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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorKevin Lin
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

This question might be really easy.

So a ($k$-)linear category with a single object is the same as an algebra – namely the endomorphism algebra of the single object.

I want to figure out what a (($k$-)linear) monoidal category with a single object is. However, the definition of monoidal category requires a unit object $1$, and moreover we require $\operatorname{End}(1) = k$ (at least according to Definition 1.1.7(v) on page 12 of Bakalov-Kirillov), which I think makes the situation rather trivial and uninteresting.

So, maybe it’s better to ask, what is a monoidal category with two objects?

Or, what is a monoidal category with a single object when we don’t require $\operatorname{End}(1) = k$? We get the composition $\operatorname{End}(1) \otimes \operatorname{End}(1) \stackrel{\circ}{\to} \operatorname{End}(1)$ and we get the tensor product $\operatorname{End}(1) \otimes \operatorname{End}(1) \stackrel{\otimes}{\to} \operatorname{End}(1)$. Both are associative. Are they the same, by some kind of Eckmann-Hilton argument? Maybe I am being stupid but I don’t see it. In particular, for the Eckmann-Hilton argument to work I need the identity endomorphism of the unit object $1$ to be the unit for the tensor product $\otimes$, but I don’t see how or whether this is true.

• CommentRowNumber2.
• CommentAuthorHarry Gindi
• CommentTimeJul 24th 2010

@Kevin: What do you need a monoidal category with a single object for? It seems like a rather unnatural construction.

• CommentRowNumber3.
• CommentAuthorKevin Lin
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

I’ve sometimes found it helpful, when trying to study something about categories, to first understand the situation for algebras. Right now I want to understand a few things about monoidal categories, so I’d like to start by looking at monoidal-categories-with-a-single-object.

• CommentRowNumber4.
• CommentAuthorHarry Gindi
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

Monoidal categories with only a single object or even with two objects is still going to be degenerate. It’s like asking about groups with one or two elements.

A better idea is to think in terms of monoidal categories generated by some class of objects.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeJul 24th 2010

I think that’s kind of an odd point of view around here, Harry, in view of the delooping hypothesis and stabilization hypothesis and horizontal categorification. One-object things can be very interesting.

• CommentRowNumber6.
• CommentAuthorKevin Lin
• CommentTimeJul 24th 2010

@Harry: What do you mean by “degenerate”?

Algebras (= categories with one object) can certainly be pretty nontrivial things.

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

A monoidal category with only one object is the same as a bicategory with only one object and only one 1-arrow. By Eckmann-Hilton, we get that this is the same as a commutative monoid (and yes, the two compositions agree, and the unit of the monoid is the identity arrow of the given object, which is necessary the tensor unit). The coherence conditions on $\otimes$ take care of the interaction between ). Actually, by Cheng-Gurski’s result on the low end of the periodic tabke, this monoid comes with a chosen invertible element (actually they attribute it to Leinster (math.CT/9901139)), so it is a bit more than just a monoid.

As far as two objects ($I$ and $x$, say) go, it depends on what $Hom(I,x)$ looks like, and if they are isomorphic or not. In this case there may be some interesting interplay for the $k$-linear version, but I haven’t thought about it before.

• CommentRowNumber8.
• CommentAuthorHarry Gindi
• CommentTimeJul 24th 2010

My mistake. I feel like I’m off my game tonight (look at the question I just asked!).

• CommentRowNumber9.
• CommentAuthorKevin Lin
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

@DavidRoberts: Thank you. And yeah, I think I figured out why the unit of the tensor product is the identity morphism of the unique object. It’s because of the unitors.

Follow-up question: What is a braided monoidal category with one object?

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2010

It should again give a commutative monoid, again with some distinguished invertible elements. Intuitively this comes from a tricategory with only one k-cell for k=0,1,2, and in Cheng-Gurski’s arXiv:0706.2307 they show this is indeed a commutative monoid with eight (count them, eight!) distinguished invertible elements. But, and this is important, a braided monoidal category with one object is perhaps not quite a tricategory with only one k-cell for k=0,1,2, because there is likely some coherence information required by the tricategory definition not present in the braided monoidal category.

• CommentRowNumber11.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2010

I might add that this is clear, because a braided monoidal category is a monoidal category after all, just with more structure morphisms. This give a couple more distinguished elements in the one-object case.

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeJul 24th 2010

In my opinion, the distinguished elements should not really be taken seriously here – they’re just an artifact of the particular way we’ve chosen to present a (braided) monoidal category, and they go away when you consider things up to the right notion of equivalence. E.g. the 2-category of pointed one-object monoidal categories is equivalent to the category of commutative monoids, as are the 2-category of pointed one-object braided monoidal categories and the tricategory of pointed one-object one-morphism bicategories.

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeJul 24th 2010

Also, the tetracategory of pointed one-object one-morphism tricategories should be equivalent to that of pointed one-object one-morphism Gray-categories, by coherence for tricategories, and unless I’m mistaken a one-object one-morphism Gray-category is precisely a category with two strict monoidal structures satisfying an interchange law up to coherent isomorphism – and these are known to be equivalent to braided monoidal categories. So I think it really does work at all known levels, when you consider things up to the right kind of equivalence.

• CommentRowNumber14.
• CommentAuthorKevin Lin
• CommentTimeJul 24th 2010

What does “pointed” mean here?

• CommentRowNumber15.
• CommentAuthorTobyBartels
• CommentTimeJul 25th 2010

The adjective ‘pointed’ means that it is equipped with a point (in this case, equipped with an object); see pointed object.

But it’s also important that the morphisms (and higher morphisms) preserve the chosen point in an appropriate way; that’s what makes the extra distinguished elements end up irrelevant.

There is discussion at k-tuply monoidal n-category.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeAug 16th 2010

(From Mike #5)

One-object things can be very interesting.

Indeed. Certainly they can encode a lot of complexity: representations of one-object things are equivalent to representations of their Cauchy completions, and such Cauchy completions can be fairlyy complicated. For example, the Cauchy completion of the monoid of endomorphisms on $X$ is equivalent to the category of nonempty sets whose cardinality is that of $X$ or less. In fact, I wonder which categories arise as Cauchy completions of monoids?

• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeAug 17th 2010

I wonder which categories arise as Cauchy completions of monoids?

Those which are Cauchy complete and contain an object of which every other object is a retract?

• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeAug 17th 2010

Well, yes. I guess you’re really suggesting there’s probably no good answer besides “it is what it is”?

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeAug 18th 2010

I suppose there might be a better answer, but it seems unlikely to me.