Singular Vector Fields as Basis Elements in Contact Planes.

Question

I'm trying to generate specific examples of contact structures in $\mathbb R^3$, but I'm running into something strange. It seems like the contact planes associated to the structure (using cylindrical coordinates in $\mathbb R^3$)

EDIT/CORRECTION: My apologies to all who read this for wasting your time; the actual contact structure is, in cylindrical coordinates $( r, \theta, z)$: $$Ker[\cos( \pi r)dz+ \sin(\pi r) d \theta], $$ ( the correction: original was: $Ker (cos(\pi r) dr + sin (\pi r) d\theta).$ To be more rigorous this time, will compute $ w \wedge dw$: we have $dw=-\pi sin (\pi r)dzdr+\pi cos(\pi r)d\theta dr$. We see in expanding $w \wedge dw$ that we have two coefficients for $d\theta dz dr$ which do not cancel each other out; we actually get $\pi dzd\theta dr \neq 0$.)

Still, the issue remains after the correction:

Set $$(\cos( \pi r)dz+ \sin(\pi r) d \theta )(a_z \partial z+a_{\theta} \partial \theta ):=0.$$ (I got rid of the $\pi$ )

Then we get $a_z cos( \pi r)+ a_{\theta} sin (\pi r)=0$.

How do we avoid dividing here when we find a basis? We can choose, e.g., $a_{\theta}=1$ , then we have

$a_z cos (\pi r)+sin(\pi r)=0 $ .

And, now, solving for $a_z$ , we still have to divide , to get $a_z =-tan (\pi r )$,

and a basis {$(\partial r,0,0),(0,\partial \theta, -tan(\pi r) \partial z)$}

And we still have singularities whenever $\pi r= \pi/2+k \pi$ ; $k $ in $\mathbb Z$ , i.e., whenever $r=1/2+k$

How do we deal with this?

Answer 1

The contact structure you are considering is defined in $\mathbb R^3$ only away from the $z$-axis ($r=0$) and cannot be extended to all of $\mathbb R^3$.

One way to "see" this (and also prove it), is to consider the field of lines perpendicular to the contact planes (kernels of $\alpha$). Consider $N=r\partial_\theta +\partial_z$. You can check that this $N$ has constant norm (square root of 2), and is perpendicular to the contact planes, up to the addition of a smooth vector field on $\mathbb R^3$ which vanishes along the $z$-axis. This means that near the $z$-axis, your contact planes "look like" the planes perpendicular to $N$. Then it is quite easy to get convinced, both intuitively and formally, that the field of lines that $N$ generates cannot be extended to the $z$-axis, hence neither your field of contact planes.

As for finding a pair of vector fields in $\mathbb R^3$, away from the $z$-axis, that span at each point the contact plane of your contact structure, you basically have it. Just multiply your second vector field by $\cos \theta$. You get the pair of vector fields $\partial_r,$ $ \cos(\pi r)\partial_\theta-\sin(\pi r)\partial_z$. Note that neither of them can be extended to the $z$-axis.

As for the general problem of finding a pair of vector fields on some open subset $U\subset\mathbb R^3$, spanning at each point the contact plane of a given contact structure on $U$ (or, more generally, on some 3-dimensional manifold), it cannot always be done. It depends on the topological properties of $U$ and the specific contact structure. This is a (mild) topological problem, covered in many textbooks, concerning the triviality of the plane-bundle defined by the contact structure (see for example Milnor+Stashef).

For $\mathbb R^3$, being a contractible space, any vector bundle on it is trivial, hence the contact planes of any contact structure can be spanned by a pair of globally defined vector fields.

For the $U$ of your problem (complement of the $z$-axis), unlike your example of contact structure, there are contact structures whose contact planes cannot be spanned by a pair of vector fields; take for example the contact form $\cos(2\theta)dr-\sin(2\theta)dz$. Note that this contact form is not globally well-defined on $U$, yet the distribution of contact planes defined by it is well-defined.

I hope this explanation is helpful to you.