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Differential Geometry

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Differential Geometry 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: Velocity: α0(t) = (x0(t); y0(t); z0(t)). Speed: jα0(t)j = px0(t)2 + y0(t)2 + z0(t)2. 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 Arc length: jα0(t)j dt. It is typical to use ds for arc length integrals. a Definition. If a curve α(t) has constant speed jα0(t)j = 1, we say α is parametrized by arc length. Examples. 1 Consider α(t) = (cos t; sin t; t), −∞ < t < 1, the standard helix in R3. Velocity and speed are calculated to be p p α0(t) = (− sin t; cos t; 1) and jα0(t)j = 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 2πp p jα0(t)j dt = 2 dt = 2 2π: 0 0 p 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 jα0(t)j = 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 jα0(t)j 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 p 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 < 1. 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 jα(t0)j 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 = jα(t)j = 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 2 I.
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