Understanding the definition of Lie bracket of vector fields as $[X,Y](f)=X(Y(f))-Y(X(f))$

Question

The Jacobi-Lie bracket or simply Lie bracket, $[X,Y]$, of two vector fields $X$ and $Y$ is the vector field such that $[X,Y](f) = X(Y(f))-Y(X(f)) \,.$ (http://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields)

So for the right-hand side, for $X(Y(f))$, do we evaluate $Y(f)$ first, then evaluate $X$ at position $Y(f)$? I can only think this way. If it is wrong, please tell me.

Also, for $\left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} (\mathrm{d}\Phi^X_{-t}) Y_{\Phi^X_t(x)}$, does this mean that $(\mathrm{d}\Phi^X_{-t})$ is evaluated with vector $Y_{\Phi^X_t(x)}$ and then differentiate with regard to $t=0$?

Answer 1

Given a function $f : M \rightarrow \Bbb R$ and a vector field $X$, we can define a new function $X f : M \rightarrow \Bbb R$ by $$(X f)(p) = \frac{d}{dt} (f \circ \gamma)(t)|_{t=0}$$ where $\gamma(t)$ is any curve such that $\gamma(0) = p$ and $\gamma'(0) = X_p$

So given a function $f$, we can define a new function $X f$ from the manifold into the reals (thought of as the rate of change of $f$ along $X$). If we now have a second vector field $Y$, we can compute $Y (X f)$ which is another function from $M$ into $\Bbb R$.

Edit: Regarding your comment of evaluating $X$ at position $Y(f)$, this doesn't really make sense. $Y(f)$ is a value in $T \Bbb R = \Bbb R$ and $X$ is evaluated at positions on the manifold so we can't really talk about $X$ evaluated at $Y f$. We can however apply $X$ to $Y f$ since vector fields can act on functions as a differential operator.

Answer 2

Let $X,Y\in\Gamma(TM)$ be smooth vector fields on a smooth manifold $M$.

This definition of Lie bracket hinges on the bijection between vector fields and derivations on $C^\infty(M)$. More precisely, if a vector field is a map $X:M\to TM$, we can define the corresponding derivation as $\nabla_X:C^\infty(M)\to C^\infty(M)$, such that $$\nabla_X f \equiv \eta\circ\mathrm df \circ X \in C^\infty(M)\equiv C^\infty(M,\mathbb R),$$ where $\eta$ projects onto the second component (this notation is from Terry Tao's notes). Note that here $\mathrm df:TM\to T\mathbb R$ and thus $$\mathrm df\circ X: p\mapsto (f(p), \mathrm df_p(\eta\circ X(p))), \qquad \eta\circ X(p)\in T_p M.$$

In other words, $\nabla$ maps vector fields into derivations: $\nabla:\Gamma(TM)\to\mathrm{Der}(C^\infty(M))$.

This bijection can be leveraged to define the commutator of vector fields, because there is a natural notion of commutator on derivations. We can then define $[X,Y]$ to be that vector field such that $$\nabla_{[X,Y]} = [\nabla_X,\nabla_Y] \equiv \nabla_X\circ\nabla_Y - \nabla_Y\circ\nabla_X.$$ Again, being $X\mapsto \nabla_X$ a bijection, the above can be safely used to define $[X,Y]$.

To more directly connect with the notation $[X,Y](f)=X(Y(f))-Y(X(f))$, we need simply observe that $X(f)\equiv \nabla_X f$ etc.