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Let $D$ be a knot diagram and $Q$ a quandle. We use $c$ to denote a fixed coloring of $D$ with $Q$. If $D'$ is another knot diagram of the same knot, and $R_1$ is a sequence of Reidemeister moves connecting $D$ and $D'$. Obviously $R_1$ decides a unique $Q$-coloring of $D'$.

My question is, if $R_2$ is another sequence of Reidemeister moves connecting $D$ and $D'$, then $R_2$ will aslo bring a $Q$-coloring to $D'$. Are these two colorings equivalent?

Here we say two colorings of a knot diagram are equivalent if one can be transformed into the other one by some isotopy on the plane, i.e. no Reidemeister moves. For example let us consider the Fox 3-colorings on the standard diagram of trefoil knot, then there are five different colorings: three trivial colorings and two nontrivial colorings.

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No.

For example, form a small loop with an R1 move on a strand coloured $x$, imagine the whole rest of the knot inside a small ball, and pass that ball on a loop-the-loop through the R1 loop. Then perform another R1 move to reverse your first.

I drew a diagram of this ambient isotopy in:

Moskovich, D. 2006 Surgery untying of coloured knots, Algebr. Geom. Topol. 6, 673-697, arXiv:math.GT/0506541

You obtain the same diagram you started with, but with every colour acted on by $x$. If the quandle action of $x$ is non-trivial, in general this colouring with be inequivalent to the one which you started with.

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  • $\begingroup$ Thanks a lot. This is really an illuminating counterexample. $\endgroup$ – Zhiyun Cheng Sep 15 '14 at 14:11

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