Chapter 4

Tangent Space

• The set of all tangent vectors (to all curves) at some point P ∈ 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 ∈ TP M , (4.1)

aXP ∈ TP M , (4.2) ∀XP ,YP ∈ TP M, a ∈ 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 ˙ ∞ λP (f) = XP (f) + YP (f)∀f ∈ C (M). n Define λ : I → R in some chart ϕ around P by λ = ϕ ◦ γ + ϕ ◦ n µ − ϕ(P ). Then λ is a curve in R , and

λ = ϕ−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 ϕ with i coordinates x . Then take n curves λk such that

 1 k n  ϕ ◦ 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) = ∀f ∈ C∞(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 ◦ ϕ ≡ k (4.7) ∂x P ∂x ϕ(P ) ∂x ϕ(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 λk|P as a basis.

Take any vector vP ∈ 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 ◦ ϕ ) ◦ (ϕ ◦ γ)) . (4.9) dt ϕ(P ) t=0

n 1 n Note that ϕ ◦ 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 ◦ ϕ ) (x ◦ γ) (4.10) ∂x ϕ(P ) dt t=0

∂ −1 i = i (f ◦ ϕ ) vP (x ) . (4.11) ∂x ϕ(P ) 19

The first factor is exactly as shown in Eq. (4.7), so we can write   ∂ i ∞ vP (f) = k fvP (x ) ∀f ∈ C (M) (4.12) ∂x P i.e., we can write  ∂  v = vi ∀v ∈ 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 ϕ. 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 ∀f, g ∈ C (M) and α ∈ R (4.16)