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](http://books.google.com.sg/books?id=MWTuMJOvwzgC&pg=PA96&lpg=PA96&dq=prime+decomposition+of+knots&source=bl&ots=M-bzlGAbuR&sig=edG6JtWWUXdRAT23ejE7trhA84k&hl=en&sa=X&ei=wE9qUYzcHo7prQeU4YHYDA&ved=0CDUQ6AEwAQ#v=onepage&q=prime%20decomposition%20of%20knots&f=false). There is also a smooth proof-- see *e.g.* [Ryan Budney's paper](http://arxiv.org/abs/math/0309427v6). [Recent work of Korablev](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=timm&paperid=761&option_lang=eng) 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.



Diagrammatic proof of unique prime decomposition of knots
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](http://www.ams.org/mathscinet-getitem?mr=2900614) by Ozawa, as well as Problem 10.6 in [this book](http://www.springer.com/mathematics/geometry/book/978-3-642-33301-9).) 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.


