Riemannian Geometry Dr Emma Carberry Semester 2, 2015 Lecture 15 [Riemannian Geometry – Lecture 15]Riemannian Geometry – Lecture 15 Riemann Curvature Tensor Dr. Emma Carberry September 14, 2015 Sectional curvature Recall that as a (1; 3)-tensor, the Riemann curvature endomorphism of the Levi-Civita connection r is R(X; Y )Z = rX (rY Z) − rY (rX Z) − r[X;Y ]Z: It measures the failure of the manifold to be locally isometric to Euclidean space. Starting with the Riemann curvature tensor, there are various simplifications of this tensor one can define. An important one is sectional curvature, because it • is the natural generalisation of Gauss curvature of the surface • completely determines the Riemann curvature tensor A two-dimensional subspace π of TpM is called a tangent 2-plane to M at p. Sectional curvature is defined on tangent 2-planes π ⊂ TpM. We will define the sectional curvature of a tangent 2-plane π in terms of a basis for π. In order to obtain an expression which is independent of this choice of basis, we will need a normalisation factor. Definition 15.1. For linearly independent vectors X; Y 2 TpM, let Q(X; Y ) = (area of the parallelogram with sides X and Y )2 = hX; Xi hY; Y i − hX; Y i2 : Definition 15.2. Given any point p 2 M and any tangent 2-plane π to M at p, the sectional curvature of π is the real number hR(X; Y )Y; Xi Rm(X; Y; Y; X) sec(π) := = ; Q(X; Y ) Q(X; Y ) where fX; Y g denotes a basis of π. In components Kij := sec (spanf@i;@jg) Independence of basis choice Proposition 15.3. Sectional curvature sec(π) is well-defined, independent of the choice of basis fX; Y g for π. Proof: If fX;~ Y~ g is another basis for π, then X~ = aX + bY; Y~ = cX + dY; 1 with non-zero determinant ad − bc. An easy computation gives Q(X;~ Y~ ) = (ad − bc)2Q(X; Y ): Using symmetry of Rm, Rm(X;~ Y;~ Y;~ X~) = Rm(aX + bY; cX + dY; cX + dY; aX + bY ) = (ad − bc) Rm(X; Y; cX + dY; aX + bY ) = (ad − bc)2Rm(X; Y; Y; X) so sec(π) is independent of the choice of basis fX; Y g. We have seen that the sectional curvature of M is a real-valued function defined on the set of all tangent 2-planes of M. Although sectional curvature seems simpler than the curvature tensor Rm, it encapsulates the same information. Lemma 15.4. Given sec(π) for all tangent 2-planes π ⊂ TpM, we can algebraically determine the curvature tensor Rm at p. Proof: The idea is to use the symmetry of the Riemann curvature tensor. Set X; Y; Z; W 2 TpM, and consider the polynomial in t defined by f(t) = Rm(X + tW; Y + tZ; Y + tZ; X + tW ) − t2(Rm(X; Z; Z; X) + Rm(W; Y; Y; W )) Due to the symmetries in each term, we can write f in terms of sectional curvatures (and the function Q which is given by the inner product). The coefficient of t2 in f(t) is Rm(X; Y; Z; W ) + Rm(X; Z; Y; W ) + Rm(W; Z; Y; X) + Rm(W; Y; Z; X): Recall that Rm is skew-symmetric in each of the first and last 2 entries ( Rm(X; Y; Z; W ) = −Rm(Y; X; Z; W ); Rm(X; Y; Z; W ) = −Rm(X; Y; W; Z) and it is symmetric between the first pair and last pair Rm(X; Y; Z; W ) = Rm(Z; W; X; Y ): Hence the coefficient of t2 in f(t) is 2Rm(X; Y; Z; W ) − 2Rm(Z; X; Y; W ): (1) Interchanging X and Y , we similarly obtain that the expression 2Rm(Y; X; Z; W ) − 2Rm(Z; Y; X; W ); is determined by the sectional curvatures of M at p. Again using symmetry of Rm, this is − 2Rm(X; Y; Z; W ) + 2Rm(Y; Z; X; W ): (2) Recall also the algebraic Bianchi identity: X Rm(X; Y; Z; W ) = 0: all cyclic permutations of (X;Y;Z) Using this, (1) − (2) = 6Rm(X; Y; Z; W ); so the Riemann curvature tensor is determined by the sectional curvature. 2 Definition 15.5. A Riemannian manifold M has constant curvature if its sectional curvature function is constant, that is there exists a constant k such that secp(π) = k for all p 2 M and all tangent planes π ⊂ TpM. Remark 15.6. Let p 2 M. If the sectional curvature function at p is zero (that is, secp = 0), then the curvature Rm = 0 at p. Hence, a Riemannian manifold (M; g) is flat if and only if the sectional curvature is identically zero. Let us consider the special case when our Riemannian manifold is a surface. In that case we had already an intrinsic notion of curvature, namely the Gauss curvature. Proposition 15.7. For a Riemannian surface (M; g), the sectional curvature of M is equal to its Gauss curvature. Proof: By definition, for p 2 M 2, Rm2112 sec(TpM) = 2 (g11g22 − g12) h(r (r @ ) − r (r @ ));@ i = @2 @1 1 @1 @2 1 2 ; det g which is a formula for the Gauss curvature. Exercise 15.8. Prove this expression, using • the definition of Gauss curvature K(p) as the determinant of the shape operator −dNp, where N is the Gauss map det(II) • the resulting expression K = det I where I and II denote the 1st and 2nd fundamental forms. (You should remind yourself how to obtain this expression.) • the fact (which you should make sure you understand) that k r@i (@j) = Γij@k + hijN where hij denotes the coefficients of the 2nd fundamental form. We will next study some spaces which have constant sectional curvature, for which the following remark will prove helpful. Remark 15.9. Let (M; g) be a Riemannian manifold and consider the (0; 4)-tensor Q defined by Q(X; Y; Z; W ) = hX; W i hY; Zi − hY; W i hX; Zi for X; Y; W; Z 2 X (M). Then M has constant sectional curvature equal to K0 if and only if Rm = K0Q, where Rm denotes the Riemann curvature tensor. Proof: Suppose M has constant sectional curvature K0 and take p 2 M and linearly independent X; Y 2 TpM. Then 2 2 2 Rm(X; Y; Y; X) = K0(jXj jY j − hX; Y i ) = K0Q(X; Y; Y; X): As in the above proof, Lemma 15.4, this then implies that Rm(X; Y; Z; W ) = K0Q(X; Y; Z; W ) for all X; Y; Z; W 2 X (M), as required. The converse is trivial. Corollary 15.10 (Constant Sectional Curvature). Let (M; g) be a Riemannian manifold, p 2 M and fe1; : : : ; eng an orthonormal basis of TpM. Then the following are equivalent: 1. sec(π) = K0 for all 2-planes π ⊂ TpM 2. Rmijkl = K0(δilδjk − δikδjl): 3. Rmijji = −Rmijij = K0 for all i 6= j and Rmijkl = 0 otherwise. 3 Lecture 16 [Riemannian Geometry – Lecture 16]Riemannian Geometry – Lecture 16 Computing Sectional Curvatures Dr. Emma Carberry September 14, 2015 Stereographic projection of the sphere Example 16.1. We write Sn or Sn(1) for the unit sphere in Rn+1, and Sn(r) for the sphere of radius r > 0. Stereographic projection φ from the north pole N = (0;:::; 0; r) gives local coordinates on Sn(r) n fNg Example 16.1 (Continued). Example 16.1 (Continued). Example 16.1 (Continued). We have for each i = 1; : : : ; n that ui = cxi (3) and using the above similar triangles, we see that r c = : (4) r − xn+1 Hence stereographic projection φ : Sn n fNg ! Rn is given by rx1 rxn φ(x1; : : : ; xn; xn+1) = ;:::; r − xn+1 r − xn+1 The inverse φ−1 of stereographic projection is given by (exercise) 2r2ui xi = ; i = 1; : : : ; n juj2 + r2 2r3 xn+1 = r − : juj2 + r2 and so we see that stereographic projection is a diffeomorphism. 4 Stereographic projection is conformal Definition 16.2. A local coordinate chart (U; φ) on Riemannian manifold (M; g) is conformal if for any p 2 U and X; Y 2 TpM, 2 gp(X; Y ) = F (p) hdφp(X); dφp(Y )i where F : U ! R is a nowhere vanishing smooth function and on the right-hand side we are using the usual Euclidean inner product. Exercise 16.3. Prove that this is equivalent to: 1. dφp preserving angles, where the angle between X and Y is defined (up to adding multiples of 2π) by g(X; Y ) cos θ = pg(X; X)pg(Y; Y ) and also to 2 2. gij = F δij for some nowhere vanishing smooth function F on U. Example 16.1 (Continued). We now proceed to show that stereographic projection is a conformal map. Now −1 −1 gij(u) = hdφu (ei); dφu (ej)i and @xj @ dφ−1(e ) = (u) u i @ui @xj ! 2r2 4r2(ui)2 @ 4r3ui @ = − + juj2 + r2 (juj2 + r2)2 @xi (juj2 + r2)2 @xn+1 | {z } | {z } A(i) B(i) so gij = A(i)A(j)δij + B(i)B(j) and simplifying gives 4r4 gij = δij: (5) (juj2 + r2)2 Pre-composing with the rotation of the sphere is another coordinate chart which includes the point N = (0;:::; 0; 1) and since rotation is an isometry, (5) holds for this coordinate chart also and hence we have covered the sphere with an atlas satisfying 4r4 gij = δij: (juj2 + r2)2 We next use this to compute the sectional curvature of Sn(r).