Math 120B Discussion Session Week 2 Notes April 11, 2019 Today we'll begin with an example of parallel transport before proving a couple of curvature results. An instance of holonomy 3 Throughout this section, let M ⊂ R be a smooth surface. We begin by recalling two definitions. Definition. A differentiable vector field along a curve γ :[a; b] ! M is a differentiable function 3 X:[a; b] ! R with the property that X(t) is a tangent vector to M at γ(t), for each t 2 [a; b]. Remark. Later in the quarter we should be able to give a more satisfying definition for the differ- entiability of a vector field. Definition. We say that a differentiable vector field X is parallel along γ if X_ is perpendicular to M. Remark. We might also get to see a more satisfying notion of parallelism later on! If we start with a curve γ and a vector tangent to M at some point on γ, it's natural to ask whether we can extend this vector to a parallel vector field along γ | that is, whether or not we can translate the vector along γ in a parallel way. Indeed, we can do this in a unique way, and today we want to see an explicit example. In constructing our example, we'll use the following proposition, which you'll prove for home- work. Proposition 1. Let x: U! M be a coordinate patch and let γ(t) = x(γ1(t); γ2(t)) be a regular P i curve. If X is a differentiable vector field along γ with X = X xi, then X is parallel along γ if and only if 2 dXk X dγj 0 = + Γk Xi ; (1) dt ij dt i;j=1 for k = 1; 2. The existence and uniqueness of parallel translates follows from applying Picard's theorem to this differential equation. Example 1. We have the familiar coordinate patch x(φ, θ) = (sin φ cos θ; sin φ sin θ; cos φ); (φ, θ) 2 (0; π) × (−π; π) for M = S2. Notice that this parametrizes M as a surface of revolution, with r(φ) = sin φ and z(φ) = cos φ. So we can use last week's computations to see that 1 0 (g ) = : ij 0 sin2 φ 1 Figure 1: Parallel transport along a circle of latitude. Here we have φ0 = π=3, so the angle between X(0) and X(2π) is π. So the only nonzero Christoffel symbols are 2 2 1 Γ12 = Γ21 = cot φ and Γ22 = − sin φ cos φ. Now consider a circle of latitude φ = φ0, along with the vector x1(φ0; 0). That is, we have γ(θ) = x(φ0; θ); 1 2 so γ (θ) = φ0 and γ (θ) = θ. Substituting these quantities into (1), we see that a differentiable P i vector field X(θ) = X (θ)xi along γ will be parallel iff dX1 dX2 0 = − sin φ cos φ X2 and 0 = + cot φ X1: dθ 0 0 dθ 0 1 If we want X(0) = x1(φ0; 0), the initial values for this system of differential equations are X (0) = 2 φ0 and X (0) = 0. Now, we see from these equations that d2Xk = − cos φ Xk; dθ2 0 1 2 for k = 1; 2. This suggests, for instance, that X (θ) = cos((cos φ0)θ). We may then solve for X to 2 find X (θ) = − sin((cos φ0)θ)= sin φ0. That is, sin((cos φ0)θ) X(θ) = cos((cos φ0)θ)x1(φ0; θ) − x2(φ0; θ): sin φ0 We see from Figure1 that parallel translation around a closed loop need not return a vector to its original position. Indeed, X(0) = x1(φ0; 0), while sin(2π cos φ0) X(2π) = cos(2π cos φ0)x1(φ0; 0) − x2(φ0; 0): sin φ0 (We're conveniently ignoring the domain of definition for x.) So we see that hX(0); X(2π)i = cos(2π cos φ0) and, since both X(0) and X(2π) are unit vectors, the angle between X(0) and X(2π) 2 is 2π cos φ0. This is an example of holonomy, and one can show that even for a more general Reimannian manifold M, the holonomy map hp : TpM ! TpM along a closed, piecewise smooth curve is always a linear map. Indeed, the Riemann curvature tensor gains some of its geometric meaning from such maps. Holonomy maps are also related to curvature forms by the Ambrose-Singer theorem. Umbilic surfaces Next we'll prove a cute curvature result which is intuitively obvious: if the normal curvatures of our surface are the same in every direction, then our surface is a sphere (perhaps of infinite radius). This result is attributed to Meusnier, and is related to but distinct from another result known as Meusnier's theorem. Theorem 2. If every point in the image of a coordinate patch is umbilic, then the coordinate patch is contained in a plane or a sphere. Proof. First, recall that a point p 2 M in a surface is umbilic if its principal curvatures are equal. Another way of saying this is that the Weingarten map is a scalar multiple of the identity, in which case we can write L = κId. Notice that, a priori, the scalar κ could vary as we move around in our coordinate patch. Let's show that this can't happen. First, recall that L(xi) = −ni in a coordinate patch. Then ni = −κxi, so @κ @κ n = − x + κx and n = − x + κx ; ij @uj i ij ji @ui j ij with the second formula following by symmetry. The equality of mixed partials then tells us that nij = nji, so @κ @κ x = x : @uj i @ui j 1 2 But x1 and x2 are linearly independent, so this can only be the case if @κ=@u = 0 and @κ/∂u = 0. That is, κ is constant in the patch. Because κ is constant, we have @ (n + κx) = n + κx = 0 @ui i i for i = 1; 2. So n = −κx + k for some constant vector k. If κ = 0, this means that n is constant 3 and therefore our patch is contained in a plane. If κ 6= 0, we can then choose a vector c 2 R for which we may write n = −κ(x − c); meaning that our patch is contained in the sphere of radius 1/κ centered at c. 3 Surfaces of constant curvature If a surface has constant curvature, it seems reasonable that any two points on the surface would have geometrically indistinguishable neighborhoods. That is, the surface would have the same local appearance from any point on the surface. Moreover, if some other surface has the same constant curvature, any of its neighborhoods should look like any of ours. This is what we learn from the following theorem. Theorem 3. Any two surfaces of the same constant Gaussian curvature are locally isometric. Recall the definition of locally isometric: Definition. Two surfaces M and N are locally isometric if for each point p 2 M there is an open set U ⊂ M containing p, along with an isometry f : U ! V for some open subset V ⊂ N. We call f a local isometry. You probably proved the following characterization of local isometry in in 120A: Proposition 4. Two surfaces M and N are locally isometric if and only if for each p 2 M there is 2 an open set U ⊂ R , a coordinate patch x: U ! M with p 2 x(U), and another coordinate patch y : U ! N such that x and y have the same metric coefficients. Our proof of Theorem3 (which is stolen from the textbook) will also use geodesic coordinate patches, which you defined in lecture this week. Definition. A coordinate patch x: U ! M is said to be a geodesic coordinate patch if g11 ≡ 1 and g12 ≡ 0. We say that x is a geodesic coordinate patch along γ if γ :[a; b] ! x(U) is a curve with the property that the u2-curve through any point on γ is γ itself. In lecture you showed that there exists a geodesic coordinate patch along any simple, regular curve in M. Also notice that the metric coefficient matrix of a geodesic coordinate patch has the form 1 0 (g ) = ; ij 0 h2 1 2 1 2 where h > 0 is the function h(u ; u ) = kx2(u ; u )k. Before proving Theorem3, we need the following lemma. Lemma 5. Suppose that x is a coordinate patch along a geodesic curve, and let h = kx2k as above. 2 2 2 Then h(0; u ) = 1, K = −h11=h, and h1(0; u ) = 0, for any u in the appropriate range. Proof. We'll prove the first and third claims, leaving the middle for homework. For the first claim we have 2 2 2 h(0; u ) = kx2(0; u )k = kγ_ (u )k = 1; where γ is the geodesic along which we've defined x. Because γ is a geodesic, we know thatγ ¨ is 1 perpendicular to the surface. In particular, x22 is perpendicular to the surface whenever u = 0. Now in general we have @g @g 22 = 2hx ; x i = 2 12 − hx ; x i = −2hx ; x i; @u1 12 2 @u2 1 22 1 22 4 @ 2 1 so @u1 (h ) = 0 whenever u = 0. But then @ 0 = (h2) = 2hh @u1 1 1 along u = 0. Since h = 1 along this curve, h1 = 0 along this curve.