6
$\begingroup$

Given an (oriented) knot projection, we can create a quandle by assigning each curve segment a generator, and a relation $a \triangleright b = c$ for each intersection. If two projections are related by Reidemeister moves, then we can easily construct an isomorphism between the two corresponding quandles, by looking at the local behavior on Reidemeister moves. This can be rephrased to use fundamental groups equipped with some suitable additional data.

However, this just establishes that there exists an isomorphism, since there may be two different paths of Reidemeister moves that give different isomorphisms. And establishing that these isomorphisms commute seems more difficult. There are probably no MacLane-style coherence theorems that reduce the problem to checking finitely many diagrams, see the question Are there moves between Reidemeister moves? But in this particular case, it seems that the isomorphisms should commute. Is it possible to prove this (preferably without using powerful tools in 3-dimensional manifold theory, since this doesn't look like a question of that depth)?

$\endgroup$
9
  • $\begingroup$ I'm a little unsure what your question is. Are you asking for the automorphism group of the quandle to be abelian? $\endgroup$ Commented May 28 at 14:49
  • 1
    $\begingroup$ @RyanBudney No. Consider a sequence of Reidemeister move that starts and ends with the same knot projection. This induces a specific isomorphism of the quandle. Is this isomorphism always the identity? $\endgroup$
    – Trebor
    Commented May 28 at 15:54
  • $\begingroup$ Your usage of the word "commute" in your question misled me. That isomorphism is generally not the identity. You get a faithful representation of the fundamental group of the space of knots into this automorphism group. $\endgroup$ Commented May 28 at 16:57
  • $\begingroup$ @RyanBudney I'm using it in the categorist sense. But also I don't understand how the automorphism group helps, because two knot diagrams give two different presentations of a quandle which are not canonically isomorphic (that is exactly the question being asked), so given an isomorphism of these quandles I don't know how to interpret that as an element of the automorphism group. $\endgroup$
    – Trebor
    Commented May 28 at 17:31
  • $\begingroup$ I don't understand your source of confusion. The answer to your question is no. $\endgroup$ Commented May 28 at 17:37

1 Answer 1

3
$\begingroup$

As in Ryan’s comments, in general the knot group will act on the fundamental quandle by conjugation. I think it’s easier to work on the 2-sphere, where one may pass strands over “infinity” (note that this doesn’t change the presentation of the fundamental quandle as you have defined it). The fundamental group is generated by meridians, so if we can describe how meridians act, then we get the action of the full fundamental group by composition. To see how a meridian acts, move the strand to the outer part of the diagram. Then perform a “sweep-around move” in the terminology of this paper (see diagram 3.1).

enter image description here

As the outer strand is dragged across the diagram, each quandle generator corresponding to another strand gets conjugated by the outer strand.

Note that in the case of torus knots, there is non-trivial center so the inner automorphism group is a quotient of the fundamental group by the center.

The example shown by Ryan Budney in the link is an example of a non-trivial mapping class giving an outer automorphism of the knot group (and hence quandle). More generally, these exist by Dehn twist around any torus: if you have a satellite, then the pattern may be cycled around the satellite torus. Also, the geometric components of knots can have finite outer automorphism/mapping class group which induces other automorphisms of the quandle (as in Lawson’s example of the trefoil knot in the comments). I’m not sure how one would prove these things without 3-manifold tools (and I haven’t thought deeply about quandles, so I may be missing some issues).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .