On Birman-Wenzlyfying the B2 spider

Prelude: First of all, let "S matrix" denote "an abstract 4D tensor satisfying the usual isotypy rules (with no arrows!)". I'm busy trying to classify all possible S matrices (paper pending) - it's just computation and no topology :-)

Lately, I was playing with Kuperbergs G2/B2 constructions. As standalones (defining some reduction rules for trivalent graphs and prove they define an invariant) they might be one of the few possible, but I'm working the reverse way: Again I start with a 3D tensor representing a trivalent node and check the rules by stupid computing.

But I said Birman-Wenzl style: All rules I write using 2+2 tensors (see my "Braid*Temperley-Lieb=?" thread), so we now have S (the braid generator), H (the Temperley-Lieb generator) and U, the spider generator. It looks this way: |=| The double line just signifies it's something different from a single line :-) Technically, U is the tensor product of two trivalent nodes: |=| = >= * =<. (Only the second = is an equal sign :-) Rotate U by 90 deg to get T, now our elements are complete and we can list the basic rules: Pic (Completed by the whirl type rules - like H1H2T1=H1U2 - and the Reidemeister 3 type rules - S1S2U1=U2S1S2 etc.) You have (for now) the free parameters o,z,w,W,a,A,B.

As you might have noted, in this approach there are no natural rules for reducing 4- and 5-gons. But again, since I start with the tensors, I could work out backwards what the rules would be. But the nice thing is: I can take almost (see below!) any working S matrix, and for each of the eigenvalues of S I can find a working U. This gives tons of B2 spider analogues (of course "the" B2 spider is among them).

But here comes my question. One type of S matrix refuses to be augmented to a spider. I hypothesize all below characterizations are equivalent: - Let t be a matrix you can "pull a line over": t1S1S2=S1S2t2. Then t must NOT necessarily be the representation of some actual tangle (one you can actually cut out of a link). - Not for every 2+2 tangle t (cut out of a link) S*t=t*S. - Or: The S matrix can distinguish mutants. - More algebraically: The infinity tangle H can not be written as a sum of finite many powers of S. Does any of the characterizations ring a bell? Here is an example S: S={ {-c^(-1),0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0},
{0,0,0,c*(-c^(-2)+c^2),0,0,0,0,0,0,0,0,-c,0,0,0},
{0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,c^(-3),0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,c^3,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0},
{0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,c^3,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,c^(-3),0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0},
{0,0,0,-c,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-c^(-1)}};

Hauke

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I have spent a lot of time studying skein relations. Your discussion looks very much like things I have thought about. However I don't know what the question is. – Bruce Westbury Apr 12 '11 at 16:01
@Bruce: <blush> Frankly, I don't know it myself, or at least much less that I'd like...In nuce, I want to know what's so special about S matrices that can distinguish mutants. (At least one mathematician said to me that in the framework of the "abstract tensor" construction he wouldn't have expected that they could exist at all!) Here's yet another classification that might be equivalent to all of the above, and is directly computable: Since S.H=w*H, w (the writhe fudge factor) is always an eigenvalue of S. Usually the multiplicity of w is 1. Here it is 2. Hauke – Hauke Reddmann Apr 13 '11 at 12:24
If you start with a representation $V$ of a quantum group so $S\in End(V\otimes V)$ then if $S$ is diagonalisable and generates $End(V\otimes V)$ (as an algebra) then you won't distinguish mutants. This means that if you want to distinguish mutants you had better take a $V$ so that some representation appears in $V\otimes V$ with multiplicity at least two. – Bruce Westbury Apr 13 '11 at 17:30
Could you somehow sharpen the statement from "some representation" to "the representation belonging to w"? E.g. the most simple and natural occuring matrix (that of the Jones polynomial) has one eigenvalue w=z^3 and three -1/z. (Every S matrix I know has multiple eigenvalues.) This would then answer my question (except for the part why exactly this type doesn't allow the spider construction I described - if that's indeed the case). – Hauke Reddmann Apr 14 '11 at 14:38
These multiplicities are accounted for by the representation theory. For the Jones polynomial you take the natural two dimensional representation of SL(2). The second tensor power decomposes (as always) into the symmetric square (of dimension three) and the exterior square (of dimension one). In this case these are irreducible. – Bruce Westbury Apr 20 '11 at 13:34