4
$\begingroup$

Consider $\mathbb{R}^3$ with the standard contact structure $\ker(dz-y\,dx)$. According to the contact version of Weinstein's theorem, any Legendrian knot $L\subset \mathbb{R}^3$ has a tubular neighborhood $U\supset L$ that is contactomorphic to a neighborhood of the zero section $\{0\}\times 0_{T^*L}$ in the 1-jet bundle $J^1(L,\mathbb{R})\approx \mathbb{R}\times T^*L$ with contact structure $\ker (ds - p\,dq)$.

I am wondering what is known regarding how big these tubular neighborhoods can be. Specifically, I have the following

Question. Does there exist a Legendrian knot $L\subset \mathbb{R}^3$ and a Legendrian tubular neighborhood $U\supset L$ such that $\mathbb{R}^3\setminus U$ is a stratified subset of positive codimension, or more generally, a measure zero set?

$\endgroup$
4
  • $\begingroup$ There's work by Etnyre and Honda (arxiv.org/abs/math/0306330) which might be relevant. They have a property called "uniform thickness", but it might be a different/unrelated measure of thickness. $\endgroup$ Commented Jun 11 at 8:36
  • $\begingroup$ You'll have better luck with transverse knots than Legendrian ones. The transverse $sl=-1$ unknot in standard $S^3$ has an (open) n'hood whose complement is an unknot (the two knots are two fibres of the Hopf fibration). But complements of standard n'hoods of Legendrians are contactomorphic, so "small" vs "large" doesn't net you anything. $\endgroup$
    – magicker72
    Commented Jun 11 at 12:35
  • $\begingroup$ @MarcoGolla thank you. I will look into this notion and try to understand if it is relevant. $\endgroup$ Commented Jun 12 at 0:08
  • $\begingroup$ @magicker72 If the complement of a tubular neighborhood as in my question happens to be a stratified subset of positive codimension, or more generally a measure zero set, what would it mean for that complement to be contactomorphic to…anything? I edited the question to emphasize that I am specifically asking about the existence of such complements. I am not sure if your comment is relevant to this. Maybe I am misunderstanding. $\endgroup$ Commented Jun 12 at 0:11

0

You must log in to answer this question.