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OK, I have to ask a dumb question again: Where do Lie groups enter in the construction of the Reshitikhin-Turaev invariant? The parts of the proof I understand are that 6j symbols take care of themselves. So why not define 6j symbols axiomatically (Biedenharn-Elliott plus symmetry plus function value at a 0 argument should suffice)? You also need an integer-valued triangle function f(a,b,c) giving the number of "irrep" c in the tensor product of a and b for a start (to construct an infinite version of a fusion category, or whatchamacallit), so {abc|def}=0 if f(a,b,c)=0. Not even talking of multiplicity hell. But these only look like "technical" difficulties to me, but not impossible. Of course, with a Lie group you get the 6j symbols "for free" (eh, semisimple? Otherwise I don't see why you can't use e.g. Vogels general Lie group).

In fact, the last months I constructed general 6j symbols for the E7 series, just for fun, here is one:

{JJA|VVV}=-I*Sqrt[Q20]*Sqrt[Q11]*Sqrt[Q10]*Sqrt[Q02]*Q11/Q32/Q30/Q43/Q22

(V defining, J adjoint, A antisymmetric, Qxy means QuantumInteger[x*m/2+y], where m is the parameter in Westburys "Magic" paper. Insert m=-2/3 to get the SU2 6j. If not, there is a typo...). Unfortunately, with higher irreps involved the 6j symbols lost such "pretty" form, I landed in phase choice hell and stopped.

At the moment I construct 6j symbols without resorting to Lie groups at all, just with diagrams. (I'm not even past the Clebsch-Gordan series of V*V, since the technical difficulties are enormous.) So again, is there any reason why the Reshitikhin-Turaev construction with 6j symbols, but without Lie groups, shouldn't work? A non-constructive existence proof would completely suffice, and I could lay myself to rest finally after 20 years :-)

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1 Answer 1

up vote 8 down vote accepted

Indeed any ribbon tensor category gives knot invariants. There's no reason why the ribbon tensor category has to come from a Lie group. However, we just don't know many other examples of ribbon tensor categories.

It's certainly plausible that you can get knot invariants from quantum versions of other points on the Vogel plane. This is closely related to the theory of finite type invariants and to Bar Natan's program to generalize Etingof-Kazhdan quantization. However, showing that such a braided tensor category exists is harder than just showing that the corresponding Lie algebra object exists (which is itself completely open and hard).

It is an interesting question to try to work out what knot invariants you would get from a quantized version of the E7 series. I don't know of any known results in that direction.

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THX for explaining the "technicalities" - I probably read these names before, but I have already enough trouble with Lie algebras :-) Just for fun again, here are a few values for the q-E7 series I computed (just copypasta): e3=-1/e1^3; (this specializes to E7 series) u=e1*e2+e1*e3+e2*e3;v=1/e1/e2+1/e1/e3+1/e2/e3;w=e1*e2*e3; o=(-1+u*w^2-v*w^6+w^8)/((u-v)*w^4);(unlink) z=(-u+u^2*w^2-u*v*w^2+v^2*w^2-u^2*w^6+u*v*w^6-v^2*w^6+v*w^8)/((-u+v)*w^4);(Hopf‌​) p3=(-1+2*u*v-v^2+u*w^2-u^3*w^2+u^2*v*w^2-u*v^2*w^2+u^3*w^6-v*w^6-u^2*v*w^6+u*v^2‌​*w^6+w^8-2*u*v*w^8+v^2*w^8)/((u-v)*w^5);(trefoil) –  Hauke Reddmann Sep 20 '12 at 10:33
    
Interesting. I would have guessed that the trefoil would already have been too hard. –  Noah Snyder Sep 20 '12 at 13:22
    
I'm a little confused about your notation. In the end you should be getting functions of two variables, one is q and the other is the parameter for the E7 family. –  Noah Snyder Sep 20 '12 at 13:24

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