August 13, 2007

Posted by Urs Schreiber Hendryk Pfeiffer asked me to forward the following question to the Café.

Dear $n$-category people,

I have a question about tensor categories on which I would appreciate comments and references. As probably several people are interested in this, I decided to ask this question here.

The short version of my question is:

Are there examples of $k$-linear additive spherical categories that are non-degenerate, but not semisimple?

In more detail:

I am interested in $k$-linear additive spherical categories where $k$ is a field. The notion of a spherical category was defined in

 J. W. Barrett, B. W. Westbury, Spherical categories, Adv Math 143 (1999) 357, hep-th/9310164

A spherical category is a pivotal category in which left- and right-traces agree. A pivotal category is, roughly speaking, a monoidal category in which each object X has a specified dual $X^*$ and in which $X^{**}$ is naturally isomorphic to $X$. The details are in

 P. Freyd, D. N. Yetter, Coherence theorems via knot theory, J Pure Appl Alg 78 (1992) 49

In a $k$-linear spherical category, the trace defines bilinear maps

\begin{aligned} \mathrm{tr}: Hom(X,Y) \otimes Hom(Y,X) &\to k \\ f \otimes g &\mapsto \mathrm{tr}_Y (f\circ g) \end{aligned}

A $k$-linear spherical category is called non-degenerate if all these traces are non-degenerate, i.e. if for any $f:X \to Y$ the following holds:

$\mathrm{tr}_Y (f\circ g)=0 \;\mathrm{for}\; \mathrm{all}\; g:Y \to X \; \mathrm{implies} f=0 \,.$

As usual, an object $X$ is called simple if $\mathrm{Hom}(X,X)=k$, and the category is called [finite] semisimple if every object is isomorphic to a finite direct sum of simple objects [and if the set of isomorphism classes of simple objects is finite].

The following implication is known to hold:

If a $k$-linear additive spherical category is finite semisimple, then it is non-degenerate.

In

 J. W. Barrett, B. W. Westbury, Invariants of piecewise-linear 3-manifolds, Trans AMS 348 (1996) 3997, hep-th/9311155

the property of being non-degenerate is part of the definition of semisimple. With the more standard definitions I used above, however, one has to prove this. The proof is completely analogous to the one for ribbon categories in Section II.4.2 of

 V. G. Turaev, Quantum invariant of knots and 3-manifolds, de Gruyter, 1994.

In order to understand why the converse implication fails, I am interested in learning about examples of k-linear additive spherical categories that are

(1) non-degenerate and not finite semisimple

(2) non-degenerate and not semisimple

(3) non-degenerate, not semisimple, and which are of the form A-mod for some finite-dimensional k-algebra A

I should be grateful for any sort of comments.

Hendryk Pfeiffer

Posted at August 13, 2007 4:57 PM UTC

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I feel like (3) cannot be possible. If you look at the endomorphism induced by an element of the Jacobson radical on any representation, there’s no way that could have non-zero trace, since it induces the 0 map on the associated graded by any filtration with simple quotients.

More generally, this should imply that if your category is non-degenerate, then the endomorphism rings of all objects with a finite composition series must have semi-simple endomorphism rings. I think this should imply that all objects of finite length must be semi-simple (if you have an object of finite length $M$ that isn’t semi-simple, then there’s a non-split injection of a simple object $N\to M$. Consider the endomorphism algebra $\mathrm{End}(M\oplus N)$. The inclusion of $N$ into $M$ considered as an element of this algebra should be in the Jacobson radical. Q.E.D.).

Of course, if you’re willing to leave Artinian categories, you might get some more interesting stuff.

Posted by: Ben Webster on August 13, 2007 7:45 PM | Permalink | Reply to this

You seem to be thinking about matrix traces.

Consider a finite dimensional commutative algebra. Then any linear functional is a trace.

Posted by: Bruce Westbury on August 14, 2007 6:19 AM | Permalink | Reply to this

But can that trace extend to a trace on the category of representations of said algebra?

The only assumption I made about the traces on endomorphism algebras induced by a spherical structure is the following (this could be false. I’ve sketched a proof in my head, but am fairly willing to believe that proof is wrong):

Fact?: If we have an exact sequence $M'\to M \to M''$ and $f:M\to M$ is a morphism which preserves $M'$, then $\mathrm{tr}(f,M)=\mathrm{tr}(f,M')+\mathrm{tr}(f,M'').$

If this true, then I stand by my previous post. If it’s not, then I guess I’ll just have to accept that traces on spherical categories are weirder than I thought.

Posted by: Ben Webster on August 15, 2007 3:44 AM | Permalink | Reply to this

Can we have some clarification on the wording of the question?

When you say $k$-linear do you require that each Hom-set is finitely generated as a $k$-module?

Do you want to assume that you have taken the idempotent completion (if necessary)?

Do you want to say that $k$ has characteristic zero?

Posted by: Bruce Westbury on August 13, 2007 8:05 PM | Permalink | Reply to this

Yes, I’d like all Hom-sets to be finitely generated k-modules. Idempotent completion is not required and k may be any field.

Posted by: Hendryk Pfeiffer on August 14, 2007 7:58 PM | Permalink | Reply to this

A simple example of a spherical category which is non-degenerate and which does not satisfy your definition of being semi-simple is the category of even dimensional vector spaces with the one dimensional vector space. What has gone wrong here is that there are objects missing.

It seems plausible to me that you could start with a finite semisimple example; choose a simple V, add 2V and remove V. This should then give finite counterexamples.

There also seems to me to be another way this can fail but I have not checked this.

I suspect that there are Frobenius algebras which are not semisimple.

Given a finite non-degenerate spherical category then you can construct an algebra in the usual way by taking the direct sum of the Hom spaces. Then (maybe) this algebra is a Frobenius algebra. Also (maybe) this algebra is semisimple if and only if the category is semisimple.

Posted by: Bruce Westbury on August 15, 2007 6:51 PM | Permalink | Reply to this

Bruce wrote:

I suspect that there are Frobenius algebras which are not semisimple.

Yes, even commutative ones. For example, take

$A = \mathbb{C}[x]/\langle x^n \rangle$

to be the algebra of polynomials in $x$ modulo the ideal generated by $x^n$. Equip it with the linear functional $tr: A \to \mathbb{C}$ for which $tr(x^i) = 0$ for $i \ne n$ and $tr(x^n) = \lambda$ for some nonzero $\lambda \in \mathbb{C}$. It’s easy to see that this gives a commutative Frobenius algebra.

Since examples of this sort can be manufactured ad nauseum, it’s hopeless to classify commutative Frobenius algebras over $\mathbb{C}$ unless we demand that they’re semisimple.

But I urge that you look at page 6 of Steve Sawin’s paper on direct sum decompositions of TQFTs, where Proposition 2 ‘classifies’ the indecomposable commutative Frobenius algebras over $\mathbb{C}$ — at least modulo the classification of commutative algebras with one-dimensional socle.

(Don’t be scared of ‘socles’. In the example I gave, the ‘socle’ of $A$ is the 1d subspace spanned by $x^n$. In general, it’s the space of all guys $a$ such that $a b = 0$ for all nilpotent $b$. Intuitively speaking, they’re the guys at the very ‘top’ of $A$, which vanish into oblivion if you try to push them any higher.)

Posted by: John Baez on September 29, 2009 10:54 PM | Permalink | Reply to this

I believe examples of both 1 and 2 given in Wenzl-Tuba Section 9.1. The example comes from the Kauffman polynomial and can be thought of as the “quantum group” U_q(O_t) where q is a root of unity but t is not an integer.

Was question 3 ever settled? Like Ben my intuition here was that there’d be no examples of 3.

Also it’s worth noting the importance of working in characteristic 0 everywhere here, otherwise semisimple does not imply nondegenerate.

Posted by: Noah Snyder on September 29, 2009 7:44 PM | Permalink | Reply to this