differential forms, cylindrical coordinates, geometric interpretation

Question

Consider a differential $1$-form $\beta$ which in cylindrical coordinates $(r, \theta, z)$ has the form $$\beta = f(r)\,dz + g(r)\,d\theta,$$where $g'(0) = 0$. Find a condition when $\beta \wedge d\beta$ is a volume form, i.e. it does not vanish anywhere. Interpret this condition geometrically in terms of the properties of the curve given in $\mathbb{R}^2$ with Cartesian coordinates $(u, v)$ by parametric equations $$u = f(r),\text{ }v = g(r)\text{ for }r \in [0, \infty).$$

Answer 1

We have$$d\beta = f'(r)\,dr \wedge dz + g'(r)\,dr \wedge d\theta,$$and$$\beta \wedge d\beta = (f'g - g'f)\,dr \wedge dz \wedge d\theta.$$Hence the required condition reads$$f'(r)g(r) - g'(r)f(r) \neq 0$$for all $r \neq 0$.

$($Here, it is worthwhile to remark that if $r=0$ we cannot make computations in cylindrical coordinates. The condition $g'(0) = g(0) = 0$ together with the condition $f'(0) = 0$ allows us to extend the form smoothly to $r = 0$. The condition that $\beta \wedge d\beta \neq 0$ along the $z$-axis then reads: $f(0) \neq 0$, $g''(0) \neq 0$.$)$

The condition $f'g - gf' \neq 0$ means that the velocity vector $(f', g')$ of the curve$$u = f(r),\,v = g(r)\text{ for }r \in [0, \infty)$$is never collinear with the radius vector $(f, g)$ of the curve. If we re-express this condition in polar coordinates $(\rho, \theta)$ in the $(u, v)$-plane, it then reads that $\phi' \neq 0$, i.e. when $r \to \infty$ the point $(f(r), g(r))$ keeps rotating around the origin in the same direction.