A pivotal monoidal category is called non-degenerate if the inner product $\left(x,y\right) = Tr\left(xy^{*}\right)$ (where $y^{*}$ is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension $0$). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

  • $\begingroup$ I should mention that abelian is crucial here. There are plenty of examples of non-degenerate idempotent complete pivotal monoidal categories over the complex numbers which aren't semisimple. Quantum groups at very small roots of unity tend to do the trick. Also according to Wenzl-Tuba, U_q(O_t) where q is a root of unity but t is not an integer is another example. $\endgroup$ – Noah Snyder Oct 9 '09 at 23:16

I believe this is Proposition 5.7 in Deligne's “La Categorie des Representations du Groupe Symetrique S_t, lorsque t n’est pas un Entier Naturel”. See also this question.

  • $\begingroup$ Amusingly, I was reading this because Noah told me it might answer one of my MO questions. $\endgroup$ – David E Speyer Jan 5 '10 at 21:42
  • $\begingroup$ Actually I think it was Victor Ostrik who pointed you to that source? $\endgroup$ – Noah Snyder Jan 5 '10 at 23:32
  • $\begingroup$ Oh, you're right. Last night, I read Halverson and Ram, which was your suggestion. $\endgroup$ – David E Speyer Jan 5 '10 at 23:53

To my understanding, the answer is "yes"; at least if everything is sufficiently linear over a decent field k. Isn't this proved in Ulrieke Tillmann's great paper, "S-structures for k-linear categories and the definition of a modular functor"? By the way, one needs to obtain the journal version of that paper; the arXiv version doesn't have the pictures :-(


OK, let me try again. First, a general construction. Let V be an object in an abelian category, A = End(V) and J a finitely generated two-sided ideal of A, with generators (j_1, ..., j_r). Define V/JV to be the cokernel of V^{r} --> V, where the map is multiplication by j_i on the ith coordinate. We need to check that this depends only on J and not on the choice of generators; I haven't actually done this. Define JV to be the kernel of V --> V/JV. So we have a short exact sequence

0 --> JV --> V --> V/JV --> 0

I claim that the action of A on V passes to an action of A/J on V/JV. Also, A acts on JV. (One can say more than this, but we don't need to.)

Using the action of A on JV, we can repeat this construction to get JV/J^2V, J^2 V/J^3 V, etcetera. All of these come with actions of A/J.

Now, suppose that all of our Hom spaces are finite dimensional. A finite dimensional algebra is semi-simple if and only if it has no nontrivial nilpotent two-sided ideal. So, suppose for the sake of contradiction that there is some (V,A,J) as above with J nilpotent. Then J^k V is eventually zero. Trace is additive in short exact sequences, so

Tr(f: V --> V) = \sum Tr(f: J^k V/J^{k+1} V --> J^k V/J^{k+1} V).

If f is in J, the right hand side is 0. Also, if f is in J, so is fg for any g in A because J is an ideal. So J is in the kernel of the trace pairing, and we deduce that J=0.

So all endomorphism rings are semi-simple and, by the lemma cited above, so is the category.

  • $\begingroup$ One obvious potential flaw: am I right that trace is additive in short exact sequences? $\endgroup$ – David E Speyer Oct 12 '09 at 14:47
  • $\begingroup$ Hrm, I'm not sure, I'll think it over tomorrow. $\endgroup$ – Noah Snyder Oct 13 '09 at 6:26
  • $\begingroup$ So I haven't fully checked this, I'm waiting to get my hands on a copy of Bakalov and Kirilov, but it seems likely that trace would be additive in short exact sequences because the pivotal structure should be exact. Thanks, I'll look into it more. $\endgroup$ – Noah Snyder Oct 13 '09 at 19:25

Noah: I believe so. There is a standard lemma: if you have an additive category, in which every idempotent has an image, and in which every endomorphism ring is semi-simple, then the category is semisimple abelian. See, for example, www.ams.org/mathscinet-getitem?mr=1150598

It shouldn't be too hard to show that a ring with nondegenerate trace is semisimple.

  • $\begingroup$ I think this argument doesn't work as stated. Semisimplicity is the same as the "obvious trace" (trace of left multiplication) being nondegenerate, but you can find a non-semisimple ring with a different trace which is non-degenerate (there's some examples in Wenzl-Tuba, but they're not stated in an elementary way). But I'm pretty sure their examples aren't abelian. $\endgroup$ – Noah Snyder Oct 11 '09 at 14:27
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    $\begingroup$ So consider C[x]/x^2 with the trace tr(a+bx) = b. Then (a+bx, c+dx) = tr(ac+(ad+bc)x) = ad+bc. That's nondegenerate. So you definitely need to use the tensor structure somewhere here, not just the individual endomorphism spaces. $\endgroup$ – Noah Snyder Oct 11 '09 at 19:04
  • $\begingroup$ Hmmm, that's irritating. I'll back off of any specific claims then, and just suggest that this lemma about semisimple endomorphism rings might be useful. $\endgroup$ – David E Speyer Oct 12 '09 at 5:06

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