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M435: Introduction to Differential Geometry

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M435: Introduction to Differential Geometry M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY MARK POWELL Contents 1. Introduction 1 2. Geometry of Curves 2 2.1. Tangent vectors and arc length 3 2.2. Curvature of plane curves 5 2.3. Curvature of curves in R3 8 2.4. The wedge product 9 2.5. Formula for curvature without computing arc length 10 2.6. The normal and binormal vectors and the torsion 10 2.7. The Serret-Frenet formulae 11 2.8. Spherical curves 14 References 17 1. Introduction The plan is as follows, although this is the maximum we will do, we may cover less, depending on time. Extra reading can be found in the bibliography, [DoCa] and [Hit] are recommended sources. (1) Geometry of curves. (2) Topology of surfaces. (3) Geometry of surfaces. (4) Intrinsic vs. extrinsic. Theorema Egregium. (5) Geodesics. (6) Gauss-Bonnet theorem, Poincar´e’s theorem (Morse theory on surfaces). (7) Hyperbolic Geometry. (8) Back to curves - Fenchel and Fary-Milnor theorems, Four Vertex theorem. After a warm up on curves, in topology of surfaces we will construct surfaces using glueing methods, and we will classify all possible closed surfaces up to home- omorphism. In the geometry of surfaces we will work on understanding the effect of intrinsic geometry, measured in particular by the curvature, for example on whether the angles of a triangle add up to π, or more than π, or less than π. The remarkable theorem, or Theorema Egregium of Gauss says that the curvature, which at first glance appears to depend on the embedding of a surface in 3-space, is in fact an intrinsic property. That is, we don’t need pictures of the Earth from space in order 1 M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 2 to tell that it is a ball. We will be particularly interested in surfaces with constant curvature. The highlight of the course is undoubtedly the Gauss-Bonnet theorem, which is the archetypal local-global theorem. It is shown that for a surface a global topo- logical quantity, the Euler characteristic, can be computed from local information at each point, namely the Gaussian curvature of the surface. Now that your appetite is whetted, let us begin. 2. Geometry of Curves Definition 2.1. A differentiable curve in the plane is a differentiable map α: I R2 where I = (a, b) R is an interval. (We can have a = or b = , or both.)→ A differentiable curve⊆ in R3 is a differentiable map α: I −∞R3 where∞I = (a, b) R is an interval. (We can have a = or b = , or both.)→ ⊆ Here differentiable means that α(−∞t) = (x(t),y∞(t), z(t)), where x,y,z are differen- tiable functions (a, b) R. A curve α is regular→ if α′(t) = 0 for all t I. 6 ∈ Curves in the plane can also be given as the level sets of two variable functions f(x,y) = c. This is implicit form. A function whose image coincides with such a level set is sometimes called the parametric form. Example 2.2. (i) α(t) = a + bt, t R. A straight line through the point determined by the vector a parallel∈ to the vector b. (ii) α(t) = (cos t, sin t) t R. The unit circle in R2 R3. (iii) α(t) = (a cos t, a sin t,∈ bt) t R, a, b > 0 constants.⊂ A helix, lying on a cylinder of radius a. ∈ (iv) α(t) = (t3,t2). A non-regular differentiable curve. The derivative vanishes at t = 0. Exercise: sketch the curve. (v) Non-example. α(t) = (t, t ), t R. Not a differentiable curve since y(t) is not a differentiable function.| | ∈ (vi) β(t) = (cos 2t, sin 2t) t R. Same circle as example (i), i.e. same subset of R3, but different parametrisation.∈ In both cases the image is (x,y) R2 x2 = y2 = 1 , however β(t) is twice as fast as α(t). { ∈ | } β′(t) = ( 2 sin 2t)2 + (2 cos 2t)2 = 2; | | − α′(t) p= ( sin t)2 + (cos t)2 = 1. | | − R R R2 (vii) Given a function f : p, its graph is (x,y) y f(x) = 0 . We can parametrise this set as→ a differntiable curve{ by α∈(t) =| (t,f− (t)). } Exercise 2.3. Find a parametrisation for the curves in the plane given as the level sets of the following equations (i) Ellipse x2/a2 + y2/b2 = 1. (ii) Hyperbola x2/a2 y2/b2 = 1. (iii) A figure 8 curve,− or an symbol, with the self intersection point at the origin of R2. Here is one∞ implicit form x4 + 2x2y2 + y4 x2 + y2 = 0. Here − is one method: deduce this equation from the polar equation r = √cos 2θ, M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 3 and then using x = r cos θ and y = r sin θ make θ the parameter. This can be made to give an α which is piecewise defined. Another implicit formula which describes a curve with the right qualitative shape is x4/16 + y2 x2 = 0. Can you find parametric equations for this implicit curve? Try typing− the equations into Wolfram Alpha, it will draw them for you. You can also look up “lemniscate” on wikipedia, if you want to cheat. Exercise 2.4. Show that the following formula gives a parametrisation of the subset of R2 from Example (v) above which is differentiable but which is not regular. (t2, t2) t 0 α: t 2 −2 ≤ 7→ ((t ,t ) t> 0 In particular you need to check that this function is differentiable at t = 0. The vector α′(t) is the tangent vector to α at t I. α′(t) is the speed of α at t (think of t as time). We consider regular curves only∈ from| now| on. Points where α′(t) = 0 are called singular points. A regular curve is also often called an immersion. 2.1. Tangent vectors and arc length. For a curve without singular points we have the unit tangent vector α′(t) T(t) := . α′(t) | | Note that we do not have to have any special notation to indicate a vector in this course. It is allowed but not required to over- or underline or bold-face vector functions, but it will be apparent what are vector functions because we will always specify clearly the domain and codomain of all functions. f : A B, A is the domain and B is the codomain. The latter is not to be confused with→ f(A), which is the image or range of f. Definition 2.5. The Tangent line to a curve α at a point p = α(t0) is the line through p which is parallel to α′(t0). Example 2.6. Suppose α(t) = (cos t, sin t,t). Then α′(t) = ( sin t, cos t, 1). So − α′(t) = sin2 t + cos2 t +1= √2. At p = ( 1, 0,π) we have t = π and α′(π) = (0| , 1, 1).| Therefore the tangent line to α at−p is given by γ(t) = ( 1, 0,π) + p t(0, 1, 1). − − Exercise 2.7. Find the tangent line to the curve α(t) = (t,t2, log t) at the point (1, 1, 0). Using the tangent vector we can compute the arc length of a curve between two of its points. We know that in a constant speed situation the distance covered is the speed multiplied by the time for which this speed is maintained. When the speed is variable with time we take an integral instead of multiplying. M435: INTRODUCTION TO DIFFERENTIAL GEOMETRY 4 Definition 2.8. For a differentiable curve α: I = (a, b) R3 (or R2) the arc → length between t = t0 and t = t1, with a t0 t1 b is: ≤ ≤ ≤ t1 α′(u) du. | | Zt0 We can reparametrise a curve by arc length. We can always do this in theory, although in practice it is often not worth the effort. It will be useful for the theory to assume that we have a curve parametrised by arc length. The point is that parametrisations are highly nonunique. By fixing a starting point and an orientation of a curve, namely a choice of preferred direction, we obtain a canonical parametrisation — the arc length one. If we switch the orientation, then we switch the canonical parametrisation s s. 3→− Definition 2.9. Let α: I R be a curve. Fix a starting point t0 I. The arc length function is → ∈ t s(t) := α′(u) du. ± | | Zt0 The sign is decided according to whether increasing t from t0 moves along the orientation± of α or against it. In order to reparametrise a regular curve by arc length first write s = f(t). Since α′(t) > 0 for all t, f is a one-one function (mean value theorem). Therefore f is | | a bijection between (t0, b) and f(t0, b), which is another interval. so f admits an inverse. Substitute t = f −1(s) in α(t) to obtain the parametrisation by arc length. Note that ds = α′(t) . If t = s, then ds = 1 and T(t)= α′(t)= α′(s). dt | | dt Example 2.10. Suppose that α(t) = (cos t, sin t,t), t R. We already saw that ∈ α′(t) = √2. Therefore the arc length function, starting at t0 = 0, is | | t t ′ t s(t) := α (u) du = √2du = [√2u]0 = √2t. 0 | | 0 Z Z Inverting this function we have t = s/√2= s√2/2. This gives the reparametrisa- tion by arc length of α: α(s) = (cos(s√2/2), sin(s√2/2),s√2/2).
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