Finding all $2$-forms in the right half-plane that are invariant under glide transformations I'm trying to find all 2-forms $\omega$ that are invariant under glide transformations in the right half-plane, $X = \{ (x,y) \in \mathbb{R}^2 : x > 0\}$.
To do this, we can write the vector field corresponding to such transformations as $V = x \frac{\partial}{\partial x} - y \frac{\partial}{\partial y}$. Now I'd like to solve $\mathcal{L}_X\omega = 0$ for $\omega$.
We can write an arbitrary $2$-form on $X$ as $$\omega = a(x,y) \textrm{d}x \wedge \textrm{d}y$$ and then we can use the usual rules of the Lie derivative to find that the equation becomes $$x\frac{\partial a}{\partial x} = y\frac{\partial a}{\partial y}.$$ The only solution I can find to this equation (by looking for a separable solution) is $a(x,y) = c(\log(x)+\log(y))$.
My difficulty is, $log(y)$ is not even defined everywhere on this manifold so how can this be the solution?
 A: Meditating on the p.d.e. for a minute, we can see that any function of the form $a(x, y) = b(xy)$ satisfies $x a_x = y a_y$, and this is (I think) the general solution, but I don't see offhand how to prove this.
Instead, here's another strategy for finding the glide-invariant $2$-forms that avoids solving a p.d.e.:
Computing gives $\mathcal L_V (dx \wedge dy) = 0$ (which reflects the fact that glides are area-preserving), so the Leibniz rule for the Lie derivative gives $$\mathcal L_V \omega = (V \cdot f) \,dx \wedge dy,$$
and thus the solutions of of $\mathcal L_V \omega = 0$ are the $2$-forms $f(x, y) \,dx \wedge dy$ for which the (smooth) function $f(x, y)$ is constant on the flow lines of $V$. This is an improvement over the original method in the question in the sense that it involves solving an o.d.e., rather than a p.d.e. (This is a real advantage for us, as in this setting we're guaranteed uniqueness of solutions of o.d.e.s satisfying a given initial condition.) In particular, if $(x(t), y(t))$ parametrizes a flow line of $V$, we get the simple system
$$
\left\{
\begin{array}{rcl}
x' &=& \phantom{-}x \\
y' &=& -y
\end{array}
\right. .
$$

This has the general solution $(x(t), y(t)) = (x_0 e^t, y_0 e^{-t})$; we see immediately that for any solution, $x(t) y(t) = x_0 y_0$, and it's not hard to check that any solution traces over the entire hyperbolic arc $\{x y = k, x > 0\}$, where $k = x_0 y_0$. Thus, the functions $f$ constant on the flow lines are the functions of the form $f(x, y) = g(xy)$, and the general solution $\omega$ to $\mathcal L_V \omega = 0$ is $$\boxed{\omega = g(x y) \,dx \wedge dy} .$$

(My original answer addressed a previous version of the question; I'll leave it here for posterity. In fact, one can adapt the method there to give another solution to this problem, using hyperbolic coordinates in place of polar coordinates.)

The given formula for the infinitesimal generator $V$ of the rotations is incorrect; it should be, e.g.,
$$V := x \partial_y - y \partial_x .$$ (As it's written now, $V$ generates glides, which are hyperbolic analogues of rotation.)
Hint Any rotationally invariant $2$-form on the half-plane extends uniquely to a rotationally invariant $2$-form on the punctured plane, $\Bbb R^2 - \{ 0 \}$, and since we're interested in rotationally symmetric objects, we may as well work in polar coordinates. In any such coordinates, we can write $$\omega = f(r, \theta) \, dr \wedge d\theta,$$ and the rotations are generated by $\partial_{\theta}$.
So, the invariance condition becomes $$0 = \mathcal L_{\partial_{\theta}} f = \frac{\partial f}{\partial \theta} ,$$ and so the rotationally invariant $2$-forms are those of the form $f(r) \,dr \wedge d\theta$.
