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4 the Tangent Bundle 4 The Tangent Bundle 4.1 Tangent spaces n For embedded submanifolds M R , the tangent space TpM at p M can be defined ✓ 2 as the set of all velocity vectors v = g˙(0), where g : J M is a smooth curve with ! g(0)=p; here J R is an open interval around 0. ✓ n It turns out (not entirely obvious!) that TpM becomes a vector subspace of R . (Warn- ing: In pictures we tend to draw the tangent space as an affine subspace, where the origin has been moved to p.) Example 4.1. Consider the sphere Sn Rn+1, given as the set of x such that x 2 = 1. ✓ || || A curve g(t) lies in Sn if and only if g(t) = 1. Taking the derivative of the equation || || g(t) g(t)=1 at t = 0, we obtain (after dividing by 2, and using g(0)=p) · p g˙(0)=0. · n+1 3 That is, TpM consists of vectors v R that are orthogonal to p R 0 . It is not 2 2 \{ } hard to see that every such vector v is of the form g˙(0),⇤ hence that n TpS =(Rp)?, the hyperplane orthogonal to the line through p. Given v, take g(t)=(p +tv)/ p +tv . ⇤ || || 70 4 The Tangent Bundle 2 n n Exercise 45. What is T0R ? What is T0R ? How about TpR for some other p Rn? 2 To extend this idea to general manifolds, note that the vector v = g˙(0) defines a “directional derivative” C•(M) R: ! v : f d f (g(t)). 7! dt |t=0 Exercise 46. Fix some p Rn Show that the directional derivative in Rn 2 f (p +tv) f (p) v( f )=lim − t 0 t ! n n • n (for v TpR = R and f C (R )) has the following properties. For any 2 ⇠ • n 2 v TpM and f ,g C (R ): 2 2 a) v : C• R is well-defined (that is, the limit exists for any f C•). ! 2 b) v : C• R is linear: v(l f + µg)=lv( f )+µv(g). ! c) v : C• R is a derivation: v( fg)=v( f )g + fv(g). ! For a general manifold, we will define TpM as a set of directional derivatives. Definition 4.1 (Tangent spaces – first definition). Let M be a manifold, p M. The • 2 tangent space TpM is the set of all linear maps v : C (M) R of the form ! v( f )= d f (g(t)) dt |t=0 for some smooth curve g C•(J,M) with g(0)=p. 2 The elements v T M are called the tangent vectors to M at p. 2 p The following local coordinate description makes it clear that TpM is a linear subspace of the vector space L(C•(M),R) of linear maps C•(M) R, of dimension ! equal to the dimension of M. Theorem 4.1. Let (U,j) be a coordinate chart around p. A linear map v : C•(M) ! R is in TpM if and only if it has the form, m ∂( f j 1) v( f )= ai ◦ −  ∂ui u=j(p) i=1 for some a =(a1,...,am) Rm. 2 Proof. Given a linear map v of this form, let g˜ : R j(U) be a curve with g˜(t)= ! j(p)+ta for t sufficiently small. Let g = j 1 g˜. Then | | − ◦ d d 1 f (g(t)) = ( f j− )(j(p)+ta) dt t=0 dt t=0 ◦ m 1 ∂( f j− ) = ai ◦ ,  ∂ui u=j(p) i=1 4.1 Tangent spaces 71 by the chain rule. Conversely, given any curve g with g(0)=p, let g˜ = j g be the ◦ corresponding curve in j(U) (defined for small t ). Then g(0)=j(p), and | | d d 1 f (g(t)) = ( f j− )(g˜(te)) dt t=0 dt t=0 ◦ m 1 i ∂( f j− ) = Âa ◦ i u=g(p), i=1 ∂u | where a = dg˜ . dt t=0 ut We can use this result as an alternative definition of the tangent space, namely: Definition 4.2 (Tangent spaces – second definition). Let (U,j) be a chart around • p. The tangent space TpM is the set of all linear maps v : C (M) R of the form ! m ∂( f j 1) v( f )= ai ◦ − (4.1)  ∂ui u=j(p) i=1 for some a =(a1,...,am) Rm. 2 Remark 4.1. From this version of the definition, it is immediate that TpM is an m- dimensional vector space. It is not immediately obvious from this second definition that TpM is independent of the choice of coordinate chart, but this follows from the equivalence with the first definition. Alternatively, one may check directly that the subspace of L(C•(M),R) characterized by (4.1) does not depend on the chart, by studying the effect of a change of coordinates. According to (4.1), any choice of coordinate chart (U,j) around p defines a vector m 1 m space isomorphism TpM = R , taking v to a =(a ,...,a ). In particular, we see that m ⇠ if U R is an open subset, and p U, then TpU is the subspace of the space of ✓ • 2 linear maps C (M) R spanned by the partial derivatives at p. That is, TpU has a ! basis ∂ ∂ ,..., ∂x1 |p ∂xm |p m identifying TpU R . Given ⌘ ∂ v = ai  ∂xi |p the coefficients ai are obtained by applying v to the coordinate functions x1,...,xm : U R, that is, ai = v(xi). ! We now describe yet another approach to tangent spaces which again charac- terizes “directional derivatives” in a coordinate-free way, but without reference to curves g. Note first that every tangent vector satisfies the product rule, also called the Leibniz rule: Lemma 4.1. Let v T M be a tangent vector at p M. Then 2 p 2 v( fg)= f (p)v(g)+v( f )g(p) (4.2) for all f ,g C•(M). 2 72 4 The Tangent Bundle Proof. Letting v be represented by a curve g, this follows from d d d f g(t) g g(t) = f (p) g g(t) + f g(t) g(p). dt t=0 dt t=0 dt t=0 ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ ut Alternatively, in local coordinates it is just the product rule for partial derivatives. It turns out that the product rule completely characterizes tangent vectors: Exercise 47. Suppose that v : C•(M) R is a linear map satisfying the product ! rule (4.2). Prove the following two facts, which will be used in the proof of Theorem 4.2 below: a) v vanishes on constants. That is, if f C•(M) is the constant map, then 2 v( f )=0. b) Suppose f ,g C•(M) with f (p)=g(p)=0. Then v( fg)=0. 2 • Theorem 4.2. A linear map v : C (M) R defines an element of TpM if and only if ! it satisfies the product rule (4.2). The proof of this result will require the following fact from multivariable calcu- lus: m Lemma 4.2 (Hadamard Lemma). Let U = BR(0) R be an open ball of radius ✓ R > 0 and h C•(U) a smooth function. Then there exist smooth functions h 2 i 2 C•(U) with m i h(u)=h(0)+ u hi(u) i=1 for all u U. Here h (0)= ∂h (0). 2 i ∂ui Proof. Let h be the functions defined for u =(u1,...,um) U by i 2 1 h(u1,...,ui,0,...,0) h(u1,...,ui 1,0,0,...,0) if ui = 0 ui − − 6 h (u)= i 8 ∂h ( 1,..., i 1, , ,..., ) i = < ∂ui u u − 0 0 0 if u 0 Using Taylor’s: formula with remainder, one sees that these functions are smooth†. i m i If all u = 0, then the sum  u hi(u) is a telescoping sum, equal to h(u) h(0). By 6 i=1 − continuity, this result extends to all u. Finally, evaluating the derivative m ∂h ∂ i k ∂hk i = i h(0)+ u hi(u) = hi(u)+Âu i ∂u ∂u i=1 ! k ∂u at u = 0, we see that ∂h = h (0). ∂ui u=0 i ut † It is a well-known fact from calculus (proved e.g. by using Taylor’s theorem with remain- der) that if f is a smooth function of a real variable x, then the function g, defined as g(x)=x 1( f (x) f (0)) for x = 0 and g(0)= f (0), is smooth. − − 6 0 4.1 Tangent spaces 73 Proof (Theorem 4.2). Let v : C•(M) R be a linear map satisfying the product rule ! (4.2). Step 1: If f1 = f2 on some open neighborhood U of p, then v( f1)=v( f2). Equivalently, letting f = f1 f2, we show that v( f )=0 if f = 0 on U. Choose • − a ‘bump function’ c C (M) with c(p)=1, with c M U = 0. Then f c = 0. The 2 | \ product rule tells us that 0 = v( f c)=v( f )c(p)+v(c) f (p)=v( f ). Step 2: Let (U,j) be a chart around p, with image U = j(U). Then there is unique linear map v : C•(U) R such that v( f )=v( f ) whenever f agrees with ! f j 1 on some neighborhood of p. e ◦ − Given f , we cane alwayse find a function fesuche that f agrees withe f j 1 on ◦ − some neighborhood of p.
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