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Stream: learning: questions

Topic: Morita equivalent categories need not be equivalent

John Baez (Mar 30 2025 at 16:45):

Similarly, I would not be surprised if two categories are equivalent if and only if their category algebras are Morita equivalent, i.e. have equivalent categories of representations. I don't actually know if this is true, but if it were true maybe one could argue that the 'category algebra' construction should be thought of, not as a mere functor from the 1-category Cat to the usual 1-category of algebras, but from the 2-category Cat to the 2-category of algebras, bimodules and bimodule homomorphisms.

So I think these issues can be debated, with the 'pro-evil' and 'anti-evil' attitudes to category theory both trying to prove their merit in these applications to fields where people treat categories as humble algebraic objects much like groups or rings.

I'd like to see these arguments be judged, not just by category theorists already committed to the principle of equivalence, but by 'practioners', like people who use category algebras in the study of quivers, or groupoid algebras in noncommutative algebra.

The category theorists can nudge the practitioners, of course, but it would be very interesting to discover if the 'inner logic' of these subjects pushes practitioners toward the principle of equivalence, or not.

Chris Grossack (she/they) (Mar 30 2025 at 18:41):

@John Baez -- It's definitely true that two equivalent categories $\mathcal{C}_1$ and $\mathcal{C}_2$ have morita equivalent category algebras, since their module categories will be equivalent to $\text{Fun}(\mathcal{C}_i, \mathsf{Vect})$, and these functor categories will be equivalent if $\mathcal{C}_1$ and $\mathcal{C}_2$ are. The converse is false since if $G$ and $H$ are two nonisomorphic groups with the same number of conjugacy classes (regarded as inequivalent one object categories) then their categories of representations are nonetheless equivalent (as abelian categories, since they're 2-vector spaces of the same dimension).

Chris Grossack (she/they) (Mar 30 2025 at 18:43):

That said, I remember reading somewhere that for categories that look like posets (I forget exactly what I mean by this) then morita equivalence implies the categories you started with are actually isomorphic! I'll try to find a reference for this if I remember, or maybe someone else will beat me to it.

Chris Grossack (she/they) (Mar 30 2025 at 18:49):

I think it's still an interesting question whether you get a (bi)functor from Cat to the Morita Bicategory, even though we know it won't be faithful...

You would first want to know that a functor $\mathcal{C} \to \mathcal{D}$ gives you a bimodule. I don't quite see how to do this directly (I'm sure it's easy, though), but you can cheat since a functor $F : \mathcal{C} \to \mathcal{D}$ gives you a functor on module categories by left kan extension, and this is probably cocontinuous and additive so that you win by eilenberg-watts.

Then you would want to say that a natural transformation between these functors gives you a bimodule hom, and without having a concrete sense of what these bimodules look like, I'm not sure how to see this... And (embarrassingly) I don't know enough about kan extensions to say for sure that a natural transformation $F \to G$ induces a natural transformation on their left kan extensions... But this is probably true, which would mean that there probably is a 2-functor Cat --> Morita

Chris Grossack (she/they) (Mar 30 2025 at 18:50):

Anyways, this is somewhat off topic, so I'll stop now, haha. If people are interested in saying more about this we can start a new thread ^_^ (Edit: that's exactly what happened, haha)

John Baez (Mar 30 2025 at 19:25):

Chris Grossack (she/they) said:

John Baez -- It's definitely true that two equivalent categories $\mathcal{C}_1$ and $\mathcal{C}_2$ have morita equivalent category algebras, since their module categories will be equivalent to $\text{Fun}(\mathcal{C}_i, \mathsf{Vect})$, and these functor categories will be equivalent if $\mathcal{C}_1$ and $\mathcal{C}_2$ are. The converse is false since if $G$ and $H$ are two nonisomorphic groups with the same number of conjugacy classes (regarded as inequivalent one object categories) then their categories of representations are nonetheless equivalent (as abelian categories, since they're 2-vector spaces of the same dimension).

Thanks for all that. It should have been obvious to me, but it clearly wasn't! But as soon as you said it, it was: it hit my brain like butter melting on a hot pancake.

The concept of two finite groups being equivalent if their categories are equivalent is fairly pathetic, since having the same number of conjugacy classes is so weak. One could instead ask that there be a homomorphism inducing an equivalence of representation categories, and all of a sudden it gets a lot harder.

Someday I want to compile a list of related concepts:

• The groups $\mathbb{Z}/2$ and $\mathbb{Z}/3$ both have 2 conjugacy classes, so their categories of complex representations are isomorphic.
• The quaternion group $Q_8$ and the dihedral group $D_8$ may be nonisomorphic groups with isomorphic character tables, and may be the smallest.
• There are supposedly two nonisomorphic 64-element groups whose categories of complex representations are equivalent as monoidal categories, but not as symmetric monoidal categories.
• The above paper claims two finite groups whose categories of complex representations are equivalent as 'symmetric tensor categories' are necessarily isomorphic. It cites a paper of Deligne and Milne, but proving this result seems to require a version of the Doplicher-Roberts theorem, so I'd like to check into this.
• Two finite groups whose categories of complex representations are equivalent as symmetric monoidal categories over $\mathsf{Vect}$ are necessarily isomorphic by Tannaka duality.
• I have vague clues hinting that the quaternion group $Q_8$ and the dihedral group $D_8$ may be nonisomorphic groups whose categories of representations have isomorphic Grothendieck rings. These are called the [[representation rings]] of those groups.
• We could also look for nonisomorphic groups whose representation rings are isomorphic not just as rings, but as $\lambda$-rings! The $\lambda$-ring depends on some of the symmetric monoidal structure of the representation category, while the ring only depends on the monoidal structure.

Notification Bot (Mar 30 2025 at 20:34):

A message was moved here from #learning: questions > Interrogating equivalence by Chris Grossack (she/they).

Notification Bot (Mar 30 2025 at 20:34):

6 messages were moved here from #learning: questions > Interrogating equivalence by Chris Grossack (she/they).

Simon Burton (Mar 31 2025 at 01:04):

Regarding the two nonisomorphic 64-element groups I'm not sure how you got 64 elements...? i think these groups are well known in quantum computing... The so-called "affine pseudosymplectic group" over $\mathbb{F}_2$ is a quotient of the Clifford group by it's center (roughly speaking), i wrote more about this over here, section 2.3.1 where its called the affine metaplectic group. These are big groups, for 3 qubits (dim(Y)=3 in Theorem 1.2 in the Etingof-Gelaki paper) I'm getting a group of order 92897280, which is the affine part of order $2^6$ (oh maybe this is where the 64 comes from) times the symplectic part which is $Sp(6,\mathbb{F}_2)$ of order 1451520.

David Michael Roberts (Mar 31 2025 at 02:53):

Chris Grossack (she/they) said:

You would first want to know that a functor $\mathcal{C} \to \mathcal{D}$ gives you a bimodule.

Another way to do it might be to factor the functor as an injective on objects functor, and a retract that is an equivalence. This gives a cospan consisting the inj-on-obj functor and a fairly canonical section of the retract that is a quasi-inverse, both of which are injective on objects. You might at least under some summability conditions for the left leg get a cospan of algebras, and from that cook up a bimodule as a direct summand of the middle algebra.

I've only thought about the groupoid case before, though. The 'summability conditions' might be moot, if one is just taking the plain algebraic category algebra, where elements are finite sums. I was caring about the C*-algebra version, so things were nastier

Chris Grossack (she/they) (Mar 31 2025 at 05:51):

I also decided to look into that memory, and I found this mse post which indeed says for finite posets and free categories on graphs morita equivalence implies isomorphism. This is almost certainly what I was remembering

Chris Grossack (she/they) (Mar 31 2025 at 05:52):

Interestingly this looks like some mild generalization of this is false, see the associated mo post.... But I'm too sleepy right now to try and work out what's going on

John Baez (Mar 31 2025 at 05:59):

Simon Burton said:

Regarding the two nonisomorphic 64-element groups I'm not sure how you got 64 elements...?

I got the number 64 from Pasquale Zito's comment on MathOverflow, which also linked to the Etinghof-Gelaki paper we're linking to... but you're right, it's not at all obvious how a 64-element group shows up there, so that comment could even be mistaken.

These are big groups, for 3 qubits (dim(Y)=3 in Theorem 1.2 in the Etingof-Gelaki paper) I'm getting a group of order 92897280, which is the affine part of order $2^6$ (oh maybe this is where the 64 comes from) times the symplectic part which is $Sp(6,\mathbb{F}_2)$ of order 1451520.

That suggests Pasquale Zito got confused. I will query him on MathOverflow.

Kevin Carlson (Mar 31 2025 at 07:03):

Morita equivalence is a phenomenon that has its instantiations in lots of places. The simplest case is set-theoretic Morita equivalence: two small categories are set-theoretically Morita equivalent if they have the same presheaf categories. This turns out to be the same as having the same free completions under splitting of idempotents. In particular, since free categories and posets have no nontrivial idempotents, set-theoretic Morita equivalence for them is (also) the same as category-theoretic equivalence, and since they're skeletal, this is the same as isomorphism.

You can ask about Morita equivalence of $V$-enriched categories for any enrichment base $V$ in just the same way, and John's original question is related to this one for the free $k$-linear (i.e. $k$-vector space enriched) category generated by a category $C$ , which you might call the category algebroid of $C.$ ("Related:" The category algebroid is much more natural than the category algebra, but for some reason normal people think representations should only have one module in them, and actions of the category algebroid on a single vector space are, I think, the same as modules of the category algebra.)

Anyway, $V$-enriched Morita equivalence over any $V$ amounts to equivalence after freely completing under absolute $V$-colimits. Over sets, the "only" absolute colimits are splitting of idempotents, but over linear categories, biproducts are also absolute (that covers all of them over $\mathbb Z,$ and I think also over $k$, though I'm not entirely sure. So you're more or less wondering when the category algebroids of two categories have the same closures under biproducts and idempotent splittings, which amounts to looking at the category of finitely presented free representations. Thus in this case Morita equivalence looks a whole lot closer to equivalence of the full representation categories than it does in the case of set-theoretic Morita equivalence--there are too many $k$-linear absolute colimits!

John Baez (Mar 31 2025 at 16:18):

Yes, for $k$-linear categories where $k$ is a field you can get all absolute colimits by splitting idempotents and taking biproducts. I bet this is true for any commutative ring $k$ but I would want to check with @Todd Trimble on that.

Thus in this case Morita equivalence looks a whole lot closer to equivalence of the full representation categories than it does in the case of set-theoretic Morita equivalence [....]

I know and love this when we have a finite group $G$ and a field $k$. If we think of $G$ as a one-object category, the 'category algebra' construction we'd been talking about is the same as Kevin's preferred 'category algebroid', because there's just one object. People usually call the result of applying this construction the [[group algebra]] $k[G]$, but since we're hotshots we can think of $k[G]$ as a one-object $k$-linear category.

Then we can complete it under absolute colimits. What do we get?

Well, when $k$ has characteristic zero, we get the category of all finite-dimensional representations of $G$!

That's because in this case every representation of $G$ is a biproduct of irreducible representations, and every irreducible representation can obtained from the 'regular representation' $k[G]$ by splitting idempotents.

But we may not get the category of all representations when $k$ has nonzero characteristic! To see this we need to dig into some actual calculations, and see that my previous paragraph works because we can average over the action of $G$ on any representation, which involves dividing by $|G|.$

Indeed, if take the one-object algebroid $k[G]$, split idempotents, and freely throw in biproducts, we get the category of all finite-dimensional representations of $G$ whenever $|G|$ has an inverse in $k$. That means it's either zero or not divisible by the characteristic of $k$.

John Baez (Mar 31 2025 at 16:27):

When we stray away from these cases life gets very scary - to me, anyway. Nobody even knows all the representations of symmetric groups $S_n$ over $\mathbb{F}_p$ when $n$ is divisible by $p$. :scared:

Todd Trimble (Mar 31 2025 at 16:43):

Yes, this is true. For a commutative ring $R$, its Cauchy completion is the category of finitely generated projective modules.

John Baez (Mar 31 2025 at 17:25):

Thanks! I wish the nLab had a list of what Cauchy completions of $V$-enriched categories for various $V$. I suppose I should start it, but right now my list is very short.

Kevin Carlson (Mar 31 2025 at 17:40):

About the only other one I know is that cones and suspensions are absolute in differential graded categories.

John Baez (Mar 31 2025 at 20:23):

I posted a question about this stuff on MathOverflow:

• To what extent can you reconstruct groups from their representations?

and I'm learning some interesting things. Here's the wildest: there exist nonisomorphic finite groups $G, H$ such that their group rings are isomorphic: $\mathbb{Z}[G] \cong \mathbb{Z}[H]$. However, the paper that proves this uses a group $G$ with $2^{21} \cdot 97^{28}$ elements!

This makes Simon Burton's group of order $92897280$ look downright puny. I really want to know why the number $97$ is used here. That's not a number I've ever seen used before.

Kevin Carlson (Mar 31 2025 at 20:32):

It's just the largest two-digit prime, clearly.

Kevin Carlson (Mar 31 2025 at 20:45):

I looked and Hertweck actually explains that $97$ is the smallest prime $p$ for which you can solve $(3+2\sqrt 2)^k=pr+8s\sqrt 2$ in integers. This arithmetic condition helps you produce lots of interesting units in the group ring of the order-$8$ cyclic group, which...helps in some way eventually.

John Baez (Mar 31 2025 at 21:00):

Oh, wow, you beat me to it. This is a crazy-sounding construction!

John Baez (Mar 31 2025 at 21:02):

I like how the question "does the free functor from groups to rings reflect isomorphicness, at least for finite groups?" leads one into the clutches of math that would terrify most category theorists - including me, and I'm barely a category theorist!

Chris Grossack (she/they) (Apr 01 2025 at 00:07):

Kevin Carlson said:

About the only other one I know is that cones and suspensions are absolute in differential graded categories.

Actually, this is a subtlety I've been confused by in the past, which maybe someone here can clarify. By analogy with the $k$-linear case (where cauchy completion gives you fg modules), I assumed that the dg-cauchy-completion would give you perfect complexes... I still suspect this is true in case you're cauchy completing $k$, the dga you started with, but I remember reading somewhere that if you cauchy complete more general dg categories there's actually some ~bonus~ absolute colimits besides cones and suspension that you have to worry about... Does anyone have intuition for what those bonus absolute colimits are?

Chris Grossack (she/they) (Apr 01 2025 at 00:08):

I might ask this in a third, even newer thread if it sounds like people have a lot to say

Kevin Carlson (Apr 01 2025 at 02:09):

That motivated me to dig up the real answer to this question, which is known. The bonus secret weighted colimits, which were seemingly not explicitly described until 2021 by Nikolić, Street (who must be mildly annoyed that nobody else got this problem half a century after he first mentioned it!), and Tendas, are cokernels of protosplit chain maps. These are the chain maps that have a splitting as graded abelian group maps, but not necessarily as chain maps. Their coequalizers are absolute, you can have all the other differential graded absolute colimits without having protosplit cokernels, and every differential graded Cauchy module can be built out of protosplit cokernels, finite biproducts, and cones.

Kevin Carlson (Apr 01 2025 at 02:10):

In particular, I think you do always get your "perfect complexes" description of the Cauchy completion, because a bounded chain complex of finitely generated projectives must be a proto-split quotient of a bounded complex of f.g. frees, but not necessarily an actually split one! These proto-split maps are all over homological algebra.

John Baez (Apr 01 2025 at 03:18):

What's an example?

Kevin Carlson (Apr 01 2025 at 07:03):

Of what, a protosplit map? So for instance the two standard model structures on chain complexes have, respectively, cofibrations that are level wise monos or fibrations that are levelwise epis. If monos (epis) split, then that means every (co)fibration is one of these protosplits. You won’t think that’s a very good example, but that’s a reasonably faithful explanation of why I said “everywhere.”

Kevin Carlson (Apr 01 2025 at 07:05):

More narrowly, when the complex is of projectives, every epi into it splits levelwise, and dually for injectives. And every complex of projectives will have a levelwise (split) epi from frees. That brings us back to the perfect complexes from before.

John Baez (Apr 01 2025 at 16:26):

Thanks, yes I was asking for examples of protosplit maps.