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{∂}{∂x})_{p}) Sometimes we will drop the p and write s(p) = (\frac{∂}{∂x})_{p} Let f ∈ C^{∞}(R). For all p ∈ R, s(p)(f) = (\frac{∂f}{∂x})_{p} = \frac{∂f}{∂x}(p) Define s_{f} : R → R, s_{f}(p) = \frac{∂f}{∂x}(p). I.e., s_{f} = \frac{∂f}{∂x} 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{∂(fg)}{∂x}(p) = f(p)\frac{∂g}{∂x}(p) + \frac{∂f}{∂x}(p)g(p) Thus 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{∂f}{∂x}. I.e., S = \frac{∂}{∂x} Ex: If M = R, then s(p) = a(p)(\frac{∂}{∂x})_{p} where a : R → R is a smooth function. Let f ∈ C^{∞}(R). For all p ∈ R, s(p)(f) = a(p)(\frac{∂f}{∂x})_{p} = a(p)\frac{∂f}{∂x}(p) Define s_{f} : R → R, s_{f}(p) = a(p)\frac{∂f}{∂x}(p). I.e., s_{f} = a\frac{∂f}{∂x} 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{a(p)∂(fg)}{∂x}(p) = a(p)f(p)\frac{∂g}{∂x}(p) + a(p)\frac{∂f}{∂x}(p)g(p) Thus 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{∂f}{∂x} I.e., S = a\frac{∂}{∂x} We can also write this as: 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.