Contact form in Polar Coordinates

contact-geometrydifferential-formsdifferential-geometry

While going through Etnyres Lectures on Contact Topology (which can be found here) Example 2.8 two questions came up

  1. He uses cylindrcal coordinates $(r,\theta, z)$ to define a 1-form $\alpha_2 = dz +r^2d\theta$ on $\mathbb{R}^3$ and it is not clear to me how this is well defined. I would interpret this as follows: $\theta$ should be a function $\mathbb{R}^3\to\mathbb{R}$ sending a point $p$ to the angle of the projection onto the xy-plane of p with respect to the ray along the x-axis going out to + $\infty$, $d\theta$ would then be the exterior derivative. However, $\theta$ does not define a smooth map on all of $\mathbb{R}^3$ because of problems at the origin. My question is then, how should one interpret this definition of $\alpha_2$.

  2. He claims that $\alpha_2$ defines a contact form, that is $\alpha_2\wedge d\alpha_2\neq 0$,
    but his computation shows that
    $$\alpha_2\wedge d\alpha_2 = 2rdr\wedge d\theta\wedge dz$$
    which gives the trivial form at the origin since $r(0,0,0) = 0$. I thought the contact condition implies that the form is non-trivial at every point $p\in\mathbb{R}^3$, or is it enough that the form is non-trivial at a single point?

Best Answer

Your point is well-taken. However, note that $$r^2\,d\theta = (x^2+y^2)\left(\frac{-y\,dx+x\,dy}{x^2+y^2}\right) = -y\,dx + x\,dy$$ is in fact globally defined, even though the polar coordinate system breaks down at the origin.

Then $\alpha_2 = dz-y\,dx+x\,dy$ and $$\alpha_2\wedge d\alpha_2 = 2\,dz\wedge dx\wedge dy = 2(dx\wedge dy)\wedge dz = 2(r\,dr\wedge d\theta)\wedge dz,$$ as promised.

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