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Let $\mathcal C$ be a [added later: semi-simple] tensor category, and let $A=(A,m:A\otimes A\to A,i:1\to A)$ be an algebra object in $\mathcal C$. The algebra is...

Separable if there is an $A$-$A$-bimodule map $\Delta:A\to A\otimes A$ such that $m\circ\Delta=\mathrm{id}_A$

Frobenius if there is an $A$-$A$-bimodule map $\Delta:A\to A\otimes A$ that is coassociative and counital.

I'm wondering whether separable implies Frobenius.

I can show that separable implies coassociative but I suspect that separable does not imply counital. I'm having a hard time finding counterexamples.



PS: by a "tensor category", I mean a category that is monoidal (not necessarily symmetric) and linear over some field $k$.

PS2: In a previous version of this question, I had added the condition that $\mathcal C$ be rigid. I'd be actually more interested to have counterexamples where the ambient category $\mathcal C$ is semi-simple. I understand that there are many things one might mean by "semisimple" (e.g., is the category of all vector spaces over $k$ semisimple?), I'm deliberately keeping things vague to allow for more counterexamples.

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  • $\begingroup$ Your question might be a duplicate: mathoverflow.net/questions/1939/… $\endgroup$ – Chris Schommer-Pries Aug 25 '15 at 15:04
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    $\begingroup$ My question is not a duplicate. The question you linked shows that ($A$ is separable) + (extra assumptions on $A$) + (extra assumptions on $\mathcal C$) implies ($A$ is Frobenius). $\endgroup$ – André Henriques Aug 25 '15 at 15:51
  • $\begingroup$ I miss-read your question as being about classical algebras in R-modules. You didn't specify what you are assuming about the ground ring and naturally I assumed you were working over a field in which case those extra assumptions are automatically satisfied. Now I see you want to work in an arbitrary tensor category. I wold be surprised if the answer were known in this generality. $\endgroup$ – Chris Schommer-Pries Aug 25 '15 at 16:49
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    $\begingroup$ For example what precisely do you mean by a tensor category? Different schools use the term differently. Do you mean just a monoidal category? A rigid monoidal category? Is is $k$-linear? is it Abelian? is the tensor product assumed to be exact? etc, etc. $\endgroup$ – Chris Schommer-Pries Aug 25 '15 at 20:01
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    $\begingroup$ I'm trying to understand under what conditions one can expect separable to imply Frobenius. The example of Gabriel Drummond-Cole indicates that, unless one assumes $\mathcal C$ to be rigid (= have duals), there is little hope to get such a result. So I'll add rigid to my assumptions. $\endgroup$ – André Henriques Aug 25 '15 at 21:40
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You should be able to get a (perhaps unnatural) counterexample by constructing a tensor category (or algebra object) with no maps to $1$. Let's try to make the most trivial such counterexample possible.

Let $C$ have two objects $A$ and $1$ with one non-identity morphism $i:1\to A$. Equip this with a structure where $1$ is the unit and $A\otimes A=A$. I believe these make $C$ unambiguously into a tensor category.

There are then unique maps $m:A\otimes A\to A$ and $\Delta:A\to A\otimes A$, both of which are the identity on $A$. Then $m$ and $i$ seem to make $A$ an algebra object in $C$ which trivially satisfies the separability condition. But this can't be counital because there are no maps $A\to 1$.

This is obviously a bit unnatural but the only "natural" tensor categories I know offhand where objects don't map to the unit are ones with disjoint union... But they can't supply any counterexamples because multiplication is forced to be the fold map but then you can't make a bimodule map $A\to A\sqcup A$ unless $A$ is empty.

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    $\begingroup$ One way to construct this is to look at the subsemigroup $\{0,1\}$ of the multiplicative semigroup of $\mathbb R$ with its induced order, turning it into a category as a poset and as a monoidal category using the semigroup structure. $\endgroup$ – Mariano Suárez-Álvarez Aug 25 '15 at 18:46
  • $\begingroup$ Wonderful. This is very instructive. I would like, however, to have some less "degenerate" counterexamples... $\endgroup$ – André Henriques Aug 25 '15 at 21:42

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