Introduction to Differential Geometry 1 - Resolution to Homework #9 | MATH 6230, Assignments of Mathematics

Download Introduction to Differential Geometry 1 - Resolution to Homework #9 | MATH 6230 and more Assignments Mathematics in PDF only on Docsity! Math 6230 Homework #9 Solutions 1. Textbook problem, Chapter 7: 4 For v ∈ V and ω ∈ Ωk(V ), we define the contraction vy ω ∈ Ωk−1(V ) by (vy ω)(v1, . . . , vk−1) = ω(v, v1, . . . , vk−1). This is sometimes also called the inner product and the notation ivω is also used. (a) Show that vy (wy ω) = −wy (vy ω). Solution: We apply both sides to (k − 2) vectors:( vy (wy ω) ) (v1, . . . , vk−2) = (wy ω)(v, v1, . . . , vk−2) = ω(w, v, v1, . . . , vk−2) = −ω(v, w, v1, . . . , vk−2) = ( wy (vy ω) ) (v1, . . . , vk−2). Since this is true for any vectors v1, . . . , vk−2, we have the desired equation. (b) Show that if v1, . . . , vn is a basis of V with dual basis φ1, . . . , φn, then vjy (φ i1 ∧ · · · ∧ φik) = { 0 j 6= any iα (−1)α−1φi1 ∧ · · · ∧ φ̂iα ∧ · · · ∧ φik if j = iα. Solution: Applying the left side to any (k − 1) vectors in the basis, we get( vjy (φ i1 ∧ · · · ∧ φik) ) (vm1 , . . . , vmk−1) = (φ i1 ∧ · · · ∧ φik)(vj, vm1 , . . . , vmk−1). This is nonzero iff {i1, . . . , ik} = {j, m1, . . . ,mk−1}. Of course, if j is not equal to any iα, then these sets cannot be equal no matter what the m’s are. So we must get the zero form. If j = iα, then we can reorder the m’s so that m1 = i1, . . . ,mα−1 = iα−1, mα = iα+1, . . . ,mk−1 = ik. Then we have( vjy (φ i1 ∧ · · · ∧ φik) ) (vm1 , . . . , vmk−1) = (φ i1 ∧ · · · ∧ φik)(viα , vi1 , . . . , viα−1 , viα+1 , vik) = (−1)α−1(φi1 ∧ · · · ∧ φik)(vi1 , . . . , viα−1 , viα , viα+1 , vik) = (−1)α−1. Since this value is nonzero iff {m1, . . . ,mk−1} = {i1, . . . , iα−1, iα+1, . . . , ik}, we must have vjy (φ i1 ∧ · · · ∧ φik) = ±φi1 ∧ · · · ∧ φ̂iα ∧ · · · ∧ φik , with the sign determined to be (−1)α−1 by the computation above. 1 (c) Show that for ω1 ∈ Ωk(V ) and ω2 ∈ Ωl(V ) we have vy (ω1 ∧ ω2) = (vy ω1) ∧ ω2 + (−1)kω1 ∧ (vy ω2). (Use (b) and linearity of everything.) Solution: By linearity of the wedge product, it is enough to prove this formula when ω1 = φi1∧· · ·∧φik and ω2 = φm1∧· · ·∧φml , for basis elements φi ∈ Ω1(V ). Furthermore, by linearity of the operation on v, it is enough to prove this when v = vj, a basis element. Then we have vy (ω1 ∧ ω2) = vjy (φi1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φml). There are now a few cases to consider. If j is one of the i’s but not one of the m’s, say j = iα, then we get vjy (φ i1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φml) = (−1)α−1(φi1 ∧ · · · ∧ φiα−1 ∧ φiα+1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φml) = (vy ω1) ∧ ω2. Since vy ω2 = 0 in this case, this proves the formula in this case. Now if j is one of the m’s but not one of the i’s, say j = mβ, then we get vjy (φ i1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φml) = (−1)k+β−1(φi1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φmβ−1 ∧ φmβ+1 ∧ φml) = (−1)kω1 ∧ (vy ω2). This proves the formula for this case, since vy ω1 = 0 in this case. If j is neither one of the i’s nor one of the m’s, then all terms in both sides are zero, so there is nothing to prove. Finally, if j is both one of the i’s and one of the m’s, then the left side is zero (since ω1 and ω2 have a repeated 1-form in common). So we just have to prove the right side is also zero. Suppose j = iα and j = mβ. We get (vy ω1) ∧ ω2 + (−1)kω1 ∧ (vy ω2) = (−1)α−1(φi1 ∧ · · · ∧ φiα−1 ∧ φiα+1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φml) + (−1)k+β−1(φi1 ∧ · · · ∧ φik ∧ φm1 ∧ · · · ∧ φmβ−1 ∧ φmβ+1 ∧ · · · ∧ φml = (−1)α−1(−1)k−1+β−1φj ∧ (φi1 ∧ · · · ∧ φiα−1 ∧ φiα+1 ∧ · · · ∧ φik)∧ ∧ (φm1 ∧ · · · ∧ φmβ−1 ∧ φmβ+1 ∧ · · · ∧ φml) + (−1)k+β−1(−1)α−1φj ∧ (φi1 ∧ · · · ∧ φiα−1 ∧ φiα+1 ∧ · · · ∧ φik)∧ ∧ (φm1 ∧ · · · ∧ φmβ−1 ∧ φmβ+1 ∧ · · · ∧ φml) = 0. Hence we have proved the formula in all cases. 2 by the antisymmetry of the wedge product on 1-forms. However, for this ω we have ω ∧ ω = 2φ1 ∧ φ2 ∧ φ3 ∧ φ4, which is not zero since the φ’s are linearly independent. 3. Textbook problem, Chapter 8: 7 Let ω be a 1-form on a manifold M . Suppose that ∫ c ω = 0 for every closed curve c in M . Show that ω is exact. Hint : If we do have ω = df , then for any curve c we have∫ c ω = f ( c(1) ) − f ( c(0) ) . Solution: Fix a base point p. Define a function f : M → R by f(q) = ∫ c ω, where c is any (nonclosed) curve with c : [0, 1] → M with c(0) = p and c(1) = q. This f does not depend on the choice of curve c, since if c1 and c2 are two such curves, then c1 − c2 is a closed curve, so that ∫ c1 ω = ∫ c2 ω. We want to prove that df = ω. To do this, take a small coordinate neighborhood near any point q ∈ M . Write ω = ∑ ωi(x)dx i in coordinates; we will prove that ωi|q = df ( ∂ ∂xi |q ) , i.e., that ωi|q = ∂f∂xi |q. Suppose without loss of generality that the coordinates are chosen so that q = (0, . . . , 0). For any i ∈ {1, . . . , n}, take a curve c1 from p to the inverse image of (0, . . . , 1, . . . , 0) (where there is a 1 in the ith place only). For any h with 0 < h < 1, let c2 be the coordinate curve c2(t) = (0, . . . , 1− (1− h)t, . . . , 0) and let c3 be the coordinate curve c3(t) = (0, . . . , h(1− t), . . . , 0). Then for any h we have f(q) = ∫ c1 ω + ∫ c2 ω + ∫ c3 ω = f(0, . . . , h, . . . , 0) + ∫ c3 ω = f(0, . . . , h, . . . , 0) + ∫ 1 0 ωi(0, . . . , h(1− t), . . . , 0) dt. Therefore, rescaling the integral, we have f(0, . . . , h, . . . , 0)− f(0, . . . , 0) = ∫ h 0 ωi(0, . . . , t, . . . , 0) dt. So by the fundamental theorem of calculus, we get ∂f ∂xi ∣∣∣ (0,...,0) = ωi(0, . . . , 0). Since this is true for every i, we have shown that df = ω at q. Since q was arbitrary, we do indeed have df = ω everywhere. 5 4. Textbook problem, Chapter 8: 13 Suppose M is a compact orientable n-manifold (with no boundary) and θ is an (n−1)- form on M . Show that dθ is 0 at some point. Solution: Of course it’s just Stokes’ Theorem. Take any nowhere-zero n-form µ on M ; then dθ = fµ. By Stokes’ Theorem we have∫ M fµ = ∫ M dθ = ∫ ∂M θ = 0. Since µ is nowhere zero, the only way the integral of f can be zero is if f is zero somewhere. Hence dθ is zero somewhere. 5. Textbook problem, Chapter 8: 27 Let {X t} be a C∞ family of C∞ vector fields on a compact manifold M . (To be more precise, suppose X is a C∞ vector field on M× [0, 1]; then X t(p) will denote πM ∗X(p,t).) From the addendum to Chapter 5, and the argument which was used in the proof of Theorem 5-6, it follows that there is a C∞ family {φt} of diffeomorphisms of M [not necessarily a 1-parameter group], with φ0 = identity, which is generated by {X t}, i.e., for any C∞ function f : M → R we have (X tf)(p) = lim h→0 f ( φt+h(p) ) − f ( φt(p) ) h . For a family ωt of k-forms on M we define the k-form ω̇t = lim h→0 ωt+h − ωt h . (a) Show that for η(t) = φ∗t ωt we have η̇t = φ ∗ t (LXtωt + ω̇t). Solution: We have (in any particular tangent space) η̇t = lim h→0 ηt+h − ηt h = lim h→0 η∗t+hωt+h − η∗t ωt h = lim h→0 η∗t+hωt+h − η∗t+hωt + η∗t+hωt − η∗t ωt h = lim h→0 η∗t+h ωt+h − ωt h + lim h→0 η∗t+h − η∗t h ωt = η∗t ω̇t + lim h→0 η∗t+hωt − η∗t ωt h . To figure out the second limit, we observe that ηt+h = ηt + hX t + O(h2) (this is a bit imprecise, but can be made precise in a coordinate system). We also know 6 that ηh ◦ ηt = ηt + hX t + O(h2). So while it is not true that ηt+h = ηh ◦ ηt, the error in using this as an approximation is only O(h2). Hence dividing by h and taking the limit gives an error approaching zero. So we can say that lim h→0 η∗t+hωt − η∗t ωt h = LXtφ∗t ωt. Finally we observe that LXtφ∗t ωt = φ∗tLXtωt since X t generates φt. (b) Let ω0 and ω1 be nowhere zero n-forms on a compact oriented n-manifold M , and define ωt = (1− t)ω0 + tω1. Show that the family φt of diffeomorphisms generated by {X t} satisfies φ∗t ωt = ω0 for all t if and only if LXtωt = ω0 − ω1. Solution: Easy enough from the previous formula. We have d dt φ∗t ωt = φ ∗ t (LXtωt + ω̇t) = φ∗t (LXtωt + ω1 − ω0). This derivative is zero for all t if and only if LXtωt = ω0 − ω1. (c) Using Problem 7-18, show that this holds if and only if d(X ty ωt) = ω0 − ω1. Solution: Problem 7-18 says that X ty dωt = LXtωt − d(Xty ωt). Hence we will be done if we know that X ty dωt = 0 for all t automatically. Of course this is true, since ωt is an n-form for every t, and hence dωt = 0 for all t. (d) Suppose that ∫ M ω0 = ∫ M ω1, so that ω0 − ω1 = dλ for some λ. Show that there is a diffeomorphism f1 : M → M such that ω0 = f ∗1 ω1. Solution: That ω0 − ω1 = dλ for some λ follows from Spivak’s Theorem 8.9. Hence ωt = ω0 − tdλ, so the condition is satisfied if d(X ty ωt) = dλ, which is satisfied in particular if X ty ωt = λ. Now the fact that M is orientable means that there is some nowhere zero n-form µ, and since ω0 and ω1 are also nowhere-zero n-forms, we have ω0 = fµ and 7