vi which is independent of any parametrization of S at p. IIp takes tangent vectors v to S (at p) and assigns to them the normal curvature of a regular curve on S passing through p with tangent vector v. Notice that if ~x(u, v) is a parametrization of S at p then IIp(~xu) = e and IIp(~xv) = g so our version of the second fundamental form, the triple (e, f, g), describes IIp in some coordinate system (u, v).

Definition. Let p ∈ S. A vector v ∈ Tp(S) is an asymptotic direction at p if kN (v) = 0. Moreover, curve α lying on S is an asymptotic curve if for each α(s), s ∈ I, the tangent vector t(s) is an asymptotic direction at α(s). Question 3.3.2. Under what conditions are coordinate curves asymptotic? Consider the coordinate curve α(s) = ~x(s, v) for a fixed v. Then α is asymptotic if and 0 0 0 0 only if hdN ·α , α i = 0, but α = ~xu so the second fundamental form is IIp(α ) = e. Therefore α is asymptotic if and only if e = 0. Similarly, the other coordinate curve β(s) = ~x(u, s) for fixed u is asymptotic if and only if g = 0. Further, notice that a general direction w ∈ Tp(S) is asymptotic if and only if IIp(w) = 0, but this can be written as a differential equation: e(u0)2 + 2fu0v0 + g(v0)2 = 0 where ~x(u, v) is a parametrization about p. Question 3.3.3. Under what conditions are coordinate curves lines of curvature? Similar calculations show that a coordinate curve α(s) = ~x(s, v) is a line of curvature (i.e. a principal direction) if and only if the first and second fundamental forms satisfy F = f = 0. We saw in the last example that this is indeed the case for surfaces of revolution, so the coordinate curves in those cases are lines of curvature.

49 3.4 Gaussian Curvature and Area 3 Curvature of Surfaces

3.4 Gaussian Curvature and Area

The Gaussian curvature of a surface at a point p has a nice geometric interpretation which was Gauss’s original motivation for describing K. Consider the Gauss map

S R N p p

We want to compute the ratio of the area of a patch on S to the area of its image under the Gauss map. Specifically, we will show that

A(N(R)) ≈ |K|, A(R) where K is Gaussian curvature at p. Since R and N(R) are both parametrized by some ~x(u, v), let U = ~x−1(R) ⊂ R2 which also equals (N ◦~x)−1(N(R)). The areas of these patches can be expressed as integrals: ZZ ZZ A(R) = dA = |~xu × ~xv| du dv R U ZZ and A(N(R)) = |Nu × Nv| du dv. U

We saw in Section 3.3 that |Nu × Nv| = |K| |~xu × ~xv| so RR A(N(R)) U |K| |~xu × ~xv| du dv = RR . A(R) U |~xu × ~xv| du dv Unfortunately we cannot just factor out K since curvature depends on (u, v), i.e. the point on the surface. However, by the mean value theorem for (double) integrals and the continuity of K, as (u, v) → p, K → Kp, the Gaussian curvature at p. Therefore as (u, v) → p, ZZ ZZ |K| |~xu × ~xv| du dv −→ |Kp| |~xu × ~xv| du dv U U which implies

A(N(R)) |Kp| A(R) lim = lim = lim |Kp| = |Kp|. A(R)→0 A(R) A(R)→0 A(R) A(R)→0

50 4 Intrinsic Geometry

4 Intrinsic Geometry

So far we have seen many attributes of curves and surfaces that require a specific parametriza- tion to understand. For example, the second fundamental form (Section 3.3) was defined in terms of a coordinate system, and the computational version of the first fundamental form (E,F,G) is likewise determined in terms of a parametrization ~x(u, v) of the surface. Such characteristics are extrinsic to the surface, in that they depend on how the surface embeds in space. On the other hand, some of the more theoretical concepts studied so far are intrinsic geometric properties: they only depend on the geometry of the surface itself and not on any particular parametrization. The first fundamental form Ip(w) = hw, wi is perhaps the most important tool in intrinsic geometry as it allows us to calculate area, angles, arc length and more. Such computations can be made without leaving the surface, so they are especially desirable. In this chapter we will study many other intrinsic properties of surfaces, which are in most cases related to the first fundamental form.

4.1 Isometries

Recall from Example 2.5.3 that R2 and the unit cylinder have the same first fundamental form: E = G = 1,F = 0. There is something interesting going on, as one can see by slicing the cylinder vertically and unwrapping it to resemble the plane. Intrinsically, these surfaces behave the same way.

Definition. A diffeomorphism ϕ : S → S beween surfaces is an isometry if for every p ∈ S and w1, w2 ∈ Tp(S), hw1, w2ip = hdϕp ·w1, dϕp ·w2iϕ(p). In other words, an isometry preserves the inner product on S.

The notation hw1, w2ip indicates that the inner product is taken at p, i.e. in the tangent space Tp(S), and likewise hdϕp · w1, dϕp · w2iϕ(p) is taken at ϕ(p).

S S dϕ(~w2) ~w1 ϕ

p ϕ(p)

dϕ(~w1) ~w2

Isometry is the fundamental notion of equivalence in differential geometry, much like isomorphism in abstract algebra, isotopy in knot theory or homeomorphism in topology. The definition of isometry implies that it is metric-preserving; the next proposition shows that when checking ϕ is an isometry, it suffices to check that the first fundamental form Ip is preserved.

51 4.1 Isometries 4 Intrinsic Geometry

Proposition 4.1.1. Let S, S be regular surfaces. Then ϕ : S → S is an isometry if and only if Ip(w) = Iϕ(p)(dϕ · w).

Proof. ( =⇒ ) is obvious since Ip(w) is an inner product on Tp(S). ( ⇒ = ) Assume hw, wip = hdϕ(w), dϕ(w)iϕ(p) for all w ∈ Tp(S). Consider hw1, w2ip for w1, w2 ∈ Tp(S). We can realize this inner product as part of a quadratic expression:

2hw1, w2ip = hw1 + w2, w1 + w2ip − hw1, w1ip − hw2, w2ip.

Then everything on the right side is a value of Ip so

2hdϕ(w1), dϕ(w2)iϕ(p) = hdϕ(w1 + w2), dϕ(w1 + w2)iϕ(p)

− hdϕ(w1), dϕ(w1)iϕ(p) − hdϕ(w2), dϕ(w2)iϕ(p)

= hw1 + w2, w1 + w2ip − hw1, w1ip − hw2, w2ip

= 2hw1, w2ip. Dividing through by 2 gives the desired equality. Definition. Let S, S be surfaces, p ∈ S and V a neighborhood of p. We say the map ϕ : V → S is a local isometry provided there exists a neighborhood V ⊂ S so that ϕ : V → V is an isometry. In general, a global isometry is a rather strong condition on two surfaces. Local isometries allow us to study the intrinsic properties of surfaces that are not diffeomorphic but still have characteristics in common.

Example 4.1.2. The unit cylinder C = {(x, y, z) ∈ R3 : x2 + y2 = 1} is locally isometric to the plane R2 at every point. However, they are not isometric since they are not diffeomorphic (e.g. their fundamental groups are distinct).

R2 ϕ

ϕ(p) p

~y = id ~x(u, v) v

~x−1(p) u

52 4.1 Isometries 4 Intrinsic Geometry

Consider the parametrizations ~x(u, v) = (cos u, sin u, v), 0 < u < 2π, −∞ < v < ∞ for C and ~y the identity for R2. This describes a neighborhood V of the cylinder and a corresponding rectangle ~x−1(V ) = {(u, v) : 0 < u < 2π, −∞ < v < ∞} in the plane. In Example 2.5.3 we computed the first fundamental form of C in this parametrization as E = G = 1,F = 0. In other words, Ip is the identity operator on Tp(S). The first fundamental form of R2 for ~y is also E = G = 1, F = 0. Define ϕ = ~y ◦ ~x−1. To see that ϕ : V → ϕ(V ) is an isometry, take w = a~xu + b~xb ∈ Tp(S) and consider

2 2 Ip(w) = a E + 2abF + b G = a2E + 2abF = b2G

= Iϕ(p)(dϕp(a~xu + b~xv))

= Iϕ(p)(dϕp · w). Hence ϕ is a local isometry between the cylinder and the plane. Notice that we didn’t even use the values of E,F,G or E, F, G. This is a result of the more general fact described below. Proposition 4.1.3. If there exists parametrizations ~x : U → S and ~y : U → S whose first −1 fundamental forms satisfy E~x = E~y,F~x = F~y,G~x = G~y then ϕ : ~y ◦ ~x : ~x(U) → S is a local isometry. Proof. The proof is exactly the same as the computation above. Examples. 1 Consider the curve parametrized by (a cosh v, av), called the catenary.

z

(a cosh v, av)

x

Physically, a (horizontal) catenary is formed by extending a piece of rope without any tension over a gap such as a ravine and letting gravity act on it. Bridges that are only suspended by their two points of contact are examples of catenaries. In our case, we form a surface of revolution, called the catenoid, by revolving the catenary about the z-axis. This can be parametrized by ~x(u, v) = (a cosh v cos u, a cosh v sin u, av).

53 4.1 Isometries 4 Intrinsic Geometry

It turns out that catenoids are examples of minimal surfaces (surfaces that assume a minimal surface area in R3). A piece of the catenoid is shown below.

~x(u, v)

The partials for this parametrization are ~xu = (−a cosh v sin u, a cosh v cos u, 0) and ~xv = (a sinh v cos u, a sinh v sin u, a) so the first fundamental form of the catenoid is E = a2 cosh2 v, F = 0 and G = a2(sinh2 v + 1) = a2 cosh2 v.

There is another minimal surface which turns out to be locally isometric to the catenoid. Recall the helix from Section 1.2. Varying the radius of the helix smoothly gives a regular surface called the helicoid, pictured below.

54 4.1 Isometries 4 Intrinsic Geometry

The helicoid is parametrized by ~y(¯u, v¯) = (¯v cosu, ¯ v¯ sinu, ¯ au¯), 0 < u¯ < 2π, 0 < v¯ < ∞. To compare this to the catenoid, setu ¯ = u andv ¯ = a sinh v, so that ~y can be written

~y(u, v) = (a sinh v cos u, a sinh v sin u, au).

The partial derivatives for this parametrization are ~yu = (−a sinh v sin u, a sinh v cos u, a) and ~yv = (a cosh v cos u, a cosh v sin u, 0), the first fundamental form is

E = a2 cosh v, F = 0 and G = a2 cosh2 v.

Therefore by Proposition 4.1.3, there is a local isometry ϕ : ~y ◦ ~x−1 between the parametrized regions of the catenoid and the helicoid. Clearly these surfaces are not globally isometric, as they are not homeomorphic.

2 Recall the cone from Section 2.2.

α

This corresponds to the function z2 = k2(x2 + y2) where k = cot α and α is the angle made by the cone with the z-axis. For simplicity, let us assume z > 0 so that we avoid the singularity at the origin and we have a connected, regular surface C. Solving for z, we have z = kpx2 + y2 so C may be parametrized as a surface of revolution:

  θ   θ   ~x(ρ, θ) = ρ sin α cos , ρ sin α sin , ρ cos α , 0 < ρ < ∞, 0 < θ < 2π sin α. sin α sin α

Consider the following subset U of the plane:

U

2π sin α

55 4.1 Isometries 4 Intrinsic Geometry

Then U is a regular surface parametrized by

~y(ρ, θ) = (ρ cos θ, ρ sin θ, 0), 0 < ρ < ∞, 0 < θ < 2π sin α.

Notice that C and U have the same parameters coming from the same subset of R2. Thus in order to check that ϕ = ~y ◦~x−1 is an isometry on the associated neighborhoods of C and U, it suffices to show that their first fundamental forms are the same. For U, consider ~yρ = (cos θ, sin θ, 0) and ~yθ = (−ρ sin θ, ρ cos θ, 0). So E = 1,F = 0 and G = ρ2. On the other hand, for C we have

  θ   θ   ~x = sin α cos , sin α sin , cos α ρ sin α sin α   θ   θ   and ~x = −ρ sin , ρ cos , 0 . θ sin α sin α

Thus E = 1,F = 0 and G = ρ2 as well. Hence ϕ is a local isometry between the upper half of the double cone (with a seam) and the open subset U ⊂ R2. The parametrization of C can easily be shifted to derive local isometries at every point of the cone.

3 Let F : U ⊂ R2 → R3 be defined by F (u, v) = (u sin α cos v, u sin α sin v, u cos α),

where α is a positive constant and U = {(u, v) ∈ R2 : u > 0}. We first show that F is a diffeomorphism from U to the cone C with its vertex at the origin that makes an angle of α radians with the z-axis, the same as in the last example.

α

Note that if we were to parametrize C from scratch, one option is to recognize C as the surface obtained from revolving the line α(t) = (t sin α, t cos α) about the z- axis. This clearly generates the cone, as the side of the cone lying in the xz-plane cos α is precisely the line with slope sin α . Then the parametrization for the surface of revolution is (u sin α cos v, u sin α sin v, u cos α), which is precisely F (u, v). Therefore F is a diffeomorphism. 2 Is F a local isometry? Consider the line segments {(u, v) ∈ R : u = c1, 0 < v < 2π} 2 and {(u, v) ∈ R : u = c2, 0 < v < 2π} for some positive constants c1, c2.

56 4.1 Isometries 4 Intrinsic Geometry

v

F (u, v) 2π

u u = c1 u = c2

By construction these segments each have length 2π, but the corresponding curves on the surface of the cone are two circles with radius c1 sin α and c2 sin α, respectively. Assuming c1 6= c2, the circumferences of the circles on the right must be different, so F cannot possibly be an isometry. This shows that just because R2 and C are locally isometric, not every parametrization we choose has to preserve distances. In the last example, we used the fact that an isometry preserves arc lengths. The converse is also true; we prove these statements below. Proposition 4.1.4. Suppose S and S are regular surfaces. A map ϕ : S → S is an isometry if and only if arc length is invariant under ϕ. Proof. ( =⇒ ) Assume ϕ : S → S is an isometry. If ~x(u, v) is a parametrization of a point p ∈ S then ϕ(u, v) = ϕ ◦ ~x(u, v) is the induced parametrization for ϕ(p) ∈ S.

S S ϕ

p ϕ(p)

ϕ(u, v) = ϕ ◦ ~x(u, v) ~x(u, v) v

u

Let E,F,G be the first fundamental form for ~x(u, v) and let E, F, G be the first fundamental form for ϕ(u, v). Then Proposition 4.1.3 shows that since ϕ is an isometry, E = E,F = F

57 4.2 Christoffel Symbols and Gauss’ Theorema Egregium 4 Intrinsic Geometry

and G = G. Let α(t), a ≤ t ≤ b, be a curve on the surface S and let β(t) = ϕ ◦ α(t) be the corresponding curve on S. Then

0 0 0 0 |β (t)| = |dϕ ◦ α | = |u dϕ · ~xu + v dϕ · ~xv| 0 0 = |u ~xu + v ~xv| by isometry = |α0(t)|. Since arc length is determined by Z b Z b length(α) = |α0(t)| dt and length(β) = |β0(t)| dt, a a we see that these are equal. Hence ϕ preserves arc length. ( ⇒ = ) Conversely, if arc length is preserved under ϕ, it is preserved for coordinate curves. In particular, consider the coordinate curve α with v constant and u = t the parameter. By the arc length formula in Section 2.5, we have Z b Z b√ length(α) = p(u0)2E + 2u0v0F + (v0)2G = E. a a Z bp Likewise, length(ϕ ◦ α) = E. We may assume a < b so that if arc length is equal, a 1 Z b√ Z bp  E − E = 0. b − a a a Taking the limit as b → a, this can be written 1 Z b√ Z bp  √ p 0 = lim E − E = E − E b→a b − a a a by the fundamental theorem of calculus. Therefore E = E. The proofs that F = F and G = G are similar, using different curves. Therefore by Proposition 4.1.3, ϕ is an isometry.

4.2 Christoffel Symbols and Gauss’ Theorema Egregium

Suppose ~x : U ⊂ R2 → S parametrizes an oriented, regular surface S. Like the Fr´enet trihedron for curves in Chapter 1, we associate a trihedron {~xu, ~xv,N} to every point in the neighborhood ~x(U) of S. To understand this trihedron, we express the following partial derivatives as linear combinations of ~xu, ~xv and N: 1 2 ~xuu = Γ11~xu + Γ11~xv + L1N 1 2 ~xuv = Γ12~xu + Γ12~xv + L2N 1 2 ~xvu = Γ21~xu + Γ21~xv + L2N 1 2 ~xvv = Γ22~xu + Γ22~xv + L3N

Nu = a11~xu + a21~xv

Nv = a12~xu + a22~xv.

58 4.2 Christoffel Symbols and Gauss’ Theorema Egregium 4 Intrinsic Geometry

The aij in the last two lines of this sytem are the coefficients from the Weingarten equations in Section 3.3. i Definition. The coefficients Γjk, 1 ≤ i, j, k ≤ 2, are called the Christoffel symbols for the parametrization ~x(u, v) of S.

Notice that by equality of mixed partials, ~xuv = ~xvu so we see that the Christoffel symbols i i are symmetric in j, k:Γjk = Γkj for all 1 ≤ i, j, k ≤ 2. Moreover, taking inner products with N allows us to express the Li in terms of the second fundamental form:

L1 = h~xuu,Ni = e, L2 = L2 = h~xuv,Ni = f and L3 = h~xvv,Ni = g. Taking a series of inner products of the above system gives us the following linear system: 1 2 1 h~xuu, ~xui = Γ11E + Γ11F = 2 Eu 1 2 1 h~xuu, ~xvi = Γ11F + Γ11G = Fu − 2 Ev 1 2 1 h~xuv, ~xui = Γ12E + Γ12F = 2 Ev 1 2 1 h~xuv, ~xvi = Γ12F + Γ12G = 2 Gu 1 2 1 h~xvv, ~xui = Γ22E + Γ22F = Fv − 2 Gu 1 2 1 h~xvv, ~xvi = Γ22F + Γ22G = 2 Gv. Notice that each pair of equations has determinant EG − F 2 6= 0, so unique solutions exist for the Christoffel symbols for any parametrization of a regular surface S. Example 4.2.1. In this example we compute the Christoffel symbols for a surface of revo- lution formed by rotating a regular curve α = (f, g) about the z-axis. As we’ve seen before, this can be parametrized by ~x(u, v) = (f(v) cos u, f(v) sin u, g(v)).

z

α = (f, g)

x

The partials and first fundamental form are:

~xu = (−f(v) sin u, f(v) cos u, 0) 0 0 0 ~xv = (f (v) cos u, f (v) sin u, g (v)) E = f(v)2 F = 0 G = f 0(v)2 + g0(v)2.

59 4.2 Christoffel Symbols and Gauss’ Theorema Egregium 4 Intrinsic Geometry

To examine the system in the previous paragraph, we need the following partials of the first fundamental form:

Eu = 0 Fu = 0 Gu = 0 0 0 00 0 00 Ev = 2f(v)f (v) Fv = 0 Gv = 2f (v)f (v) + 2g (v)g (v)

We can use these values to obtain solutions for the Christoffel symbols:

1 2 2 1 Γ11f + Γ11(0) = 0 =⇒ Γ11 = 0 −ff 0 Γ1 (0) + Γ2 ((f 0)2 + (g0)2) = 0 − 1 (2ff 0) =⇒ Γ2 = 11 11 2 11 (f 0)2 + (g0)2 f 0 Γ1 f 2 + Γ2 (0) = 1 (2ff 0) =⇒ Γ1 = 12 12 2 12 f 1 2 0 2 0 2 1 2 Γ12(0) + Γ12((f ) + (g ) ) = 2 (0) =⇒ Γ12 = 0 1 2 2 1 1 Γ22f + Γ22(0) = 0 − 2 (0) =⇒ Γ22 = 0 f 0f 00 + g0g00 Γ1 (0) + Γ2 ((f 0)2 + (g0)2) = 1 (2f 0f 00 + 2g0g00) =⇒ Γ2 = . 22 22 2 22 (f 0)2 + (g0)2

To summarize, the Christoffel symbols for a surface of revolution generated by a regular plane curve α = (f, g) are

−ff 0 Γ1 = 0 Γ2 = 11 11 (f 0)2 + (g0)2 f 0 Γ1 = Γ2 = 0 12 f 12 f 0f 00 + g0g00 Γ1 = 0 Γ2 = . 22 22 (f 0)2 + (g0)2

i Christoffel symbols are intrinsic: the preceding discussion shows that the set of Γjk can be computed in terms of the first fundamental form, which we know is intrinsic. These symbols are also useful for describing the differential of the Gauss map dNp : Tp(S) → Tp(S). As we know, this differential has a matrix form   a11 a12 dNp = a21 a22 where the aij are the coefficients of the Weingarten equations: fF − eG gF − fG a = a = 11 EG − F 2 12 EG − F 2 eF − fE fF − gE a = a = . 21 EG − F 2 22 EG − F 2 Question 4.2.2. To what extent are the E, F, G, e, f and g from the first and second fun- damental forms independent? Or, if they are not independent, how do they relate?

60 4.2 Christoffel Symbols and Gauss’ Theorema Egregium 4 Intrinsic Geometry

Recall the formulas for the partials of ~x and N:

1 2 ~xuu = Γ11~xu + Γ11~xv + eN 1 2 ~xuv = Γ12~xu + Γ12~xv + fN 1 2 ~xvv = Γ22~xu + Γ22~xv + gN

Nu = a11~xu + a21~xv

Nv = a12~xu + a22~xv.

We will use the fact that the second and third mixed partials are equal to write

~xuuv − ~xuvu = 0

~xvvu − ~xuvv = 0

and Nuv − Nvu = 0.

Combining these two sets of equations, we have a system

Ai~xu + Bi~xv + CiN = 0 for i = 1, 2, 3.

Since ~xu, ~xv and N are linearly independent, we obtain nine relations: Ai = 0,Bi = 0,Ci = 0 for i = 1, 2, 3. For example, let’s compute the relation given by B1 = 0. This comes from setting the coefficients for ~xv in ~xuuv − ~xuvu = 0:

1 2 2 2 2 1 2 2 2 2 0 = Γ11Γ12 + Γ11Γ22 + ea22 + (Γ11)v − Γ12Γ11 − Γ12Γ12 + fa21 + (Γ12)u.

Then by the Weingarten equations,

eg − f 2 (Γ2 ) − (Γ2 ) + Γ1 Γ2 + Γ2 Γ2 − Γ2 Γ2 − Γ1 Γ2 = −E = −EK, 12 u 11 v 12 11 12 12 11 22 11 12 EG − F 2 where K is Gaussian curvature. This is a surprising result: K only depends on the Christoffel symbols so it is intrinsic! This was important enough that Gauss called it the Theorema Egregium (Remarkable Theorem):

Theorem 4.2.3 (Gauss). The Gaussian curvature of a surface is intrinsic.

Proof. By the work above, there is a formula for K in terms of the Christoffel symbols and i we showed that the Γjk are intrinsic. Therefore K is intrinsic. Examples. The following examples highlight the broad power of the Theorema Egregium.

1 Since K is intrinsic to surfaces, there is no isometry from the sphere (positive Gaussian curvature) to the plane (zero Gaussian curvature); there isn’t even a local isometry! This illustrates why it is impossible to create a perfect flat map of the Earth.

2 The sphere (K > 0), cylinder (K = 0) and saddle graphed by z2 = x2 − y2 (K < 0) are not locally isometric.

61 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

3 One may wonder if there exists a regular surface for any possible combination of first and second fundamental forms. In fact, the Theorema Egregium rules out many options. For instance, suppose S has E = G = 1,F = 0, e = 1, f = 0 and g = −1. Then the Gaussian curvature of S is eg − f 2 K = = −1. EG − F 2 However we saw in Example 4.1.2 that the plane also has first fundamental form E = G = 1 and F = 0. Therefore such a surface S would have a local isometry with a patch of the plane, but since the plane has zero Gaussian curvature, the Theorema Egregium shows this to be impossible. Therefore no surface exists with E = G = 1,F = 0, e = 1, f = 0, g = −1.

4.3 Geodesics and the Covariant Derivative

What does it mean for a curve to be a straight line on a surface? Similarly, what does parallel translate to on surfaces? To study these and related questions, we introduce an object called the covariant derivative. Assume α is a differentiable curve lying on a surface S. Our definition will work even if α is not regular.

Definition. Let w(t) be a differentiable vector field defined on α(t) such that w(t) ∈ Tα(t)(S) for all t. The covariant derivative of w is defined to be the projection of w0(t) onto the tangent space Tα(t)(S): Dw dw = proj . dt Tα(t)(S) dt

0 Using the trihedron {~xu, ~xv,N}, for each t we can write w (t) = a~xu + b~xv + cN; notice that a, b, c, ~xu, ~xv and N all depend on t. Then the covariant derivative of w may be written Dw = w0(t) − hw0(t), ~x i~x − hw0(t), ~x i~x . dt u u v v Example 4.3.1. The most useful choice of vector field on α(t) is simply the curve’s tangent vectors: w(t) = α0(t). Other choices of vector field include the tangent vectors in the coordinate directions, e.g. w(t) = ~xu(α(t)) or ~xv(α(t)); or tangents in the principal directions: e1(α(t)) and e2(α(t)).

Dw Definition. We say a vector field w(t) is parallel to α(t) if dt = 0. Definition. A curve α(t) on S is a geodesic if the tangent vector field w(t) = α0(t) is parallel.

62 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

Theorem 4.3.2 (Properties of Covariant Derivatives). Let S be a regular surface and let α(t) be a differentiable curve on S. Dw (1) For any vector field w(t) on α, dt is intrinsic. d (2) If w1 and w2 are parallel vector fields on α then dt hw1, w2i = 0. (3) If w is a parallel vector field on α, then |w| is constant.

Proof. (1) Since w(t) ∈ Tα(t)(S), we may write w(t) = a~xu + b~xv. Then dw = a0~x + a~x0 + b0~x + b~x dt u u v v 0 0 0 0 0 0 = a ~xu + b ~xu + a(~xuuu + ~xuvv ) + b(~xvuu + ~xvvv ) 0 0 1 0 1 0 1 0 1 = (a + au Γ11 + av Γ12 + bu Γ12 + bv Γ22)~xu 0 0 2 0 2 0 2 0 2 + (b + au Γ11 + av Γ12 + bu Γ12 + bv Γ22)~xv + (au0e + av0f + bu0f + bv0g)N.

dw Notice that the coefficients for ~xu and ~xv in this description of dt are intrinsic – they only depend on a, b, u, v and the Christoffel symbols for S. Since the covariant derivative is defined as a projection, Dw dw = proj , dt Tα(t)(S) dt Dw we see that dt will only depend on the coefficients of ~xu and ~xv, which in turn are in terms i Dw of a, b, u, v and the Γjk. Therefore dt is intrinsic. (2) By the product rule for inner products, d dw   dw  hw , w i = 1 , w + w , 2 . dt 1 2 dt 2 1 dt

dw1 Dw1 Note that dt = (a~xu + b~xv) + cN = dt + cN for some a, b and c; and since w2 ∈ Tα(t)(S), dw1 Dw1 dw2 hN, w2i = 0 so we can replace dt , w2 with dt . Similarly, we can replace hw1, dt i with Dw2 hw1, dt i to obtain d Dw   Dw  hw , w i = 1 , w + w , 2 dt 1 2 dt 2 1 dt

= h0, w2i + hw1, 0i since w1, w2 are parallel to α = 0 + 0 = 0. (3) follows immediately from (2), since |w|2 = hwi.

Theorem 4.3.3. Given a curve α(t) on S and a tangent vector w0 ∈ Tα(t0)(S), there is a unique parallel vector field w(t) on α(t) such that w(t0) = w0. Proof. This comes down to solving the initial value problem Dw = 0, w(t ) = w . dt 0 0 We may apply the existence and uniqueness theorem from ODEs to obtain the result.

63 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

Definition. The vector field w(t) satisfying Theorem 4.3.3 is called the parallel transport of w0 along α.

S

α

w0

Example 4.3.4. Consider the curve α with constant height on the sphere S:

Dw dv N

dw α dv

This curve can be parametrized by α(v) = (cos u cos v, cos u sin v, sin u). Attach to α the vector field of tangents: w(v) = α0(v). To determine if α is a geodesic, we need to decide if w(v) is parallel along α. Recall the parametrization ~x(u, v) = (cos u cos v, cos u sin v, sin u) of the sphere has partial derivatives

~xu = (− sin u cos v, − sin u sin v, cos u) and ~xv = (− cos u sin v, cos u cos v, 0). To determine the covariant derivative of w(v), we first compute dw = α00(v) = (− cos u cos v, − cos u sin v, 0) dv dw  dw  dw  = , ~x ~x + , ~x ~x + ,N N dv u u dv v v dv 2 = (sin u cos u)~xu + (0)~xv + (− cos u)N. Then the covariant derivative is the projection of this vector in the tangent plane, that is, the plane spanned by ~xu and ~xv: Dw = (sin u cos u)~x . dv u We can see from this expression that α is a geodesic of the sphere precisely when u = 0, i.e. when α is the equator.

64 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

Suppose w(t) is a vector field on a curve α(t). If |w(t)| = 1 for all t, what can we say about its covariant derivative?

N N × w α w

dw In this case dt is still perpendicular to w since, by a trick we’ve used before, |w| = 1 implies d dw  hw, wi = 0 =⇒ , w = 0. dt dt

dw Therefore we can express dt as a linear combination of the trihedron {N, w, N × w}: dw = αN + βw + λ(N × w). dt

dw Since dt , w = 0, β = 0 so the covariant derivative of such a vector field is Dw = βw + λ(N × w) = λ(N × w). dt

Dw Definition. The coefficient λ ∈ R such that dt = λ(N × w) is called the algebraic value Dw  Dw  of dt , denoted λ = dt . Example 4.3.5. Consider a curve α(s) = (x(s), y(s), 0) in the plane S = R2. z

N = (0, 0, 1)

y

α x

65 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

Let w be the vector field of tangents: w(s) = α0(s) = (x0, y0, 0). Using the Fr´enetequations from Chapter 1, we compute dw = (x00, y00, 0) = k(−y0, x0, 0), ds where k is the curvature of α in the plane. Now N × w = (0, 0, 1) × (x0, y0, 0) = (−y0, x0, 0) so  Dw  we see that the algebraic value of the covariant derivative of w is ds = k. This motivates the following definition for any surface S.

Definition. Let α(s) be a curve on a surface S with orientation N. Then the geodesic curvature of α is the algebraic value

Dw k (s) = , g ds

where w(s) = α0(s).

Example 4.3.6. (Curves on the torus) Consider three curves formed by slicing a torus T with different horizontal planes.

α3 α2

α1

Are any of these geodesics? Asymptotic? First, α1 and α3 are geodesics, since in both cases dw = kn = ckN for w(t) = α0(t), i = 1, 2, dt i so w(t) is parallel to the normal vector N to the surface and therefore its covariant derivative dw is zero. On the other hand, α2 is not a geodesic since all its dt lie in the tangent spaces to the surface. α2 is asymptotic though, since n and N are orthogonal and kN = khn, Ni = 0. The other two curves are not asymptotic, as n and N are not orthogonal in these cases.

Theorem 4.3.7. Given an orthogonal parametrization ~x(u, v) (i.e. one with F = 0) of a surface S and a vector field w(t) defined on S with |w| = 1 everywhere, the algebraic value of w is   Dw 1 0 0 0 = √ (Guv − Evu ) + ϕ , dt 2 EG

where ϕ is the angle between ~xu and w.

66 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

Proof. We may obtain an orthonormal basis {e1, e2,N} for the tangent space at any point of S by setting ~xu ~xv e1 = √ , e2 = √ and N = e1 × e2. E G For any t, w(t) is tangent to S and thus may be written w(t) = ae1 + be2 for some a, b that depend on t. The requirement |w| = 1 implies a2 + b2 = 1, and even better, we know a = cos ϕ and b = sin ϕ as the figure below illustrates.

e2

w ϕ

e1

Then we have dw = a0e + ae0 + b0e + be0 dt 1 1 2 2 De  De  = a0e + b0e + a 1 + γN + b 2 + δN 1 2 dt dt for some values γ, δ. Projecting on the tangent plane, i.e. dropping the terms in the normal direction, gives the following expression for the covariant derivative of w: Dw De De = a0e + b0e + a 1 + b 2 . dt 1 2 dt dt Note that the definition of algebraic value easily leads to Dw Dw  = ,N × w , dt dt so to produce the desired formula, we may compute N × w and take this inner product. Consider N × w = N × (ae1 + be2) = a(N × e1) + b(N × e2) = ae2 − be1. Then Dw   De De  ,N × w = a0e + b0e + a 1 + b 2 · (−be + ae ) dt 1 2 dt dt 1 2 De  De  De  De  = −a0b − ab 1 , e − b2 2 , e + ab0 + a2 1 , e + ab 2 , e . dt 1 dt 1 dt 2 dt 2

To evaluate these inner products, recall that he1, e2i = 0, so that differentiating yields de  de  de  de  2 , e + 1 , e = 0 =⇒ 2 , e = − 1 , e dt 1 dt 2 dt 1 dt 2 De  De  =⇒ 2 + cN, e = − 1 + dN, e dt 1 dt 2 De  De  =⇒ 2 , e = − 1 , e . dt 1 dt 2

67 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

De1 De2 Similarly, he1, e1i = 1 and he2, e2i = 1 imply dt , e1 = dt , e2 = 0. This simplifies the computation dramatically:

Dw De  De  = −a0b + (a2 + b2) 1 , e + ab0 = ab0 − a0b + 1 , e dt dt 2 dt 2 since a2 + b2 = 1. Now De  de  1 , e = 1 , e as above dt 2 dt 2 0 0 = h(e1)uu + (e1)vv , e2i 0 0 = h(e1)uu , e2i + h(e1)vv , e2i.

The first term can be written * √ √ + ~x E − ~x ( E) h(e ) u0, e i = uu u u , e u0 1 u 2 E 2 * √ + 1 ( E)u 1 0 = √ ~xuu − ~xu, √ ~xv u E E G

1 0 = √ h~xuu, ~xviu EG   1 1 0 = √ Fu − Ev u by the formulas from Section 4.2 EG 2 −1 0 = √ Evu . 2 EG Similarly, the second term can be expanded:

0 1 0 h(e1)vv , e2i = √ Guv . 2 EG Lastly, recall that a = cos ϕ and b = sin ϕ. Then

ab0 − a0b = (cos ϕ)(cos ϕ · ϕ0) − (− sin ϕ · ϕ0)(sin ϕ) = (cos2 ϕ + sin2 ϕ)ϕ0 = ϕ0.

Putting everything together gives the desired formula:   Dw 1 0 0 0 = √ (Guv − Evu ) + ϕ . dt 2 EG

68 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

Examples.

1 Let S be the unit sphere with parametrization ~x(u, v) = (cos u cos v, cos u sin v, sin u). Recall that the first fundamental form for S is E = 1,F = 0,G = cos2 u. For the horizontal curve α(t) = ~x(u, t) with vector field wu(t) = ~xu(u, t), we want to compute  Dwu  the algebraic value dt . By Theorem 4.3.7, Dw  1 u = (−2 cos u sin u − 0) + 0 = − sin u. dt 2 cos u

 Dwv  Similarly, wv(t) = ~xv(u, t) has algebraic value dt = − sin u as well. 2 Suppose ~x(u, v) is an orthogonal parametrization of a surface S.

S

α(s)

v = c1

u = c2

0 Consider α(s) = ~x(u(s), c1) where c1 is a constant. Note that v = 0 for this curve and 0 0 0 0 0 α (s) = ~xuu (s) + ~xvv (s) = ~xuu (s). Since α is regular, we may assume |α | = 1 so we see that u0 = 1 = √1 . Then by Theorem 4.3.7, |~xu| E

 0    Dα 1 1 −Ev kg = = √ Gu(0) − Ev √ = √ . ds 2 EG E 2E G In particular, we see that the coordinate curves on the sphere with v held constant are all geodesics:

69 4.3 Geodesics and the Covariant Derivative 4 Intrinsic Geometry

3 Suppose α is a curve on a surface S that is both a line of curvature and a geodesic. Let N be the normal vector to S at an arbitrary point lying on α and let (t, n, b) denote the Fr´enet trihedron for α, as in Chapter 1. A line of curvature satisfies N 0 = λα0 and a geodesic satisfies N = n. Differentiating and writing these in terms of the Fr´enet equations gives us N 0 = λt and N 0 = −kt − τb. Since t and b are linearly independent, we deduce that λ = −k and τ = 0. From the characterization of curves in Section 1.5, we know that a curve with zero torsion must be a plane curve. This reflects the earlier discovery that only certain lines of curvature on the torus are geodesics.

Let (kg)1 and (kg)2 denote respectively the geodesic curvatures of the first and second coordinate curves of S, i.e. the curves with v or u held constant. Recall that we proved

−Ev Gu (kg)1 = √ and (kg)2 = √ . 2E G 2G E The following formula says that any geodesic curvature may be written as a combination of (kg)1 and (kg)2. Theorem 4.3.8 (Liouville’s Formula). Let ~x(u, v) be an orthogonal parametrization of a regular surface S at a point p and suppose α is a curve on S passing through p. Let ϕ be the 0 angle between ~xu and α . Then the geodesic curvature at p is dϕ k = (k ) cos ϕ + (k ) sin ϕ + . g g 1 g 2 ds

 Dα0  Proof. Using the general formula for ds and the expressions for (kg)1 and (kg)2 above, we may write Dα0  √ du √ dv dϕ k = = (k ) E + (k ) G + . g ds g 1 ds g 2 ds ds In addition, we have   √   √ 0 0 0 ~xu 0 ~xv 0 α (s) = ~xuu + ~xvv = √ Eu + √ Gv E G  ~x   ~x  = √u cos ϕ + √v sin ϕ. E G Combining these gives Liouville’s Formula.

70 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

4.4 The Gauss-Bonnet Theorem

One of the most important theorems in differential geometry in the last 200 years is the Gauss-Bonnet theorem. We begin with the simpler notion of triangles in the plane, which is reminiscent of Euclid’s studies of plane geometry. Consider a triangle in the plane:

θ2

θ1

θ3

Label its exterior angles θ1, θ2 and θ3, as pictured. Then θ1 + θ2 + θ3 = 2π. Notice that a triangle is just the region formed by the intersection of three nonparallel lines in the plane. Gauss realized that this could be generalized to the sphere and later to a regular surface. Consider a triangle R on the unit sphere:

θ1 θ2 R

θ3

We consider R to be a triangle if its three sides are geodesics. Again label the exterior angles by θ1, θ2 and θ3. Then Gauss proved that the sum of the angles satisfies

3 X A(R) + θ3 = 2π, i=1 where A(R) is the area of the triangle. Bonnet extended this result to regular surfaces:

Theorem 4.4.1 (Bonnet). Let S be a regular surface with Gaussian curvature K. If R is a triangle whose sides are geodesic curves αi meeting at exterior angles θ1, θ2, θ3 then

3 3 X Z si+1 ZZ X kg ds + K dσ + θi = 2π, i=1 si R i=1 where s1 < s2 < s3 < s4 satisfy α1(s2) = α2(s2, α2(s3) = α3(s3) and α3(s4) = α1(s1) and kg denotes geodesic curvature.

71 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

S

θ1 θ2 R

θ3

The full generalization of these results is called the local Gauss-Bonnet Theorem, which gives a similar statement to those above for any figure in S whose sides are curves lying in S, not necessarily even geodesics. Theorem 4.4.2 (Local Gauss-Bonnet Theorem). Let ~x be an orthogonal parametrization of a regular surface S and suppose R is a simply-connected region parametrized by ~x whose boundary ∂R is parametrized by a piecewise regular curve α. Denote the vertices of R by α(s1), α(s2), . . . , α(sn+1) = α1(s1); and denote the exterior angles by θ1, . . . , θn. Then n n X Z si+1 ZZ X kg ds + K dσ + θi = 2π. i=1 si R i=1 Proof. Assume α is positively-oriented. Recall the formula for geodesic curvature: 1  dv du dϕ kg = √ Gu − Ev + 2 EG ds ds ds 0 where ϕ is the angle between ~xu and α (s) at any point on ∂R. We begin by computing the sum of the line integrals: n n Z si+1 Z si+1      Z si+1  X X Gu dv Ev du dϕ k ds = √ − √ ds + . g ds ds ds i=1 si i=1 si 2 EG 2 EG si dϕ We can evaluate each ds integral using the fundamental theorem of calculus and sum over the vertices to obtain: n n X Z si+1 dϕ X = (ϕ(s ) − ϕ(s )). ds i+1 i i=1 si i=1 Set this aside for now. The rest of the expression is a line integral over ∂R: n Z si+1      X Gu dv Ev du √ − √ ds ds ds i=1 si 2 EG 2 EG Z  E G  = − √ v du + √ u dv ∂R 2 EG 2 EG ZZ  G   E   = √ u + √ v du dv R 2 EG u 2 EG v

72 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

by Green’s Theorem. (We have abused notation slightly: the line integral is really being taken over ∂~x−1(R) in the uv-plane and likewise the double integral is over ~x−1(R).) Now the integrand of the double integral can be evaluated as: ZZ  G   E   ZZ √ √ u + √ v du dv = − K EG du dv. R 2 EG u 2 EG v R √ √ 2 Recall that |~xu × ~xv| = EG − F ; since ~x is orthogonal, F = 0 so this just equals EG. We can therefore write this term as a surface integral over the surface proper: ZZ ZZ − K|~xu × ~xv| du dv = − K dσ. ~x−1(R) R

Putting everything together so far, we have (after rearranging terms):

n n X Z si+1 ZZ X kg ds + K dσ = (ϕ(si+1) − ϕ(si)) i=1 si R i=1 n X = 2π − θi by the turning tangents lemma below. i=1 We have thus derived the local version of the Gauss-Bonnet Theorem.

In the last step of the proof, we needed to evaluate the sum of each ϕ(si+1) − ϕ(si) at the vertices. To do so, we have the following lemma. Lemma 4.4.3 (Turning Tangents). For a positively-oriented curve α(s) on a surface S, with vertices α(s1), . . . , α(sn),

n n X X (ϕ(si+1) − ϕ(si)) + θi = 2π, i=1 i=1

where ϕ denotes the angle from ~xu to α(s) at each point on the curve. Proof. See Hopf’s Compositio Mathematica vol. 2 for the case when n ≥ 2 and Section 5.7 of do Carmo for the case with no vertices. There is an even more general statement of Gauss-Bonnet that describes the global geometry of S: Theorem 4.4.4 (Global Gauss-Bonnet Theorem). Suppose R is a regular region of a surface S whose boundary consists of simple, closed, regular curves C1,...,Cn. Denote by θ1, . . . , θp the exterior angles of the meetings of the Ci at the vertices of R. Then

n p X Z ZZ X kg ds + K dσ + θj = 2πχ(R), i=1 Ci R j=1

where χ(R) is the Euler characteristic of the region R.

73 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

Proof. Start with a triangle R1 and add a second triangle R2 along one edge, with the same orientation as R1:

S

R2

R1

The local case of Gauss-Bonnet for the two triangles is

3 3 Z si+1 ZZ X X 1 R1 : kg ds + K dσ + θi = 2π i=1 si R1 i=1 3 3 Z sj+1 ZZ X X 2 R2 : kg ds + K dσ + θj = 2π. j=1 sj R2 j=1

Add the two expressions together and notice that two of the kg integrals cancel since R1 and R2 have the same orientation: 4 3 3 Z si+1 ZZ X X 1 X 2 kg ds + K dσ + θi + θj = 4π, i=1 si R1∪R2 i=1 j=1

after relabeling the si to match the new figure. Consider one of the angles made by the meeting of R1 and R2:

ψ2 ψi θi ψ1 θ2

θ1

If ψi denotes the corresponding interior angle at θi, then θi = π − ψi. Consider the labeled figure on the right. We have θ1 + θ2 = 2π − (ψ1 + ψ2) = π + θ where θ is the new exterior angle of R1 ∪ R2. Thus the summation can be written as the sum of the four exterior angles of R1 ∪ R2 but with an extra 2π radians. Subtracting this extra 2π, we have the following Gauss-Bonnet formula for R = R1 ∪ R2: 4 4 X Z si+1 ZZ X kg ds + K dσ + θi = 2π. i=1 si R i=1 This shows that gluing triangles together along one edge does not change the G-B formula. What if, after repeating this process several times, we have situation in which we wish to add a triangle along two edges?:

74 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

It’s easy to check that the angles will work out again so that still sums to 2π, as we did above. So gluing along one or two edges doesn’t the G-B sum. Finally, we see what happens when we glue a triangle along all three edges:

One can check that there is cancellation of the θ at each vertex of the triangle to be added, but there are no new factors of 2π. This decreases the right side of the G-B formula by 2π, resulting in an expression that sums to 0. This reflects the fact that adding a triangle along three edges changes the Euler characteristic of the figure. To see this another way, let F be the figure obtained after adding the triangle R above; that is F has no holes. Since this can be constructed by gluing triangles one/two edges at a time, we must have

n n X Z si+1 ZZ X kg ds + K dσ + θi = 2π. i=1 si F i=1 The G-B expression for the triangle contributes the 2π to the expression for F , so it must be that m m X Z sj+1 ZZ X kg ds + K dσ + θj = 0. j=1 sj F rR j=1 At this point we have essentially reduced the global case to the piecing together of the local case for triangles. This is all we need though, since from topology we know that every surface (and in particular every regular surface) has a triangulation. Putting everything together then, we have the desired statement of the global Gauss-Bonnet Theorem: If R is a region of a surface S with boundary components Ci meeting at angles θ1, . . . , θp, then

n p X Z ZZ X kg ds + K dσ + = 2πχ(R). i=1 Ci R j=1

75 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

The following are relevant properties of the Euler characteristic χ(S) of a surface S: ˆ χ(S) = V − E + F , where V,E,F denote respectively the number of vertices, edges and faces in a legal triangulation of S. Implicit to this property is the fact that χ is well-defined on a surface, i.e. it doesn’t change with different choices of triangulation. ˆ As mentioned above, every surface is triangulable. ˆ χ(S) is a topological invariant, meaning it is invariant under homeomorphism. ˆ The genus of a surface is, informally, the number g of ‘holes’ in the surface. For example, a torus with a single hole has genus 1, a torus with two holes has genus 2, etc. The Euler characteristic is related to the genus of a surface S in the following way: χ(S) = 2 − 2g.

The Gauss-Bonnet Theorem is highly useful for deducing several results about surfaces that may be difficult to prove otherwise. We include several of these consequences here. Proposition 4.4.5. Every compact, orientable surface with positive Gaussian curvature is homeomorphic to a sphere. Proof. Let R be such a surface. Then R is a region with no edges or angles, so the Gauss- Bonnet formula for R reads ZZ K dσ = 2πχ(R). R If K > 0 then the right hand side must also be positive. However, by the classification of compact surfaces, the only positive Euler characteristic for an orientable surface is χ(R) = 2 which is the Euler characteristic of the sphere. Thus R ∼= S2. Examples. 1 What can be said about a surface with negative curvature everywhere? Consider for example the catenoid from Section 4.1:

R

76 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

If R is a region with two boundary components that don’t intersect, with orientations as pictured above, can they be geodesics? If we assume they are geodesics, then P R kg ds = 0 in the Gauss-Bonnet formula. Moreover, there are no angles so the formula can be written ZZ K dσ = 2πχ(R). R However the Euler characteristic for R is zero (count V,E and F for any triangulation) so the right side of the formula is 0. This contradicts the assumption that K < 0 on the left, so this is impossible. Therefore the boundary components of R cannot be geodesics. A similar proof shows that two closed curves on a sphere that don’t intersect cannot be both geodesics.

α2 R α1

Assuming α1 and α2 are both geodesics and computing the Gauss-Bonnet formula for the region R between them will produce a contradiction. The way this is often stated is: every pair of geodesics on a sphere must intersect.

2 Consider the torus T .

Here χ(T ) = 0 so by the global Gauss-Bonnet theorem (4.4.4), we obtain a nice expression for the Gaussian curvature over the entire torus: ZZ K dσ = 0. T RR It is common to view T K dσ as the signed area of the image of the Gauss map N : T → S2. Dividing the torus into the ‘outer’ region with positive curvature and the ‘inner’ region with negative curvature, we see that N produces two copies of S2 – one with positive signed area and one with negative signed area. These fit together to RR produce the identity T K dσ = 0.

77 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

3 The theory in this section is best applied to bounded (better, compact) surfaces, like the torus. However we can obtain interesting results for infinite surfaces using the Gauss-Bonnet theorem. Consider the paraboloid S. If we choose an orientation, e.g. with normal vectors pointing inwards as shown, then the image of the Gauss map over the whole surface is the top hemisphere of S2.

N

This seems paradoxical: an infinite surface corresponds to some finite signed area in the same way as a compact surface like the torus. The Gauss-Bonnet theorem suggests the following formula to describe the global Gaussian curvature of the paraboloid: ZZ K dσ = π. S We will see that this remarkable formula does indeed hold for the entire paraboloid. To start, S may be parametrized by

~x(u, v) = (u cos v, u sin v, u2), 0 < u < ∞, 0 < v < 2π.

Let α be a coordinate curve for a fixed value of u: α(v) = (u cos v, u sin v, u2). Also let R be the region of S bounded by α; this is illustrated in the figure below.

α(v)

R

The Gauss-Bonnet formula for R is Z ZZ kg ds + K dσ = 2π, α R since there is only one boundary component (α), there are no angles and χ(R) = 1. For this parametrization of S, we have

~xu = (cos v, sin v, 2u) and ~xv = (−u sin v, u cos v, 0).

78 4.4 The Gauss-Bonnet Theorem 4 Intrinsic Geometry

The first fundamental form for S is then calculated to be

E = 1 + 4u2,F = 0 and G = u2.

0 In particular, ~x is an orthogonal parametrization so the angle ϕ between ~xu and ~xv = α dϕ is constant and thus dv = 0. Now we use the algebraic value formula from Theo- rem 4.3.7 to write

1 0 0 dϕ 1 1 kg = √ (Guv − Evu ) + = √ (2u − 0) + 0 = √ . 2 EG dv 2u 1 + 4u2 1 + 4u2 To evaluate the geodesic curvature integral, first realize that α(v) is not parametrized 0 by arc length. Therefore we must change variables by multiplying by |α | = |~xv| = u, which together with the above calculations gives us

Z Z 2π Z 2π u u kg ds = kgu dv = √ dv = 2π √ . 2 2 α 0 0 1 + 4u 1 + 4u Now as we take u → ∞, the region below α approaches the entire paraboloid and the Gauss-Bonnet formula becomes Z ZZ  lim kg ds + K dσ = 2π u→∞ α R  u  ZZ lim 2π √ + K dσ = 2π 2 u→∞ 1 + 4u S ZZ π + K dσ = 2π S ZZ K dσ = π. S We have derived the proposed formula for the total curvature of the paraboloid. This suggests that, under certain conditions, we are able to piece together the differential geometry of compact surfaces into similarly powerful statements for infinite surfaces.

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