Finding all $2$-forms in the right half-plane that are invariant under glide transformations

Question

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?

Answer 1

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.)

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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 $f(r) \,dr \wedge d\theta$.