C^{∞}(M) = {g | g^{smooth} : M → R} D is a derivation iff D : C^{∞}(p) → R and D is linear and satisfies the Leibniz rule. That is D is a derivation if D(f) ∈ R, D(cf) = cD(f), D(f + g) = D(f) + D(g), D(fg) = f(p)Dg + g(p)Df Defn: A vector field or section of the tangent bundle T M is a smooth function s: M → T M so that π ∘ s = id [i.e., s(p) = (p, v_{p})]. Ex: If M = R, let s(p) = (p, (\frac{d}{dx})_{p}) Sometimes we will drop the p and write s(p) = (\frac{d}{dx})_{p} Let f ∈ C^{∞}(R). For all p ∈ R, s(p)(f) = (\frac{df}{dx})_{p} = \frac{df}{dx}(p) Define s_{f} : R → R, s_{f}(p) = \frac{df}{dx}(p). I.e., s_{f} = \frac{df}{dx} Note s_{f} is smooth. Lemma 3.4.1: For any vector field s and smooth functions f and g on M, we have

s_{fg}(p) = f(p) · s_{g}(p) + s_{f}(p) · g(p)

Proof: \frac{d(fg)}{dx}(p) = f(p)\frac{dg}{dx}(p) + \frac{df}{dx}(p)g(p) We can think of a vector field as a function S : C^{∞}(M) → C^{∞}(M), S(f) = s_{f} Ex: S : C^{∞}(R) → C^{∞}(R), S(f) = \frac{df}{dx}. I.e., S = \frac{d}{dx} Ex: If M = R, then s(p) = a(p)(\frac{d}{dx})_{p} where a : R → R is a smooth function.

Let f ∈ C^{∞}(R). For all p ∈ R, s(p)(f) = a(p)(\frac{df}{dx})_{p} = a(p)\frac{df}{dx}(p)

Define s_{f} : R → R, s_{f}(p) = a(p)\frac{df}{dx}(p). I.e., s_{f} = a\frac{df}{dx}

Note s_{f} is smooth.

Lemma 3.4.1: For any vector field s and smooth functions f and g on M, we have

s_{fg}(p) = f(p) · s_{g}(p) + s_{f}(p) · g(p)

Proof: a(p)\frac{d(fg)}{dx}(p) = a(p)f(p)\frac{dg}{dx}(p) + a(p)\frac{df}{dx}(p)g(p)

We can think of a vector field as a function S : C^{∞}(M) → C^{∞}(M), S(f) = s_{f}

Ex: S : C^{∞}(R) → C^{∞}(R), S(f) = a\frac{df}{dx} I.e., S = a\frac{d}{dx} In the above we used the charts ϕ_{p} : R → R, ϕ_{p}(x) = x − p. Thus \frac{d(g(ϕ_{p}^{−1}(x)))}{dx}|_{x=0} = \frac{d(g(x+p))}{dx}|_{x=0} = \frac{dg}{dx}(p) Note ϕ_{0}(x) = ϕ_{p}(x + p). Thus \frac{d(ϕ_{p}(ϕ_{0}^{−1}(x))}{dx}|_{x=0} = \frac{d(ϕ_{p}(ϕ_{p}^{−1}(x+p)))}{dx}|_{x=0} = \frac{d(x+p)}{dx}|_{x=0} = 1

If we use the chart ψ_{q} : R → R, ψ_{q}(x) = q − x. Then \frac{d(g(ψ_{p}^{−1}(x)))}{dx}|_{x=0} = \frac{d(g(p−x))}{dx}|_{x=0} = \frac{−dg}{dx} (p) Note \frac{d(ψ_{q}(x+p))}{dx}|_{x=0} = \frac{dψ_{q}}{dx} |_{p} = \frac{d(q−x)}{dx}|_{p} = −1 Example of a non-smooth vector field on R: If p ≥ 0, let s(p) = (p, (\frac{d}{dx})_{p}) [i.e., the basis element of T_{p}(R) from ϕ_{p}] If p < 0, let s(p) = (p, (−\frac{d}{dx})_{p}) [i.e., the basis element of T_{p}(R) from ψ_{p}] Ex: If M = R^{2}, then s(p) = a(p)(\frac{∂}{∂x})_{p} + b(p)(\frac{∂}{∂y})_{p} where a, b : R^{2} → R are smooth functions.

Ex: Let \{(\frac{∂}{∂x_{1}} )_{p}, ..., (\frac{∂}{∂x_{m}})_{p}\} be a basis for T_{p}(M).

Let s : M → T M, s(p) = (p, Σ_{i=1}^{m}a_{i}(p)(\frac{∂}{∂x_{i}})_{p})