The short of it:
Concerning Lie brackets of vector fields: Who's right with the sign of the "gap vector"? Does $[V=d/d\mu, W=d/d\lambda]$ point from $A$ to $B$ or from $B$ to $A$?
Adapted from Schutz's "Geometrical methods in mathematical physics":
But in other sources the sign differs, e.g. Hehl 2007:
Who is right?
The long of it: I'm reading Bernard Schutz's "Geometrical methods in mathematical physics", which -- being a physicist myself -- strikes a great balance for me between explanatory detail and rigor.
I'd like to prefix my question with some things that I think I understand, so that it's easier for others to point to where I might have erred. I added two figures to illustrate what I'm trying to get to.
I cannot find an answer to my specific question, though I have seen similar arguments as the following posted.
I can understand how vector fields can be seen as operators acting on functions. The coordinate representation of the Lie bracket
$[V, W] = \left(V^j \frac{W^i}{\partial x_j} - W^j \frac{V^i}{\partial x_j} \right)\partial_i$
is clear to me and using the product rule, I can derive the "product rule"
$[fX, gY] = fg[X, Y] + fX(Y) - gY(f)X$.
Schutz describes the geometric interpretation of the Lie bracket by using the exponentiation of vector fields that describe finite motions along integral curves: Let $V = d / d\lambda$ and $W = d / d \mu$ be vector fields. Starting at a point $P$, moving $\Delta \lambda = \epsilon$ along the V curve through $P$ to a point $R$, and then moving $\Delta \mu = \epsilon$ along a W curve, ending at a point $A$. Then start again at $P$, go first $\Delta \mu$ along $W$ and then $\Delta \lambda$ along $V$ ending at a point $B \neq A$. The vector from $A$ to be $B$ is $\epsilon^2[V, W]$, to lowest order in $\varepsilon$, since
$x^i(R) = \exp\left(\epsilon \frac{d}{d\lambda}\right)x^i|_P \\ x^i(A) = \exp\left(\epsilon \frac{d}{d\mu}\right) \exp\left(\epsilon \frac{d}{d\lambda}\right)x^i|_P$
and analogously
$x^i(B) = \exp\left(\epsilon \frac{d}{d\lambda}\right) \exp\left(\epsilon \frac{d}{d\mu}\right)x^i|_P$.
Using the Taylor series of the exponentials yields
$x^i(B) - x^i(A) = \left[\exp\left(\epsilon \frac{d}{d\lambda}\right) \exp\left(\epsilon \frac{d}{d\mu}\right), \exp\left(\epsilon \frac{d}{d\mu}\right) \exp\left(\epsilon \frac{d}{d\lambda}\right)\right]x^i|_P \\ = \epsilon^2\left[V, W\right]|_P$.
This makes sense to me, the exponentials act from the left on $x^i$ so the rightmost exponential acts first. You from $P$ to $A$ via $WV$ and you get from $P$ to $B$ via $VW$.
So all of this should be easy to verify with an example, right?
Lets take polar coordinates $(r, \phi)$ in 2d and three vector fields:
$V := \cos(\phi) \partial_x + \sin(\phi) \partial_y = \frac{x}{r} \partial_x + \frac{y}{r} = \partial_r\\ W := -r\sin(\phi) \partial_x + r \cos(\phi) \partial_y = -y \partial_x + x \partial_y = \partial_{\phi}\\ Z := -\sin(\phi) \partial_x + \cos(\phi) \partial_y = \frac{1}{r} \partial_{\phi}$
The vector fields $V$ and $W$ are the usual ones for polar coordinates and form a coordinate basis, since they commute with each other: $[V, W] = 0$ (just plug in the definition of the Lie bracket and use the product rule for the Lie bracket together with commuting partial derivatives $[\partial_x, \partial_y] = 0$. Ok.
Using the product rule for Lie brackets yet again yields straightforwardly:
$[V, Z] = [V, \frac{1}{r} W] = \frac{1}{r} [V, W] + \frac{\partial r^{-1}}{\partial_r} \partial_\phi = -\frac{1}{r^2} \partial_\phi= -\frac{1}{r} Z \neq 0$
I get the same result by separately computing $V(Z) = ... = 0$ and $Z(V) = \frac{1}{r^2}\partial_\phi$ and again $[V, Z] = -\frac{1}{r^2} \partial_\phi$.
So the vector field $[V, Z]$ points opposite to $\partial_\phi$.
Finally my question: How does this connect to $[V, Z]$ pointing from $A$ to $B$? Following the above reasoning of following the flows, I get to $A$ by following first $V$ and then $Z$, so as an operator it's $ZV$; and I get to $B$ by following first $Z$ and then $V$, so as an operator it's $VZ$. Sketching this for small $\epsilon$ (or large radii so that the circle segments look almost like straight lines), starting at $P = (x=1, y=0)$ results in $B$ being almost directly above $A$, so $\vec B - \vec A = \epsilon^2 [V, Z] + O(\epsilon^4)$ points almost completely in positive $y$ direction. But from my calculations I see that $[V, Z] = - \frac{1}{r^2}\partial_\phi$ so evaluated at $P = (1, 0)$ this yields $-\partial_y$. Exactly opposite of what I expected!
I can follow the derivation of why the Lie bracket points from $A$ to $B$, but the example yields the opposite!
Where am I going wrong?
Sorry for the rather long post, but I think my question may be unclear without context.
Thanks for your help!