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Curvatures of Riemannian Manifolds

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Curvatures of Riemannian Manifolds Chapter 10 Curvatures of Riemannian Manifolds 10.1. Riemann Curvature Tensor 10.1.1. Motivations and Definitions. Recall that Gauss’s Theorema Egregium is an immediate consequence of the Gauss’s equation, which can be written using covariant derivatives as: @ @ ql @ rirj − rjri = g (hjkhli − hikhlj) : @uk @uk @uq Taking the inner product with @ on both sides, we get @up @ @ @ ql g rirj − rjri ; = g (hjkhli − hikhlj) gpq @uk @uk @up l = δp (hjkhli − hikhlj) = hjkhpi − hikhpj: In dimension 2, we get 2 det[h] = h11h22 − h12 = g(r1r2@2 − r2r1@2;@1): It motivates us to define a (3; 1)-tensor T and a (4; 0)-tensor S so that T (@1;@2;@2) = r1r2@2 − r2r1@2 and S(@1;@2;@2;@1) = g(r1r2@2 − r2r1@2;@1); then the Gauss curvature in the 2-dimension case can be expressed in tensor notations as q T122g1q S1221 K = 2 = 2 : g11g22 − g12 g11g22 − g12 Naturally, one may attempt: T (X; Y; Z) = rX (rY Z) − rY (rX Z) for the (3; 1)-tensor. However, even though it indeed gives T122 = r1r2@2 − r2r1@2, one can easily verify that such T is not tensorial. We modify such an T a bit and define the following very important tensors in Riemannian geometry: 239 240 10. Curvatures of Riemannian Manifolds Definition 10.1 (Riemann Curvature Tensors). Let (M; g) be a Riemannian manifold, then its Riemann curvature (3; 1)-tensor is defined as (3;1) Rm (X; Y )Z := rX (rY Z) − rY (rX Z) − r[X;Y ]Z; and the Riemann curvature (4; 0)-tensor is defined as: Rm(4;0)(X; Y; Z; W ) := gRm(3;0)(X; Y )Z; W : If the tensor type is clear from the context, we may simply call it the Riemann curvature tensor and simply denote it by Rm. Alternative notations include R, R, etc. When X = @ and Y = @ , we have [X; Y ] = 0, so that it still gives @ui @uj (3;1) Rm (@1;@2)@2 = r1r2@2 − r2r1@2; and at the same time Rm(3;1) is tensorial. Exercise 10.1. Verify that Rm(3;1) is tensorial, i.e. Rm(3;1)(fX; Y )Z = Rm(3;1)(X; fY )Z = Rm(3;1)(X; Y )fZ = fRm(3;1)(X; Y )Z: It is clear from the definition that Rm(3;1)(X; Y )Z = −Rm(3;1)(Y; X)Z. That explains why we intentionally write its input vectors by (X; Y )Z instead of (X; Y; Z), so as to emphasize that X and Y are alternating. (3;1) l We express the local components of Rm by Rijk, so that @ Rm(3;1) = Rl dui ⊗ duj ⊗ duk ⊗ ijk @ul where l @ (3;1) @ @ @ Rijk = Rm ; @ul @ui @uj @uk = ri rj@k − rj ri@k − r[@i;@j ]@k ! @Γl l jk @Γik p l p l @ = − + ΓjkΓip − ΓikΓjp : @ui @uj @ul Therefore, the local components of Rm(3;1) are given by @Γl l l jk @Γik p l p l Rijk = − + ΓjkΓip − ΓikΓjp: @ui @uj For the (4; 0)-tensor, the local components are given by: (4;0) (3;1) p p Rijkl = Rm (@i;@j;@k;@l) = g Rm (@i;@j)@k;@l = g(Rijk@p;@l = gplRijk: 10.1.2. Geometric Meaning of the Riemann Curvature (3; 1)-Tensor. While l Rijk and Rijkl are related to the Gauss curvature in the two dimension case, it is not obvious what are their geometric meanings in higher dimensions at the first glance. In fact, the Riemann curvature (3; 1)-tensor has a non-trivial relation with parallel transports. (3;1) Given tangent vectors X1;X2;Y at p, the vector Rm (X1;X2)Y at p in fact measures the defect between the parallel transport of Y along two different paths. For simplicity, consider local coordinates fuig such that p has coordinates (0; ··· ; 0), and let’s only consider the form Rm(@1;@2)Y . Take an arbitrary Y (p) 2 TpM. We are going to transport Y (p) in two different paths. One path is first along the u1-direction, then along the u2-direction; another is the opposite: along u2-direction first, followed by the u1-direction. 10.1. Riemann Curvature Tensor 241 k @ Let’s consider the former path: first u1, then u2. Let Y = Y be the transported @uk k vector, then along the first u1-segment (u1; u2) = (u1; 0), the components Y satisfy the equation: k @Y j k r@1 Y = 0 =) + Y Γ1j = 0 for any k: @u1 Provided that we are sufficiently close to p, we have the Taylor expansion: k 2 k k k @Y 1 @ Y 2 3 Y (u1; 0) = Y (0; 0) + (u1 − 0) + 2 (u1 − 0) + O(u1): @u1 (0;0) 2 @u1 (0;0) Using the parallel transport equation, both the first and second derivatives of Y k can be expressed in terms of Christoffel’s symbols. k @Y j k = −Y Γ1j @u1 2 k j k @ Y @Y k j @Γ1j 2 = − Γ1j − Y @u1 @u1 @u1 k l j k j @Γ1j = Y Γ1lΓ1j − Y @u1 k These give the local expression of Y along the first segment (u1; 0): k k k j k 1 l j k j @Γ1j 2 3 Y (u1; 0) = Y (0; 0) − Y Γ1j u1 + Y Γ1lΓ1j − Y u1 + O(u1): (0;0) 2 @u1 (0;0) Next, we consider the second segment: transporting Y (u1; 0) along the u2-curve. By a similar Taylor expansion, we get: k 2 k k k @Y 1 @ Y 2 3 Y (u1; u2) = Y (u1; 0) + u2 + u + O(u ): @u 2 @u2 2 2 2 (u1;0) 2 (u1;0) As before, we next rewrite the derivative terms using Christoffel’s symbols. The parallel transport equation along the u2-curve gives: k @Y j k = −Y Γ2j @u2 2 k k @ Y l j k j @Γ2j 2 = Y Γ2lΓ2j − Y : @u2 @u2 Plugging these in, we get: k k k j k 1 l j k j @Γ2j 2 3 Y (u1; u2) = Y (u1; 0) − Y Γ u2 + Y Γ Γ − Y u + O(u ): 2j 2 2l 2j @u 2 2 (u1;0) 2 (u1;0) k k k Each term Y (u1; 0) above can then be expressed in terms of Y (0; 0) and Γij(0; 0) using our previous calculations. It seems a bit tedious, but if we just keep terms up to 242 10. Curvatures of Riemannian Manifolds second-order, then we can easily see that: k k k j k 1 l j k j @Γ1j 2 Y (u1; u2) = Y (0; 0) − Y Γ1j u1 + Y Γ1lΓ1j − Y u1 (0;0) 2 @u1 (0;0) j l j k − Y (0; 0) − Y (0; 0)Γ1l(0; 0)u1 Γ2j(u1; 0) u2 k ! 1 l j k j @Γ2j 2 3 + Y Γ Γ − Y u + O j(u1; u2)j 2 2l 2j @u 2 2 (u1;0) k k j k 1 l j k j @Γ1j 2 = Y (0; 0) − Y Γ1j u1 + Y Γ1lΓ1j − Y u1 (0;0) 2 @u1 (0;0) k j l j k j @Γ2j − Y (0; 0) − Y (0; 0)Γ1l(0; 0)u1 Γ2j(0; 0) u2 − Y (0; 0) u1u2 @u1 (0;0) k ! 1 l j k j @Γ2j 2 3 + Y Γ2lΓ2j − Y u2 + O j(u1; u2)j : 2 @u2 (0;0) Next, we consider the parallel transport in another path, first along u2, then along k @ u1. Let Y = Y be the local expression of the transported vector, then the local e e @uk k k expression of Ye at (u1; u2) can be obtained by switching 1 and 2 of that of Y (u1; u2): k k k j k 1 l j k j @Γ2j 2 Ye (u1; u2) = Ye (0; 0) − Ye Γ2j u2 + Ye Γ2lΓ2j − Ye u2 (0;0) 2 @u2 (0;0) k j l j k j @Γ1j − Ye (0; 0) − Ye (0; 0)Γ2l(0; 0)u2 Γ1j(0; 0) u1 − Ye (0; 0) u1u2 @u2 (0;0) k ! 1 l j k j @Γ1j 2 3 + Ye Γ1lΓ1j − Ye u1 + O j(u1; u2)j : 2 @u1 (0;0) Recall that Y (0; 0) = Ye(0; 0). By comparing Y (u1; u2) and Ye(u1; u2), we get: k k Y (u1; u2) − Ye (u1; u2) k k l j k j k @Γ1l @Γ2l 3 = Y Γ1lΓ2j − Γ2lΓ1j + − u1u2 + O j(u1; u2)j @u2 @u1 (0;0) k l 3 = R21lY u1u2 + O j(u1; u2)j : (0;0) In other words, we have: 3 Y (u1; u2) − Ye(u1; u2) = Rm(@2;@1)Y u1u2 + O j(u1; u2)j : (0;0) Therefore, the vector Rm(@2;@1)Y at p measures the difference between the parallel transports of those two different paths. 10.1.3. Symmetric properties and Bianchi identities. We will explore more about the geometric meaning of the Rm(4;0)-tensor in the next section. Meanwhile, let’s discuss some nice algebraic properties of this tensor. The Riemann curvature (4; 0)-tensor satisfies some nice symmetric properties. The first two indices ij, and the last two indices kl of the components Rijkl, and is symmetric if one swap the whole ij with the whole kl. Precisely, we have: 10.1. Riemann Curvature Tensor 243 Proposition 10.2 (Symmetric Properties of Rm(4;0)). The local components of Rm(4;0) satisfy the following properties: • Rijkl = −Rjikl = −Rijlk, and • Rijkl = Rklij.
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