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Consider a knot to be a diagram in a plane--- i.e. a drawing of a finite connected planar graph (loops and multiple edges allowed) whose vertices are 4-valent with cyclic ordering for the incident edges of each vertex, two non-consecutive of which are called "over" and two "under" (with an additional connectivity condition to ensure there is one component)--- modulo Reidemeister moves. If we can find a disc in the plane such that a representative for the knot K intersects the boundary of the disc at two points, and intersecting K with a disc closes to give a non-trivial knot (and the same for intersecting K with the complement of the disc), then the resulting sub-knot is called a connect summand of K. A knot with no connect-summands is called prime.

Schubert (1949) proved that every knot has a unique prime decomposition, up to permutation of the factors. The proof is geometric and uses Seifert surfaces. See for example Theorem 7.12 of Burde-Zieschang. There is also a smooth proof-- see e.g. Ryan Budney's paper. Recent work of Korablev concerns prime decomposition of virtual knots, where there is no Seifert surface in sight of course, but the proof is also geometric, and involves modifications of thickened surfaces in which the virtual knot is embedded.

Question: Is there a purely diagrammatic proof for unique prime decomposition of knots? (no topology or geometry, definitely no Seifert surfaces, diagrammatic algebra and combinatorics only). Who is it due to?

Unique prime decomposition is a fairly elementary property of knots, so it feels like there should be a diagrammatic proof- but it is also a global property, whereas the diagrammatic approach is local/quantum, so I don't know. There is a weakly analogous problem which I'm looking at for coloured knots, in which there is an easy diagrammatic proof (but the statement is also much weaker in that context).

Note that quantum knot invariants are diagrammatic (Jones polynomial, HOMFLYPT, Kontsevich invariant), and we know that these see a lot, so a direct proof of unique prime decomposition using the Jones polynomial, for example, would qualify.

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Schubert's original proof was in the language of PL manifolds but it is readily adapted to the smooth world. Basically the only technology he was using was the Alexander theorem, that a 2-sphere in $\mathbb R^3$ bounds a disc, and the corresponding theorem for tori in $S^3$, that on one side it bounds a solid torus. It's hard for me to imagine a diagrammatic proof that doesn't go through an argument like the additivity of crossing numbers. – Ryan Budney Apr 14 '13 at 18:17
Thanks for your answer. The statement of unique prime decomposition has purely combinatorial content- one way or reformulating it would be that there is a unique way to partition chords in a Gauss diagram into disjoint connected sets, modulo Reidemeister moves. And we can get a lot of mileage out of Gauss diagrams (e.g. the Homflypt polynomial), so I'm surprised if we strictly need topology and things like crossing numbers. – Daniel Moskovich Apr 15 '13 at 2:10
Incidentally, grid position (or arc index) behaves better with respect to connect sums, since the arc index is additive ($-1$). But the proof of this (by Dynnikov) uses 3-dimensions. – Ian Agol Apr 24 '13 at 22:44
One may also try to reformulated the decomposition problem in terms of braids. – Reza Rezazadegan Oct 7 '14 at 11:34
up vote 5 down vote accepted

I am fairly certain that there's no known diagrammatic proof of the uniqueness of prime decompositions. That answers the "who is it due to?" question.

Of course, a diagrammatic argument may still be out there, waiting to be discovered. But, echoing Ryan's comment, I expect that line of argument to be very difficult, for the following reason.

Suppose you're looking at a diagram $D(K)$ of some composite knot. Perhaps the diagram even shows $K$ to be a connected sum in some fashion. The crux of what you need to show is that there is no alternate prime decomposition besides the one you see. In other words, you would need to show that any prime decomposition of $K$ is visible in some (suitably nice) diagram.

At present, the Jones polynomial and its relatives are only known to place strong restrictions on diagrams for certain classes of knots and links. These include alternating links (where everything is easiest) and, to a smaller extent, adequate and semi-adequate links. It's conjectured that if a semi-adequate knot is composite, every semi-adequate diagram must also be composite -- and that conjecture is probably within reach. (See this paper by Ozawa, as well as Problem 10.6 in this book.) However, any solution is likely to involve essential surfaces, which you want to avoid. Furthermore, the semi-adequate setting is essentially the limit of where current knowledge about Jones-type invariants can reach diagrammatic information.

Without these invariants, you find yourself looking at problems like the additivity of crossing number, which are known to be devilishly hard. In particular, if the additivity conjecture does get solved, the solution would surely involve something beyond purely diagrammatic methods.

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Alternatively, Marc Lackenby has a formalism for normal surfaces that one could view as being almost diagrammatic. He customizes a decomposition of the knot complement to a diagram in a way that allows for normal surface theory. This appears in his recent paper on the additivity of crossing number problem. – Ryan Budney Apr 24 '13 at 19:00
Ryan, that's a good point. In fact, it brings up the grey and amorphous boundary of the "diagrammatic" concept. For instance, take Menasco's proof that prime decompositions of alternating knots must be visible in the diagram. That argument is diagrammatic in flavor, but cut-and-paste topology (of the kind that Daniel seems to want to rule out) is also present. So is this in or out? – Dave Futer Apr 24 '13 at 19:23
Thank you for this answer! So the answer, essentially, is that no such proof exists and that any such proof seems out of reach. So, for now anyway, prime factorization of knots and links is essentially a topological rather than a combinatorial fact. – Daniel Moskovich Apr 25 '13 at 2:18

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