4402 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam


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1 4402 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP n x α 0}, we defined for ech d Z line bundle E d by the trnsition functions g αβ : U α U β R, g αβ = ( xβ Let L RP 1 be ny line bundle. It is fct (esy to prove) tht L cn lwys be trivilized over ech U α, nd therefore L is given by trnsition functions h αβ : U α U β R, with respect to the cnonicl covering of RP 1. x α ) d. () Show tht if the imge of ll h αβ lies in R >0, then L is isomorphic to the trivil bundle. (b) Show tht if the imge of ll h αβ lies in R <0, then L is isomorphic to the trivil bundle. (c) Show tht if neither of the two previous situtions occur, then L E 1 is isomorphic to the trivil bundle. (d) Conclude tht the set of isomorphism clsses of line bundles over RP 1 is isomorphic to Z/2Z. Solution. () By ssumption, h 01 : U 0 U 1 R is lwys strictly positive. Note tht U 0 U 1 = R. By shrinking the chrts U 0, U 1 little bit, i.e. replcing them by U 0 = φ 1 0 (( A, A)), U 1 = φ 1 1 (( A, A)), for some A > 0 lrge, we my ssume tht C 1 < h 01 < C holds on U 0 U 1 = [ A, 1/A] [1/A, A], for some C > 1. Let then s 0 : U 0 R be identiclly equl to 1 nd s 1 : U 1 R be smooth function which is equl to 1 h 01 on φ 1 1 (( A, 1/A)) nd on φ 1 1 ((1/A, A)), nd it interpoltes smoothly between these in the region φ 1 1 ([ 1/A, 1/A]), stying lwys strictly positive. This is possible becuse C 1 < h 01 < C on U 0 U 1. By construction we hve s 0 = h 01 s 1, 1
2 on U 0 U 1, nd so we get globl smooth section s of L which is never vnishing. This mens tht L is isomorphic to the trivil bundle. (b) The sme proof s in () pplies to this cse, with strictly positive replced by strictly negtive. (c) Let [x 0 : x 1 ] be the usul homogeneous coordintes on RP 1, nd cover it by the two chrts U α = {x α 0} = R, α = 0, 1. The chrt mps re φ 0 ([x 0 : x 1 ]) = x 1 /x 0 nd φ 1 ([x 0 : x 1 ]) = x 0 /x 1, nd the trnsition functions re φ 0 φ 1 1 (x) = 1/x = φ 1 φ 1 0 (x), which re functions from R = U 0 U 1 to itself. The trnsition functions of the bundle E 1 re g αβ = x β /x α, α, β = 0, 1. In prticulr, in these chrts, we cn write g 01 (φ 1 1 (x))(v) = g 01([x : 1])(v) = v x. This mens tht g 01 hs imge in both components of R, with g 01 > 0 on one component of U 0 U 1 = R nd g 01 < 0 in the other component. On the other hnd, by ssumption, the trnsition functions h αβ of L re > 0 on one component of U 0 U 1 = R nd < 0 on the other. Since the trnsition functions of L E 1 re h αβ g αβ, we see tht these hve imge lying ll in R >0 or R <0. Thnks to prts () nd (b), we conclude tht L E 1 is isomorphic to the trivil bundle. (d) If L is ny line bundle over RP 1, then one of options (),(b) or (c) pplies to L. Therefore, either L or L E 1 is isomorphic to trivil. We lso know from previous homework tht E 1 is not isomorphic to trivil. This mens tht the set of isomorphism clsses of line bundles over RP 1 is isomorphic to Z/2Z, with the trivil bundle corresponding to 0 Z/2Z nd E 1 corresponding to 1. 2) Let M be smooth mnifold (s usul connected). Show tht the ction of the diffeomorphism group of M on M is trnsitive, i.e. show tht given ny two points p, q M there is diffeomorphism F : M M with F (p) = q. Solution. Let us first do the cse when M = B, the unit bll in R n. Given two points p, q B (we my ssume tht p q, otherwise F = Id works), let m = mx( p, q ) < 1. Let χ be smooth cutoff function supported in B, with χ 1 on B m (0). Let X 0 be the constnt vector field on R n, equl to the vector q p everywhere, nd let X = χx 0. This is smooth vector field on R n, compctly supported in B, nd its flow Θ stisfies Θ(1, p) = q. 2
3 Therefore F = θ 1 : B B is the desired diffeomorphism. Note tht F is the identity outside B. For generl M, ssuming gin w.l.o.g. tht p q, we choose smooth curve γ : [0, 1] M with γ(0) = p, γ(1) = q. Cover γ([0, 1]) by finitely mny coordinte chrts (U 1, φ 1 ),..., (U N, φ n ), with φ j (U j ) ll equl to the unit bll B in R n. Since {γ 1 (U j )} cover [0, 1], we cn choose points p 0 = p, p 1,..., p N = q with p j = γ(t j ) U j U j+1 for j = 1,..., N 1. We pply the construction in the first prt of the solution, nd produce F 1,..., F N diffeomorphisms of M (with F j equl to the identity outside U j ), with F j (p j 1 ) = p j. Then F = F N F N 1 F 1 is the desired diffeomorphism. 3) Let M be smooth nmnifold nd X 1,..., X k be smooth vector fields on M which re linerly independent t ech point. Let D = Spn(X 1,..., X k ) be the smooth kdistribution on M spnned by them. () Given ny p M show tht there is p U open, nd there exist smooth 1forms ω 1,..., ω n k on U such tht smooth vector field X on U lies in D iff ω j (X) = 0 for ll j = 1,..., n k. Fix now such n open set U nd 1forms ω j. (b) Show tht D is integrble over U iff dω j (X, Y ) = 0 for ll j = 1,..., n k nd for ll X, Y D over U. (c) Show tht D is integrble over U iff there exist smooth 1forms α ij, 1 i, j n k, on U such tht holds on U for ll 1 i n k. n k dω i = α ij ω j, Solution. () Given ny point p there is n open set p U nd smooth vector fields X k+1,..., X n on U such tht {X 1,..., X n } form bsis of T q M for ll q U. Let {α 1,..., α n } be the dul bsis of smooth 1forms on U, so by definition α i (X j ) = δ ij. Every vector field X on U cn be written s X = i X i, for some smooth functions i, nd i = ω i (X). This shows tht X lies in D iff α j (X) = 0 for ll k + 1 j n. Then ω j = α j+k, 1 j n k do the job. (b) For ech 1 j n k nd ny X, Y vector fields on U we hve j=1 dω j (X, Y ) = X(ω j (Y )) Y (ω j (X)) ω j ([X, Y ]). 3
4 Thnks to prt (), if X, Y D we see tht dω j (X, Y ) = ω j ([X, Y ]), nd so [X, Y ] D iff dω j (X, Y ) = 0 for ll 1 j n k. (c) Going bck to the nottion from (), we hve tht {α i α j } 1 i<j n form bsis of 2forms on U, nd so for ech 1 l n we cn write dα l = c l ijα i α j, 1 i<j n for some smooth functions c l ij on U. Thus, for ny vector fields X, Y on U we hve dα l (X, Y ) = c l ij(α i (X)α j (Y ) α i (Y )α j (X)). 1 i<j n Thnks to prt (b) we know tht D is integrble iff for ll k + 1 l n nd ll 1 < b k we hve which is equivlent to dα l = n n i=1 j=k+1 0 = dα l (X, X b ) = c l b, c l ijα i α j = n j=k+1 ( n ) c l ijα i α j. If we cll α ij = n p=1 ci+k p(j+k) α p, 1 i, j n k, nd go bck to the nottion with ω j, this reds exctly i=1 for ll 1 i n k. n k dω i = α ij ω j, j=1 4) Consider the vector field X = x y y x, in R 3 with the stndrd coordintes (x, y, z). Find locl coordintes (u, v, w) in neighborhood of (x, y, z) = (1, 0, 0), such tht in these coordintes we hve X = u. 4
5 Solution. Since X is never zero in neighborhood of (1, 0, 0), we know (s shown in clss) tht such coordintes exist. To find them, compute the flow Θ : R R 3 R 3 s in problem 6 (it is lmost the sme vector field), nd get Θ(t, x, y, z) = (x cos t y sin t, y cos t + x sin t, z). If we restrict the flow to the hypersurfce y = 0 in R 3, we obtin the mp Ψ(t, x, z) = (x cos t, x sin t, z). From the definition of the flow, we see tht ( ) dψ = X. t Thnks to this clcultion we define (x, y, z) = F (u, v, w) = (v cos u, v sin u, w). Note tht F (0, 1, 0) = (1, 0, 0). The jcobin of F is v sin u cos u 0 df = v cos u sin u 0, which t (u, v, w) = (0, 1, 0) is invertible, nd hence (u, v, w) re new coordintes ner (x, y, z) = (1, 0, 0). We hve ( ) df = ( v sin u) u x + (v cos u) y = y x + x y = X, which is wht we wnt. 5) Consider the vector fields X = x y y x, Y = y z, in R 3 with the stndrd coordintes (x, y, z). Cn you find locl coordintes (u, v, w) in neighborhood of (x, y, z) = (1, 0, 0), such tht in these coordintes we hve X = u, Y = v? 5
6 Solution. Since [ u, ] = 0, v necessry condition is tht [X, Y ] vnishes identiclly in neighborhood of (1, 0, 0). We hve ( [X, Y ] = x y y ) ( y ) ( y ) ( x x z z y y ) = x x z. Since this vector field is not identiclly zero in neighborhood of (1, 0, 0), we conclude tht there re no such coordintes. 6) Determine explicitly the flow Θ of the vector field on R 2. X = y x x y, Solution. Let γ(t) = (f(t), g(t)) be smooth curve in R 2. The condition tht γ be flow line for X is γ (t) = X γ(t), i.e., f (t) = g(t), Substituting, we obtin f (t) = f(t), g (t) = f(t). g (t) = g(t). From elementry ODE theory we know tht the generl solutions of these re f(t) = cos t + b sin t, g(t) = c cos t + d sin t, for, b, c, d R. If we set γ(0) = (x 0, y 0 ), then we hve f(0) = = x 0, g(0) = c = y 0, nd nd so f (0) = g(0) = y 0, g (0) = f(0) = x 0, f(t) = x 0 cos t + y 0 sin t, g(t) = y 0 cos t x 0 sin t. 6
7 In prticulr we see tht X is complete, nd its flow Θ : R R 2 R 2 is given by Θ(t, x, y) = (x cos t + y sin t, y cos t x sin t). 7) Let D be the distribution on R 3 spnned by the vector fields X = ye x y z, Y = x. () Write down n integrl submnifold for D pssing through (1, 0, 0) (b) Is D integrble? (c) Cn you write down n integrl submnifold for D pssing through (1, 1, 1)? Solution. () The xzplne S = {y = 0} is n integrl submnifold for D pssing through (1, 0, 0). Indeed, t ny point (x, 0, z) S we hve X = z, Y = x, which re clerly tngent to S nd spn its tngent plne. (b) We compute ( [X, Y ] = ye x y ) ( ) ( ) ( ye x z x x y ) = ye x z y. Since this cnnot be written s liner combintion of the form fx + gy, with f, g smooth functions, we conclude tht D is not involutive, nd so not integrble either. (c) There is no such submnifold, from the sme proof tht we sw in clss tht D integrble implies D involutive. Briefly, if we hve n integrl submnifold S then X, Y would be tngent to S nd therefore so would be [X, Y ]. This would imply tht for ny p S, [X, Y ] p = X p + by p, for some rel numbers, b (which depend on p). But t the point p = (1, 1, 1) we hve X p = e y z, Y p = x, [X, Y ] p = e y, nd there re no such, b. 8) Let (M, g) be Riemnnin mnifold nd γ : [, b] M smooth curve, with length Define the energy of γ to be L(γ) = E(γ) = b b 7 γ (t) g dt. γ (t) 2 gdt.
8 Show tht we hve L(γ) 2 (b )E(γ), nd tht equlity holds iff γ (t) g is constnt. Solution. We hve L(γ) = b ( b γ (t) g dt ) 1 ( 2 b ) 1 2 γ (t) 2 gdt 1dt = (b ) 1 2 E(γ) 1 2, using the Hölder inequlity. Squring this, we get the desired inequlity. Equlity holds in the Hölder inequlity iff γ (t) g = c 1 holds for some constnt c nd for.e. t [, b]. Since γ (t) g is continuous, we see tht equlity holds iff γ (t) g is constnt. 9) Let f : R 2 R be smooth function, nd consider its grph M = {(x, y, z) R 3 z = f(x, y)}. () Show tht M is submnifold of R 3. (b) Let ι : M R 3 be the inclusion. Using (x, y) s globl coordintes on M, write down the metric g = ι g Eucl induced from the Eucliden metric on R 3. (c) Write down the sme metric explicitly when f(x, y) = x 2 + y 2. Solution. () Consider the smooth function F : R 3 R given by F (x, y, z) = f(x, y) z. Then M = F 1 (0). We hve ( df = x, ) y, 1, which is never zero, hence 0 R is regulr vlue of F, nd so M is submnifold of R 3. (b) We hve tht ι(x, y) = (x, y, f(x, y)), nd Then g Eucl = dx dx + dy dy + dz dz. ι g Eucl = dx dx + dy dy + df df ( ( ) ) ( 2 = 1 + dx dx x 8 ( ) ) 2 dy dy + y x dx dy + y x dy dx. y
9 (c) Since we obtin x = 2x, y = 2y, ι g Eucl = ( 1 + 4x 2) dx dx + ( 1 + 4y 2) dy dy + 4xydx dy + 4xydy dx. 10) Let M be smooth mnifold nd be connection on T M, with torsion T (X, Y ) = X Y Y X [X, Y ]. Show tht for ny smooth functions f, g on M nd ny X, Y T (M), we hve T (fx, gy ) = fgt (X, Y ). Solution. Just compute T (fx, gy ) = fx (gy ) gy (fx) [fx, gy ] = f X (gy ) g Y (fx) fg[x, Y ] fx(g)y + gy (f)x = fg X Y + fx(g)y fg Y X gy (f)x fg[x, Y ] fx(g)y + gy (f)x = fg( X Y Y X [X, Y ]) = fgt (X, Y ). 11) Let G be connected compct Lie group, nd F : G R be Lie group homomorphism. Show tht F is constnt equl to 1. Solution. We hve tht F (G) is compct nd connected subgroup of R. Since F (e) = 1, it follows tht F (g) > 0 for ll g G. If F is not constnt, there exist g G nd t R >0, t 1, such tht F (g) = t. Up to replcing t by t 1, we my ssume tht t < 1. Then F (g n ) = t n for ll n N, nd t n 0 s n. Since G is compct, subsequence of {g n } converges to g G, nd by continuity we get F (g ) = 0, bsurd. 9
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