Chapter 4 Tangent Space • The set of all tangent vectors (to all curves) at some point P 2 M is the tangent space TP M at P: 2 Proposition: TP M is a vector space with the same dimension- ality n as the manifold M. Proof: We need to show that TP M is a vector space, i.e. XP + YP 2 TP M ; (4.1) aXP 2 TP M ; (4.2) 8XP ;YP 2 TP M; a 2 R: That is, given curves γ; µ passing through P such that XP = γ_ P ;YP =µ _ P ; we need a curve λ passing through P such that _ 1 λP (f) = XP (f) + YP (f)8f 2 C (M): n Define λ : I ! R in some chart ' around P by λ = ' ◦ γ + ' ◦ n µ − '(P ): Then λ is a curve in R ; and λ = '−1 ◦ λ : I ! M (4.3) is a curve with the desired property. 2 Note: we cannot define λ = γ + µ − P because addition does not make sense on the right hand side. The proof of the other part works similarly. (Exercise!) To see that TP M has n basis vectors, we consider a chart ' with i coordinates x : Then take n curves λk such that 1 k n ' ◦ λk(t) = x (P ); ··· ; x (P ) + t; ··· ; x (P ) ; (4.4) 17 18 CHAPTER 4. TANGENT SPACE i.e., only the k-th coordinate varies along t: So λk is like the axis of the k-th coordinate (but only in some open neighbourhood of P ). @ Now denote the tangent vector to λk at P by k ; i.e., @x P @ d f = λ_ (f) = (f ◦ λ ) : (4.5) k k k @x P P dt P This notation makes sense when we remember Eq. (3.9). Using it we can write @f λ_ (f) = 8f 2 C1(M): (4.6) k k P @x P @ Note that k is notation. We should understand this as @x P @ @ −1 @f k f = k f ◦ ' ≡ k (4.7) @x P @x '(P ) @x '(P ) @ in a chart around P . The are defined only when this chart @xk P n is given, but these are vectors on the manifold at P , not on R . _ Let us now show that the tangent space at P has λkjP as a basis. Take any vector vP 2 TP M ; which is the tangent vector to some curve γ at P . (We may sometimes refer to P as γ(0) or as t = 0.) Then d vP (f) = (f ◦ γ) (4.8) dt t=0 d −1 = ((f ◦ ' ) ◦ (' ◦ γ)) : (4.9) dt '(P ) t=0 n 1 n Note that ' ◦ γ : I ! R ; t 7! (x (γ(t)); ··· ; x (γ(t))) are the coor- dinates of the curve γ; so we can use the chain rule of differentiation to write @ −1 d i vP (f) = i (f ◦ ' ) (x ◦ γ) (4.10) @x '(P ) dt t=0 @ −1 i = i (f ◦ ' ) vP (x ) : (4.11) @x '(P ) 19 The first factor is exactly as shown in Eq. (4.7), so we can write @ i 1 vP (f) = k fvP (x ) 8f 2 C (M) (4.12) @x P i.e., we can write @ v = vi 8v 2 T M (4.13) P P k P P @x P where vi = v (xi). Thus the vectors @ span T M : These are P P @xk P P to be thought of as tangents to the coordinate curves in ': These can be shown to be linearly independent as well, so @ form a @xk P basis of T M and vi are the components of v in that basis. P P P @ The k are called coordinate basis vectors and the set n o@x P @ is called the coordinate basis. @xk P It can be shown quite easily that for any smooth (actually C1) function f a vector vP defines a derivation f 7! vP (f) , i.e., satisfies linearity and Leibniz rule, vP (f + αg) = vP (f) + αvP (g) (4.14) vP (fg) = vP (f)g(P ) + f(P )vP (g) (4.15) 1 8f; g 2 C (M) and α 2 R (4.16).