Distinguishing Legendrian knots with trivial orientation-preserving symmetry group

TL;DR

This paper simplifies the process of distinguishing Legendrian knots with trivial symmetry groups, enabling comparisons where algebraic invariants are ineffective, and disproves a specific equivalence conjecture for boundary components of certain annuli.

Contribution

It introduces a simplified method for distinguishing Legendrian knots with trivial symmetry groups, removing the need for exhaustive surface diagram searches, and proves non-equivalence in a specific annulus case.

Findings

Simplified method for Legendrian knot comparison when symmetry group is trivial

Disproved the conjecture about boundary components of certain annuli being Legendrian equivalent

Established algorithmic solvability for comparing Legendrian knots with trivial symmetry groups

Abstract

In a recent work of I.\,Dynnikov and M.\,Prasolov a new method of comparing Legendrian knots is proposed. In general, to apply the method requires a lot of technical work. In particular, one needs to search all rectangular diagrams of surfaces realizing certain dividing configurations. In this paper, it is shown that, in the case when the orientation-preserving symmetry group of the knot is trivial, this exhaustive search is not needed, which simplifies the procedure considerably. This allows one to distinguish Legendrian knots in certain cases when the computation of the known algebraic invariants is infeasible or is not informative. In particular, it is disproved here that when~$A\subset\mathbb R^3$ is an annulus tangent to the standard contact structure along~$\partial A$, then the two components of~$\partial A$ are always equivalent Legendrian knots. A candidate counterexample was proposed recently by I.\,Dynnikov and M.\,Prasolov, but the proof of the fact that the two components of~$\partial A$ are not Legendrian equivalent was not given. Now this work is accomplished. It is also shown here that the problem of comparing two Legendrian knots having the same topological type is algorithmically solvable provided that the orientation-preserving symmetry group of these knots is trivial.