Randell 2.1 Let p ∈ M. Suppose g_{i} : U_{i} → N, where p ⊂ U_{i}^{open} ⊂ M. g_{1} ∼ g_{2} if ∃V such that p ∈ V ⊂ U_{1} ∩U_{2} and g_{1}(x) = g_{2}(x) ∀x ∈ V. The equivalence class [g] is a germ. G(p, N) = {[g] | g^{smooth} : U → N, for some U^{open} such that p ∈ U ⊂ M} G(p) = G(p, R) G(p) is an algebra over R. Let α : I → M where I = an interval ⊂ R, α(0) = p. Note [α] ∈ G[0, M] Directional derivative of [g] in direction [α] =

D_{α}g = \frac{d(g ∘ α)}{dt}|_{t=0} ∈ R

Note D_{α} : G(p) → R is linear and satisfies the Leibniz rule.

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 T_{p}(M) = {v : G(p) → R | v is linear and satisfies the Leibniz rule } If v ∈ T_{p}(M) is called a derivation D_{α} ∈ T_{p}(M) T_{p}(M) is closed under addition and scalar multiplication and hence is a vector space over R Find a basis for T_{p}(M): Let (U, ϕ) be a chart at p such that ϕ : U^{open} → ϕ(U)^{open} ⊂ R^{m} is a homeomorphism, ϕ(p) = 0 ∈ R^{m} 0 ∈ ϕ(U)^{open} implies ∃ϵ > 0 such that B_{ϵ}(0) ⊂ ϕ(U) Thus if t ∈ (−ϵ, ϵ), then (0, ..., t, ..., 0) ⊂ ϕ(U). Define α_{i} : (−ϵ, ϵ) → M, α_{i}(t) = ϕ^{−1}(0, ..., t, ..., 0) Let v_{i} = D_{α_{i}}. Claim: {v_{1}, ..., v_{m}} is a basis for T_{p}(M).

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},