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Stream: deprecated: algebraic geometry

Topic: Morita equivalence

Jade Master (May 13 2020 at 03:25):

Why are those two versions of Morita equivalence the same?

Jade Master (May 13 2020 at 03:47):

John Baez said:

Puzzle. Can you guess how a lot of what I just said can be abstracted a bit so it's not about commutative algebras over a field $k$?

I'm not sure if anyone has answered this puzzle. This is all reminding me of the bicategory of categories, profunctors, and natural transformations. I think this is the Set-enriched case of what you're describing? More generally for an enriching category $V$, there is a bicategory of V-enriched categories, V-profunctors, and V-transformations. This is still too general because you need the $V$-categories to have one object. If V is $\mathsf{Vect}_k$, then V-categories with one object are k-algebras, V-profunctors are bimodules, and V-transformations are bimodule homomorphisms.

John Baez (May 13 2020 at 05:03):

Jade Master said:

Why are those two versions of Morita equivalence the same?

John Baez (May 13 2020 at 05:06):

I can get you started on this. Suppose A and B are algebras. Suppose we have bimodules M: A -> B and N: B -> A that are inverses up to isomorphism. We want to show that $\mathsf{Mod}(A) \simeq \mathsf{Mod}(B)$.

John Baez (May 13 2020 at 05:07):

This is the easy direction of the equivalence between the two concepts of Morita equivalence.

John Baez (May 13 2020 at 05:07):

I hope you can guess what I mean by all this, like "a bimodule M: A -> B". If not, ask questions.

Joe Moeller (May 13 2020 at 05:09):

I'll copy some stuff out of Weibel's K-book. Maybe somebody can help me understand this:

We say that two rings $R$ and $S$ are Morita equivalent if Rmod and Smod are equivalent as abelian categories, that is, if there exist additive functors $T \colon Rmod \to Smod$ and $U \colon Smod \to Rmod$ such that $UT \cong id_R$ and $TU \cong id_S$. This implies that $T$ and $U$ preserve filtered colimits. Set P=T(R) and Q=U(S); P is an RS bimodule and Q is an SR bimodule via the maps $R = End_R(R) \to End_S(P)$ and $S = End_S(S) \to End_R(Q)$. Since $T(\oplus R) = \oplus P$ and $U(\oplus S) = \oplus Q$, it follows that we have $T(M) \cong M \otimes_R P$ and $U(N) \cong N \otimes_S Q$ for all M,N. Both $UT(R) \cong P \otimes_S Q \cong R$ and $TU(S) \cong Q \otimes_R P \cong S$ are bimodule isomorphisms.

John Baez (May 13 2020 at 05:09):

Assuming you know what I mean by the question, let's get started on the answer. Can you think of a way to use our bimodule M: A -> B to get a functor from $\mathsf{Mod}(A)$ to $\mathsf{Mod}(B)$, or maybe the other way around? If you can, that'll be a big step.

John Baez (May 13 2020 at 05:10):

I find that stuff from Weibel to be more confusing than how I'd talk about this (obviously: everyone finds themselves clearer than everyone else).

John Baez (May 13 2020 at 05:10):

So I think it's best if you just think about my puzzle instead of reading Weibel. It's not very hard.

John Baez (May 13 2020 at 05:12):

All this stuff is very abstract and simple, I think; I don't think "abelian categories" or "filtered colimits" or "additive functors" have much to do with it.

John Baez (May 13 2020 at 05:13):

Jade Master said:

John Baez said:

Puzzle. Can you guess how a lot of what I just said can be abstracted a bit so it's not about commutative algebras over a field $k$?

I'm not sure if anyone has answered this puzzle. This is all reminding me of the bicategory of categories, profunctors, and natural transformations. I think this is the Set-enriched case of what you're describing? More generally for an enriching category $V$, there is a bicategory of V-enriched categories, V-profunctors, and V-transformations. This is still too general because you need the $V$-categories to have one object. If V is $\mathsf{Vect}_k$, then V-categories with one object are k-algebras, V-profunctors are bimodules, and V-transformations are bimodule homomorphisms.

John Baez (May 13 2020 at 05:15):

Yes! A lot of the Brauer 3-group stuff can be generalized to this. I think we can replace "rings" with "one-object V-enriched categories" (or in other words "monoids in V") in all this stuff, where V is any sufficiently nice category.

John Baez (May 13 2020 at 05:15):

Similarly, "commutative rings" could be replaced with "commutative monoids in V".

John Baez (May 13 2020 at 05:17):

So I'm saying that for any commutative monoid $k$ in a sufficiently nice category V, you'll get a Brauer 3-group of

John Baez (May 13 2020 at 05:17):

[ $k$-algebras, bimodules, bimodule homomorphisms ]

John Baez (May 13 2020 at 05:18):

A $k$-algebra is an object on which $k$ acts, which is also a monoid, where the monoid multiplication commutes with the action of $k$.

Joe Moeller (May 13 2020 at 05:19):

I think the stuff I copied from Weibel goes in the more difficult direction. But I'll try out the puzzle.

John Baez (May 13 2020 at 05:20):

Okay. Some of the difficulty may come from going in what I consider the more difficult direction, but I bet some of it comes him from knowing too much stuff.

John Baez (May 13 2020 at 05:20):

I think if we do the easy direction first, the other direction will seem easier later.

John Baez (May 13 2020 at 05:22):

I'll post the puzzle again:

John Baez (May 13 2020 at 05:22):

Suppose A and B are algebras. Suppose we have bimodules $M: A \to B$ and $N: B \to A$ that are inverses up to isomorphism. We want to show that $\mathsf{Mod}(A) \simeq \mathsf{Mod}(B)$.

I hope you can guess what I mean by all this, like "a bimodule $M : A \to B$". If not, ask questions.

Assuming you know what I mean by the question, let's get started on the answer. Can you think of a way to use our bimodule $M: A \to B$ to get a functor from $\mathsf{Mod}(A)$ to $\mathsf{Mod}(B)$, or maybe the other way around? If you can, that'll be a big step.

Joe Moeller (May 13 2020 at 05:25):

A bimodule M: A -> B is invertible if there is another bimodule N : B -> A such that $M \otimes_B N \cong A$ and $N \otimes_A M \cong B$, right?

John Baez (May 13 2020 at 05:25):

Right!

John Baez (May 13 2020 at 05:26):

There's some conventions involved here. It looks you're saying a bimodule $M: A \to B$ is a left A, right B bimodule. I like that convention.

Joe Moeller (May 13 2020 at 05:29):

You get a functor $\mathsf{Mod}(A) \to \mathsf{Mod}(B)$ by tensoring... it seems you could do it with either M or N...

Joe Moeller (May 13 2020 at 05:30):

I guess it's another convention: whether Mod(A) is left or right A-modules.

John Baez (May 13 2020 at 05:30):

I don't think you could do it with either one.

John Baez (May 13 2020 at 05:30):

Yeah, we have to decide what Mod(A) means.

John Baez (May 13 2020 at 05:31):

I think I'm gonna be old-fashioned and take left A-modules as the default option.

John Baez (May 13 2020 at 05:31):

So how do you get a functor Mod(A) $\to$ Mod(B), exactly?

Joe Moeller (May 13 2020 at 05:32):

Ok, if its left A-modules, then you want to tensor on the left with a B,A-bimodule, which is N.

John Baez (May 13 2020 at 05:33):

Right.

John Baez (May 13 2020 at 05:33):

So let's see, I'm saying tensoring with $N: B \to A$ gives a functor Mod(A) $\to$ Mod(B).

John Baez (May 13 2020 at 05:34):

I think our conventions suck.

John Baez (May 13 2020 at 05:34):

Let's change our minds about something, eh?

John Baez (May 13 2020 at 05:34):

Let's say Mod(A) means right A-modules.

John Baez (May 13 2020 at 05:35):

Now it should be that tensoring with $M : A \to B$ gives a functor Mod(A) $\to$ Mod(B).

John Baez (May 13 2020 at 05:35):

I think this will save us a lot of suffering.

Joe Moeller (May 13 2020 at 05:36):

This is where the pre- and post- subscripting becomes useful, I think.

John Baez (May 13 2020 at 05:36):

Well, I think if $M: A \to B$ means ${}_A M {}_B$, which makes sense, we are committed to reading from left to right.

John Baez (May 13 2020 at 05:37):

And this makes right modules a bit more fundamental!

Joe Moeller (May 13 2020 at 05:37):

$(-)_A \otimes_A {}_A M_B \colon \mathsf{Mod}(A) \to \mathsf{Mod}(B)$ is our functor.

John Baez (May 13 2020 at 05:37):

Right, that's nice.

John Baez (May 13 2020 at 05:38):

I don't think I want to keep writing so many subscripts, using TeX here.

John Baez (May 13 2020 at 05:38):

I can compose bimodules $M$ and $N$ just by writing stuff like $MN$, and it's less stressful to write, though less clear.

John Baez (May 13 2020 at 05:39):

Okay, see if you can do this now:

Suppose A and B are algebras. Suppose we have bimodules $M: A \to B$ and $N: B \to A$ that are inverses up to isomorphism. We want to show that $\mathsf{Mod}(A) \simeq \mathsf{Mod}(B)$.

Joe Moeller (May 13 2020 at 05:40):

John Baez said:

I don't think I want to keep writing so many subscripts, using TeX here.

No, just at critical moments when we're clarifying things, I think.

John Baez (May 13 2020 at 05:41):

Yup.

Joe Moeller (May 13 2020 at 05:42):

So we have MN=1 and NM=1. Our functors are (-)M and (-)N.

Joe Moeller (May 13 2020 at 05:42):

So that's it, these functors form an equivalence.

John Baez (May 13 2020 at 05:43):

Yes! Btw, what thing you just wrote was a white lie?

Joe Moeller (May 13 2020 at 05:44):

=?

John Baez (May 13 2020 at 05:44):

Right. Just checking.

John Baez (May 13 2020 at 05:44):

So the proof of this direction is super-easy and sweet.

John Baez (May 13 2020 at 05:45):

Here's one thing you sorta glossed over, which is still easy.

John Baez (May 13 2020 at 05:45):

It's that when we turn bimodules into functors as we've just done, composition of bimodules goes to composing functors.

Joe Moeller (May 13 2020 at 05:49):

Now we can try going in the other direction.

Joe Moeller (May 13 2020 at 05:50):

That passage from Weibel gives a big hint I think.

John Baez (May 13 2020 at 05:50):

Yeah.

John Baez (May 13 2020 at 05:50):

Sorry, my "Yeah" was to going in the other direction.

Joe Moeller (May 13 2020 at 05:51):

If we have an equivalence of the categories of modules, I think he wants us to produce the representing bimodule by hitting the ring as a module with the equivalence.

John Baez (May 13 2020 at 05:51):

Oh good, that's what I figured we should do.

John Baez (May 13 2020 at 05:51):

My reasoning: now we gotta turn a functor into a bimodule.

John Baez (May 13 2020 at 05:52):

Given a functor Mod(A) -> Mod(B) we gotta create an A,B-bimodule out of thin air.

John Baez (May 13 2020 at 05:52):

We don't have much to work with!

John Baez (May 13 2020 at 05:53):

All we know is that A is an A,A-bimodule and B is a B,B-bimodule.

John Baez (May 13 2020 at 05:53):

So we probably have to use these, and our functor, to dream up an A,B-bimodule.

John Baez (May 13 2020 at 05:54):

How are we gonna do it?

John Baez (May 13 2020 at 05:56):

I'm getting nervous. It's possible this direction won't even work at all unless our functor is a bit "nice".

Joe Moeller (May 13 2020 at 05:58):

Weibel suggests they have to be additive.

John Baez (May 13 2020 at 05:59):

Yeah. I'm not sure. It's sorta sad that the other direction is so easy and this one seems harder.

John Baez (May 13 2020 at 05:59):

But we have to try the obvious thing: take $A \in \mathsf{Mod}(A)$ and hit it with our functor and see what we can get out of it.

John Baez (May 13 2020 at 06:01):

I guess it's important to think about how $A$ is better than your average guy in $\mathsf{Mod}(A)$.

John Baez (May 13 2020 at 06:01):

What are all the extra special things we can say about A as an A-module?

Joe Moeller (May 13 2020 at 06:04):

Tensoring over A with A is equivalent to the identity functor.

John Baez (May 13 2020 at 06:04):

Yes.

John Baez (May 13 2020 at 06:05):

Let's see, that's what's special about A as an A,A-bimodule, right?

John Baez (May 13 2020 at 06:06):

I actually asked what's special about A as a (right) A-module, i.e. as an object of Mod(A).

John Baez (May 13 2020 at 06:06):

The reason I asked is that then, when we hit A with our functor Mod(A) $\to$ Mod(B), we may get an object In Mod(B) that's "special" in some ways.

John Baez (May 13 2020 at 06:07):

Ideally it would be so special that it would turn out to be an A,B-bimodule!

John Baez (May 13 2020 at 06:08):

I'm actually getting a bit more optimistic.

John Baez (May 13 2020 at 06:09):

But I gotta sleep now! Or at least head in that direction.

John Baez (May 13 2020 at 06:09):

This has been lots of fun....

John Baez (May 13 2020 at 18:46):

So I feel pretty optimistic about this question. Let me remind us what we're trying to do. Say we have algebras A and B over some field k. We get categories of right modules Mod(A) and Mod(B). We've already seen that an (A,B)-bimodule M gives rise to a functor

Mod(A) $\rightarrow$ Mod(B)

John Baez (May 13 2020 at 18:47):

namely the one that takes a right A-module X and sends it to X $\otimes_A$ M.

John Baez (May 13 2020 at 18:47):

Now we're trying to reverse this process.

John Baez (May 13 2020 at 18:48):

Given a functor

Mod(A) $\rightarrow$ Mod(B)

we want to find an (A,B)-bimodule that gives it as above (up to natural isomorphism).

John Baez (May 13 2020 at 18:48):

We use the method of "what could possibly work?"

John Baez (May 13 2020 at 18:49):

I think we'll succeed only if our functor is $\mathsf{Vect}_{k}$-enriched.

John Baez (May 13 2020 at 18:51):

But anyway, we're trying to get an (A,B)-bimodule. All we have at our disposal is the god-given (A,A)-bimodule A and the (B,B)-bimodule B and the functor from Mod(A) to Mod(B). So this is what we must use!

How can we do it?

Morgan Rogers (he/him) (May 13 2020 at 19:18):

Well, the left A-module structure on A is encoded in the category of right A-modules in a way that is preserved by any functor!

Morgan Rogers (he/him) (May 13 2020 at 19:19):

Because for each $a \in A$, multiplying on the left by $a$ is a right-module homomorphism

Morgan Rogers (he/him) (May 13 2020 at 19:20):

so F(A), a right B-module, gets a left A-action for free (with a acting by F(a)) :tada:

John Baez (May 13 2020 at 19:37):

Right! Here's how I'd say it: since A is an (A,A)-bimodule, it's a left A-module object in Mod(A) (which recall is the category of right A-modules).

John Baez (May 13 2020 at 19:37):

If we assume F is Vect-enriched, it follows that F(A) $\in$ Mod(B) is a left A-module object in F(B).

John Baez (May 13 2020 at 19:38):

We need the functor to be Vect-enriched to map A-module objects to A-module objects.

John Baez (May 13 2020 at 19:39):

So assume this; then F(A) is a left A-module object in F(B), which is the same as an (A,B)-bimodule.

So my guess is that this is how we cook up our desired (A,B)-bimodule.

Paolo Capriotti (May 13 2020 at 19:44):

I wonder if one could organise the argument like this: First prove that from a single (enriched) equivalence $\mathsf{Mod}(R) \to \mathsf{Mod}(S)$ one can construct a whole family of equivalences between $T$-$R$-bimodules and $T$-$S$-bimodules, pseudonatural in the $k$-algebra $T$. Then invoke the appropriate higher categorical Yoneda lemma to conclude that $R$ and $S$ are equivalent, since their representables are.

John Baez (May 13 2020 at 19:47):

That sounds elegant.

John Baez (May 13 2020 at 19:49):

I was trying to proceed in a more elementary way: directly cook up from any enriched functor $\mathsf{Mod}(R) \to \mathsf{Mod}(S)$ an (R,S)-bimodule.

Paolo Capriotti (May 13 2020 at 19:50):

Yes, I think that would be applying the argument just to $T = R$, because that's what you need to prove the "easy" direction of the Yoneda lemma.

John Baez (May 13 2020 at 19:53):

The dream here would be to get an equivalence of bicategories

[$k$-algebras, bimodules, bimodule homomorphisms]

and

[nice $\mathsf{Vect}_k$-enriched categories, nice enriched functors, nice natural transformations]

where a "nice" category is one that's of the form Mod(A) for some algebra A, and so on. Ideally one would have a good intrinsic characterization of these nice categories, etc. And ideally one could do this for enriched categorie more general than $\mathsf{Vect}_k$-enriched categories.

Then the equivalence of the two definitions of "Morita equivalence" would fall out as a puny corollary. :upside_down:

John Baez (May 13 2020 at 19:54):

I probably don't have the energy to do all this, esp. since someone has probably already done it!

John Baez (May 13 2020 at 19:55):

But it's good for me to think about it, because it's helping me understand Morita equivalence a bit better.

John Baez (May 14 2020 at 20:07):

Here's something that'd be really helpful to understand:

Morita’s Theorem. Let $O$ be an object in an abelian category $\mathcal{A}$, and let $R:=\mathrm{hom}_\mathcal{A}(O,O)$. Then $\mathrm{hom}_\mathcal{A}(O,-)$ is an equivalence of categories between $\mathcal{A}$ and $\mathsf{Mod}(R)$ with inverse functor $O\otimes_R-$ if and only if $O$ is a compact projective generator (‘progenerator’) in $\mathcal{A}$. Furthermore, every equivalence between $\mathcal{A}$ and a category of modules arises in this way.

The proof is explained here. What's nice is that it settles the question "which categories are categories of modules of a ring?", which is good if you want to take the "Morita attitude" and say rings are Morita equivalent iff they have equivalent categories of modules.

John Baez (May 14 2020 at 20:09):

It shows that for any abelian category and any object in it you can get a ring $R$ and a functor from $\mathsf{Mod}(R)$ to your abelian category. Then the question becomes: when is this functor an equivalence? And the theorem gives an answer!

John Baez (May 15 2020 at 00:18):

I would like to prove something like this. Maybe someone has.

Fix a commutative ring $k$. There's a bicategory $\mathsf{Alg}_k$ of

[$k$-algebras, bimodules, bimodule homomorphisms]

and a category $\mathsf{Cat}_{\ast, k}$ of

[ $\mathsf{Mod}(k)$-enriched categories with a chosen object, $\mathsf{Mod}(k)$-enriched functors, $\mathsf{Mod}(k)$-enriched natural transformations]

where $\mathsf{Mod}(k)$ is the symmetric monoidal category of $\mathsf{k}$-modules with its usual tensor product.

John Baez (May 15 2020 at 00:20):

Conjecture. There is a map of bicategories

$\mathrm{Mod} \colon \mathsf{Alg}_k \to \mathsf{Cat}_{*,k}$

sending any $k$-algebra $A$ to $\mathsf{Mod}(A)$, which is $\mathsf{Mod}(k)$-enriched automatically and has as its chosen object $A$ itself.

John Baez (May 15 2020 at 00:23):

And, there's a map of bicategories going back, say

$\mathrm{Alg} \colon \mathsf{Cat}_{*,k} \to \mathsf{Alg}_k$

sending any $\mathsf{Vect}_k$-enriched category $V$ with chosen object $x$ to the $k$-algebra $\mathrm{hom}(x,x)$.

John Baez (May 15 2020 at 00:23):

Furthermore, these maps are adjoints (more precisely pseudoadjoints).

John Baez (May 15 2020 at 00:25):

$\;$

Some of the details may be wrong, or the statement might even require a lot of fixing. But this seems to be idea behind Morita theory: going back and forth between algebras and their categories of representations.

For example, Morita's theorem above says which objects in are the essential image of $\mathrm{Mod}$: that is, which categories are categories of modules of something.

John Baez (May 15 2020 at 00:28):

I wouldn't be surprised if someone had already proved a (corrected) version of this conjecture. Does anyone here know?

Tobias Fritz (May 15 2020 at 00:33):

At the $1$-morphism level, the relevant counterpart of Morita's theorem is the Eilenberg-Watts theorem. The statement is that the relevant $Mod(k)$-enriched functors are (up to isomorphism) exactly those that preserve colimits.

I've worked out the details a couple of years ago as a fun extended exercise for $k = \mathbb{Z}$. If I remember correctly, at the 2-morphism level nothing surprising happens: that 2-functor is a bijection between bimodule homomorphisms and mere natural transformations.

Perhaps The formal theory of Tannaka duality is relevant here? (I haven't read it though.)

John Baez (May 15 2020 at 00:39):

Nice, thanks! I didn't know that theorem.

What I'm trying to do is set up a (pseudo)adjunction between two bicategories that are pretty simple to describe. This adjunction should restrict to an equivalence on some smaller bicategories in some obvious way - namely the sub-bicategories where going forwards and back gets you back where you started (up to equivalence at the object level, or isomorphism at the morphisms level). Then characterizing these sub-bicategories should involve Morita's theorem and the Eilenberg-Watts theorem.

John Baez (May 15 2020 at 00:42):

The technical conditions that show up in this process - the technical conditions in Morita's theorem and the Eilenberg-Watts theorem - are then revealed as the price you pay when you restrict your adjunction to an equivalence.

Reid Barton (May 15 2020 at 12:27):

John Baez said:

and a category $\mathsf{Cat}_{\ast, k}$ of

[ $\mathsf{Mod}(k)$-enriched categories with a chosen object, $\mathsf{Mod}(k)$-enriched functors, $\mathsf{Mod}(k)$-enriched natural transformations]

I'm a bit confused about this setup: you didn't impose any structure relating to the chosen object on the 1- and 2-morphisms, which means that changing the chosen object results in an equivalent object of the 2-category. But then "endomorphisms of the chosen object" can't be a functor back to k-algebras.

Reid Barton (May 15 2020 at 12:29):

Yes, I think it needs some or possibly a lot of fixing :upside_down:

Reid Barton (May 15 2020 at 12:45):

That page tells me how to recover a single ring from a single category that looks like it could be its category of modules, but John wanted to extend this to a functor, ideally part of an adjunction with {k-algebras and bimodules}, and I'm not sure how to choose the 1-morphisms to make that work out.

Reid Barton (May 15 2020 at 12:46):

If you wanted {k-algebras and k-algebra homomorphisms} on the other side, then I think you could just take the 1-morphisms to be functors which preserve the chosen object (up to specified isomorphism).

John Baez (May 15 2020 at 19:25):

Rongmin Lu said:

John Baez said:

Nice! Did you see my "conjecture" above? This sort of thing should interact with your program somehow.

I reckon it's probably Tannaka duality you had in mind, as Tobias Fritz has suggested above.

Tannaka duality relies on a fiber functor, and I don't see that as fundamental in this Morita stuff. Maybe I'm completely wrong! Is the choice of the "progenerator" in Morita's theorem secretly a choice of a fiber functor???

John Baez (May 15 2020 at 19:26):

For convenience, I'll repeat it:

Morita’s Theorem. Let $O$ be an object in an abelian category $\mathcal{A}$, and let $R:=\mathrm{hom}_\mathcal{A}(O,O)$. Then $\mathrm{hom}_\mathcal{A}(O,-)$ is an equivalence of categories between $\mathcal{A}$ and $\mathsf{Mod}(R)$ with inverse functor $O\otimes_R-$ if and only if $O$ is a compact projective generator (‘progenerator’) in $\mathcal{A}$. Furthermore, every equivalence between $\mathcal{A}$ and a category of modules arises in this way.

John Baez (May 15 2020 at 19:28):

In Tannaka duality we'd do something like choose a fiber functor $p: \mathcal{A} \to \mathsf{AbGp}$ sending each object to an underlying abelian group, and then try to reconstruct $R$ as endo-natural-transformations of $p$.

Reid Barton (May 15 2020 at 19:30):

I think the fiber functor is Hom(O, -) (the AbGp-valued Hom).

Reid Barton (May 15 2020 at 19:30):

By Yoneda, its natural transformations should be the same as those of O

John Baez (May 15 2020 at 19:36):

Okay, great - that sounds believable. If $\mathcal{A} = \mathsf{Mod}(R)$ and we take $O = R$, then it seems to work: for any $R$-module $M$ the abelian group $\mathsf{hom}(R, M)$ is isomorphic to the underlying abelian group of $M$.

John Baez (May 15 2020 at 19:38):

So apparently the Morita crowd prefer to talk about "compact progenerators" - specially nice objects in some abelian category $\mathcal{A}$ - rather than specially nice functors from $\mathcal{A}$ to $\mathsf{AbGp}$. The Tannaka crowd prefers to talk about specially nice functors!

John Baez (May 15 2020 at 19:39):

By the way: did Morita and Tannaka know each other? Did one of them influence the other a lot?

John Baez (May 15 2020 at 19:45):

Morgan Rogers said:

Is there a reason why you call $p,q$ projections rather than idempotents? I suppose projections are always idempotents, maybe it's just a matter of personal preference.

I'll just wildly guess it's because David has been working with C-algebras a lot, and there they like self-adjoint idempotents, which are usually called 'projections'. (The orthogonal projection onto a closed subspace of a Hilbert space is the motivating example, where our C-algebra consists of all bounded operators on that Hilbert space.)

John Baez (May 15 2020 at 19:49):

Reid Barton said:

That page tells me how to recover a single ring from a single category that looks like it could be its category of modules, but John wanted to extend this to a functor, ideally part of an adjunction with {k-algebras and bimodules}, and I'm not sure how to choose the 1-morphisms to make that work out.

Yeah, I'm not quite sure. The Eilenberg-Watts theorem must be part of the answer.

John Baez (May 16 2020 at 23:16):

We've been talking about how for any commutative ring $k$ there is a bicategory with

• algebras over $k$ as objects,
• bimodules as morphisms,
• bimodule homomorphisms as 2-morphisms.

This is a monoidal bicategory, since we can take the tensor product of algebras, and everything else gets along nicely with that.

John Baez (May 16 2020 at 23:16):

Given any monoidal bicategory we can take its core: that is, the sub-monoidal bicategory where we only keep invertible objects (invertible up to equivalence), invertible morphisms (invertible up to 2-isomorphism), and invertible 2-morphisms. The core is a monoidal bicategory where everything is invertible in a suitably weakened sense, so it's called a 3-group.

I call the particular 3-group we get from a commutative ring $k$ its Brauer 3-group, and denote it as $\mathbf{Br}(k)$.

John Baez (May 16 2020 at 23:17):

I just wrote a bunch about it here:

• The Brauer 3-group.

I had a question and Jacob Lurie answered it... but it turns out Noether had already thought about this in 1929.