# Are there moves between Reidemeister moves?

### Background

Knots are typically written in 2 dimensions as a loop in the plane with normal crossings. One then asks when two such diagrams describe the same knot. Two diagrams describe the same knot when one can be made into the other by a sequence of Reidemeister moves. These are three simple manipulations of a diagram which obviously don't change the underlying knot.

### The Reidemeister Graph

One may consider the 'Reidemeister graph' of a knot diagram (probably not an official term), which consists of every diagram which is equivalent to the original, and an edge for every Reidemeister move between them. Since every Reidemeister move can be undone by the same move again, this is an undirected graph.

Two diagrams may be connected by a many different sequences of Reidemeister moves. It is not hard to find a sequence of moves which takes a knot to itself, and is not trivial in the sense that involves doing and immediately undoing the same move. As a consequence, the Reidemeister graph is infinite and not simply connected.

### Question

I can think of a loop in the Reidemeister graph as a kind of 'relation between relations' (where the moves are the relations). I would like to find a finite list of relations between relations, such that every loop is built from these relations.

I'll be more specific. Define a higher Reidemeister move to be a locally-defined sequence of Reidemeister moves which relate a given diagram to itself. I would like a finite list of higher Reidemeister moves, such that if one fills in the corresponding loops in the Reidemeister graph with a 2-cell, the resulting space is simply connected.

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Your requirement that "if one fills in the corresponding loops in the Reidemeister graph with a 2-cell, the resulting space is simply connected" seems too strong. I think you probably want to require instead that the inclusion of your 2-complex into the space of knots induces an isomorphism on the fundamental group of each component. – Michael Hutchings Nov 29 '10 at 3:52
As Michael mentions, attaching 2-cells won't give you a simply connected space. For one, 2-cells don't change the path-components of a space so regardless of which 2-cells you attach you'll still have countably-many components (the knot types). If you want the components to be simply-connected you can of course do that but it won't be all that interesting a space. – Ryan Budney Nov 29 '10 at 4:00
I was only consider the Reidemeister graph of a single diagram; ie, a connected component of the space of ALL diagrams and moves. And certainly I can make this simply connected by filling in cells, the question is whether I can do it with a finite number of kinds of cells. – Greg Muller Nov 29 '10 at 4:03
The space of smooth embeddings $S^1 \to \mathbb R^3$ has a fairly delicate homotopy-type. It sounds like you want to move towards a cellular decomposition of this space that is compatible with projections and knot diagrams -- in particular the 1-skeleton being the Reidemeister move graph. The 2-cells for this space I believe were worked out quite a while ago, likely not in this exact context but it follows from fairly elementary singularity theory, the work of Thom and Mather, etc. Take a look at some of the recent papers of Fiedler: front.math.ucdavis.edu/0606.5381 – Ryan Budney Nov 29 '10 at 4:06
Have you looked at the Carter-Saito movie moves? books.google.com/… – Ben Webster Nov 29 '10 at 4:29

In some strict sense I think the answer to your question is no, there are likely no finite collection of 2-cells doing what you want. If you were to ask the more natural question where you're looking for a 2-complex whose inclusion into $Emb(S^1,\mathbb R^3)$ is an isomorphism on $\pi_1$ (component-by-component) the answer I'm near-certain is yes (via Thom-Mather singularity theory).

For example, here is a non-trivial loop in the space of knots which you could imagine as a loop in your Reidemeister graph once you refine things suitably. This loop isn't a problem if you only want the map (2-complex) $\to Emb(S^1,\mathbb R^3)$ to be an isomorphism on $\pi_1$. But for the complex you want, these loops are a problem, as they're very much global things and can't be described readily in terms of local diagram moves.

The loop described in this picture can be done for any combination of summands -- as long as the summand knots are non-trivial this is a non-trivial loop. So how are you going to construct a finite collection of 2-cells that kill off all these loops?

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What if you restrict to prime knots? – Jim Conant Nov 29 '10 at 11:23
Or, probably better, hyperbolic knots... – Jim Conant Nov 29 '10 at 11:24
This is an excellent point. I don't want to kill off loops that are genuinely non-trivial. I suppose what I want is to consider maps $S^1\times S^1\rightarrow \mathbb{R}^3$ which are embeddings in the first factor, and which can be extended to maps $S^1\times B^2\rightarrow \mathbb{R}^3$ (where $B^2$ is the ball). Those are the 2-cells to fill in, which won't leave a simply connected space as you say. – Greg Muller Nov 29 '10 at 22:33
Okay, so it sounds like you're after what Hutchings and I first suggested. The Fiedler paper I cited above: front.math.ucdavis.edu/0606.5381 is what you want. The moves they work out are for 2-parameter families of knot/link diagrams in $S^1 \times D^2$ but it's a nice example of how the Thom-Mather theory works and I think there's little you have to do to adapt it for links in $\mathbb R^3$ -- I believe Fiedler essentially explains this in his paper. – Ryan Budney Nov 30 '10 at 3:48
@JimConant: prime knots have similar problems, but the diagrams tend to be a little more complicated. For example, take a cable of the connect sum above. The above loop can be made to also be a loop for the cable knot, which is prime. The space of the figure-8 knot similarly has non-trivial loops. and on and on... – Ryan Budney Jan 27 at 6:18

Take a look at page 180 of Low dimensional topology by Tomasz Mrowka, Peter Steven Ozsvát, (following Ben Webster's comment about Movie Moves elucidated by Baez and Langford and their 30 basic movie moves, and by Carter and Saito who describe a 31st basic movie move.) A movie move is a sequence of frames of a braid (or subregion of a knot, I suppose).

Carter and Saito have a theorem that

two movies represent the same tangle cobordism iff they can be related by a sequence of movie moves

If you take the subset of Movie Moves where each movie is a composition of a sequence of Reidemeister moves, it seems like that would be similar or equivalent to what you are calling "Higher Reidemeister moves." Am I understanding you correctly?

I would point you out to the appropriate page on that wiki o' info, but "Movie moves" does not even show up on their search page.

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I think that may represent a good answer, although I'm not familiar with these movie moves myself (surely there must be a better name!) To offer a complete answer I think we need to know the homotopy type of the configuration space of the knot. It's not simply connected, is it aspherical? What is its fundamental group? – James Griffin Nov 29 '10 at 17:33
Perhaps a relevant link: math.cornell.edu/~hatcher/Papers/knotspaces.pdf The action of SO(4) means that we probably do have spheres, but maybe they can be filled in in a coherent way. – James Griffin Nov 29 '10 at 17:37
Another relevant link: arxiv.org/abs/math/0506524 (see end of p4). – James Griffin Nov 29 '10 at 17:50
The three spaces $Emb(S^1,\mathbb R^3)$, $Emb(S^1, S^3)$ and $Emb(\mathbb R,\mathbb R^3)$ "long knots", all have closely related homotopy-types. To see the precise relationship, check out Theorem 2.1 and Prop 2.2 in: front.math.ucdavis.edu/0605.5069 In particular $Emb(\mathbb R, \mathbb R^3)$, denoted $\mathcal K_{3,1}$ in the above paper, all the path components are $K(\pi,1)$-spaces. The fundamental groups of the components are described in the 2nd paper James cites, and in the paper I cite. My new preprint on the "splicing operad" has a more uniform and geometric description. – Ryan Budney Nov 30 '10 at 4:51
... of the homotopy-equivalences. – Ryan Budney Nov 30 '10 at 4:52