1 Geometry of surfaces

Transcription

1 1 Geometry of surfaces Let M E 3 be a smooth surface. M can be represented locally by a R 3 -valued function r : D R 3 where D is an open set on R 2 : r : D r(d) =: U R 3 (u, v) r(u, v) = (x(u, v), y(u, v), z(u, v)). For any fixed v 0, r(u, v 0 ) is a curve on the surface M. Similarly we can fix u 0 and get a curve r(u 0, v) depending on v. Assume r is differentiable. Define the tangent vectors: r u (u 0, v 0 ) = r, r v (u 0, v 0 ) = r. (1) u v (u0,v 0) (u0,v 0) For (u, v) to give a coordinate on r(d), we require r u and r v are linearly independent. In other words, we assume: r u r v 0 for any (u, v) D. (2) In this case, we say that r is a regular parametrization of M on U and we say that M is regular on U. From now on, we assume r(u, v) is a C 3 -function of (u, v). Most of the time, we will assume that r(u, v) is smooth. Proposition 1.1. Any regular surface locally is isomorphic to the graph of a function. Proof. Because r u r v = = ( (y, z) (u, v) i j k x u y u z u x v y v z v (z, x) (x, y),, (u, v) (u, v) If ( r u r v )(u 0, v 0 ) 0, then we can assume without loss of generality that (y,z) (u,v) (u 0, v 0 ) 0. Then (u, v) (x, y) is a map there is locally invertible near p. By inverse function theorem, locally we can solve u = u(y, z), v = v(y, z). So the surface is represented as (x(y, z), y, z) where x(y, z) = x(u(y, z), v(y, z)). ). Regular parameterizations are not unique. We can always change the parametrization. Consider a map: φ : D D (ũ, ṽ) u = u(ũ, ṽ), v = v(ũ, ṽ) Assume (i) φ is C 3 -differentiable; (ii) (u, v)/ (ũ, ṽ) 0. Then we get a regular parametrization r(ũ, ṽ) = r(u(ũ, ṽ), v(ũ, ṽ)). By the chain rule: u rũ = r u ũ + r v v ũ, r u ṽ = r u ṽ + r v v ṽ. (3) so that: (u, v) rũ rṽ = ( r u r v ) (ũ, ṽ). (4) When we are give a regular parametrization r = r(u, v), then r u r v determines an orientation of M. By (4), the coordinate change preserves the orientation if and only if (u, v)/ (ũ, ṽ) > 0. 1

2 1.1 Tangent Space, First Fundamental Form If C is a smooth curve on M, we can parametrize it by p : ( ɛ, ɛ) M The tangent vector of p(t) at t = 0 is given by: t r(u(t), v(t)). p (0) = r u u (0) + r v v (0). Notice that different curves may give the same tangent vector. For any p M, the tangent space T p M consists of all tangent vectors of a curve passing through p: T p M = {a r u + b r v ; a, b R}. For any X p = a r u + b r v T p M, we can consider its length using the inner product of E 3 : where we have denoted: X p 2 = a r u + b r v 2 = a 2 r u 2 + 2ab( r u, r v ) + b 2 r v 2 =: Ea 2 + 2F ab + Gb 2, E = E(u, v) = r u 2, F = F (u, v) = ( r u, r v ), G = G(u, v) = r v 2. To encode these information, we define two linear function and a quadratic form on T p M. Denote T p M = Hom R (T p M, R) the dual space of T p M. Let du and dv denote the basis of T p M that is dual to T p M. In other words, du and dv are linear functions on T p M satisfying: du( r u ) = 1, du( r v ) = 0; dv( r u ) = 0, dv( r v ) = 1. Definition 1.2. For any p M, the first fundamental form I is a quadratic form on T p M defined as: I(X, X) = Edu(X) 2 + 2F du(x)dv(x) + Gdv(X) 2 =: (Edu du + F (du dv + dv du) + Gdv dv) (X, X). Theorem 1.3. On a regular surface, locally we can always find an orthogonal coordinate chart. Proof. Fix any p M. Assume (u, v) is a coordinate chart around p such that u(p) = v(p) = 0. Consider the coordinate change of the type ũ = u, ṽ = f(u, v) for f(u, v) that satisfies f v (0, 0) 0. We have the identity: and ( uũ uṽ vũ vṽ So F = rũ rṽ = F f 1 v ) ( ) 1 ũu ũ = v = ṽ u ṽ v ( ) 1 ( 1 0 = f u f v 1 0 ) fv 1 f u fv 1 rũ = r u uũ + r v vũ = r u r v f 1 v f u, rṽ = r u uṽ + r v vṽ = r v f 1 v. Gfv 2 f u. So we just need find f(u, v) such that f u = f v G 1 F or equivalently (f u, f v )//(F, G). (6) (5) 2

3 So we see that the level set of f is tangent to the vector field ( G, F ). Note that G < 0 is non-vanishing. So the level curve of f is the graph of a function v = v(u) satisfying the ordinary differential equation: dv du = (G 1 F )(u, v), v(0) = w, (7) for w close to 0. Then v = v(u, w) is a differentiable function of (u, w) satisfying v w (0, w) = 1 0. By Inverse Function Theorem, we can solve w in terms of v: w = f(u, v). Now it s easy to see that f(u, v) satisfies f u = f v G 1 F. To see this, notice that we have the identity: w = f(u, v(u, w)). Both sides are considered as functions of (u, w). So we can take partial derivative with respect to u on both sides and get: 0 = w u = f u + f v dv du = f u + f v ( G 1 F ). 1.2 Unit Normal Vector, Shape Operator, Second Fundamental Form For a regular parametrization, we can define the unit normal vector: Then it s clear that: n p = r u r v (u(p), v(p)). r u r v T p M = n p. (8) Assume p(t) is a curve on M with p(0) = p and ṗ = X p T p M. n(p(t)) 2 = 1. Taking derivative with respect to t we get: 0 = d ( ) d n dt ( n(t), n(t)) = 2 dt, n. (9) This means that n(p(t)) n p(t) = T p(t)m. Theorem 1.4. The vector d n/dt t=0 depends only on X p and not on the curve p(t) chosen. Denote: S(X p ) = d n dt. (10) t=0 Then X p S(X p ) is a linear map of T p M T p M. Proof. n(t) = n(u(t), v(t)) : ( ɛ, ɛ) R 3. By the chain rule, we have: d dt n(t) = n n (u(0), v(0)) u(0) + (u(0), v(0)) v(0). t=0 u v So S(X p ) only depends on the components of X p = u(0) r u + v(0) r v. Theorem 1.5. S : T p M T p M is a symmetric linear map, in the sense that: for any X, Y T p M, we have the identity: (S(X), Y ) = (X, S(Y )) 3

4 Proof. We will calculate the matrix of S under the basis r u, r v. For any curve p = p(t) with p(0) = p and p (0) = u(0) r u + v(0) r v, we have: (S(X), r u ) = ( n, r u ) = d ( dt ( n, r u) + n, d ) dt ( r u) = (n, r uu u + r uv v) = ul + vm; (S(X), r v ) = (ṅ, r v ) = d ( dt (n, r v) + n, d ) dt ( r v) = (n, r vu u + r vv v) = um + vn; where we have used the identity r uv = r vu and introduced the notations: L = ( r uu, n), M = ( r uv, n), N = ( r vv, n). (11) So if X = u r u + v r v = a r u + b r v and Y = c r u + d r v, then (S(X), Y ) = (S(X), c r u + d r v ) = c(s(x), r u ) + d(s(x), r v ) = c(al + bm) + d(am + bn) = Lac + M(bc + ad) + Nbd. This is clearly symmetric with respect to X and Y : (S(X), Y ) = (S(Y ), X). It s easy to see that, equivalently, the matrix of S : T p M T p M under the basis r u, r v is equal to: ( ) 1 ( ) E F L M S =. F G M N Definition 1.6. For any p M, the second fundamental form II is a quadratic form on T p M defined as: II(X, X) = Ldu(X) 2 + 2Mdu(X)dv(X) + Ndv(X) 2 =: (Ldu du + M(du dv + dv du) + Ndv dv) (X, X). 1.3 Normal curvatures and Principle Curvatures A smooth curve p(t) is parametrized by arc-length if and only if p (t) = 1. In this case we usually write the parameter t as s. Fix a unit tangent vector X T p M (i.e. X = 1). We can choose a smooth curve p = p(s) parametrized by the arc-length satisfying with p(0) = p and p (0) = X. Definition 1.7. The normal curvature of p(t) at p is defined to be: κ n (X) = (p (0), n) = d dt (p (t), n) (p (0), n(0)) = (S(X), X). (12) t=0 We see that κ n (X) is well-defined: it does not depend on the chosen curve. By standard linear algebra, we know that there exist an orthonormal basis { e 1, e 2 } of T p Mand two real numbers κ 1 κ 2 such that I and II are diagonalized as: I = e 1 e 1 + e 2 e 2, II = κ 1 e 1 e 1 + κ 2 e 2 e 2. Definition 1.8. κ 1 and κ 2 are called principle curvatures of the surface M at p. Exercise 1. Show that κ 1 and κ 2 are the eigenvalues of S : T p M T p M. Moreover, we have: II(X, X) κ 1 = min X 0 I(X, X), κ II(X, X) 2 = max X 0 I(X, X). 4

5 Definition 1.9. The mean curvature H and Gauss curvature K of M at p is defined to be: H = 1 2 (κ 1 + κ 2 ), K = κ 1 κ 2. Exercise 2. Show the following two identities hold: H = Examples EN 2F M + GL tr(s) = 2(EG F 2, K = det(s) = LN M 2 ) EG F 2. (13) Example (Euclidean space) r(u, v) = (u, v, 0). I = du 2 + dv 2, II = 0. H = K = 0. Example (Unit sphere) r(u, v) = (sin u cos v, sin u sin v, cos u), 0 < u < π, 0 < v < 2π.. r u = (cos u cos v, cos u sin v, sin u), r v = ( sin u sin v, sin u cos v, 0). E = r u r u = 1, F = r u r v = 0, G = r v r v = sin 2 u. r u r v = (sin 2 u cos v, sin 2 u sin v, cos u sin u) = sin u r n(u, v) = r(u, v) L = n u r u = 1, M = r u r v = 0, N = n v r v = sin 2 u. κ 1 = κ 2 = 1, H = 2, K = 1. Example (Helicoid) r(u, v) = (u cos v, u sin v, av). r u = (cos v, sin v, 0), r v = ( u sin v, u cos v, a).. E = 1, F = 0, G = u 2 + a 2 r u r v = (a sin v, a cos v, u), n = (a sin v, a cos v, u) a2 + u 2. r uu = (0, 0, 0), r uv = ( sin v, cos v, 0), r vv = ( u cos v, u sin v, 0). a L = r uu n = 0, M = r uv r = a2 + u, r 2 vv n = 0. H = LG 2F M + EN EG F 2 = 0, K = LN M 2 EG F 2 = a 2 (a 2 + u 2 ) 2. Example (Rotation surface) r(u, v) = (u cos v, u sin v, f(u)), u 0, 0 v < 2π. r u = (cos v, sin v, f (u)), r v = ( u sin v, u cos v, 0). E = 1 + f (u) 2, F = 0, G = u 2. r u r v = ( f (u)u cos v, f (u)u sin v, u), n = ( f (u) cos v, f (u) sin v, 1) 1 + f (u) 2. r uu = (0, 0, f (u)), r uv = ( sin v, cos v, 0), r vv = ( u cos v, u sin v, 0). H = L = r uu n = f (u), M = r uv n = 0, N = r vv n = uf (u). 1 + f f 2 LG 2F M + EN EG F 2 = uf + f (1 + f 2 ) 2u(1 + f 2 ) 3/2, K = LN M 2 EG F 2 = f f u(1 + f 2 ) 2. 5

6 Exercise A tractrix curve is obtained as follows. Suppose an object is placed at (a, 0) and a puller at origin starts to move along the y-axis in the positive direction. At every moment, the thread is tangent to the curve y = y(x) described by the object. The movement will be described by the differential equation: dy dx = a2 x 2. (14) x Solve for the equation of the tractrix curve y = f(x), x > Consider the rotational surface r(x, θ) = (x cos θ, x sin θ, f(x)). (15) Calculate the mean curvature and Gaussian curvature of this surface. Exercise 4. A surface with zero mean curvature is called minimal. For example, the Helicoid is a minimal surface. In this exercise, we will find rotational surfaces that are minimal (these surfaces are called Catenoid). By Example 1.13, we know that a rotational surface r(u, v) = (u cos v, u sin v, f(u)) has zero mean curvature if and only if the following equation is satisfied: uf (u) + f (u)(1 + f (u) 2 ) = 0. (16) Solve this differential equation to get the parametrization of the minimal surface. Hint: first integrate to get: f 2 a2 = 1 + f 2 u 2. Exercise 5. Assume the graph of z = f(x, y) is a minimal surface. Derive the equation satisfied by f. Exercise 6. Prove the graph of z = arctan y x curvatures. is a minimal surface. Calculate its principle 1.5 Gauss Theorem Egregium Theorem The Gauss curvature depends only on the first fundamental form of the surface. Same fundamental form implies the tangent vectors have the same length. So the two surfaces with the same fundamental form are isometric to each other. Gauss s theorem means that the Gauss curvature depends only on the metric structure of the surface and does NOT depend on how the surface is embedded into R 3. The following are examples of isometric surfaces: Example The plane and r 1 (u, v) = (u, v, 0) and the cylinder r 1 (u, v) = (cos u, sin u, v) have the same fundamental form I = du 2 + dv 2 and both have Gauss curvature 0. Example and the The Helicoid: r 1 (u, v) = (u cos v, sin v, v) ( Catenoid: r 2 (u, v) = 1 + u2 cos v, ( 1 + u 2 sin v, log u + )) u 2 + 1). have the same first fundamental form and have the same Gauss curvature. 6

7 We will use tensorial notation and the Einstein summation convention. The index 1 denotes the variable u and 2 denotes the variable v. If an index i appear as upper and lower index in a term, we mean the summation 2 i=1. Moreover, we will denote: g 11 = r u r u = E, g 12 = g 21 = r u r v = F, g 22 = r v r v = G; b 11 = r uu n = L, b 12 = r uv n = F, b 22 = r vv n = G. g = (g ij ) is a 2 2 positive definite matrix and g 1 = (g ij ) is the inverse of g. In other words, we have: g ij g jk = 2 g ij g jk = δi k. j=1 Proof of Theorem Assume r : D r(d) M is a regular parametrization. At any point p = r(u, v) M, we can write down the following structural equation (notice that r i T p M = span R { r u, r v } = span R { r 1, r 2 }): { rij = Γ k ij r k + b ij n n i = b j i r j. (17) Notice that b ij = b ji and Γ k ij = Γk ji. Moreover: b j i g jk = n i r k = i ( n r k ) + n r ki = b ki = b ik. Taking derivatives on both sides, we get: Using r ijl = r ilj, we get: So we have: r ijl = ( l Γ k ij) r k + Γ k ij r kl + ( l b ij ) n + b ij n l = ( l Γ k ij) r k + Γ k ij(γ m kl r m + b kl n) + ( l b ij ) n + b ij ( b m l r m ) = ( l Γ k ij + Γ m ij Γ k ml b ij b k l ) r k + ( l b ij + Γ k ijb kl ) n. R k i lj := l Γ k ij j Γ k il + Γ m ij Γ k ml Γ m il Γ k mj = b ij b k l b il b k j. (18) R imlj := g km R k i lj = b ij b lm b il b jm. (19) Hence we have the following identity (called the Gauss equation): R 1221 = b 11 b 22 b 2 12 = LN M 2. (20) So it s sufficient to show that Γ k ij depends only on I. Γ k ijg kl = r ij r l = j ( r i r l ) r i r lj = j g il r i r jl = j g jl l g ij + r il r j = j g il l g ij + i ( r l r j ) r l r ji = j g il l g ij + i g lj Γ k jig kl. Since Γ k ij = Γk ji, we get: Γ k ij = 1 2 gkl ( j g il + i g jl l g ij ). (21) The coefficient Γ k ij in the above proof are called Christoffel symbols. 7

8 Exercise 7. Prove the following identities for any indices i, j, k, l: 1. R k i lj = R k i jl ; 2. R k i lj + R k l ji + R k j il = R ijkl = R klij (hint: use Gauss equation). Exercise 8. In general, calculating Gauss curvature involves long computations. This computation is greatly simplified by using the orthogonal coordinate system, which is guaranteed to exist by Theorem 1.3. In this exercise, we will obtain the Gauss curvature under orthogonal coordinate system: g 12 = g 21 = 0, i.e. F = Prove the following expression of Christoffel symbols: Γ 1 11 = 1 2 E 1 u E, Γ 2 11 = 1 2 G 1 v E; Γ 1 12 = 1 2 E 1 v E, Γ 2 21 = 1 2 G 1 u G; Γ 1 22 = 1 2 E 1 u G Γ 2 22 = 1 2 G 1 v G. 2. Prove the following expression for R 1221 = g 22 R : R 1221 = 1 ( u G + G 1 ( u G) 2 + E 1 ( u E)( u G) 2 ve 2 + E 1 ( v E) 2 + G 1 ( v E)( v G) ) ( ) ( )] v E = [ v u G + u (EG) 1/2. G E So we get the expression of Gauss curvature under orthogonal coordinate system: [( K = R 1221 EG = 1 ( ) ( G) u ( ) ] E) v +. (22) EG E G u v On a regular surface, a coordinate chart {D, (u, v)} is called an isothermal coordinate, if E = G and F = 0. It s a deep result that near any point p M, one can always find an isothermal coordinate around p. Under isothermal coordinate, the formula for Gauss curvature becomes very simple: 1.6 Gauss-Bonnet formula K = 1 2E ((log E) uu + (log E) vv ) = 1 log E. (23) 2E Theorem If M is a closed regular surface, then we have the identity: Kdσ = 2πχ(M). (24) M 8

9 The rest of this section is denoted to a proof of (24). dσ = EG F 2 dudv is the area element on M. χ(m) is the Euler number of M and can be calculated by the identity χ(m) = v e + f for any polytope partition of M such that v, e and f are the numbers of vertex, edges and faces respectively. Assume P is a domain on M with piece wisely smooth boundary curves and is contained in an orthogonal coordinate chart and has piecewisely smooth boundary. We want to calculate the integral: Kdσ. Using equation (22) and Green s formula we get: P [( 1 ( ) ( G) Kdσ = u ( ) ] E) v EGdudv + P P EG E u G v ( E) v = du ( G) u dv. (25) G E P Assume γ(s) = r(u(s), v(s)) is a smooth segment of P that is parametrized by the arclength. Denote e 1 = r u / E, e 2 = r v / G. Then we have: γ (s) = r u u (s) + r v v (s) = Eu e 1 + Gv e 2. Since γ(s) is parametrized by arc-length, we have γ (s) 2 = u 2 E + v 2 G = 1. Let cos θ = u E and sin θ = v G so that γ (s) = cos θ(s) e 1 + sin θ(s) e 2. So we have: γ (s) = θ ( sin θ e 1 + cos θ e 2 ) + cos θ e 1 + sin θ e 2. Denote by J : T M T M the linear homomorphism of counter-clockwise rotation by 90. Then Jγ = sin θ e 1 + cos θ e 2. The geodesic curvature of γ = γ(s) is by definition equal to: κ g (γ(s)) := γ Jγ = θ + cos 2 θ e 1 e 2 sin 2 θ e 2 e 1 = θ + e 1 e 2. (26) The second identity is because e 1 is perpendicular to e 2 : e2 e 1 = d ds ( e 2 e 1 ) e 2 e 1 = e 2 e 1. To calculate the e 1 e 2, we note that: ( ) d r u e 1 e 2 = e 2 ds E = r uuu + r uv v ( ) d e 2 + E ds (E 1/2 ) r u r v / G = ve 2 EG u + ug 2 EG v where we have used the following two identities: = ( E) v u + ( G) u v, (27) G E r uu e 2 = r uu r v = 1 ( u ( r u r v ) r u r vu ) G G and r uv e 2 = r vu = v( r u r u ) 2 G = ve 2 G ; r v = 1 u ( r v r v ) = ug G 2 G 2 G. 9

10 Combining (25), (26) and (27), we get: Kdσ = (θ κ g (s))ds P γ P γ = 2π + i (α i π) κ g ds. (28) P In the above, α i are inner angles of the polytope at the vertices. We used the formula: dθ + (π α i ) = 2π. γ i γ P Remark Notice that dθ = d arctan y x = xdy ydx x 2 +y. If P is smooth, then we would get: 2 xdy ydx dθ = x 2 + y 2 = 2π. P P Now partition the surface M into polytopes such that each polytope is contained in an orthonormal coordinate system. Then we can use the above formula (28) for each polytope in the partition and then sum them up. Notice that the terms P κ g are cancelled out so we get: M Kdσ = 2πf + m f j (α ij π) j=1 i j=1 = 2πf + 2πv π2e = 2π(v e + f) = 2πχ(M). Exercise 9. In the above proof, we defined geodesic curvature of a curve on the surface to be k g (γ(s)) = γ(x) J γ(s) where s is the arc length parameter. A curve with zero geodesic curvature is called a geodesic. Prove that an latitude on the sphere is a geodesic if and only if it s the equator. By symmetry, we also know that the big circles (circles cutting out by planes through the center) are geodesics. Exercise 10. For any orthogonal coordinate chart (u, v) D, we have the moving orthonormal frame: e 1 = ru E, e 2 = rv G. Consider the set: S 1 T D := {cos θ e 1 + sin θ e 2 ; θ [0, 2π), (u, v) D} = D S 1. (29) This is called the unit circle bundle over D. Consider the following differential on S 1 T D: η := (( e 1 ) u e 2 ) du + ( e 1 ) v e 2 ) dv. (30) 1. Show that η = ( E) v G du + ( G) u dv. E 2. Using the chain rule, show that for any other coordinate (ũ, ṽ) on D (not necessarily orthogonal), we have: η = (( e 1 )ũ e 2 )dũ + (( e 1 )ṽ e 2 )dṽ. (31) So η just depends on the moving orthogonal frame and does not depend on the coordinate. 10

11 3. Assume we have another moving orthonormal frame: f1, f 2. The two orthonormal frames differ by a rotation: e 1 = (cos α) f 1 + (sin α) f 2, e 2 = (sin α) f 1 + (cos α) f 2. (32) Note any element in S 1 T D can be expressed under { e 1, e 2 } and { f 1, f 2 } as: (cos θ) e 1 + (sin θ) e 2 = cos(θ + α) f 1 + sin(θ + α) f 2 = cos( θ) f 1 + sin( θ) f 2. (33) where θ := θ + α. Define Prove the following identity: η = (( f 1 ) u f 2 )du + (( f 1 ) v f 2 )dv. η = η + dα. Note that this is equivalent to dθ + η = d θ + η. The above calculation thus shows that dθ + η is a globally well-defined form on the unit circle bundle S 1 T M. Choose any smooth vector field V with isolated zero points {p i }. Let B r (p i ) denote the ball of radius r around p i. Then X := V V : M \ ib r (p i ) ST M is a smooth map. We can then calculate: Kdσ = lim Kdσ M r 0 M\ ib r(p i) = lim dη r 0 M\ ib r(p i) = lim r 0 M\ ib r(p i) = lim r 0 i = lim r 0 i p V (p) V (p) X ( d(η + dθ)) X η + B r(p i) i B r(p i) X dθ = 2π i B r(p i) X dθ ind pi (V ). Here ind p V is the index or the winding number of V at p. The last sum is equal to the Euler characteristic (this is called Poincare-Hopf theorem). The above calculation can be generalized to higher dimensions, which leads to S.S.Chern s intrinsic proof of Gauss-Bonnet formula for higher dimensional Riemannian manifolds. Exercise 11. The above calculation can actually be made very concrete in the case of surfaces. Assume V is a smooth vector field V with isolated zero points {p i }. Triangulate the surface such that each triangle contains at most one zero point. Let X := denote the associated unit vector field away from the punctures {p i }. Then under an orthogonal coordinate chart, we can write (using the same notation as above): X = (cos ϕ) e 1 + (sin ϕ) e 2. V V 11

12 1. Prove that along any part of boundary curve γ P which is parametrized using arclength parameter γ = γ(s) : I R 3, we have the similar identity as in (26) (dot means derivative w.r.t. s): Ẋ JX = ϕ + e 1 e One can show the following purely topological facts (assume these two facts): If an triangle P contains an isolated zero point (a puncture) p i, then define: ind pi (V ) := 1 dϕ. (34) 2π One can show that this is an integer and does not depend on the triangle containing the point p i. If P does not contain any punctures, then P dϕ = 0 Sum up over the triangles to prove the identity (you want to use Green s formula as before): Kdσ = 2π ind pi (V ). (35) p i M Combine this with the Gauss-Bonnet theorem to prove the Hopf-Poincaré formula: p i ind pi (V ) = χ(m). (36) 3. Deduce from the Hopf-Poincaré theorem that at any time, there must be a point on the earth where the velocity of wind is 0. 2 Differentiable manifolds A topological manifold of dimension n is a Hausdorff space with a countable basis of open sets and the further property that each point has a neighborhood homeomorphic to an open subset of R n. Recall the basic concepts from point-set topology: Definition 2.1. (i) A topological space X is an ordered pair (X, τ), where M is a set and τ is a collection of subsets of M satisfying (1):, M belong to τ; (2): Any union of members of τ still belongs to τ. (3): The intersection of any finite number of members of τ still belong to τ. If p U X with U τ, then U is called a neighborhood of p. (ii) X is Hausdorff if for any p, q X, there exist neighborhood U, V of p, q respectively such that U V =. M is of countable basis of open sets if there exists a countable collection U = {U i } i=1 of open sets of M such that any open subset of M can be written as a union of elements of some subfamily of U. (iii) A map f : X Y between two topological spaces is continuous if f 1 (U) is open for any open set U Y. Each pair (U, ϕ), where U is an open set of M and ϕ : U ϕ(u) R is a homeomorphism, is called a coordinate neighborhood. For each q U, we denote the i-th coordinate function x i (q) = x i (ϕ(q)). Assume (V, ψ) is a second coordinate neighborhood and it has coordinates (y 1 (q),, y n (q)). Since ϕ and ψ are both homeomorphisms, we get a homeomorphism: P ψ ϕ 1 : ϕ(u V ) ψ(u V ), (37) 12

13 which is represented by continuous functions: y i = h i (x 1,..., x n ), i = 1,..., n; (38) giving the y-coordinates of each q U V in terms of its x-coordinates. Conversely we also have: ϕ ψ 1 : ψ(u V ) ϕ(u V ), x j = g j (y 1,..., y n ). (39) Definition 2.2. We say that two coordinate neighborhoods (U, ϕ) and (V, ψ) are C -compatible (resp. C r -compatible) if U V nonempty implies that the function h i (x) and g j (y) giving the change of coordinates are C (C r ); this is equivalent to ϕ ψ 1 and ψ ϕ 1 being diffeomorphisms of the open subsets ϕ(u V ) and ψ(u V ) of R n. Definition 2.3. A C (resp. C r ) structure on a topological manifold M is a family U = {(U α, ϕ α )} of coordinate neighborhoods such that (1) M = α U α, i.e. {U α } cover M; (2) for any α, β, the neighborhoods (U α, ϕ α ) and (U β, ϕ β ) are C -compatible (resp. C r - compatible); (3) any coordinate neighborhood (V, ψ) C -compatible (resp. C r -compatible) with every (U α, ϕ α ) U is itself in U. A C (resp. C r ) manifold is a topological manifold together with a C structure. Theorem 2.4. Let M be a Hausdorff space with a countable basis of open sets. If {V β, ψ β } is a covering of M by C -compatible coordinate neighborhoods, then there is a unique C structure on M containing these coordinate neighborhoods. Fact: Every 2 and 3-dimensional topological manifolds have a unique smooth structure. This is not true for higher dimensional manifolds. There are topological 4-manifolds that admit no smooth structure. 2.1 Examples 1: (Euclidean spaces) M = R n. U = M, ϕ : U R n is the identity. 2: (regular surfaces) If M R 3 is a C r (r 1) regular surface, then any p M has a neighborhood with a regular parametrization r : D r(d) M. Using inverse function theorem, one can show that the transition functions are C r. 3: (n-dimensional Sphere) M = S n = {z = (z 1,..., z n+1 ) R n+1 ; z 2 := i z2 i = 1}. {(U i, ϕ i ), (V i, ψ i )}. U i = {z i > 0}, ϕ i (z) = (z 1,..., ẑ i,..., z n+1 ) =: (x 1 i,..., xn i ). V i = {z i < 0}, ψ(z) = (z 1,... ẑ i,..., z n+1 ) =: (yi 1,..., yn i ). Change of coordinates: ϕ 2 ϕ 1 1 : ϕ 1 (U 1 U 2 ) ϕ 2 (U 1 U 2 ) is given by: ϕ 2 ϕ 1 1 (x1 1,..., x n 1 ) = ϕ 1 ( 1 x 2, x 1 1,..., x n 1 ) = 1 (x i 1 )2, x 2 1,..., x n 1. (40) i 4: (open submanifold of a smooth manifold). If M is an smooth submanifold and N M is an open set of M. Assume U = (U α, ϕ α ) is an atlas of M. Then N is a smooth strucutre with an atlas given by {(U α N, ϕ α )}. For example, assume m n and denote by Mat m n (R) the set of m n real matrices. Let N denote that the set of rank m matrices in Mat m n (R) = R mn is an open set of M and hence is a smooth manifold. In particular, the set of non-singular 13

14 m m matrices is a smooth manifold. Denote this set by GL(m, R). Note that GL(m, R) is naturally a group. Examples of Quotient manifolds Let X be a topological space and an equivalence relation on X. Denote by [x] = {y X; y X} X the equivalence class of x, and for a subset A X, denote [A] = x A [x]. X/ denotes the set of equivalence classes and π : X X/ is the natural projection. The quotient topology on X/ is defined as follows: a set U X/ is open if and only if π 1 (U) is open. This topology makes π continuous. Definition 2.5. The equivalence relation is open if [A] is open whenever A is open. This is equivalent to π : X X/ being an open mapping: Lemma 2.6. is open if and only if π is an open mapping (i.e. it maps open sets to open sets). If X has a countable basis of open sets and is open, then X/ is a topological space with countable basis of open sets. Lemma 2.7. Let be an open equivalence relation on a topological space X. Then R = {(x, y); x y} is a closed subset of X X if and only if the quotient topology on X/ is Hausdorff. Exercise 12 (Bo III.2.2). Let X = R {±1} with the Cartesian topology and C structure. Define an equivalence relation on X by (x, i) (y, j) if either (i): x = y < 0 or (ii): x = y and i = j. Show that X/ is locally Euclidean and has a countable basis of open sets but is not Hausdorff. 5: (real projective space) M = RP n = S n /Z 2 ; z z. U i = {z S n ; z i 0}/Z 2, ϕ i : U i R n, [(z 1,..., z n+1 )] sign(z i ) (z 1,..., ẑ i,..., z n+1 ) where z i > 0. is an open equivalence relation since [A] = A ( A). Define F : S n S n R, F (x, y) = x y 2 x+y 2. Then R = {(x, y); x y} = F 1 (0) is a closed subset of S n S n. So RP n = S n / is a Hausdorff topological space with countable basis of open sets. Exercise 13. Show that RP n = (R n+1 {0})/, where z z if and only if z = λz for λ 0 R. We have {[ ] } z U i = {[z]; z i 0} = ; z i 0. z Denote ϕ i : U i R n by: ϕ i ([z]) = ( z 1 (,..., ) zi z i z i,..., z n+1 z i ). Then ϕ 1 ϕ 1 1 (x1,..., x n ) = ϕ 1 ϕ 1 1 (y1,..., y n ) = ( ( x 1 1 x 2,..., y y 2,..., ) x n ; 1 x 2 ) y n. 1 + y 2 6: (Grassmannian manifold) Fix m n Z. Denote by M = Gr(m, R n ) the set of subspaces of dimension m inside R n. Let M 1 denote the set of m n matrices of rank m. The group GL(m, R) acts on M 1 on the left. Then M is the quotient manifold M 1 /GL(m, R). Any element p M is represented by a basis b = { v 1,..., v m } of p. Extend p to a basis { v 1,..., v m, v m+1,..., v n } of R n. 14

15 Exercise 14. Prove that there is an open set U p of M containing p such that each q U is generated by a basis of the form: n n v 1 + a 1j v j,..., v m + a mj v j. j=m+1 j=m+1 Moreover, such basis is uniquely determined by q. This implies {a ij ; 1 i m, m + 1 j n m} is a coordinate on U p. Hence M is a smooth manifold of dimension m(n m). 7: (Product manifolds) Assume M i, i = 1,..., k are k smooth manifolds. Show that M 1 M k is a smooth manifold. 8: (torus) (1-dimensional) M = {(x, y) R 2 ; x 2 + y 2 = 1} = S 1. U 1 = {(cos α, sin α); 0 < α < 2π}, ϕ 1 (x, y) = α. U 2 = {(cos β, sin β); π < α < π}, ϕ 2 (x, y) = β. (n-dimensional torus) M = torus. 2.2 Differentiable mapping and its rank n times {}}{ S 1 S 1 is a smooth manifold and is called the n-dimensional Assume M and N are two smooth manifolds. A map f : M N is smooth, if for any p M, there exist coordinate neighborhoods (U, ϕ = (x 1,..., x m )) of p and (V, ψ = (y 1,..., y n )) of f(p) such that f is represented by smooth functions: y i = y i (x 1,..., x m ), i = 1,..., n. The rank of f at p M is defined to be the rank of the Jacobian matrix: J(f)(x 1,..., x m ) = (y1,..., y n ( ) ) y i (x 1,..., x m ) = x j f is called an immersion, if rk(f) m.. 1 i n,1 j m Definition 2.8. An embedding (or imbedding) is a one-to-one immersion F : N M which is a homeomorphism of N into M, that is, F is a homeomorphism of N onto its image, Ñ = F (N), with its topology as a subspace of M. The image of an embedding is called an embedded submanifold. Theorem 2.9. If F : N M is an immersion. Then each p N has a neighborhood U such that F U is an embedding of U into M. Theorem 2.9 follows immediately from the following important result about smooth mappings on open sets of Euclidean spaces: Theorem 2.10 (Rank Theorem). Let A 0 R n, B 0 R m be open sets, F : A 0 B 0 be a C r mapping, and suppose the rank of F on A 0 is equal to k. If a A 0 and b = F (a) B 0, then there exist open sets A A 0 and B B 0 with a A and b B, and there exist C r diffeomorphism G : A U R n, H : B V R m such that H F G 1 (U) V and this map has the simple form: H F G 1 (x 1,..., x n ) = (x 1,..., x k, 0,..., 0). 15

16 Proof. F is locally represented by y i = y i (x 1,..., x m ). Without the loss of generality, we can assume (y 1,..., y k )/ (x 1,..., x k ) 0 and all bigger size sub-determinants of J(F ) = (y 1,..., y n )/ (x 1,..., x m ) are zero. Then consider the map G(x 1,..., x m ) = (y 1 (x),..., y k (x), x k+1,..., x m ). It s straight forward to verify that Rank(DG) = m. By IFT, we get an inverse G 1 represented by: x j = x j (y 1,..., y k, x k+1,..., x m ), j = 1,..., k so that F G 1 (y 1,..., y k, x k+1,..., x m ) = (y 1,..., y k, F 1 (y, x ),..., F n (y, x )). Moreover, we have: Rank(D(F g)) = k. This implies F i / x j = 0 for i = 1,..., n; j = k + 1,..., m. In other words, F i only depends on y = (y 1,..., y k ). Next consider the map: H(y 1,..., y n ) = (y 1,..., y k, y 1 F 1 (y ), y n F (y )). Then Rank(DH) = n and hence H is a local diffeomorphism and we have: H F G 1 (y 1,..., y n ; x k+1,..., x m ) = (y 1,..., y k, 0,..., 0). The Rank Theorem can be proved via the Inverse Function Theorem: Theorem 2.11 (Inverse Function Theorem). Let W be an open subset of R n and F : W R n be a C r mapping with r 2. If a W and DF (a) is nonsingular, then there exists an open neighborhood U of a in W such that V = F (U) is open and F : U V is a C r diffeomorphism. If x U and y = F (x), then we have the following formula for the derivatives of F 1 at y: DF 1 (y) = (DF (x)) 1. Proof of the existence of inverse. We want to find small balls B r (a) W and B ɛ (b) R n such that for each y B ɛ (b), there exists a unique solution x B r (a) such that F (x) = y. Assume DF (a) = A. Because F C r for r 2, we can find an r sufficiently small such that for any x B r (a), we have: DF (a) 1 DF (x ) I 1 4. (41) Then by using Newton-Leibniz formula, we get, if r is sufficiently small, then for any two points x 1, x 2 B r (a) 1 A 1 (F (x 2 ) F (x 1 )) (x 2 x 1 ) = (DF (a) 1 DF (x 1 + t(x 2 x 1 )) I) (x 2 x 1 )dt For any y R n, consider the map Φ : B r (a) R n x 2 x 1. x A 1 (F (x) y) + x. Define the sequence x 0 = 0, x k+1 = Φ(x k ), k 0. Then we want to show that for y sufficiently close to b, {x k } is a Cauchy sequence and hence converges to a limit x = x (y) satisfying Φ(x ) = x or equivalently F (x ) = y. 16

17 1. Φ maps small ball around a to itself: Φ(x) a A 1 (F (x) y) + x a = A 1 (F (x) F (a)) A 1 (F (a) b) + A 1 (y b) + (x a) A 1 (F (x) F (a)) (x a) + A 1 (y b) 1 4 x a + A 1 y b. So if y b A 1 1 r 4 =: ɛ and x a r, then Φ(x) a 1 2r < r. In other words, for any y B ɛ (b), Φ(B r (a)) B r (a)) with the choice of r made earlier. 2. The map Φ is contracting: For any x 1, x 2 B r (a) W we have: Φ(x 2 ) Φ(x 1 ) (A 1 (F (x 2 ) F (x 1 )) (x 2 x 1 ) 1 2 x 2 x 1. Now it s straight forward to verify that {x k } converges to a solution x to Φ(x) = x. This solution is actually unique: if Φ(x i ) = x i, then the inequality implies x 1 = x 2. x 2 x 1 = Φ(x 2 ) Φ(x 1 ) 1 2 (x 2 x 1 ) Exercise 15. Prove that the inverse map obtained in the above proof is differentiable and we have DF 1 (y) = DF (x) if F (x) = y. Exercise 16. Let A 0 be an open set of R n. Assume that a C r mapping F : U R m := R k R m k satisfies for any a F 1 (0 R m k ) we have Image(J(F )) + (0 R m k ) = R m where J(F ) : R n R m = R k R m k is the Jacobian of F. (In this case, F is called transversal to 0 R m k ) Show that there exists a diffeomorphism G : A U such that F G 1 (x) = (x 1,..., x k, f k+1 (x 1,..., x n ),..., f m (x 1,..., x n )). Examples of immersed submanifolds which are not embedded submanifolds: 1. Curve spiraling to (0, 0): F (t) = (t 1 cos(t), t 1 sin t). 2. Curve spiraling towards the unit circle: F (t) = ((t + 1)t 1 cos t, (t + 1)t 1 sin t). 3. Figure 8 curves 4. dense curves of irrational slopes on a torus 5. curves that have are not locally connected under the subspace topology. Theorem 2.12 (Whitney). Any smooth manifold M m can be immersed into R 2m. Remark The dimension 2m in Whitney s theorem is sharp. For example, S 1 can not be smoothly immersed into R Smale discovered that the 2-sphere can be turned from inside to outside by a family of smooth immersions. This can be done only for spheres of dimension 0, 2 and 6. 17

18 2.3 Submanifolds Definition Let M be an m-dimensional smooth manifold. A subset N M has an n-submanifold property if for any p N, there exists a coordinate neighborhood (U, ϕ = {x 1,..., x m }) of p M such that N U = {x n+1 = 0,..., x m = 0}. Such a coordinate neighborhood is called preferred. Remark We can also require the coordinate neighborhood (U, ϕ) to satisfy the refined property: (i) ϕ(p) = 0; (ii) ϕ(u) = C m ɛ ; (iii) ϕ(u N) = {x C m ɛ (0); x n+1 = = x m = 0}. This can be achieved by shrinking the neighborhood U. Lemma Let N M have the n-submanifold property. Then N with the relative topology is a topological manfold n-manifold with a C structure defined by the preferred coordinate neighborhoods. Definition A regular submanifold of a C manifold M is any subspace N with the submanifold property and with the C structure that the corresponding preferred coordinate neighborhoods determine on it. Theorem Let F : N M be an embedding of C manifolds. Then F (N) is a regular submanifold of M and F : N F (N) is a diffeomorphism. Proof. Use the coordinate system constructed by the Rank Theorem. Use the assumption that N is homeomorphic to F (N) with the relative topology to conclude that a neighborhood of F (U) for p U N does not contain other branches of F (N) as long as U is sufficiently small. Theorem If F : N M is a one-to-one immersion and N is compact, then F is an embedding and F (N) is a regular submanifold of M. Proof. Proof 1: Image of compact set under a continuous map is compact. For any p N, there exist coordinate neighborhoods (U, ϕ) of p and (V, ψ) of F (p) M such that F (x 1,..., x n ) = (x 1,..., x n, 0,..., 0). Since F (N \ 1 2 (U)) is a closed compact set, we can choose V sufficiently small such that V F (N \ 1 2 (U)) =. Proof 2: Show that F maps closed subsets of N to closed subsets F (N) (with respect to the subspace topology). Being one-to-one implies that F also maps open subsets of N to open subsets of F (N). So F is a homemorphism of N to F (N) and therefore is an imbedding. So F (N) is a regular submanifold. Theorem Let F : N n M m be a smooth mapping. Suppose F has constant rank k and q F (N). Then F 1 (q) is a closed, regular submanifold of N of dimension m k. Proof. Choose any p f 1 (q). Using the Rank Theorem, we know that near p and q, locally F looks like ψ V F ϕ U : (x 1,..., x n ) (x 1,..., x k, 0,..., 0). We have assumed ϕ(p) = 0 R n and ψ(q) = 0 R m. Then F 1 (q) U = {x ϕ(u); x 1 = = x k = 0}. So F 1 (q) satisfies the submanifold property. Hence F 1 (q) is a submanifold of dimension m k. Corollary If the rank of F is equal to n for every point of A = F 1 (a), then A is a closed, regular submanifold of dimension n m of N. F : N M is called a submersion if rk(f ) m. Example F : R n R, F (x 1,..., x n ) = n i=1 (xi ) 2 has rank 1 on R n {0}. So S n 1 = F 1 (1) is an (n 1)-dimensional submanifold of R n. More generally, if F is a degree d polynomial such that {F = 0, F/ x i = 0} is empty, then F 1 (0) is a submanifold of R n. 18

19 Exercise 17 (Bo III.4.2). Show that a submersion is an open mapping and that if both manifolds have the same dimension, it is locally-but not necessarily globally- a diffeomorphism on to its image. Exercise 18 (Bo III.5.7). Prove that F : S 2 R 4 given by F (x 1, x 2, x 3 ) = ((x 1 ) 2 (x 2 ) 2, x 1 x 2, x 1 x 3, x 2 x 3 ) induces an embedding RP 2 R Lie group Definition A topological group G is a Lie group if it is a smooth manifold and the group operations are smooth: (x, y) xy and x x 1 are both C maps. Example Discrete groups, GL(n, R), (S 1 ) r, C, GL(n, C). G 1 G 2 if G 1 and G 2 are Lie groups. Theorem Let G be a Lie group and H a subgroup which is also a regular submanifold. Then H is a Lie group. This follows easily from the following lemma. Lemma Let F : A M be a C mapping of C manifolds and suppose F (A) N, N being a regular submanifold of M. Then F is C as a mapping into N. Proof. Use the adapted coordinate neighborhoods of N as a regular submanifold of M. Definition Let F : G 1 G 2 be an algebraic homomorphism of Lie groups. F is called a homomorphism of Lie groups if F is also a C mapping. Theorem If F is a homomorphism of Lie groups, then the rank of F is constant. As a consequence, the kernel is a closed regular submanifold and thus a Lie subgroup; dim KerF = dim G 1 dim G 2. Proof. Fix any point a G 1. We have: F (x) = F (xa 1 a) = F (xa 1 )F (a) So we have: F = R F (a) F R a 1. Since R F (a) and R a 1 rk(f )(a) = rk(f )(e) for any a G. are diffeomorphisms, we have: Example SL(n, R) = {X GL(n, R); det(x) = 1}; SL(n, C) O(n) = {X GL(n, R); X t X = I}; SO(n) = {X O(n); det(x) = 1}. Proof that O(n) is a Lie group. Consider the map F : GL(n, R) GL(n, R), X X t X. The derivative of F at X is given by DF (A) = A t X + X t A = (X t A) t + X t A. So Im(DF ) = {symmetric matrices} and rk(f ) = dim{symmetric matrices} = n2 +n 2 is constant. So F 1 (I) = O(n) is a regular submanifold and is a Lie group. SO(n) is the connected component of O(n) and is a Lie subgroup. Alternatively, we have F (X) = X t X = X 0 (X0 1 X t XX0 1 0 = X0(XX t 0 1 (XX0 1 0 = L X t 0 R X0 F R X 1(X). 0 Because the left and right multiplications on Lie groups are diffeomorphism, rk(f )(X 0 ) = rk(f )(I) is independent of X 0 GL(n, R). 19

20 U(n) = {X GL(n, C); X X = I}; SU(n) = {X U(n); det(x) = 1}. Exercise 19 (Boothby III.7.1). Show that given any neighborhood U of e, the identity of a Lie group G, then there exists a neighborhood V of e such that V V 1 = {g 1 g2 1 ; g 1, g 2 V } U, and a neighborhood W of e such that W 2 = W W = {g 1 g 2 ; g 1, g 2 W } U. Exercise 20 (Boothby III.7.2). Show that the collection {xu} over all neighborhoods U of e, is a base of neighborhoods for x (similarly for {Ux}). Definition A Lie subgroup of a Lie group is an algebraic subgroup which is also a submanifold with its C structure as an immersed submanifold. Example H : R T 2 = S 1 S 1 given by t (exp(2πit), exp(2πiαt)) is a immersed Lie subgroup of T 2 = S 1 S 1. If α is an irrational number. Then H(R) is a immersed submanifold of T 2 which is diffeomorphic to R and is dense as a subset of T 2.. Exercise 21 (Boothby III.7.9). Show that H : R T 2 = S 1 S 1 given by t (exp(2πit), exp(2πiαt)) is a Lie subgroup of T 2 = S 1 S 1. Assume α is an rational number. Then H(R) is a regular submanifold of T 2 diffeomorphic to S 1. Theorem If H is a regular submanifold and subgroup of a Lie group G, then H is closed as a subset of G. Proof. Assume H h k g G. We want to show that g H. Fix a neighborhood V of e such that V H is closed. Because the maps (g 1, g 2 ) g 1 g 2 and g 1 g1 1 are both continuous, we can choose a neighborhood U of e such that UU 1 V. For k, l sufficiently large, g 1 h k U and g 1 h l U. Then we have h 1 k h l = (g 1 h k ) 1 (g 1 h l ) V H. Letting l +, we get h 1 k g V H. So g H. Group action on smooth manifolds: A group G acts on M (on the left) if there is a homomorphism θ : G S M where S M is the group of permutations on the set of points of M. Equivalently, we have a map: θ : G M M, (g, x) θ(g, x) =: g x = gx satisfying the following properties: 1. e x = x for any x M; 2. g 1 g 2 x = (g 1 g 2 )x for any x M. The action is effective if θ is injective. The action is free if for any g e G, g does not have fixed points: g x x for any x M. The stabilizer (stability or isotropy group) G x of a point x M is defined to be the following subgroup: The orbit of x M under G is given by: {g G; g x = x}. (42) Gx = {g x; g G}. Exercise Show that there is a bijection between Gx and the quotient G/G x. 2. Show that if Gx = Gy, then G x is conjugate to G y, i.e. there exists g 1 G such that G y = g 1 1 G xg 1. 20

21 The collection {Gx; x M} is a partition of M which defines an equivalence relation on M. Each equivalence class is an orbit of the G-action. The action is called transitive if there is only one orbit under G, or equivalently, for any x, y M, there exists g G such that g x = y. In this case, M = Gx = G/G x for any x M. The quotient space M/G consists of all the orbits and is equipped with the quotient topology. If G is a Lie (resp. topological) group, then the action is called smooth (resp. continuous) if the map θ is smooth (resp. continuous) mapping. Under suitable assumptions, M/G is also a smooth manifold. Exercise 23. Assume that G is a discrete subgroup that acts smoothly on M. The action is called properly discontinuous if it satisfies the following two properties: (i) For any point x M, there is a neighborhood U of x such that {g G; gu U } is finite; (ii) If Gx Gy, then there exist neighborhood U of x and V of y, such that GU V =. Proving the following statements: 1. Show that the equivalence relation defined by a smooth group action is always open. 2. Show that the property (ii) is equivalent to the property that M/G is Hausdorff. 3. Assume furthermore that G acts freely on M. Show that (i) allows one to construct coordinate neighborhoods on M/G such that they define a smooth structure on M/G. For example RP n = S n /Z 2 is a quotient manifold obtained by a properly discontinuous group action. Exercise 24. Fix any α, β R 2 >0. Show that the following action of Z 2 on R 2 is properly discontinuous: (m, n) (x, y) = (x + αm, y + βn). (43) What is the quotient manifold R 2 /Z 2? Exercise If H is a Lie subgroup of G and is a regular submanifold, then G acts transitively on the set of cosets G/H. 2. What is the stabilizer of G at any point gh G/H? 3. If a Lie subgroup H is a closed set of G (which is true if H is a regular submanifold by Theorem 2.32), then G/H is a Hausdorff space. 3 Vector fields 3.1 Tangent space and cotangent space Let M be a smooth manifold. Let C (p) denote the germ of smooth functions at p. Fix any point p M. A tangent vector at p is a derivative at p: Definition 3.1. A tangent vector V at p is a linear map V : C (p) R satisfying: (i) V is R-linear: for any f, g C (p) and λ R, we have: V (f + λg) = V (f) + λv (g). (ii) (Leibniz rule) For any f, g C (p), we have: V (fg) = V (f)g(p) + f(p) V (g). 21

22 In (ii) above, if we let f = g = 1, then we get: V (1) = 2V (1) which implies V (1) = 0. As a consequence, V (λ) = λv (1) = 0 for any λ R considered as the constant function on M. Choose a coordinate neighborhood (U, ϕ = (x 1,..., x m )) of p M. In the following, we will just f(x 1,..., x m ) for f ϕ 1 if it s not confusing. Proposition 3.2. Denote a i = V (x i ). Then for any f C (p), we have: V (f)(p) = m i=1 a i x i (f ϕ 1 )(ϕ(p)). (44) Proof. With out the loss of generality we can assume x = 0. By Newton-Leibniz formula, we can write 1 d 1 f f(x) = f(0) + f(tx)dt = f(0) + 0 dt x i (tx)xi dt = f(0) + i x i g i (x). 0 i So we have the identity: V (f(x)) x=0 = i = i V (x i )g i (x) + x i V (g i (x)) V (x i ) f x i (0) x=0 The set of tangent vectors at p is called the tangent space at p M, which is usually denoted by T p M. From Proposition 3.2, T p M is a vector space which is spanned by a basis given by { x }. We call this to be the coordinate frame (associated to the coordinate neighborhood i (U, ϕ U )). If we choose a different coordinate neighborhood (V, ϕ V = (y 1,..., y m )). Then by the chain rule we have: (f ϕ 1 yi V ) = (f ϕ 1 yi U ) (ϕ U ϕ 1 V ) m [ ] = (f ϕ 1 xj U ) (ϕ U ϕ 1 V ) y i (ϕ U ϕ 1 = j=1 m j=1 f x j (x(p)) xj y i (y(p)). V ) So we can just write: m y i = j=1 x j y i x j. (45) Definition 3.3. Assume F : M N is a smooth mapping. p M and q = f(p) N. Then the differential of F at p is the map F : T p M T q N, satisfying for any g C (q), we have: F (V )(g) = V (g F ). (46) 22

23 Under local coordinate chart, F is given by (x 1,..., x m ) (y 1,..., y n ). Then F ( x i )(g) = x i (g F ) = n j=1 y j g(y(q)) yj x i (x(p)). So under the frame { x i ; i = 1,..., m} and { y j ; j = 1,..., n}, we have: F ( x i ) = n j=1 y j x i (x(p)) y j If γ : ( ɛ, ɛ) M is a smooth curve passing through p : γ(0) = p, then γ (p) = γ ( t (0)). Exercise 26. Prove the following equivalence: T p M = {smooth curves passing through p M}/, where γ γ if and only if γ (0) = γ (0), or equivalently, γ γ iff for any function f C (p), we have: d (f γ)(t) dt = d t=0 dt t=0 (f γ)(t). Moreover if F : M N is a smooth mapping, then we have F (γ (p)) = (F γ) (F (p)). Exercise 27. Let F : M R be a smooth function. If t denotes the coordinate on R. Then F (V ) = V (F ) t for any V T pm. Definition 3.4. The cotangent space at p M, denoted by T p M is the dual space of T p M. An element in T p M is called a cotangent vector. For any f C (p), V V (f) is a linear function on T p M. So f determines an element in T p M which will be denote by df. From these definition, it s easy to see Proposition 3.5. For any coordinate neighborhood (U, ϕ U = (x 1,..., x m )), the dual basis of the coordinate frame { x ; i = 1,..., m} is {dx i ; i = 1,..., m}. In other words, dx i ( i x j ) = δ ij. Exercise 28. Show that for two coordinate neighborhoods (U, ϕ U = (x 1,..., x m )) and (V, ϕ V = (y 1,..., y m )), we have: m dy i y i = x j dxj. j=1 3.2 Tangent bundle and vector fields Let M be a smooth manifold and (U, ϕ U = (x 1,..., x n )) be a coordinate neighborhood on M. Then we have a globally defined frame of tangent vectors x, i = 1,..., n on U. Any i vector V at p U can be written as V = i ai x. The tangent bundle of A is defined as i { T U = (p, i a i x i); p A, {a i } R n } = U R n = ϕu (U) R n. The tangent bundle T M is a smooth manifold constructed as follows. Assume {(U α, ϕ Uα )} is an atlas of coordinate neighborhoods covering M. Then T M is a smooth manifold obtained 23

24 by gluing T U α by the transition map: F β Fα 1 : (U α U β ) R n (U α U β ) R n ( p, ) a i α x i p, x j a i β α i α x i j α There is a smooth mapping π : T M M such that for any coordinate neighborhood (U α, ϕ Uα ) we have a diffeomorphism F α : π 1 (U α ) = T U α U α R n. As a set, we have T M = p M T p M. Definition 3.6. A smooth vector field V on M is a smooth mapping V : M T M such that π V = id M. In particular for any p M, V (p) T p M. Moreover, if (U, ϕ U ) is a coordinate neighborhood on M. Then V = i ai (x) x with a i (x) C (ϕ i U (U)). If (V, ϕ V = (y 1,..., y m )) is another coordinate neighborhood, then over the intersection U V, we have: x i α. V = i a i (x) x i = i,j a i (x(y)) yj x i (x(y)) y j. Definition 3.7. If F : M N is a smooth mapping. X and Y are smooth vector fields on M and N. X and Y are F -related if F (X p ) = Y F (p) for any p M. If F : M M is a smooth mapping and F X = X then X is said to be F -invariant. Theorem 3.8. Let G be a Lie group and T e (G) be the tangent space at the identity. Then each X e T e G determines uniquely a C -vector field X on G which is invariant under left translations. In particular G is parallelizable. Proof. The vector field X defined by X g = (L g ) X e is left invariant: (L g1 ) X g2 = (L g1 ) (L g2 ) X e = (L g1g 2 ) X e = X g1g 2. We just need to show its smoothness. For any g 1 G, choose a coordinate neighborhood (U, ϕ U ) = (x 1,..., x n ) of g 1 G. Because the product operation is continuous, there exist coordinate neighborhood (V, ϕ V ) = (y 1,..., y n ) of g 1 and a neighborhood (W, ϕ W ) = (z 1,..., z n ) of e G such that V W U and the product mapping G G G is represented by a smooth function x = Φ(y, z). Assume X e = i ai z T i e G. Then X g = (L g ) X e = j a i xj (y, 0) zi x j T gg. X is a smooth vector field because the coefficient functions are smooth. Exercise Prove that SU(2) = {A GL(2, C); A A = I 2, det(a) = 1} is diffeomorphic to S 3 by showing that: {( ) } α β SU(2) = β ; α 2 + β 2 = 1. ᾱ 2. Prove that T e SU(2) = {B GL(2, C); B + B = 0, tr(b) = 0}. 3. Find the explicit expression of 3 vector fields which are left invariant and linearly independent every where on S 3. 24

25 3.3 One parameter and local one-parameter groups A one parameter group action is a smooth mapping θ : R M M satisfying the following properties: 1. θ 0 is the identity transformation; 2. θ t θ s = θ t+s. Equivalently a one parameter group is an action of R on M. In particular θ : R Diff(M) is a homomorphism. Definition 3.9. The infinitesimal generator of θ is the C vector field X defined as: X p = d dt θ t (p). (47) t=0 Example (Translation) M = R n, θ t (x) = (x i + a i t), X(x) = i ai x. i ( ) ( ) cos t sin t 2. (Rotation) M = R 2 x1, θ t (x) =. X sin t cos t x p (x) = x 2 x x 1 x. 2 Exercise 30. For any A = (a ij ) Mat n n (R), we have the exponential e ta : R GL(n, R) defined as: e ta = I + ta + (ta)2 2! + + (ta)k k! +. (48) Prove that θ t (x) = e ta x defines a one-parameter subgroup with the infinitesimal generator given by: X = a ij x j x i. i,j Proposition If θ : R M M is a C action of R on M, then the infinitesimal generator X is invariant under this action: θ t (X p ) = X θt(p) for all t R. Definition Given a vector field X on a manifold M, a curve t γ(t) defined on an open interval J of R is called an integral curve of X if dγ dt = X γ(t) on J. By definition, an integral curve is connected. Let M be a smooth manifold and W R M be an open set which satisfies: For every p M there exist real numbers α(p) < 0 < β(p) such that W (R {p}) = {(t, p) α(p) < t < β(p)}. Definition A local one-parameter group action or flow on a smooth manifold M is a C map θ : W M which satisfies the following two conditions: 1. θ 0 (p) = p for all p M; 2. If (s, p) W, then α(θ s (p)) = α(p) s, β(θ s (p)) = β(p) s, and moverover for any t such that α(p) s < t < β(p) s, θ t+s is defined and θ t θ s (p) = θ t+s (p). Example M = R 2 {0}, X = r = 1 r (x x + y y ). Then W = {(t, x); x < t < + } and θ : W M is given by (t, x) = x + t x x. The domain of definition of θ t is defined as V t = {p M; (t, p) W }. 25