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)).

m b) Let Y : B1(0) → R be a vector ﬁeld with isolated non-degenerate zeros. Suppose that Y (x) 6= 0 for x ∈ ∂B1(0) and P ι(Y, p) = 0. Then there exists a nowhere vanishing vector ﬁeld Z : B (0) → m which agrees with x∈Y −1(0) 1 R Y near the boundary.

c) Prove that if χ(M) = 0 then there is a vector field X ∈ Vect(M) with no zeroes. Hint: For a): Start with any vector field X with isolated non-degenerate zeros. Use the homogeneity lemma of the previous exercise to move all the zeros into a ball inside a coordinate chart of X. m−1 For b): The Gauss map ∂B1(0) → S defined by x 7→ Y (x)/kY (x)k has degree zero under the given assumptions. Hence it is homotopic to a constant map by the Hopf degree theorem. For c): Combine a) and b) and make sure that all your modifications produce a smooth vector field. Solution:

−1 a) Let X be a vector ﬁeld with isolated non-degenerate zeroes X (0) = {p1, . . . , pk} m and ϕ : U → R be a chart deﬁned on U ⊂ M with B1(0) ⊂ ϕ(U). Let ` = −1 −1 −1 #X (0)∩ϕ (B1(0)). If ` = k, we are done. Else pick q ∈ ϕ (B1(0)) with X(q) 6= −1 0, i ∈ {1, . . . , k} such that pi ∈/ ϕ (B1(0)) and choose small open neighbourhoods pj ∈ Vj for j 6= i such that

K := M\(V1 ∪ · · · ∪ Vj−1 ∪ Vj+1 ∪ · · · ∪ Vk) ◦ is a compact connected subset with pj, q ∈ K . By the previous exercise, there exists an isotoply {ψt} supported in K with ψ1(pj) = q. Then Y = ψ∗X is a vector field on −1 −1 −1 M such that Y (0) = {p1, . . . , pk, q}\{pi}. Hence, #Y (0) ∩ ϕ (B1(0)) = ` + 1. −1 We can repeat this argument until all the zeroes are in ϕ (B1(0)). b) Since Y has only isolated zeros, it follows from the Hopf lemma that the signed count P m−1 x∈Y −1(0) ι(Y, p) agrees with the degree of the Gauss map f : ∂B1(0) = S → Sm−1 defined by x 7→ Y (x)/kY (x)k (see Exercise 2 on Exercise Sheet 4). Hence this has degree zero and it is homotopic to a constant map by the Hopf degree theorem. m−1 m−1 Say Ht : S → S such that H1 = f and H0 ≡ pN . Take > 0 such that −1 B1(0) \ B1−(0) ∩ Y (0) = ∅. Choose a cut off function ρ : R → R such that ρ = 0 near 1 − , ρ = 1 near 1 and ρ is strictly increasing. Define Z ∈ Vect(B1(0)) by  Z(p) = pN for p ∈ B1−(0),

Z(rσ) = (ρ(r)|Y (σ)| + (1 − ρ(r)))Hρ(r)(σ) for rσ ∈ B1(0) \ B1−(0), where r ∈ [1 − , 1), σ ∈ Sm−1. This is a smooth vector ﬁeld which is non vanishing and agrees with Y near the boundary.

4 D-Math FS 18 Diﬀerential Geometry II Prof. Dr. D.A. Salamon March 27, 2018

m c) Let ϕ : U → R be a chart defined on U ⊂ M with B1(0) ⊂ ϕ(U). Take as in a) −1 X ∈ Vect(M) which does not vanish outside of ϕ (B1(0)) and has only isolated non-degenerate zeros in ϕ−1(B (0)). Then Y = ϕ X| fulfills the assumptions in 1 ∗ B1(0) b) due to Poincaré–Hopf theorem and χ(M) = 0. Therefore we can find a nowhere m vanishing vector field Z : B1(0) → R which agrees with Y near the boundary. Now define the vector X˜ ∈ Vect(M) by  X(p), for p∈ / ϕ−1(B (0)), X˜(p) = 1 −1 −1 (ϕ )∗Z, for p ∈ ϕ (B1(0)).

By construction, X˜ is smooth and has no zeroes.

5. Let f : M → R be a Morse function as in Exercise 2, Sheet 3. The Morse Lemma asserts that for every p ∈ Crit(f), m there is a chart ϕp : Up ⊂ M → R with ϕp(p) = 0 such that

−1 2 2 2 2 f ◦ ϕp (x) = f(p) − x1 − · · · − xk + xk+1 + ··· + xm.

Here k := µ(f, p) is determined as the dimension of the negative eigenspace of the Hessian Hpf and it is called the Morse index of f at p.

Let g be a Riemannian metric on M. The gradient vector ﬁeld X = ∇gf of a smooth function f : M → R is deﬁned by

g(X(p), pˆ) = df(p)ˆp, for all p ∈ M and pˆ ∈ TpM.

Prove the following properties for the gradient ﬁeld of a Morse function f.

a) For every metric g the gradiant X = ∇gf is a vector ﬁeld with isolated, non-degenerate zeroes.

b) For every p ∈ Crit(f) and any two Riemannian metrics g1, g2 the indices ι(∇g0 f, p) = ι(∇g1 f, p) agree. µ(f,p) c) For every p ∈ Crit(f) it holds ι(∇gf, p) = (−1) . Solution:

a) The linearization of the gradient vector ﬁeld ∇f at a critical point p is the linear map ∇(∇f)(p): TpM → TpM, pˆ 7→ ∇pˆ(∇f)(p).

We verify that h∇pˆ(∇f), ·i = Hpf(ˆp, ·) is related to the Hessian deﬁned in Exercise 2 of Exercise Sheet 3. Indeed,

Hpf(X,Y ) = LX df(p)Y − df(p)∇X Y

= LX h∇f(p),Y (p)i − h∇f(p), ∇X Y (p)i

= h∇X ∇f(p),Y (p)i.

Here we used that the Levi-Civita connection ∇ is Riemannian. Hence Hpf is nondegenerate if and only if ∇(∇f)(p) is an isomorphism. This shows that every zero is non-degenerate. Non-degeneracy implies isolated. This follows from the following more general statement. Surjectivity of ∇(∇f)(p) at every point p ∈ M with ∇f(p) = 0 is equivalent to the statement that ∇f : M → TM intersects the zero section M × {0} ⊂ TM transversely. Therefore the intersection {p ∈ M | ∇f(p) = 0} is a 0-dimensional manifold and hence a discrete set of points.

5 D-Math FS 18 Diﬀerential Geometry II Prof. Dr. D.A. Salamon March 27, 2018

b) The set of Riemannian metrics is convex. Hence gt := (1−t)g0 +tg1 is a Riemannian metric for 0 ≤ t ≤ 1. This defines a homotopy of the gradient vector fields defined

by ∇gt f. The zeros for each of these vector ﬁeld are the same, since ∇gt f(p) = 0 if and only if df(p) = 0 and the later statement does not depend on the metric. It follows from the homotopy invariance of the degree, that all the zeros have the same

index for ∇g0 f and ∇g1 f.

c) Deﬁne h : Rm → R by 2 2 2 2 h(x) = f(p) − x1 − · · · − xk + xk+1 + ··· + xm. Its gradient for the standard metric is the vector ﬁeld

m m ∇h(x): R → R : ∇h(x) := 2(−x1,..., −xk, xk+1, . . . , xm). Using the Morse Lemma, the index of ∇f at a critical point p of index k := µ(f, p) agrees with the index of ∇h at the origin. This is deﬁned as degree of the Gauss map Sm−1 → Sm−1 given by

(x1, . . . , xm) 7→ (−x1,..., −xk, xk+1, . . . , xm). This is a composition of k reﬂections and hence has degree (−1)k. Hence, it follows µ(f,p) ι(∇gf, p) = (−1) for any metric g by part b).

n 6. a) Show that fn : RP → R deﬁned by

Pn jx2 j=1 j fn ([x0 : x1 : ... : xn]) := Pn x2 j=0 j

n is a Morse function on RP . Determine all the critical points of fn and their Morse index. n n b) Show that χ(RP ) = 0 when n is odd and χ(RP ) = 1 when n is even. n c) Show that gn : CP → R deﬁned by

Pn j|z |2 j=1 j gn ([z0 : z1 : ... : zn]) := . Pn |z |2 j=0 j

n is a Morse function on CP . Determine all the critical points of gn and their Morse index. n d) Show that χ(CP ) = n + 1 for n ≥ 1. Hint: Use the gradient vector ﬁelds of the Morse functions to compute the Euler characteristic in part b) and d) (see Exercise 5 above) Solution:

n n a) We write fn in the standard charts for RP . Deﬁne hi : R → R by

hi(x1, . . . , xn) := fn([x1 : ··· : xi−1 : 1 : xi : ··· : xn]) Pi 2 P 2 j=1(j − 1)xj + i + j=i+1 jxj = Pn 2 1 + j=1 xj

The partial derivatives ∂khi is for k ≤ i given by

Pn 2 hPi 2 P 2i 1 + j=1 xj 2(k − 1)xj − 2xk j=1(j − 1)xj + i + j=i+1 jxj ∂khi(x) = Pn 2 2 (1 + j=1 xj )

6 D-Math FS 18 Diﬀerential Geometry II Prof. Dr. D.A. Salamon March 27, 2018

and for k > i by

Pn 2 hPi 2 P 2i 1 + j=1 xj 2kxj − 2xk j=1(j − 1)xj + i + j=i+1 jxj ∂khi(x) = Pn 2 2 . (1 + j=1 xj )

It follows directly that 0 is a critical value of hi. Inductively, one can show that ∂nhi(x) = 0 implies xn = 0 and then ∂n−1hi(x) = 0 yields xn−1 = 0, etc. Therefore 0 is the only critical value of hi. When calculating the second partial derivatives, we see that most terms vanish at the origin, and get as ﬁnal expression ( 2(k − 1)δk` − 2iδk` for k ≤ i, ∂`∂khi(0) = 2kδk` − 2iδk` for k > i.

Hence H0(hi) = diag (−2i, −2i + 2,..., −2, 2,..., 2n − 2j). This shows that the origin is a nondegenerate critical point for hi with Morse index j − 1. n For the function fn : RP → R it follows that the critical points are

p0 = [1 : 0 : ... : 0], p1 = [0 : 1 : 0 : ... : 0], ..., pn = [0 : ... : 0 : 1].

They are all non-degenerate and have Morse index µ(fn, pi) = i. b) It follows from Exercise 5 and part a) that

m m ( X X 1 for n even, χ( P n) = (−1)µ(fn,pi) = (−1)i = R 0 for n odd. i=0 i=0 Note that this is half the Euler characteristic of Sn, which hints at another proof using the Poincaré–Hopf theorem.

c) We proceed similarly to part a). To make things more explicit, we write zj = xj + iyj and use real notation. In local coordinates the function gn is represented by ˜ 2n n hi : R → R with Pi (j − 1)(x2 + y2) + i + P j(x2 + y2) ˜ j=1 j j j=i+1 j j hi(x, y) = Pn 2 2 1 + j=1 xj + yj q q ˜ 2 2 2 2 Since hi(x, y) = hi x1 + y1,..., xn + yn , it follows from the chain rule and part ˜ a) that the only critical point of hi is the origin. The second partial derivatives are: ( 2(k − 1) − 2i for k ≤ i, ∂ h (0) = ∂ h (0) = xkxk i ykyk i 2k − 2i for k > i. and all other mixed partial derivative vanish. Hence the Hessian is again a diagonal matrix with precisely 2i negative entries. n It follows that gn : CP → R has the critical points

p0 = [1 : 0 : ... : 0], p1 = [0 : 1 : 0 : ... : 0], ..., pn = [0 : ... : 0 : 1].

They are all non-degenerate and have Morse index µ(fn, pi) = 2i.

7 D-Math FS 18 Diﬀerential Geometry II Prof. Dr. D.A. Salamon March 27, 2018

d) It follows from Exercise 5 and part c) that

m m n X µ(g ,p ) X 2i χ(CP ) = (−1) n i = (−1) = n + 1. i=0 i=0