Exponentiating the Lie Derivative: Which direction does it rotate? I've been working through the book Functional Differential Geometry by Sussman and Wisdom and am having trouble with an example they give.
We can exponentiate a Lie derivative
$$
e^{\mathsf{t\mathcal{L}_v}}\mathsf{y} = \mathsf{y + t \mathcal{L}_v y + \frac{t^2}{2!} \mathcal{L}^2_v y + \dots }
$$
which evolves $\mathsf{y}$ along the integral curves of $\mathsf{v}$.  As a concrete example they evolve the coordinate basis vector $\frac{\partial}{\partial \mathsf{y}}$ along $\mathsf{J}_z = x\frac{\partial}{\partial \mathsf{y}} - y\frac{\partial}{\partial \mathsf{x}}$  (a counter clockwise circular field or z-angular momentum generator) and give as an answer
$$\exp(a \mathcal{L}_{\mathsf{J}_z})\tfrac{\partial}{\partial \mathsf{y}} = -\sin(a)\tfrac{\partial}{\partial \mathsf{x}} + \cos(a)\tfrac{\partial}{\partial \mathsf{y}}.$$
This agrees at $a=0$ and indicates that the evolution maintains the orientation of $\mathsf{v}$ and $\mathsf{y}$ along the flow.  $\tfrac{\partial}{\partial\mathsf{y}}$ just rotates along $\mathsf{J}_z$.
If I try to calculate the expansion, for the first term I get
$$
a\mathcal{L}_{x\frac{\partial}{\partial \mathsf{y}} - y\frac{\partial}{\partial \mathsf{x}}}\tfrac{\partial}{\partial \mathsf{y}} = 
a\mathcal{L}_{- y\tfrac{\partial}{\partial \mathsf{x}}}\tfrac{\partial}{\partial \mathsf{y}} = -a[y\tfrac{\partial}{\partial \mathsf{x}}, \tfrac{\partial}{\partial \mathsf{y}}] = a\tfrac{\partial}{\partial \mathsf{x}} 
$$
which disagrees with the expansion of the answer by a sign, giving rotation in the opposite direction.
The intuitive notions of the Lie Derivative I have also say it should be $a\tfrac{\partial}{\partial \mathsf{x}}$.  For example, beginning at $x=1, y=0$ we can travel along $\epsilon\mathsf{J}_z$ and then  $\epsilon \tfrac{\partial}{\partial \mathsf{y}} $, or begin along $\epsilon \tfrac{\partial}{\partial \mathsf{y}} $ and then along $\epsilon\mathsf{J}_z$ (points slightly to the left).  The difference between these paths will be $\epsilon^2 \tfrac{\partial}{\partial \mathsf{x}} $.  The same goes for beginning at $x=0,y=1$ or if I do it using a pushforward.  I'm tempted to say the book is mistaken, and that to get a rotation operator which coincides with the direction of $\mathsf{J}_z$ we should use $e^{-t\mathcal{L}_{\mathsf{J}_z}}$ but would appreciate some confirmation.
 A: You are right. I believe it is a small typo in the book. Indeed, here is a geometric interpretation of $exp(a \mathcal{L}_{J_z})\frac{\partial}{\partial y}$. You flow by $J_z$ for some time $a$, evaluate the vector field $\frac{\partial}{\partial y}$ at that new point, then take $d\varPhi_{-a}$ of that vector. As you can see, we are making use of $d\varPhi_{-a}$ (note the minus sign). Thus your calculation was right, and I believe it was a small typo in the book. I think it should be:
$\exp(a \mathcal{L}_{\mathsf{J}_z})\tfrac{\partial}{\partial \mathsf{y}} = \sin(a)\tfrac{\partial}{\partial \mathsf{x}} + \cos(a)\tfrac{\partial}{\partial \mathsf{y}}$.
One way to check would be to do the calculation explicitly (though I did not do that).
A: To add to @Malkoun: I would definitely say a typo. It seems the book is based upon a technical report by Sussman and Wisdom which can be found here. In that report (on page 48) they have the same calculation but the result this time is $+\sin(a)\frac{\partial}{\partial x}$, as you have calculated. 
It does makes me chuckle that they preface the result of the calculation with "apparently the result is" - maybe they were unsure of the answer too! 
