Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Jacobi-Lie bracket or simply Lie bracket, $[X,Y]$, of two vector fields $X$ and $Y$ is the vector field such that $[X,Y](f) = X(Y(f))-Y(X(f)) \,.$ (

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 $t=0$?

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
I got it. Thanks! – Differential Apr 16 '13 at 9:22
Oh wait. What about the last part? Is that correct? – Differential Apr 16 '13 at 9:47
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
Thanks! I appreciate your help. – Differential Apr 16 '13 at 10:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.