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Stream: theory: category theory

Topic: Looking for a compact closed category with some properties

Jean-Baptiste Vienney (Feb 21 2024 at 21:06):

I'm looking for a category $\mathcal{C}$ which is:

1. compact closed
2. has binary products (and thus binary coproducts by self-duality)

such that:

3. products distribute over coproducts (i.e. it is like a distributive category but without requiring initial and terminal objects) (note that it implies that coproducts distribute over products by self-duality)
4. theses products are not biproducts (i.e. it is false that for every objects $a,b$, $a \times b \simeq a + b$)
5. $\mathcal{C}$ is not thin (i.e. it is not a preoder)

Note some non-examples:

• every non-trivial abelian lattice-ordered group, considered as a poset, verifies $1.$, $2.$, $3.$ and $4.$ but not $5.$
• the category of finite-dimensional vector spaces over a field $k$ verifies $1.$, $2.$ and $5.$ but neither $3$ nor $4.$

Jean-Baptiste Vienney (Feb 21 2024 at 21:06):

Note also that if a category verifies $1.$, $2.$ and has an initial object, then it can't verify $4.$ because finite products are biproducts in a compact closed category.

Jean-Baptiste Vienney (Feb 21 2024 at 21:07):

Of course, I'm as much interested in a proof that such a category doesn't exist in case it doesn't exist.

Jean-Baptiste Vienney (Feb 21 2024 at 21:11):

I've been trying to somehow categorify an abelian lattice-ordered group to find an example, but I didn't succeed.

Mike Shulman (Feb 21 2024 at 21:13):

I don't think vector spaces satisfy (3). Since the products and coproducts are biproducts, distributivity would mean $a\oplus (b\oplus c) \cong (a\oplus b) \oplus (a\oplus c)$ which can't be true unless $a=0$.

Jean-Baptiste Vienney (Feb 21 2024 at 21:13):

Oh yes, sorry I meant finite-dimensional ones

Mike Shulman (Feb 21 2024 at 21:13):

So did I.

Jean-Baptiste Vienney (Feb 21 2024 at 21:14):

Ok, what you say is true. I corrected this.

Jean-Baptiste Vienney (Feb 21 2024 at 21:16):

I made the mistake because I was thinking to the distributivity of the tensor product over binary products $\simeq$ binary coproducts.

Mike Shulman (Feb 21 2024 at 21:19):

But in your example, you do want the cartesian products to distribute over the cartesian coproducts?

Jean-Baptiste Vienney (Feb 21 2024 at 21:21):

Yes, exactly.

Jean-Baptiste Vienney (Feb 21 2024 at 21:22):

This is like the distributivity of the $gcd$ over the $lcm$.

Jean-Baptiste Vienney (Feb 21 2024 at 21:23):

Or like in any other distributive lattice.

Morgan Rogers (he/him) (Feb 21 2024 at 21:45):

So you have both $a \times (b+c) \cong (a \times b) + (a \times c)$ and $a + (b \times c) \cong (a+b) \times (a+c)$? Taking $a = b=c$ that means that $a \times (a+a) \cong (a \times a) + (a \times a) \cong ((a \times a) + a) \times ((a \times a) + a) \cong ((a+a) \times (a+a)) \times ((a+a) \times (a+a)) \cong \cdots$

Jean-Baptiste Vienney (Feb 21 2024 at 21:47):

It looks correct. This is not so surprising because in a lattice both the product and the coproduct are idempotent.

Morgan Rogers (he/him) (Feb 21 2024 at 21:49):

Do you have any restrictions on how the monoidal product interacts with products/coproducts?

Jean-Baptiste Vienney (Feb 21 2024 at 21:50):

Yes, indirectly. These restrictions come from asking for a compact closed category. In a compact closed category, the monoidal product distributes both over products and coproducts because $a \otimes -$ is both a left and a right adjoint.

Morgan Rogers (he/him) (Feb 21 2024 at 21:56):

I have a feeling that an Eckmann-Hilton type argument will show that the product and coproduct do have to coincide, although that argument does rely on having units...

John Baez (Feb 21 2024 at 21:58):

There's this paper:

• Robin Houston, Finite products are biproducts in a compact closed category.

Jean-Baptiste Vienney (Feb 21 2024 at 21:59):

Of course :) I talked about it me too a few lines above :sweat_smile:

John Baez (Feb 21 2024 at 22:00):

Okay, sorry, I didn't read everything.

Jean-Baptiste Vienney (Feb 21 2024 at 22:02):

Morgan Rogers (he/him) said:

I have a feeling that an Eckmann-Hilton type argument will show that the product and coproduct do have to coincide, although that argument does rely on having units...

This is interesting but be aware that there are examples which verify all the conditions except the one of not being thin to your proof should use the non-thinness.

Morgan Rogers (he/him) (Feb 21 2024 at 22:06):

I noticed that the sketch proof that products and coproducts coincide in Robin Houston's paper seems to use the monoidal unit rather than a terminal object to prove it!

Morgan Rogers (he/him) (Feb 21 2024 at 22:07):

Are you assuming that there is no monoidal unit either?

Jean-Baptiste Vienney (Feb 21 2024 at 22:08):

Sorry, I'm back in 20 minutes or so

Jean-Baptiste Vienney (Feb 21 2024 at 22:18):

I assume that there is a monoidal unit!

Jean-Baptiste Vienney (Feb 21 2024 at 22:19):

Morgan Rogers (he/him) said:

I noticed that the sketch proof that products and coproducts coincide in Robin Houston's paper seems to use the monoidal unit rather than a terminal object to prove it!

I don't think so :sweat_smile:

Jean-Baptiste Vienney (Feb 21 2024 at 22:28):

The counterexample he gives of a compact closed category with binary products but without terminal object, such that the products are not biproducts is the poset $\mathbb{Z}$ with the usual order $\le$ (which happens to also have the distributivity of products over coproducts), $-$ ("minus") for the compact closed structure, with $min$ as the product and $max$ as the coproduct.

Jean-Baptiste Vienney (Feb 21 2024 at 22:30):

The example I was thinking to which is similar is $\mathbb{Q}_{>0}$ with the divisibility, multiplicative inverse for the compact closed structure, gcd and lcm where you extend divisibility, gcd and lcm in the "natural" way from $\mathbb{N}_{>0}$ to $\mathbb{Q}_{>0}$ (I'm thinking to this by using the "prime power factorization" of rational numbers, where you have powers $p^n$ with $n \in \mathbb{Z}$)

Jean-Baptiste Vienney (Feb 21 2024 at 22:32):

In fact the two examples are very related. You have morphisms from the first to the second by making $n \mapsto p^n$ for some prime number $p$

Jean-Baptiste Vienney (Feb 21 2024 at 22:33):

But they are posets!

Mike Shulman (Feb 21 2024 at 22:49):

It looks like in the absence of initial/terminal objects, the part of Houston's argument still goes through showing that there is a canonical isomorphism

$$((A_1\otimes B_1) \times (A_1\otimes B_2)) + ((A_2\otimes B_1) \times (A_2\otimes B_2)) \cong ((A_1\otimes B_2) + (A_2\otimes B_1)) \times ((A_1\otimes B_2) + (A_2\otimes B_2))$$

and therefore also

$$(A\times A) + (B\times B) \cong (A+B) \times (A+B)$$

which we could write as

$$A^2 + B^2 \cong (A+B)^2.$$

If $\times$ also distributes over $+$, then we also have

$$(A+B)^2 \cong A^2 + 2\cdot A\times B + B^2.$$

If the resulting isomorphism

$$A^2 + B^2 \cong A^2 + B^2 + 2\cdot A\times B$$

coincides with the coproduct injection (which seems like not an unreasonable expectation, although I haven't tried to check it), it would follow that if $X$ is any object such that there exists at least one morphism $A^2+B^2 \to X$ (for which it suffices that there exists a morphism $A\to X$ and a morphism $B\to X$), then there exists a unique morphism $2\cdot A\times B \to X$. In particular, therefore, if $\mathcal{C}$ had a weakly initial object $W$, then $2\cdot W\times W$ would be an initial object.

Mike Shulman (Feb 21 2024 at 22:55):

Ah, and since there exists a morphism $A\to A$, there exists a unique morphism $2\cdot A\times A\to A$. But $2\cdot A\times A = A^2 + A^2$, and there are four morphisms $A^2+A^2 \to A$ determined by the two projections $A^2 \to A$. Thus these four morphisms are all equal, and therefore the two projections $A^2 \to A$ are equal, which is to say that $A$ is subterminal. Since $A$ was arbitrary, the category is thin.

Jean-Baptiste Vienney (Feb 21 2024 at 22:58):

Congrats, it looks like you have solved the problem :) It means that there doesn't exist any categorical model of arithmetic in the sense I was thinking to which is not a pre-ordered set.

Mike Shulman (Feb 21 2024 at 23:02):

Well, my argument still depends on verifying that that isomorphism is the coproduct injection.

Jean-Baptiste Vienney (Feb 21 2024 at 23:04):

Oh yes, ok.

Jean-Baptiste Vienney (Feb 21 2024 at 23:27):

If, it works, it would be somehow a coherence theorem: in this class of categories (those which verify $(1)$, $(2)$, $(3)$ if I understand), "all the diagrams commute"... Also, second thing is that, still if it is true, I may be able to find a proof-theoretic proof given that it looks like I have a logic for this class of categories and that it also looks like it has cut-elimination.

Mike Shulman (Feb 22 2024 at 03:57):

Ok, I checked that the isomorphism is the coproduct injection. First, Houston's isomorphism $A^2 + B^2 \cong (A+B)^2$ has component $A^2 \to (A+B)^2$ given by "squaring" the injection $A\to A+B$ and dually. And the distributivity isomorphism $A^2 + 2AB + B^2 \cong (A+B)^2$ is the left-to-right map also defined on $A^2$ and $B^2$ by squaring the injections and on $2AB$ by multiplying the two injections. Thus, the composite

$$A^2+B^2 \to A^2+2AB+B^2 \to (A+B)^2$$

with the injection on the left is also given by squaring the two injections, so it is Houston's map. Inverting the distributivity isomorphism we get the result.

Mike Shulman (Feb 22 2024 at 04:00):

This fact kind of reminds me of the Cockett-Seely observation that a category with finite products and coproducts is both distributive and linearly distributive (with the same products and coproducts) iff it is thin. Obviously it's not the same, since you have a noncartesian monoidal structure as well, but it feels like a sort of similar result about the incompatibility of combining linear structure, classical structure, and proof-relevance.

Jean-Baptiste Vienney (Feb 22 2024 at 04:01):

I didn't know that. And what does "proof relevance" mean?

Mike Shulman (Feb 22 2024 at 04:13):

Roughly, non-thin categories.

Mike Shulman (Feb 22 2024 at 04:13):

(In this context)

Mike Shulman (Feb 22 2024 at 04:14):

If you think of a category as providing semantics for a logic, with objects corresponding to propositions (= types) and morphisms to proofs (= terms), then "proof relevance" means that the category is not thin and therefore can semantically "distinguish" between different proofs of the same proposition.

Jean-Baptiste Vienney (Feb 22 2024 at 04:15):

Hmm ok. This word is a bit dismissive. I don't think the proofs in logics whose semantics is in a poset are uninteresting even if we call them irrelevant. Else the proofs in classical propositional logic would be uninteresting.

Jean-Baptiste Vienney (Feb 22 2024 at 04:15):

But I get the idea.

Mike Shulman (Feb 22 2024 at 04:24):

Yeah, "irrelevant" is one of those words you have to learn to strip away the negative connotations of in your head when it's used with a precise mathematical meaning.

James Deikun (Feb 22 2024 at 04:49):

"connotation-irrelevant semantics"

Morgan Rogers (he/him) (Feb 22 2024 at 07:40):

Jean-Baptiste Vienney said:

Morgan Rogers (he/him) said:

I noticed that the sketch proof that products and coproducts coincide in Robin Houston's paper seems to use the monoidal unit rather than a terminal object to prove it!

I don't think so :sweat_smile:

The sketch proof in the introduction doesn't mention initial or terminal objects in the part about binary products and coproducts, and indeed they don't show up until Lemma 5 (hence Mike's argument).

Jean-Baptiste Vienney (Feb 22 2024 at 14:52):

Morgan Rogers (he/him) said:

Jean-Baptiste Vienney said:

Morgan Rogers (he/him) said:

I noticed that the sketch proof that products and coproducts coincide in Robin Houston's paper seems to use the monoidal unit rather than a terminal object to prove it!

I don't think so :sweat_smile:

The sketch proof in the introduction doesn't mention initial or terminal objects in the part about binary products and coproducts, and indeed they don't show up until Lemma 5 (hence Mike's argument).

My apologies, you were absolutely right.

Jean-Baptiste Vienney (Feb 22 2024 at 16:48):

I would be interested to know if @Mike Shulman, you would want to write your proof in some paper. There is probably a day when I will like to be able to cite a reference for it. I can always cite this Zulip conversation also but it is probably not the best practice.

Jean-Baptiste Vienney (Feb 22 2024 at 16:50):

I guess this is sufficiently interesting to be written in a paper, right?

John Baez (Feb 22 2024 at 16:51):

Could you state the theorem that's been obtained, Jean-Baptiste? I haven't had time to follow all this.

John Baez (Feb 22 2024 at 16:52):

Something much better than citing a Zulip conversation is putting the theorem and its proof on the nLab, and citing that.

Jean-Baptiste Vienney (Feb 22 2024 at 16:54):

If I understand well, the theorem is: "Every compact closed category with binary products such that products distribute over coproducts is thin". (Note that the binary coproducts exist by self-duality. )

Nathanael Arkor (Feb 22 2024 at 17:07):

Are there nontrivial examples of categories with biproducts, for which the biproduct distributes over itself?

Nathanael Arkor (Feb 22 2024 at 17:09):

Presumably the associator for the biproduct exhibit linear distributivity, from which Cockett and Seely's result may be applied?

Mike Shulman (Feb 22 2024 at 17:47):

@Nathanael Arkor Yes, that sounds right to me.

Mike Shulman (Feb 22 2024 at 17:48):

I agree the result should be recorded. I'll think about whether to put it on the nLab or in a paper.

Mike Shulman (Feb 22 2024 at 17:52):

Returning to the original list of criteria, we have:

• An example satisfying 1,2,3,4, but not 5.
• An example satisfying 1,2,5 but not 3 or 4.
• A proof that 1,2,3,5 are inconsistent.
• Of course, examples of 2,3,4,5 abound, like Set.

It doesn't make sense to talk of omitting 2 while satisfying 3 or 4, but how about omitting 3 and satisfying all the others? Can we have a non-thin compact closed category with binary products (and coproducts) that are not biproducts, without asking for distributivity? By Houston's result such a category can't have an initial or terminal object.

Mike Shulman (Feb 22 2024 at 18:04):

I admit I don't have any idea where to go looking for such a thing. Naturally-occurring examples of categories with binary products and coproducts but not an initial or terminal object are hard enough to come by, even without also asking for compact closure. Of course we can't just take a compact closed category that has all finite products and remove the zero object, since then the products would still be biproducts by Houston's result in the larger categroy.

Nathanael Arkor (Feb 22 2024 at 18:12):

(It's probably also worth mentioning that Houston's result is generalised in Garner and Schäppi's When coproducts are biproducts and Zekić's Biproducts in monoidal categories. Perhaps Mike's proof can be strengthened by weakening compact closure similarly.)

Mike Shulman (Feb 22 2024 at 18:33):

Good suggestion. I don't see a way to use their approaches exactly; they seem to use the initial object earlier on. But we can get some mileage by inspecting exactly how much is needed for the particular instance of Houston's Lemma 3 that is used in my proof, which has $A_1=A_2=A$ and $B_1=B_2=I$. I think that means that instead of compact closure it suffices to assume that each functor $A\otimes (-)$ preserves the product $I\times I$ and that the functor $(-)\otimes (I\times I)$ preserves each coproduct $A+A$.

Jean-Baptiste Vienney (Feb 22 2024 at 18:55):

Mike Shulman said:

Can we have a non-thin compact closed category with binary products (and coproducts) that are not biproducts, without asking for distributivity? By Houston's result such a category can't have an initial or terminal object.

I can say that a thin compact closed category with binary products has automatically products which distribute over coproducts in case it is skeletal. It is usually stated as: "every lattice-ordered abelian group is distributive". I think that we can remove the assumption of being skeletal since the distributivity property is preserved by equivalence of categories. So, I think that a compact closed category with binary products without the distributivity is automatically non-thin.

Jean-Baptiste Vienney (Feb 22 2024 at 19:12):

Also every lattice-ordered abelian group with an initial object is automatically trivial. So, almost all the examples don't have an initial object! It is very easy to prove this. Without any category theory, a lattice-ordered abelian group is an abelian group which is also a poset, such that if $x \le y$, then $xz \le yz$ and such that for every two elements, $x,y$ the sup of $\{x,y\}$ exists (and is denoted $x \vee y$). From, there you obtain that $x \wedge y := (x^{-1} \vee y^{-1})^{-1}$ is the inf of $\{x,y\}$ and other properties such that $x \le y \Rightarrow y^{-1} \le x^{-1}$, $(x \vee y).z=(xz) \vee (yx)$ or $(x \wedge y).z=(xz) \wedge (yz)$. It is very easy if I remember to prove that the existence of a smallest element makes all elements equal to it. It is similar to the impossibility of inverting $0$.

Jean-Baptiste Vienney (Feb 22 2024 at 19:13):

If $\bot$ is a smallest element, then I think the main factor in the proof is that $\frac{\bot}{\bot}=1$.

Jean-Baptiste Vienney (Feb 22 2024 at 19:18):

The reason why I started the topic was because I wanted to know how much we can categorify lattice-ordered abelian groups. But the answer is that we can't...

Jean-Baptiste Vienney (Feb 22 2024 at 19:24):

They are very interesting. There is theorem named Krull-Kaplansky-Jaίfard-Ohm theorem which states that "any lattice-ordered abelian group is the group of divisibility of a Bezout domain". Apparently it is a well-known result.

Jean-Baptiste Vienney (Feb 22 2024 at 19:24):

So, they are intrinsically linked to arithmetic.

Jean-Baptiste Vienney (Feb 22 2024 at 19:26):

And now, we have learned they could maybe be studied by purely category-theoretic methods because they are captured by a purely category-theoretic definition (up to equivalence of categories, 1,2,3).

Jean-Baptiste Vienney (Feb 22 2024 at 19:36):

I'm not sure there is anything to do with this category-theoretic definition but at least it is cool

Morgan Rogers (he/him) (Feb 22 2024 at 20:49):

Compact closed is the strongest categorification you could consider. A more relaxed version would be to have an adjunction between the category and its opposite (the power-object adjunction between a topos and its opposite springs to mind as an example of this kind of relationship). It's totally unclear to me what ingredients would be important to maintain with regard to the arithmetic connection, though.

Jean-Baptiste Vienney (Feb 22 2024 at 21:07):

I've just realized something. As I said I'm pretty sure that a compact closed category with binary products has the distributivity in case it is thin. It would mean that:

A compact closed category with binary products has the distributivity of products over coproducts if and only if it is thin.

Jean-Baptiste Vienney (Feb 22 2024 at 21:13):

Morgan Rogers (he/him) said:

Compact closed is the strongest categorification you could consider. A more relaxed version would be to have an adjunction between the category and its opposite (the power-object adjunction between a topos and its opposite springs to mind as an example of this kind of relationship). It's totally unclear to me what ingredients would be important to maintain with regard to the arithmetic connection, though.

Oh, I have an idea. Replacing compact closed by traced symmetric monoidal doesn't change much in the context of arithmetic. A lattice-ordered commutative monoid is traced if and only if it is cancellative. So a traced lattice-ordered commutative monoid is potentially still something very arithmetic.

Jean-Baptiste Vienney (Feb 22 2024 at 21:15):

And a categorification of traced lattice-ordered commutative monoid would be: traced symmetric monoidal category with binary products and coproducts such that products distribute over coproducts.

Jean-Baptiste Vienney (Feb 22 2024 at 21:16):

So a non-thin example without biproducts of this would be great if it exists.

Mike Shulman (Feb 22 2024 at 22:22):

Doesn't $x\vee y$ usually denote the sup, and $x\wedge y$ the inf?

Jean-Baptiste Vienney (Feb 22 2024 at 22:38):

Yes, you're right, I corrected this.

John Baez (Feb 23 2024 at 00:25):

It always annoys me that $\vee$ looks like the Hasse diagram of an inf: one thing that's less than two others.

Jean-Baptiste Vienney (Feb 23 2024 at 00:37):

Yeah, it's true. I have a lot of trouble to remember which one corresponds to what. What I use when trying to find the good one is that $\vee$ is a disjunction and $\wedge$ a conjunction in logic.

Mike Shulman (Feb 23 2024 at 00:43):

I'm so used to it that I don't need a mnemonic any more, but when I was learning I used that $\vee$ looks like $\cup$ for union, and $\wedge$ looks like $\cap$ for intersection.

John Baez (Feb 23 2024 at 00:51):

When I was a kid I had no trouble remembering $\vee$ meant "or", because I knew how "or" was related to $\cup$. But once I learned about Hasse diagrams and that "or" was a sup, life got harder.

Mike Shulman (Feb 23 2024 at 00:51):

Heh.

Mike Shulman (Feb 23 2024 at 00:51):

I guess maybe you can know too much...

John Baez (Feb 23 2024 at 00:53):

Any sort of convention involve mirror images tends to throw me, probably because I'm left handed. If I try to quickly say something like "turn left!" I usually get it backwards, but if I engage my brain a little I get it right. I mean left.

Mike Shulman (Feb 23 2024 at 00:59):

If I were to add this result to the nLab, anyone have an opinion on where it should go? [[distributive category]]?

John Baez (Feb 23 2024 at 01:06):

I think of it as expressing a tension between [[distributive categories]] and [[compact closed categories]], saying it's hard for something to be both. So I'd put it in one of those with proof, and mention it on the other. Which one? Well, Houston's paper is already referred to in [[compact closed categories]], so maybe that one.

Mike Shulman (Feb 23 2024 at 01:43):

Ok, I did that.

Jean-Baptiste Vienney (Feb 23 2024 at 02:09):

Awesome, now I will be able to talk about my little sequent calculus and say that the denotational semantics is in pre-lattice-ordered abelian groups aka "compact closed categories with binary products which distribute over coproducts" by the result on the nLab.