Topology III Home Assignment 8 Subhadip Chowdhury

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1 Topology Home Assignment 8 Subhadip Chowdhury Problem 1 Suppose {X 1,..., X n } is an orthonormal basis of T x M for some point x M. Then by definition, the volume form Ω is defined to be the n form such that Ω x (X 1,..., X n ) = 1 Now suppose x 1,..., x n is a local coordinate system on M and let i = A j i X j. Then Ω x ( 1,..., n ) = det(a)ω x (X 1,..., X n ) where A is the n n matrix with (i, j) th entry as A i j. Now g ij = g( i, j ) = A ki X k, A ljx l = A k i A k j = (A T A) ij det(g) = (det(a)) 2 Thus we have Ω( 1,..., n ) = det g = Ω = det gdx 1 dx n Problem 2 Recall that for any vector field X X(M), we have, X = ι X ω where ω is the volume form. Hence div(x) = d ι X ω = ( 1) n 1 dι X ω dι X ω = div(x) = div(x)ω Thus for given in our problem we can write div(x)dvol = div(x)ω = dι X ω By Stokes theorem we then have div(x)dvol = ι X ω 1

2 So it is enough to prove that, on we have, ι X ω = X, ν darea Suppose { 1,..., n 1 } form an positively oriented o.n.b. of T x ( ) for x. Then {ν, 1,..., n 1 } forms an oriented o.n.b. for T x (). Suppose the corresponding dual basis of T x is {dx 0, dx 1,..., dx n } so that dx 0 (ν) = 1 and dx i ( j ) = δ ij. Hence Hence Problem 3 ι X ω = ι X (dx 0 dx 1 dx n 1 ) = dx 0 (X)(dx 1 dx n 1 ) ( = X, ν dx 0 (ν) + i = X, ν darea div(x)dvol = X, i dx 0 ( i ) X, ν darea (i) Take any p M. Observe that since α p dα p is nonzero, dα p ξp ) darea is non-degenerate and hence ker(dα p ) ξ p = {0}. Since dim T p M = 3 = dim ξ p + 1, it follows that the dimension of ker(dα p ) is zero or one. Suppose ker(dα p ) is trivial. Note that by Cartan s formula, dα(x, Y ) = Xα(Y ) Y α(x) α([x, Y ]) = dα(y, X) Hence dα p is a symplectic form on T p M. But then dim T p M must be even which it is not. Hence ker(dα) ξ = T M i.e. ker(dα) p is a one-dimensional subspace of T p M for all p M. t follows that ker(dα) is a line bundle over M. Thus we can find a unique vector field X X(M) which is a continuous section of ker(dα) and α(x) 0 everywhere. So we can normalize X at each point to make sure that α(x) = 1. Thus by construction, X is the unique Reeb vector field associated to α. (ii) Suppose ϕ t (p) is the flow generated by X so that X(ϕ t (p)) = d dt ϕ t(p) and hence Now and Thus for all t. ϕ t d dt ϕ t (α dα) = L X (α dα) = d(ι X (α dα)) + ι X (d(α dα)) d(ι X (α dα)) = d(ι X α dα) d(α ι X dα) = d(1 dα) d(α 0) = 0 ι X (d(α dα)) = ι X (dα dα) ι X (α d 2 α) = 0 d dt ϕ t (α dα) = ϕ t (0) = 0 = ϕ t (α dα) = α dα 2

3 (iii) We are given that X is the Reeb flow associated to the contact form α. ker(dα) ξ = T M, any vector field V X(M) can be uniquely written as Note that since V = fx + Y where Y is a smooth section ξ and f is a smooth function. We define the Riemannian metric g =.,. on T p M as fx p + Y p, hx p + Z p = fh + g(y p, Z p ) where g is the Euclidean metric derived from the identification of T p M = R 3. Then X, X = 1. Clearly g is symmetric, nondegenarate and positive definite. Suppose X = X i i. Then X = X i g ij dx j where g ij = i, j. Thus 1 = X, X = X i X j g ij = X (X) Hence using α(x) = 1 and comparing coefficients we get α = X. To show that the flow lines generated by X are geodesic, we need to show that γ γ = X X = 0 where γ (t) = X(γ(t)). Recall that we proved in first part that ker(dα) ξ = T M. Hence we can find vector fields Y, Z X(M) such that X p, Y p, Z p forms an orthonormal basis of T p M and Y p, Z p form a basis of ξ p. Then it is enough to show that X X, X = X X, Y = X X, Z = 0 ˆ 2 X X, X = X X, X = X(1) = 0. ˆ By Koszul formula, to see that X X, Y = 0; it is enough to prove that [X, Y ], X since other terms are trivially zero. Thus we need to show that [X, Y ] p ξ p or equivalently α p ([X, Y ] p ) = 0. Now by Cartan s formula, since ι X dα = 0. ˆ Similarly, X X, Z = 0. α([x, Y ]) = Xα(Y ) Y α(x) dα(x, Y ) = 0 Y (1) dα(x, Y ) = 0 = 0, 3

4 (iv) Let X be a Beltrami field. Suppose α = X. Then we have ( dx ) = fx = ( dα) = fx = fα = dα = ( 1) 2 f( α) = f( α) = α dα = f(α α) = f α 2 ω 0 since f does not vanish anywhere. Thus α is a contact form. Also we have ι X dα = ι X (f( α)) = fι X ( X ) = fι X ι X ω = 0 since (ι X ) 2 = 0. Thus rescaling X to make sure α(x) = 1, we can get a Reeb flow of α. Problem 4 Suppose dim M = n. Let N = {1, 2,..., n} Suppose at a point q, and let {dz 1,..., dz n } form the orthonormal basis of Tq M such that {,..., } forms an o.n.b of R n. Then we have smooth functions z 1 z n A j i such that dx i = A j i dzj We will use the notation A J for, J N, = J = p, to mean j A J := A j 1 i1 A j 2 i2... A jp Note that in above definition, we require J = p but it is not necessary that j 1 < j 2 <... < j p. However we have i 1 < i 2 <... <. Now suppose dz J = dz j 1 dz jp where j 1 < j 2 <... < j p. Then, dx = J A σ(j) σ S p dz J = J B J dz J (,let) Let α = f dx = J f J dz J Then by definition, f A σ(j) σ = g J = (C J )f where C is inverse of B. Note that B and C are both matrices of dimension ( ( n p) n p). Now we claim that α = ( g )dz Let ω = ω dz for some fixed = {i 1 < i 2 <... < } N. Choose j 1,..., j n p such that { z i 1,..., z ip, z j 1,..., 4 z j n p }

5 forms a positive orthonormal basis of R n. Henceforth we will always consider Now ω dω = dz j k dz z j k l {1,..., p}, k {1,..., n p} dω = ( 1) p+k 1 ω dz j 1 z dz j j k dz j n p k d dω = + d dω = + ( 1) p+k 1 p 2 ω ( z j k ) 2 dzj k dz j 1 dz j k dz j n p ( 1) p+k 1 ( 1) p+p(n p) p ( 1) pd+l Recall that δ = ( 1) n(p+1)+1 d. Hence δdω = ( 1) 2 ω ( z j k ) 2 dz + Analogously, we can derive, dδω = p ( 1) 2 ω ( z i l ) 2 dz + 2 ω dz j l dz j 1 z dz j j k k z i dz j n p l 2 ω ( z j k ) 2 dzi 1 dz ip 2 ω dz j k dz i 1 z dz i j l k z i dz l p ( 1) 1+l 2 ω dz j k dz i 1 z dz i j l k z i dz l p ( 1) l 2 ω dz j k dz i 1 z dz i i l l z j dz k ω = dδω + δdω = ( 1) Thus by linearity of d and δ it follows that n 2 ω ( z m ) 2 dz = ( ω )dz α = J ( g J )dz J 5

6 We now consider the change of basis {dz j } {dx i }. Let B J α = J = = = n n n 2 g J ( z m ) 2 dzj 2 (B Jf ) ( x m ) 2 (CJ )dx = σ f A σ(j) 2 (f ) ( x m ) 2 dx + lower order terms ( f )dx + lower order terms Then we have Problem 5 Since α is closed, it is orthogonal to δω p+1. Suppose the 1 forms {β 1,..., β k } form a R basis of H 1 = H 1 (M, R). Then by Hodge decomposition theorem, we can find real numbers r i, 1 i k and g Ω 0 (M) i.e. a real valued function g such that α = k r i β i + dg We will think of S 1 as R/Z. Recall that since S 1 = K(Z, 1), an Eilenberg-MacLane space, we have a bijection between the first singular cohomology group of M with Z coefficients and the homotopy class of maps from M to S 1. Thus we have a bijection [M, S 1 ] H 1 (M, Z) given by f f ([S 1 ]) for any f Hom(M, S 1 ). Here [S 1 ] is the generator of H 1 (S 1 ) = Z. Since H 1 (M, R) = H 1 (M, Z) R, {β i 1} form a basis of H 1 (M, R). Hence we can find differentiable functions f i : M S 1 such that [f i (dt)] = [β i ] H 1 (M, R) where dt is the obvious 1 form on S 1 derived from dx on R. Thus we can write for some real valued functions g i. Thus we have α = f i (dt) = β i dg i k r i fi (dt) + k dg i + dg Now we claim that wlog we may take the last two summands to be zero. Suppose we have a real valued map ϕ from a Riemannian manifold M ϕ : M R Let dx be the canonical 1 form on R. Then for any point p M we have an induced map f : T ϕ(p) R T p M such that for v T p M, f (dx)(v) = dx(f (v)) 6

7 But we can identify T ϕ(p) R with R so that for a choice of local coordinate dx ϕ(p) = d. Thus f (dx)(v) = f (v) = df(v) f (dx) = df Thus we can write f1 (dt) + dh = (f 1 + Π h) (dt) where Π : R S 1 is the natural projection and h is a real valued function. Thus we assume α = k r i fi (dt) Note that since r i R, given ɛ > 0, we can find rational numbers n i Q, n m i Z, m Z such that α 1 k n i fi (dt) m ɛ where the norm is derived from some Riemannian metric on M.Note that since n i Z, ni f i (dt) H 1 (M, R) and since α 0 everywhere, taking ɛ 0 we get that n i fi (dt) 0 everywhere. Consider the function F : M S 1 defined by F = k n if i. Then we have shown that F (dt) = α. Hence F (dt) is nonvanishing and F is a submersion. Since M is closed, hence compact, F is a fiber map. Thus M is a fiber bundle over S 1. 7

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