There are several equivalent definitions of the tangent space TpM at p∈M:
Definition (Directional Derivatives). A tangent vector at p is a derivation at p: a linear map v:C∞(M)→R satisfying the Leibniz rule:
v(fg)=f(p)⋅v(g)+v(f)⋅g(p)
Definition (Equivalence Classes of Curves). A tangent vector is an equivalence class of smooth curves γ:(−ε,ε)→M with γ(0)=p, where γ1∼γ2 if (φ∘γ1)′(0)=(φ∘γ2)′(0) in some (hence every) chart.
Proposition 2.1.TpM is a vector space of dimension n=dimM.
The dual space Tp∗M=(TpM)∗ is called the cotangent space at p. Its elements are covectors (linear functionals on TpM). The basis dual to {∂/∂xi} is denoted {dxi∣p}, where:
dxi(∂xj∂)=δji
Proposition 2.3. For a smooth function f:M→R, the differential dfp∈Tp∗M is given in coordinates by:
The tangent bundle TM is a special case of a vector bundle: a smooth manifold E with a surjective submersion π:E→M such that each fiber π−1(p) is a vector space, and local trivializations exist.
Definition. A vector bundle of rank k over M is a smooth manifold E with a smooth map π:E→M such that for every p∈M there exists a neighborhood U and a diffeomorphism Φ:π−1(U)→U×Rk with π=pr1∘Φ and each fiber π−1(p) maps linearly to {p}×Rk.
Example 2.4. The tangent bundle TM is a rank n vector bundle over M. The cotangent bundle T∗M is also a rank n vector bundle, dual to TM.
Example 2.5. The trivial bundle M×Rk is a rank k vector bundle. A manifold M is parallelizable if TM≅M×Rn. For example, S1 is parallelizable but S2 is not (by the hairy ball theorem).
Problem 1. Let M=R2 with coordinates (x,y) and let f:R2→R be f(x,y)=x2+y2. Compute df(1,0) in coordinates.
Solution.∂f/∂x=2x, ∂f/∂y=2y. At (1,0), df(1,0)=2xdx+2ydy∣(1,0)=2dx. So df(1,0)(v)=2v1. ■
Problem 2. Show that TpS2 is isomorphic to {v∈R3:p⋅v=0}.
Solution. Consider the embedding S2⊆R3. A curve γ(t) on S2 satisfies γ(t)⋅γ(t)=1. Differentiating: γ′(0)⋅p=0, so every tangent vector is orthogonal to p. Conversely, any v⊥p is tangent to the great circle in the p-v-plane. Thus TpS2≅p⊥. ■