Consider for example the approach to the Kuperberg G2 or the Yamada polynomial. I don't know whether relations of graph (in contrast to knot) theory are also usually called "skein" but I simply carry over the term since the gist is the same: You have a set of diagrams thought to be identical outside an area and different inside, and a linear relation between them that "mops up" infinitely many equations and defines an invariant.
To be able to recursively compute the invariant, it must be defined by the skein relation(s) at all. I.e., they can simplify all graphs somewhere. This is a no-brainer for the G2: relations are given for 1-,...,5-gons and by an Euler e+f=k+2 argument any (trivalent) graph must have one. (I omit technicalities.)
And of course the relations must be self-consistent, which boils down finitely many polynomic equations which you solve once and for all.

The obvious problem is that the more complicated the relations look, the more troublesome gets the part with the "always reduce". E.g. HOMFLYPT already lacks a no-brainer proof (my brain taken as gold standard :-).

I have a lot of working S matrices and tricks of "reverse engineering" them into tensors usable for trivalent graphs. For example, one Yabc tensor fulfils skein-relations for reducing 1-, 2- and 3-gons. Also two adjacent 4-gons can be reduced to a single 4-gon (and lower debris). Obviously, this isn't enough for an Euler argument, but everything involving 5-gons was a total computational mess (for now).
But finally the question: To make it general, suppose I have an i-,j- and k-gon with a common corner, or an i- and j-gon with a common edge or just an i-gon as "input". That conformation might be reducable to simpler graphs (with the same number of open ends, of course) with some skein relation, which I denote with ijk+/ij+/i+, or maybe not, which gets a -. So G2 could be labeled 1+2+3+4+5+, and my own example 1+2+3+4-44+.

How can you decide whether a certain labeling fulfils the "all graphs are reducible" condition just by using the Euler argument? (1+2+3+4-44+45+46+55+56+ for example looks like a reducing set, but I'm not too sure about it, since you can liberally speckle the graph with 7-gons.) After all, e+f=k+2 is a global argument and doesn't consider what n-gons are adjacent.


The short answer is you can't. The long answer is that you are working with a presentation. Then a standard approach is that you should first make your relations confluent. You can make them locally confluent by the Knuth-Bendix aka Groebner basis algorithm. This may not terminate so you need some luck here. Once you have a confluent set of relations then you are asking about irreducible graphs. You are then asking if your set of forbidden graphs is unavoidable in the sense this is used in the four colour theorem. The method for establishing this uses discharging.

  • $\begingroup$ Thanks. I vaguely heard from discharging in the framework of the 4 color theorem. (And had to google "confluent", but in any case it's enough to know that my attempted task doesn't belong to the no-brain category :-). Small addendum: Kuperbergs proof for the B2 spider uses an argument about interior angles to prove reduceableness. Is this essentially a variant of "Euler logic"? $\endgroup$ – Hauke Reddmann May 9 '11 at 15:30

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