This problem may sound a little strange at first, but read on to the end before you pass judgement. Consider a theta graph (a graph with two vertices and three edges, each edge from one vertex to the other) embedded in the three sphere. Removing any edge from the graph leaves two edges that form a knot. Thus the embedding defines three knots. The question is: Can one embed a theta graph so that all three knots are isotopic to the same non-trivial knot. (The non-trivial condition is important because it’s easy to embed it so that you get three unknots. Also note that unlike a theta graph that comes from an unknotting tunnel, I’m not requiring that the complement of the theta graph is a handlebody.)

This question should sound suspiciously similar to problems about intrinsically knotted graphs. There, the idea is that given a sufficiently complex graph, no matter how you embed it, there will be a subgraph that defines a non-trivial knot or a linked link or more generally a knot or link with some measure of complexity bounded from below. In other words, given a graph, there is some set of knots and/or links such that any embedding contains a knot or link that is not in that set. Let’s call this the unnecessary set (though if you have a better name for it, feel free to suggest it.) As mentioned above, the results about intrinsically knotted or linked graphs have focused on highly complex graphs and looked at unnecessary sets that contain the unknot and perhaps all knots/links below some complexity.

But why focus on big graphs and unnecessary sets that contain the unknot? For example, it seems reasonably likely that the theta graph should have a non-trivial unnecessary set. This would be the case if and only if there is a non-trivial knot type such that it’s impossible to embed the theta graph so that it contains three copies of this knot. In fact, I’m willing to conjecture that the theta graph has an unnecessary set containing the trefoil (hence the title – intrinsically untrefoiled sort of like intrinsically un unknotted = intrinsically knotted, get it?).

A simple exercise using Zorn’s Lemma shows that there is always a maximal unnecessary set, though I would imagine constructing a maximal one is pretty hard. Also, maximal unnecessary sets aren’t necessarily unique, and probably aren’t ever unique. This may be a frivolous generalization of the theory of intrinsically knotted graphs, but I think at least for the theta graph, it makes an interesting problem.

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I’m interested in a question having a similar flavor: Take a genus two handlebody W and embed in in S^3 in some complicated way. (Require, for instance, that the exterior be boundary-irreducible.) Given a non-separating disc A in W, cutting W along A gives the regular neighborhood of a knot K_A in the 3-sphere. (Scharleman math.GT/0603705 calls this “refilling the meridian A”.) So we can ask, given two non-separating discs A and B in W which can’t be isotoped to be disjoint, can both K_A and K_B be the unknot? If so, what is the relationship between A and B in W?

To phrase this like the problems in the post: in addition to considering isotopy of the theta graph, allow edge slides. You might then also want to restrict the embeddings of the theta graph (eg requiring the exterior to be boundary irreducible) or the discs A and B. In the post, A and B were required to be disjoint, but as in the question above, you might require them to intersect (essentially). The question is then, can you get the unknot in two ways? Scharlemann’s paper and (at the risk of self-promotion) my paper 0709.4051 answer some related questions, but the question (with my initial hypotheses) about K_A and K_B both being unknots is still open. I’m open to suggestions!

It’s maybe worth noting that a number of classical knot theory operations (eg. changing a crossing) can be modeled by the operation described in the first paragraph. It’s also a special case of “two-handle addition” which is similar to Dehn surgery. So the questions aren’t all that strange!

Comment by Scott Taylor — January 29, 2008 @ 12:58 am |