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D-Math Prof. Dr. DA Salamon Differential Geometry II FS 18 D-Math FS 18 Differential Geometry II Prof. Dr. D.A. Salamon March 27, 2018 Solution 6 m 1. A parallelizable manifold M is one where the tangent bundle TM is isomorphic to the trivial bundle M × R . In other words, M carries m vector fields that are linearly independent at every point of M. a) Prove that S3 is parallelizable. b) Prove that S7 is parallelizable. c) Prove that S2n is not parallelizable for all n ≥ 1. Hint: For a) and b), see S3,S7 as the unit quaternions resp. unit octonions. It is a hard theorem that S0, S1, S3 and S7 are the only parallelizable spheres. Solution: a) We can see S3 as the set of unit quaternions: Define 4 H = {z = x + yi + vj + wk :(x, y, v, w) ∈ R } with relations ij = k , i2 = j2 = k2 = −1 and 3 2 2 2 2 S = {z = x + yi + vj + wk ∈ H : zz¯ = x + y + v + w = 1}. Here the conjugate is defined as z¯ = x − yi − vj − wk). We can define three linearly independent vector fields X, Y, Z on S3 by X(z) = iz, Y (z) = jz and Z(z) = kz. This statement is implied by proving that {z, X(z),Y (z),Z(z)} is an orthonormal basis of R4 for every z ∈ S3. In the standard basis of R4 we get x −y −v −w y x w −v M(z) := (z, X(z),Y (z),Z(z)) = . v −w x y w v −y x > Then we see that M M = 1 which indeed means that M ∈ O(4). (As M(pN ) = 1, this even gives SO(4).) Note that the quaternion multiplication gives S3 the structure of a Lie group and every Lie group is parallelizable. (Simply take a basis of left invariant vector fields. This is exactly what we did by taking X, Y, Z.) b) We can see S7 as the set of unit octonions: The octonians are defined as O = {z = z1 + z2l : z1, z2 ∈ H} with relations l2 = −1, il = −li, jl = −lj and kl = −lk. Then 7 2 2 S = {z = z1 + z2l ∈ O : zz = (z1 + z2l)(¯z1 − z2l) = |z1| + |z2| = 1}. 7 (Note that z2l = lz¯2.) We can define seven linearly independent vector fields on S by X1(z) = iz, X2(z) = jz, X3(z) = kz, X4(z) = lz, X5(z) = ilz, X6(z) = jlz, and X7(z) = klz. 1 D-Math FS 18 Differential Geometry II Prof. Dr. D.A. Salamon March 27, 2018 This statement is implied by proving that {z, X1(z).X2(z),X3(z),X4(z),X5(z),X6(z),X7(z)} is an orthonormal basis for every z ∈ S7. In the standard basis of R8 we get M(z) := (z, X1(z),X2(z),X3(z),X4(z),X5(z),X6(z),X7(z)) x1 −y1 −v1 −w1 −x2 −y2 −v2 −w2 y1 x1 w1 −v1 −y2 x2 w2 −v2 v1 −w1 x1 y1 −v2 −w2 x2 y2 w v −y x −w v −y x 1 1 1 1 2 2 2 2 = . x2 y2 v2 w2 x1 −y1 −v1 −w1 y −x w −v y x −w v 2 2 2 2 1 1 1 1 v2 −w2 −x2 y2 v1 w1 x1 −y1 w2 v2 −y2 −x2 w1 −v1 y1 x1 > Then we see that M M = 1 which indeed means that M ∈ O(8). (As M(pN ) = 1, this even gives SO(8).) Note that S7 with octonian multiplication is not a Lie group since associativity fails. c) We already know χ(S2n) = 2. So by Poincaré–Hopf, there are no non-vanishing vector fields. Therefore S2n cannot be parallelizable. 2. Let M be a compact manifold. a) Prove that there exist a finite collection X1,X2,...,Xk ∈ Vect(M) such that {X1(p),X2(p), ··· ,Xk(p)} spans TpM at every p ∈ M. b) Prove that the number k from a) can be chosen such that k ≤ 2m, where m = dim(M). k Hint: For b): Start with any large number of vector fields X1,...,Xk and consider the map F : TpM → R whose j-th coordinate is given by Fj (p, v) := hv, Xj (p)i. What does Sard’s theorem tell us when k > 2m and how can you use this to construct k − 1 vector fields Y1,...,Yk−1 which still span TpM at every p ∈ M? Solution: a) We show the following local statement first: For every p ∈ M exists an open neigh- (p) (p) borhood p ∈ Up and vector field X1 ,...,Xm ∈ Vect(M) such that (p) (p) X1 (q),...Xm (q) is a basis for TqM for every q ∈ Up. Indeed, let ϕp : Vp → Ωp be a chart defined on an open neighborhood p ∈ Vp ⊂ M m with image Ωp ⊂ R . Choose a bounded open subset ϕp(p) ∈ Wp ⊂ Ωp with W p ⊂ Ωp and a cut-off function ρp : Wp → [0, 1] with ρ(x) = 1 for all x ∈ Wp and such that supp(ρp) ⊂ Ωp is compact (i.e. ρp vanishes near the boundary). Define m Yj :Ωp → R ,Yj(x) := ρ(x)ej m (p) where e1, . , em denotes the standard basis of R . We define the vector fields Xj as the pullback of Yj: ( (p) 0 q∈ / Vp Xj (q) := −1 dϕ(q) Yj(ϕ(q)) q ∈ Vp 2 D-Math FS 18 Differential Geometry II Prof. Dr. D.A. Salamon March 27, 2018 These are clearly smooth vector fields on M and yield a basis of TqM for all q ∈ −1 Up := ϕ (Wp). By compactness of M, there exists finitely many points p1, . , pN such that Up1 ,...,UpN (pi) cover M. Then {Xj } with j = 1, . , m and i = 1,...,N is a finite collection of vector fields spanning TpM at every p ∈ M. b) Let X1,...,Xk ∈ Vect(M) be any finite number of vector fields which span TpM at every point p ∈ M. Choose a Riemannian metric (or embedding M ⊂ Rn) to define the map k F : TM → R ,F (p, v) = (hv, X1(p)i, hv, X2(p)i,..., hv, Xk(p)i) . Suppose k > 2m. Then it follows from Sard’s theorem that F is not surjective, since dim(TM) = 2m. Choose ξ ∈ Rk\Image(F ). Note that F is homogeneous in the sense that F (p, tv) = tF (p, v) for all (p, v) ∈ TM and t ∈ R. This yields the stronger assertion Rξ ∩ Image(F ) = {0}. We may assume (after relabelling if necessary) that ξk 6= 0 and define Yj(p) := ξkXj(p) − ξiXk(p), for j = 1, . , k − 1. We claim that Y1,...,Yk−1 span TpM at every point p ∈ M. If not, then there exists 0 6= v ∈ TpM such that 0 = hYj(p), vi = ξkhXi(p), vi − ξihXk(p), vi = ξkFi(p, v) − ξiFk(p, v) and hence ! ξ1 ξ2 ξk−1 F (p, v) = Fk(p, v), Fk(p, v),..., Fk(p, v),Fk(p, v) ξk ξk ξk Fk(p, v) = (ξ1, . , ξk) . ξk Since Rξ ∩ Image(F ) = {0}, it follows F (p, v) = 0 and hence v = 0. This contradic- tions our assumption and shows that Y1,...,Yk−1 indeed span TpM. Repeating the procedure described above, we can decrease the number of vector fields successively until k ≤ 2m. For k = 2m the argument ceases to work, since F might be surjective. Nevertheless, in many examples one can do better, e.g. for any closed hypersurface M m ⊂ Rm+1 the coordinate vector fields project to k = m + 1 vector fields spanning TpM at every point. Can you find an example where k = 2m vector fields are needed? 3. Let M be a connected manifold without boundary and let p, q ∈ M. Let K ⊂ M be a compact set containing p, q ∈ K◦ in its non empty, connected interior. Then there exists an isotopy ψt : M → M for t ∈ [0, 1] such that ψ0 = id, ψ1(p) = q and [ supp({ψt}t∈[0,1]) := {x ∈ M : ψt(x) 6= x} ⊂ K. 0≤t≤1 We call this an isotopy with compact support in K. Hint: This is a slight generalization of the homogenisation lemma from the lecture. Have a careful look at the proof and check how it can be adopted to encompass this case. 3 D-Math FS 18 Differential Geometry II Prof. Dr. D.A. Salamon March 27, 2018 Solution: Look at the proof of the homogenisation lemma given in class or in Milnor’s book at page 22. In this proof, we constructed isotopies Ft supported on a small ball in a chart around any point which sends the origin to any desired point in this ball. Thus say that two interior points x, y ∈ K◦ are ‘isotopic’ in K if there is an isotopy supported in K which sends x to y. By the previous construction in the chart, we see that all points near y ∈ K◦ are ‘isotopic’ to y in K as long as we choose the small ball in the interior of K. "In other words, each isotopy class of points in the interior of K is an open set and the interior of K is partitioned into disjoint open isotopy classes. But the interior of K is connected; hence there can only be one isotopy class." (quote from p:23 of Milnor’s book.) 4. Let M m be a compact, connected manifold without boundary where m ≥ 2. m a) Let ϕ : U → R be a chart defined on U ⊂ M with B1(0) ⊂ ϕ(U). Prove that there exists X ∈ Vect(M) which does −1 −1 not vanish outside of ϕ (B1(0)) and has only isolated non-degenerate zeros in ϕ (B1(0)).
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