DIFFERENTIAL GEOMETRY HW 12

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1 DIFFERENTIAL GEOMETRY HW 1 CLAY SHONKWILER 3 Find the Lie algebra so(n) of the special orthogonal group SO(n), and the explicit formula for the Lie bracket there. Proof. Since SO(n) is a subgroup of GL(n), we know that so(n) gl(n) M(n), the n n matrices. Consider the matrix I + ta, where I is the n n identity matrix and A so(n). If I + ta SO(n), then I = (I+tA)(I+tA) T = (I+tA)(I T +ta T ) = (I+tA)(I+tA T ) = I +ta+ta T +t AA T. Hence, differentiating both sides with respect to t, 0 = A + A T + taa T. For t = 0, I + ta SO(n), so, plugging in t = 0 to the above equation, 0 = A + A T. Therefore, since our choice of A so(n) was arbitrary, we see that so(n) consists of skew-symmetric matrices, where [A, B] = AB BA is just regular matrix multiplication (as it is in gl(n)). In fact, since the dimension of the space of skew-symmetric matrices is n(n 1), which is equal to the dimension of SO(n) and, therefore, so(n), so(n) consists precisely of the skew-symmetric n n matrices. Show that x : R R 4 given by x(θ, φ) = 1 (cos θ, sin θ, cos φ, sin φ), (θ, φ) R is an immersion of R into the unit sphere S 3 (1) R 4, whose image x(r ) is a torus T with sectional curvature zero in the induced metric. Proof. First, note that dx = 1 sin θ 0 1 cos θ sin φ 1 0 cos φ 1,

2 CLAY SHONKWILER which is certainly of rank, so x is an immersion. Moreover, x(θ, φ) = 1 (cos θ + sin θ + cos φ + sin φ) = 1 () = 1, so the image of x is contained in S 3 (1). Since (cos θ, sin θ) and (cos φ, sin φ) parameterize the unit circle, the image of x is S 1 S 1 = T. Now, if X i = x i, θ = 1 ( sin θx 1 + cos θx ) and φ = 1 ( sin φx 3 + cos φx 4 ). ( ) Note that K θ, φ = 0 since all sectional curvatures of R 4 are zero. Hence, using the Gauss Theorem, ( ) ( ) ( ) K θ, = K φ θ, K φ θ, φ ( = B θ, ) ( ) ( ), B θ φ, φ B θ, φ (1) = /θ θ /θ θ, /φ φ /φ φ /θ φ /θ φ = /θ θ, /φ φ /θ φ since /θ θ = 0, /φ φ = 0 and /θ φ give coordinates on R ). Now, extend θ V (x 1, x, x 3, x 4 ) = 1 ( x, x 1, x 4, x 3 ). Then /θ θ = V V = 1 x X 1 +x 1 X ( x X 1 + x 1 X ) = 0 (because θ and φ to the vector field V where = 1 [ x X1 ( x X 1 + x 1 X ) + x 1 X ( x X 1 + x 1 X )] = 1 [ x X 1 (x 1 )X + x 1 X ( x )X 1 ] = 1 [ x 1X 1 x X ], since all the other terms in the expansion of the third line are zero. A similar argument shows that /φ φ = 1 [ x 3X 3 x 4 X 4 ] and that /θ φ = 0. Therefore, ( ) K θ, = φ /θ θ, /φ φ /θ φ = 1 4 x 1X 1 x X, x 3 X 3 x 4 X 4 0 = 0.

3 DIFFERENTIAL GEOMETRY HW 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. Show that h is a totally geodesic immersion. Proof. Since the metric H inherits from G is bi-invariant on H, we know that the geodesics through the identity in H are one-parameter subgroups of H. However, a one-parameter subgroup of H is also a one-parameter subgroup of G. Since the one-parameter subgroups of G are just the geodesics through the identity, this implies that the geodesics through the identity in H are geodesics in G, which demonstrates that H is a geodesic immersion at the identity. Since left-translation is an isometry of H and isometries take geodesics to geodesics, this in turn means that H is a geodesic immersion at every p H, so H is a totally geodesic immersion. The metric inherited from G on H is simply u, v H = (dh) g u, (dh) g v G, where u, v T g H. To see that the metric H inherits is actually bi-invariant on H, suppose u, v T g H. Then u, v H = (dh g )u, (dh) g v G = (dl h(g )) g (dh g )u, (dl h(g )) g (dh) g v G = d(l h(g ) h) g u, d(l h(g ) h) g v G = d(h h 1 g L h(g ) h) g u, d(h h 1 g L h(g ) h) g G = d(h L g ) g u, d(h L g ) g v G = (dh) g (dl g ) g u, (dh) g (dl g ) g v G = (dl g ) g u, (dl g ) g v H. A similar calculation shows that this inherited metric is right-invariant, so we see that it is, in fact, bi-invariant. Consider the immersion x : R R 4 given in Exercise. (a): Show that the vectors 8 e 1 = ( sin θ, cos θ, 0, 0), e = (0, 0, sin φ, cos φ) form an orthonormal basis of the tangent space, and that the vectors n 1 = 1 (cos θ, sin θ, cos φ, sin φ), n = 1 ( cos θ, sin θ, cos φ, sin φ) form an orthonormal basis of the normal space. Proof. First, and e 1, e = 0, e 1, e 1 = sin θ + cos θ = 1 e, e = sin φ + cos φ = 1,

4 4 CLAY SHONKWILER so e 1 and e form an orthonormal basis of the tangent space. Moreover, n 1, n 1 = 1 (cos θ + sin θ + cos φ + sin φ) = 1, and n, n = 1 (cos θ + sin θ + cos φ + sin φ) = 1 n 1, n = 1 ( cos θ sin θ + cos φ + sin φ) = 0. Also, e 1, n 1 = 1 ( sin θ cos θ + sin θ cos θ) = 0, e 1, n = 1 (sin θ cos θ sin θ cos θ) = 0 and, similarly, e, n 1 = 0 and e, n = 0. Therefore, n 1 and n form an orthonormal basis of the normal space. (b): Use the fact that S nk (e i ), e j = ei n k, e j = ei e j, n k, where is the covariant derivative of R 4, and i, j, k = 1,, to establish that the matrices of S n1 and S n with respect to the basis {e 1, e } are S n1 = 1 ( S n = 1 ( Proof. Note that, using what we showed in problem, e 1 = θ and e = φ. Since S n k (e i ), e j = ei e j, n k, we have that S n1 (e 1 ), e 1 = e1 e 1, n 1 = = /θ θ, n 1 cos θx 1 sin θx, = 1 ( cos θ sin θ) = 1. ) ) 1 (cos θx 1 + sin θx + cos φx 3 + sin φx 4 ) The other 7 calculations are tedious but similar and allow us to conclude that S n1 and S n are as above.

5 DIFFERENTIAL GEOMETRY HW 1 5 (c): From Exercise, x is an immersion of the torus T into S 3 (1) (the Clifford torus). Show that x is a minimal immersion. Proof. Note that 1 (cos θ, sin θ, cos φ, sin φ) = 1 (cos θ + sin θ + cos φ + sin φ) = 1, so n 1, thought of as a point, lies on the sphere S 3 (1), so n 1, thought of as a vector, is orthogonal to S 3 (1). Therefore, the subspace of T p S 3 normal to T p T (where T is the image of x), is spanned by n. Therefore, the condition of being minimal boils down to the condition that trace S n = 0, which is clearly does, given the computation in part (b) above. 11 Let f : M n+1 R be a differentiable function. Define the Hessian, Hess f of f at p M as the linear operator Hess f : T p M T p M, (Hess f)y = Y grad f, Y T p M, where is the Riemannian connection of M. Let a be a regular value of f and let M n M n+1 be the hypersurface in M defined by M = {p M; f(p) = a}. Prove that: (a): The Laplacian f is given by f = trace Hess f. Proof. By definition, if Y T p M, f(y ) = div grad f(y ) = Trace Y grad f = Trace Hess f(y ). since our choice of Y was arbitrary, we see that Deltaf = Hess f. (b): If X, Y X(M), then (Hess f)y, X = Y, (Hess fx. Conclude that Hess f is self-adjoint, hence determines a symmetric bilinear form on T p M, p M, given by (Hess f)(x, Y ) = (Hess f)x, Y, X, Y T p M.

6 6 CLAY SHONKWILER Proof. Let p M and let E 1,..., E n+1 be a geodesic frame at the point p. Then, for i, j {1,..., n + 1}, () (Hess f)e i, E j = Ei grad f, E j n+1 = Ei (E k (f))e k, E j = = n+1 [E k (f) Ei E k + E i (E k (f))e k ], E j n+1 E i (E k (f))e k, E j = E i (E j (f)). On the other hand, (3) E i, (Hess f)e j = E i, Ej grad f n+1 = E i, Ej (E k (f))e k n+1 = E i, [E k (f) Ej E k + E j (E k (f))e k ] = E j (E i (f)). Now, 0 = Ei E j Ej E i = [E i, E j ] = E i E j E j E i, so we see that E i (E j (f)) = E j (E i (f)). Hence, the expressions in () and (3) are equal, so (Hess f)e i, E j = E i, (Hess f)e j. Since this holds for all i, j {1,..., n + 1}, we conclude that the Hessian is self-adjoint. (c): The mean curvature H of M M is given by ( ) grad f nh = div. grad f Proof. Let p M M. Let E 1,..., E n, E n+1 = orthonormal frame about p. Note that η η, η = 1 ηη, η = 1 η(1) = 0. grad f grad f = η be an

7 DIFFERENTIAL GEOMETRY HW 1 7 Thus, by definition of divergence and of the mean curvature, n nh = Trace S η = S η (E i ), E i i=1 = n Ei η, E i η η, η i=1 n+1 = Ei η, E i i=1 = div M η ( ) grad f = div. grad f (d): Observe that every embedded hypersurface M n M n+1 is locally the inverse image of a regular value. Conclude from (c) that the mean curvature H of such a hypersurface is given by H = 1 div N, n where N is an appropriate local extension of the unit normal vector field on M n M n+1. Proof. Let M be an embedded hypersurface. Then, locally, M = f 1 (r) for some regular value r. Hence, for each point p M, grad f(p) 0, so the expression derived in (c) is well-defined on all of M. Hence, H = 1 div N, n where N is some extension of the unit normal vector field on M n M n+1. 1 Let X be a Killing vector field on a Riemannian manifold M. Let N = {p M; X(p) = 0}. Prove that: (a): If p N, and V M is a normal neighborhood of p, with q N V, then the radial geodesic segment γ joining p to q is contained in N. Conclude that γ V N. Proof. Let φ t denote the flow of X. If p N and V M is a normal neighborhood of p with q N V, then let γ be the radial geodesic joining p to q. Since q N, φ t (q) = q for all t where the flow is defined. Now, since φ t is an isometry on V, φ t maps geodesics to geodesics; specifically, φ t (γ) is a geodesic. Since φ t fixes p and q, φ t (γ) is a geodesic passing through p and q in V ; by uniqueness of

8 8 CLAY SHONKWILER geodesics, then, φ t (γ) = γ. Therefore, for any q γ V, φ t (q ) = q, so X(q ) = 0. Thus, γ V N. (b): If p N, there exists a neighborhood V M of p such that V N is a submanifold of M (this implies that every connected component of N is a submanifold of M). Proof. We prove this by induction. If p is an isolated point, then {p} is a neighborhood of p and N {p} = {p} is a 0-submanifold of M. Otherwise, let V be a normal neighborhood of p such that there exists q 1 V N not equal to p. Let γ 1 be the radial geodesic from p to q 1. By part (a), γ 1 V N. If V N = γ 1 V, then we re done, since γ 1 V is a 1-submanifold of M. If V N γ 1 V, then let q V N {γ 1 }. Let γ be the radial geodesic joining p to q. Note that, by the same argument as in (a), γ V N. Now, let Q T p M be the subspace generated by exp 1 p v N. Hence, (q 1 ) and exp 1 p Then v = a exp 1 p (q ); let N = exp p (Q exp 1 p (V )). Suppose (q 1 ) + b exp 1 p (q ) for some a, b R. (dφ t ) p (v) = (dφ t ) p (a exp 1 p (q 1 ) + b exp 1 p (q )) = a(dφ t ) p (exp 1 p (q 1 )) + b(dφ t ) p (exp 1 p (q )) = a exp 1 p (q 1 ) + b exp 1 p (q ) = v since q 1 and q are fixed by φ t. Therefore, we see that (dφ t ) p is the identity on Q. Therefore, if q N, φ t (q) = exp p ((dφ t ) p (exp 1 (q))) = exp p (exp 1 p (q)) = q, so we see that N V N. If V N = N, then we re done, since N is a -submanifold of M. Otherwise, if V N N, we pick q 3 V N N and iterate the above procedure. At each stage, either this algorithm terminates with N i, an i-submanifold of M, or we proceed to the next stage. Since we have only n dimensions to work with, we see that this procedure must terminate, and so V N is a submanifold of M. (c): The codimension, as a submanifold of M, of a connected component N k of N is even. Assume the following fact: if a sphere has a non-vanishing differentiable vector field on it then its dimension must be odd. Proof. Let p N k M. Let E p = (T p N k ) and let V M be a normal neighborhood of p. Let Nk = exp p(e p exp 1 p (V )). Now, suppose v E p. Then (dφ t ) p (v) E p ; otherwise, we would have that (dφ t ) p (v) T p N k and so exp p ((dφ t ) p v) N k V, meaning that φ t (exp p ((dφ t ) p v)) = exp p ((dφ t ) p v) = φ t (exp p (v)). However, since

9 DIFFERENTIAL GEOMETRY HW 1 9 exp p (v) / N k, it can never flow to a point in N k. Thus, we can conclude that (dφ t ) p : E p E p, so X is tangent to Nk. On the other hand, since p is the unique point in Nk where X vanishes, we know (from HW 8 Problem 5(b)) that X 0 is tangent to the geodesic spheres of Nk centered at p. Thus, X is a non-vanishing vector field on the geodesic spheres of Nk, which are homeomorphic to plain old spheres. Therefore, by the Hairy Ball Theorem, these geodesic spheres must be odd-dimensional, so Nk must be even dimensional. Since dim Nk = codim N k, we conclude that N k has even codimension in M. DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu

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