> [!quote]- Introduction > > Let $X$ be a [[Manifold|manifold]] of class $C^p$, and $\xi, \eta \in \vf(X)$ be [[Vector Field|vector fields]] of class $C^{p - 1}$. As vector fields act on functions as [[Vector Fields as Derivations|derivations]], for any $f \in C^{p - 1}(X)$, $\xi f, \eta \xi f \in C^{p - 1}(X)$. However, adding $g \in C^{p - 1}(X)$ into the equation and after some computations, > $ > \eta \xi (fg) = \eta f \cdot \xi g + f \cdot \eta \xi g + \xi g \cdot \eta f + g \cdot \eta \xi f > $ > the map $f \mapsto \eta \xi f$ does not satisfy the [[Product Rule|product rule]] and cannot be a [[Derivation|derivation]]. > > Let $(U, \varphi)$ be a [[Atlas|chart]] at $p$, and let $\hat \xi$ be the following coordinate representation > $ > \begin{CD} > \widehat U @>{\varphi^{-1}}>> U @>{\xi}>> TU @>{\tilde \varphi}>> \widehat{U} \times E @>{\proj_2}>> E > \end{CD} > $ > Let $\hat f = f \circ \varphi^{-1}$ be its coordinate representation with respect to $(U, \varphi)$ and $(\real, \text{Id})$. Then $\xi f$ is evaluated as > $ > \xi f(p) = \xi(p) \cdot f = D(\hat f)(\hat p)(\hat \xi (\hat p)) > $ > and expanding using the product rule > $ > \begin{align*} > D\braks{D(\hat f)\cdot \hat \xi}(\hat p)(h) &= D^2(\hat f)(\hat p)(\hat \xi(p), h) + > D(\hat f)(\hat p) \cdot D(\hat \xi)(\hat p)(h) > \end{align*} > $ > Obtaining a coordinate representation for $\eta$ in the same way, > $ > \begin{align*} > (\eta \xi f)(p) &= D(\widehat{\xi f})(\hat p)(\hat \eta (\hat p)) \\ > &= D^2(\hat f)(\hat p)(\hat \xi(p), \hat \eta(\hat p)) + > D(\hat f)(\hat p) \cdot D(\hat \xi)(\hat p)(\hat \eta(\hat p)) > \end{align*} > $ > Evaluating the composition on $f$ yields an expression involving a second derivative, meaning that $\eta \xi f$ cannot necessarily be written as the action of a vector field on $f$. However, there is a way of composing vector fields that does give us a new vector field. > [!definition] > > Let $\xi, \eta \in \vf(X)$ be vector fields of class $C^{p - 1}$. Then there exists a unique vector field $[\xi, \eta]$ of class $C^{p - 2}$ such that for any $f \in C^{p - 1}(X)$, > $ > [\xi, \eta]f = \xi\eta f - \eta \xi f > $ > known as the **bracket product** of $\xi$ and $\eta$. > > *Proof*. Let $(U, \varphi)$ be a chart at $p$, then we have local representations > $ > \begin{align*} > \xi \eta f(p) &= D^2(\hat f)(\hat p)(\hat \eta(p), \hat \xi(\hat p)) + > D(\hat f)(\hat p) \cdot D(\hat \eta)(\hat p)(\hat \xi(\hat p))\\ > \eta \xi f(p) &= > D^2(\hat f)(\hat p)(\hat \xi(p), \hat \eta(\hat p)) + > D(\hat f)(\hat p) \cdot D(\hat \xi)(\hat p)(\hat \eta(\hat p)) > \end{align*} > $ > As the [[Higher Derivatives|second derivative]] is symmetric, subtracting the two yields > $ > \xi\eta f(p) - \eta \xi f(p) = D(\hat f)(\hat p) \cdot D(\hat \eta)(\hat p)(\hat \xi(\hat p)) - D(\hat f)(\hat p) \cdot D(\hat \xi)(\hat p)(\hat \eta(\hat p)) > $ > Since the derivative is linear, we can combine the terms to obtain a local formula > $ > \begin{align*} > \xi\eta f(p) - \eta \xi f(p) &= D(\hat f)(\hat p) \cdot \braks{D(\hat \eta)(\hat p)(\hat \xi(\hat p)) - D(\hat \xi)(\hat p)(\hat \eta(\hat p))} \\ > \widehat{[\xi, \eta]}(\hat p) &= D(\hat \eta)(\hat p)(\hat \xi(\hat p)) - D(\hat \xi)(\hat p)(\hat \eta(\hat p)) > \end{align*} > $ > and $[\xi, \eta]$ is the unique vector field with the above local representation. > [!theorem] > > Let $\xi, \eta, \zeta \in \vf(X)$. Then the bracket product satisfies the following properties: > 1. **Antisymmetric:** $[\xi, \eta] = -[\eta, \xi]$ > 2. **Bilinear**: $[a\xi + \zeta, \eta] = a[\xi, \eta] + [\zeta, \eta]$ > 3. **Jacobi Identity:** $[\xi, [\eta, \zeta]] + [\eta, [\zeta, \xi]] + [\zeta, [\xi, \eta]] = 0$ > 4. For any $f, g \in C^\infty(X)$, $[f\xi, g\eta] = fg[\xi, \eta] + (f\xi g) \eta - (g \eta f) \xi$. > > In other words, smooth vector fields form a [[Lie Algebra|Lie algebra]]. > > *Proof*. Antisymmetry and linearity comes from the definition of the bracket product and the bilinearity of applying a linear map to a vector. > > For the Jacobi identity, let $f \in C^\infty(X)$, > $ > \begin{align*} > [\xi, [\eta, \zeta]]f &= \xi[\eta, \zeta]f - [\eta, \zeta]\xi f \\ > &= \xi\eta\zeta f - \xi\zeta\eta f - \eta\zeta\xi f + \zeta\eta\xi f \\ > &= \underbrace{\xi\eta\zeta f - \eta\zeta\xi f}_{L} + \underbrace{\zeta\eta\xi f - \xi\zeta\eta f}_{R} > \end{align*} > $ > Adding all three terms up, > $ > \begin{align*} > &[\xi, [\eta, \zeta]]f + [\eta, [\zeta, \xi]]f + [\zeta, [\xi, \eta]]f \\ > &= \underbrace{\pb{\xi\eta\zeta f} \po{- \eta\zeta\xi f + \eta\zeta\xi f} \pg{- \zeta\xi\eta f + \zeta\xi\eta f} \pb{- \xi\eta\zeta f}}_{L} \\ > &\underbrace{\pb{+ \zeta\eta\xi f} \po{- \xi\zeta\eta f + \xi\zeta\eta f} \pg{- \eta\xi\zeta f + \eta\xi\zeta f} \pb{- \zeta\eta\xi f}}_{R} \\ > &= 0 > \end{align*} > $ > > Lastly, let $f, g, h \in C^\infty(X)$, then by expanding through the product rule > $ > \begin{align*} > [f\xi, g\eta]h &= (f\xi)(g\eta)h - (g\eta)(f\xi)h \\ > &= g(f\xi)\eta h + \eta h(f\xi)g - f(g\eta)\xi h - \xi h(g\eta)f \\ > &= (gf)\xi\eta h- (fg)\eta\xi h + (f\xi g)\eta h - (g\eta f)\xi h \\ > &= fg[\xi, \eta]h + (f\xi g)\eta h - (g\eta f)\xi h \\ > [f\xi, g\eta]&= fg[\xi, \eta] + (f\xi g)\eta - (g\eta f)\xi > \end{align*} > $ > [!theorem] > > Let $X$ be a [[n-Manifold|n-manifold]], and $\xi, \eta \in \vf(X)$ with coordinate representations > $ > \xi = \xi^i \ppi \quad \eta = \eta^i\ppi > $ > around $p$. Then their bracket product has coordinate representation > $ > [\xi, \eta] = \paren{\xi^i\frac{\partial \eta^j}{\partial x^i} - \eta^i\frac{\partial \xi^j}{\partial x^i}} \cdot \frac{\partial}{\partial x^j} > $ > *Proof*. Since > $ > \widehat{[\xi, \eta]}(\hat p) = D(\hat \eta)(\hat p)(\hat \xi(\hat p)) - D(\hat \xi)(\hat p)(\hat \eta(\hat p)) > $ > writing out both the vector fields and the derivatives in coordinates yields > $ > \begin{align*} > \widehat{[\xi, \eta]} &= D(\hat \eta)(\hat p)\paren{\xi^ie_i} - D(\hat \xi)(\hat p)\paren{\xi^ie_i} \\ > &= \paren{\xi^i\frac{\partial \eta^j}{\partial x^i}e_j} - \paren{\eta^i\frac{\partial \xi^j}{\partial x^i}e_j} \\ > [\xi, \eta] &= \braks{\xi^i\frac{\partial \eta^j}{\partial x^i} - \eta^i\frac{\partial \xi^j}{\partial x^i}} \frac{\partial}{\partial x^i} > \end{align*} > $ > [!theorem] > > Let $X, Y$ be manifolds, $F: X \to Y$ be a [[Manifold Morphism|morphism]], $\xi_1, \xi_2 \in \vf(X)$, and $\eta_1, \eta_2 \in \vf(Y)$. If $\xi_1$ and $\xi_2$ are [[Related Vector Fields|related]] to $\eta_1$ and $\eta_2$ by $F$. Then $[\xi_1, \xi_2]$ and $[\eta_1, \eta_2]$ are related by $F$. > > As a corollary, if $F$ is a diffeomorphism, then the [[Pushforward of Vector Field|pushforward]] commutes with the bracket product: > $ > F_*[\xi_1, \xi_2] = [F_*{\xi_1}, F_*{\xi_2}] > $ > > *Proof*. Let $f \in C^\infty(X)$, then we can substitute each vector field with its related vector field > $ > \begin{align*} > [\eta_1, \eta_2](f \circ F) &= \eta_1\eta_2(f \circ F) - \eta_2\eta_1(f \circ F) \\ > &= \eta_1(\xi_2 f \circ F) - \eta_2(\xi_1 f \circ F) \\ > &= (\xi_1\xi_2f) \circ F - (\xi_2\xi_1f)\circ F \\ > &= [\xi_1, \xi_2]f \circ F > \end{align*} > $ > Hence $[\eta_1, \eta_2]$ and $[\xi_1, \xi_2]$ are $F$-related.