Differential Geometry. Homework 9. Due May 10Th. Professor: Luis Fernández

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, ϕ) = (cos θ, sin θ, cos ϕ, sin ϕ) 2 is an immersion of R2 into the unit sphere S3 ∈ 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 ϕ, cos ϕ) form an orthonormal basis of the tangent space, and that the vectors 1 1 n1 = √ (cos θ, sin θ, cos ϕ, sin ϕ), n2 = √ (− cos θ, − sin θ, cos ϕ, sin ϕ) 2 2 form an orthonormal basis of the normal space. b) Use the fact that

hSnk (ei), eji = −h∇ei nk, eji = h∇ei ej, nki,

where ∇ 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 {e1, e2} 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 ϕ(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) ∈ 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) (∇X 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  P(0,0),(t,0)(Yα(0,0)) − Yα(t,0) P(0,s),(t,s)(Yα(0,s)) − Yα(t,s)  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)  P (Y ) − Y P (Y ) − Y  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) = ∇ ∇ Y − ∇ ∇ Y, st X1 X2 X2 X1

where X1 = ∂/∂t and X2 = ∂/∂s. (Note that [X1,X2] = 0 so the ∇[X1,X2]Y term is zero.)