4
$\begingroup$

Suppose that C is a ribbon monoidal category with dominant ribbon functors F_1: C->D_1 and F_2: C->D_2 such that D_1 and D_2 are modular tensor categories, does it follow that D_1 and D_2 are equivalent as MTCs? Here dominant means that every object in the target is a summand of an object in the image of the functor.

This is certainly true if C is premodular (semisimple with finitely many simple objects) as was proved by Bruguieres. What if C is not premodular? I haven't been able to locate a more general statement in the literature.

The particular case I have in mind is where C is the Kuperberg G_2-spider specialized to q a particular root of unity. After semisimplification C is in fact premodular, but actually proving that is likely to be a lot of work (it would require writing down inductive formulas for simples, etc.).

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Noah, this is a comment to your answer to your question: unfortunately your functor is not braided. Indeed the braiding on the square V_2 has 2 eigenvalues and the braiding on
the square of F_2(V_2) has 4 different eigenvalues.. Also, I think that TL_{-1} is related with third (or sixth) root of 1.

Improve this answer
$\endgroup$
0
0
$\begingroup$

I think this is a counterexample to the result I was looking for. Let C be the idempotent completion of TL_{-1}, the Temperley-Lieb category with loop value -1. (Or equivalently, the category of finite dimensional representations of U_q(sl_2) where q is an third root of unity.)

Let D_1 be the quotient of C by negligibles. This category is just Rep(Z/2) but with the dimension of the nontrivial rep being -1.

Let B be the Fibonacci category and let B' be its Galois conjugate, let x and x' be the nontrivial objects. Let D_2 be the Deligne tensor product of B and B'. Let F_2 be the functor sending V_2 to $x \boxtimes x'$, this exists by the universal property of Temperley-Lieb (i.e. $x \boxtimes x'$ is symmetrically self-dual and has quantum dimension -1). This functor is dominant because $F_2(V_2 \otimes V_2)$ has every object of D_2 as a summand.

As Victor Ostrik points out F_2 is not a ribbon functor, so this is not a counterexample.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Despite this not being a counterexample to my question, it still suggests that the result is unlikely to be true. What this counterexample shows is that there are functors of pivotal categories C->D with D semisimple which don't factor through the "semisimplification" (quotient by all negligibles) of C. This stops the most obvious route to a positive answer to my question. $\endgroup$
    – Noah Snyder
    Commented Jan 12, 2010 at 18:48

You must log in to answer this question.

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