Boothby 2.4 T_{a}(R^{n}) = {(a, x) | x ∈ R^{n}}, ϕ(ax) = x − a canonical basis {E_{ia} = ϕ^{−1}(e_{i}) | i = 1, ..., n} Let a ∈ R^{n} Suppose f : X ⊂ R^{n} → R where a ⊂ X^{open} ⊂ M. f ∼ g if ∃U^{open} s.t. a ∈ U and f(x) = g(x)∀x ∈ U. Let C^{∞}(a) = {f : X ⊂ R^{n} → R ∈ C^{∞} | a ∈ domf} f_{i} : U_{i} → R ∈ C^{∞}(a) implies f_{1} + f_{2} : U_{1} ∩ U_{2} → R ∈ C^{∞}(a), αf_{i} : U_{i} → R ∈ C^{∞}(a), and f_{1}f_{2} : U_{1} ∩ U_{2} → R ∈ C^{∞}(a), Thus C^{∞}(a) is an algebra over R Let X_{a} = Σ_{i=1}^{n}ξ_{i}E_{ia} X_{a}^{∗} : C^{∞}(a) → R X_{a}^{∗}(f) = Σ_{i=1}^{n}ξ_{i} \frac{∂f}{∂x_{i}} |_{a} = directional derivative of f at a in the direction of X_{a}. Let x_{j} : R^{n} → R, x_{j}(x) = x_{j} X_{a}^{∗}(x_{j}) = Σ_{i=1}^{n}ξ_{i} \frac{∂x_{j}}{∂x_{i}} |_{a} = ξ_{i} X_{a}^{∗} is linear and satisfies the Leibniz rule.

Let D(a) = {D : C^{∞}(a) → R | D is linear and satisfies the Leibniz rule } D ∈ D(a) is called a derivation X_{a}^{∗} ∈ D(a) D(a) is closed under addition and scalar multiplication and hence is a vector space over R Let j : T_{a}(R^{n}) → D(a), j(X_{a}) = X_{a}^{∗} Claim: j is an isomorphism. Let X_{a} = Σ_{i=1}^{n}ξ_{i}E_{ia} and Z_{a} = Σ_{i=1}^{n}ζ_{i}E_{ia} j is a homomorphism. j is 1-1: If j(X_{a}) = j(Z_{a}), then X_{a}^{∗}(x_{j}) = Σ_{i=1}^{n}ξ_{i} \frac{∂x_{j}}{∂x_{i}} |_{a} = ξ_{i} = ζ_{i} = Z_{a}^{∗}(x_{j}) j is onto: Let D be a derivation. Suppose f(x) = 1. Then Df = 0 by product rule. Suppose g(x) = c. Then Dg = D(cf) = cDf = 0 Let h_{i}(x) = x_{i}. Let ξ_{i} = Dh_{i}. Then D = X_{a}^{∗} where X_{a} = Σ_{i=1}^{n}ξ_{i}E_{ia} (proof: long calculation, see Boothby). Note since X_{a}^{∗}(f) = Σ_{i=1}^{n}ξ_{i} \frac{∂f}{∂x_{i}} |_{a}, j(E_{ia}) = E_{ia}∗ = \frac{∂}{∂x_{i}} |_{a},