[Exercise 1.18] Let M Be a Topological Manifold. Then Any Two Smooth Atlases for M Determine the Same Smooth Structure If and Only If Their Union Is a Smooth Atlas

Chapter 1. Smooth Manifolds

Theorem 1. [Exercise 1.18] Let M be a topological manifold. Then any two smooth atlases for M determine the same smooth structure if and only if their union is a smooth atlas.

Proof. Suppose A1 and A2 are two smooth atlases for M that determine the same smooth structure A. Then A1, A2 ⊆ A, so A1 ∪ A2 must be a smooth atlas since every chart in A1 is compatible with every chart in A2. Conversely, if A1 ∪ A2 is a smooth atlas then the smooth structures determined by A1 and A2 both contain A1 ∪ A2. But there is exactly one smooth structure containing A1 ∪ A2, so A1 and A2 determine the same smooth structure. Theorem 2. [Exercise 1.44] Let M be a smooth n-manifold with boundary and let U be an open subset of M. (1) U is a topological n-manifold with boundary, and the atlas consisting of all smooth charts (V, ϕ) for M such that V ⊆ U deﬁnes a smooth structure on U. With this topology and smooth structure, U is called an open submanifold with boundary. (2) If U ⊆ Int M, then U is actually a smooth manifold (without boundary); in this case we call it an open submanifold of M. (3) Int M is an open submanifold of M (without boundary).

Proof. Parts (1) and (2) are obvious. Part (3) follows from (2) and the fact that Int M is an open subset of M. Theorem 3. [Problem 1-6] Let M be a nonempty topological manifold of dimension n ≥ 1. If M has a smooth structure, then it has uncountably many distinct ones.

s−1 Proof. We use the fact that for any s > 0, Fs(x) = |x| x defines a homeomorphism from Bn to itself, which is a diffeomorphism if and only if s = 1. Let A be a smooth structure for M and choose some coordinate map ϕ : U → Bn centered at some p ∈ U. Let A0 be the smooth atlas obtained by replacing every coordinate map ψ : V → R in 0 A with ψ : V ∩ (M \{p}) → R, except when ψ = ϕ. For any s > 0, let As be the 0 smooth atlas obtained from A by replacing ϕ with Fs ◦ϕ. Since Fs is a diffeomorphism if and only if s = 1, the smooth structures determined by As and At are the same if and only if s = t. This shows that there are uncountably many distinct smooth structures on M. Theorem 4. [Problem 1-8] An angle function on a subset U ⊆ S1 is a continuous function θ : U → R such that eiθ(z) = z for all z ∈ U. There exists an angle function θ on an open subset U ⊆ S1 if and only if U 6= S1. For any such angle function, (U, θ) is a smooth coordinate chart for S1 with its standard smooth structure. 1 2

Proof. The ﬁrst part follows from the lifting criterion (Proposition A.78). By rotating R2 appropriately, we may assume that N = (0, 1) ∈/ U. Let σ : S1 \{N} → R be the stereographic projection given by σ(x1, x2) = x1/(1 − x2). We can compute cos α (σ ◦ θ−1)(α) = , 1 − sin α which is a diﬀeomorphism on θ(U). Theorem 5. [Problem 1-10] Let k and n be integers satisfying 0 < k < n, and let n P,Q ⊆ R be the linear subspaces spanned by (e1, . . . , ek) and (ek+1, . . . , en), respec- n tively, where ei is the ith standard basis vector for R . For any k-dimensional subspace S ⊆ Rn that has trivial intersection with Q, the coordinate representation ϕ(S) constructed in Example 1.36 is the unique (n − k) × k matrix B such that S is spanned by I the columns of the matrix k , where I denotes the k × k identity matrix. B k

Proof. Let B = {e1, . . . , ek}. The matrix of ϕ(S) represents the linear map (πQ|S) ◦ −1 −1 (πP |S) . Since πP |S is an isomorphism, the vectors (πP |S) (B) form a basis for S. But then −1 −1 (IdRn |S ◦ (πP |S) )(B) = ((πP |S + πQ|S) ◦ (πP |S) )(B) −1 = (IdP +πQ|S ◦ (πP |S) )(B) is a basis for S. This is precisely the result desired.

Theorem 6. [Problem 1-12] Suppose M1,...,Mk are smooth manifolds and N is a smooth manifold with boundary. Then M1 × · · · × Mk × N is a smooth manifold with boundary, and ∂(M1 × · · · × Mk × N) = M1 × · · · × Mk × ∂N.

Proof. Denote the dimensions of M1,...,Mk,N by n1, . . . , nk, d. Let p = (p1, . . . , pk, q) ∈ ni M1 × · · · × Mk × N, let ϕi : Mi → R be coordinate maps around pi for i = 1, . . . , k, d and let ψ : N → H be a coordinate map around q. Then ϕ1 × · · · × ϕk × ψ : M1 × · · · × n1+···+nk+d Mk × N → H is a coordinate map, and (ϕ1 × · · · × ϕk × ψ)(p1, . . . , pk, q) ∈ n +···+n +d d ∂H 1 k if and only if ψ(q) ∈ ∂H , which is true if and only if q ∈ ∂N.

Chapter 2. Smooth Maps

Theorem 7. [Exercise 2.1] Let M be a smooth manifold with or without boundary. Then pointwise multiplication turns C∞(M) into a commutative ring and a commutative and associative algebra over R.

Proof. The product of two smooth functions is also smooth. 3

Theorem 8. [Exercise 2.2] Let U be an open submanifold of Rn with its standard smooth manifold structure. Then a function f : U → Rk is smooth in the sense of smooth manifolds if and only if it is smooth in the sense of ordinary calculus.

n Proof. Obvious since the single chart IdRn covers R . Theorem 9. [Exercise 2.3] Let M be a smooth manifold with or without boundary, and suppose f : M → Rk is a smooth function. Then f ◦ ϕ−1 : ϕ(U) → Rk is smooth for every smooth chart (U, ϕ) for M.

Proof. If p ∈ U then there is a smooth chart (V, ψ) for M such that f ◦ ψ−1 is smooth and p ∈ V . But on ϕ(U ∩ V ) we have f ◦ ϕ−1 = f ◦ ψ−1 ◦ ψ ◦ ϕ−1, which is smooth since ψ ◦ ϕ−1 is smooth. This shows that f ◦ ϕ−1 is smooth in a −1 neighborhood of every point in ϕ(U), so f ◦ ϕ is smooth. Theorem 10. [Exercise 2.7(1)] Suppose M and N are smooth manifolds with or without boundary, and F : M → N is a map. Then F is smooth if and only if either of the following conditions is satisﬁed: (1) For every p ∈ M, there exist smooth charts (U, ϕ) containing p and (V, ψ) containing F (p) such that U ∩ F −1(V ) is open in M and the composite map ψ ◦ F ◦ ϕ−1 is smooth from ϕ (U ∩ F −1(V )) to ψ(V ). (2) F is continuous and there exist smooth atlases {(Uα, ϕα)} and {(Vβ, ψβ)} for M −1 and N, respectively, such that for each α and β, ψβ ◦ F ◦ ϕα is a smooth map −1 from ϕα (Uα ∩ F (Vβ)) to ψβ(Vβ).

Proof. (1) is obvious, as is (2) ⇒ (1). If (1) holds then F is continuous by Proposition 2.4, and we can choose the smooth atlases to be the smooth structures for M and N. −1 For every α and β the map ψβ ◦ F ◦ ϕα is smooth since for all p ∈ M we can ﬁnd smooth coordinate maps ϕ containing p and ψ containing F (p) such that −1 −1 −1 −1 ψβ ◦ F ◦ ϕα = ψβ ◦ ψ ◦ ψ ◦ F ◦ ϕ ◦ ϕ ◦ ϕα on a small neighborhood around ϕα(p); this map is smooth since the above charts are all smoothly compatible. Theorem 11. [Exercise 2.7(2)] Let M and N be smooth manifolds with or without boundary, and let F : M → N be a map.

(1) If every point p ∈ M has a neighborhood U such that the restriction F |U is smooth, then F is smooth. (2) Conversely, if F is smooth, then its restriction to every open subset is smooth. 4

Proof. (1) is obvious. For (2), let E be an open subset of M. Let p ∈ E and choose smooth charts (U, ϕ) containing p and (V, ψ) containing F (p) such that F (U) ⊆ V and −1 ψ ◦ F ◦ ϕ is smooth. By restricting ϕ to U ∩ E, we ﬁnd that F |E is also smooth. Theorem 12. [Exercise 2.9] Suppose F : M → N is a smooth map between smooth manifolds with or without boundary. Then the coordinate representation of F with respect to every pair of smooth charts for M and N is smooth.

Proof. Almost identical to Theorem 9. Theorem 13. [Exercise 2.11] Let M, N and P be smooth manifolds with or without boundary. (1) Every constant map c : M → N is smooth. (2) The identity map of M is smooth. (3) If U ⊆ M is an open submanifold with or without boundary, then the inclusion map U,→ M is smooth.

Proof. If M 6= ∅ then let y be the constant value that c assumes and let (V, ψ) be a smooth chart containing y. If x ∈ M and (U, ϕ) is any smooth chart containing x then ψ ◦ c ◦ ϕ−1 is constant and therefore smooth. This proves (1). Part (2) follows from the fact that any two coordinate maps for M are smoothly compatible. For (3), if p ∈ U −1 and (E, ϕ) is any smooth chart in U containing p then ϕ ◦ ϕ = Idϕ(E) is smooth. Theorem 14. [Exercise 2.16] (1) Every composition of diffeomorphisms is a diffeomorphism. (2) Every finite product of diffeomorphisms between smooth manifolds is a diffeomorphism. (3) Every diffeomorphism is a homeomorphism and an open map. (4) The restriction of a diffeomorphism to an open submanifold with or without boundary is a diffeomorphism onto its image. (5) “Diffeomorphic” is an equivalence relation on the class of all smooth manifolds with or without boundary.

Proof. If f : M → N and g : N → P are diffeomorphisms then f −1 ◦ g−1 is a smooth inverse to g ◦ f, which proves (1). If fi : Mi → Ni are diffeomorphisms for i = 1, . . . , n −1 −1 then f1 ×· · ·×fn is a smooth inverse to f1×· · ·×fn, which proves (2). Part (3) follows from the fact that smooth maps are continuous, and part (4) follows immediately from (3) and Theorem 11. Part (5) follows from (1). Theorem 15. [Exercise 2.19] Suppose M and N are smooth manifolds with boundary and F : M → N is a diffeomorphism. Then F (∂M) = ∂N, and F restricts to a diffeomorphism from Int M to Int N. 5

Proof. It suffices to show that F (∂M) ⊆ ∂N, for then F (Int M) = Int N by Theorem 1.46 and F |Int M is a diffeomorphism by Theorem 14. Let x ∈ ∂M and choose smooth charts (U, ϕ) containing x and (V, ψ) containing F (x) such that ϕ(x) ∈ ∂Hn, F (U) = V and ψ ◦ F ◦ ϕ−1 is a diffeomorphism. Then ϕ ◦ F −1 is a smooth coordinate map sending n F (x) to ϕ(x) ∈ ∂H , so F (x) ∈ ∂N by definition. Example 16. [Exercise 2.27] Give a counterexample to show that the conclusion of the extension lemma can be false if A is not closed. Take f(x) = 1/x defined on R \{0}. Example 17. [Problem 2-1] Define f : R → R by ( 1, x ≥ 0, f(x) = 0, x < 0.

Show that for every x ∈ R, there are smooth coordinate charts (U, ϕ) containing x and (V, ψ) containing f(x) such that ψ ◦ f ◦ ϕ−1 is smooth as a map from ϕ (U ∩ f −1(V )) to ψ(V ), but f is not smooth as deﬁned in this chapter. f is clearly smooth on R\{0}. If x = 0 then ϕ = Id(−1,1) is a coordinate map containing −1 x and ψ = Id(1/2,3/2) is a coordinate map containing f(x) = 1 such that ψ ◦ f ◦ ϕ is smooth on ϕ ((−1, 1) ∩ f −1 ((1/2, 3/2))) = ϕ ([0, 1)) = [0, 1). However, f is clearly not smooth since it is not continuous at 0.

Theorem 18. Let M1,...,Mk−1 be smooth manifolds without boundary and let Mk be a smooth manifold with boundary. Then each projection map πi : M1 × · · · × Mk → Mi is smooth.

Proof. Let ni be the dimension of Mi. Let x = (x1, . . . , xk) ∈ M1 × · · · × Mk. For each i = 1, . . . , k, choose a smooth chart (Ui, ϕi) containing xi. Write U = U1 × · · · × Uk and −1 ϕ = ϕ1 × · · · × ϕk. Then (U, ϕ) is a smooth chart containing x and ϕi ◦ πi ◦ ϕ = pi n1 nk ni ni is smooth, where pi is a projection from R × · · · × H to R (or H if i = k). This shows that πi is smooth.

Theorem 19. [Problem 2-2] Suppose M1,...,Mk and N are smooth manifolds with or without boundary, such that at most one of M1,...,Mk has nonempty boundary. For each i, let πi : M1 × · · · × Mk → Mi denote the projection onto the Mi factor. A map F : N → M1 × · · · × Mk is smooth if and only if each of the component maps Fi = πi ◦ F : N → Mi is smooth.

Proof. For simplicity we assume that Mk is a smooth manifold with boundary. From Proposition 2.10(d) it is clear that every Fi is smooth if F is smooth. Suppose that each Fi is smooth and let x ∈ N. For each i, choose smooth charts (Ui, ϕi) containing x −1 −1 and (Vi, ψi) containing Fi(x) such that ψi ◦ Fi ◦ ϕi = ψi ◦ πi ◦ F ◦ ϕi . Replace U1 with 6

−1 U = U1 ∩ · · · ∩ Uk and replace ϕ1 with ϕ1|U ; by Theorem 12, each map ψi ◦ πi ◦ F ◦ ϕ1 is smooth. Write V = V1 × · · · × Vk and ψ = ψ1 × · · · × ψk so that (V, ψ) is a smooth −1 −1 chart containing F (x). Since the ith component of ψ ◦ F ◦ ϕ1 is just ψi ◦ πi ◦ F ◦ ϕ1 , −1 the map ψ ◦ F ◦ ϕ1 is smooth. This shows that F is smooth. Theorem 20. [Problem 2-4] The inclusion map f : Bn ,→ Rn is smooth when Bn is regarded as a smooth manifold with boundary.

Proof. It is clear that f is smooth on Int Bn = Bn. For points on ∂Bn, the map provided by Problem 3-4 in [[1]] is a suitable coordinate map. Theorem 21. [Problem 2-7] Let M be a nonempty smooth n-manifold with or without boundary, and suppose n ≥ 1. Then the vector space C∞(M) is inﬁnite-dimensional.

∞ Proof. We ﬁrst show that any collection {fα} from C (M) with nonempty disjoint supports is linearly independent. Suppose there exist r1, . . . , rk ∈ R such that

r1fα1 + ··· + rkfαk = 0 where α1, . . . , αk are distinct. For each i, choose some xi ∈ supp fαi such that fαi (xi) 6= 0. Then

rifαi (xi) = (r1fα1 + ··· + rkfαk )(xi) = 0, so ri = 0 since fαi (xi) 6= 0. Choose some coordinate cube U ⊆ M and a homeomorphism ϕ : U → (0, 1)n. For each n−1 i = 1, 2,... , let Bi = (0, 1) × (1/(i + 1), 1/i) and let Ai be any nonempty closed −1 subset of Bi. By Proposition 2.25, there is a smooth bump function fi for ϕ (Ai) −1 −1 supported in ϕ (Bi). Furthermore, since the sets {ϕ (Bi)} are disjoint, the functions ∞ {fi} are linearly independent. This shows that C (M) is infinite-dimensional. Theorem 22. [Problem 2-8] Define F : Rn → RPn by F (x1, . . . , xn) = [x1, . . . , xn, 1]. Then F is a diffeomorphism onto a dense open subset of RPn. A similar statement holds for G : Cn → CPn defined by G(z1, . . . , zn) = [z1, . . . , zn, 1]. Proof. The inverse to F is given by F −1[y1, . . . , yn, yn+1] = (y1/yn+1, . . . , yn/yn+1); it is easy to check that the two maps are smooth since F −1 itself is a coordinate map for RPn. The image of F is dense in RPn since any point [x1, . . . , xn, 0] not in the image 1 n of F can be approximated by some sequence [x , . . . , x , ak] where ak → 0. Theorem 23. [Problem 2-9] Given a polynomial p in one variable with complex coef- 1 1 ficients, not identically zero, there is a unique smooth map pe : CP → CP that makes the following diagram commute, where CP1 is 1-dimensional complex projective space and G : C → CP1 is the map of Theorem 22: 7

G C CP1 p pe

1 C G CP

Proof. We can assume that p is not constant, for otherwise the result is clear. For any 1 z = [z1, z2] ∈ CP , deﬁne ( [p(z1/z2), 1], z2 6= 0, pe(z) = [1, 0], z2 = 0.

The map pe clearly satisfies the given diagram and is smooth on the image of G, so it remains to show that pe is smooth in a neighborhood of [1, 0]. Define a diffeomorphism H : C → CP1 by H(z) = [1, z]; by definition, (H(C),H−1) is a smooth chart for CP1. We can compute −1 −1 (H ◦ pe◦ H)(z) = (H ◦ pe)[1, z] ( H−1[p(1/z), 1], z 6= 0, = H−1[1, 0], z = 0, ( 1/p(1/z), z 6= 0, = 0, z = 0, and it is easy to check that this map is smooth in a neighborhood of 0 whenever p is non-constant (since 1/p(1/z) is a rational function in z). This shows that pe is smooth. Theorem 24. [Problem 2-10] For any topological space M, let C(M) denote the algebra of continuous functions f : M → R. Given a continuous map F : M → N, define F ∗ : C(N) → C(M) by F ∗(f) = f ◦ F .

(1) F ∗ is a linear map. (2) Suppose M and N are smooth manifolds. Then F : M → N is smooth if and only if F ∗(C∞(N)) ⊆ C∞(M). (3) Suppose F : M → N is a homeomorphism between smooth manifolds. Then it is a diﬀeomorphism if and only if F ∗ restricts to an isomorphism from C∞(N) to C∞(M). 8

Proof. (1) is obvious. For (2), if F is smooth and f ∈ C∞(N) then F ∗(f) = f ◦ F ∈ C∞(M) by Proposition 2.10(d). Conversely, suppose that F ∗(C∞(N)) ⊆ C∞(M). Let x ∈ M and let n be the dimension of N. Choose smooth charts (U, ϕ) containing x and (V, ψ) containing F (x) such that F (U) ⊆ V . For each i = 1, . . . , n the map ∗ n πi ◦ ψ ◦ F = F (πi ◦ ψ) is smooth, where πi : R → R is the projection onto the ith −1 −1 coordinate. By 9, πi ◦ ψ ◦ F ◦ ϕ is smooth for i = 1, . . . , n, so ψ ◦ F ◦ ϕ must be smooth. This shows that F is smooth. Part (3) follows directly from part (2): if F is a diffeomorphism then (F −1)∗ : C∞(M) → C∞(N) is clearly an inverse to F ∗, and ∗ −1 conversely if F |C∞(N) is an isomorphism then F and F are both smooth. Theorem 25. [Problem 2-11] Suppose V is a real vector space of dimension n ≥ 1. Define the projectivization of V , denoted by P(V ), to be the set of 1-dimensional linear subspaces of V , with the quotient topology induced by the map π : V \{0} → P(V ) that sends x to its span. Then P(V ) is a topological (n − 1)-manifold, and has a unique smooth structure with the property that for each basis (E1,...,En) for V , the map n−1 1 n i E : RP → P(V ) defined by E[v , . . . , v ] = [v Ei] is a diffeomorphism.

Proof. Obvious. Theorem 26. [Problem 2-14] Suppose A and B are disjoint closed subsets of a smooth manifold M. There exists f ∈ C∞(M) such that 0 ≤ f(x) ≤ 1 for all x ∈ M, f −1(0) = A, and f −1(1) = B.

−1 Proof. By Theorem 2.29, there are functions fA, fB : M → R such that fA ({0}) = A −1 and fB ({0}) = B; the function deﬁned by f (x) f(x) = A fA(x) + fB(x) satisﬁes the required properties.

Chapter 3. Tangent Vectors

Theorem 27. [Exercise 3.7] Let M, N and P be smooth manifolds with or without boundary, let F : M → N and G : N → P be smooth maps, and let p ∈ M.

(1) dFp : TpM → TF (p)N is linear. (2) d(G ◦ F )p = dGF (p) ◦ dFp : TpM → TG◦F (p)P .

(3) d(IdM )p = IdTpM : TpM → TpM. (4) If F is a diﬀeomorphism, then dFp : TpM → TF (p)N is an isomorphism, and −1 −1 (dFp) = d(F )F (p). 9

Proof. If v, w ∈ TpM and c ∈ R then

dFp(v + w)(f) = (v + w)(f ◦ F ) = v(f ◦ F ) + w(f ◦ F ) = dFp(v)(f) + dFp(w)(f) and

dFp(cv)(f) = (cv)(f ◦ F ) = cdFp(v)(f) for all f ∈ C∞(N), which proves (1). To prove (2), we have

d(G ◦ F )p(v)(f) = v(f ◦ G ◦ F ) = dFp(v)(f ◦ G) = dGF (p)(dFp(v))(f) ∞ for all f ∈ C (N) and v ∈ TpM, so d(G ◦ F )p = dGF (p) ◦ dFp. For (3),

d(IdM )p(v)(f) = v(f ◦ IdM ) = v(f), so d(IdM )p(v) = v for all v ∈ TpM and therefore d(IdM )p = IdTpM . Part (4) follows immediately, since

−1 −1 d(F )F (p) ◦ dFp = d(F ◦ F )p = d(IdM )p = IdTpM −1 and similarly dFp ◦ d(F )F (p) = IdTF (p)N . Example 28. [Exercise 3.17] Let (x, y) denote the standard coordinates on R2. Verify 2 that (x,e ye) are global smooth coordinates on R , where 3 xe = x, ye = y + x . Let p be the point (1, 0) ∈ R2 (in standard coordinates), and show that

∂ ∂ 6= , ∂x p ∂xe p even though the coordinate functions x and xe are identically equal. 3 3 Since an inverse to (x, y) 7→ (x, y + x ) is given by (x,e ye) 7→ (x,e ye− xe ), the coordinates are smooth and global. Now

∂ ∂ ∂ = + 3 , ∂x p ∂xe p ∂ye p

∂ ∂ = , ∂y p ∂ye p so ∂/∂x|p 6= ∂/∂xe|p. Theorem 29. Let X be a connected topological space and let ∼ be an equivalence relation on X. If every x ∈ X has a neighborhood U such that p ∼ q for every p, q ∈ U, then p ∼ q for every p, q ∈ X. 10

Proof. Let p ∈ X and let S = {q ∈ X : p ∼ q}. If q ∈ S then there is a neighborhood U of q such that q1 ∼ q2 for every q1, q2 ∈ U. In particular, for every r ∈ U we have p ∼ q and q ∼ r which implies that p ∼ r, and U ⊆ S. This shows that S is open. If q ∈ X \ S then there is a neighborhood U of q such that q1 ∼ q2 for every q1, q2 ∈ U. If p ∼ r for some r ∈ U then p ∼ q since q ∼ r, which contradicts the fact that q ∈ X \ S. Therefore U ⊆ X \ S, which shows that S is closed. Since X is connected, S = X. Corollary 30. Let X be a connected topological space and let f : X → Y be a continuous map. If every x ∈ X has a neighborhood U on which f is constant, then f is constant on X. Theorem 31. [Problem 3-1] Suppose M and N are smooth manifolds with or without boundary, and F : M → N is a smooth map. Then dFp : TpM → TF (p)N is the zero map for each p ∈ M if and only if F is constant on each component of M.

Proof. Suppose F is constant on each component of M and let p ∈ M. For all v ∈ TpM we have dFp(v)(f) = v(f ◦ F ) = 0 since f ◦ F is constant on a neighborhood of p (see Proposition 3.8), so dFp = 0. Conversely, suppose that dF = 0, let p ∈ M, and choose smooth charts (U, ϕ) containing p and (V, ψ) containing F (p) such that ϕ(U) is an open ball and F (U) ⊆ V . Let Fb = ψ ◦ F ◦ ϕ−1 be the coordinate representation of F . From equation (3.10), the derivative of Fb is zero since dFp = 0 and therefore Fb is constant on ϕ(U). This implies that F = ψ−1 ◦ Fb ◦ ϕ is constant on U. By Corollary 30, F is constant on each component of M.

Theorem 32. [Problem 3-2] Let M1,...,Mk be smooth manifolds, and for each j, let πj : M1 × · · · × Mk → Mj be the projection onto the Mj factor. For any point p = (p1, . . . , pk) ∈ M1 × · · · × Mk, the map

α : Tp(M1 × · · · × Mk) → Tp1 M1 ⊕ · · · ⊕ Tpk Mk deﬁned by

α(v) = (d(π1)p(v), . . . , d(πk)p(v)) is an isomorphism. The same is true if one of the spaces Mi is a smooth manifold with boundary.

Proof. It is clear that α is a linear map. For each j = 1, . . . , k, deﬁne ej : Mj → M1 × · · · × Mk by ej(x) = (p1, . . . , x, . . . , pk) where x is in the jth position. Each ej induces a linear map d(ej)pj : Tpj Mj → Tp(M1 × · · · × Mk), so we can deﬁne a linear map

β : Tp1 M1 ⊕ · · · ⊕ Tpk Mk → Tp(M1 × · · · × Mk)

(v1, . . . , vk) 7→ d(e1)p1 (v1) + ··· + d(ek)pk (vk). 11

The jth component of (α ◦ β)(v1, . . . , vk) is then given by k ! k X X d(πj)p d(ei)pi (vi) = d(πj ◦ ei)pi (vi) i=1 i=1

= d(πj ◦ ej)pj (vj)

= vj since πj ◦ ei is constant when i 6= j, and πj ◦ ej = IdMj . This shows that α is surjective. But

dim(Tp(M1 × · · · × Mk)) = dim(Tp1 M1 ⊕ · · · ⊕ Tpk Mk), so α is an isomorphism.

Theorem 33. If (Uj, ϕj) are smooth charts containing pj as deﬁned in Theorem 32, then ϕ = ϕ1 ×· · ·×ϕk is a smooth coordinate map containing p. Let nj be the dimension 1 nj of Mj, and write ϕj = (xj , . . . , xj ) for each j so that

1 n1 1 nk ϕ = ϕ1 × · · · × ϕk = (xe1,..., xe1 ,..., xek,..., xek ) i i where xej = xj ◦ πj. (1) The collection ( ) ∂ : 1 ≤ j ≤ k, 1 ≤ i ≤ nj ∂xi ej p

forms a basis for Tp(M1 × · · · × Mk). (2) Let v ∈ T (M × · · · × M ) and write v = P vji ∂/∂xi . Then the jth p 1 k i,j ej p

component of α(v) is P vji ∂/∂xi (the variable j is ﬁxed). i j pj

n 1 nj o −1 Proof. Since Bj = ∂/∂x , . . . , ∂/∂x is a basis for Tpj Mj, the set α (B1 ∪ j pj j pj · · · ∪ Bk) = β(B1 ∪ · · · ∪ Bk) is a basis for Tp(M1 × · · · × Mk). But ! ! ∂ ∂ β (f) = d(ej)pj (f) ∂xi ∂xi j pj j pj ! −1 ∂ = d(ej ◦ ϕ )ϕ (p ) (f) j j j ∂xi j ϕj (pj ) −1 ∂(f ◦ ej ◦ ϕj ) = i (ϕj(pj)) ∂xj ∂(f ◦ ϕ−1) = i (ϕ(p)) ∂xej 12 ! −1 ∂ = d(ϕ )ϕ(p) (f) ∂xi ej ϕ(p)

∂ = (f), ∂xi ej p which proves (1). For (2), we compute the jth component of α(v) as ! ! X sr ∂ X sr ∂ d(πj)p v (f) = v d(πj)p (f) ∂xr ∂xr r,s es p r,s es p −1 X ∂(f ◦ πj ◦ ϕ ) = vjr (ϕ(p)) ∂xr r ej −1 X ∂(f ◦ ϕj ) = vjr (ϕ (p )) ∂xr j j r j ! X ji ∂ = v (f). ∂xi i j pj Theorem 34. [Problem 3-3] If M and N are smooth manifolds, then T (M × N) is diﬀeomorphic to TM × TN.

Proof. Let m and n denote the dimensions of M and N respectively. For each (p, q) ∈ M × N, there is an isomorphism α(p,q) : T(p,q)(M × N) → TpM ⊕ TqN as in Theorem 32. Deﬁne a bijection F : T (M × N) → TM × TN by sending each tangent vector v ∈ T(p,q)(M × N) to α(p,q)(v). For any (p, q) ∈ M × N, choose smooth charts (U, ϕ) containing p and (V, ψ) containing q. Let π : T (M × N) → M × N be the natural −1 projection map. Then ρ = ϕ × ψ is a smooth coordinate map for M × N, ρe : π (U × 2(m+n) −1 −1 V ) → R is a smooth coordinate map for T (M×N), and ϕe×ψe : π (U)×π (V ) → R2(m+n) is a smooth coordinate map for TM × TN. By Theorem 33 the coordinate representation of F is F (x1, . . . , xm, y1, . . . , yn, v1, . . . , vm+n) = (x1, . . . , xm, v1, . . . , vm, y1, . . . , yn, vm+1, . . . , vm+n) −1 which is clearly smooth. Similarly, F is also smooth. Theorem 35. [Problem 3-4] T S1 is diﬀeomorphic to S1 × R.

Proof. We begin by showing that for each z ∈ S1 we can choose an associated tangent 1 vector vz ∈ TzS . Let θ : U → R be any angle function deﬁned on a neighborhood of 13

−1 0 0 z, and set vz = d/dx|z = (dθ)z d/dx|θ(z). If θ : U → R is any other angle function defined around z then θ − θ0 is constant on U ∩ U 0 and (θ0 ◦ θ−1)(t) = t + c for some 0 −1 0 −1 0 c ∈ R. Therefore (dθ )z d/dx|θ0(z) = (dθ )z d/dx |θ0(z), and vz is well-defined. 1 1 Define a map F : S × R → T S by sending (z, r) to rvz. If θ : U → R is an angle function defined around z then the coordinate representation of F is F (z, r) = (θ(z), r), 1 so F is smooth. An inverse to F is defined by sending v ∈ TzS to (z, r), where v = rvz (since T S1 is one-dimensional). The coordinate representation of F −1 is F −1(x, r) = −1 −1 (θ (x), r), so F is also smooth. Theorem 36. [Problem 3-5] Let S1 ⊆ R2 be the unit circle, and let K ⊆ R2 be the boundary of the square of side 2 centered at the origin: K = {(x, y) : max(|x| , |y|) = 1}. There is a homeomorphism F : R2 → R2 such that F (S1) = K, but there is no diffeomorphism with the same property.

Proof. The map defined by px2 + y2 F (x, y) = (x, y) max(|x| , |y|) is a homeomorphism such that F (S1) = K. Suppose there is a diffeomorphism F : R2 → R2 such that F (S1) = K. Let γ :(−1, 1) → S1 be given by s 7→ exp(2πi(s + c)), where c is chosen so that (F ◦ γ)(0) = (1, 1). We can assume without loss of generality that (F ◦γ)((−ε, 0)) lies on the right edge of K and that (F ◦γ)((0, ε)) lies on the top edge of K, for small ε. Then there are smooth functions γ1, γ2 such that (F ◦ γ)(t) = (1, γ1(t)) 0 0 for t ∈ (−ε, 0) and (F ◦ γ)(t) = (γ2(t), 1) for t ∈ (0, ε). Since (F ◦ γ) (t) = (0, γ1(t)) 0 0 0 0 and (F ◦ γ) (t) = (γ2(t), 0), taking t → 0 shows that (F ◦ γ) (0) = 0. But dF (γ (0)) 6= 0 0 since γ (0) 6= 0 and dF is an isomorphism, so this is a contradiction. Theorem 37. [Problem 3-7] Let M be a smooth manifold with or without boundary ∞ and p be a point of M. Let Cp (M) denote the algebra of germs of smooth real-valued ∞ functions at p, and let DpM denote the vector space of derivations of Cp (M). Define a map φ : DpM → TpM by (φv)f = v([f]p). Then φ is an isomorphism.

Proof. It is clear that φ is a homomorphism. If φv = 0 then v([f]p) = 0 for all ∞ f ∈ C (M), so v = 0 and therefore φ is injective. If v ∈ TpM then we can define 0 0 v ∈ DpM by setting v ([f]p) = v(f). This is well-defined due to Proposition 3.8. Then 0 φv = v, so φ is surjective. Theorem 38. [Problem 3-8] Let M be a smooth manifold with or without boundary and p ∈ M. Let VpM denote the set of equivalence classes of smooth curves starting 0 0 at p under the relation γ1 ∼ γ2 if (f ◦ γ1) (0) = (f ◦ γ2) (0) for every smooth real- valued function f defined in a neighborhood of p. The map ψ : VpM → TpM defined by ψ[γ] = γ0(0) is well-defined and bijective. 14

0 0 0 0 Proof. If γ1, γ2 ∈ [γ] then γ1(0)(f) = (f ◦ γ1) (0) = (f ◦ γ2) (0) = γ2(0)(f), which shows that ψ is well-deﬁned. The same argument shows that ψ is injective. Finally, Proposition 3.23 shows that ψ is surjective.

Chapter 4. Submersions, Immersions, and Embeddings

Theorem 39. [Exercise 4.4] (1) The composition of two smooth submersions is a smooth submersion. (2) The composition of two smooth immersions is a smooth immersion. (3) The composition of two maps of constant rank need not have constant rank.

Proof. Let F : M → N and G : N → P be smooth maps; parts (1) and (2) follow from 2 2 the fact that d(G ◦ F )x = dGF (x) ◦ dFx. For (3), define F : R → R and G : R → R by F (x) = (1, x),G(x, y) = x + y2 so that 0 DF (x) = ,DG(x, y) = 1 2y ,D(G ◦ F )(x) = 2x. 1 Theorem 40. [Exercise 4.7] (1) Every composition of local diffeomorphisms is a local diffeomorphism. (2) Every finite product of local diffeomorphisms between smooth manifolds is a local diffeomorphism. (3) Every local diffeomorphism is a local homeomorphism and an open map. (4) The restriction of a local diffeomorphism to an open submanifold with or without boundary is a local diffeomorphism. (5) Every diffeomorphism is a local diffeomorphism. (6) Every bijective local diffeomorphism is a diffeomorphism. (7) A map between smooth manifolds with or without boundary is a local diffeomorphism if and only if in a neighborhood of each point of its domain, it has a coordinate representation that is a local diffeomorphism.

Proof. Let F : M → N and G : N → P be local diffeomorphisms. If x ∈ M then there are neighborhoods U of x and V of F (x) such that F |U : U → V is a diffeomorphism, 0 0 and there are neighborhoods V of F (x) and W of (G ◦ F )(x) such that G|V 0 : V → W is a diffeomorphism. Then (G ◦ F )|F −1(V ∩V 0) is a diffeomorphism onto its image, which proves (1). Parts (2)-(5) are obvious. Suppose F : M → N is a bijective local diffeomorphism. If y ∈ N then there are neighborhoods U of F −1(y) and V of y such −1 that F |U : U → V is a diffeomorphism, so (F )|V is smooth. Theorem 11 then shows −1 that F is smooth, which proves (6). Part (7) is obvious. 15

Theorem 41. [Exercise 4.10] Suppose M, N, P are smooth manifolds with or without boundary, and F : M → N is a local diﬀeomorphism. (1) If G : P → M is continuous, then G is smooth if and only if F ◦ G is smooth. (2) If in addition F is surjective and G : N → P is any map, then G is smooth if and only if G ◦ F is smooth.

Proof. For (1), if F ◦ G is smooth and x ∈ P then there are neighborhoods U of G(x) and V of (F ◦ G)(x) such that F |U : U → V is a diﬀeomorphism, so G|G−1(U) = −1 (F |U ) ◦ (F ◦ G)|G−1(U) is smooth. By Theorem 11, G is smooth. For (2), if G ◦ F is smooth and x ∈ N then there is a point p ∈ M such that x = F (p), and there are neighborhoods U of p and V of x such that F |U : U → V is a diﬀeomorphism. Then −1 G|V = G ◦ F ◦ (F |U ) is smooth, so G is smooth by Theorem 11. Example 42. [Exercise 4.24] Give an example of a smooth embedding that is neither an open nor a closed map. Let X = [0, 1) and Y = [−1, 1], and let f : X → Y be the identity map on X. Then f is a smooth embedding, X is both open and closed, but f(X) is neither open nor closed. Example 43. [Exercise 4.27] Give an example of a smooth map that is a topological submersion but not a smooth submersion. Consider f(x) = x3 at x = 0.

Theorem 44. [Exercise 4.32] Suppose that M, N1, and N2 are smooth manifolds, and π1 : M → N1 and π2 : M → N2 are surjective submersions that are constant on each other’s ﬁbers. Then there exists a unique diﬀeomorphism F : N1 → N2 such that F ◦ π1 = π2.

Proof. Theorem 4.30 shows that there are unique smooth maps πe1 : N2 → N1 and πe2 : N1 → N2 such that π1 = πe1 ◦ π2 and π2 = πe2 ◦ π1. Then πe2 ◦ πe1 ◦ π2 = πe2 ◦ π1 = π2, so π ◦ π = Id by the uniqueness part of Theorem 4.30. Similarly, π ◦ π = Id . e2 e1 N2 e1 e2 N1 Theorem 45. [Exercise 4.33] (1) Every smooth covering map is a local diffeomorphism, a smooth submersion, an open map, and a quotient map. (2) An injective smooth covering map is a diffeomorphism. (3) A topological covering map is a smooth covering map if and only if it is a local diffeomorphism.

Proof. Let π : E → M be a smooth covering map. Let e ∈ E and x = π(e), let U be an evenly covered neighborhood of x, and let Ue be the component of π−1(U) containing 16 e. By definition π| : U → U is a diffeomorphism, so π is a local diffeomorphism and Ue e the other properties in (1) follow. If π is injective then π is a bijection. Every bijective local diffeomorphism is a diffeomorphism, proving (2). Part (3) is obvious. Theorem 46. [Exercise 4.37] Suppose π : E → M is a smooth covering map. Every local section of π is smooth.

Proof. Let τ : V → E be a local section of π where V is open in M. Let x ∈ V and let U be an evenly covered neighborhood of x contained in V . By Proposition 4.36, there is a unique smooth local section σ : U → E such that σ(x) = τ(x). Therefore τ|U = σ, and τ is smooth by Theorem 11.

Theorem 47. [Exercise 4.38] Suppose E1,...,Ek and M1,...,Mk are smooth manifolds (without boundary), and πi : Ei → Mi is a smooth covering map for each i = 1, . . . , k. Then π1 × · · · × πk : E1 × · · · × Ek → M1 × · · · × Mk is a smooth covering map.

Proof. This is true for topological covering maps. But any finite product of local diffeomorphisms is a local diffeomorphism, so the result follows from Theorem 45. Theorem 48. Suppose π : E → M is a smooth covering map. If U ⊆ M is evenly covered in the topological sense, then π maps each component of π−1(U) diffeomorphically onto U.

Proof. Let U be a component of π−1(U) and suppose π| : U → U is a homeomorphism e Ue e (or merely a bijection). Since π| is a local diﬀeomorphism and it is bijective, it is a Ue diﬀeomorphism. Theorem 49. [Exercise 4.42] Suppose M is a connected smooth n-manifold with boundary, and π : E → M is a topological covering map. Then E is a topological n-manifold with boundary such that ∂E = π−1(∂M), and it has a unique smooth structure such that π is a smooth covering map.

Proof. Identical to Proposition 4.40. Theorem 50. [Exercise 4.44] If M is a connected smooth manifold, there exists a simply connected smooth manifold Mf, called the universal covering manifold of M, and a smooth covering map π : Mf → M. The universal covering manifold is unique in the following sense: if Mf0 is any other simply connected smooth manifold that admits a smooth covering map π0 : Mf0 → M, then there exists a diﬀeomorphism Φ: Mf → Mf0 such that π0 ◦ Φ = π. 17

Proof. Every manifold has a universal covering space, so let π : Mf → M be a universal covering. Proposition 4.40 shows that Mf is a smooth manifold with a unique smooth structure such that π is a smooth covering map. If π0 : Mf0 → M is another universal smooth covering, then there is a homeomorphism Φ : Mf → Mf0 such that π0 ◦ Φ = π. Let e ∈ Mf, let x = π(e) and choose some e0 ∈ π−1({x}). Let U be an evenly covered neighborhood of x with respect to both π and π0, let Ue be the component of π−1(U) containing e, and let Ue0 be the component of (π0)−1(U) containing e0. Then Φ| = (π0| )−1 ◦ π| , so Φ| is a diffeomorphism. This shows that Φ is a bijective local Ue Ue0 Ue Ue diffeomorphism and therefore Φ is a diffeomorphism. Example 51. [Problem 4-1] Use the inclusion map ι : Hn ,→ Rn to show that Theorem 4.5 does not extend to the case in which M is a manifold with boundary.

Lemma 3.11 shows that dι0 is invertible. If Theorem 4.5 holds, then there are neigh- n n borhoods U ⊆ H of 0 and a neighborhood V ⊆ R of 0 such that ι|U → V is a diﬀeomorphism. Since ι is a bijection, we must have U = V . This is impossible, because U cannot be open in Rn. Theorem 52. [Problem 4-2] Suppose M is a smooth manifold (without boundary), N is a smooth manifold with boundary, and F : M → N is smooth. If p ∈ M is a point such that dFp is nonsingular, then F (p) ∈ Int N.

Proof. Let n be the dimension of M, which is the same as the dimension of N. Suppose F (p) ∈ ∂N. Choose smooth charts (U, ϕ) centered at p and (V, ψ) centered at F (p) n −1 such that ψ(V ) ⊆ H , F (U) ⊆ V and ψ(F (p)) = 0. Let Fe = ψ ◦ F ◦ ϕ . Then dFe0 is invertible, so the inverse function theorem shows that there are neighborhoods Ue0 of 0 and V of 0 open in n such that F | : U → V is a diﬀeomorphism. But V is a e0 R e Ue0 e0 e0 e0 n n neighborhood of 0 open in R and contained in H , which is impossible. Example 53. [Problem 4-4] Let γ : R → T2 be the curve of Example 4.20. The image set γ(R) is dense in T2.

2 2πis1 2πis2 Let (z1, z2) ∈ T and choose s1, s2 ∈ R so that e = z1 and e = z2. Then 2πis1 2πiαs1 2πiαn γ(s1 + n) = (e , e e ), so it remains to show that 2πiαn S = e : n ∈ Z is dense in S1 whenever α is irrational. Let ε > 0. By Lemma 4.21, there exist integers n, m such that |nα − m| < ε. Let θ = nα − m so that e2πiθ = e2πiαn ∈ S and θ 6= 0 since α is irrational. Any integer power of e2πiθ is also a member of S, and since ε was arbitrary, we can approximate any point on S1 by taking powers of e2πiθ. 18

3 2 2 2 Example 54. We identify S with the set (z, w) ∈ C : |z| + |w| = 1 . Let p : C → S2 \{N} be the inverse of the stereographic projection given by 1 p(x + yi) = (2x, 2y, x2 + y2 − 1), x2 + y2 + 1 or 1 p(z) = (z +z, ¯ −i(z − z), |z|2 − 1). |z|2 + 1 Deﬁne q : S3 \{S1 × {0}} → C by (z, w) = z/w. Then we have a smooth map p ◦ q : S3 \{S1 × {0}} → S2 \{N} given by 1 z z¯ z z¯ z 2 (p ◦ q)(z, w) = + , −i − , − 1 |z/w|2 + 1 w w¯ w w¯ w = (zw¯ +zw, ¯ −i(zw¯ − zw¯ ), zz¯ − ww¯). It is easy to see that we can extend p ◦ q so that it maps S3 into S2 by using the formula above. Theorem 55. [Problem 4-5] Let CPn be n-dimensional complex projective space. (1) The quotient map π : Cn+1 \{0} → CPn is a surjective smooth submersion. (2) CP1 is diﬀeomorphic to S2.

n+1 Proof. Let (z1, . . . , zn+1) ∈ C \{0} and assume without loss of generality that zn+1 6= 0. We have a coordinate map ϕ given by w1 wn [w1, . . . , wn, wn+1] 7→ ,..., . wn+1 wn+1 Then  −1 −2  wn+1 0 ··· 0 −w1wn+1 0 w−1 ··· 0 −w w−2 0  n+1 2 n+1 (ϕ ◦ π) (w1, . . . , wn+1) =  . . . . .  ,  ......  −1 −2 0 0 ··· wn+1 −wnwn+1 which clearly has full rank when wk = zk for k = 1, . . . , n + 1 (since the first n columns form an invertible matrix). This proves (1). The map π : C2 \{0} → CP1 is a surjective smooth submersion. Define a map ψ : C2 \{0} → S2 by (z1, z2) ψ(z1, z2) = (p ◦ q) , |(z1, z2)| where p ◦ q is the map in 54. By Theorem 4.30, ψ descends to a unique smooth map 1 2 ψe : CP → S satisfying ψe ◦ π = ψ. This map is a diffeomorphism. Theorem 56. [Problem 4-6] Let M be a nonempty smooth compact manifold. There is no smooth submersion F : M → Rk for any k > 0. 19

Proof. By Proposition 4.28, F (M) is open in Rk. Also, F (M) is compact because M is compact. Therefore F (M) is open, closed and bounded. But Rk is connected and M k is nonempty, so F (M) = R . This contradicts the boundedness of F (M). Theorem 57. [Problem 4-7] Suppose M and N are smooth manifolds, and π : M → N is a surjective smooth submersion. There is no other smooth manifold structure on N that satisfies the conclusion of Theorem 4.29; in other words, assuming that Ne represents the same set as N with a possibly different topology and smooth structure, and that for every smooth manifold P with or without boundary, a map F : Ne → P is smooth if and only if F ◦ π is smooth, then IdN is a diffeomorphism between N and Ne.

Proof. Write ι for the identity map from N to N, and ι for the identity map from N e Ne N to N. Then ι is smooth if and only if ι ◦ π = π is smooth and ι is smooth if and e N N Ne only if ι ◦ π = π is smooth. Both of these conditions hold, and since ι and ι are Ne N Ne inverses, Id = ι = ι is a diﬀeomorphism. N N Ne Theorem 58. [Problem 4-8] Let π : R2 → R be deﬁned by π(x, y) = xy. Then π is surjective and smooth, and for each smooth manifold P , a map F : R → P is smooth if and only if F ◦ π is smooth; but π is not a smooth submersion (cf. Theorem 4.29).

Proof. It is clear that F ◦ π is smooth if F is smooth. If F ◦ π is smooth then the map x 7→ (F ◦ π)(x, 1) is smooth and identical to F , so F is smooth. However, we can compute Dπ(x, y) = y x , which does not have full rank when (x, y) = (0, 0). Theorem 59. [Problem 4-9] Let M be a connected smooth manifold and let π : E → M be a topological covering map. There is only one smooth structure on E such that π is a smooth covering map (cf. Proposition 4.40).

Proof. Let M be the smooth structure constructed in Proposition 4.40. Suppose there is another smooth structure M0 for which π is a smooth covering map. Let p ∈ E and let 0 (U, ϕe) be a smooth chart in M containing p. Let V be an evenly covered neighborhood of ϕ(p) and let V be the component of π−1(V ) containing p so that π| : V → V is e e Ve e a diﬀeomorphism (with respect to M0). By shrinking V and Ve if necessary, we may assume that there is a smooth chart (V, ψ) containing ϕ(p). Then ψ ◦ π| is a smooth Ve coordinate map in M, and ϕ ◦ (ψ ◦ π| )−1 = ϕ ◦ (π| )−1 ◦ ψ−1 e Ve e Ve is smooth. This shows that ϕ and ψ ◦ π| are smoothly compatible, and since p was Ve 0 e arbitrary, M = M . 20

Example 60. [Problem 4-10] Show that the map q : Sn → RPn deﬁned in Example 2.13(f) is a smooth covering map.

Since Sn is compact, it is clear that q is proper. For each i = 1, . . . , n + 1, let + n Si = {(x1, . . . , xn+1) ∈ S : xi > 0} , − n Si = {(x1, . . . , xn+1) ∈ S : xi < 0} .

Then q| + and q| − are diﬀeomorphisms for i = 1, . . . , n + 1, which shows that q is a Si Si local diﬀeomorphism. By Proposition 4.46, q is a smooth covering map.

Example 61. [Problem 4-13] Deﬁne a map F : S2 → R4 by F (x, y, z) = (x2 − y2, xy, xz, yz). Using the smooth covering map of Example 2.13(f) and Problem 4-10, show that F descends to a smooth embedding of RP2 into R4. Let q : S2 → RP2 be the map of Example 60. Since F (−x, −y, −z) = F (x, y, z), the map F descends to a unique smooth map Fe : RP2 → R4. It is easy to check that Fe−1 is smooth and given by ruv ruw rvw Fe−1(t, u, v, w) = q , , . w v u Note that we take uv/w = 0 if w = 0, since w = yz = 0 implies that either u = 0 or v = 0. This is true for the other coordinates. Therefore Fe is a homeomorphism onto its image. It remains to show that F is a smooth immersion. Let

2 2 2 ϕ : RP \ [x, y, z] ∈ RP : z 6= 0 → R x y [x, y, z] 7→ , z z be coordinate map. Then

(Fe ◦ ϕ−1)(x, y) = Fe[x, y, 1] = (x2 − y2, xy, x, y) so that 2x −2y −1  y x  D(Fe ◦ ϕ )(x, y) =   ,  1 0  0 1 which clearly has full rank. We can perform similar computations for the other coordinate maps, showing that dFe[x,y,z] is always surjective. 21

Chapter 5. Submanifolds

Example 62. [Exercise 5.10] Show that spherical coordinates (Example C.38) form a slice chart for S2 in R3 on any open subset where they are deﬁned. Spherical coordinates are given by (ρ, ϕ, θ) 7→ (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ). For the image to be a subset of S2, we must have ρ = 1. Theorem 63. [Exercise 5.24] Suppose M is a smooth manifold with or without boundary, S ⊆ M is an immersed k-submanifold, ι : S,→ M is the inclusion map, and U is an open subset of Rk. A map X : U → M is a smooth local parametrization of S if and only if there is a smooth coordinate chart (V, ϕ) for S such that X = ι ◦ ϕ−1. Therefore, every point of S is in the image of some local parametrization.

Proof. If X is a smooth local parametrization of S then X−1 : X(U) → U is a diﬀeo- morphism, so (X(U),X−1) is a smooth chart such that X = ι ◦ (X−1)−1. Conversely, suppose that (V, ϕ) is a smooth chart such that X = ι ◦ ϕ−1. Then X(ϕ(V )) = V is open in S and X is a diﬀeomorphism onto V , which shows that X is a smooth local parametrization. Theorem 64. [Exercise 5.36] Suppose M is a smooth manifold with or without boundary, S ⊆ M is an immersed or embedded submanifold, and p ∈ S. A vector v ∈ TpM is in TpS if and only if there is a smooth curve γ : J → M whose image is contained in S, and which is also smooth as a map into S, such that 0 ∈ J, γ(0) = p, and γ0(0) = v.

Proof. Let ι : S,→ M be the inclusion map. If v ∈ TpS then by Proposition 3.23 there 0 is a smooth curve γ0 : J → S such that 0 ∈ J, γ0(0) = p and γ0(0) = v. Therefore γ = ι ◦ γ0 is the desired smooth curve. Conversely, if there is a smooth curve γ with the speciﬁed properties then γ0 : J → S given by t 7→ γ(t) is smooth. Therefore 0 0 0 γ0(0) ∈ TpS and v = γ (0) = dιp(γ0(0)), so v is in TpS (as a subspace of TpM). Theorem 65. [Exercise 5.40] Suppose S ⊆ M is a level set of a smooth map Φ: M → N with constant rank. Then TpS = ker dΦp for each p ∈ S.

Proof. Follows from Theorem 5.12 and Proposition 5.38. Theorem 66. [Exercise 5.42] Suppose M is a smooth n-dimensional manifold with boundary, p ∈ ∂M, and (xi) are any smooth boundary coordinates deﬁned on a neighborhood of p. The inward-pointing vectors in TpM are precisely those with positive xn-component, the outward-pointing ones are those with negative xn-component, and n the ones tangent to ∂M are those with zero x -component. Thus, TpM is the disjoint union of Tp∂M, the set of inward-pointing vectors, and the set of outward-pointing vectors, and v ∈ TpM is inward-pointing if and only if −v is outward-pointing. 22

Proof. As in Proposition 3.23. Theorem 67. [Exercise 5.44] Suppose M is a smooth manifold with boundary, f is a boundary deﬁning function, and p ∈ ∂M. A vector v ∈ TpM is inward-pointing if and only if vf > 0, outward-pointing if and only if vf < 0, and tangent to ∂M if and only if vf = 0.

Proof. Let (x1, . . . , xn) be smooth coordinates defined on a neighborhood of p. Since f is a boundary defining function, (∂f/∂xi)(p) = 0 for i = 1, . . . , n−1 and (∂f/∂xn)(p) > 0. Pn i i Write v = i=1 v (∂/∂x |p) so that n X ∂f ∂f vf = vi (p) = vn (p); ∂xi ∂xn i=1 then the result follows from Theorem 66. Example 68. [Problem 5-1] Consider the map Φ : R4 → R2 defined by Φ(x, y, s, t) = (x2 + y, x2 + y2 + s2 + t2 + y). Show that (0, 1) is a regular value of Φ, and that the level set S = Φ−1({(0, 1)}) is diffeomorphic to S2. We compute 2x 1 0 0 DΦ(x, y, s, t) = . 2x 2y + 1 2s 2t This matrix clearly has full rank when s 6= 0 or t 6= 0, so suppose that s = t = 0. The determinant of the submatrix formed by the first two columns is 2x(2y + 1) − 2x = 4xy, which is zero if and only if x = 0 or y = 0. In either case, we cannot have Φ(x, y, s, t) = 4 4 (0, 1). This shows that dΦp is surjective for all p ∈ S. Define ϕ : R → R by ϕ(x, y, s, t) = (x, x2 + y, s, t); then ϕ is a diffeomorphism and Φe = (Φ ◦ ϕ−1)(u, v, s, t) = (v, v + (v − u2)2 + s2 + t2). It is easy to see that 4 4 2 2 ϕ(S) = (u, 0, s, t) ∈ R : u + s + t = 1 . This set can be considered as an embedded submanifold of R4 diffeomorphic to S2, such that ϕ|S is a diffeomorphism. Therefore S is diffeomorphic to ϕ(S). Theorem 69. [Problem 5-2] If M is a smooth n-manifold with boundary, then with the subspace topology, ∂M is a topological (n−1)-dimensional manifold (without boundary), and has a smooth structure such that it is a properly embedded submanifold of M. 23

Proof. Let x ∈ ∂M and choose a smooth boundary chart (U, ϕ) containing x. Then ϕ(∂M ∩ U) = (x1, . . . , xn) ∈ ϕ(U): xn = 0 , which shows that ∂M satisﬁes the local (n − 1)-slice condition. The result follows from Theorem 5.8 and Proposition 5.5. Theorem 70. [Problem 5-3] Suppose M is a smooth manifold with or without boundary, and S ⊆ M is an immersed submanifold. If any of the following holds, then S is embedded.

(1) S has codimension 0 in M. (2) The inclusion map S ⊆ M is proper. (3) S is compact.

Proof. If (1) holds then Theorem 4.5 shows that the inclusion map ι : S,→ M is an open map. The result follows from Proposition 4.22. Example 71. [Problem 5-4] Show that the image of the curve β :(−π, π) → R2 of Example 4.19 given by β(t) = (sin 2t, sin t) is not an embedded submanifold of R2. It is well-known that the image of β with the subset topology is not a topological manifold at all. (This can be seen by considering the point (0, 0), around which there is no coordinate map into R.) Example 72. [Problem 5-5] Let γ : R → T2 be the curve of Example 4.20. Show that γ(R) is not an embedded submanifold of the torus. Again, γ(R) with the subset topology is not a topological manifold because it is not locally Euclidean.

Theorem 73. [Problem 5-6] Suppose M ⊆ Rn is an embedded m-dimensional submanifold, and let UM ⊆ T Rn be the set of all unit tangent vectors to M: n UM = {(x, v) ∈ T R : x ∈ M, v ∈ TxM, |v| = 1} . It is called the unit tangent bundle of M. Prove that UM is an embedded (2m − 1)- dimensional submanifold of T Rn ≈ Rn × Rn.

Proof. Let (x, v) ∈ UM. Since M is an embedded submanifold of Rn, we can choose a smooth chart (U, ϕ) for Rn containing x such that ϕ(M ∩ U) = (x1, . . . , xn) ∈ ϕ(U): xm+1 = ··· = xn = 0 . 24

Similarly, Sm−1 is an embedded submanifold of Rm, so we can choose a smooth chart (V, ψ) for Rm containing v such that m−1 1 m m ψ(S ∩ V ) = (x , . . . , x ) ∈ ψ(V ): x = 0 . 1 n −1 2n Write ϕ = (x , . . . , x ) and Ue = ϕ (U). By deﬁnition, the map ϕe : Ue → R given by

i ∂ 1 n 1 n v i 7→ (x (p), . . . , x (p), v , . . . , v ) ∂x p n is a coordinate map for T R . By shrinking Ue, we can assume that V ⊆ π(ϕe(Ue)), where π : R2n → Rm is the projection onto the coordinates n + 1, . . . , n + m. Define θ(x1, . . . , xn, v1, . . . , vn) = (x1, . . . , xn, ψ(v1, . . . , vm), vm+1, . . . , vn); 2n then θ is a diffeomorphism onto its image, and θ ◦ ϕe : Ue → R is still a coordinate map for T Rn. Furthermore, (θ ◦ ϕe)(UM ∩ Ue) n 1 n 1 n m+1 n m n o = (x , . . . , x , v , . . . , v ) ∈ (θ ◦ ϕe)(Ue): x = ··· = x = v = ··· = v = 0 , so UM satisfies the local (2m−1)-slice condition. By Theorem 5.8, UM is an embedded n (2m − 1)-dimensional submanifold of T R . Example 74. [Problem 5-7] Let F : R2 → R be defined by F (x, y) = x3 + xy + y3. Which level sets of F are embedded submanifolds of R2? For each level set, prove either that it is or that it is not an embedded submanifold. We compute DF (x, y) = 3x2 + y x + 3y2 , which has full rank unless 3x2 + y = 0 and x + 3y2 = 0. In this case, we have y = −3x2 = −3(−3y2)2 = −27y4 and 0 = y + 27y4 = y(3y + 1)(9y2 − 3y + 1), 1 1 so y = 0 or y = − 3 . By symmetry, we have x = 0 or x = − 3 . From this, it is clear 1 1 that DF (x, y) has full rank if and only if (x, y) 6= (0, 0) and (x, y) 6= (− 3 , − 3 ). Note 1 1 1 −1 that F (0, 0) = 0 and F (− 3 , − 3 ) = 27 . We first examine the level set S1 = F ({0}). The point p = (0, 0) is a saddle point of F , so S1 is a self-intersecting curve at p. Next, −1 1 we examine the level set S2 = F ( 27 ). Since F has a strict local maximum at 1 1 p = (− 3 , − 3 ), the point p is isolated in S2. In both cases, the level sets fail to be locally Euclidean. Therefore, we can conclude that F −1({c}) is an embedded 1-dimensional 2 1 submanifold of R if and only if c 6= 0 and c 6= 27 . Theorem 75. Any closed ball in Rn is an n-dimensional manifold with boundary. 25

n n Proof. Let B = B1(0) be the closed unit ball in R , let N = (0,..., 0, 1) be the “north pole”, and let T = {0}n−1 × [0, 1] be the vertical line connecting 0 and N. Deﬁne σ : Bn \ T → Hn given by x1 xn−1 −1 σ(x1, . . . , xn) = ,..., , kxk − 1 , kxk − xn kxk − xn with its inverse given by

−1 1 2 σ (u1, . . . , un) = 2 (2u1,..., 2un−1, kuek − 1) (un + 1)(kuek + 1) n where ue = (u1, . . . , un−1). Since both maps are continuous, this proves that B \ T is homeomorphic to Hn. Furthermore, we can repeat the same argument by taking the vertical line T = {0}n−1 × [−1, 0] instead, giving us Euclidean neighborhoods of every point in Bn except for 0. But the identity map on the open unit ball Bn is a suitable coordinate chart around 0, so this proves that Bn is an n-manifold with boundary. Theorem 76. [Problem 5-8] Suppose M is a smooth n-manifold and B ⊆ M is a regular coordinate ball. Then M \ B is a smooth manifold with boundary, whose boundary is diﬀeomorphic to Sn−1.

Proof. We have coordinate balls around every point of M \B, so it remains to show that we have coordinate half-balls around each point of ∂B. Since B is a regular coordinate ball, this reduces to showing that Br(0) \ Bs(0) is a smooth manifold with boundary whenever s < r. But this follows easily from the stereographic projection described in Theorem 75. Theorem 77. [Problem 5-9] Let S ⊆ R2 be the boundary of the square of side 2 centered at the origin (see Theorem 36). Then S does not have a topology and smooth structure in which it is an immersed submanifold of R2.

Proof. Suppose S has a topology and smooth structure for which the inclusion map ι : S,→ R2 is a smooth immersion. By Theorem 4.25, there is a neighborhood U of (1, 1) such that ι|U is a smooth embedding. We can then apply the argument of Theorem 36 to derive a contradiction. 2 Example 78. [Problem 5-10] For each a ∈ R, let Ma be the subset of R deﬁned by 2 Ma = (x, y): y = x(x − 1)(x − a) . 2 For which values of a is Ma an embedded submanifold of R ? For which values can Ma be given a topology and smooth structure making it into an immersed submanifold? Let F (x, y) = y2 − x(x − 1)(x − a); then DF (x, y) = −(3x2 − (2a + 2)x + a) 2y . 26

Therefore 0 is a regular value of F unless there is a point (x, y) such that y = 0, 3x2 − (2a + 2)x + a = 0 and F (x, y) = −x(x − 1)(x − a) = 0. We have the following cases to consider: a = 0 and (x, y) = (0, 0), a = 1 and (x, y) = (1, 0). When a = 0 the point (0, 0) is a local minimum of F , so (0, 0) is an isolated point of M0. Therefore M0 cannot be an embedded or immersed submanifold. When a = 1 the curve M1 is self-intersecting at (1, 0). By giving M1 an appropriate topology in which 2 it is disconnected, we can make M1 an immersed submanifold of R . Theorem 79. [Problem 5-12] Suppose E and M are smooth manifolds with boundary, and π : E → M is a smooth covering map. The restriction of π to each connected component of ∂E is a smooth covering map onto a component of ∂M.

Proof. Let F be a connected component of ∂E. Let x ∈ π(F ), let U be an evenly covered neighborhood of x, choose e so that x = π(e), and let Ue be the component of π−1(U) containing e. Then π| : U → U is a diﬀeomorphism. It is clear that Ue e π(Ue ∩ F ) = U ∩ ∂M. Since Ue ∩ F is an embedded submanifold of Ue and π(Ue ∩ F ) is an embedded submanifold of U by Theorem 5.11, π| is a diﬀeomorphism. This shows Ue∩F that U ∩∂M is an evenly covered neighborhood of x, and that the image of π|F is open. Let x∈ / π(F ), let U be an evenly covered neighborhood of x, and let V = U ∩ ∂M. By shrinking V if necessary, we may assume that V is a connected neighborhood of x in U ∩ ∂M. Suppose there is a f ∈ F such that y = π(f) ∈ V . Let Ue be the component of π−1(U) containing f and choose e ∈ Ue so that x = π(e). Let γ be a path from x to y in V . Then (π| )−1 ◦ γ is a path from e to f in U ∩ ∂E, so e ∈ F since f ∈ F Ue e and F is a connected component of ∂E. This contradicts the fact that x∈ / π(F ), so V ⊆ ∂M \ π(F ). This shows that π(F ) is both open and closed in ∂M, so it is a component of ∂M. Theorem 80. [Problem 5-14] Suppose M is a smooth manifold and S ⊆ M is an immersed submanifold. For the given topology on S, there is only one smooth structure making S into an immersed submanifold.

Proof. Let ι : S,→ M be the inclusion map, which is a smooth immersion. Let S0 denote S with some other smooth structure such that ι0 : S0 ,→ M is a smooth immersion. Since S and S0 have the same topology, the maps ι : S → S0 and ι0 : S0 → S are both continuous, and Theorem 5.29 shows that the maps are inverses of each other. 0 Therefore S is diﬀeomorphic to S . 27

Theorem 81. [Problem 5-16] If M is a smooth manifold and S ⊆ M is a weakly embedded submanifold, then S has only one topology and smooth structure with respect to which it is an immersed submanifold.

Proof. Almost identical to Theorem 80. Theorem 82. [Problem 5-17] Suppose M is a smooth manifold and S ⊆ M is a smooth submanifold. (1) S is embedded only if every f ∈ C∞(S) has a smooth extension to a neighborhood of S in M. (2) S is properly embedded only if every f ∈ C∞(S) has a smooth extension to all of M.

Proof. First suppose that S is embedded. Let p ∈ S and choose a smooth slice chart (Up, ϕp) for S in M containing p. Then S ∩Up is closed in Up, so Proposition 2.25 shows that there is smooth function fp : Up → R such that fp|S∩Up = f|S∩Up . Let {ψp}p∈S be a partition of unity subordinate to {Up}p∈S; then X fe(x) = ψp(x)fp(x) p∈S S is a smooth function deﬁned on U = p∈S Up such that fe|S = f, taking fp to be zero on U \supp fp. For (2), Proposition 5.5 shows that S is closed in M. Therefore we have a partition of unity {ψp} ∪ {ψ0} for M subordinate to {Up} ∪ {M \ S}, allowing fe to be deﬁned on all of M. Theorem 83. [Problem 5-19] Suppose S ⊆ M is an embedded submanifold and γ : J → M is a smooth curve whose image happens to lie in S. Then γ0(t) is in the subspace Tγ(t)S of Tγ(t)M for all t ∈ J. This need not be true if S is not embedded.

Proof. Corollary 5.30 shows that γ is smooth as a map into S. Denote this map by 0 0 γ0 : J → S and let ι : S,→ M. Then γ = ι ◦ γ0, so γ (t) = dιγ(t)(γ0(t)) by Proposition 0 3.24 and therefore γ (t) ∈ Tγ(t)S ⊆ Tγ(t)M. This need not hold if S is merely an immersed submanifold. For example, consider a curve γ that crosses the point of self- intersection in the ﬁgure-eight curve β of Example 4.19, such that γ is not continuous as a map into the image of β. Theorem 84. [Problem 5-21] Suppose M is a smooth manifold and f ∈ C∞(M). (1) For each regular value b of f, the sublevel set f −1 ((−∞, b]) is a regular domain in M. (2) If a and b are two regular values of f with a < b, then f −1 ([a, b]) is a regular domain in M. 28

Proof. Let m be the dimension of M. It is clear that f −1 ((−∞, b)) satisfies the local m-slice condition. Furthermore, there is a boundary slice chart around each point of f −1({b}). We can see this by applying the argument in Corollary 5.14, with the following modification to Theorem 5.12: for each p ∈ f −1({b}), the rank theorem shows that there are smooth charts (U, ϕ) centered at p and (V, ψ) containing b for which the coordinate representation fb of f has the form fb(x1, . . . , xm) = x1. By modifying fband ψ appropriately, U becomes a boundary slice chart for f −1 ((−∞, b]) in M. Theorem 5.51 then shows that f −1 ((−∞, b]) is an embedded submanifold with boundary in M, and it is properly embedded since it is closed in M. Part (2) is similar. Theorem 85. [Problem 5-22] If M is a smooth manifold and D ⊆ M is a regular domain, then there exists a defining function for D. If D is compact, then f can be taken to be a smooth exhaustion function for M.

Proof. Let U = {(Up, ϕp)}p∈D be a collection of interior or boundary slice charts for D in M. Then U ∪ {M \ D} covers M, and we can repeat the argument of Proposition

5.43 by constructing functions {fp}p∈D ∪ {f0} as follows: set fp = −1 when p ∈ Int D, 1 n n set fp(x , . . . , x ) = −x when p ∈ ∂D, and set f0 = 1 for f0 : M \ D → R. If D is compact then we can modify f by using the argument of Proposition 2.28, taking a countable open cover of M \ D by precompact open subsets.

Chapter 6. Sard’s Theorem

Theorem 86. [Exercise 6.7] Let M be a smooth manifold with or without boundary. Any countable union of sets of measure zero in M has measure zero.

S∞ Proof. Let A = n=1 An where each An has measure zero. Let (U, ϕ) be a smooth chart; we want to show that ∞ [ A ∩ U = An ∩ U n=1 has measure zero. But this follows from the fact that ∞ ! ∞ [ [ ϕ An ∩ U = ϕ(An ∩ U) n=1 n=1 n has measure zero (in R ) since each ϕ(An ∩ U) has measure zero. 29

Theorem 87. [Problem 6-1] Suppose M and N are smooth manifolds with or without boundary, and F : M → N is a smooth map. If dim M < dim N, then F (M) has measure zero in N.

Proof. Let n = dim N. Deﬁne Fe : M × Rk → N by (x, y) 7→ F (x), where k = dim N − dim M. This map is smooth because it can be written as the composition k π ◦ (F × IdRk ), where π : N × R → N is the canonical projection. It suﬃces to show that the image of Fe has measure zero in N. Let (U1 ×U2, ϕ1 ×ϕ2) and (V, ψ) be smooth k k charts for M ×R and N respectively such that U1 ⊆ M and U2 ⊆ R . Let U = U1 ×U2 and ϕ = ϕ1 × ϕ2. Then ψ(Fe(U) ∩ V ) = ψ(Fe(U ∩ Fe−1(V ))) = (ψ ◦ Fe ◦ ϕ−1)(ϕ(U ∩ Fe−1(V ))) −1 −1 ⊆ (ψ ◦ Fe ◦ ϕ )(ϕ(U) ∩ (ϕ1(F (V )) × {0})), which has measure zero in Rn by Proposition 6.5. Therefore Fe(U) has measure zero. But Fe(M × Rk) can be written as a union of countably many sets of this form, so k Fe(M × R ) = F (M) has measure zero in N. Theorem 88. [Problem 6-2] Every smooth n-manifold (without boundary) admits a smooth immersion into R2n.

Proof. Let M be a smooth n-manifold; by Theorem 6.15, we can assume that M is an embedded submanifold of R2n+1. Let UM ⊆ T R2n+1 be the unit tangent bundle of M (see Theorem 73) and deﬁne G : UM → RP2n by (x, v) 7→ [v]. This map is smooth π because it is the composition UM → R2n+1 \{0} −→ RP2n. Since dim UM < dim RP2n, Sard’s theorem shows that the image of G has measure zero in RP2n, and since π(R2n) has measure zero in RP2n, there is a v ∈ R2n+1 \ R2n such that [v] is not in the image 2n+1 2n of G. Let πv : R → R be the projection with kernel Rv; then πv|M is smooth. Furthermore, if p ∈ M then v∈ / TpM, so d(πv|M )p is injective. Theorem 89. [Problem 6-3] Let M be a smooth manifold, let B ⊆ M be a closed subset, and let δ : M → R be a positive continuous function. There is a smooth function δe : M → R that is zero on B, positive on M \ B, and satisﬁes δe(x) < δ(x) everywhere.

Proof. By Theorem 2.29, there is a smooth nonnegative function f : M → R such that f −1({0}) = B. By replacing f with f/(f + 1), we can assume that f = 0 on B and 0 < f ≤ 1 on M \ B. By Corollary 6.22, there is a smooth e : M → R such that 0 < e(x) < δ(x) for all x ∈ M. Then δe = fe is the desired function. 30

Example 90. [Problem 6-7] By considering the map F : R → H2 given by F (t) = (t, |t|) and the subset A = [0, ∞) ⊆ R, show that the conclusions of Theorem 6.26 and Corollary 6.27 can be false when M has nonempty boundary.

F is smooth on the closed subset [0, ∞), but there is no smooth map G : R → H2 such that G|A = F |A. Theorem 91. [Problem 6-8] Every proper continuous map between smooth manifolds is homotopic to a proper smooth map.

Proof. Since the retraction r : U → M is proper, it suﬃces to show that if F is proper in Theorem 6.21, then Fe is also proper. Let E ⊆ Rk be compact. Since δ is continuous, it attains some maximum δ(x0) on E. Let B be a closed ball containing E such that for every x ∈ E, the closed ball of radius δ(x) around x is contained in B. Since |F (x) − Fe(x)| < δ(x), it is clear that Fe−1(E) is a closed subset of F −1(B). But F is −1 −1 proper, so F (B) is compact and therefore Fe (E) is compact. Example 92. [Problem 6-9] Let F : R2 → R3 be the map F (x, y) = (ey cos x, ey sin x, e−y).

3 For which positive numbers r is F transverse to the sphere Sr(0) ⊆ R ? For which −1 2 positive numbers r is F (Sr(0)) an embedded submanifold of R ? We compute −ey sin x ey cos x DF (x, y) =  ey cos x ey sin x . 0 −e−y We have kF (x, y)k = r if and only if r2 = e2y cos2 x + e2y sin2 x + e−2y = e2y + e−2y, i.e. 1 1 √ y = log (r2 ± r4 − 4) . 2 2 −1 2 √Thus F (Sr(0)) is√ always an embedded submanifold√ of R , being either empty (r < 2 2), one line (r =√ 2) or two lines in R (r > 2).√ However, F is transverse to Sr(0) if and only if r > 2. Let us examine the case r = 2: we have y = 0, so the image of dF(x,y) is spanned by the columns of − sin x cos x DF (x, 0) =  cos x sin x . 0 −1 31

√ The corresponding point in S 2(0) is v(x) = (cos x, sin x, 1), so the tangent space consists of all vectors orthogonal to v(x). But v(x) is orthogonal to the columns of DF (x, 0), √ 3 so dF(x,y) + Tv(x)S 2(0) is a proper subset of Tv(x)R . Theorem 93. [Problem 6-15] Suppose M and N are smooth manifolds and S ⊆ M ×N is an immersed submanifold. Let πM and πN denote the projections from M × N onto M and N, respectively. The following are equivalent: (1) S is the graph of a smooth map f : M → N. (2) πM |S is a diﬀeomorphism from S onto M. (3) For each p ∈ M, the submanifolds S and {p} × N intersect transversely in exactly one point. If these conditions hold, then S is the graph of the map f : M → N deﬁned by f = −1 πN ◦ (πM |S) .

Proof. Let m = dim M and n = dim N. Suppose S = {(x, f(x)) : x ∈ M} for some smooth map f : M → N. Then S is the image of the map g : M → S given by x 7→ (x, f(x)), so πM |S is smooth with inverse g. Conversely, if πM |S is a diﬀeomorphism −1 then we can take f = πN ◦ (πM |S) in (1). This proves that (1) ⇔ (2). If (1) holds then d(πM )g(p) ◦ dgp is always nonzero, so TpS + Tp({p} × N) = Tp(M × N). It is also clear that S ∩ ({p} × N) consists of exactly one point. If (3) holds then πM |S is a bijection. For every p ∈ M, the submanifolds S and {p} × N intersect transversely. Since dim({p} × N) = n < dim(M × N), we must have dim d(πM |S)p(TpS) ≥ m. Therefore d(πM |S)p is invertible for every p ∈ M, and πM |S is a diﬀeomorphism. This proves that (1) ⇔ (3).

Chapter 7. Lie Groups

Theorem 94. [Exercise 7.2] If G is a smooth manifold with a group structure such that the map t : G × G → G given by (g, h) 7→ gh−1 is smooth, then G is a Lie group.

Proof. The inversion map i(g) = t(e, g) is smooth, and so is the multiplication map m(g, h) = t(g, i(h)) = t(g, t(e, h)). Theorem 95. [Problem 7-1] For any Lie group G, the multiplication map m : G×G → G is a smooth submersion.

−1 Proof. For each g ∈ G, the map σg : G → G × G given by x 7→ (g, g x) is a smooth −1 local section of m since m(σg(x)) = gg x = x. Every point (g1, g2) ∈ G × G is in the image of σg1 , so Theorem 4.26 shows that m is a smooth submersion. Theorem 96. [Problem 7-2] Let G be a Lie group. 32

(1) Let m : G × G → G denote the multiplication map. Using Proposition 3.14 to identify T(e,e)(G×G) with TeG⊕TeG, the diﬀerential dm(e,e) : TeG⊕TeG → TeG is given by dm(e,e)(X,Y ) = X + Y.

(2) Let i : G → G denote the inversion map. Then die : TeG → TeG is given by

die(X) = −X. Proof. We have

dm(e,e)(X,Y ) = dm(e,e)(X, 0) + dm(e,e)(0,Y ) (1) (2) = d(m )e(X) + d(m )e(Y ), where m(1) : G → G is given by x 7→ m(x, e) and m(2) : G → G is given by y 7→ m(e, y). (1) (2) But m = m = IdG, so dm(e,e)(X,Y ) = X + Y . This proves (1). Let n = m ◦ p ◦ s where s : G → G × G is given by x 7→ (x, x) and p : G × G → G × G is given by (x, y) 7→ (x, i(y)); then n is constant, so

0 = dne(X)

= dm(e,e)(dp(e,e)(dse(X)))

= dm(e,e)(dp(e,e)(X,X))

= dm(e,e)(X, die(X))

= X + die(X).

Therefore die(X) = −X. Theorem 97. [Problem 7-3] If G is a smooth manifold with a group structure such that the multiplication map m : G × G → G is smooth, then G is a Lie group.

Proof. It suﬃces to show that the inversion map i : G → G is smooth. Deﬁne F : G × G → G × G by (g, h) 7→ (g, gh), which is smooth and bijective. We have

dF(e,e)(X,Y ) = (X,X + Y ), so dF(e,e) is an isomorphism. Write Lg and Rg for the translation maps. If (x, y) ∈ G×G then it is easy to check that

F = ((Lxy ◦ Ry−1 ) × IdG) ◦ F ◦ ((Ly−1x−1 ◦ Ry) × Ly−1 ), so dF(x,y) = ϕ ◦ dF(e,e) ◦ ψ where ϕ, ψ are isomorphisms. Therefore dF(x,y) is an isomorphism and F is a bijective local diﬀeomorphism, i.e. a diﬀeomorphism. Since i(g) is −1 the second component of F (g, e), the inversion map i is smooth. Theorem 98. [Problem 7-4] Let det : GL(n, R) → R denote the determinant function. Then −1 d(det)X (B) = (det X) tr(X B). 33

Proof. Let A ∈ M(n, R) write Ai,j for the (i, j) entry of A. Then

d d det(In + tA) = (1 + tr(A)t + ··· ) dt t=0 dt t=0 = tr(A).

If X ∈ GL(n, R) and B ∈ TX GL(n, R) we have −1 det(X + tB) = det(X) det(In + tX B), so

d d(det)X (B) = det(X + tB) dt t=0

d −1 = det(X) det(In + tX B) dt t=0 = det(X) tr(X−1B).

Theorem 99. [Problem 7-5] For any connected Lie group G, the universal covering group is unique in the following sense: if Ge and Ge0 are simply connected Lie groups that admit smooth covering maps π : Ge → G and π0 : Ge0 → G that are also Lie group homomorphisms, then there exists a Lie group isomorphism Φ: Ge → Ge0 such that π0 ◦ Φ = π.

Proof. Let e be the identity in Ge and let e0 be the identity in Ge0. By Theorem 50 there is a diﬀeomorphism Φ : Ge → Ge0 such that π0 ◦ Φ = π, so it remains to show that Φ is a −1 group homomorphism. By replacing Φ with Φ(e) Φ = LΦ(e)−1 ◦ Φ, we can assume that Φ(e) = e0. (Since π0(Φ(e)−1Φ(g)) = π0(Φ(e))−1π(g) = π(g), we still have π0 ◦ Φ = π.) If g, h ∈ Ge then π0(Φ(gh)) = π(g)π(h) = π0(Φ(g)Φ(h)), so (g, h) 7→ Φ(gh) and (g, h) 7→ Φ(g)Φ(h) are both lifts of (g, h) 7→ π(gh). Since the maps all agree at the point (e, e), we have Φ(gh) = Φ(g)Φ(h). Theorem 100. [Problem 7-6] Suppose G is a Lie group and U is any neighborhood of the identity. There exists a neighborhood V of the identity such that V ⊆ U and gh−1 ∈ U whenever g, h ∈ V .

Proof. Deﬁne f(g, h) = gh−1 and let W = f −1(U). Since (e, e) ∈ W , there are neighborhoods W1,W2 of e such that (e, e) ∈ W1 × W2 ⊆ W . Then V = W1 ∩ W2 is the desired neighborhood of the identity. 34

Theorem 101. [Problem 7-7] Let G be a Lie group and let G0 be its identity component. Then G0 is a normal subgroup of G, and is the only connected open subgroup. Every connected component of G is diﬀeomorphic to G0.

Proof. The subgroup H generated by G0 is a connected open subgroup of G, so H ⊆ G0 since G0 is a connected component. Therefore H = G0. For each g ∈ G, deﬁne the −1 conjugation map Cg : G → G by h 7→ ghg . This map is a Lie group isomorphism, so Cg(G0) is a connected open subgroup of G and again we have Cg(G0) = G0. Every left coset gG0 is the image of G0 under the diﬀeomorphism Lg, so gG0 is a connected open subset of G. But G is the union of all such cosets, so the connected components of G are precisely the cosets of G0. Theorem 102. [Problem 7-8] Suppose a connected topological group G acts continu- ously on a discrete space K. Then the action is trivial.

Proof. Let θ : G × K → K be the group action. For any x ∈ K the set θ(G × {x}) is connected and therefore consists of one point. But θ(e, x) = x, so θ(g, x) = x for every g ∈ G. Theorem 103. [Problem 7-9,7-10] The formula A · [x] = [Ax] deﬁnes a smooth, transitive left action of GL(n + 1, R) on RPn. The same formula deﬁnes a smooth, transitive left action of GL(n + 1, C) on CPn.

Proof. The smooth map θe : GL(n + 1, R) × Rn+1 \{0} → RPn given by (A, x) 7→ [Ax] is constant on each line in Rn+1 since A is linear, so it descends to a smooth map θ : GL(n + 1, R) × RPn → RPn given by (A, [x]) 7→ [Ax]. It is easy to see that θ is a group action. Furthermore, for any two points v, w ∈ Rn+1 there is an invertible linear map that takes v to w, so θ is transitive. Example 104. [Problem 7-11] Considering S2n+1 as the unit sphere in Cn+1, deﬁne an action of S1 on S2n+1, called the Hopf action, by z · (w1, . . . , wn+1) = (zw1, . . . , zwn+1). This action is smooth and its orbits are disjoint unit circles in Cn+1 whose union is S2n+1. Theorem 105. [Problem 7-12] Every Lie group homomorphism has constant rank.

Proof. If F : G → H is a Lie group homomorphism, then we can apply Theorem 7.25 with the left translation action on G and the action g · h = F (g)h on H. 35

Theorem 106. [Problem 7-13,7-14] For each n ≥ 1, U(n) is a properly embedded n2-dimensional Lie subgroup of GL(n, C) and SU(n) is a properly embedded (n2 − 1)- dimensional Lie subgroup of U(n).

Proof. We can use the argument in Example 7.27, replacing the transpose operator with the conjugate transpose operator. Theorem 107. [Problem 7-15] SO(2), U(1), and S1 are all isomorphic as Lie groups.

Proof. U(1) is obviously identical to S1. We also have a Lie group isomorphism f : S1 → SO(2) given by Re(z) −Im(z) cos θ − sin θ f(z) = = , Im(z) Re(z) sin θ cos θ where θ = arg(z). Theorem 108. [Problem 7-16] SU(2) is diﬀeomorphic to S3.

3 2 2 2 Proof. Identifying S with the subspace (z, w) ∈ C : |z| + |w| = 1 , we have a diffeomorphism z −w¯ (z, w) 7→ . w z¯ Example 109. [Problem 7-17] Determine which of the following Lie groups are compact: GL(n, R), SL(n, R), GL(n, C), SL(n, C), U(n), SU(n). Clearly GL(n, R) and GL(n, C) are not compact. SL(n, R) and SL(n, C) are not compact, since the operator norm of r  −1  r  ∈ SL(n, R) I is unbounded as r → ∞. However, U(n) is compact since every unitary matrix has operator norm 1, and SU(n) is compact because it is a properly embedded subgroup of U(n). Theorem 110. [Problem 7-18] Suppose G is a Lie group, and N,H ⊆ G are Lie subgroups such that N is normal, N ∩ H = {e}, and NH = G. Then the map (n, h) 7→ nh is a Lie group isomorphism between N oθ H and G, where θ : H × N → N is the −1 action by conjugation: θh(n) = hnh . Furthermore, N and H are closed in G. 36

Proof. Let ϕ be the map (n, h) 7→ nh. If nh = e then n = h−1 ∈ N ∩ H, so n = h = e. Also, ϕ is surjective since NH = G. Therefore ϕ is a bijection. We have ϕ((n, h)(n0, h0)) = ϕ(nhn0h−1, hh0) = nhn0h−1hh0 = nhn0h0 = ϕ(n, h)ϕ(n0, h0), so ϕ is an isomorphism. Since ϕ is smooth, it is a Lie group isomorphism. Theorem 111. [Problem 7-19] Suppose G, N, and H are Lie groups. Then G is isomorphic to a semidirect product NoH if and only if there are Lie group homomorphisms ∼ ϕ : G → H and ψ : H → G such that ϕ ◦ ψ = IdH and ker ϕ = N.

Proof. Suppose there is an isomorphism f : G → N oθ H. Let π : N oθ H → H be the projection (n, h) 7→ h and let ι : H → N oθ H be the injection h 7→ (e, h). ∼ Take ϕ = π ◦ f and ψ = ι; then ϕ ◦ ψ = IdH and ker ϕ = N. Conversely, suppose the homomorphisms ϕ and ψ exist. Let N = ker ϕ and H1 = ψ(H). Then N is normal in G and N ∩ H1 = {e}, since x = ψ(h) and x ∈ ker ϕ implies that h = ϕ(ψ(h)) = ϕ(x) = e. For any g ∈ G we have ϕ(ψ(ϕ(g))) = ϕ(g), so g = nψ(ϕ(g)) for some n ∈ ker ϕ. Therefore NH1 = G. By Theorem 110, G is a semidirect product. Example 112. [Problem 7-20] Prove that the following Lie groups are isomorphic to semidirect products as shown. ∼ (1)O( n) = SO(n) o O(1). ∼ (2)U( n) = SU(n) o U(1). ∼ ∗ (3) GL(n, R) = SL(n, R) o R . ∼ ∗ (4) GL(n, C) = SL(n, C) o C . For (1) to (4), deﬁne ϕ by M 7→ det(M) and ψ by z z 7→ In−1 where In−1 is the (n − 1) × (n − 1) identity matrix, and apply Theorem 111. Theorem 113. [Problem 7-23] Let H∗ be the Lie group of nonzero quaternions, and let S ⊆ H∗ be the set of unit quaternions. Then S is a properly embedded Lie subgroup of H∗, isomorphic to SU(2).

Proof. The norm |·| : H∗ → R∗ is a smooth homomorphism, since q |(a, b)| = p(a, b)(¯a, −b) = |a|2 + |b|2 37 and |pq| = |p| |q|. Therefore ker |·| = S is a properly embedded Lie subgroup of H∗. Furthermore, we have a Lie group isomorphism a −¯b (a, b) 7→ b a¯ into SU(2).

Chapter 8. Vector Fields

Theorem 114. [Exercise 8.29] For X,Y ∈ X(M) and f, g ∈ C∞(M), we have [fX, gY ] = fg[X,Y ] + (fXg)Y − (gY f)X.

Proof. We compute [fX, gY ]h = fX(gY h) − gY (fXh) = gfXY h + Y hfXg − fgY Xh − XhgY f = fg[X,Y ]h + (fXg)Y h − (gY f)Xh. Theorem 115. [Exercise 8.34] Let A : g → h be a Lie algebra homomorphism. Then ker A and im A are Lie subalgebras.

Proof. It suﬃces to show that ker A and im A are closed under brackets. If X,Y ∈ ker A then A[X,Y ] = [AX, AY ] = 0, so [X,Y ] ∈ ker A. Similarly, if X,Y ∈ im A then X = AX0 and Y = AY 0 for some X0,Y 0 ∈ g, so A[X0,Y 0] = [AX0, AY 0] = [X,Y ] and [X,Y ] ∈ im A. Theorem 116. [Exercise 8.35] Suppose g and h are ﬁnite-dimensional Lie algebras and A : g → h is a linear map. Then A is a Lie algebra homomorphism if and only if A[Ei,Ej] = [AEi, AEj] for some basis (E1,...,En) of g.

Proof. One direction is obvious. Suppose that (E1,...,En) is a basis of g and A[Ei,Ej] = Pn Pn [AEi, AEj]. Let X,Y ∈ g and write X = i=1 xiEi and Y = i=1 yiEi; then " n n # X X A[X,Y ] = A xiEi, yjEj i=1 j=1 n n X X = xi yjA [Ei,Ej] i=1 j=1 n n X X = xi yj [AEi, AEj] i=1 j=1 38

" n n # X X = A xiEi,A yjEj i=1 j=1 = [AX, AY ].

Theorem 117. [Exercise 8.43] If V is any ﬁnite-dimensional real vector space, the composition of canonical isomorphisms

Lie(GL(V )) → TId GL(V ) → gl(V ) yields a Lie algebra isomorphism between Lie(GL(V )) and gl(V ).

Proof. Choose a basis B for V and let ϕ : GL(V ) → GL(n, R) be the associated Lie group isomorphism. It is also a Lie algebra isomorphism gl(V ) → gl(n, R) since it is linear. By Corollary 8.31 there is a Lie algebra isomorphism ρ : Lie(GL(V )) → Lie(GL(n, R)) induced by the diffeomorphism ϕ, and by Proposition 8.41 there is a Lie algebra isomorphism ψ : Lie(GL(n, R)) → gl(n, R). Then ϕ−1 ◦ ψ ◦ ρ is the desired Lie algebra isomorphism. Theorem 118. [Problem 8-1] Let M be a smooth manifold with or without boundary, and let A ⊆ M be a closed subset. Suppose X is a smooth vector field along A. Given any open subset U containing A, there exists a smooth global vector field Xe on M such that Xe|A = X and supp Xe ⊆ U.

Proof. For each p ∈ A, choose a neighborhood Wp of p and a smooth vector ﬁeld Xep on Wp that agrees with X on Wp ∩ A. Replacing Wp by Wp ∩ U we may assume that Wp ⊆ U. The family of sets {Wp : p ∈ A} ∪ {M \ A} is an open cover of M. Let {ψp : p ∈ A} ∪ {ψ0} be a smooth partition of unity subordinate to this cover, with supp ψp ⊆ Wp and supp ψ0 ⊆ M \ A.

For each p ∈ A, the product ψpXep is smooth on Wp by Proposition 8.8, and has a smooth extension to all of M if we interpret it to be zero on M \supp ψp. (The extended function is smooth because the two deﬁnitions agree on the open subset Wp \supp ψp where they overlap.) Thus we can deﬁne Xe : M → TM by X Xex = ψp(x)Xep|x. p∈A

Because the collection of supports {supp ψp} is locally finite, this sum actually has only a finite number of nonzero terms in a neighborhood of any point of M, and therefore 39 defines a smooth function. If x ∈ A, then ψ0(x) = 0 and Xep|x = Xx for each p such that ψp(x) 6= 0, so ! X X Xex = ψp(x)Xx = ψ0(x) + ψp(x) Xx = Xx, p∈A p∈A so Xe is indeed an extension of X. It follows from Lemma 1.13(b) that [ [ supp Xe = supp ψp = supp ψp ⊆ U. p∈A p∈A

Theorem 119. [Problem 8-2] Let c be a real number, and let f : Rn \{0} → R be a smooth function that is positively homogeneous of degree c, meaning that f(λx) = λcf(x) for all λ > 0 and x ∈ Rn \{0}. Then V f = cf, where V is the Euler vector ﬁeld deﬁned in Example 8.3.

Proof. We want to compute n X ∂f (V f)(x) = V f = xi (x). x ∂xi i=1 Diﬀerentiating λcf(x) = f(λx) with respect to λ on (0, ∞), we have

n X ∂f cλc−1f(x) = Df(λx)(x) = xi (λx). ∂xi i=1 Replacing x with λ−1x gives n X ∂f cλc−1f(λ−1x) = λ−1 xi (x) ∂xi i=1 n X ∂f ⇒cλc−1λ−cf(x) = λ−1 xi (x) ∂xi i=1 n X ∂f ⇒cf(x) = xi (x) = (V f)(x). ∂xi i=1 Theorem 120. [Problem 8-3] Let M be a nonempty positive-dimensional smooth manifold with or without boundary. Then X(M) is inﬁnite-dimensional. 40

Proof. Let n = dim M. If n ≥ 2 then the result is clear from Proposition 8.7 since there are infinitely many lines in R2, i.e. RP1 is infinite. Suppose n = 1 and choose any diffeomorphism ϕ :(−ε, 1 + ε) → U where ε > 0 and U is open in M. For each m = 1, 2,... , define a vector field Xm along {1/k : k = 1, . . . , m} by setting ( 0, k < m, Xm|1/k = d dt 1/k , k = m.

For each m, we have an extension Xem deﬁned on (−ε, 1 + ε) by Lemma 8.6. We will show that {Xem : m = 1, 2,... } is linearly independent by induction. Since Xem is nonzero, the set {Xem0 } is linearly independent for any m0. Now suppose

a1Xem1 + ··· + arXemr = 0 with m1 < ··· < mr. Then

a1Xem1 |1/m1 + ··· + arXemr |1/m1 = a1Xem1 |1/m1 = 0, so a1 = 0 since Xem1 |1/m1 6= 0, and a2 = ··· = ar = 0 by induction. Therefore {ϕ∗Xem : m = 1, 2,... } is a set of linearly independent vector fields defined on U. We can extend these vector fields to M by restricting each ϕ∗Xem to the closed set ϕ([0, 1]) and applying Lemma 8.6. This proves that X(M) is infinite-dimensional. Theorem 121. [Problem 8-4] Let M be a smooth manifold with boundary. There exists a global smooth vector field on M whose restriction to ∂M is everywhere inward- pointing, and one whose restriction to ∂M is everywhere outward-pointing.

Proof. Let n = dim M. Let {(Uα, ϕα)} be a collection of smooth boundary charts whose domains cover ∂M. For each α, let Xα be the nth coordinate vector field. Let {ψα} S P be a partition of unity of U = α Uα subordinate to {Uα}; then X = α ψαXα is a smooth vector field defined on U. For any boundary defining function f : M → [0, ∞) we have X Xpf = ψαXα|pf > 0 α since each Xα|p is inward-pointing, so Xp is always inward-pointing. By restricting X to ∂M and applying Lemma 8.6, we have a global smooth vector field Xe whose restriction to ∂M is everywhere inward-pointing. Also, the restriction of −Xe to ∂M is everywhere outward-pointing. Theorem 122. [Problem 8-5] Let M be a smooth n-manifold with or without boundary.

(1) If (X1,...,Xk) is a linearly independent k-tuple of smooth vector ﬁelds on an open subset U ⊆ M, with 1 ≤ k < n, then for each p ∈ U there exist smooth 41

vector ﬁelds Xk+1,...,Xn in a neighborhood V of p such that (X1,...,Xn) is a smooth local frame for M on U ∩ V . (2) If (v1, . . . , vk) is a linearly independent k-tuple of vectors in TpM for some p ∈ M, with 1 ≤ k ≤ n, then there exists a smooth local frame (Xi) on a neighborhood of p such that Xi|p = vi for i = 1, . . . , k. (3) If (X1,...,Xn) is a linearly independent n-tuple of smooth vector ﬁelds along a closed subset A ⊆ M, then there exists a smooth local frame (Xe1,..., Xen) on some neighborhood of A such that Xei|A = Xi for i = 1, . . . , n.

Proof. See Theorem 162. Theorem 123. [Problem 8-6] Let H be the algebra of quaternions and let S ⊆ H be the group of unit quaternions.

(1) If p ∈ H is imaginary, then qp is tangent to S at each q ∈ S. (2) Deﬁne vector ﬁelds X1,X2,X3 on H by

X1|q = qi, X2|q = qj, X3|q = qk. These vector ﬁelds restrict to a smooth left-invariant global frame on S. (3) Under the isomorphism (x1, x2, x3, x4) ↔ x11 + x2i + x3j + x4k between R4 and H, these vector ﬁelds have the following coordinate representations: ∂ ∂ ∂ ∂ X = −x2 + x1 + x4 − x3 , 1 ∂x1 ∂x2 ∂x3 ∂x4 ∂ ∂ ∂ ∂ X = −x3 − x4 + x1 + x2 , 2 ∂x1 ∂x2 ∂x3 ∂x4 ∂ ∂ ∂ ∂ X = −x4 + x3 − x2 + x1 . 3 ∂x1 ∂x2 ∂x3 ∂x4

2 Proof. S is a level set of the norm squared |·| : H∗ → (0, ∞), which is given by |(a + bi, c + di)|2 = a2 + b2 + c2 + d2 and has derivative 2a 2b 2c 2d .

If q = (a, b, c, d) then by Theorem 65, TqS is the kernel of this matrix. If p ∈ H is imaginary then p has the form (ei, f + gi), so qp = (−be − cf − dg + (ae + cg − df)i, af − bg + de + (ag + bf − ce)i) and a(−be − cf − dg) + b(ae + cg − df) + c(af − bg + de) + d(ag + bf − ce) = 0. 42

0 Therefore qp ∈ TqS. This proves (1). For (2), let g = (a, b, c, d) and g = (e+fi, g +hi). We compute a −b −c −d 0 b a −d c  DLg(g ) =   . c d a −b d −c b a Then a −b −c −d −f −(be + af − dg + ch) b a −d c   e   ae − bf − cg − dh      =   , c d a −b  h   de − cf + bg + ah  d −c b a −g −(ce + df + ag − bh) which shows that d(Lg)g0 (X1|g0 ) = X1|gg0 , i.e. X1 is left-invariant. Similar calculations show that X2 and X3 are left-invariant. Example 124. [Problem 8-9] Show by finding a counterexample that Proposition 8.19 is false if we replace the assumption that F is a diffeomorphism by the weaker assumption that it is smooth and bijective. Consider the smooth bijection ω : [0, 1) → S1 given by s 7→ e2πis and consider the smooth vector field X given by x 7→ (1 − 2x) d/dt|x. There is no way of defining an 1 F -related smooth vector field Y on S , since the sign of Y1 is ambiguous. Example 125. [Problem 8-10] Let M be the open submanifold of R2 where both x and y are positive, and let F : M → M be the map F (x, y) = (xy, y/x). Show that F is a diffeomorphism, and compute F∗X and F∗Y , where ∂ ∂ ∂ X = x + y ,Y = y . ∂x ∂y ∂x The inverse ru √ F −1(u, v) = , uv v is smooth, so F is a diffeomorphism. We compute y x DF (x, y) = −y/x2 1/x so that √ uv pu/v DF (F −1(u, v)) = . −vpv/u pv/u

Therefore the coordinates of F∗X are given by √ p uv u/v pu/v 2u √ = , −vpv/u pv/u uv 0 43 and the coordinates of F∗Y are given by √ √ uv pu/v uv uv = . −vpv/u pv/u 0 −v2 So ∂ ∂ ∂ F X = 2u ,F Y = uv − v2 . ∗ ∂u ∗ ∂u ∂v Example 126. [Problem 8-11] For each of the following vector fields on the plane, compute its coordinate representation in polar coordinates on the right half-plane {(x, y): x > 0}. ∂ ∂ (1) X = x + y . ∂x ∂y ∂ ∂ (2) Y = x − y . ∂x ∂y ∂ (3) Z = (x2 + y2) . ∂x Let M be the right half-plane. We have a diffeomorphism F : M → (0, ∞) × (−π, π) given by F (x, y) = (px2 + y2, arctan(y/x)), with inverse F −1(r, θ) = (r cos θ, r sin θ). Then x/px2 + y2 y/px2 + y2 DF (x, y) = , −y/(x2 + y2) x/(x2 + y2) so cos θ sin θ DF (F −1(r, θ)) = . −r−1 sin θ r−1 cos θ We have cos θ sin θ r cos θ r cos θ r2 r r cos 2θ r2 cos θ = . −r−1 sin θ r−1 cos θ r sin θ −r sin θ 0 0 − sin 2θ −r sin θ Therefore ∂ X = r , ∂r ∂ ∂ Y = r cos 2θ − sin 2θ , ∂r ∂θ ∂ ∂ Z = r2 cos θ − r sin θ . ∂r ∂θ Example 127. [Problem 8-12] Let F : R2 → RP2 be the smooth map F (x, y) = [x, y, 1], and let X ∈ X(R2) be defined by X = x∂/∂y − y∂/∂x. Prove that there is a vector field Y ∈ X(RP2) that is F -related to X, and compute its coordinate representation in 44 terms of each of the charts defined in Example 1.5. Let π : R3 \{0} → RP2 be the quotient map and define Xe : R3 \{0} → T RP2 by ! ∂ ∂ Xe(x,y,z) = dπ x − y . ∂y (x,y,z) ∂x (x,y,z)

Since Xe is constant on the ﬁbers of π, it descends to a vector ﬁeld Y : RP2 → T RP2. 3 2 n Let Uei ⊆ R \{0}, Ui ⊆ RP and ϕi : Ui → R be as in Example 1.5: 1 2 3 1 i 2 i 3 i ϕi[x , x , x ] = (x /x , x /x , x /x ), −1 1 2 3 1 i−1 i+1 3 ϕi (u , u , u ) = [u , . . . , u , 1, u , . . . , u ].

The coordinate representations of dπ and Y with respect to ϕ3 are given by u1 u2 v1 u1v3 v2 u2v3 dπ(u1, u2, u3, v1, v2, v3) = , , 1, − , − , 0 u3 u3 u3 (u3)2 u3 (u3)2 and ! 1 2 3 1 ∂ 2 ∂ Y (u , u , u ) = dπ u − u ∂y (u1,u2,1) ∂x (u1,u2,1) = (u1, u2, 1, −u2, u1, 0). On this chart, the coordinate representation of dF is dF (u1, u2, v1, v2) = (u1, u2, 1, v1, v2, 0), so for all (x, y) ∈ R2, ! ∂ ∂ dF(x,y)(X(x,y)) = dF x − y ∂y (x,y,z) ∂x (x,y,z) = (x, y, 1, −y, x, 0) = Y (x, y, 1).

The computations for ϕ1 and ϕ2 are similar. This shows that Y is F -related to X. Theorem 128. [Problem 8-13] There is a smooth vector ﬁeld on S2 that vanishes at exactly one point.

Proof. Let N = (0, 0, 1) be the north pole and let σ : S2\{N} → R2 be the stereographic projection given by σ(x1, x2, x3) = (x1, x2)/(1 − x3), σ−1(u1, u2) = (2u1, 2u2, |u|2 − 1)/(|u|2 + 1). 45

Let X be the 1st coordinate vector field on R2 and define a vector field Y on S2 by ( 0, p = N, Y (p) = −1 ((σ )∗X)p p 6= N. This vector field vanishes at exactly one point. Let S = (0, 0, −1) be the south pole and let τ : S2 \{S} → R2 be the stereographic projection given by τ(x1, x2, x3) = (x1, x2)/(1 + x3), τ −1(u1, u2) = (2u1, 2u2, 1 − |u|2)/(|u|2 + 1). Then (σ ◦ τ −1)(u1, u2) = (2u1, 2u2)/(2 |u|2), and the coordinate representation of Y for (u1, u2) 6= (0, 0) with respect to τ is given by 1 2 −1 Y (u , u ) = d(σ )(σ◦τ −1)(u1,u2)X(σ◦τ −1)(u1,u2) ! −1 d = d(σ )(σ◦τ −1)(u1,u2) dx (σ◦τ −1)(u1,u2) 2(1 − (u1)2 + (u2)2) 4u1u2 4u1 = τ −1(u1, u2), , − , − , (1 + |u|2)2 (1 + |u|2)2 (1 + |u|2)2 2 which shows that Y is smooth on S . Theorem 129. [Problem 8-14] Let M be a smooth manifold with or without boundary, let N be a smooth manifold, and let f : M → N be a smooth map. Define F : M → M × N by F (x) = (x, f(x)). For every X ∈ X(M), there is a smooth vector field on M × N that is F -related to X.

Proof. Let m = dim M and n = dim N. By Lemma 8.6, it suffices to show that there is a smooth vector field along F (M) that is F -related to X. By Proposition 5.4, F (M) is an embedded m-submanifold of M ×N. Let p ∈ F (M), let (U, ϕ) be a slice chart for F (M) m+n containing p, let pe = ϕ(p), and let Ue = ϕ(U) ⊆ R . By shrinking U, we may assume m+n that Ue is an open cube containing pe, and that ϕ(F (M) ∩ U) = {(a, b) ∈ R : b = 0}. We identify T U with U × m+n. Define a vector field Y on U by assigning Y the pe e e R e (a,b) coefficients of dϕ(Xπ(ϕ−1(a,0))), where π : M ×N → M is the canonical projection. Then −1 Y is smooth, and (ϕ )∗Y is a smooth vector field defined on U that agrees with X on F (M) ∩ U. Theorem 130. [Problem 8-15] Suppose M is a smooth manifold and S ⊆ M is an embedded submanifold with or without boundary. Given X ∈ X(S), there is a smooth vector field Y on a neighborhood of S in M such that X = Y |S. Every such vector field extends to all of M if S is properly embedded. 46

Proof. As in Theorem 82. Example 131. [Problem 8-16] For each of the following pairs of vector fields X,Y defined on R3, compute the Lie bracket [X,Y ]. ∂ ∂ ∂ (1) X = y − 2xy2 ; Y = . ∂z ∂y ∂y ∂ ∂ ∂ ∂ (2) X = x − y ; Y = y − z . ∂y ∂x ∂z ∂y ∂ ∂ ∂ ∂ (3) X = x − y ; Y = x + y . ∂y ∂x ∂y ∂x For (1), we have ∂X2 ∂X2 ∂X3 = −2y2, = −4xy, = 1, ∂x ∂y ∂y so ∂ ∂ [X,Y ] = 4xy − . ∂y ∂z For (2), we have ∂X1 ∂X2 ∂Y 2 ∂Y 3 = −1, = 1, = −1, = −1, ∂y ∂x ∂z ∂y so ∂ ∂ [X,Y ] = z − x . ∂x ∂z For (3), we have ∂X1 ∂X2 ∂Y 1 ∂Y 2 = −1, = 1, = 1, = 1, ∂y ∂x ∂y ∂x so ∂ ∂ [X,Y ] = 2x − 2y . ∂x ∂y Theorem 132. [Problem 8-17] Let M and N be smooth manifolds. Given vector fields X ∈ X(M) and Y ∈ X(N), we can define a vector field X ⊕ Y on M × N by

(X ⊕ Y )(p,q) = (Xp,Yq), where we think of the right-hand side as an element of TpM ⊕ TqN, which is naturally identiﬁed with Tp,q(M × N) as in Proposition 3.14. Then X ⊕ Y is smooth if X and Y are smooth, and [X1 ⊕ Y1,X2 ⊕ Y2] = [X1,X2] ⊕ [Y1,Y2].

Proof. The smoothness of X ⊕Y follows easily from Proposition 8.1. If f ∈ C∞(M ×N) then

[X1 ⊕ Y1,X2 ⊕ Y2](p,q)f = (X1 ⊕ Y1)(p,q)(X2 ⊕ Y2)f − (X2 ⊕ Y2)(p,q)(X1 ⊕ Y1)f 47

= (X1|pX2fq,Y1|qY2fp) − (X2|pX1fq,Y2|qY1fp)

= ([X1,X2]fq, [Y1,Y2]fp)

= ([X1,X2] ⊕ [Y1,Y2])f,

∞ ∞ where fq ∈ C (M) is the map p 7→ f(p, q) and fp ∈ C (N) is the map q 7→ f(p, q). Theorem 133. [Problem 8-18] Suppose F : M → N is a smooth submersion, where M and N are positive-dimensional smooth manifolds. Given X ∈ X(M) and Y ∈ X(N), we say that X is a lift of Y if X and Y are F -related. A vector field V ∈ X(M) is said to be vertical if V is everywhere tangent to the fibers of F (or, equivalently, if V is F -related to the zero vector field on N).

(1) If dim M = dim N, then every smooth vector field on N has a unique lift. (2) If dim M 6= dim N, then every smooth vector field on N has a lift, but it is not unique. (3) Assume in addition that F is surjective. Given X ∈ X(M), X is a lift of a smooth vector field on N if and only if dFp(Xp) = dFq(Xq) whenever F (p) = F (q). If this is the case, then X is a lift of a unique smooth vector field.

Proof. Let m = dim M and n = dim N. If m = n then F is a local diffeomorphism, and if Y ∈ X(N) then we can define a smooth vector field X ∈ X(M) by

−1 Xp = (dFp) (YF (p)).

This is clearly the only vector field that satisfies dFp(Xp) = YF (p), and smoothness follows from Theorem 41 and the fact that dF is a local diffeomorphism. This proves (1). Now suppose Y ∈ X(N) and m 6= n, i.e. m > n. Let p ∈ M, let q = F (p), and choose smooth coordinates (xi) centered at p and (yi) centered at q in which F has the coordinate representation π(x1, . . . , xm) = (x1, . . . , xn). If ε is a sufficiently small number, the coordinate cube i Cp = x : x < ε for i = 1, . . . , m is a neighborhood of p whose image under F is the cube 0 i Cp = y : y < ε for i = 1, . . . , n .

i i Define a smooth vector field Xp on Cp by setting Xp(x) = Y (F (x)) for i = 1, . . . , n i and Xp(x) = 0 for i = n + 1, . . . m. Then dFx(Xp|x) = YF (x) for all x ∈ Cp. Choose a partition of unity {ψp} subordinate to the open cover {Cp}. Then X X = ψpXp p∈M 48 is a smooth vector field on M (taking ψpXp = 0 on M \ supp ψp), and ! X dFx(Xx) = dFx ψp(x)Xp|x p∈M X = ψp(x)dFx(Xp|x) p∈M X = ψp(x)YF (x) p∈M

= YF (x) for all x ∈ M. This proves (2). For (3), F is a quotient map. If X is a lift of some Y ∈ X(N) then for all p, q ∈ M,

dFp(Xp) = YF (p) = YF (q) = dFq(Xq).

Conversely, if dFp(Xp) = dFq(Xq) whenever F (p) = F (q) then dF ◦X is constant on the ﬁbers of F , so it descends to a smooth vector ﬁeld Y on N satisfying dF ◦ X = Y ◦ F , i.e. dFp(Xp) = YF (p) for all p ∈ M. Theorem 134. [Problem 8-20] Let A ⊆ X(R3) be the subspace spanned by {X,Y,Z}, where ∂ ∂ ∂ ∂ ∂ ∂ X = y − z ,Y = z − x ,Z = x − y . ∂z ∂y ∂x ∂z ∂y ∂x Then A is a Lie subalgebra of X(R3) which is isomorphic to R3 with the cross product.

Proof. Since {X,Y,Z} is linearly independent, it is a basis for A. Simple computations similar to those in Example 131 show that A is closed under brackets. Deﬁne an isomorphism ϕ : A → R3 by setting ϕX = (1, 0, 0), ϕY = (0, 1, 0), ϕZ = (0, 0, 1), and extending linearly. It is easy to check that ϕ is a Lie algebra isomorphism. For example, ϕ[Z,X] = ϕY = [(0, 0, 1), (1, 0, 0)] = [ϕZ, ϕX]. Theorem 135. [Problem 8-21] Up to isomorphism, there are exactly one 1-dimensional Lie algebra and two 2-dimensional Lie algebras. All three algebras are isomorphic to Lie subalgebras of gl(2, R).

Proof. Any 1-dimensional Lie algebra is abelian, and is isomorphic to the subalgebra {rI2 : r ∈ R} of gl(2, R). Let A be a 2-dimensional Lie algebra and let {x, y} be a basis for A. If A is abelian then it is isomorphic to the subalgebra r1 0 , r1, r2 ∈ R 0 r2 49 of gl(2, R). Otherwise, we have [x, y] = ax + by for some a, b ∈ R, not both zero. Assume without loss of generality that b 6= 0. (If b = 0, then we can swap x and y.) Deﬁne an homomorphism ϕ : A → gl(2, R) by setting 1 0 −ab−1 0 ϕx = , ϕy = 0 1 + b 1 −ab−1(1 + b) and extending linearly. Then ϕ is an isomorphism onto its image, and it is easy to verify that ϕ[x, y] = [ϕx, ϕy]. Theorem 136. [Problem 8-22] Let A be any algebra over R.A derivation of A is a linear map D : A → A satisfying D(xy) = (Dx)y + x(Dy) for all x, y ∈ A. If D1 and D2 are derivations of A, then [D1,D2] = D1 ◦ D2 − D2 ◦ D1 is also a derivation. The set of derivations of A is a Lie algebra with this bracket operation.

Proof. Identical to Lemma 8.25. Theorem 137. [Problem 8-23] (1) Given Lie algebras g and h, the direct sum g⊕h is a Lie algebra with the bracket deﬁned by [(X,Y ), (X,Y 0)] = ([X,X0], [Y,Y 0]). (2) Suppose G and H are Lie groups. Then Lie(G × H) is isomorphic to Lie(G) ⊕ Lie(H).

Proof. We have isomorphisms ε : Lie(G × H) → Te(G × H), ε1 : Lie(G) → Te(G) and ε2 : Lie(H) → Te(H). Thus we have an isomorphism 0 ∼ ε : Lie(G) ⊕ Lie(H) → Te(G) ⊕ Te(H) = Te(G × H), −1 0 and it remains to show that ε ◦ ε respects brackets. Let (X1,X2), (Y1,Y2) ∈ Lie(G) ⊕ Lie(H); then −1 0 −1 0 (ε ◦ ε )[(X1,X2), (Y1,Y2)](g,h) = ((ε ◦ ε )([X1,Y1], [X2,Y2]))(g,h) −1 = (ε ([X1,Y1]e, [X2,Y2]e))(g,h)

= d(L(g,h))(e,e)([X1,Y1] ⊕ [X2,Y2])(e,e)

= d(L(g,h))(e,e)[X1 ⊕ X2,Y1 ⊕ Y2](g,h)

= [X1 ⊕ X2,Y1 ⊕ Y2](g,h) −1 0 −1 0 = [(ε ◦ ε )(X1,X2), (ε ◦ ε )(Y1,Y2)](g,h) since [X1 ⊕ X2,Y1 ⊕ Y2] is left-invariant (see Theorem 132). Theorem 138. [Problem 8-24] Suppose G is a Lie group and g is its Lie algebra. A vector ﬁeld X ∈ X(G) is said to be right-invariant if it is invariant under all right translations. 50

(1) The set g¯ of right-invariant vector ﬁelds on G is a Lie subalgebra of X(G). −1 (2) Let i : G → G denote the inversion map i(g) = g . The pushforward i∗ : X(G) → X(G) restricts to a Lie algebra isomorphism from g to g¯.

Proof. Part (1) follows from Corollary 8.31. If X ∈ g then for all g, g0 ∈ G we have

d(Rg)g0 ((i∗X)g0 ) = (d(Rg)g0 ◦ dii−1(g0))(Xi−1(g0))

= d(Rg ◦ i)(g0)−1 (X(g0)−1 )

= d(i ◦ Lg−1 )(g0)−1 (X(g0)−1 )

= di(g0g)−1 (d(Lg−1 )(g0)−1 X(g0)−1 )

= di(g0g)−1 (X(g0g)−1 )

= (i∗X)g0g.

−1 Therefore i∗(g) ⊆ g¯. A similar calculation shows that i∗(g¯) ⊆ g, so (i∗) = i∗ and i∗ is an isomorphism. Corollary 8.31 shows that i∗ is a Lie algebra isomorphism. Theorem 139. [Problem 8-25] If G is an abelian Lie group, then Lie(G) is abelian.

Proof. If G is abelian then the inversion map i : G → G is a Lie group isomorphism. The induced Lie algebra isomorphism is given by

(i∗X)g = d(Lg)e(die(Xe))

= d(Lg)e(−Xe)

= −d(Lg)eXe

= −Xg, so i∗X = −X for all X ∈ Lie(G). Then

[X,Y ] = [−X, −Y ] = [i∗X, i∗Y ] = i∗[X,Y ] = −[X,Y ], so [X,Y ] = 0 for all X,Y ∈ Lie(G). This proves that Lie(G) is abelian. Theorem 140. [Problem 8-26] Suppose F : G → H is a Lie group homomorphism. The kernel of F∗ : Lie(G) → Lie(H) is the Lie algebra of ker F (under the identiﬁcation of the Lie algebra of a subgroup with a Lie subalgebra as in Theorem 8.46).

Proof. Consider ι∗X for some X ∈ ker F , where ι : ker F,→ G is the inclusion map. Then F∗ι∗X = (F ◦ ι)∗X = 0 since F ◦ ι = 0. This shows that ι∗(Lie(ker F )) ⊆ ker F∗. Conversely, suppose F∗Y = 0 for some Y ∈ Lie(G). By deﬁnition we have dFe(Ye) = 0, so Ye ∈ ker(dFe) = Te(ker F ) by Theorem 65. By Theorem 8.46, this shows that ker F∗ ⊆ ι∗(Lie(ker F )). 51

Theorem 141. [Problem 8-27] Let G and H be Lie groups, and suppose F : G → H is a Lie group homomorphism that is also a local diﬀeomorphism. The induced homomorphism F∗ : Lie(G) → Lie(H) is an isomorphism of Lie algebras.

Proof. Since F is a local diﬀeomorphism, dFe is bijective. If F∗X = 0 then dFe(Xe) = 0, so Xe = 0 and X = 0. Suppose Y ∈ Lie(H). Let ε : Lie(G) → TeG be the canonical isomorphism. Then −1 −1 −1 −1 (F∗ε (dFe) Ye)e = dFe((ε (dFe) Ye)e) −1 = dFe((dFe) Ye)

= Ye,

−1 −1 so F∗(ε (dFe) Ye) = Y . This proves that F∗ is a bijection. Theorem 142. [Problem 8-28] Considering det : GL(n, R) → R∗ as a Lie group homomorphism, its induced Lie algebra homomorphism is tr : gl(n, R) → R.

Proof. This follows immediately from Theorem 98. Example 143. [Problem 8-29] Theorem 8.46 implies that the Lie algebra of any Lie subgroup of GL(n, R) is canonically isomorphic to a subalgebra of gl(n, R), with a similar statement for Lie subgroups of GL(n, C). Under this isomorphism, show that ∼ Lie(SL(n, R)) = sl(n, R), Lie(SO(n)) ∼= o(n), ∼ Lie(SL(n, C)) = sl(n, C), Lie(U(n)) ∼= u(n), Lie(SU(n)) ∼= su(n), where

sl(n, R) = {A ∈ gl(n, R) : tr A = 0}, T o(n) = {A ∈ gl(n, R): A + A = 0}, sl(n, C) = {A ∈ gl(n, C) : tr A = 0}, ∗ u(n) = {A ∈ gl(n, C): A + A = 0}, su(n) = u(n) ∩ sl(n, C). ∼ We have SL(n, R) = ker det, so Lie(SL(n, R)) = ker det∗ = ker tr = sl(n, R). The same computation holds for SL(n, C). The others follow by considering the kernel of the map A 7→ AT A (or A 7→ A∗A in the complex case). 52

Example 144. [Problem 8-30] Show by giving an explicit isomorphism that su(2) and o(3) are isomorphic Lie algebras, and that both are isomorphic to R3 with the cross product. We have an isomorphism from R3 to o(3) given by a  0 c −b b 7→ −c 0 a  . c b −a 0 We also have an isomorphism from R3 to su(2) given by a 1 ai b + ci b 7→ √ .   −b + ci −ai c 2 Theorem 145. [Problem 8-31] Let g be a Lie algebra. A linear subspace h ⊆ g is called an ideal in g if [X,Y ] ∈ h whenever X ∈ h and Y ∈ g. (1) If h is an ideal in g, then the quotient space g/h has a unique Lie algebra structure such that the projection π : g → g/h is a Lie algebra homomorphism. (2) A subspace h ⊆ g is an ideal if and only if it is the kernel of a Lie algebra homomorphism.

Proof. Deﬁne a bracket on g/h by [X + h,Y + h] = [X,Y ] + h. If X0 + h = X + h and Y 0 + h = Y + h then [X0,Y 0] = [X0 + X − X0,Y 0] = [X,Y 0 + Y − Y 0] = [X,Y ] since X − X0 ∈ h and Y − Y 0 ∈ h, which shows that the bracket is well-deﬁned. It is clearly the unique bracket making π a Lie algebra homomorphism. If h ⊆ g is an ideal then it is the kernel of the projection π : g → g/h. Conversely, if f is a Lie algebra homomorphism then ker f is an ideal since X ∈ ker f implies that f[X,Y ] = [fX, fY ] = 0.

Chapter 9. Integral Curves and Flows

Theorem 146. [Exercise 9.37] Suppose v ∈ Rn and W is a smooth vector ﬁeld on an open subset of Rn. The directional derivative

d DvW (p) = Wp+tv dt t=0 i i is equal to (LV W )p, where V is the vector ﬁeld V = v ∂/∂x with constant coeﬃcients in standard coordinates. 53

Proof. The ﬂow of V is given by θ(t, x) = x + tv. Therefore

d (LV W )p = d(θ−t)θt(p)(Wθt(p)) dt t=0

d = d(θ−t)p+tv(Wp+tv) dt t=0

d = Wp+tv dt t=0 since θ−t(x) = x − tv is a translation of x. Theorem 147. [Problem 9-1] Suppose M is a smooth manifold, X ∈ X(M), and γ is a maximal integral curve of X. (1) We say that γ is periodic if there is a number T > 0 such that γ(t + T ) = γ(t) for all t ∈ R. Exactly one of the following holds: (a) γ is constant. (b) γ is injective. (c) γ is periodic and nonconstant. (2) If γ is periodic and nonconstant, then there exists a unique positive number T (called the period of γ) such that γ(t) = γ(t0) if and only if t − t0 = kT for some k ∈ Z. (3) The image of γ is an immersed submanifold of M, diﬀeomorphic to R, S1, or R0.

Proof. If γ is constant then (b) and (c) cannot hold, so we assume that γ is nonconstant. If γ is not injective then γ(t0) = γ(t1) for some points t0 < t1, and γ is defined on at least (t0 − ε, t1 + ε) for some ε > 0. Let T = t1 − t0. Then γ and t 7→ γ(t1 + t) are both integral curves of X starting at γ(t0), so Theorem 9.12 shows that γ must be defined on at least (t0 − ε, t1 + T + ε). By induction, t0 + kT is in the domain of γ for all k ∈ Z, so γ is defined on R and has period T . This proves (1). For (2), let A be the set of all T > 0 such that γ(t + T ) = γ(t) for all t ∈ R, which is nonempty since γ is periodic. If we can show that A is closed, then T0 = inf A ∈ A is the period of γ. But if T/∈ A then γ(t + T ) 6= γ(t) for some t ∈ R, so γ−1(M \{γ(t)}) − t = {s − t : γ(s) 6= γ(t)} is a neighborhood of T contained in R\A. For (3), if γ is constant then the image of γ is diffeomorphic to R0. Otherwise, Proposition 9.21 shows that γ is a smooth immersion. If γ is injective then Proposition 5.18 shows that the image of γ is diffeomorphic to R, and if γ is periodic then it descends to a smooth injective immersion from S1 to M (using a suitable smooth covering of S1 by R), which is a diffeomorphism onto its image by Proposition 5.18. 54

Theorem 148. [Problem 9-2] Suppose M is a smooth manifold, S ⊆ M is an immersed submanifold, and V is a smooth vector ﬁeld on M that is tangent to S.

(1) For any integral curve γ of V such that γ(t0) ∈ S, there exists ε > 0 such that γ ((t0 − ε, t0 + ε)) ⊆ S. (2) If S is properly embedded, then every integral curve that intersects S is contained in S. (3) (2) need not hold if S is not closed.

Proof. Let x = γ(t0) ∈ S. By Proposition 8.23, there is a smooth vector field V |S on S that is ι-related to V . Let α : I → S be a maximal integral curve of V |S starting at x, where I contains (−ε, ε) for some ε > 0. Then 0 0 (ι ◦ α) (0) = dιx(α (0)) = dιx((V |S)x) = Vx, which shows that ι ◦ α is an integral curve of V starting at x. By Theorem 9.12, γ is defined on J = (t0 − ε, t0 + ε) and γ(t) = (ι ◦ α)(t − t0) ∈ S for t ∈ J. This proves (1). Now suppose that S is properly embedded and let γ be an integral curve that intersects S, i.e. γ(t0) ∈ S for some t0. Let α be defined as above. By Theorem 9.12, γ is defined on at least I + t0 = (a, b) and γ(t) ∈ S for all t ∈ (a, b). Suppose γ can be extended to an open interval J larger than (a, b). Since S is closed, γ(b) ∈ S by continuity. Then (1) shows that γ ((b − ε, b + ε)) ⊆ S for some ε > 0, which contradicts the maximality of α. This proves (2). For (3), let S be the open ball of radius 1 in R and let V be the 1st coordinate vector field on R. Then θ : R × R → R given by θ(t, x) = x + t is an integral curve of V that intersects S, but θ is not contained in S. Example 149. [Problem 9-3] Compute the flow of each of the following vector fields on R2: ∂ ∂ (1) V = y + . ∂x ∂y ∂ ∂ (2) W = x + 2y . ∂x ∂y ∂ ∂ (3) X = x − y . ∂x ∂y ∂ ∂ (4) Y = y + x . ∂x ∂y Let γ(t) = (x(t), y(t)) be a curve. We have the differential equations x0(t) = y(t), y0(t) = 1, x0(t) = x(t), y0(t) = 2y(t), x0(t) = x(t), y0(t) = −y(t) x0(t) = y(t), y0(t) = x(t). 55

Therefore the flows (in this order) are 1 θ (x, y) = x + ty + t2, y + t , t 2 t 2t θt(x, y) = (xe , ye ), t −t θt(x, y) = (xe , ye ), 1 θ (x, y) = et(x + y) + e−t(x − y), et(x + y) − e−t(x − y) . t 2 Theorem 150. [Problem 9-4] For any integer n ≥ 1, define a flow on the odd- dimensional sphere S2n−1 ⊆ Cn by θ(t, z) = eitz. The infinitesimal generator of θ is a smooth nonvanishing vector field on S2n−1.

Proof. This is obvious. Theorem 151. [Problem 9-5] Suppose M is a smooth, compact manifold that admits a nowhere vanishing smooth vector ﬁeld. There exists a smooth map F : M → M that is homotopic to the identity and has no ﬁxed points.

Proof. Let V be a nowhere vanishing smooth vector field on M. Since M is compact, Corollary 9.17 shows that there is a global flow θ : R × M → M of V . By Theorem 9.22, each p has a neighborhood Up in which V has the coordinate representation 1 ∂/∂s . Choosing a sufficiently small neighborhood Vp of p, there is some Tp > 0 such that θt(x) = x + (t, 0,..., 0) in local coordinates for all 0 ≤ t ≤ Tp and x ∈ Vp.

Since {Vp : p ∈ M} is an open cover of M, there is a ﬁnite subcover {Vp1 ,...,Vpn }. Let

T = min {Tp1 ,...,Tpn }. Then θT has no ﬁxed points, and the map H : M × I → M given by (x, t) 7→ θ(tT, x) is a homotopy from the identity to θT . Theorem 152. [Problem 9-6] Suppose M is a smooth manifold and V ∈ X(M). If γ : J → M is a maximal integral curve of V whose domain J has a ﬁnite least upper bound b, then for any t0 ∈ J, γ ([t0, b)) is not contained in any compact subset of M.

Proof. Let θ be the flow of V . Let {bn} be an increasing sequence contained in [t0, b) and converging to b. Since γ ([t0, b)) is contained in a compact set E ⊆ M, there is a subsequence {γ(bnk )} of {γ(bn)} that converges to a point p in E. Choose some ε > 0 and a neighborhood U of p such that θ is defined on (−2ε, 2ε) × U, and choose an integer m such that bm ∈ (b − ε, b) and γ(bm) ∈ U. Define γ1 :[t0, b + ε) → M by ( γ(t), t0 ≤ t < b, γ1(t) = (γ(bm)) θ (t − bm) b ≤ t < b + ε.

For all t ∈ (bm, b) we have

(γ(bm)) θ (t − bm) = θt−bm (γ(bm)) = γ(t), 56

(γ(bm)) so γ1 is smooth because γ and t 7→ θ (t − bm) agree where they overlap. But γ1 extends γ at b, which contradicts the maximality of γ. Theorem 153. [Problem 9-7] Let M be a connected smooth manifold. The group of diﬀeomorphisms of M acts transitively on M: that is, for any p, q ∈ M, there is a diﬀeomorphism F : M → M such that F (p) = q.

Proof. First assume that M = Bn. By applying Lemma 8.6 to the map x 7→ q − p defined on the line segment between p and q, we have a compactly supported smooth n vector field on B whose flow θ satisfies θ1(p) = q. Theorem 9.16 shows that θ1 is a diffeomorphism on Bn, which proves the result for M = Bn. The general case follows from Theorem 29 and the fact that every point of M is contained in a coordinate ball. Theorem 154. [Problem 9-8] Let M be a smooth manifold and let S ⊆ M be a compact embedded submanifold. Suppose V ∈ X(M) is a smooth vector field that is nowhere tangent to S. There exists ε > 0 such that the flow of V restricts to a smooth embedding Φ:(−ε, ε) × S → M.

Proof. Let θ : D → M be the flow of V and let Φ = θO where O = (R × S) ∩ D. Let δ : S → R be as in Theorem 9.20. Since S is compact, δ attains a minimum α on S, and Φ:(−α, α)×S → M is a smooth immersion. Let ε = α/2. Then [−ε, ε]×S is compact, so Φ : [−ε, ε] × S → M is a smooth embedding. Therefore Φ : (−ε, ε) × S → M is also a smooth embedding. Theorem 155. [Problem 9-9] Suppose M is a smooth manifold and S ⊆ M is an embedded hypersurface (not necessarily compact). Suppose further that there is a smooth vector field V defined on a neighborhood of S and nowhere tangent to S. Then S has a neighborhood in M diffeomorphic to (−1, 1) × S, under a diffeomorphism that restricts to the obvious identification {0} × S ≈ S.

Proof. Using Theorem 9.20, we have that Φ|Oδ is a diffeomorphism onto an open submanifold of M. So it remains to show that Oδ = {(t, p) ∈ O : |t| < δ(p)} is diffeomorphic to (−1, 1) × S. Let ϕ :(−1, 1) × S → Oδ be given by (t, p) 7→ (tδ(p), p); then ϕ is −1 smooth, and its inverse ϕ (t, p) = (t/δ(p), p) is also smooth. Example 156. [Problem 9-10] For each vector field in Example 149, find smooth coordinates in a neighborhood of (1, 0) for which the given vector field is a coordinate vector field.

We only consider the ﬁrst vector ﬁeld. We have V(1,0) = ∂/∂y, so we can take S to be the x-axis, parametrized by X(s) = (s, 0). Therefore 1 Ψ(t, s) = θ (s, 0) = s + t2, t , t 2 57 and 1 Ψ−1(x, y) = x − y2, y ; 2 1 −y DΨ−1(x, y) = . 0 1 In these coordinates, we have ∂ ∂ V = y + ∂x ∂y ∂ ∂ ∂ = s + −s + ∂s ∂s ∂t ∂ = . ∂t

Chapter 10. Vector Bundles

Theorem 157. [Exercise 10.1] Suppose E is a smooth vector bundle over M. The projection map π : E → M is a surjective smooth submersion.

Proof. Let v ∈ E and let Φ : π−1(U) → U × Rk be a smooth local trivialization of E such that π(v) ∈ U. Then πU ◦ Φ = π, so dπv = d(πU )π(v) ◦ dΦv is surjective since dΦv and d(πU )π(v) are both surjective. Theorem 158. [Exercise 10.9] The zero section ζ of every vector bundle π : E → M is continuous, and the zero section of every smooth vector bundle is smooth.

Proof. Let p ∈ M and let Φ : π−1(U) → U × Rk be a local trivialization of E such that p ∈ U. We have (Φ ◦ ζ)(x) = (x, 0) for all x ∈ U, so Φ ◦ ζ|U is continuous and ζ|U is continuous since Φ is a homeomorphism. But p was arbitrary, so ζ is continuous. A similar argument holds if π is a smooth vector bundle. Theorem 159. [Exercise 10.11] Let π : E → M be a smooth vector bundle.

(1) If σ, τ ∈ Γ(E) and f, g ∈ C∞(M), then fσ + gτ ∈ Γ(E). (2)Γ( E) is a module over the ring C∞(M).

Proof. Let Φ : π−1(U) → U × Rk be a local trivialization of E; then

(Φ ◦ (fσ + gτ))(x) = (x, f(x)πRk (σ(x)) + g(x)πRk (τ(x))) , so fσ + gτ is smooth on U. Therefore fσ + gτ is smooth. 58

Theorem 160. [Exercise 10.13] Let π : E → M be a smooth vector bundle over a smooth manifold M with or without boundary. Suppose A is a closed subset of M, and σ : A → E is a section of E|A that is smooth in the sense that σ extends to a smooth local section of E in a neighborhood of each point. For each open subset U ⊆ M containing A, there exists a global smooth section σe ∈ Γ(E) such that σe|A = σ and supp σe ⊆ U.

Proof. For each p ∈ A, choose a neighborhood Wp of p and a local section σep : Wp → E of E that agrees with σ on Wp ∩ A. Replacing Wp by Wp ∩ U we may assume that Wp ⊆ U. The family of sets {Wp : p ∈ A} ∪ {M \ A} is an open cover of M. Let {ψp : p ∈ A} ∪ {ψ0} be a smooth partition of unity subordinate to this cover, with supp ψp ⊆ Wp and supp ψ0 ⊆ M \ A.

For each p ∈ A, the product ψpσep is smooth on Wp by Theorem 159, and has a smooth extension to all of M if we interpret it to be zero on M \ supp ψp. (The extended function is smooth because the two deﬁnitions agree on the open subset Wp \ supp ψp where they overlap.) Thus we can deﬁne σe : M → E by X σe(x) = ψp(x)σep(x). p∈A

Because the collection of supports {supp ψp} is locally finite, this sum actually has only a finite number of nonzero terms in a neighborhood of any point of M, and therefore defines a smooth function. If x ∈ A, then ψ0(x) = 0 and σep(x) = σ(x) for each p such that ψp(x) 6= 0, so ! X X σe(x) = ψp(x)σ(x) = ψ0(x) + ψp(x) σ(x) = σ(x), p∈A p∈A so σe is indeed an extension of σ. It follows from Lemma 1.13(b) that [ [ supp σe = supp ψp = supp ψp ⊆ U. p∈A p∈A Theorem 161. [Exercise 10.14] Let π : E → M be a smooth vector bundle. Every element of E is in the image of a smooth global section.

Proof. Let v ∈ E and let p = π(v). The assignment p 7→ v is a section of E|{p}, which is easily seen to be smooth in the sense of Theorem 160. Theorem 162. [Exercise 10.16] Suppose π : E → M is a smooth vector bundle of rank k. 59

(1) If (σ1, . . . , σm) is a linearly independent m-tuple of smooth local sections of E over an open subset U ⊆ M, with 1 ≤ m < k, then for each p ∈ U there exist smooth sections σm+1, . . . , σk deﬁned on some neighborhood V of p such that (σ1, . . . , σk) is a smooth local frame for E over U ∩ V . (2) If (v1, . . . , vm) is a linearly independent m-tuple of elements of Ep for some p ∈ M, with 1 ≤ m ≤ k, then there exists a smooth local frame (σi) for E over some neighborhood of p such that σi(p) = vi for i = 1, . . . , m. (3) If A ⊆ M is a closed subset and (τ1, . . . , τk) is a linearly independent k-tuple of sections of E|A that are smooth in the sense described in Theorem 160, then there exists a smooth local frame (σ1, . . . , σk) for E over some neighborhood of A such that σi|A = τi for i = 1, . . . , k.

Proof. Choose a smooth local trivialization Φ : ϕ−1(V ) → V × Rk such that p ∈ V . By shrinking V , we may assume there is a coordinate map ϕ : V → Rn and that V ⊆ U; i let (x ) be the coordinates. Let vi = (πRk ◦ Φ ◦ σi)(p) for i = 1, . . . , m, and complete k {v1, . . . , vm} to a basis {v1, . . . , vk} of R . For each i = m + 1, . . . , k, let σi : V → E −1 k be the local section of E defined by σi(x) = Φ (x, vi). Let W be the subspace of R −1 n −1 spanned by {vm+1, . . . , vk}; then F = Φ (V × (R \ W )) is open, so Vi = σi (F ) is a neighborhood of p for each i = 1, . . . , m. Let Ve = V ∩ V1 ∩ · · · ∩ Vm. Then {σ1, . . . , σk} is a smooth local frame for E on Ve since (πRk ◦ Φ ◦ σi)(x) ∈/ W for all x ∈ Ve and i = 1, . . . , k. This proves (1). Part (2) follows by choosing constant coefficient local sections σi such that σi(p) = vi on a sufficiently small neighborhood of p. For (3), apply Theorem 160 to each τi and restrict to a sufficiently small neighborhood of A. Theorem 163. [Exercise 10.27] A smooth rank-k vector bundle over M is smoothly trivial if and only if it is smoothly isomorphic over M to the product bundle M × Rk.

Proof. If π : E → M is smoothly trivial then there is a trivialization Φ : E → M × Rk. k This map satisﬁes πM ◦ Φ = π, so E and M × R are smoothly isomorphic over M. The converse is similar. Theorem 164. [Exercise 10.31] Given a smooth vector bundle π : E → M and a smooth subbundle D ⊆ E, the inclusion map ι : D,→ E is a smooth bundle homomorphism over M.

Proof. This is simply the condition π ◦ ι = π|D. Theorem 165. [Problem 10-1] The M¨obiusbundle E is not trivial.

Proof. E is not homeomorphic to S1 × R. Removing the center circle S1 × {0} from S1 × R leaves a disconnected space, but removing the center circle from E leaves a connected space. 60

Theorem 166. [Problem 10-2] Let E be a vector bundle over a topological space M. The projection map π : E → M is a homotopy equivalence.

Proof. Let ζ : M → E be the zero section. Then π ◦ ζ = IdM , and H : E × I → E defined by H(v, t) = t IdE is a homotopy from ζ ◦ π to IdE. This shows that π is a homotopy equivalence. Theorem 167. [Problem 10-3] Let VB denote the category whose objects are smooth vector bundles and whose morphisms are smooth bundle homomorphisms, and let Diff denote the category whose objects are smooth manifolds and whose morphisms are smooth maps. The assignment M 7→ TM, F 7→ dF defines a covariant functor from Diff to VB, called the tangent functor.

Proof. This follows immediately from Corollary 3.22. Theorem 168. [Problem 10-4] The map τ : U ∩ V → GL(k, R) in Lemma 10.5 is smooth.

Proof. We know that σ :(U ∩ V ) × Rk → Rk given by σ(p, v) = τ(p)v is smooth. k Let {e1, . . . , ek} be the standard basis for R . The (i, j) entry of τ(p) is πi(τ(p)ej) = k πi(σ(p, ej)), where πi : R → R is the projection onto the ith coordinate. This map is smooth as a function of p, so τ is smooth. Theorem 169. [Problem 10-5] Let π : E → M be a smooth vector bundle of rank k over a smooth manifold M with or without boundary. Suppose that {Uα}α∈A is an open cover −1 of M, and for each α ∈ A we are given a smooth local trivialization Φα : π (Uα) → k Uα × R of E. For each α, β ∈ A such that Uα ∩ Uβ 6= ∅, let ταβ : Uα ∩ Uβ → GL(k, R) be the transition function deﬁned by (10.3). The following identity is satisﬁed for all α, β, γ ∈ A: ταβ(p)τβγ(p) = ταγ(p), p ∈ Uα ∩ Uβ ∩ Uγ.

Proof. This follows from the fact that −1 (p, ταγ(p)v) = (Φα ◦ Φγ )(p, v) −1 −1 = (Φα ◦ Φβ ◦ Φβ ◦ Φγ )(p, v)

= (p, ταβ(p)τβγ(p)v). Theorem 170. [Problem 10-6] Let M be a smooth manifold with or without boundary, and let {Uα}α∈A be an open cover of M. Suppose for each α, β ∈ A we are given a smooth map ταβ : Uα ∩ Uβ → GL(k, R) such that

ταβ(p)τβγ(p) = ταγ(p), p ∈ Uα ∩ Uβ ∩ Uγ 61 is satisﬁed for all α, β, γ ∈ A. Then there is a smooth rank-k vector bundle E → M −1 k with smooth local trivializations Φα : π (Uα) → Uα × R whose transition functions are the given maps ταβ.

` k k Proof. Note that ταα = IdRk for all α ∈ A. Let F = α∈A Uα×R and let iα : Uα×R → F be the canonical injection for each α ∈ A. Deﬁne an equivalence relation ∼ on F by 0 0 0 0 declaring that (p, v)α ∼ (p , v )β if and only if p = p and ταβ(p)v = v. This relation is symmetric because

ταβ(p)τβα(p) = ταα(p) = IdRk , 0 0 so ταβ(p)v = v implies that v = τβα(p)v. Similarly, the relation is transitive because 0 00 0 ταβ(p)v = v and τβγ(p)v = v implies that 00 00 0 ταγ(p)v = ταβ(p)τβγ(p)v = ταβ(p)v = v. k Let E = F/ ∼ and let q : F → E be the quotient map. Let Ep = q {p} × R for each −1 p ∈ M; we want to give Ep a vector space structure. Define m : R × q (Ep) → Ep 0 e0 0 by m(r, (p, v)α) = q((p, rv)α). If (p, v)α ∼ (p, v )β then ταβ(p)v = v and ταβ(p)rv = 0 rταβ(p)v = rv. Therefore me descends to a continuous map m : R × Ep → Ep, which is scalar multiplication in Ep. A similar construction shows that vector addition can be defined on Ep. Let π : F → M be given by (p, v) 7→ p; then π descends to a continuous map π : E → M e e −1 k satisfying π ◦ q = πe. For each α ∈ A, define Φα : π (Uα) → Uα × R as the unique k bijection satisfying Φα ◦ q ◦ iα = IdUα×R . We now check conditions (ii) and (iii) in

Lemma 10.6. For (ii), it is clear that Φα|Ep is linear, so it is an isomorphism. For (iii), we have −1 (Φα ◦ Φβ )(p, v) = Φα(q((p, v)β))

= Φα(q((p, ταβ(p)v)α))

= (p, ταβ(p)v). Lemma 10.6 now shows that π : E → M is a smooth rank-k vector bundle, with {(Uα, Φα)} as smooth local trivializations. Example 171. [Problem 10-7] Compute the transition function for T S2 associated with the two local trivializations determined by stereographic coordinates. See Theorem 128.

Theorem 172. [Problem 10-8] Let Vec1 be the category whose objects are ﬁnite-dimensional real vector spaces and whose morphisms are linear isomorphisms. If F is a covariant functor from Vec1 to itself, for each ﬁnite-dimensional vector space V we get a map F : GL(V ) → GL(F(V )) sending each isomorphism A : V → V to the induced isomorphism F(A): F(V ) → F(V ). We say F is a smooth functor if this map is smooth for every V . Given a smooth vector bundle π : E → M and a smooth functor 62

F : Vec1 → Vec1, there is a smooth vector bundle πe : F(E) → M whose fiber at each point p ∈ M is F(Ep). ` Proof. Let k be the rank of π. Define F(E) = p∈M F(Ep) and let πe : F(E) → M be the projection. Let {Up}p∈M be an open cover of M where each Up contains p and −1 k has an associated local trivialization Φp : π (Up) → Up × R . Choose an arbitrary k ∼ n k isomorphism ϕ : F(R ) = R , where n is the dimension of F(R ). For each p ∈ M, −1 n −1 define Φep : πe (Up) → Up × R as follows: for each x ∈ Up and v ∈ Eex = πe ({x}), set

Φep(v) = (x, (ϕ ◦ F(Φp|Ex ))(v)) where k ∼ k F(Φp|Ex ): F(Ex) → F({x} × R ) = F(R ) is an isomorphism. For each p, q ∈ M with Up ∩ Uq 6= ∅, we have −1 −1 −1 (Φep ◦ Φeq )(x, v) = (x, (ϕ ◦ F(Φp|Ex ) ◦ F(Φq|Ex ) ◦ ϕ )(v)) −1 −1 = (x, (ϕ ◦ F(Φp|Ex ◦ (Φq|Ex ) ) ◦ ϕ )(v))

= (x, τpq(x)v) for some smooth τpq since F is a smooth functor. Applying Lemma 10.6 completes the proof. Theorem 173. [Problem 10-9] Suppose M is a smooth manifold, E → M is a smooth vector bundle, and S ⊆ M is an embedded submanifold with or without boundary. For any smooth section σ of the restricted bundle E|S → S, there exist a neighborhood U of S in M and a smooth section σe of E|U such that σ = σe|S. If E has positive rank, then every smooth section of E|S extends smoothly to all of M if and only if S is properly embedded.

Proof. As in Theorem 82. Theorem 174. [Problem 10-11] Suppose E and E0 are smooth vector bundles over a smooth manifold M with or without boundary, and F : E → E0 is a bijective smooth bundle homomorphism over M. Then F is a smooth bundle isomorphism.

Proof. Let k be the rank of E and let k0 be the rank of E0; let π : E → M and π0 : E0 → M be the projections. We want to show that F −1 is smooth. Let v ∈ E and let Φ : π−1(U) → U × Rk be a smooth local trivialization of E satisfying π(v) ∈ U 0 0 −1 0 0 k0 0 and πU ◦ Φ = π. Let Φ :(π ) (U ) → U × R be a smooth local trivialization of E 0 0 0 0 0 −1 satisfying π (F (v)) ∈ U and πU 0 ◦ Φ = π . Consider the smooth map f = Φ ◦ F ◦ Φ with its inverse f −1 = Φ ◦ F −1 ◦ (Φ0)−1 deﬁned on a suﬃciently small neighborhood 0 k0 0 Ue ×R of Φ (F (v)). For each x ∈ U the map f|{x}×Rk is an isomorphism whose inverse −1 0 is f | k0 , where y(x) = π (F (ζ(x))) and ζ : M → E is the zero section. Since {y(x)}×R 63

0 operator inversion is smooth and y is smooth, f −1 is smooth on Ue0 ×Rk . Therefore F −1 −1 is smooth on a neighborhood of F (v), and since v was arbitrary, F is smooth.

Theorem 175. [Problem 10-12] Let π : E → M and πe : Ee → M be two smooth rank-k vector bundles over a smooth manifold M with or without boundary. Suppose {Uα}α∈A is an open cover of M such that both E and Ee admit smooth local trivializations over each Uα. Let {ταβ} and {τeαβ} denote the transition functions determined by the given local trivializations of E and Ee, respectively. Then E and Ee are smoothly isomorphic over M if and only if for each α ∈ A there exists a smooth map σα : Uα → GL(k, R) such that −1 (*) τeαβ(p) = σα(p)ταβ(p)σβ(p) , p ∈ Uα ∩ Uβ.

Proof. Suppose there is a smooth bundle isomorphism F : E → Ee. Let α ∈ A and let −1 k −1 k Φα : π (Uα) → Uα × R and Φeα : πe (Uα) → Uα × R be the associated smooth local trivializations. Since π ◦ Φ ◦ F = π ◦ F = π, the map Φ ◦ F : π−1(U ) → U × k Uα eα e eα α α R is a smooth local trivialization of E. Therefore −1 (Φeα ◦ F ◦ Φα )(p, v) = (p, σα(p)v) for some smooth linear map σα : Uα → GL(k, R). Now let α, β ∈ A. We have −1 (p, τeαβ(p)v) = (Φeα ◦ Φeβ )(p, v) −1 −1 −1 −1 = (Φeα ◦ F ◦ Φα ◦ Φα ◦ Φβ ◦ Φβ ◦ F ◦ Φeβ )(p, v) −1 = (p, σα(p)ταβ(p)σβ(p) v), −1 so τeαβ(p) = σα(p)ταβ(p)σβ(p) . Conversely, suppose that (*) holds. For each α ∈ A, −1 −1 −1 deﬁne a map Fα : π (Uα) → πe (Uα) by Fα = Φeα ◦ ϕα ◦ Φα, where ϕα(p, v) = (p, σα(p)v). If α, β ∈ A and Uα ∩ Uβ 6= ∅ then −1 Fβ = Φeβ ◦ ϕβ ◦ Φβ −1 −1 −1 = Φeα ◦ Φeα ◦ Φeβ ◦ ϕβ ◦ Φβ ◦ Φα ◦ Φα −1 = Φeα ◦ ϕα ◦ Φα

= Fα −1 −1 on π (Uα) ∩ π (Uβ) by (*), so the gluing lemma shows that there is a smooth map

−1 F : E → Ee such that F |π (Uα) = Fα for all α ∈ A. The restriction of F to each ﬁber −1 is clearly linear. On any π (Uα) we have −1 πe ◦ F = πe ◦ Φeα ◦ ϕα ◦ Φα

= πUα ◦ ϕα ◦ Φα

= πUα ◦ Φα 64

= π, so πe ◦ F = π on E. By Theorem 174, F is a smooth bundle isomorphism. Theorem 176. [Problem 10-14] Suppose M is a smooth manifold with or without boundary, and S ⊆ M is an immersed submanifold with or without boundary. Iden- tifying TpS as a subspace of TpM for each p ∈ S in the usual way, TS is a smooth subbundle of TM|S.

Proof. Let ι : S,→ M be the inclusion map. For each p ∈ S, let Dp = dιp(TpS). Let S D = p∈S Dp = TS ⊆ TM|S. Let p ∈ S, let V be a neighborhood of p in S that is an embedded submanifold of M, and let (U, ϕ) be a slice chart for V containing p. Write ϕ = (x1, . . . , xn) and let k be the dimension of S. Then V ∩ U is neighborhood of p in i S, and the maps σi : V ∩ U → TM|S given by σi(q) = dϕq(∂/∂x |q) for i = 1, . . . , k satisfy the condition in Lemma 10.32. Theorem 177. [Problem 10-17] Suppose M ⊆ Rn is an immersed submanifold. The n ambient tangent bundle T R |M is isomorphic to the Whitney sum TM ⊕ NM, where NM → M is the normal bundle.

Proof. Obvious. Theorem 178. [Problem 10-19] Suppose π : E → M is a ﬁber bundle with ﬁber F .

(1) π is an open quotient map. (2) If the bundle is smooth, then π is a smooth submersion. (3) π is a proper map if and only if F is compact. (4) E is compact if and only if both M and F are compact.

Proof. For (1), let V be open in E. Let x ∈ π(V ) and let Φ : π−1(U) → U × F be a local trivialization of E such that x ∈ U. Choose any f ∈ F and a neighborhood Ve ⊆ π−1(U) of Φ−1(x, f). Then Φ(Ve) is open in U × F since Φ is a homeomorphism, and πU (Φ(Ve)) = π(Ve) is a neighborhood of x contained in π(V ). This shows that π is open, and π is a quotient map since it is surjective. For (2), let v ∈ E and let Φ: π−1(U) → U × F be a smooth local trivialization of E such that π(v) ∈ U. Then πU ◦ Φ = π, so dπv = d(πU )π(v) ◦ dΦv is surjective since dΦv and d(πU )π(v) are both surjective. This shows that π is a smooth submersion. For (3), suppose that π is proper. If x ∈ M then π−1({x}) ≈ {x} × F ≈ F is compact. Conversely, suppose that F is compact and A ⊆ M is compact. For each p ∈ A, choose a local trivialization −1 Φ: π (Up) → Up × F such that p ∈ Up, and choose a precompact neighborhood Vp of p such that Vp ⊆ Up. Since A is compact, the open cover {Vp} has a ﬁnite subcover 65

−1 {Vp1 ,...,Vpk }. Then π (A) is a closed subset of k k [ −1 ∼ [ π (Vpi ) = Vpi × F, i=1 i=1 which is compact. Therefore π−1(A) is compact. For (4), if E is compact then M = π(E) is compact and F is compact by (3) since π is proper. Conversely, if M and F −1 are compact then π is proper by (3), so E = π (M) is compact.

Chapter 11. The Cotangent Bundle

Theorem 179. [Exercise 11.10] Suppose M is a smooth manifold and E → M is a smooth vector bundle over M. Deﬁne the dual bundle to E to be the bundle E∗ → M ∗ ` ∗ ∗ whose total space is the disjoint union E = p∈M Ep , where Ep is the dual space to Ep, with the obvious projection. Then E∗ → M is a smooth vector bundle, whose transition functions are given by τ ∗(p) = (τ(p)−1)T for any transition function τ : U → GL(k, R) of E.

Proof. Since ∗ ∗ −1 T −1 T ταβ(p)τβγ(p) = (ταβ(p) ) (τβγ(p) ) −1 T = ((ταβ(p)τβγ(p)) ) ∗ = ταγ(p), we can apply Theorem 170. Theorem 180. [Exercise 11.12] Let M be a smooth manifold with or without boundary, and let ω : M → T ∗M be a rough covector field. (1) ω is smooth. (2) In every smooth coordinate chart, the component functions of ω are smooth. (3) Each point of M is contained in some coordinate chart in which ω has smooth component functions. (4) For every smooth vector field X ∈ X(M), the function ω(X) is smooth on M. (5) For every open subset U ⊆ M and every smooth vector field X on U, the function ω(X): U → R is smooth on U. Proof. (1) ⇒ (2) ⇒ (3) is clear. If (3) holds for a coordinate chart (U, (xi)) and the i −1 ∗ natural coordinates (x , ξi) on π (U) ⊆ T M then the coordinate representation of ω on U is 1 n ωb(x) = (x , . . . , x , ω1(x), . . . , ωn(x)), which is smooth if ω1, . . . , ωn are smooth. This proves (3) ⇒ (1). This also implies (3) ⇒ (4). Suppose that (4) holds, U ⊆ M is open and X is a smooth vector field on U. 66

For each p ∈ U, choose a smooth bump function ψ that is equal to 1 on a neighborhood of p and supported in U. Deﬁne Xe = ψX, extended to be zero on M \ supp ψ. Then ω(Xe) is smooth and is equal to ω(X) in a neighborhood of p. So ω(X) is smooth in a neighborhood of every point in U, which proves (4) ⇒ (5). If (5) holds then (2) holds, i i since ωi = ω(∂/∂x ) for any smooth chart (U, (x )). Theorem 181. [Exercise 11.16] Let M be a smooth manifold with or without boundary, and let ω be a rough covector ﬁeld on M. If (εi) is a smooth coframe on an open subset U ⊆ M, then ω is smooth on U if and only if its component functions with respect to (εi) are smooth.

Proof. This follows from Proposition 10.22. Example 182. [Exercise 11.17] Let f(x, y) = x2 on R2, and let X be the vector ﬁeld ∂ X = grad f = 2x . ∂x Compute the coordinate expression for X in polar coordinates (on some open subset on which they are deﬁned) and show that it is not equal to ∂f ∂ ∂f ∂ + . ∂r ∂r ∂θ ∂θ Set x = r cos θ and y = r sin θ; then f(r, θ) = r2 cos2 θ, so ∂f ∂ ∂f ∂ ∂ ∂ + = 2r cos2 θ − 2r2 sin θ cos θ . ∂r ∂r ∂θ ∂θ ∂r ∂θ The derivative of the transition map ϕ(r, θ) = (r cos θ, r sin θ) is cos θ −r sin θ . sin θ r cos θ Therefore ∂ ∂ ∂f ∂ ∂f ∂ X = 2r cos θ cos θ + sin θ 6= + . ∂r ∂θ ∂r ∂r ∂θ ∂θ Theorem 183. [Exercise 11.21] Let M be a smooth manifold with or without boundary, and let f, g ∈ C∞(M). (1) If a and b are constants, then d(af + bg) = a df + b dg. (2) d(fg) = f dg + g df. (3) d(f/g) = (g df − f dg)/g2 on the set where g 6= 0. (4) If J ⊆ R is an interval containing the image of f, and h : J → R is a smooth function, then d(h ◦ f) = (h0 ◦ f) df. (5) If f is constant, then df = 0. 67

Proof. For (1), we have

d(af + bg)p(v) = v(af + bg) = avf + bvg = (a dfp + b dgp)(v). For (2), we have

d(fg)p(v) = v(fg) = f(p)vg + g(p)vf = (f dgp + g dfp)(v). For (3), if p ∈ M and (U, (xi)) is a smooth chart containing p then on U, n X ∂(f/g) d(f/g) = (x) dxi| x ∂xi x i=1 n 1 X ∂f ∂g = g(x) (x) − f(x) (x) dxi| g(x)2 ∂xi ∂xi x i=1 2 = (g(x) dfx − f(x) dgx)/g(x) . For (4), we have 0 d(h ◦ f)p(v) = dhf(p)(dfp(v)) = h (f(p))dfp(v). For (5), if (U, (xi)) is a smooth chart and fb is the coordinate representation of f on U then Dfb = 0, so fbis constant and f is constant on U. The result follows from Theorem 29. Theorem 184. [Exercise 11.24] For a smooth real-valued function f : M → R, p ∈ M is a critical point of f if and only if dfp = 0.

Proof. Obvious. Theorem 185. [Exercise 11.35] Let M be a smooth manifold with or without boundary. ∗ Suppose γ :[a, b] → M is a piecewise smooth curve segment, and ω, ω1, ω2 ∈ X (M).

(1) For any c1, c2 ∈ R,

(c1ω1 + c2ω2) = c1 ω1 + c2 ω2. ˆγ ˆγ ˆγ

(2) If γ is a constant map, then γ ω = 0. (3) If γ1 = γ|[a,c] and γ2 = γ|[c,b] with´ a < c < b, then

ω = ω + ω. ˆγ ˆγ1 ˆγ2 (4) If F : M → N is any smooth map and η ∈ X∗(N), then

F ∗η = η. ˆγ ˆF ◦γ 68

Proof. We can assume that γ is smooth rather than piecewise smooth. For (1),

∗ (c1ω1 + c2ω2) = γ (c1ω1 + c2ω2) ˆγ ˆ[a,b]

∗ ∗ = c1 γ ω1 + c2 γ ω2 ˆ[a,b] ˆ[a,b]

= c1 ω1 + c2 ω2 ˆγ ˆγ since γ∗ is linear. For (2), if γ is constant then dγ = 0 and γ∗ = 0. For (3),

ω = γ∗ω ˆγ ˆ[a,b] = γ∗ω + γ∗ω ˆ[a,c] ˆ[c,b] = ω + ω. ˆγ1 ˆγ2 For (4), F ∗η = γ∗F ∗η = (F ◦ γ)∗η = η. ˆγ ˆ[a,b] ˆ[a,b] ˆF ◦γ Theorem 186. [Exercise 11.41] A smooth covector ﬁeld ω is conservative if and only if its line integrals are path-independent, in the sense that ω = ω whenever γ and γ γe γe are piecewise smooth curve segments with the same starting´ and´ ending points.

Proof. Suppose that ω is conservative, γ is defined on [a, b], and γe is defined on [c, d]. Define γ − γe :[a, b + d − c] → M by ( γ(t), 0 ≤ t ≤ b, (γ − γe)(t) = γe(b + d − t), b ≤ t ≤ b + d − c. Then γ − γe is piecewise smooth, so 0 = ω = ω − ω ˆ ˆ ˆ γ−γe γ γe by Theorem 185. The converse is obvious since any piecewise smooth closed curve segment has the same starting and ending points as a constant curve segment. Theorem 187. [Problem 11-1] (1) Suppose V and W are finite-dimensional vector spaces and A : V → W is any linear map. The following diagram commutes: 69

A V W

ξV ξW

V ∗∗ W ∗∗ (A∗)∗

where ξV and ξW denote the natural isomorphisms. (2) There does not exist a way to assign to each ﬁnite-dimensional vector space V ∗ an isomorphism βV : V → V such that for every linear map A : V → W , the following diagram commutes: A V W

βV βW

V ∗ W ∗ A∗

Proof. For (1), we have ∗ ∗ ∗ ∗ ((A ) (ξV v))ω = (ξV v)(A ω) = (A ω)v = ω(Av) = (ξW (Av))ω.

For (2), suppose that there are such isomorphisms βV . Take V = W = R. The diagram commutes when A = 0, so

βR = 0 ◦ βR ◦ 0 = 0.

This contradicts the fact that βR is an isomorphism.

Theorem 188. [Problem 11-3] Let Vec1 be the category of ﬁnite-dimensional vector spaces and linear isomorphisms as in Theorem 172. Deﬁne a functor F : Vec1 → Vec1 by setting F(V ) = V ∗ for a vector space V , and F(A) = (A−1)∗ for an isomorphism A. Then F is a smooth covariant functor, and for every M, F(TM) and T ∗M are canonically smoothly isomorphic vector bundles.

Proof. F is covariant because ((A ◦ B)−1)∗ = (B−1 ◦ A−1)∗ = (A−1)∗ ◦ (B−1)∗, and it is smooth because matrix inversion and transposition are smooth. So Theorem 172 shows that F(TM) exists, and we can deﬁne a smooth vector bundle isomorphism ∗ ∗ ∗ ϕ : F(TM) → T M by setting ϕ|Tp M = IdTp M for each p ∈ M. Theorem 189. [Problem 11-4] Let M be a smooth manifold with or without boundary ∞ and let p be a point of M. Let Jp denote the subspace of C (M) consisting of smooth 70

2 functions that vanish at p, and let Jp be the subspace of Jp spanned by functions of the form fg for some f, g ∈ Jp. 2 (1) f ∈ Jp if and only if in any smooth local coordinates, its ﬁrst-order Taylor polynomial at p is zero. ∗ 2 (2) Deﬁne a map Φ: Jp → Tp M by setting Φ(f) = dfp. The restriction of Φ to Jp 2 ∗ is zero, and Φ descends to a vector space isomorphism from Jp/Jp to Tp M.

2 Pk Proof. If f ∈ Jp then f = i=1 αigihi for some αi ∈ R and gi, hi ∈ Jp. If in smooth local coordinates we have Df1(p) = Df2(p) = 0, then

D(gihi)(p)u = gi(p)[Dhi(p)u] + [Dgi(p)u]hi(p) = 0, so the ﬁrst-order Taylor polynomial of f is zero by linearity. Conversely, if Df(p) = 0 then Theorem C.15 shows that locally we can write n X 1 ∂2f f(x) = (xi − pi)(xj − pj) (1 − t) (p + t(x − p)) dt, ˆ ∂xi∂xj i,j=1 0 2 i i j j which is an element of Jp since x − p and x − p times the integral are elements of Pk Jp. Therefore f = i=1 αigihi for some αi ∈ R and gi, hi ∈ Jp in a neighborhood U of p. Choose a precompact neighborhood V of p with V ⊆ U and choose a partition of unity {ψ0, ψ1} subordinate to the open cover U, M \ V . We take gi = hi = 0 on M \ supp ψ0. Then ψ1, f ∈ Jp and k X f = αi(ψ0gi)hi + ψ1f, i=1 which proves (1). Part (2) follows immediately. Theorem 190. [Problem 11-5] For any smooth manifold M, T ∗M is a trivial vector bundle if and only if TM is trivial.

Proof. If TM is trivial then it admits a smooth global frame; Lemma 11.14 shows that its dual coframe is a smooth global coframe for M, so T ∗M is trivial. The converse is similar. Theorem 191. [Problem 11-6] Suppose M is a smooth n-manifold, p ∈ M, and y1, . . . , yk are smooth real-valued functions deﬁned on a neighborhood of p in M.

1 n ∗ 1 n (1) If k = n and (dy |p, . . . , dy |p) is a basis for Tp M then (y , . . . , y ) are smooth coordinates for M in some neighborhood of p. 1 k (2) If (dy |p, . . . , dy |p) is a linearly independent k-tuple of covectors and k < n, then there are smooth functions yk+1, . . . , yn such that (y1, . . . , yn) are smooth coordinates for M in a neighborhood of p. 71

1 k ∗ i1 in (3) If (dy |p, . . . , dy |p) span Tp M, there are indices i1, . . . , in such that (y , . . . , y ) are smooth coordinates for M in a neighborhood of p.

1 k Proof. Part (1) follows from the inverse function theorem. For (2), complete (dy |p, . . . , dy |p) 1 k k+1 n i i to a basis (dy |p, . . . , dy |p, ω , . . . , ω ), choose smooth functions y such that dy |p = i ω for i = k + 1, . . . , n, and apply (1). For (3), choose indices i1, . . . , in such that i1 in ∗ dy |p, . . . , dy |p is a basis for Tp M and apply (1). Example 192. [Problem 11-7] In the following problems, M and N are smooth manifolds, F : M → N is a smooth map, and ω ∈ X∗(N). Compute F ∗ω in each case.

(1) M = N = R2; F (s, t) = (st, et); ω = x dy − y dx. (2) M = R2 and N = R3; F (θ, ϕ) = ((cos ϕ + 2) cos θ, (cos ϕ + 2) sin θ, sin ϕ); 2 ω = z dx. √ (3) M = {(s, t) ∈ R2 : s2 + t2 < 1} and N = R3 \{0}; F (s, t) = (s, t, 1 − s2 − t2); ω = (1 − x2 − y2) dz.

For (1), dx = t ds + s dt, dy = et dt, so ω = st(et dt) − et(t ds + s dt) = −tet ds + set(t − 1) dt. For (2), dx = −(cos ϕ + 2) sin θ dθ − sin ϕ cos θ dϕ, so ω = sin2 ϕ(−(cos ϕ + 2) sin θ dθ − sin ϕ cos θ dϕ). For (3), dx = ds, dy = dt, −s −t dz = √ ds + √ dt, 1 − s2 − t2 1 − s2 − t2 so √ √ ω = −s 1 − s2 − t2 ds − t 1 − s2 − t2 dt. Theorem 193. [Problem 11-8]

(1) Suppose F : M → N is a diﬀeomorphism, and let dF ∗ : T ∗N → T ∗M be the ∗ ∗ map whose restriction to each cotangent space Tq N is equal to dFF −1(q). Then dF ∗ is a smooth bundle homomorphism. 72

(2) Let Diff1 be the category whose objects are smooth manifolds, but whose only morphisms are diffeomorphisms; and let VB be the category whose objects are smooth vector bundles and whose morphisms are smooth bundle homomorphisms. ∗ ∗ The assignment M 7→ T M, F 7→ dF defines a contravariant functor from Diff1 to VB, called the cotangent functor.

Proof. It is clear that dF ∗ is a smooth bundle homomorphism covering F −1. Since ∗ ∗ ∗ ∗ d(F ◦ G)p = (dFG(p) ◦ dGp) = dGp ◦ dFG(p) ∗ ∗ ∗ ∗ ∗ we have d(F ◦ G) = dG ◦ dF . Also, d(IdM ) = IdT ∗M . This shows that M 7→ T M, ∗ F 7→ dF deﬁnes a contravariant functor. Example 194. [Problem 11-9] Let f : R3 → R be the function f(x, y, z) = x2 +y2 +z2, and let F : R2 → R3 be the following map (the inverse of the stereographic projection): 2u 2v u2 + v2 − 1 F (u, v) = , , . u2 + v2 + 1 u2 + v2 + 1 u2 + v2 + 1 Compute F ∗df and d(f ◦ F ) separately, and verify that they are equal. We have df = 2x dx + 2y dy + 2z dz and 2(u2 + v2 + 1) − 4u2 −4uv dx = du + dv, (u2 + v2 + 1)2 (u2 + v2 + 1)2 −4uv 2(u2 + v2 + 1) − 4v2 dy = du + dv, (u2 + v2 + 1)2 (u2 + v2 + 1)2 4u 4v dz = du + dv, (u2 + v2 + 1)2 (u2 + v2 + 1)2 so df = 0. Alternatively we have f ◦ F = 1, so d(f ◦ F ) = 0. Example 195. [Problem 11-10] In each of the cases below, M is a smooth manifold and f : M → R is a smooth function. Compute the coordinate representation for df, and determine the set of all points p ∈ M at which dfp = 0. (1) M = {(x, y) ∈ R2 : x > 0}; f(x, y) = x/(x2 + y2). Use standard coordinates (x, y). (2) M and f are as in (1); this time use polar coordinates (r, θ). (3) M = S2 ⊆ R3; f(p) = z(p) (the z-coordinate of p as a point in R3). Use north and south stereographic coordinates. 2 (4) M = Rn; f(x) = |x| . Use standard coordinates. 73

For (1), −x2 + y2 −2xy df = dx + dy, (x2 + y2)2 (x2 + y2)2 which is zero when x2 = y2 and xy = 0, i.e. never on M. For (2) we have cos θ f(r, θ) = , r so cos θ sin θ df = − dr − dθ, r2 r which is zero when cos θ = sin θ = 0, i.e. never on M. For (3) we have u2 + v2 − 1 f(u, v) = u2 + v2 + 1 on S2 \ N using the north stereographic projection, so 4u 4v df = du + dv, (u2 + v2 + 1)2 (u2 + v2 + 1)2 which is zero when u = v = 0, i.e. when p is the south pole (0, 0, −1). A similar calculation using the south stereographic projection shows that dfp = 0 when p is the north pole (0, 0, 1). For (4) we have df = 2x1 dx1 + ··· + 2xn dxn, which is zero when p = 0. Theorem 196. [Problem 11-11] Let M be a smooth manifold, and C ⊆ M be an embedded submanifold. Let f ∈ C∞(M), and suppose p ∈ C is a point at which f attains a local maximum or minimum value among points in C. Given a smooth local deﬁning function Φ: U → Rk for C on a neighborhood U of p in M, there are real numbers λ1, . . . , λk (called Lagrange multipliers) such that 1 k dfp = λ1dΦ |p + ··· + λkdΦ |p.

Proof. Let n be the dimension of M; then C has dimension n − k. Let (V, ϕ) be a smooth slice chart for C in M centered at p such that V ⊆ U, let fe = f ◦ ϕ−1, let −1 Φe = Φ◦ϕ , and let pe = ϕ(p). It suﬃces to show that there are real numbers λ1, . . . , λk such that 1 k Dfe(p) = λ1DΦe (p) + ··· + λkDΦe (p). −1 n Since Φ (0) ⊆ V ∩ C, this follows from the method of Lagrange multipliers on R . Theorem 197. [Problem 11-12] Any two points in a connected smooth manifold can be joined by a smooth curve segment. 74

Proof. Let M be a connected smooth n-manifold. Define an equivalence relation ∼ on M by setting p ∼ q if and only if there is a smooth curve segment from p to q. This relation is clearly reflexive and symmetric; if we can show that it is transitive, then the result follows from Theorem 29. Suppose p ∼ q and q ∼ r, let γ :[−1, 0] → M be a smooth curve segment from p to q, and let γ0 : [0, 1] → M be a smooth curve segment from q to r. Let (U, ϕ) be a smooth chart centered at q; by shrinking U, we may assume that ϕ(U) = Bn. Choose 0 < ε < 1 so that γ((−ε, 0]) ⊆ U and γ0([0, ε)) ⊆ U. 0 0 n Let γe = ϕ ◦ γ :(−ε, 0] → U and γe = ϕ ◦ γ : [0, ε) → U. Let ` :(−ε, ε) → B be given by t 7→ (t/ε, 0,..., 0). By Proposition 2.25, there is a smooth bump function n ψ :(−ε, ε) → R for [−ε/4, ε/4] supported in (−ε/2, ε/2). Define αe :(−ε, ε) → B by 0 t 7→ ψ(t)`(t) + (1 − ψ(t))(γe(t) + γe (t)), taking γ = 0 on (0, ε) and γ0 = 0 on (−ε, 0). It is easy to check that α is smooth on e e −1 e (−ε, 0), (−ε/4, ε/4) and (0, ε), so α is smooth. Let α = ϕ ◦ α; then α, γ|[−1,ε/2), and 0 e e γ |(ε/2,1] agree on their overlaps, so we have a smooth curve segment defined on [−1, 1] from p to r. Theorem 198. [Problem 11-13] The length of a smooth curve segment γ :[a, b] → Rn is defined to be the value of the (ordinary) integral b L(γ) = |γ0(t)| dt. â There is no smooth covector field ω ∈ X∗(Rn) with the property that ω = L(γ) ˆγ for every smooth curve γ.

Proof. Let −γ denote the curve t 7→ γ(b − t + a). Then L(−γ) = L(γ), but

ω = − ω = −L(γ). ˆ−γ ˆγ Example 199. [Problem 11-14] Consider the following two covector fields on R3: 4z dx 2y dy 2x dz ω = − + + , (x2 + 1)2 y2 + 1 x2 + 1 4xz dx 2y dy 2 dz η = − + + . (x2 + 1)2 y2 + 1 x2 + 1 (1) Set up and evaluate the line integral of each covector field along the straight line segment from (0, 0, 0) to (1, 1, 1). (2) Determine whether either of these covector fields is exact. 75

(3) For each one that is exact, ﬁnd a potential function and use it to recompute the line integral.

Let γ : [0, 1] → R3 be given by t 7→ (t, t, t). Then 1 4t 2t 2t ω = − 2 2 + 2 + 2 dt ˆγ ˆ0 (t + 1) t + 1 t + 1 1 4t3 = 2 2 dt ˆ0 (1 + t ) = 2 log 2 − 1, and 1 4t2 2t 2 η = − 2 2 + 2 + 2 dt ˆγ ˆ0 (t + 1) t + 1 t + 1 1 2(t3 − t2 + t + 1) = 2 2 dt ˆ0 (1 + t ) = log 2 + 1. It is easy to check that ∂ ∂ ω 6= ω , ∂z 1 ∂x 3 so ω is not closed and not exact. It is similarly easy to check that η is closed, so Theorem 11.49 shows that η is exact. Suppose f is a potential for η; it must satisfy ∂f 4xz ∂f 2y ∂f 2 = − , = , = . ∂x (x2 + 1)2 ∂y y2 + 1 ∂z x2 + 1 Integrating the first equation, we have 2z f(x, y, z) = + C (y, z). x2 + 1 1 Then ∂C 2y 1 = , ∂y y2 + 1 so 2 C1(y, z) = log(y + 1) + C2(z). Finally we have 2 ∂C 2 + 2 = , x2 + 1 ∂z x2 + 1 so C2 is constant. It is now easy to check that for any C ∈ R, 2z f(x, y, z) = + log(y2 + 1) + C x2 + 1 76 is a potential for η. Therefore 2 η = df = + log(2) − 0 = log 2 + 1. ˆγ ˆγ 2 Theorem 200. [Problem 11-15] Let X be a smooth vector field on an open subset U ⊆ Rn. Given a piecewise smooth curve segment γ :[a, b] → U, define the line integral of X over γ, denoted by γ X · ds, as ´ b 0 X · ds = Xγ(t) · γ (t) dt, ˆγ â where the dot on the right-hand side denotes the Euclidean dot product between tangent vectors at γ(t), identified with elements of Rn.A conservative vector field is one whose line integral around every piecewise smooth closed curve is zero. (1) X is conservative if and only if there exists a smooth function f ∈ C∞(U) such that X = grad f. (2) Suppose n = 3. If X is conservative then curl X = 0, where ∂X3 ∂X2 ∂ ∂X1 ∂X3 ∂ ∂X2 ∂X1 ∂ curl X = − + − + − . ∂x2 ∂x3 ∂x1 ∂x3 ∂x1 ∂x2 ∂x1 ∂x2 ∂x3 (3) If U ⊆ R3 is star-shaped, then X is conservative on U if and only if curl X = 0.

Proof. For any X ∈ X(U), define a smooth covector field ω by ωx(v) = Xx · v; then b 0 X · ds = ωγ(t)(γ (t)) dt = ω. ˆγ â ˆγ Part (1) follows immediately from Theorem 11.42, and part (2) follows from Proposition 11.44. Part (3) follows from Theorem 11.49. Theorem 201. [Problem 11-16] Let M be a compact manifold of positive dimension. Every exact covector field on M vanishes at least at two points in each component of M.

Proof. Let ω = df be an exact covector ﬁeld. Let U be a component of M. Since U is compact, f attains a maximum at some x ∈ U and a minimum at some y ∈ U. If x = y then f is constant on U, so df = 0 on U. Otherwise, df vanishes at the distinct points x and y. Theorem 202. [Problem 11-17] Let Tn = S1 × · · · × S1 ⊆ Cn denote the n-torus. For n each j = 1, . . . , n, let γj : [0, 1] → T be the curve segment 2πit γj(t) = (1, . . . , e ,..., 1). A closed covector ﬁeld ω on n is exact if and only if ω = 0 for j = 1, . . . , n. T γj ´ 77

1 n Proof. Let εn : Rn → Tn be the smooth covering map εn(x1, . . . , xn) = (e2πix , . . . , e2πix ). Note that εn is a local diﬀeomorphism, so Corollary 11.46 shows that (εn)∗ω is closed and Theorem 11.49 shows that (εn)∗ω is exact. If ω is exact then ω = 0 since each γj γj is a closed curve segment. Conversely, suppose that ω = 0 for´ j = 1, . . . , n. Let γj n ´ n γ : [0, 1] → T be a piecewise smooth closed curve segment and let γe : [0, 1] → R be a lift of γ such that γ = εn ◦ γ; Theorem 41 shows that γ is smooth. Write γ(0) = 1 n 1 e n e e (x , . . . , x ) and γ(1) = (x +m1, . . . , x +mn) for integers m1, . . . , mn. Assume without e 1 n n loss of generality that x = ··· = x = 0. For each i = 1, . . . , n, let αi : [0, 1] → R be the straight line path from (m1, . . . , mi−1, 0,..., 0) to (m1, . . . , mi,..., 0). Let α be the concatenation of α1, . . . , αn so that α(0) = 0 and α(1) = (m1, . . . , mn). Then Theorem 186 shows that

ω = ω ˆ ˆ n γ ε ◦γe = (εn)∗ω ˆ γe = (εn)∗ω ˆα = (εn)∗ω + ··· + (εn)∗ω, ˆα1 ˆαn = ω + ··· + ω n n ˆε ◦α1 ˆε ◦αn = 0

n since each ε ◦ αi is the concatenation of zero or more copies of γi. Theorem 203. [Problem 11-18] Suppose C and D are categories, and F, G are (covariant or contravariant) functors from C to D.A natural transformation λ from F to G is a rule that assigns to each object X ∈ Ob(C) a morphism λX ∈ HomD(F(X), G(X)) in such a way that for every pair of objects X,Y ∈ Ob(C) and every morphism f ∈ HomC(X,Y ), the following diagram commutes:

F(f) F(X) F(Y )

λX λY

G(X) G(Y ). G(f) 78

(If either F or G is contravariant, the corresponding horizontal arrow should be re- versed.)

(1) Let VecR denote the category of real vector spaces and linear maps, and let D be the contravariant functor from VecR to itself that sends each vector space to its dual space and each linear map to its dual map. The assignment V 7→ ξV , where ∗∗ ξV : V → V is the map defined by ξV (v)ω = ω(v), is a natural transformation from the identity functor of VecR to D ◦ D. (2) There does not exist a natural transformation from the identity functor of VecR to D. (3) Let Diff1 be the category of smooth manifolds and diffeomorphisms and VB the category of smooth vector bundles and smooth bundle homomorphisms, and let ∗ T,T : Diff1 → VB be the tangent and cotangent functors, respectively. There does not exist a natural transformation from T to T ∗.

(4) Let X : Diﬀ1 → VecR be the covariant functor given by M 7→ X(M), F 7→ F∗; and let X × X : Diﬀ1 → VecR be the covariant functor given by M 7→ X(M) × X(M), F 7→ F∗ × F∗. The Lie bracket is a natural transformation from X × X to X.

Proof. Theorem 187 proves (1) and (2). If there is a natural transformation from T to T ∗, then taking the one-point space {∗} produces a natural transformation from the identity functor of VecR to D, contradicting (2). This proves (3). Corollary 8.31 proves (4).

Chapter 12. Tensors

Theorem 204. [Exercise 12.18] T kT ∗M, T kTM, and T (k,l)TM have natural structures as smooth vector bundles over M.

Proof. Let n = dim M. Let (U, ϕ) be a smooth chart for M with coordinate functions (xi). Deﬁne Φ : π−1(U) → U × Rn(k+l) by ! i1,...,ik ∂ ∂ j j Φ a ⊗ · · · ⊗ ⊗ dx 1 ⊗ · · · ⊗ dx l j1,...,jl i1 ik p p ∂x p ∂x p 1,...,1 n,...,n = (p, (a1,...,1, . . . , an...,n )).

i Suppose (U,e ϕe) is another smooth chart with coordinate functions (xe ), and let Φ:e π−1(Ue) → Ue × Rn be deﬁned analogously. It follows from (11.6) and (11.7) that ∂ ∂xi ∂ ∂xj = (p) , dxj| = e (p) dxi| , j j i e p i p ∂xe p ∂xe ∂x p ∂x 79

−1 −1 −1 where xe = ϕe ◦ ϕ and x = ϕ ◦ ϕe = xe , and p denotes either a point in M or its coordinate representation as appropriate. Therefore on π−1(U ∩ Ue) we have −1 1,...,1 n,...,n (Φ ◦ Φe )(p, (ea1,...,1,..., ean...,n )) ∂x1 ∂x1 ∂xj1 ∂xjl = p, (p) ··· (p) e (p) ··· e (p)ai1,...,ik , i1 i 1 1 ej1,...,jl ∂xe ∂xe k ∂x ∂x ∂xn ∂xn ∂xj1 ∂xjl ..., (p) ··· (p) e (p) ··· e (p)ai1,...,ik . i1 i n n ej1,...,jl ∂xe ∂xe k ∂x ∂x It follows from the vector bundle chart lemma that T ∗M has a smooth structure making it into a smooth vector bundle for which the maps Φ are smooth local trivializations. Theorem 205. [Proposition 12.25] Suppose F : M → N and G : N → P are smooth maps, A and B are covariant tensor ﬁelds on N, and f is a real-valued function on N. (1) F ∗(fB) = (f ◦ F )F ∗B. (2) F ∗(A ⊗ B) = F ∗A ⊗ F ∗B. (3) F ∗(A + B) = F ∗A + F ∗B. (4) F ∗B is a (continuous) tensor ﬁeld, and is smooth if B is smooth. (5)( G ◦ F )∗B = F ∗(G∗B). ∗ (6) (IdN ) B = B.

Proof. For (1), ∗ ∗ ∗ ∗ (F (fB))p = dFp ((fB)F (p)) = dFp (f(F (p))BF (p)) = (f ◦ F )(p)(F B)p ∗ since dFp is linear. For (2), ∗ (F (A ⊗ B))p(v1, . . . , vk+l) = (A ⊗ B)p(dFp(v1), . . . , dFp(vk+l))

= Ap(dFp(v1), . . . , dFp(vk))Bp(dFp(vk+1), . . . , dFp(vk+l)) ∗ ∗ = (F A)p(v1, . . . , vk)(F B)p(vk+1, . . . , vk+l). ∗ Part (3) follows from the fact that dFp is linear. For (4), let p ∈ M be arbitrary, and choose smooth coordinates (yj) for N in a neighborhood V of F (p). Let U = F −1(V ), which is a neighborhood of p. Writing B in coordinates as

j1 jk B = Bj1,...,jk dy ⊗ · · · ⊗ dy for continuous functions Bj1,...,jk on V , we have

∗ ∗ j1 jk F B = F (Bj1,...,jk dy ⊗ · · · ⊗ dy )

∗ j1 jk = (Bj1,...,jk ◦ F )F (dy ⊗ · · · ⊗ dy )

∗ j1 ∗ jk = (Bj1,...,jk ◦ F )F dy ⊗ · · · ⊗ F dy

j1 jk = (Bj1,...,jk ◦ F )d(y ◦ F ) ⊗ · · · ⊗ d(y ◦ F ), 80 which is continuous, and is smooth if B is smooth. Part (5) follows from the fact that ∗ d(G ◦ F )p(α)(v1, . . . , vk) = α(dGF (p)(dFp(v1)), . . . , dGF (p)(dFp(vk))) ∗ = dGF (p)(α)(dFp(v1), . . . , dFp(vk)) ∗ ∗ = dFp (dGF (p)(α))(v1, . . . , vk). Part (6) is obvious. Example 206. [Problem 12-1] Give an example of finite-dimensional vector spaces V and W and a specific element α ∈ V ⊗ W that cannot be expressed as v ⊗ w for v ∈ V and w ∈ W . 2 Take V = W = R , let e1 = (1, 0) and e2 = (0, 1), and let α = e1 ⊗e1 +e2 ⊗e2. Suppose 2 i j i j α = v ⊗ w for v, w ∈ R , and write v = v ei and w = w ej. Then α = v w ei ⊗ ej, so v1w1 = 1, v1w2 = 0, v2w1 = 0, v2w2 = 1. This implies that w1 6= 0, so v2 = 0, contradicting v2w2 = 1. Theorem 207. [Problem 12-2] For any finite-dimensional real vector space V , there ∼ ∼ are canonical isomorphisms R ⊗ V = V = V ⊗ R.

Proof. We can deﬁne a multilinear map f : R × V → V by (r, v) 7→ rv, which induces a linear map fe : R ⊗ V → V satisfying fe(r ⊗ v) = rv. It is easy to check that the linear map v 7→ 1 ⊗ v is an inverse to fe, so fe is an isomorphism. A similar argument shows ∼ that V = V ⊗ R. Theorem 208. [Problem 12-3] Let V and W be ﬁnite-dimensional real vector spaces. The tensor product space V ⊗ W is uniquely determined up to canonical isomorphism by its characteristic property. More precisely, suppose πe : V ×W → Z is a bilinear map into a vector space Z with the following property: for any bilinear map B : V ×W → Y , there is a unique linear map Be : Z → Y such that the following diagram commutes: B V × W Y.

πe Be Z

Then there is a unique isomorphism Φ: V ⊗ W → Z such that πe = Φ ◦ π, where π : V × W → V ⊗ W is the canonical projection.

Proof. Since πe is bilinear, there exists a unique linear map Φ : V ⊗ W → Z such that πe = Φ ◦ π. Similarly, there exists a unique linear map Ψ : Z → V ⊗ W such that 81

π = Ψ ◦ πe. We have Φ ◦ Ψ ◦ πe = Φ ◦ π = πe, so Φ ◦ Ψ = IdZ by uniqueness. Similarly, Ψ ◦ Φ ◦ π = Ψ ◦ πe = π implies that Ψ ◦ Φ = IdV ⊗W by uniqueness. Therefore Φ is an isomorphism.

Theorem 209. [Problem 12-4] Let V1,...,Vk and W be ﬁnite-dimensional real vector spaces. There is a canonical isomorphism ∗ ∗ ∼ V1 ⊗ · · · ⊗ Vk ⊗ W = L(V1,...,Vk; W ).

Proof. Proposition 12.10 shows that ∗ ∗ ∼ V1 ⊗ · · · ⊗ Vk = L(V1,...,Vk; R), so it remains to show that ∼ L(V1,...,Vk; R) ⊗ W = L(V1,...,Vk; W ).

The bilinear map ϕ : L(V1,...,Vk; R) × W → L(V1,...,Vk; W ) deﬁned by

ϕ(f, w)(v1, . . . , vk) = f(v1, . . . , vk)w induces a linear map ϕe : L(V1,...,Vk; R) ⊗ W → L(V1,...,Vk; W ) satisfying ϕe(f ⊗ w)(v1, . . . , vk) = f(v1, . . . , vk)w. It is easy to check that ϕe takes a basis of L(V1,...,Vk; R)⊗W to a basis of L(V1,...,Vk; W ), so ϕe is an isomorphism. Theorem 210. [Problem 12-5] Let V be an n-dimensional real vector space. Then n + k − 1 dim Σk(V ∗) = . k

Proof. Let {ε1, . . . , εn} be a basis for V ∗; we want to show that

i1 ik B = ε ··· ε : 1 ≤ i1 ≤ · · · ≤ ik ≤ n is a basis for Σk(V ∗), from which the result follows immediately. It is clear that B is linearly independent. Let α ∈ Σk(V ∗) and write

i1 ik α = ai1,...,ik ε ⊗ · · · ⊗ ε . We have

i1 ik α = Sym α = ai1,...,ik Sym ε ⊗ · · · ⊗ ε

i1 ik = ai1,...,ik ε ··· ε ,

i1 ik which is in the span of B after reordering the indices i1, . . . , ik in each term ε ··· ε k ∗ to be in ascending order. This shows that B is a basis for Σ (V ). Theorem 211. [Problem 12-6] 82

(1) Let α be a covariant k-tensor on a finite-dimensional real vector space V . Then Sym α is the unique symmetric k-tensor satisfying (Sym α)(v, . . . , v) = α(v, . . . , v) for all v ∈ V . (2) The symmetric product is associative: for all symmetric tensors α, β, γ, (αβ)γ = α(βγ). (3) Let ω1, . . . , ωk be covectors on a finite-dimensional vector space. Their symmetric product satisfies 1 X ω1 ··· ωk = ωσ(1) ⊗ · · · ⊗ ωσ(k). k! σ∈Sk

Proof. Let {E1,...,Ek} be a basis for V . For (1), 1 X (Sym α)(v, . . . , v) = α(v, . . . , v) = α(v, . . . , v). k! σ∈Sk The uniqueness of Sym α follows from the fact that ⊗k (Ei1 + ··· + Ein ) : 1 ≤ n ≤ k, 1 ≤ i1 < ··· < in ≤ k spans V ⊗k. For (2), we have (Sym(Sym α ⊗ β) ⊗ γ)(v, . . . , v) = (Sym α ⊗ β)(v, . . . , v)γ(v, . . . , v) = α(v, . . . , v)β(v, . . . , v)γ(v, . . . , v) = α(v, . . . , v)(Sym β ⊗ γ)(v, . . . , v) = (Sym α ⊗ (Sym β ⊗ γ))(v, . . . , v), so the result follows from (1). (Note that the tuples (v, . . . , v) may have diﬀerent numbers of elements.) Part (3) is obvious. Example 212. [Problem 12-7] Let (e1, e2, e3) be the standard dual basis for (R3)∗. The tensor e1 ⊗e2 ⊗e3 is not equal to a sum of an alternating tensor and a symmetric tensor. 3 Let (e1, e2, e3) be the standard basis for R . Suppose that e1 ⊗ e2 ⊗ e3 = α + β where α is alternating and β is symmetric. We have 1 2 3 1 = (e ⊗ e ⊗ e )(e1, e2, e3) = α(e1, e2, e3) + β(e1, e2, e3) but 1 2 3 0 = (e ⊗ e ⊗ e )(e2, e3, e1) = α(e2, e3, e1) + β(e2, e3, e1)

= α(e1, e2, e3) + β(e1, e2, e3), 83 which is a contradiction.

Chapter 13. Riemannian Metrics

Theorem 213. [Exercise 13.21] Let (M, g) be a Riemannian n-manifold with or without boundary. For any immersed k-dimensional submanifold S ⊆ M with or without boundary, the normal bundle NS is a smooth rank-(n − k) subbundle of TM|S. For each p ∈ S, there is a smooth frame for NS on a neighborhood of p that is orthonormal with respect to g.

Proof. Let p ∈ M be arbitrary, and let (X1,...,Xk) be a smooth local frame for TS over some neighborhood V of p in M. Because immersed submanifolds are locally embedded, by shrinking V if necessary, we may assume that it is a single slice in some coordinate ball or half-ball U ⊆ Rn. Since V is closed in U, Theorem 122(c) shows that we can complete (X1,...,Xk) to a smooth local frame (Xe1,..., Xen) for TM|S over U, and then Proposition 13.6 yields a smooth orthonormal frame (Ej) over U such that span(E1|p,...,Ek|p) = span(X1|p,...,Xk|p) = TpS for each p ∈ U. It follows that (Ek+1,...,En) restricts to a smooth orthonormal frame for NS over V . Thus NS satisﬁes the local frame criterion in Lemma 10.32, and is therefore a smooth subbundle of TM|S.

Theorem 214. [Exercise 13.24] Suppose (M, g) and (M,f ge) are Riemannian manifolds with or without boundary, and F : M → M is a local isometry. Then L (F ◦γ) = L (γ) f ge g for every piecewise smooth curve segment γ in M.

Proof. We have b 0 Lg(F ◦ γ) = |(F ◦ γ) (t)| dt e ge â b 0 = dFγ(t)(γ (t)) g dt â e b q 0 0 = ge(F ◦γ)(t)(dFγ(t)(γ (t)), dFγ(t)(γ (t))) dt â b p ∗ 0 0 = (F ge)(γ (t), γ (t)) dt â b = pg(γ0(t), γ0(t)) dt â = Lg(γ). 84

Theorem 215. [Exercise 13.27] Suppose (M, g) and (M,f ge) are connected Riemannian manifolds and F : M → Mf is a Riemannian isometry. Then d (F (p),F (q)) = d (p, q) ge g for all p, q ∈ M.

Proof. If γ is a piecewise smooth curve segment from p to q, then F ◦ γ is a piecewise smooth curve segment from F (p) to F (q), and Theorem 214 shows that L (γ) = L (F ◦ g ge −1 γ). Similarly, if γe is a piecewise smooth curve segment from F (p) to F (q) then F ◦ γe is a piecewise smooth curve segment from p to q and L (γ) = L (F −1 ◦ γ). ge e g e Theorem 216. [Problem 13-1] If (M, g) is a Riemannian n-manifold with or without n o boundary, let UM ⊆ TM be the subset UM = (x, v) ∈ TM : |v|g = 1 , called the unit tangent bundle of M. Then UM is a smooth ﬁber bundle over M with model ﬁber Sn−1.

Proof. Define ` : TM → R by (x, v) 7→ g(x)(v, v) and for fixed x ∈ M, define `x : TxM → R by v 7→ g(x)(v, v). In any smooth local coordinates around x we have D`(v)(u) = 2g(x)(v, u) and in particular D`(v)(v) = 2g(x)(v, v) > 0 for all v 6= 0. Therefore Corollary 5.14 shows that UM = `−1({1}) is a properly embedded submanifold of TM. Let π : TM → M be the canonical projection. Let Φ : π−1(U) → U × Rn be a smooth local trivialization. Define −1 n−1 ϕ : π (U) ∩ UM → U × S (π n ◦ Φ)(x, v) (x, v) 7→ x, R k(πRn ◦ Φ)(x, v)k and n−1 −1 ψ : U × S → π (U) ∩ UM Φ−1(x, v) (x, v) 7→ −1 . |Φ (x, v)|g Then ϕ and ψ are smooth and inverses of each other, so ϕ is a local trivialization of UM over M. Theorem 217. [Problem 13-2] Suppose that E is a smooth vector bundle over a smooth manifold M with or without boundary, and V ⊆ E is an open subset with the property that for each p ∈ M, the intersection of V with the fiber Ep is convex and nonempty. By a “section of V ”, we mean a (local or global) section of E whose image lies in V .

(1) There exists a smooth global section of V . 85

(2) Suppose σ : A → V is a smooth section of V deﬁned on a closed subset A ⊆ M. (This means that σ extends to a smooth section of V in a neighborhood of each point of A.) There exists a smooth global section σe of V whose restriction to A is equal to σ. If V contains the image of the zero section of E, then σe can be chosen to be supported in any predetermined neighborhood of A.

Proof. For (1), let p ∈ M and choose a local trivialization Φ : π−1(U) → U × Rn such −1 0 that p ∈ U. Choose any v ∈ π ({p}) and choose neighborhoods Up of p and Fp of v = −1 −1 0 (πRn ◦Φ)(v) such that Up ×Fp ⊆ Φ(π (U)∩V ). Define τp : Up → V by x 7→ Φ (x, v ). The family of sets {Up : p ∈ M} is an open cover of M. Let {ψp : p ∈ M} be a smooth partition of unity subordinate to this cover, with supp ψp ⊆ Up. For each p ∈ M, the product ψpτp is smooth on Wp by Theorem 159, and has a smooth extension to all of M if we interpret it to be zero on M \ supp ψp. (The extended function is smooth because the two definitions agree on the open subset Wp \ supp ψp where they overlap.) Thus we can define τ : M → V by X τ(x) = ψp(x)τp(x). p∈M

Because the collection of supports {supp ψp} is locally finite, this sum actually has only a finite number of nonzero terms in a neighborhood of any point of M, and therefore P defines a smooth function. The fact that p∈M ψp = 1 implies that τ(x) is a convex combination of vectors in Ex for every x ∈ M, so the image of τ does indeed lie in V . For (2), let τ : M → V be a smooth global section of V . If V contains the imaage of the zero section of E, then let U be some neighborhood of A and replace τ with the zero section of E. For each p ∈ A, choose a neighborhood Wp of p and a local section σep : Wp → V of V that agrees with σ on Wp ∩A. If U was chosen, then replacing Wp by Wp ∩ U we may assume that Wp ⊆ U. The family of sets {Wp : p ∈ A} ∪ {M \ A} is an open cover of M. Let {ψp : p ∈ A} ∪ {ψ0} be a smooth partition of unity subordinate to this cover, with supp ψp ⊆ Wp and supp ψ0 ⊆ M \ A. For each p ∈ A, the product ψpσep is smooth on Wp and has a smooth extension to all of M if we interpret it to be zero on M \ supp ψp. Thus we can define σe : M → V by X σe(x) = ψ0(x)τ(x) + ψp(x)σep(x). p∈A If x ∈ A, then ψ0(x) = 0 and σep(x) = σ(x) for each p such that ψp(x) 6= 0, so ! X X σe(x) = ψp(x)σ(x) = ψ0(x) + ψp(x) σ(x) = σ(x), p∈A p∈A P so σe is indeed an extension of σ. The fact that ψ0 + p∈A ψp = 1 implies that σe(x) is a convex combination of vectors in Ex for every x ∈ M, so the image of σe does indeed 86 lie in V . If U was chosen, then it follows from Lemma 1.13(b) that [ [ supp σe = supp ψp = supp ψp ⊆ U. p∈A p∈A Theorem 218. Suppose (M, g) is a Riemannian manifold, and X is a vector field on M. Given any positive continuous function δ : M → R, there exists a smooth vector field Xe ∈ X(M) that is δ-close to X. If X is smooth on a closed subset A ⊆ M, then Xe can be chosen to be equal to X on A. (Two vector fields X, Xe are δ-close if

Xx − Xex < δ(x) for all x ∈ M. g

Proof. If X is smooth on the closed subset A, then by Lemma 8.6, there is a smooth vector ﬁeld X0 ∈ X(M) that agrees with X on A. Let

U0 = {y ∈ M : |X0|y − Xy|g < δ(y)}.

Then U0 is an open subset containing A. (If there is no such set A, we just take U0 = A = ∅ and F0 = 0.) ∞ We will show that there are countably many points {xi}i=1 in M \ A and smooth ∞ coordinate balls Ui of xi in M \ A such that {Ui}i=1 is an open cover of M \ A and

(*) |Xy − Xi|y|g < δ(y) for all y ∈ Ui, where Xi ∈ X(Ui) is a constant coeﬃcient vector ﬁeld such that Xi|xi =

Xxi . To see this, for any x ∈ M \ A, let Ux be a smooth coordinate ball around x contained in M \ A and small enough that

1 (x) 1 δ(y) > δ(x) and Xy − X |y < δ(x) 2 g 2 (x) for all y ∈ Ux, where X ∈ X(Ux) is a constant coeﬃcient vector ﬁeld such that (x) X |x = Xx. (Such a neighborhood exists by continuity of δ and X.) Then if y ∈ Ux, we have (x) 1 Xy − X |y < δ(x) < δ(y). g 2 The collection {Ux : x ∈ M \ A} is an open cover of M \ A. Choosing a countable ∞ (xi) subcover {Uxi }i=1 and setting Ui = Uxi and Xi = X , we have (*).

Let {ϕ0, ϕi} be a smooth partition of unity subordinate to the cover {U0,Ui} of M, and deﬁne Xe ∈ X(M) by X Xey = ϕ0(y)X0|y + ϕi(y)Xi|y. i≥1 87

Then clearly Xe is smooth, and is equal to X on A. For any y ∈ M, the fact that P i≥0 ϕi = 1 implies that ! X X Xey − Xy = ϕ0(y)X0|y + ϕi(y)Xi|y − ϕ0(y) + ϕi(y) Xy g i≥1 i≥1 g X ≤ ϕ0(y) |X0|y − Xy|g + ϕi(y) |Xi|y − Xy|g i≥1 X < ϕ0(y)δ(y) + ϕi(y)δ(y) = δ(y). i≥1 Lemma 219. Let V be a real inner product space and let k·k be the norm on V . Let {u1, . . . , uk} be an orthonormal set of vectors in V with kuik = 1 for i = 1, . . . , k. Let v ∈ V with kvk = 1.

(1) We have 2 2 hu1, vi + ··· + huk, vi ≤ 1,

and equality holds if and only if the set {v, u1, . . . , uk} is linearly dependent. (2) If w ∈ V and 2 2 2 kv − wk < 1 − hu1, vi − · · · − huk, vi ,

then {w, u1, . . . , uk} are linearly independent.

Proof. For (1), complete {u1, . . . , uk} to an orthonormal basis {u1, . . . , un} of V . Write v = a1u1 + ··· + anun where a1, . . . , an ∈ R. Then 2 2 2 2 hu1, vi + ··· + huk, vi = hu1, a1u1i + ··· + huk, akuki 2 2 = a1 + ··· + ak 2 2 = 1 − ak+1 − · · · − an since kvk = 1. Clearly the last expression is equal to 1 if and only if ak+1, . . . , an = 0, which is true if and only if v is in the span of {u1, . . . , uk}. For (2), suppose w = a1u1 + ··· + akuk where a1, . . . , ak ∈ R. Then 2 k k X X 2 0 > v − aiui − 1 + hui, vi i=1 i=1 * k + k 2 k

X X X 2 = 1 − 2 v, aiui + aiui − 1 + hui, vi i=1 i=1 i=1 88

k X 2 2 = (ai − 2ai hui, vi + hui, vi ) i=1 k X 2 = (ai − hui, vi) , i=1 which is a contradiction. Theorem 220. [Problem 13-3] Let M be a smooth manifold. (1) If there exists a global nonvanishing vector field on M, then there exists a global smooth nonvanishing vector field. (2) If there exists a linearly independent k-tuple of vector fields on M, then there exists such a k-tuple of smooth vector fields.

Proof. Let g be a Riemannian metric on M. For (1), let X be a global nonvanishing vector field on M. By Theorem 218, there is a smooth vector field Xe ∈ X(M) such that |Xx − Xex|g < |Xx|g for all x ∈ M. This implies that |Xex|g > 0 for all x ∈ M, so Xe is nonvanishing. For (2), let (X1,...,Xk) be a linearly independent k-tuple of vector fields on M. Using the Gram-Schmidt process, we may assume that (X1,...,Xk) is an orthonormal k-tuple of vector fields. Assume that X1,...,Xj−1 are already smooth. Define s X 2 δ(x) = 1 − hXj|x,Xi|xig; i6=j then δ : M → R is a positive continuous function by Lemma 219. By Theorem 218, there is a smooth vector field Xej ∈ X(M) such that 2 X 2 |(Xj|x − Xej|x)|g < 1 − hXj|x,Xi|xig i6=j for all x ∈ M. Then Lemma 219 shows that (X1,..., Xej,...,Xk) is linearly independent. Apply the Gram-Schmidt process to (X1,..., Xej) to obtain an orthonormal k-tuple of vector fields such that the first j vector fields are smooth. Repeating this procedure k times produces an orthonormal k-tuple of smooth vector fields. ◦ Example 221. [Problem 13-4] Let g denote the round metric on Sn. Compute the ◦ coordinate representation of g in stereographic coordinates. We have (2u1,..., 2un, |u|2 − 1) (x1, . . . , xn+1) = , |u|2 + 1 89 so 2 2(|u| − 2uj + 1) X 4uiuj dxj = duj − dui 2 2 2 2 (|u| + 1) i6=j (|u| + 1) for each j = 1, . . . , n and n X 4ui dxn+1 = dui. 2 2 i=1 (|u| + 1) Therefore ◦ g = (dx1)2 + ··· + (dxn+1)2 4 = ((du1)2 + ··· + (dun)2). (|u|2 + 1)2 Theorem 222. [Problem 13-5] Suppose (M, g) is a Riemannian manifold. A smooth 0 curve γ : J → M is said to be a unit-speed curve if |γ (t)|g = 1. Every smooth curve with nowhere-vanishing velocity has a unit-speed reparametrization.

Proof. We can assume that J is open. Choose any t0 ∈ J and define ` : J → R by t 0 `(t) = |γ (x)|g dx. ˆt0 0 Then x 7→ |γ (x)|g is smooth since v 7→ |v|g is smooth on {(x, v) ∈ TM : v 6= 0} and γ0(x) 6= 0 for all x ∈ J. Therefore ` is smooth, and since `0(t) 6= 0 for all t ∈ J, the −1 inverse function theorem shows that ` : `(J) → J is smooth. Define γ1 : `(J) → M −1 by γ1 = γ ◦ ` . We have 0 −1 0 0 −1 γ1(x) = (` ) (x)γ (` (x)) γ0(`−1(x)) = `0(`−1(x)) γ0(`−1(x)) = 0 −1 , |γ (` (x))|g so γ1 is a unit-speed reparametrization of γ. Theorem 223. [Problem 13-6] Every Riemannian 1-manifold is flat.

Proof. Let (M, g) be a Riemannian 1-manifold. Suppose x ∈ M and (U, ϕ) is a smooth chart containing x. By shrinking U, we can assume that U is connected. Therefore ϕ−1 : ϕ(U) → U is a smooth curve, and Theorem 222 shows that there is a diﬀeomorphism ψ : E → ϕ(U) such that ϕ−1 ◦ ψ is a unit-speed curve. We want to show that ϕ−1 ◦ ψ : 90

E → U is an isometry. If v, w ∈ TpE are nonzero vectors then v = r d/dt|p and w = s d/dt|p for some r, s ∈ R, so ! −1 ∗ −1 ∗ d d ((ϕ ◦ ψ) gp)(v, w) = rs((ϕ ◦ ψ) gp) , dt p dt p ! 2 −1 d = rs d(ϕ ◦ ψ)p dt p g −1 0 2 = rs (ϕ ◦ ψ) (p) g = rs

= g(ϕ−1◦ψ)(p)(v, w). Theorem 224. [Problem 13-7] A product of ﬂat metrics is ﬂat.

Proof. If (R1, g1),..., (Rk, gk) are Riemannian manifolds, then

(R1 × · · · × Rk, g1 ⊕ · · · ⊕ gk) is also a Riemannian manifold. Let (x1, . . . , xk) ∈ R1 × · · · × Rk. For each i = 1, . . . , k, let Fi : Ui → Vi be an isometry such that xi ∈ Ui. Then F1 × · · · × Fk is an isometry from the neighborhood U1 × · · · × Uk to the open set V1 × · · · × Vk, since ∗ ∗ ∗ (F1 × · · · × Fk) (g ⊕ · · · ⊕ g) = F1 g ⊕ · · · ⊕ Fk g

= g1 ⊕ · · · ⊕ gk. Corollary 225. [Problem 13-8] Let Tn = S1 × · · · × S1 ⊆ Cn, and let g be the metric on Tn induced from the Euclidean metric on Cn (identified with R2n). Then g is flat. Theorem 226. [Problem 13-10] The shortest path between two points in Euclidean space is a straight line segment. More precisely, for x, y ∈ Rn, let γ : [0, 1] → Rn be the curve segment γ(t) = x + t(y − x). Any other piecewise smooth curve segment γe from x to y satisfies Lg(γe) > Lg(γ) unless γe is a reparametrization of γ.

n Proof. First note that if γe :[a, b] → R then b b 0 0 |γe (t)| dt ≥ γe (t)dt = |γe(b) − γe(a)| . ˆa ˆa n−1 By rotating Rn, we can assume that x, y ∈ X, where X = R × {0} is the x1- n axis. Let γe :[a, b] → R be a piecewise smooth curve segment from x to y such that 91

Lg(γe) = |y − x|. Suppose that γe(c) ∈/ X for some c ∈ (a, b). Then c b 0 0 Lg(γe) = |γe (t)| dt + |γe (t)| dt â ˆc ≥ |γe(c) − γe(a)| + |γe(b) − γe(c)| > |y − x| = Lg(γ), n which is a contradiction. Therefore γe([a, b]) ⊆ X. Let π : R → X be the canonical projection. A similar argument shows that π ◦ γe is a diffeomorphism from [a, b] to the line segment from π(x) to π(y). Example 227. [Problem 13-11] Let M = R2 \{0}, and let g be the restriction to M of the Euclidean metric g. Show that there are points p, q ∈ M for which there is no piecewise smooth curve segment γ from p to q in M with Lg(γ) = dg(p, q). Let p = (−1, 0) and q = (1, 0). For each 0 < r < 1, let p0 = (−r, 0), q0 = (r, 0), p00 = (−r, r), q00 = (r, r) and define αr : [0, 5] → M by (1 − t)p + tp0, 0 ≤ t < 1,  (2 − t)p0 + (t − 1)p00, 1 ≤ t < 2,  00 00 αr(t) = (3 − t)p + (t − 2)q , 2 ≤ t < 3,  00 0 (4 − t)q + (t − 3)q , 3 ≤ t < 4,  (5 − t)q0 + (t − 4)q, 4 ≤ t ≤ 5. We have

Lg(αr) = 1 − r + 4r + 1 − r = 2 + 2r → 2 as r → 0. Therefore dg(p, q) ≤ 2. But Theorem 226 shows that dg(p, q) ≥ 2, so dg(p, q) = 2. There is no piecewise smooth curve segment γ from p to q such that Lg(γ) = 2, due to Theorem 226. Theorem 228. [Problem 13-12] Consider Rn as a Riemannian manifold with the Eu- clidean metric g.

(1) Suppose U, V ⊆ Rn are connected open sets, ϕ, ψ : U → V are Riemannian isometries, and for some p ∈ U they satisfy ϕ(p) = ψ(p) and dϕp = dψp. Then ϕ = ψ. (2) The set of maps from Rn to itself given by the action of E(n) on Rn described in Example 7.32 is the full group of Riemannian isometries of (Rn, g). 92

Proof. By considering ϕ ◦ ψ−1 : V → V , it suﬃces to show that for any connected open set U ⊆ Rn and any Riemannian isometry f : U → U satisfying f(p) = p and dfp = IdTpU for some p ∈ U, we have f = IdU . If γ : I → U is a smooth curve then Lg(γ) = Lg(f ◦ γ), so Theorem 226 implies that f takes straight line segments to straight line segments (of the same length). Let

E = {x ∈ U : f(x) = x, dfx = IdTxU } ; then E is nonempty. Let x ∈ E and let B = Br(x) be an open ball of radius r around x such that B ⊆ U. For any v ∈ Sn−1, let γ :[−1, 1] → U be the curve t 7→ x + trv and let L = γ([−1, 1]). Then f ◦ γ is a straight line segment of length 2r such that (f ◦ γ)(0) = x. But d(f ◦ γ)0 = dfx ◦ dγ0 = dγ0, so (f ◦ γ)([−1, 1]) = L. Using the fact that dg(x, y) = dg(f(x), f(y)) = dg(x, f(y)) for all y ∈ U, we conclude that f ◦γ = γ. But v was arbitrary, so f|B is the identity map on B(x). This shows that B ⊆ E, and therefore E is open. It is clear from continuity that E is closed. Since U is connected, we must have E = U. This proves (1). n n −1 Let ϕ : R → R be a Riemannian isometry. Then ϕe(x) = Dϕ(0) (ϕ(x) − ϕ(0)) defines a Riemannian isometry satisfying Dϕe(0) = I and ϕe(0) = 0, so (1) shows that ϕe = IdRn . Therefore ϕ(x) = ϕ(0) + Dϕ(0)x is an affine transformation. Theorem 229. [Problem 13-15] Let (M, g) be a Riemannian manifold, and let gb be the product metric on M × R determined by g and the Euclidean metric on R. Let X = 0 ⊕ d/dt be the product vector field on M × R determined by the zero vector field on M and the standard coordinate vector field d/dt on R. Then X is a Killing vector field for (M × R, gb).

Proof. The ﬂow of X is given by θ((x, r), t) = (x, r + t). Theorem 230. [Problem 13-16] If g = f(t)dt2 is a Riemannian metric on R, then g is complete if and only if both of the following improper integrals diverge: ∞ 0 (*) pf(t) dt, pf(t) dt. ˆ0 ˆ−∞

Proof. If x, y ∈ R and γ :[x, y] → R is the identity map then y y y 0 p dg(x, y) = |γ (t)|g dt = |1|g dt = f(t) dt. ˆx ˆx ˆx ∞ p Note that f(t) > 0 for all t ∈ R. If 0 f(t) dt converges, then {xn} = {1, 2,... } is a Cauchy sequence. Suppose that´g is complete; then n → x for some x ∈ R. n p If x ≥ 0 then dg(n, x) = x f(t) dt converges to a positive number as n → ∞, which is a contradiction. A´ similar contradiction arises if x < 0. Therefore g is not 93 complete. Conversely, suppose that both integrals in (*) diverge and let {xn} be a Cauchy sequence. Then {xn} is contained in a set K ⊆ R that is bounded with respect to the metric induced by g. Since the integrals in (*) diverge, K must also be bounded with respect to the Euclidean metric. By replacing K with a closed and bounded interval containing K, we may assume that K is compact. There are positive constants c, C such that c ≤ pf(t) ≤ C and pf(t)/C ≤ 1 ≤ pf(t)/c for all t ∈ K. Let ε > 0. Since {xn} is Cauchy, there exists an integer N such that

xn −cε < pf(t) dt < cε ˆxm for all m, n ≥ N. Then cε 1 xn p 1 xn p cε −ε < − < f(t) dt ≤ |xn − xm| ≤ f(t) dt < = ε, C C ˆxm c ˆxm c so in the Euclidean metric, {xn} is Cauchy and xn → x for some x ∈ K. We have x p f(t) dt → 0 ˆxn as n → ∞, so xn → x in the metric induced by g. Theorem 231. [Problem 13-18] Suppose (M, g) is a connected Riemannian manifold, S ⊆ M is a connected embedded submanifold, and ge is the induced Riemannian metric on S.

(1) d (p, q) ≥ d (p, q) for p, q ∈ S. ge g (2) If (M, g) is complete and S is properly embedded, then (S, ge) is complete. (3) Every connected smooth manifold admits a complete Riemannian metric.

Proof. Let ι : S,→ M be the inclusion map. If γ :[a, b] → S is a piecewise smooth curve segment from p to q in S then ι ◦ γ is a piecewise smooth curve segment from p to q in M and L (γ) = L (ι ◦ γ) ≥ d (p, q). Since this holds for all γ, we have ge g g d (p, q) ≥ d (p, q). This proves (1). Let {x } be a Cauchy sequence in S and let ε > 0. ge g n There exists an integer N such that d (x , x ) ≤ d (x , x ) < ε g m n ge m n for all m, n ≥ N, so {xn} is a Cauchy sequence in M and xn → x for some x ∈ M. But S is closed in M, so x ∈ S. This proves (2). Let N be a connected smooth manifold. Theorem 6.15 shows that there is a proper smooth embedding j : N,→ R2n+1. Since 2n+1 ∗ (R , g) is complete, (N, j g) is complete by (2). This proves (3). 94

Example 232. [Problem 13-19] The following example shows that the converse of Theorem 231(b) does not hold. Deﬁne F : R → R2 by F (t) = ((et+1) cos t, (et+1) sin t). Show that F is an embedding that is not proper, yet R is complete in the metric induced from the Euclidean metric on R2. We have et(cos t − sin t) − sin t DF (t) = , et(cos t + sin t) + cos t −1 which is never zero. Since F is an open map, it is a smooth embedding. But F B2(0) is not compact, where B2(0) is the closed ball of radius 2 around (0, 0). The metric on R induced from the Euclidean metric on R2 is given by f ∗g = d((et + 1) cos t)2 + d((et + 1) sin t)2 = (2e2t + 2et + 1)dt2. But ∞ √ 0 √ 2e2t + 2et + 1 dt and 2e2t + 2et + 1 dt ˆ0 ˆ−∞ diverge, so Theorem 230 shows that f ∗g is complete. Theorem 233. [Problem 13-21] Let (M, g) be a Riemannian manifold, let f ∈ C∞(M), and let p ∈ M be a regular point of f.

(1) Among all unit vectors v ∈ TpM, the directional derivative vf is greatest when v points in the same direction as grad f|p, and the length of grad f|p is equal to the value of the directional derivative in that direction. (2) grad f|p is normal to the level set of f through p.

Proof. We have hgrad f|p, vi = vf, so the Cauchy-Schwarz inequality shows that

| hgrad f|p, vig | is maximized when v = ± grad f|p. If S is the level set of f through p then TpS = ker dfp by Proposition 5.38. Clearly hgrad f|p, vi = vf = dfp(v) = 0 for every v ∈ ker dfp. Theorem 234. [Problem 13-22] For any smooth manifold M with or without boundary, the vector bundles TM and T ∗M are smoothly isomorphic over M.

∗ Proof. Choose any Riemannian metric g on M and let gb : TM → T M be the smooth bundle homomorphism deﬁned at each p ∈ M by gb(v)(w) = gp(v, w). Then gb is injective (and therefore bijective) at each point, so Theorem 174 shows that gb is a smooth bundle isomorphism. 95

Example 235. [Problem 13-23] Is there a smooth covector ﬁeld on S2 that vanishes at exactly one point? Yes, due to Theorem 128 and Theorem 234.

Chapter 14. Differential Forms

Theorem 236. [Exercise 14.4] Let α be a covariant tensor on a ﬁnite-dimensional vector space.

(1) Alt α is alternating. (2) Alt α = α if and only if α is alternating.

Proof. If τ ∈ Sk then 1 X (Alt α)(v , . . . , v ) = (sgn σ)α(v , . . . , v ) τ(1) τ(k) k! (σ◦τ)(1) (σ◦τ)(k) σ∈Sk 1 X = sgn(τ) sgn(σ ◦ τ)α(v , . . . , v ) k! (σ◦τ)(1) (σ◦τ)(k) σ∈Sk 1 X = sgn(τ) (sgn σ)α(v , . . . , v ) k! σ(1) σ(k) σ∈Sk

= sgn(τ)(Alt α)(v1, . . . , vk), which proves (1). If α is alternating then 1 X (Alt α)(v , . . . , v ) = (sgn σ)α(v , . . . , v ) 1 k k! σ(1) σ(k) σ∈Sk 1 X = (sgn σ)2α(v , . . . , v ) k! 1 k σ∈Sk

= α(v1, . . . , vk). Conversely, if Alt α = α then α is alternating since Alt α is alternating. This proves (2). Theorem 237. [Exercise 14.12] The wedge product is the unique associative, bilinear, and anticommutative map Λk(V ∗) × Λl(V ∗) → Λk+l(V ∗) satisfying

εi1 ∧ · · · ∧ εik = εI

i ∗ for any basis (ε ) for V and any multi-index I = (i1, . . . , ik). 96

k ∗ l ∗ Proof. Let ∧e be such a map. Let ω ∈ Λ (V ) and η ∈ Λ (V ). By Lemma 14.10,

0 ! 0 ! X I X J ω∧eη = ωI ε ∧e ηJ ε I J 0 0 X X I J = ωI ηJ ε ∧eε I J 0 0 X X i1 ik j1 jl = ωI ηJ ε ∧e · · · ∧ee ∧ee ∧e · · · ∧ee I J 0 0 X X IJ = ωI ηJ ε I J 0 0 X X I J = ωI ηJ ε ∧ ε I J 0 ! 0 ! X I X J = ωI ε ∧ ηJ ε I J = ω ∧ η.

Theorem 238. [Exercise 14.14] ΛkT ∗M is a smooth subbundle of T kT ∗M, and there- n fore is a smooth vector bundle of rank k over M.

Proof. This follows from Proposition 14.8 and Lemma 10.32. Theorem 239. [Exercise 14.28] The diagram below commutes:

grad curl div C∞(R3) X(R3) X(R3) C∞(R3)

Id [ β ∗

Ω0(R3) Ω1(R3) Ω2(R3) Ω3(R3). d d d

Therefore curl ◦ grad = 0 and div ◦ curl = 0 on R3. The analogues of the left-hand and right-hand square commute when R3 is replaced by Rn for any n. 97

Proof. In Rn, we have n ! n X ∂f ∂ X ∂f ([ ◦ grad)(f) = [ = dxi = df ∂xi ∂xi ∂xi i=1 i=1 and 1 n (d ◦ β)(X) = d(Xy(dx ∧ · · · ∧ dx )) n ! X = d (−1)i−1dxi(X)dx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxn i=1 n ! X = d (−1)i−1Xidx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxn i=1 n n X X ∂Xi = (−1)i−1 dxj ∧ dx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxn ∂xj i=1 j=1 n X ∂Xi = dx1 ∧ · · · ∧ dxn ∂xi i=1 n ! X ∂Xi = ∗ ∂xi i=1 = (∗ ◦ div)(X).

In R3 we have ∂X3 ∂X2 ∂ ∂X1 ∂X3 ∂ (β ◦ curl)(X) = β − + − ∂x2 ∂x3 ∂x1 ∂x3 ∂x1 ∂x2 ∂X2 ∂X1 ∂ + − ∂x1 ∂x2 ∂x3 ∂X2 ∂X1 ∂X3 ∂X1 = − dx1 ∧ dx2 + − dx1 ∧ dx3 ∂x1 ∂x2 ∂x1 ∂x3 ∂X3 ∂X2 + − dx2 ∧ dx3 ∂x2 ∂x3 = d X1dx1 + X2dx2 + X3dx3 = (d ◦ [)(X).

Theorem 240. [Problem 14-1] Covectors ω1, . . . , ωk ∈ V ∗ are linearly dependent if and only if ω1 ∧ · · · ∧ ωk = 0. 98

Proof. Let n = dim V . Suppose that ω1, . . . , ωk are linearly dependent and assume 1 k 1 without loss of generality that a1ω + ··· + akω = 0 with a1 6= 0. Then ω = −1 2 k −a1 (a2ω + ··· + akω ), so 1 k −1 2 k 2 k ω ∧ · · · ∧ ω = −a1 (a2ω + ··· + akω ) ∧ ω ∧ · · · ∧ ω = 0. Conversely, suppose that ω1, . . . , ωk are linearly independent. We can extend ω1, . . . , ωk to a basis {ω1, . . . , ωn} of Λn(V ∗). Then ω1 ∧ · · · ∧ ωn = (ω1 ∧ · · · ∧ ωk) ∧ (ωk+1 ∧ · · · ∧ ωn) 6= 0, 1 k so ω ∧ · · · ∧ ω 6= 0. Theorem 241. [Problem 14-5] Let M be a smooth n-manifold with or without boundary, and let (ω1, . . . , ωk) be an ordered k-tuple of smooth 1-forms on an open subset 1 k U ⊆ M such that (ω |p, . . . , ω |p) is linearly independent for each p ∈ U. Given smooth 1-forms α1, . . . , αk on U such that k X αi ∧ ωi = 0, i=1 each αi can be written as a linear combination of ω1, . . . , ωk with smooth coeﬃcients.

Proof. First note that each αi is in the span of ω1, . . . , ωk, since αi ∧ ω1 ∧ · · · ∧ ωk = (−1)i+1ω1 ∧ · · · ∧ αi ∧ ωi ∧ · · · ∧ ωk X = (−1)iω1 ∧ · · · ∧ αj ∧ ωj ∧ · · · ∧ ωk j6=i = 0.

i j Therefore we have α = rijω for some coefficients rij : U → R. But the span of ω1, . . . , ωk forms a smooth subbundle of TM, so Proposition 10.22 shows that the functions rij are smooth. Example 242. [Problem 14-6] Define a 2-form ω on R3 by ω = x dy ∧ dz + y dz ∧ dx + z dx ∧ dy. (1) Compute ω in spherical coordinates (ρ, ϕ, θ) defined by (x, y, z) = (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ). (2) Compute dω in both Cartesian and spherical coordinates and verify that both expressions represent the same 3-form. (3) Compute the pullback ι∗ ω to 2, using coordinates (ϕ, θ) on the open subset S2 S where these coordinates are defined. (4) Show that ι∗ ω is nowhere zero. S2 99

We have dx = sin ϕ cos θ dρ + ρ cos ϕ cos θ dϕ − ρ sin ϕ sin θ dθ, dy = sin ϕ sin θ dρ + ρ cos ϕ sin θ dϕ + ρ sin ϕ cos θ dθ, dz = cos ϕ dρ − ρ sin ϕ dϕ, so ω = ρ3 sin ϕ dϕ ∧ dθ. Also, dω = 3dx ∧ dy ∧ dz and dω = 3ρ2 sin ϕ dρ ∧ dϕ ∧ dθ. The pullback of ω to S2 is given by ∗ ι 2 ω = sin ϕ dϕ ∧ dθ, S which is deﬁned for (ϕ, θ) ∈ (0, π)×(0, 2π). This expression is never zero since sin ϕ > 0 for ϕ ∈ (0, π).

Chapter 15. Orientations

Theorem 243. [Exercise 15.4] Suppose M is an oriented smooth n-manifold with or without boundary, and n ≥ 1. Every local frame with connected domain is either positively oriented or negatively oriented.

Proof. Let (Ei) be a local frame with a connected domain U. Let D be the points of U at which (Ei) is positively oriented. Let p ∈ D and let (Eei) be an oriented local j frame deﬁned on a neighborhood V ⊆ U of p. Writing Ei = Bi Eej for continuous j j component functions Bi : V → R, we have det(Bi (p)) > 0. Since det is continuous, there is a neighborhood of p on which Ei is oriented. This shows that D is open. A similar argument shows that U \ D is open, so D is either empty or equal to U by the connectedness of U.

Theorem 244. [Exercise 15.8] Suppose M1,...,Mk are orientable smooth manifolds. There is a unique orientation on M1 × · · · × Mk, called the product orientation, with the following property: if for each i = 1, . . . , k, ωi is an orientation form for the ∗ ∗ given orientation on Mi, then π1ω1 ∧ · · · ∧ πkωk is an orientation form for the product orientation. 100

Proof. Let η1, . . . , ηk be orientation forms for M1,...,Mk and let O be the orientation ∗ ∗ determined by π1η1∧· · ·∧πkηk. Suppose ω1, . . . , ωk are orientation forms for M1,...,Mk. Then ωi = fiηi for some strictly positive continuous functions fi : Mi → R, so ∗ ∗ π1ω1 ∧ · · · ∧ πkωk = (f1 ◦ π1) ··· (fk ◦ πk)η1 ∧ · · · ∧ ηk. ∗ ∗ Since (f1 ◦ π1) ··· (fk ◦ πk) is strictly positive, π1ω1 ∧ · · · ∧ πkωk determines the same orientation as η1 ∧ · · · ∧ ηk. Theorem 245. [Exercise 15.10] Let M be a connected, orientable, smooth manifold with or without boundary. Then M has exactly two orientations. If two orientations of M agree at one point, they are equal.

Proof. If dim M = 0 then M consists of a single point, so the result is clear. Suppose 0 0 O, O are two orientations of M that agree at a single point p. Let D = {x ∈ M : Ox = Ox}; this set is nonempty since p ∈ D. Let x ∈ D and let (Ei) be a local frame deﬁned in a connected neighborhood of x that is positively oriented with respect to O0. Apply- ing Theorem 243 to (M, O) shows that (Ei) is positively oriented with respect to O. Therefore D is open. A similar argument shows that M \ D is open, so D = M by the connectedness of M. Since M is orientable, there is an orientation O for M. Let ω be an orientation form 0 0 for O; then −ω determines an orientation O for M with Op 6= Op for all p ∈ M. If 00 00 00 0 O is any orientation and p is any point in M then either Op = Op or Op = Op, so 00 00 0 O = O or O = O . Theorem 246. [Exercise 15.12] Suppose M is an oriented smooth manifold with or without boundary, and D ⊆ M is a smooth codimension-0 submanifold with or without boundary. The orientation of M restricts to an orientation of D. If ω is an orientation ∗ form for M, then ιDω is an orientation form for D.

∗ Proof. It suﬃces to check that ιDω is nonvanishing. But this is clear since d(ιD)p is an isomorphism for every p ∈ M. Theorem 247. [Exercise 15.13] Suppose M and N are oriented positive-dimensional smooth manifolds with or without boundary, and F : M → N is a local diﬀeomorphism. The following are equivalent:

(1) F is orientation-preserving. (2) With respect to any oriented smooth charts for M and N, the Jacobian matrix of F has positive determinant. (3) For any positively oriented orientation form ω for N, the form F ∗ω is positively oriented for M. 101

Proof. If (U, (xi)) and (V, (yi)) are oriented smooth charts for M and N respectively, i then dFp takes the oriented basis (∂/∂x |p) to an oriented basis j ! ∂F ∂ i (p) j . ∂x ∂y p i j i Since (∂/∂y |p) is also an oriented basis, we have det(∂F /∂x (p)) > 0. This proves (1) ⇒ (2). Suppose that (2) holds. By shrinking U and V , we can assume that they are connected sets and that F |U : U → V is a diﬀeomorphism. Then ω = f dy1 ∧ · · · ∧ dyn on V for some positive continuous function f, so by Proposition 14.20 we have F ∗ω = (f ◦ F )(det DF )dx1 ∧ · · · ∧ dxn. Since det DF > 0 by (2), F ∗ω is positively oriented on U. This proves (2) ⇒ (3). Finally, suppose that (3) holds and let (Ei) be an oriented basis of TpM. If ω is a positively oriented orientation form for N then ∗ (F ω)p(E1,...,En) > 0 ⇒ ωF (p)(dFp(E1), . . . , dFp(En)) > 0, so (dFp(Ei)) is an oriented basis of TF (p)N. This proves (3) ⇒ (1). Theorem 248. [Exercise 15.14] A composition of orientation-preserving maps is orientation- preserving.

Proof. Let F : M → N and G : N → P be orientation-preserving maps. If (Ei) is an oriented basis of TpM then (dFp(Ei)) is an oriented basis of TF (p)N and (d(G◦F )p(Ei)) is an oriented basis of T(G◦F )(p)P . Theorem 249. [Exercise 15.16] Suppose F : M → N and G : N → P are local diﬀeomorphisms and O is an orientation on P . Then (G ◦ F )∗O = F ∗(G∗O).

Proof. This is clear from the fact that (G ◦ F )∗ω = F ∗G∗ω for any orientation form ω. Theorem 250. [Exercise 15.20] Every Lie group has precisely two left-invariant orientations, corresponding to the two orientations of its Lie algebra.

Proof. Let G be a Lie group. Corollary 8.39 shows that there are at least two left- invariant orientations. If O is a left-invariant orientation on G then it is completely determined by Oe, so there are exactly two left-invariant orientations.

Theorem 251. [Exercise 15.30] Suppose (M, g) and (M,f ge) are positive-dimensional Riemannian manifolds with or without boundary, and F : M → Mf is a local isometry. Then F ∗ω = ω . ge g 102

Proof. Let p ∈ M and let U be a neighborhood of p for which F |U is an isometry. If (Ei) is a local oriented orthonormal frame for M then (dF (Ei)) is a local oriented orthonormal frame for Mf on a neighborhood of F (p) since

hdF (Ei), dF (Ej)i = hEi,Eji = δij. ge g Therefore (F ∗ω )(E ,...,E ) = ω (dF (E ), . . . , dF (E )) = 1, ge 1 n ge 1 n which shows that F ∗ω is the Riemannian volume form on M. ge Theorem 252. [Problem 15-1] Suppose M is a smooth manifold that is the union of two orientable open submanifolds with connected intersection. Then M is orientable.

Proof. Denote the two open submanifolds by N and P . We can assume that there is some p ∈ N ∩ P , for otherwise the result is trivial. Choose an orientation O for N, and 0 0 0 choose an orientation O for P such that Op = Op. By Theorem 245, O and O agree on N ∩ P , and it is easy to check that the orientation given by ( 00 Ox, x ∈ N, Ox = 0 Ox, x ∈ P is well-deﬁned and continuous. Theorem 253. [Problem 15-2] Suppose M and N are oriented smooth manifolds with or without boundary, and F : M → N is a local diﬀeomorphism. If M is connected, then F is either orientation-preserving or orientation-reversing.

Proof. If M and N are 0-manifolds, then M and N both consist of a single point and the result is trivial. Let D be the set of points x for which dFx takes oriented bases of TxM to oriented bases of TF (x)N. Let ω be a positively oriented orientation form for N. Let x ∈ D and let (Ei) be an oriented local frame deﬁned on a neighborhood U of ∗ x. Then (F ω)x(E1|x,...,En|x) > 0, so by continuity there is a neighborhood V ⊆ U ∗ of x on which (F ω)(E1,...,En) > 0. Therefore V ⊆ D by Theorem 247, and D is open. A similar argument shows that M \ D is open, so D = ∅ or D = M since M is connected. Theorem 254. [Problem 15-3] Suppose n ≥ 1, and let α : Sn → Sn be the antipodal map: α(x) = −x. Then α is orientation-preserving if and only if n is odd.

Proof. The outward unit normal vector ﬁeld along Sn is N = xi∂/∂xi, and Corollary 15.34 shows that the Riemannian volume form ω of n is ge S n+1 ! ∗ 1 n+1 ∗ X i−1 i 1 i n+1 ω = ι n (N (dx ∧ · · · ∧ dx )) = ι n (−1) x dx ∧ · · · ∧ dx ∧ · · · ∧ dx . ge S y S c i=1 103

We have n+1 ! ∗ ∗ X i−1 i 1 i n+1 α ω = ι n (−1) (−x )(−dx ) ∧ · · · ∧ (\−dx ) ∧ · · · ∧ (−dx ) ge S i=1 = (−1)n+1ω , ge so F is orientation-preserving if and only if n is odd. Theorem 255. [Problem 15-5] Let M be a smooth manifold with or without boundary. The total spaces of TM and T ∗M are orientable.

Proof. Let π : TM → M be the projection. Let (U, ϕ) and (V, ψ) be smooth charts −1 −1 for M, and let (π (U), ϕe) and (π (V ), ψe) be the corresponding charts for TM. The transition map is given by ∂x1 ∂xn (ψe ◦ ϕ−1)(x1, . . . , xn, v1, . . . , vn) = x1(x),..., xn(x), e (x)vj,..., e (x)vj . e e e ∂xj ∂xj The Jacobian matrix of this map at (x1, . . . , xn, v1, . . . , vn) is

i ∂xe ∂xj (x) 0 i , ∂xe ∗ ∂xj (x) which clearly has positive determinant. The result follows from Proposition 15.6. A ∗ similar computation holds for T M. Theorem 256. [Problem 15-7] Suppose M is an oriented Riemannian manifold with or without boundary, and S ⊆ M is an oriented smooth hypersurface with or without boundary. There is a unique smooth unit normal vector ﬁeld along S that determines the given orientation of S.

Proof. Let n = dim M, let ω be an orientation form for M, and let O be the orientation on S. Let NS be the normal bundle of S (see Theorem 213), which is of rank 1. Let Z be the set of all zero vectors in NS. Deﬁne a map f : NS \ Z → {−1, +1} as follows: if v ∈ NpS then choose an oriented (with respect to O) basis (Ei) for TpS and deﬁne ω(v, E ,...,E ) f(v) = 1 n−1 . |ω(v, E1,...,En−1)|

This map is well-defined because (Ei) is oriented, and is continuous because there is an oriented local frame defined on a neighborhood of every p. Let P = f −1({+1}). It is easy to see that P is convex and open in NS, so Theorem 217 shows that there is a smooth global section X : S → P . Then Xe = X/ |X|g is the desired smooth unit normal vector field along S. 104

Theorem 257. [Problem 15-8] Suppose M is an orientable Riemannian manifold, and S ⊆ M is an immersed or embedded submanifold with or without boundary.

(1) If S has trivial normal bundle, then S is orientable. (2) If S is an orientable hypersurface, then S has trivial normal bundle.

Proof. If S has trivial normal bundle then there is a global frame for NS, and Proposi- tion 15.21 shows that S is orientable. If S is an orientable hypersurface, then Theorem 256 gives a global frame for NS (since NS is of rank 1). Theorem 258. [Problem 15-9] Let S be an oriented, embedded, 2-dimensional submanifold with boundary in R3, and let C = ∂S with the induced orientation. By Theorem 256, there is a unique smooth unit normal vector field N on S that determines the orientation. Let T be the oriented unit tangent vector field on C, and let V be the unique unit vector field tangent to S along C that is orthogonal to T and inward-pointing. Then 3 (Tp,Vp,Np) is an oriented orthonormal basis for R at each p ∈ C.

Proof. It is clear that (Tp,Vp,Np) is an orthonormal basis, so it remains to check that (Tp,Vp,Np) is oriented. This is true if and only if (Np,Tp,Vp) is oriented, and by the definition of the induced orientation, this is true if and only if (Tp,Vp) is an oriented basis for TpS. This is true since (−Vp,Tp) is an oriented basis for TpS. Theorem 259. [Problem 15-10] Let M be a connected nonorientable smooth manifold with or without boundary, and let πb : Mc → M be its orientation covering. If X is any oriented smooth manifold with or without boundary, and F : X → M is any local diffeomorphism, then there exists a unique orientation-preserving local diffeomorphism Fb : X → Mc such that πb ◦ Fb = F .

Proof. Define Fb(x) = (F (x), OF (x)), where OF (x) is the orientation of TF (x)M determined by F and the orientation of X. It is clear that πb ◦ Fb = F . Let UbO be an element −1 of the basis defined in Proposition 15.40. If U is connected, then Fb (UbO) is either U or −1 empty. Therefore Fb (UbO) is open for any open set U, and Fb is continuous. By Theo- rem 41, Fb is smooth, and it is easy to check that it is orientation-preserving and a local diffeomorphism. Uniqueness follows from the fact that Fb is orientation-preserving. Theorem 260. [Problem 15-11] Let M be a nonorientable connected smooth manifold with or without boundary, and let πb : Mc → M be its orientation covering. If Mf is an oriented smooth manifold with or without boundary that admits a two-sheeted smooth covering map πe : Mf → M, then there exists a unique orientation-preserving diffeomorphism ϕ : Mf → Mc such that πb ◦ ϕ = πe. 105

Proof. By Theorem 259, there is a unique orientation-preserving local diffeomorphism ϕ : Mf → Mc such that πb ◦ ϕ = πe. Similarly, there is a unique orientation-preserving local diffeomorphism ψ : Mc → Mf such that πe ◦ ψ = πb. Then πb ◦ ϕ ◦ ψ = πe ◦ ψ = πb, so ϕ◦ψ = Id by uniqueness. Similarly, ψ◦ϕ = Id . Therefore ϕ is a diffeomorphism. Mc Mf Theorem 261. [Problem 15-22] Every orientation-reversing diffeomorphism of R has a fixed point.

Proof. Let f : R → R be such a diﬀeomorphism; then f 0(x) < 0 for all x ∈ R. Let g(x) = f(x) − x; then g0(x) = f 0(x) − 1 < −1 for all x ∈ R. If g(0) = 0, then we are done. If g(0) > 0 then by the mean value theorem, there exists a point t ∈ (0, g(0)) such that g(g(0)) − g(0) = g0(t) < −1, g(0) i.e. g(g(0)) < 0, and the intermediate value theorem shows that g(x) = 0 for some x ∈ (0, g(0)). Similarly, if g(0) < 0 then there exists a point t ∈ (g(0), 0) such that g(g(0)) − g(0) = g0(t) < −1, g(0) i.e. g(g(0)) > 0, and the intermediate value theorem shows that g(x) = 0 for some x ∈ (g(0), 0).

Chapter 16. Integration on Manifolds

Theorem 262. [Exercise 16.29] Suppose (M, g) is an oriented Riemannian manifold and f : M → R is continuous and compactly supported. Then

f dVg ≤ |f| dVg. ˆM ˆM

Proof. If f is supported in the domain of a single oriented smooth chart (U, ϕ), then

q 1 n f dVg = f(x) det(gij) dx ··· dx ˆM ˆϕ(U)

q 1 n ≤ |f(x)| det(gij) dx ··· dx ˆϕ(U)

= |f| dVg. ˆM 106

For the general case, cover supp f by ﬁnitely many domains of positively or negatively oriented smooth charts {Ui} and choose a subordinate smooth partition of unity {ψi}. Then

X f dVg = ψif dVg ˆ ˆ M i M X ≤ ψ |f| dV ˆ i g i M

= |f| dVg. ˆM Example 263. [Problem 16-2] Let T2 = S1 ×S1 ⊆ R4 denote the 2-torus, deﬁned as the set of points (w, x, y, z) such that w2 + x2 = y2 + z2 = 1, with the product orientation determined by the standard orientation on 1. Compute ω, where ω is the following S T2 2-form on R4: ´ ω = xyz dw ∧ dy.

Let D = (0, 2π) × (0, 2π) and deﬁne an orientation-preserving diﬀeomorphism F : D → T2 by F (θ, ϕ) = (cos θ, sin θ, cos ϕ, sin ϕ) so that dw = − sin θ dθ, dx = cos θ dθ, dy = − sin ϕ dϕ, dz = cos ϕ dϕ. Then

ω = sin θ cos ϕ sin ϕ(− sin θ dθ) ∧ (− sin ϕ dϕ) ˆT2 ˆT2 = sin2 θ cos ϕ sin2 ϕ dθ ∧ dϕ ˆT2 2π 2π = sin2 θ dθ cos ϕ sin2 ϕ dϕ ˆ0 ˆ0 = π · 0 = 0. Theorem 264. [Problem 16-3] Suppose E and M are smooth n-manifolds with or without boundary, and π : E → M is a smooth k-sheeted covering map or generalized covering map. 107

∗ (1) If E and M are oriented and π is orientation-preserving, then E π ω = k M ω for any compactly supported n-form ω on M. ´ ´ ∗ (2) E π µ = k M µ whenever µ is a compactly supported density on M. ´ ´ Proof. Let {Ui} be a ﬁnite open cover of supp ω by evenly covered sets, and let {ψi} be a (1) (k) subordinate smooth partition of unity. For each i, let Ui ,...,Ui be the components −1 n (j)o of π (Ui). Let ψbi be a smooth partition of unity subordinate to the open cover n (j)o ∗ Ui of supp π ω. Applying Proposition 16.6(d), we have

k X X k ω = ψ ω ˆ ˆ i M i j=1 Ui k X X ∗ = (ψi ◦ π|U (j) )π ω ˆ (j) i i j=1 Ui k X X ∗ = (ψi ◦ π|U (j) )π ω ˆ (j) i i j=1 Ui k k X X X X (j0) ∗ = ψbi0 (ψi ◦ π|U (j) )π ω ˆ (j) i i j=1 Ui i0 j0=1 k k X X (j0) X X ∗ = ψbi0 (ψi ◦ π|U (j) )π ω ˆ i E i0 j0=1 i j=1 k X X (j0) ∗ = ψb 0 π ω ˆ i E i0 j0=1

= π∗ω. ˆE This proves (1). Part (2) is similar. Theorem 265. [Problem 16-4] If M is an oriented compact smooth manifold with boundary, then there does not exist a retraction of M onto its boundary.

Proof. Suppose that there exists such a retraction r : M → ∂M. By Theorem 6.26, we can assume that r is smooth. Let ι : ∂M ,→ M be the inclusion map. Let ω be an orientation form for ∂M. By Theorem 16.11 and Proposition 14.26,

r∗ω = d(r∗ω) = r∗(dω) = 0 ˆ∂M ˆM ˆM 108 since ω is a top-degree form. By deﬁnition,

r∗ω = ι∗r∗ω = (r ◦ ι)∗ω = ω, ˆ∂M ˆ∂M ˆ∂M ˆ∂M so ω = 0. ˆ∂M This contradicts Proposition 16.6(c). Theorem 266. [Problem 16-5] Suppose M and N are oriented, compact, connected, smooth manifolds, and F,G : M → N are homotopic diﬀeomorphisms. Then F and G are either both orientation-preserving or both orientation-reversing.

Proof. Theorem 253 shows that F and G are individually orientation-preserving or orientation-reversing. By Theorem 6.29, there is a smooth homotopy H : M × I → N from F to G. Let ω be a positively oriented orientation form for N. By Theorem 16.11,

H∗ω = d(H∗ω) = H∗(dω) = 0. ˆ∂(M×I) ˆM×I ˆM×I But H∗ω = F ∗ω − G∗ω, ˆ∂(M×I) ˆM ˆM so F ∗ω = G∗ω. ˆM ˆM The result follows from Proposition 16.6(d). Theorem 267. [Problem 16-6] The following are equivalent:

(1) There exists a nowhere-vanishing vector ﬁeld on Sn. (2) There exists a continuous map V : Sn → Sn satisfying V (x) ⊥ x (with respect to the Euclidean dot product on Rn+1) for all x ∈ Sn. n n (3) The antipodal map α : S → S is homotopic to IdSn . (4) The antipodal map α : Sn → Sn is orientation-preserving. (5) n is odd.

Therefore, there exists a nowhere-vanishing vector ﬁeld on Sn if and only if n is odd.

Proof. Suppose X is a nowhere-vanishing vector field on Sn. Then taking V = X/ |X| n n proves (1) ⇒ (2). Suppose (2) holds. We can define a homotopy H1 : S × I → S from α to V by (1 − t)α(x) + tV (x) H (x, t) = , 1 |(1 − t)α(x) + tV (x)| 109 where the denominator is nonzero since (1 − t)(−x) + tV (x) = 0 contradicts the fact that V (x) ⊥ x. Similarly, we can define a homotopy H2 from V to IdSn by (1 − t)V (x) + tx H (x, t) = . 2 |(1 − t)V (x) + tx| This proves (2) ⇒ (3). Theorem 266 proves (3) ⇒ (4), and Theorem 254 proves (4) ⇒ (5). Finally, Theorem 150 proves (5) ⇒ (1). Theorem 268. [Problem 16-9] Let ω be the (n − 1)-form on Rn \{0} defined by n −n X ω = |x| (−1)i−1xi dx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxn. i=1 (1) ι∗ ω is the Riemannian volume form of n−1 with respect to the round metric Sn−1 S and the standard orientation. (2) ω is closed but not exact on Rn \{0}.

Proof. The unit normal vector ﬁeld for Sn−1 with the standard orientation is given by x 7→ x, so (1) follows from the formula in Lemma 14.13. For (2), we have n X xi dω = (−1)i−1d dx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxn |x|n i=1 n n i 2 n−2 ! X |x| − n(x ) |x| 1 n = 2n dx ∧ · · · ∧ dx i=1 |x| n X 1 n(xi)2 = − dx1 ∧ · · · ∧ dxn |x|n n+2 i=1 |x| = 0. If ω is exact then ω = 0 ˆSn−1 by Corollary 16.13, which is a contradiction. Example 269. [Problem 16-10] Let D denote the torus of revolution in R3 obtained by revolving the circle (r −2)2 +z2 = 1 around the z-axis, with its induced Riemannian metric and with the orientation determined by the outward unit normal.

(1) Compute the surface area of D. (2) Compute the integral over D of the function f(x, y, z) = z2 + 1. (3) Compute the integral over D of the 2-form ω = z dx ∧ dy. 110

Let ι : D,→ R3 be the inclusion map. Let U = (0, 2π) × (0, 2π) and deﬁne an orientation-preserving diﬀeomorphism F : U → D by F (θ, ϕ) = ((2 + cos θ) cos ϕ, (2 + cos θ) sin ϕ, sin θ). We have dx = − sin θ cos ϕ dθ − (2 + cos θ) sin ϕ dϕ, dy = − sin θ sin ϕ dθ + (2 + cos θ) cos ϕ dϕ, dz = cos θ dθ and F ∗g = dx2 + dy2 + dz2 = sin2 θ cos2 ϕ dθ2 + (2 + cos θ)2 sin2 ϕ dϕ2 + 2 sin θ sin ϕ cos ϕ(2 + cos θ) dθ dϕ + sin2 θ sin2 ϕ dθ2 + (2 + cos θ)2 cos2 ϕ dϕ2 − 2 sin θ sin ϕ cos ϕ(2 + cos θ) dθ dϕ + cos2 θ dθ2 = dθ2 + (2 + cos θ)2 dϕ2. For (1), we have

p 2 ωι∗g = (2 + cos θ) ˆD ˆ(θ,ϕ)∈U 2π = 2π (2 + cos θ) dθ ˆ0 = 8π2. For (2), we have

2 p 2 fωι∗g = (sin θ + 1) (2 + cos θ) ˆD ˆ(θ,ϕ)∈U 2π = 2π (sin2 θ + 1)(2 + cos θ) dθ ˆ0 = 12π2. For (3), we have dx ∧ dy = (− sin θ cos ϕ dθ − (2 + cos θ) sin ϕ dϕ) ∧ (− sin θ sin ϕ dθ + (2 + cos θ) cos ϕ dϕ) = −(2 + cos θ) sin θ dθ ∧ dϕ 111 so that

z dx ∧ dy = − (2 + cos θ) sin2 θ dθ ∧ dϕ ˆD ˆD 2π = −2π (2 + cos θ) sin2 θ dθ ˆ0 = −4π2. Theorem 270. [Problem 16-11] Let (M, g) be a Riemannian n-manifold with or without boundary. In any smooth local coordinates (xi) we have ∂ 1 ∂ div Xi = √ Xipdet g , ∂xi det g ∂xi where det g = det(gkl) is the determinant of the component matrix of g in these coordinates.

Proof. We have p 1 n ωg = det gdx ∧ · · · ∧ dx by Proposition 15.31, so n ∂ X p β Xi = (−1)i−1Xi det g dx1 ∧ · · · ∧ dxci ∧ · · · ∧ dxn. ∂xi i=1 Then ∂ ∂ d β Xi = Xipdet g dx1 ∧ · · · ∧ dxn ∂xi ∂xi and ∂ ∂ div Xi = ∗−1 d β Xi ∂xi ∂xi 1 ∂ = √ Xipdet g . det g ∂xi Theorem 271. [Problem 16-12] Let (M, g) be a compact Riemannian manifold with boundary, let ge denote the induced Riemannian metric on ∂M, and let N be the outward unit normal vector ﬁeld along ∂M.

(1) The divergence operator satisﬁes the following product rule for f ∈ C∞(M), X ∈ X(M):

div(fX) = f div X + hgrad f, Xig . 112

(2) We have the following “integration by parts” formula:

hgrad f, Xi dV = f hX,Ni dV − (f div X) dV . g g g ge g ˆM ˆ∂M ˆM

Proof. Theorem 270 shows that in coordinates, we have 1 ∂ div(fX) = √ fXipdet g det g ∂xi 1 ∂ ∂f = √ f Xipdet g + Xipdet g det g ∂xi ∂xi 1 ∂ ∂f = f √ Xipdet g + Xi det g ∂xi ∂xi

= f div X + hgrad f, Xig . Therefore

hgrad f, Xig dVg = div(fX) dVg − f div X dVg ˆM ˆM ˆM = f hX,Ni dV − f div X dV g ge g ˆ∂M ˆM by Theorem 16.32. Theorem 272. [Problem 16-13] Let (M, g) be a Riemannian n-manifold with or without boundary. The linear operator 4 : C∞(M) → C∞(M) deﬁned by 4u = − div(grad u) is called the (geometric) Laplacian. The Laplacian is given in any smooth local coordinates by 1 ∂ ∂u 4u = −√ gijpdet g . det g ∂xi ∂xj On Rn with the Euclidean metric and standard coordinates, n X ∂2u 4u = − . (∂xi)2 i=1

Proof. Obvious from Theorem 270. Theorem 273. [Problem 16-14] Let (M, g) be a Riemannian n-manifold with or without boundary. A function u ∈ C∞(M) is said to be harmonic if 4u = 0.

(1) If M is compact, then

u4v dV = hgrad u, grad vi dV − uNv dV , g g g ge ˆM ˆM ˆ∂M 113

(u4v − v4u) dV = (vNu − uNv)dV , g ge ˆM ˆ∂M where N and ge are as in Theorem 271. These are known as Green’s identities. (2) If M is compact and connected and ∂M = ∅, the only harmonic functions on M are the constants. (3) If M is compact and connected, ∂M 6= ∅, and u, v are harmonic functions on M whose restrictions to ∂M agree, then u = v.

Proof. (1) follows by putting f = u and X = grad v in Theorem 271. For (2), if u is harmonic then 2 |grad u|g dVg = u4u dVg = 0 ˆM ˆM by part (1). Therefore grad u = 0, and u is constant since M is connected. For (3), we have |grad(u − v)|2 dV = (u − v)N(u − v) dV = 0 g g ge ˆM ˆ∂M by part (1). Therefore grad(u − v) = 0, and u − v is constant since M is connected. Since u and v agree on ∂M, we have u = v. Theorem 274. [Problem 16-15] Let (M, g) be a compact connected Riemannian manifold without boundary, and let 4 be its geometric Laplacian. A real number λ is called an eigenvalue of 4 if there exists a smooth real-valued function u on M, not identically zero, such that 4u = λu. In this case, u is called an eigenfunction corresponding to λ. (1)0 is an eigenvalue of 4, and all other eigenvalues are strictly positive.

(2) If u and v are eigenfunctions corresponding to distinct eigenvalues, then M uv dVg = 0. ´

Proof. Any constant function is an eigenfunction with eigenvalue 0. If 4u = λu and u is not constant then Theorem 273 shows that

2 2 0 < |grad u|g dVg = u4u dVg = λ u dVg, ˆM ˆM ˆM so λ > 0. Suppose 4u = λu and 4v = µv where λ 6= µ. By Theorem 273,

0 = (u4v − v4u) dVg = (µ − λ) uv dVg, ˆM ˆM so

uv dVg = 0 ˆM since µ − λ 6= 0. 114

Theorem 275. [Problem 16-16] Let M be a compact connected Riemannian n-manifold with nonempty boundary. A number λ ∈ R is called a Dirichlet eigenvalue for M if there exists a smooth real-valued function u on M, not identically zero, such that 4u = λu and u|∂M = 0. Similarly, λ is called a Neumann eigenvalue if there exists such a u satisfying 4u = λu and Nu|∂M = 0, where N is the outward unit normal. (1) Every Dirichlet eigenvalue is strictly positive. (2)0 is a Neumann eigenvalue, and all other Neumann eigenvalues are strictly positive.

Proof. As in Theorem 274. Theorem 276. [Problem 16-17] Suppose M is a compact connected Riemannian n- manifold with nonempty boundary. A function u ∈ C∞(M) is harmonic if and only 2 if it minimizes M |grad u|g dVg among all smooth functions with the same boundary values. ´

Proof. Suppose f ∈ C∞(M) vanishes on ∂M. For all ε, we have

2 2 (f4u) dVg + 2ε |grad f|g dVg ˆM ˆM 115 must be zero at ε = 0. That is,

(f4u) dVg = 0 ˆM ∞ for all f ∈ C (M) such that f|∂M = 0. Therefore 4u = 0. Theorem 277. [Problem 16-18] Let (M, g) be an oriented Riemannian n-manifold.

k ∗ (1) For each k = 1, . . . , n, g determines a unique inner product on Λ (Tp M) (denoted by h·, ·ig, just like the inner product on TpM) satisfying

1 k 1 k i # j # ω ∧ · · · ∧ ω , η ∧ · · · ∧ η g = det (ω ) , (η ) g whenever ω1, . . . , ωk, η1, . . . , ηk are covectors at p. (2) The Riemannian volume form dVg is the unique positively oriented n-form that has unit norm with respect to this inner product. (3) For each k = 0, . . . , n, there is a unique smooth bundle homomorphism ∗ : ΛkT ∗M → Λn−kT ∗M satisfying

ω ∧ ∗η = hω, ηig dVg for all smooth k-forms ω, η. (For k = 0, interpret the inner product as ordinary multiplication.) This map is called the Hodge star operator. 0 ∗ n ∗ (4) ∗ :Λ T M → Λ T M is given by ∗f = f dVg. (5) ∗ ∗ ω = (−1)k(n−k)ω if ω is a k-form.

Proof. For (1), any such inner product must satisfy

 i1 # j1 # i1 # jk #  (ε ) , (ε ) g ··· (ε ) , (ε ) g εI , εJ = det  . .. .  g  . . .  ik # j1 # ik # jk # (ε ) , (ε ) g ··· (ε ) , (ε ) g gi1j1 ··· gi1jk  . . . = det  . .. .  gikj1 ··· gikjk I (*) = δJ for all increasing multi-indices I and J, whenever (εi) is the coframe dual to a local ij orthonormal frame (Ei). (Here (g ) is the inverse of gij(p), which is the matrix of g in these coordinates.) This proves that the inner product is uniquely determined. To prove existence, we will deﬁne h·, ·ig by (*) and extend bilinearly. To check that the 116

i deﬁnition is independent of the choice of orthonormal frame, let (εe ) be another coframe dual to a local orthonormal frame (Eei). Then 0 0 I X L J X M εe = αLε and εe = βM ε L M where α = εI (E ,...,E ) and β = εJ (E ,...,E ), so L e l1 lk M e m1 mk 0 0 X X εI , εJ = α β εL, εM e e g L M g L M 0 0 X X L = αLβM δM L M 0 X = αLβL L 0 X = εI (E ,...,E )εJ (E ,...,E ). e l1 lk e l1 lk L j j Since Ei = Ai Eej for some orthonormal matrix (Ai ), we have 0 X X X εI , εJ = Ap1 ··· Apk δI Aq1 ··· Aqk δJ e e g l1 lk P l1 lk Q L P Q 0 X X iσ(1) iσ(k) X jτ(1) jτ(k) = (sgn σ)A ··· A (sgn τ)A ··· A l1 lk l1 lk L σ∈Sk τ∈Sk =???????????????

For (2), if ω is any positively oriented n-form then ωp = c(dVg)p for some c > 0, and 2 hωp, ωpig = c hdVg, dVgig 2 1 n 1 n = c ε ∧ · · · ∧ ε , ε ∧ · · · ∧ ε g = c2 whenever (εi) is the coframe dual to a local orthonormal frame, by Proposition 15.29. If ω has unit norm then it is clear that c = 1. Example 278. [Problem 16-19] Consider Rn as a Riemannian manifold with the Eu- clidean metric and the standard orientation. (1) Calculate ∗dxi for i = 1, . . . , n. (2) Calculate ∗(dxi ∧ dxj) in the case n = 4. 117

TODO Theorem 279. [Problem 16-20] Let M be an oriented Riemannian 4-manifold. A 2-form ω on M is said to be self-dual if ∗ω = ω, and anti-self-dual if ∗ω = −ω.

(1) Every 2-form ω on M can be written uniquely as a sum of a self-dual form and an anti-self-dual form. (2) On M = R4 with the Euclidean metric, the self-dual and anti-self-dual forms in standard coordinates are given by: ??????????

Proof. TODO Theorem 280. [Problem 16-21] Let (M, g) be an oriented Riemannian manifold and X ∈ X(M). Then

[ XydVg = ∗X , div X = ∗d ∗ X[, and, when dim M = 3, curl X = (∗dX[)].

Proof. TODO Theorem 281. [Problem 16-22] Let (M, g) be a compact, oriented Riemannian n- manifold. For 1 ≤ k ≤ n, define a map d∗ :Ωk(M) → Ωk−1(M) by d∗ω = (−1)n(k+1)+1 ∗ d∗ω, where ∗ is the Hodge star operator defined in Theorem 277. Extend this definition to 0-forms by defining d∗ω = 0 for ω ∈ Ω0(M).

(1) d∗ ◦ d∗ = 0. (2) The formula

(ω, η) = hω, ηig dVg ˆM k deﬁnes an inner product on Ω (M) for each k, where h·, ·ig is the pointwise inner product on forms deﬁned in Theorem 277. (3)( d∗ω, η) = (ω, dη) for all ω ∈ Ωk(M) and η ∈ Ωk−1(M).

Proof. TODO 118

Chapter 17. De Rham Cohomology

Theorem 282. [Exercise 17.37] Suppose M, N, and P are compact, connected, oriented, smooth n-manifolds. (1) If F : M → N and G : N → P are both smooth maps, then deg(G ◦ F ) = (deg G)(deg F ). (2) If F : M → N is a diﬀeomorphism, then deg F = +1 if F is orientation- preserving and −1 if it is orientation-reversing. (3) If two smooth maps F0,F1 : M → N are homotopic, then they have the same degree.

Proof. For every smooth n-form ω on P , we have

(G ◦ F )∗ω = F ∗G∗ω ˆM ˆM = (deg F ) G∗ω ˆN = (deg F )(deg G) ω. ˆP

This proves (1). Part (2) follows from Proposition 16.6(d). If F0 and F1 are homotopic ∗ ∗ then Proposition 17.10 shows that the induced cohomology maps F0 and F1 are equal. ∗ ∗ For any smooth n-form ω on N, we have F0 ω −F1 ω = dη for some smooth (n−1)-form η. Then ∗ ∗ F0 ω − F1 ω = dη = 0. ˆM ˆM ˆM Theorem 283. [Problem 17-1] Let M be a smooth manifold with or without boundary, and let ω ∈ Ωp(M), η ∈ Ωq(M) be closed forms. The de Rham cohomology class of ω∧η depends only on the cohomology classes of ω and η, and thus there is a well-deﬁned p q p+q bilinear map `: HdR(M) × HdR(M) → HdR (M), called the cup product, given by [ω] ` [η] = [ω ∧ η].

Proof. TODO

References [1] John M. Lee. Introduction to Topological Manifolds. Springer, 2nd edition, 2011.