2
$\begingroup$

Let $F$ be a field $\{\otimes,\oplus,R_i\}$ with the one-element $R_1$ (the null- element $R_0$ isn't very relevant). Let $T_{ijk}$ be a "triangle condition" (symmetric in the indices and integer-valued) which is used to define ${R_i}\otimes{R_j}=\oplus_k T_{ijk}*R_k$. Clearly $T_{0jk}=0, T_{1jk}=\delta^j_k$ (although I wouldn't even guarantee for the latter when multiplicities greater than 1 occur...) and the associativity of $\otimes$ forces a lot of further conditions on the $T_{ijk}$. Call this setup a confusion category. :-)

You now have two possibilities: a) finite number of $R_i$ (giving a fusion category - my definition may be nonstandard or incomplete!); b) infinitely many $R_i$ which form the irreps of some Lie group. c): The Vogel plane shows that there might be a third possibility - infinitely many $R_i$ which are NOT the irreps of some Lie group (but maybe generalize them somehow). Question 1: Do all the irreps of the Vogel plane indeed form, provenly, a confusion category? (He only gave the lowest CG expansions, and the computations were hard enough, so it's not obvious to me the process can be continued ad infinitum.)

Question 2: It is probably easier to generalize the $BCD_n$ series in an analogous way, has this been done?

At the moment, I'm after the symmetrized generalization of the $E_7$ series. The first Clebsch-Gordan expansions are $V\otimes{V}=1+A_1+A_2+A_3, V\otimes{A_1}= V+B+C_2+C_3+D_1$ (and cyclic). (I already have computed all CG coefficients, Casimirs and dimensions with maximally one B, C or D occurring.) Since it was enormous work to get even that far and already $A_1\otimes{A_1}$ will have about 15 terms, including a double $A_1$ (aaargh!) - can this form a confusion category too? (If yes, one could do the Reshitikhine-Turaev 6j construction to get a symmetrized 3-variable knot polynomial, and after 30 years I could finally rest in peace :-)

$\endgroup$

1 Answer 1

1
$\begingroup$

For BCD what you want is called the BMW category.

For the Vogel plane, E7, exceptional, etc. everything is totally open.

$\endgroup$
1
  • $\begingroup$ I look up BMW. Open is baaaad. (At least for me, not necessarily for the math community :-) $\endgroup$ Commented Aug 2, 2013 at 11:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .