G(p, N) = {[g] | g^{smooth} : U → N, for some U^{open} such that p ∈ U ⊂ M} G(p) = G(p, R) C^{∞}(M) = {g | g^{smooth} : M → R} C^{∞}(p) = {g | g^{smooth} : U → N, for some U^{open} such that p ∈ U ⊂ M} 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 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

D_{α} : G(p) → R is a derivation. T_{p}(M) = {v : G(p) → R | v is linear and satisfies the Leibniz rule } v ∈ T_{p}(M) is called a derivation Given a chart (U, ϕ) at p where ϕ(p) = 0, the standard basis for T_{p}(M) = \{(\frac{∂}{∂x_{1}} )_{p}, ..., (\frac{∂}{∂x_{m}})_{p}\}, where (\frac{∂}{∂x_{i}})_{p} = D_{α_{i}} and for some ϵ > 0, α_{i} : (−ϵ, ϵ) → M, α_{i}(t) = ϕ^{−1}(0, ..., t, ..., 0) If v ∈ T_{p}(M), then v = Σ_{i=1}^{m}a_{i}(\frac{∂}{∂x_{1}} )_{p} where a_{i} = v([π_{i} ∘ ϕ])

T M = ∪_{p∈M}T_{p}(M) = {(p, v) | p ∈ M, v ∈ T_{p}M}, let π: T M → M be defined by π(p, v) = p. Let (ϕ, U) be a chart for M. If q ∈ U, let \{(\frac{∂}{∂x_{1}} )_{q}, ..., (\frac{∂}{∂x_{m}})_{q}\} be the standard basis (w.r.t (ϕ, U)) for T_{q}(M) = T_{q} t_{ϕ}: π^{−1}(U) → ϕ(U) × R^{m} ⊂ R^{2m}, t_{ϕ}(q, v) = (ϕ(q), a_{1}, ..., a_{m}) where v = Σ_{i=1}^{m}a_{i}(\frac{∂}{∂x_{i}})_{q} Let A be a maximal atlas for M. Basis for topology on T M : {W | ∃(ϕ, U) ∈ A s.t. W ⊂ π^{−1}(U) and t_{ϕ}(W) open in R^{2m}} Claim: T M is a 2m−manifold and C = {(t_{ϕ}, π^{−1}(U)) | (ϕ, U) ∈ A} is a pre-atlas for T M. π: T M → M, π(p, v) = p is smooth df : T M → T N defined by df(p, v) = (f(p), d_{p}f(v)) is smooth if f : M → N is smooth. Proof: See Hitchin 4.1 (in Chapter 1 of http://www2.maths.ox.ac.uk/ hitchin/hitchinnotes/hitchinnotes.html 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. 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. 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(ψ) = a(ψ)(\frac{∂}{∂x})_{ψ} + b(ψ)(\frac{∂}{∂y})_{ψ} 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})

Defn: s is never zero if s(p) ≠ (p, 0) for all p ∈ M. Prop: Let G be a Lie group. Then G admits a never-zero vector field. Note: S^{n} admits a never-zero vector field iff n odd.

Let p_{2}(s(p)) = p_{2}(p, v_{p}) = v_{p} Defn: The vector fields s_{1}, ..., s_{k} are linearly independent iff for all p ∈ M, p_{2}(s_{1}(p)), ..., p_{2}(s_{k}(p)) are linearly independent. Defn: M is parallelizable (or equivalently the “tangent bundle π: T M → M is trivial”) iff T M admits m linearly independent vector fields. Suppose M is parallelizable. Thus for each p ∈ M, let {v_{1,p}, ..., v_{m,p}} be ANY basis for T_{p}(M) such that s_{i} : M → T M, s_{i}(p) = (p, v_{i,p}) is a SMOOTH vector field. NOTE: We can form m vector fields using basis elements iff M is parallelizable. When M is parallelizable, we can define: t : T M → M × R^{m}, t(p, v) = (p, a_{1}, ..., a_{m}) where v = Σ_{i=1}^{m}a_{i}v_{i,p} Let ρ_{1} : M × R^{m} → M, ρ_{1}(p, x) = p. ρ_{2} : M × R^{m} → R^{m}, ρ_{1}(p, x) = x. ρ_{1} ∘ t : T M → M, (ρ_{1} ∘ t)(p, v) = π(p, v) = p Recall π^{−1}(p) = T_{p}(M) Prop: t|_{T_{p}(M)} : T_{p}(M) → {p} × R^{m} is a linear isomorphism for all p. or equivalently, ρ_{2} ∘ t|_{T_{p}(M)} : T_{p}(M) → R^{m} is a linear isomorphism for all p. since ρ_{2} ∘ t|_{T_{p}(M)}(Σ_{i=1}^{m}a_{i}v_{i,p}) = (a_{1}, ..., a_{m}) HENCE: t : T M → M × R^{m} is a diffeomorhism. Note that for all p ∈ M, given a basis {v_{1,p}, ..., v_{m,p}} be a basis for T_{p}(M), we can always define a linear isomorphism:

t_{p} : T_{p}(M) → R^{m}, T (Σ_{i=1}^{m}a_{i}v_{i,p}) = (a_{1}, ..., a_{m})

However, t : T M → M ×R^{m}, t(p, v) = (p, t_{p}(v)) may not be smooth (recall example of non-smooth vector field on p. 4). In general T M may not be diffeomorphic to M × R^{m}.

Randell 3.4 The bracket of two vector fields. 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})]. I.e, s takes p ∈ M to the derivation v_{p} : C^{∞}(M) → R Let f ∈ C^{∞}(M) Define s_{f} : M → R, s_{f}(p) = v_{p}([f]) where s(p) = (p, v_{p}) Note s_{f} is smooth. Thus we can think of a vector field as a function S : C^{∞}(M) → C^{∞}(M), S(f) = s_{f}

Lemma 3.4.2: Let S : C^{∞}(M) → C^{∞}(M) be linear, and suppose S(fg)(p) = f(p) · S(g)(p) + S(f)(p) · g(p). Then S is a vector field. Proof: Define s : M → T M, s(p) = (p, S_{p}) where Define S_{p} : C^{∞}(M) → R, S_{p}(f) = S(f)(p), i.e, the function S(f) evaluated at p. Claim S_{p} is a derivation. Show S_{p} is linear and satisfies the Leibniz rule. Claim s is smooth. Defn: If A, B are vector fields, let AB = A ∘ B Defn: The Lie Bracket of vector fields A and B is [A, B] = AB − BA : C^{∞}(M) → C^{∞}(M). Thm: The Lie bracket of vector fields is a vector field.

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 D_{α} : G(p) → R is a derivation. Given a chart (U, ϕ) at p where ϕ(p) = 0, the standard basis for T_{p}(M) = \{(\frac{∂}{∂x_{1}} )_{p}, ..., (\frac{∂}{∂x_{m}})_{p}\}, where (\frac{∂}{∂x_{i}})_{p} = D_{α_{i}} and for some ϵ > 0, α_{i} : (−ϵ, ϵ) → M, α_{i}(t) = ϕ^{−1}(0, ..., t, ..., 0) If v ∈ T_{p}(M), then v = Σ_{i=1}^{m}a_{i}(\frac{∂}{∂x_{1}} )_{p} where a_{i} = v([π_{i} ∘ ϕ]) Let (U, ϕ) be a chart for M such that 0 ∈ ϕ(U). Suppose q ∈ U. Choose ϵ > 0 such that B(ϕ(q), ϵ) ⊂ ϕ(U) and B(0, ϵ) ⊂ ϕ(U). Let τ_{q} : B(ϕ(q), ϵ) → B(0, ϵ), τ_{q}(x) = x − ϕ(q). the standard basis for T_{q}(M) with respect to (U, ϕ) =

the standard basis for T_{q}(M) with respect to (ϕ^{−1}(B(0, ϵ)), τ_{q} ∘ ϕ)

Hence the standard basis (w.r.t. (U, ϕ)) = \{(\frac{∂}{∂x_{1}} )_{q}, ..., (\frac{∂}{∂x_{m}})_{q}\}, where (\frac{∂}{∂x_{i}})_{q} = D_{α_{i}} α_{i} : (−ϵ, ϵ) → M, α_{i}(t) = ϕ^{−1}(τ^{−1}(0, ..., t, ..., 0))

= ϕ^{−1}(ϕ_{1}(q), ..., ϕ_{i−1}(q), ϕ_{i}(q) + t, ϕ_{i+1}(q), ..., ϕ_{m}(q)) where ϕ_{i} = π_{i} ∘ ϕ. Suppose f : M → R is smooth. Recall f is smooth iff for all p ∈ M, there exists a chart (U, ϕ) such that p ∈ U and f ∘ ϕ^{−1} : ϕ(U) ⊂ R^{m} → R is smooth. Claim: \frac{∂f}{∂x_{i}} : U → R, \frac{∂f}{∂x_{i}} (q) = (\frac{∂}{∂x_{i}})_{q}(f) is smooth. NOTE: \frac{∂f}{∂x_{i}} : U → R is only defined on U, and is NOT a globally defined function on M. We will show that \frac{∂f}{∂x_{i}} ∘ ϕ^{−1} : ϕ(U) ⊂ R^{m} → R is smooth. Let q = ϕ^{−1}(x) \frac{∂f}{∂x_{i}} ∘ ϕ^{−1}(x) = \frac{∂f}{∂x_{i}} (q) = (\frac{∂}{∂x_{i}})_{q}(f) = D_{α_{i}}(f) = \frac{d(f∘α_{i})}{dt}|_{t=0} = \frac{d((f∘ϕ^{−1})(ϕ_{1}(q),...,ϕ_{i−1}(q),ϕ_{i}(q)+t,ϕ_{i+1}(q),...,ϕ_{m}(q)))}{dt}|_{t=0} = \frac{∂(f∘ϕ^{−1})}{∂x_{i}}|_{ϕ(q)} = \frac{∂(f∘ϕ^{−1})}{∂x_{i}}|_{x} Thus \frac{∂f}{∂x_{i}} ∘ ϕ^{−1} is smooth since f ∘ ϕ^{−1} is smooth.