# Definition of lie bracket of vector fields

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 $t=0$?

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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.
The last part is correct. We take a flow for time $t$ along $X$ starting from a point $x$. We then look at the value of the vector field $Y$ there and map it back to $x$ using the derivative of $\Phi_{-t}^X$. We then have a curve in $T_x M$ with variable $t$ which we can differentiate. –  muzzlator Apr 16 '13 at 9:51