Resolution to Homework #1 - Introduction to Differential Geometry II | MATH 6240, Exams of Mathematics

Download Resolution to Homework #1 - Introduction to Differential Geometry II | MATH 6240 and more Exams Mathematics in PDF only on Docsity! Math 6240 Homework #1 Solutions 1. For a general path-connected metric space M , define the length of a continuous curve γ : [a, b] → M by the formula L(γ) = sup P n∑ k=1 d ( γ(tk−1), γ(tk) ) where P is the set of all partitions a = t0 < t1 < t2 < · · · < tn−1 < tn = b. Prove the following: (a) (Lower semicontinuity of the length function.) If γm is a sequence of curves on [a, b] which converges to γ in the supremum norm, then L(γ) ≤ lim infm→∞ L(γm). Solution: Let ε > 0. Fix a partition P = {t0, . . . , tn} of [a, b] such that L(γ) < n∑ k=1 d ( γ(tk−1), γ(tk) ) + ε. Then for any m, L(γ)− ε < n∑ k=1 d ( γ(tk−1), γm(tk−1) ) + d ( γm(tk−1), γm(tk) ) + d ( γm(tk), γ(tk) ) ≤ L(γm) + 2nD(γ, γm). Since lim m→∞ D(γm, γ) = 0, there is an m ′ such that D(γ, γm) < ε/2n for every m ≥ m′. Thus for m ≥ m′, we have L(γ) < 2ε + L(γm). Hence L(γ) ≤ 2ε + lim inf m→∞ L(γm). Since ε was arbitrary, L(γ) ≤ lim inf m→∞ L(γm). (b) If M is in addition compact, then between any two points p and q, there is a curve γ : [0, 1] → M with γ(0) = p, γ(1) = q, and γ being minimal among all curves. (Hint: use the Ascoli theorem.) Solution: Let D = inf{L(γ) | γ : [0, 1] → M continuous, γ(0) = p, γ(1) = q}. If D = ∞ then every curve joining p and q has infinite length, so there is nothing to prove. Suppose D < ∞ and choose a sequence of continuous curves γn : [0, 1] → M with γn(0) = p, γn(1) = q, and lim n→∞ L(γn) = D. We can suppose that each γn is parametrized proportionally to arc length, since for each n, the function hn : [0, 1] → R defined by hn(t) = L(γn ∣∣ [0,t] ) is strictly increasing and continuous; thus γ̃n = t 7→ γn ( h−1n ( thn(1) )) is parametrized proportionally to arc length, has γ̃n(0) = p, γ̃n(1) = q, γ̃n continuous, and L(γ̃n) = L(γn). The family {γn} is equicontinuous, since for every ε > 0, choosing δ = ε/ supk∈N L(γk) gives |t1 − t2| < δ =⇒ d ( γn(t1), γn(t2) ) ≤ L(γn ∣∣ [t1,t2] ) = |t1 − t2|L(γn) < ε L(γn) supk∈N L(γk) ≤ ε 1 for every n. By the Ascoli Theorem, since M is compact, there is a subsequence γnk which con- verges to some curve γ, which must be continuous by completeness of C([0, 1], M) and must have L(γ) ≤ limk→∞ L(γnk) = D. Since also D ≤ L(γ), we have D = L(γ). (c) Consider the plane R2 with metric d ( (x1, y1), (x2, y2) ) = |x1 − x2|+ |y1 − y2|. Find all length-minimizing curves between (0, 0) and (1, 1). Solution: Suppose we have a curve γ(t) = ( x(t), y(t) ) with γ(0) = (0, 0) and γ(1) = (1, 1). If x and y are both nondecreasing, then for any partition P , we have n∑ k=1 d ( γ(tk−1), γ(tk) ) = n∑ k=1 |x(tk)− x(tk−1)|+ n∑ k=1 |y(tk)− y(tk−1)| = n∑ k=1 x(tk)− x(tk−1) + n∑ k=1 y(tk)− y(tk−1) = x(1)− x(0) + y(1)− y(0) = 2. Thus L(γ) = 2. On the other hand, if x has x(t) < x(s) for some t > s, then choosing a partition P = {0, s, t, 1}, we have LP (γ) = 3∑ k=1 d ( γ(tk−1), γ(tk) ) = |x(s)− x(0)|+ |x(t)− x(s)|+ |x(1)− x(t)| + |y(s)− y(0)|+ |y(t)− y(s)|+ |y(1)− y(t)| ≥ |x(s)− x(0)|+ x(s)− x(t) + |x(1)− x(t)|+ |y(1)− y(0)| = 1 + |x(s)|+ x(s)− x(t) + |1− x(t)|. There are now three cases. • If x(s) ≤ 0, then x(t) < 0 and |x(s)| = −x(s) while |1− x(t)| = 1− x(t), so the sum is LP (γ) = 2− 2x(t) > 2. • If x(s) > 0 and x(t) ≤ 1, then the sum is LP (γ) = 1 + x(s) + x(s) − x(t) + 1− x(t) = 2 + 2 ( x(s)− x(t) ) > 2. • If x(s) > 0 and x(t) > 1, then the sum is LP (γ) = 1 + x(s) + x(s) − x(t) + x(t)− 1 = 2x(s) > 2. In any case, we conclude LP (γ) > 2, so that L(γ) > 2, so such a curve is not minimizing. (The case where y is nondecreasing is identical.) As a result, the infimum of lengths of curves joining (0, 0) and (1, 1) is 2, and this infimum is attained by all nondecreasing-component curves in the square. 2 Thus two coordinate patches still do not cover the torus. A third coordinate patch will work. So now let us take the same function F on R2 and restrict it to three different open squares: F1 = F ∣∣ (0,2π)×(0,2π), F2 = F ∣∣ (π 2 , 5π 2 )×(π 2 , 5π 2 ) , F3 = F ∣∣ (π,3π)×(π,3π). Then we have imF1 ∪ imF2 ∪ imF3 = T2. The transition functions are found from the fact that if F1(u1, v1) = F3(u2, v2), then sin u1 = sin u2, cos u1 = cos u2, cos v1 = cos v2, and sin v1 = sin v2. Thus u1 = u2 and v1 = v2, modulo 2π, depending on the location of u1, u2, v1, and v2. Since imF1 ∩ imF2 has four components, we have four formulas for the transition functions: F−13 ( F1(u, v) ) = (u, v) in (π, 2π)× (π, 2π) F−13 ( F1(u, v) ) = (u− 2π, v) in (2π, 3π)× (π, 2π) F−13 ( F1(u, v) ) = (u, v − 2π) in (π, 2π)× (2π, 3π) F−13 ( F1(u, v) ) = (u− 2π, v − 2π) in (2π, 3π)× (2π, 3π). All four of these functions are smooth, and since the domain is disconnected, we don’t need to worry about them joining up. The other transition functions are of course similar. 3. (a) Prove that if M is a compact manifold and f : M → Rn is an immersion which is also one-to-one, then f is an embedding of M . Give an example to show this is not true if M is not compact. Solution: Since f is one-to-one, the map f : M → imf ⊂ Rn is a bijection, so there is an inverse map f−1 : imf → M . We want to prove f−1 is continuous and smooth. For continuity, we want to prove (f−1)−1 of any closed set is closed set, i.e., f of any closed set is closed. Since M is compact, any closed subset K of M is also compact; thus f [K] is also compact, and hence closed. 5 For smoothness of f−1, we use the fact that f is an immersion; thus at any p ∈ M there is an open set U ⊂ M containing p such that f ∣∣ U is a diffeomorphism onto its image. Thus f−1 ∣∣ f [U ] is smooth. So f−1 is a smooth map. If M is not compact, then f may not be an embedding. For example, if f : R → R2 is f(t) = ( t t4 + 1 , t2 t4 + 1 ) , then f is an immersion since the two derivatives are not zero simultaneously. Also imf ⊂ [−1, 1] × [−1, 1], and imf is closed, since lim t→±∞ f(t) = (0, 0) ∈ imf . Thus imf is compact, while R is not, so f cannot be an embedding. (b) Define the Klein bottle as R2 modulo the group generated by the operations g1 = (x, y) → (x + 2π, y) and g2 = (x, y) → (2π − x, y + 2π). Prove this group action is properly discontinuous, i.e., that every point has a neighborhood U with U ∩ g(U) = ∅ for every group element g. (It is sufficient to do this at one point; why?) By Example 0.4.8 of do Carmo, the Klein bottle is a manifold. Solution: Any element of the group with these generators may be written as g = gi11 g j1 2 g i2 1 g j2 2 · · · gin1 g jn 2 for some integers ik and jk. We can simplify these expressions by noting that g22(x, y) = (x, y + 4π) so that g 2 2g1 = g1g 2 2. Similarly g−22 commutes with g1. Hence we can pull out all even powers of g2 to the end, so we can assume j1 = ±1, j2 = ±1, . . . , jn−1 = ±1. Furthermore, g1g2(x, y) = (4π − x, y + 2π) and g2g1(x, y) = (−x, y + 2π), so that g1g2 = g 2 1g2g1, which implies g −1 1 g2 = g2g1. Similarly g2g −1 1 = g1g2, g −1 2 g1 = g−11 g −1 2 , and g −1 2 g −1 1 = g1g −1 2 . We can use these equations to pull all g1’s to the left side. Hence the general element of the group is gm1 g n 2 , for integers m and n. If n is even, then gm1 g n 2 (x, y) = (x + 2mπ, y + 2nπ) so |gm1 gn2 (x, y)− (x, y)|2 = 4π2(m2 + n2) ≥ 4π2 if (m, n) 6= (0, 0). If n is odd, then gm1 g n 2 (x, y) = (2mπ + 2π − x, y + 2nπ) so |gm1 gn2 (x, y)− (x, y)|2 = 4(mπ + π − x)2 + 4π2n2 ≥ 4π2 since |n| ≥ 1. 6 Thus choosing U to be the 2π-ball around (x, y), we see that U ∩g(U) = for every g ∈ G. (c) Prove that the map F : R2 → R3 defined by F (x, y) = ( (1 + cos y 2 sin x− sin y 2 sin 2x) cos y,( 1 + cos y 2 sin x− sin y 2 sin 2x ) sin y, sin y 2 sin x + cos y 2 sin 2x ) restricts to the Klein bottle and defines an immersion of the Klein bottle into R3. Show however that this is not an embedding. Solution: (Notice that the original formula had x and y interchanged; this one is correct.) We obviously have F ◦g1(x, y) = F (x, y), and in addition F ◦g2(x, y) = F (x, y) as can easily be checked (since e.g., cos (y 2 + π) sin (2π − x) = (− cos y 2 )(− sin x) = cos y 2 sin x). So F does restrict to the Klein bottle. To prove F is an immersion, we need to prove that the vectors ∂F ∂x and ∂F ∂y are linearly independent for all x and y. This is easiest done by computing c = ∣∣∣∣∂F∂x × ∂F∂y ∣∣∣∣2 = ∣∣∣∣∂F∂x ∣∣∣∣2 ∣∣∣∣∂F∂y ∣∣∣∣2 − ∣∣∣∣∂F∂x · ∂F∂y ∣∣∣∣2 and showing that it is nonzero. This is a long but straightforward computation (helped perhaps by Maple). We obtain after some simplifications that it is c = ( 1 + cos y 2 sin x− sin y 2 sin 2x )2 (cos2 x+4 cos2 2x)+ 1 16 sin2 2x(8 cos2 x− 3)2. This is actually zero when x = π/2 and y = 2π (or x = 3π 2 and y = 0), but otherwise it’s nonzero. Not a very good immersion then, is it? Anyway, even if F were an immersion, it could not be an embedding since F is not one-to-one. To see this, we simply notice that most terms of F drop out when sin x = 0. So we have F (0, y) = (cos y, sin y, 0) and F (π, y) = (cos y, sin y, 0), so F is not one-to-one. (d) Define the projective plane P2 as the quotient of S2 ⊂ R3 by the antipodal map- ping. Prove that F : R3 → R4 defined by F (x, y, z) = (x2−y2, xy, yz, xz) restricts to a map on P2, and that this restriction is an embedding of P2 into R4. Solution: We obviously have F (−x,−y,−z) = F (x, y, z), so F restricts to a map f on P2. Now we show f is one-to-one on P2. If F (u, v, w) = F (x, y, z), then u2 − v2 = x2 − y2, uv = xy, vw = yz, and uw = xz. If u 6= 0, then v = xy u so u2 − x2 = v2 − y2 = y2(x2 − u2)/u2, 7