Assignment 6; Solutions

Assignment 6; Solutions - PMATH 465/665 Riemannian...

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PMATH 465/665: Riemannian Geometry Assignment 6; Solutions [1] Let ( M, g ) be a Riemannian manifold of dimension n and let p M . Let { e 1 , . . . , e n } be any orthonor- mal frame at p M . For example, we can take the coordinate frame { ∂x 1 , . . . , ∂x n } of a normal coordinate chart centred at p . [Recall that in normal coordinates, g ij ( p ) = g p ∂x i p , ∂x j p = δ ij .] Let B ε (0) be an open subset of the domain of the restricted exponential map. Hence exp p : B ε (0) T p M exp p ( B ε (0)) M is a diffeomorphism. In fact, the map is described as follows. If V p B ε (0), then exp p ( V p ) is the point in M obtained by following the geodesic with initial point p and initial unit velocity 1 | V p | gp V p for a time t = | V p | g p . In particular, all points in U = exp p ( B ε (0)) are connected to p by a unique unit-speed geodesic starting at p . We will define an orthonormal frame { E 1 , . . . , E n } on U as follows. Fix j . Define for q U , define E j ( q ) to be the parallel transport of e j to q by the unique geodesic from p to q . Since parallel transport with respect to a metric-compatible connection preserves the inner product, this procedure defines an orthonormal frame { E 1 , . . . , E n } on U . Each vector field E j is smooth because of the smooth dependence on initial conditions from the ODE theorem. By construction, each E j is parallel at q in the direction of the radial geodesics passing through q . That is, if γ : [0 , ε ) is the unique unit-speed radial geodesic starting at p and passing through q at time t [0 , ε ), then γ 0 ( t ) E j = 0. But for t = 0, we have γ (0) = p and every unit tangent vector U p at p is the initial velocity of a unit-speed radial geodesic starting at p . Thus U p E j = 0, which implies that ( E i E j )( p ) = 0, since E i | p = e i is a unit tangent vector at p . [2] Let be a connection on the tangent bundle TM of a manifold M . As in the case of the Levi-Civita connection, we define the curvature tensor R = R of by R ( X, Y ) Z = X ( Y Z ) - ∇ Y ( X Z ) - ∇ [ X,Y ] Z. We say that the connection is flat if R = 0. We proved in class that R = 0 if and only if there exists a parallel local frame { E 1 , . . . , E n } near every point p M . [We also saw that the domain U of the parallel frame was topologically trivial – it was an open rectangle, hence connected and simply connected.] Let be a flat connection on TM . Then for any p M , there is a connected, simply-connected, open neighbourhood U of p in M , and a parallel frame { E 1 , . . . , E n } on U . Let V p = V i E i | p T p M be a tangent vector at p . Define a smooth vector field V on U by V = V i E i . That is, V q = V i E i | q . The vector field V is smooth because its component functions with respect to a smooth local frame are smooth (in fact they are constant). Let γ be any path from p to q . Let D t be covariant differentiation along the curve γ . Then ( D t V )( t 0 ) = dV i dt t = t 0 E i | γ ( t 0 ) + V i γ 0 ( t 0 ) E i !

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• Fall '10
• N.A
• Geometry, Riemannian geometry, Riemann curvature tensor

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