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Stream: learning: questions

Topic: A tensor product of categories?

James Deikun (Oct 22 2024 at 18:27):

I kind of want there to be a tensor product of (enriched) categories like this:

• The monoidal unit is $\mathbf{Set}$ or whatever you're enriched in.
• The product of $[\mathcal C,\mathbf{Set}]$ and $\mathcal D$ is $[\mathcal C, \mathcal D]$.
• The category with one object, because of this, would be an absorbing element.

I've never heard of this product, and I think a lot of people would be interested in it, so there must be some obstruction to its existence. What keeps it from existing?

Mike Shulman (Oct 22 2024 at 19:01):

In general, it's hard to give a meaningful answer to the question of "why doesn't something with such-and-such properties exist". You didn't share any of your reasons for thinking such a product should exist or wanting it to exist, so it's hard to know what kind of answer you would find convincing. However, I think probably the best answer is "completeness and size restrictions".

Rémy Tuyéras (Oct 22 2024 at 19:02):

What about continous functors (as in limit preserving) between the opposite category to the other category?

Specifically, $A \otimes B = \mathsf{Cont}(A^{op},B)$

Mike Shulman (Oct 22 2024 at 19:08):

In the world of "ultimate category theory" where there is a set of all sets, we can talk about "ultimately complete" categories, which have both limits and colimits of all sizes (not just "small" ones, and in this case either the limits or the colimits implies the other). In this world every "ultimately continuous" functor (preserving all limits) is a right adjoint, and every ultimately cocontinuous functor is a left adjoint. Similarly, every ultimately continuous functor to Set is representable, so $\mathrm{Cont}(C,\mathrm{Set}) \cong C^{\mathrm{op}}$.

The category of ultimately complete categories and ultimately cocontinuous functors has not just one monoidal structure but two: it is [[star-autonomous]], where the duality is the ordinary opposite category, and the internal-hom is the category $\mathrm{Cocont}(C,D)$ of ultimately cocontinuous functors. The cotensor product is thus $C ⅋ D = \mathrm{Cocont}(C^{\mathrm{op}},D)$, and so the tensor product is $C \otimes D = (C^{\mathrm{op}} ⅋ D^{\mathrm{op}})^{\mathrm{op}} = (\mathrm{Cont}(C,D^{\mathrm{op}}))^{\mathrm{op}}$. The tensor unit is $\mathrm{Set}$ and the cotensor unit is $\mathrm{Set}^{\mathrm{op}}$, and the trivial category is absorbing for both. So this satisfies all your conditions except that in (2) you have to restrict to limit-preserving functor categories and use the cotensor product instead of the tensor product.

Mike Shulman (Oct 22 2024 at 19:08):

Unfortunately, this is inconsistent by Russell's paradox. As Todd (un)said, you can make sense of it when enriched over 2, so we're talking about the category of [[suplattices]].

Todd Trimble (Oct 22 2024 at 19:09):

You said basically all I was going to say, Mike (for which I'm glad -- saves me time!). I deleted my earlier message since you anticipated everything.

Mike Shulman (Oct 22 2024 at 19:11):

(-:

Mike Shulman (Oct 22 2024 at 19:11):

Another thing you can do is restrict to [[locally presentable categories]] and cocontinuous functors, in which case there is still a tensor product whose unit is Set and for which the trivial category is absorbing, and which satisfies a similar condition on the continuous hom. But the opposite of a locally presentable category is never again locally presentable, unless the category is a poset, so you lose the duality and the cotensor product.

James Deikun (Oct 22 2024 at 19:58):

Hm, I'd be interested in hearing more about the tensor product for LPCs ...

Kevin Carlson (Oct 22 2024 at 20:16):

Can’t you have some ultimately complete categories that aren’t posets if you get rid of choice? Are there not enough of them to make that hypothetical story work nicely?

Kevin Carlson (Oct 22 2024 at 20:17):

Or maybe it’s not even choice but needing to work in NF or something like that?

Rémy Tuyéras (Oct 22 2024 at 20:26):

I guess what @Mike Shulman was suggesting to you, @James Deikun, is to try to determine, among other things, whether you really want a tensor product, or more like a binary operation with properties, and how much "bias" would you accept for that operation? For example, for a tensor product, you might have a choice in how biased that tensor product can be

James Deikun (Oct 22 2024 at 21:07):

Well, I would like a tensor product on all categories, but I'd be perfectly willing to live with something that's a tensor product on "nice enough" categories and acts on "somewhat less nice" categories. If it came with a duality where $\mathbf{Set}$ and its slices were self-dual that would be even nicer.

What I actually want it for is transferring algebraic and coalgebraic structure from a category to other categories. Something like if I have a nice monad on $\mathcal{C}$ I can get a nice comonad on $\mathcal{C}^*$ or a nice monad on $\mathcal{C \otimes D}$.

James Deikun (Oct 22 2024 at 21:12):

I can already kind of do this where "nice categories" are presheaf categories, "somewhat less nice" categories are complete and cocomplete categories (or less if there are some mild conditions on the inheritance of structure), the duality is inherited from $^\mathrm{op}$ and the tensor product from Cartesian product both through $\mathrm{Psh}(-)$, and the "nice monads" are strongly Cartesian. It seems like this isn't the most general this could go on at least two out of the three axes though.

James Deikun (Oct 22 2024 at 21:15):

I should note that I don't need it to be monoidal on the category of categories and all functors -- I don't think my existing one is, for instance.

Mike Shulman (Oct 22 2024 at 21:18):

Kevin Carlson said:

Can’t you have some ultimately complete categories that aren’t posets if you get rid of choice? Are there not enough of them to make that hypothetical story work nicely?

If you get rid of not just choice but excluded middle, there can be non-posetal [[small complete categories]]. But I don't think there are enough of them to have a good theory. In particular, Set is not one of them.

Mike Shulman (Oct 22 2024 at 21:19):

Kevin Carlson said:

Or maybe it’s not even choice but needing to work in NF or something like that?

NF is not real good for category theory, since its category Set is not cartesian closed.

James Deikun (Oct 23 2024 at 16:00):

For $\mathcal E$ a complete category or thereabouts, there seems to be an equivalence between $\mathbf{RAdj}(\widehat{\mathcal C}^\mathrm{op}, \mathcal E)$ and $[\mathcal{C}^\mathrm{op},\mathcal E]$, given by an end over powers in one direction and evaluation on representables in the other.

James Deikun (Oct 24 2024 at 21:49):

Apparently this is described by Chapter 5 of Bird's PhD thesis and it has an action on left adjoint (or equivalently, on right adjoint) functors. It also has a closed structure $\mathbf{LAdj}(-,=)$ and I think the closed structure is determined by duals ($\mathbf{LAdj}(-,\mathbf{Set})$).