Differential Geometry. Homework 9. Due May 10th. Professor: Luis Fern´andez NOTE: if you need, please ask for hints. 1. (DoCarmo, Ex 5 of Chapter 6) Prove that the sectional curvature of the Riemannian manifold S2 × S2 with the product metric, where S2 is the unit sphere in R3, is non-negative. Find a totally geodesic, flat torus, embedded in S2 × S2 (see next exercise). 2. (DoCarmo, Ex 2 of Chapter 6): Show that x : R2 ! R4 given by 1 x(θ; ') = (cos θ; sin θ; cos '; sin ') 2 is an immersion of R2 into the unit sphere S3 2 R4, whose image x(R2) is a torus T 2 with sectional curvature zero in the induced metric. 3. (DoCarmo, Ex 6 of Chapter 6) Let G be a Lie group with a bi-invariant metric. Let H be a Lie group and let h : H ! G be an immersion that is also a homomorphism of groups (that is, H is a Lie subgroup of G). Show that h is a totally geodesic immersion. 4. (DoCarmo, Ex 8 of Chapter 6) (The Clifford torus). Consider the immersion x : R2 ! R4 given in Exercise 2. a) Show that the vectors e1 = (− sin θ; cos θ; 0; 0); e2 = (0; 0; − sin '; cos ') form an orthonormal basis of the tangent space, and that the vectors 1 1 n1 = p (cos θ; sin θ; cos '; sin '); n2 = p (− cos θ; − sin θ; cos '; sin ') 2 2 form an orthonormal basis of the normal space. b) Use the fact that hSnk (ei); eji = −hrei nk; eji = hrei ej; nki; where r is the covariant derivative (that is, the usual derivative) of R4, and i; j; k = 1; 2, to establish that the matrices of Sn1 and Sn2 with respect to the basis fe1; e2g are −1 0 1 0 S = S = n1 0 −1 n2 0 −1 c) From Exercise 2, x is an immersion of the torus T 2 into S3(1) (the Clifford torus). Show that x is a minimal immersion. 5. Let M ⊂ R3 be a surface of revolution (see Hw 6, Ex 3, which is DoCarmo, Chapter 3, Ex. 1). Suppose it is parametrized by '(u; v) = (f(v) cos u; f(v) sin u; g(v)). a) Show that the Gaussian curvature (i.e. the product of the principal curvatures) of M is −f 00(u)=f(u). b) Show that there is a surface of revolution in R3 that has constant Gaussian curvature equal to 1 but does not have constant mean curvature (i.e. the average of the principal curvatures). 6. Let M ⊂ R3 be the catenoid, which is the surface of revolution obtained by revolving the curve x = cosh z around the z-axis. Show that M is a minimal surface. 7. (From Lee, Ex. 8-10.) Suppose M is a connected n-dimensional Riemannian manifold, and a Lie group G acts effectively on M by isometries. (A group action is said to be effective if no element of G other than the identity acts as the identity on M .) Show that dimG ≤ n(n + 1)=2, and that equality is possible only if M has constant sectional curvature. 8. Again! This time without coordinates: The Riemann curvature tensor arises naturally when you consider parallel transport around a coordinate loop. This can be seen as follows. Let α : V ⊂ R2 ! M be a diffeomorphism over its image, where V contains a square containing (0; 0). Let p = α(0; 0) 2 M. We are going to do parallel transport along the chain of curves given by α(0; 0) −! α(t; 0) −! α(t; s) −! α(0; s) −! α(0; 0): For a vector field Y in M, let P(0;0);(t;0)(Y ), P(t;0);(t;s)(Y ), P(t;s);(0;s)(Y ), and P(0;s);(0;0)(Y ), denote parallel transport of Y along the corresponding curves above. a) Prove that for all s P(0;s);(t;s)(Yα(0;s)) − Yα(t;s) (rX Y )α(0;s) = − lim 1 t!0 t (and similarly with X2 doing the corresponding limit in s instead of t. Note that this follows easily from Exercise 6 from Hw 5, which is DoCarmo Ex. 2 of chapter 2). b) Show that (I mean, check that it all cancels properly) P(0;s);(0;0) ◦ P(t;s);(0;s) ◦ P(t;0);(t;s) ◦ P(0;0);(t;0)(Yα(0;0)) − Yα(0;0) st 8 P(0;0);(t;0)(Yα(0;0)) − Yα(t;0) P(0;s);(t;s)(Yα(0;s)) − Yα(t;s) 9 >P(t;0);(t;s) − > < t t = Yα(0;0) = P(0;s);(0;0) ◦ P(t;s);(0;s) ◦ − > s > st :> ;> 1 + P ◦ P (P (Y )) + P ◦ P (P (Y )) − P ◦ P (Y ) st (0;s);(0;0) (t;s);(0;s) (t;0);(t;s) α(t;0) (0;s);(0;0) (t;s);(0;s) (0;s);(t;s) α(0;s) (0;s);(0;0) (t;s);(0;s) α(t;s) 8 P (Y ) − Y P (Y ) − Y 9 >P (0;0);(t;0) α(0;0) α(t;0) − (0;s);(t;s) α(0;s) α(t;s) > <> (t;0);(t;s) t t => = P(0;s);(0;0) ◦ P(t;s);(0;s) ◦ > s > :> ;> 1 P (Y ) − Y 1 P (Y ) − Y + P ◦ P (t;0);(t;s) α(t;0) α(t;s) − P (0;0);(0;s) α(0;0) α(0;s) t (0;s);(0;0) (t;s);(0;s) s t (0;s);(0;0) s c) Do the limit when s; t ! 0 to obtain P ◦ P ◦ P ◦ P (Y ) − Y (0;s);(0;0) (t;s);(0;s) (t;0);(t;s) (0;0);(t;0) α(0;0) α(0;0) = r r Y − r r Y; st X1 X2 X2 X1 where X1 = @=@t and X2 = @=@s. (Note that [X1;X2] = 0 so the r[X1;X2]Y term is zero.).