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Stream: deprecated: mathematics

Topic: Azumaya algebras

John Baez (May 26 2023 at 20:53):

Yes, I'm the guy asking a lot of annoying ring theory questions on the category theory community server! :upside_down:

John Baez (May 26 2023 at 20:53):

But I want this stuff to become category theory.

John Baez (May 26 2023 at 20:56):

So, there are two apparently equivalent definitions of an [[Azumaya algebra]] $A$ over a commutative ring $R$:

• an algebra $A$ for which there's an algebra $B$ such that $B \otimes A$ is [[Morita equivalent]] to $R$.

• an algebra $A$ that's a [[separable algebra]] whose center is $R$.

John Baez (May 26 2023 at 20:57):

The first definition is conceptually nice because there's a monoidal bicategory of

• algebras over $R$
• bimodules
• bimodule homomorphisms

John Baez (May 26 2023 at 20:57):

where the monoidal structure comes from tensor product of algebras over $R$.

John Baez (May 26 2023 at 21:00):

Then the first definition is saying the object $A$ is Azumaya iff it's invertible: there's a $B$ such that $B \otimes A \simeq I$ in this monoidal bicategory.

(I believe this implies $A \otimes B \simeq I$, but I don't see why. I'd better check: if it doesn't, then there's a paper out there giving a bad definition of Azumaya algebra!)

John Baez (May 26 2023 at 21:02):

The second definition sounds less categorical, but actually a separable algebra is one where the multiplication $A \otimes_R A \to A$ has a section (that is, a right inverse) in the category of $A,A$-bimodules.

John Baez (May 26 2023 at 21:04):

And this is very nice! First, it makes sense very generally: for any monoid object $A$ in any monoidal category, we can say it's separable if its multiplication has right inverse in the category of $A,A$-biactions. Second, this right inverse makes $A$ into a [[Frobenius monoid]].

John Baez (May 26 2023 at 21:05):

Proving this is easy and fun with string diagrams.

John Baez (May 26 2023 at 21:06):

So, my question is whether there's an elegant proof that the two definitions of Azumaya algebra agree... a proof that might be generalized to other contexts.

Jean-Baptiste Vienney (May 26 2023 at 21:46):

John Baez said:

The second definition sounds less categorical, but actually a separable algebra is one where the multiplication $A \otimes A \to A$ has a section (that is, a right inverse) in the category of $A,A$-bimodules.

That's probably a very dumb question, but why isn't the map $A \rightarrow A \otimes A$ defined by $x \mapsto 1 \otimes x$ always a section of the multiplication $A \otimes A \rightarrow A$?

Reid Barton (May 26 2023 at 21:56):

I think because it is not a map of A,A-bimodules. It sends $ax$ to $1 \otimes ax$, and not $a \otimes x$.

Jean-Baptiste Vienney (May 26 2023 at 22:16):

I think I was mistaken by the fact that $A \otimes A := A \otimes_{R} A$ and not $A \otimes_{A} A$ so that if $a \not\in r$ indeed $1 \otimes ax \neq (1a) \otimes x = a \otimes x$.

John Baez (May 26 2023 at 22:30):

Yes, when I wrote $A \otimes A$ I meant $A \otimes_R A$. And yes, $x \mapsto 1 \otimes x$ is not a map of $A,A$ bimodules.

Jean-Baptiste Vienney (May 26 2023 at 22:32):

I see what you meant, I felt funny as well

John Baez (May 26 2023 at 22:33):

I decided to delete that "feel funny" business. I don't feel funny anymore. :upside_down:

John Baez (May 26 2023 at 22:35):

Say $A$ is the algebra of $n \times n$ matrices with entries in $R$. Let $e_{i j} \in A$ be the elementary matrices. Then here's a map $A \mapsto A \otimes_R A$ that's a right inverse for multiplication:

$e_{ij} \mapsto \frac{1}{n} \sum_{k=1}^n e_{ik} \otimes e_{kj}$

John Baez (May 26 2023 at 22:35):

But notice, this only works if we can divide by $n$ in $R$.

John Baez (May 26 2023 at 22:36):

So it turns out, for example, that the algebra of $n \times n$ matrices with entries in a field of characteristic $p$ is separable iff $n$ is not divisible by $p$.

Jean-Baptiste Vienney (May 26 2023 at 22:37):

Thanks, nice example!

John Baez (May 26 2023 at 22:39):

Similarly the group algebra $k[G]$ of a finite group is separable iff the order of the group $G$ is not divisible by the characteristic of the field $k$. Nasty things happen when the order of the group is divisible by the characteristic of the field... and that's why nobody knows all the irreducible representations of symmetric groups $S_n$ over finite fields.

This is a shocking hole in our knowledge, and if climate change weren't a problem and I was king of the Earth I would declare it an emergency. :upside_down:

Jean-Baptiste Vienney (May 26 2023 at 22:45):

Is a section given by $g \mapsto \frac{1}{|G|}\underset{h \in G}{\sum} h \otimes (h^{-1}g)$?

Jean-Baptiste Vienney (May 26 2023 at 22:49):

I'm still not sure if it's a map of $k[G],k[G]$-bimodule...

Jean-Baptiste Vienney (May 26 2023 at 22:49):

I'm not used at all to bimodules

Jean-Baptiste Vienney (May 26 2023 at 22:52):

No, it doesn't work.

Jean-Baptiste Vienney (May 26 2023 at 22:53):

A variant of this should work I guess.

Jean-Baptiste Vienney (May 26 2023 at 22:59):

I guess, it's more complicated than that and maybe you use the fact that a $k$-linear representation of $G$ is the same than a $k[G]$-module ?

Jean-Baptiste Vienney (May 26 2023 at 23:04):

Ok, I saw the answer on Wikipedia, and I was very close so a section is
$g \mapsto \frac{1}{|G|}\underset{h \in G}{\sum}(gh)\otimes h^{-1}$

Jean-Baptiste Vienney (May 26 2023 at 23:10):

Funny that it provides a Frobenius monoid, so $k[G]$ is at the same time a Hopf algebra and a Frobenius algebra...

Jean-Baptiste Vienney (May 26 2023 at 23:19):

I'm wondering if there is a class of finite dimensional hopf algebras bigger than such groups algebras which are separable as well because you can define $g \mapsto \underset{h \in H}{\sum}(gh)\otimes S(h)$ for every Hopf algebra $H$ with antipode $S$, now I don't know by what I should replace $|G|$.

John Baez (May 26 2023 at 23:39):

Jean-Baptiste Vienney said:

Ok, I saw the answer on Wikipedia, and I was very close so a section is
$g \mapsto \frac{1}{|G|}\underset{h \in G}{\sum}(gh)\otimes h^{-1}$

You were not merely close! Your answer

$g \mapsto \frac{1}{|G|}\underset{h \in G}{\sum} h \otimes (h^{-1}g)$

is equal to Wikipedia's, and frankly I like your way of writing it better.

John Baez (May 26 2023 at 23:39):

I like this way even better:

$g \mapsto \frac{1}{|G|} \underset{h, k \in G \text{ s.t. } hk = g}{\sum} h \otimes k$

since it doesn't involve arbitrarily choosing a trick for summing over all pairs $h,k$ that multiply to give $g$.

Jean-Baptiste Vienney (May 26 2023 at 23:40):

I thought to that at first but I wasn't sure that there are $|G|$ elements in the sum, so I wrote it in the other way

Jean-Baptiste Vienney (May 26 2023 at 23:43):

Jean-Baptiste Vienney said:

Is a section given by $g \mapsto \frac{1}{|G|}\underset{h \in G}{\sum} h \otimes (h^{-1}g)$?

But this way makes me think that it is maybe a natural map $k[G] \rightarrow k[G] \otimes k[G]$

Jean-Baptiste Vienney (May 26 2023 at 23:44):

If I define $k[f]:k[G] \rightarrow k[H]$ for a group morphism $f:G \rightarrow H$

John Baez (May 26 2023 at 23:46):

Jean-Baptiste Vienney said:

Funny that it provides a Frobenius monoid, so $k[G]$ is at the same time a Hopf algebra and a Frobenius algebra...

I love this fact! They have the same multiplication but different comultiplications. I keep wanting to study this more. I should read this:

• Siqi Zheng, Frobenius algebra structure in Hopf algebras....

Theorem 4.15 says any finite-dimensional Hopf algebra admits a Frobenius algebra structure with the same multiplication. But apparently not every (finite-dimensional) Frobenius algebra has a Hopf algebra structure with the same multiplication.

Jean-Baptiste Vienney (May 26 2023 at 23:55):

Oh, now I think that this comultiplication is $g \mapsto \frac{1}{dim(H)}\underset{h \in B}{\sum}(gh)\otimes S(h)$ for $B$ a basis of $H$

Jean-Baptiste Vienney (May 26 2023 at 23:55):

if $H$ is a finite-dimensional hopf algebra such that $dim(H)$ doesn't divide the characteristic of $k$

Jean-Baptiste Vienney (May 26 2023 at 23:56):

That would be even better if this this paper was written for Hopf monoids

Jean-Baptiste Vienney (May 26 2023 at 23:57):

So we should define what is a finite-dimensional Hopf monoid in a symmetric monoidal category

Jean-Baptiste Vienney (May 26 2023 at 23:58):

and there are good chances that most of the paper will still work

John Baez (May 26 2023 at 23:58):

If $H$ is an arbitrary Hopf algebra you should be summing over a basis, right? I'm not sure your formula is right in general, but it's definitely right for group algebras.

Jean-Baptiste Vienney (May 26 2023 at 23:58):

Oh yes, you're right

John Baez (May 26 2023 at 23:59):

You can define a finite-dimensional object in a symmetric monoidal category to be one that is a [[dualizable object]] - that works pretty well.

Jean-Baptiste Vienney (May 27 2023 at 00:04):

Ok, nice, we should try to see if some facts work for dualizable hopf monoids in a symmetric monoidal $k$-linear category so

John Baez (May 27 2023 at 00:05):

By the way, I fear that business of summing over a basis of the Hopf algebra is not the correct way to define a Frobenius comultiplication on a finite-dimensional Hopf algebra. There's a theory of "integrals" for Hopf algebras, and I think you need to choose an integral for your Hopf algebra to make it into a Frobenius algebra. That's what that paper explains.

Jean-Baptiste Vienney (May 27 2023 at 00:09):

Ok, the crux to obtain the Frobenius algebra structure seems to be their fundamental theorem of Hopf modules.

Jean-Baptiste Vienney (May 27 2023 at 00:10):

(Theorem 4.13)

John Baez (May 27 2023 at 00:11):

Nice.

John Baez (May 27 2023 at 00:36):

So, back to my original question. For any $R$-algebra $A$, the following are equivalent:

1) there's an algebra $B$ such that $B \otimes_R A$ is Morita equivalent to $R$

2) $A$ is separable and its center is $R$?

But what's the slickest proof? Can we find a proof that uses just string diagrams or other category-theoretic machinery?

To do this we should try to work in the symmetric monoidal bicategory $\mathbf{Mod}(R)$ of

• $R$-algebras,
• bimodules between such algebras, and
• bimodule homomorphisms

as much as possible, since that may let us generalize to other (sufficiently nice) symmetric monoidal bicategories.

Condition 1) is clearly phrased in terms of this monoidal bicategory $\mathbf{Mod}(R)$: it's just saying that our object $A$ has an object $B$ that's a "tensor inverse" of $A$.

For condition 2), "separability" is nicely phrased in terms of this bicategory: it's saying that multiplication $A \otimes_R A \to A$, viewed as homomorphism of $A,A$-bimodules, has a right inverse! But what about this business of the center of $A$ being $R$? That sounds hard to express abstractly.

But then I remembered: any separable algebra is a Frobenius algebra! And for any Frobenius algebra $A$ (over a field $R$, at least) there's a god-given way to project it down to its center! We comultiply $A \to A \otimes_R A$, then use the symmetry to "switch" $A \otimes_R A \to A \otimes_R A$, and then multiply $A \otimes_R A \to A$. This gives an idempotent $p: A \to A$ that projects $A$ down to its center. I showed why this works on page 84 here.

So, I think there's a nice way to say the center of $A$ is just $R$ when $A$ is separable. We say this map $p: A \to A$ equals the counit $\epsilon: A \to R$ followed by the unit $\iota : R \to A$.

John Baez (May 27 2023 at 00:39):

So now conditions 1) and 2) are things I can say in any symmetric monoidal bicategory, and I can try to show they're equivalent... or at least show one implication, perhaps.

John Baez (May 27 2023 at 00:41):

In 1) the condition that $A$ have a tensor inverse is a bit annoying because it has an existential quantifier. But in fact the tensor inverse, when it exists, is always $A^{\text{op}}$. That could help.

Jean-Baptiste Vienney (May 27 2023 at 01:13):

Remark: in 2) the condition of separability has also an existential quantifier. So I guess that we could maybe go from $B$ to the section $A \rightarrow A \otimes_{R} A$ and vice-versa.

John Baez (May 27 2023 at 01:55):

That's true. Actually the reason I usually think of "separable algebras" as not having an existential quantifier is that I usually think of the section $A \to A \otimes_R A$ as part of the structure of a separable algebra, just as I think of the comultiplication as part of the structure of a Frobenius algebra. (And indeed, every separable algebra equipped with a specific section gives a Frobenius algebra with a specific comultiplication!)

John Baez (May 27 2023 at 01:58):

But over in condition 1), I am acting like there's no extra structure required. So something is funny. Maybe the point is that instead of simply saying $A^{\text{op}}$ is a tensor inverse of $A$, I should be specifying a specific invertible bimodule from $R$ to $A \otimes A^{\text{op}}$.

John Baez (May 27 2023 at 01:59):

And that's nice because it smells a lot like the section $A \to A \otimes_R A$.

John Baez (May 27 2023 at 02:01):

So I/we should start calculating and see if I/we can get from condition 1), thus reformulated, to condition 2). Or vice versa. Going from 2) to 1) seems a bit easier.

John Baez (May 27 2023 at 02:03):

(I'm willing to work with anyone on this as long as they don't expect me to write a paper with them. It's so tiring to write papers with other people that by now I'd rather just stick their name on the paper and do all the writing myself, or not write a paper at all and let them do whatever they want.)

Jean-Baptiste Vienney (May 27 2023 at 02:13):

I'm a bit lost about the nature of the morphisms. The section $A \rightarrow A \otimes_{R} A$ is a morphism of $(R,R)$-bimodule, right? but now you say that $A \otimes A^{op} \cong R$ is a $(R,R)$-bimodule.

Jean-Baptiste Vienney (May 27 2023 at 02:15):

John Baez said:

I should be specifying a specific invertible bimodule from $R$ to $A \otimes A^{\text{op}}$.

I don't understand this phrase.

Jean-Baptiste Vienney (May 27 2023 at 02:18):

Maybe you wanted to say "a specific invertible bimodule morphism".

Jean-Baptiste Vienney (May 27 2023 at 02:20):

If it's the case, that's exactly what I wanted to suggest too.

Jean-Baptiste Vienney (May 27 2023 at 02:21):

Is there a canonical way to build this morphism?

John Baez (May 27 2023 at 02:21):

Maybe you wanted to say "a specific invertible bimodule morphism".

No, I meant what I said. It's confusing but in definition 1) we are working in the monoidal bicategory of

• algebras over $R$
• bimodules
• bimodule homomorphisms

so there, when I say $A^{\text{op}}$ is a weak inverse to $A$, I mean that there's a bimodule from the algebra $R$ to the algebra $A^{\text{op}} \otimes_R A$, which is an equivalence in this bicategory.

John Baez (May 27 2023 at 02:22):

Note that bimodules are the 1-morphisms in this bicategory!

John Baez (May 27 2023 at 02:23):

If what I'm saying sounds mysterious, it probably pays to read a little about [[Morita equivalence]].

John Baez (May 27 2023 at 02:24):

But I see the nLab starts by explaining Morita equivalence for rings, while I'm using it for $R$-algebras! So that's another way to get confused. :upside_down:

Jean-Baptiste Vienney (May 27 2023 at 02:26):

Yes, in fact if I understand two $R$-algebras $A$ and $B$ are Morita equivalent if there is an equivalence $A \otimes_{R} B \cong R$ in our bicategory, is it right? and I don't want to read more and be confused.

Jean-Baptiste Vienney (May 27 2023 at 02:28):

Now, I guess that the definition of an equivalence in a bicategory is obtained by generalizing the equivalence of categories in the bicategory $Cat$ in terms of objects, 1-morphisms and 2-morphisms in any bicategory?

Jean-Baptiste Vienney (May 27 2023 at 02:45):

Jean-Baptiste Vienney said:

Yes, in fact if I understand two $R$-algebras $A$ and $B$ are Morita equivalent if there is an equivalence $A \otimes_{R} B \cong R$ in our bicategory, is it right? and I don't want to read more and be confused.

About reading, I mean that most of the time it is incredibly more efficient when someone explains you shortly what you want to know in the terms you want to use rather than getting confused by 10 equivalent definitions that are not formulated in the language you prefer. That's not that I'm too lazy to read.

Jean-Baptiste Vienney (May 27 2023 at 04:21):

John Baez said:

(I'm willing to work with anyone on this as long as they don't expect me to write a paper with them. It's so tiring to write papers with other people that by now I'd rather just stick their name on the paper and do all the writing myself, or not write a paper at all and let them do whatever they want.)

I don't know a lot about ring and modules, and I enjoy more writing stuff alone too because that's the only way I can understand some subject. So if you think to me, I would be happy to prove a categorical version of this equivalence, if you had enough energy to answer my questions. That would be a very nice way for me to get my hands on more pure math stuff. Now, I still don't have enough information to do this work. I don't know how to define the section 2) from equivalence 1) and how to define the equivalence 1) from the section 2).

Jean-Baptiste Vienney (May 27 2023 at 04:22):

Maybe if you knew a reference where this equivalence is proved in the classical case of rings, I could understand how to do it by reading.

Jean-Baptiste Vienney (May 27 2023 at 04:25):

I think people have a hard time working with me or supervising me because I always want to have my own catch on the stuff.

Jean-Baptiste Vienney (May 27 2023 at 04:29):

I just try to get as much information from people to learn faster :grinning_face_with_smiling_eyes: and after I want to use it by myself

Jean-Baptiste Vienney (May 27 2023 at 04:36):

Well, I think you wanted somebody to help you and not help somebody and lose a lot of time with that :upside_down:

John Baez (May 27 2023 at 04:41):

Jean-Baptiste Vienney said:

Yes, in fact if I understand two $R$-algebras $A$ and $B$ are Morita equivalent if there is an equivalence $A \otimes_{R} B \cong R$ in our bicategory, is it right? and I don't want to read more and be confused.

Yes. But like lots of people doing bicategories or 2-categories I use $\sim$ to indicate the existence of a 1-morphism that's an equivalence, and $\cong$ to indicate the existence of a 2-morphism that's an isomorphism.

John Baez (May 27 2023 at 04:42):

Jean-Baptiste Vienney said:

Now, I guess that the definition of an equivalence in a bicategory is obtained by generalizing the equivalence of categories in the bicategory $Cat$ in terms of objects, 1-morphisms and 2-morphisms in any bicategory?

Right.

John Baez (May 27 2023 at 04:44):

Jean-Baptiste Vienney said:

About reading, I mean that most of the time it is incredibly more efficient when someone explains you shortly what you want to know in the terms you want to use rather than getting confused by 10 equivalent definitions that are not formulated in the language you prefer.

More efficient for you, sometimes less efficient for the person who has to explain it.

John Baez (May 27 2023 at 04:46):

Jean-Baptiste Vienney said:

Now, I still don't have enough information to do this work. I don't know how to define the section 2) from equivalence 1) and how to define the equivalence 1) from the section 2).

The problem is precisely to figure out those things - if it's possible at all. If I knew those things I wouldn't need help.

Jean-Baptiste Vienney (May 27 2023 at 05:02):

Ahah okay, but it is maybe in papers (for the classical case). I've found one of them when this equivalence is written saying it's a classical result but it doesn't give the proof. It says to look at the bibliography of one of the reference. For the moment, I still haven't found the proof but it may be doable.

Jean-Baptiste Vienney (May 27 2023 at 05:03):

Azumaya categories theorem 1.2 is your equivalence

Jean-Baptiste Vienney (May 27 2023 at 05:03):

(for commutative rings)

John Baez (May 27 2023 at 05:34):

Okay, thanks. I'd probably like to spend at least a few hours trying to figure it out myself before looking at this.

Jean-Baptiste Vienney (May 27 2023 at 05:48):

There isn't a proof just the proposition.

Jean-Baptiste Vienney (May 27 2023 at 05:49):

There is a proof of something which looks similar in the book "Théorie de la descente et Algèbres d'Azumaya", Théorème 5.1. p.93

Jean-Baptiste Vienney (May 27 2023 at 05:56):

They prove that 1') "There exists a $R$-algebra B and a faithfully projective $R$-module $P$ such that $A \otimes_{R} B \cong End_{R}(P)$ as $R$-algebras." is equivalent to 2) "$A$ is separable and its center is $R$."

Jean-Baptiste Vienney (May 27 2023 at 05:57):

I guess that 1') should be equivalent to 1) if you know a bit how to manipulate bimodules.

Jean-Baptiste Vienney (May 27 2023 at 06:39):

Just to save it for myself: they say that 1') is equivalent to 1) in Lemme 7.1 p.26 and refer to H. Bass, Algebraic K-theory, 1968 for a proof.

Nathanael Arkor (May 27 2023 at 11:06):

@John Baez: have you looked at Borceux–Vitale's Azumaya Categories? My impression is that they give the kind of general categorical proof you are looking for (see Theorem 3.4).

Nathanael Arkor (May 27 2023 at 11:08):

Oh, I see now that Jean-Baptiste also mentioned this paper.

John Baez (May 27 2023 at 16:31):

I have looked at it, but I looked at it again 10 minutes ago (before reading you comment!) and noticed that it says this:

Starting from the seventies, several categorical approaches to Azumaya algebras and the Brauer group have been proposed (see the bibliography in [23] for some references). Almost all these approaches deal with Azumaya monoids in a monoidal category V. If one confines one's attention to monoids (i.e. one-object V-categories), the basic results needed to develop the Azumaya-Brauer theory are quite simple: they reduce essentially to Morita theory in monoidal categories. Consequently, the conditions on V can be weaker than those assumed in our paper. For example, in the classical paper by Pareigis [20], the closedness of V is weakened, and in some recent papers by Van Oystaeyen and Zhang (see [22]) and by Alonso Alvarez et al. (see [1]), V is braided but not necessarily symmetric; however, some completeness and biclosedness requirements are needed on V.

John Baez (May 27 2023 at 16:32):

So I looked at [23]:

• Enrico M. Vitale, The Brauer and Brauer-Taylor groups of a symmetric monoidal category, 1996.

John Baez (May 27 2023 at 16:42):

On the top of page 4 this lists four equivalent definitions of an Azumaya algebra in classic case of $R$-algebras, including the two I'm interested in, and threatens to generalize these to "Azumaya monoids" in a pretty general symmetric monoidal category. So that's what I want!

John Baez (May 27 2023 at 16:43):

So, I just need to figure out what they're doing.

John Baez (May 27 2023 at 16:47):

They use very weak assumptions: they're looking at monoids in a symmetric monoidal category with coequalizers that are preserved by tensor products.

Jean-Baptiste Vienney (May 27 2023 at 16:50):

Note than in definition 3.1, c) they define a a splitting monoid as a (non-unital) monoid $A$ which have a section $A \rightarrow A \otimes_{A} A$ of the "canonical map" $A \otimes_{A} A \rightarrow A$ and such that this section and multiplication verify the comultiplication/multiplication diagram of a Frobenius monoid.

Jean-Baptiste Vienney (May 27 2023 at 16:51):

I'm a little bit confused by the fact that they use the tensor product $\otimes_{A}$ and I'm not sure what is this canonical map.

Jean-Baptiste Vienney (May 27 2023 at 16:52):

I'm not even sure what $\otimes_{A}$ means in a general monoidal category.

John Baez (May 27 2023 at 16:53):

This is why they need the coequalizers that I mentioned!

John Baez (May 27 2023 at 16:55):

If $A$ is a monoid in a monoidal category $\mathsf{C}$ with coequalizers, and $A$ acts on the right on $X$ and $A$ acts on the right on $Y$, you can define a thing $X \otimes_A Y$ using a coequalizer, copying the usual construction when $\mathsf{C} = \mathsf{Ab}$.

John Baez (May 27 2023 at 16:56):

Namely, you form the two obvious morphisms $X \otimes A \otimes Y \to X \otimes Y$ and you coequalize them.

Jean-Baptiste Vienney (May 27 2023 at 16:58):

Ok, and $A$ acts on the right on $X$ means that you have a morphism $X \otimes A \rightarrow X$ which verifies the same diagrams than for a set $X$ which would act on a monoid $A$, right?

John Baez (May 27 2023 at 17:02):

Yes, exactly.

John Baez (May 27 2023 at 17:06):

So in a symmetric monoidal category $\mathsf{C}$ with equalizers preserved by taking tensor products with objects, we can talk about:

• monoids in $\mathsf{C}$
• left and right actions of a monoid $A \in \mathsf{C}$
• the category of left actions of a monoid $A \in \mathsf{C}$, and similar for right actions
• $A,B$-bimodules (or 'biactions') of a pair of monoids $A,B \in \mathsf{C}$
• how to take an $A,B$-bimodule and a $B,C$-bimodule and tensor them over $B$ to get an $A,C$-bimodule

and finally the monoidal bicategory of

• monoids in $\mathsf{C}$
• bimodules between such monoids
• bimodule homomorphisms

This is a pretty general context for Azumaya--Brauer--Morita theory.

Jean-Baptiste Vienney (May 27 2023 at 17:09):

Nice, I think I understand how to define this stuff! Just a question: do you really mean "right adjunctions" or is it "right actions"?

John Baez (May 27 2023 at 17:13):

I meant actions. Interesting mistake - you can see how much I type "adjunctions"! I'll fix that, for other people reading.

John Baez (May 27 2023 at 17:14):

This abstract theory is very pretty and I'm basically glad people developed it to the point of defining "Azumaya monoids" in a bunch of equivalent ways.

John Baez (May 27 2023 at 17:19):

But I still think there should be a simple string diagram proof that these two concepts are equivalent:

1) a monoid $A$ in $\mathsf{C}$ for which there's a monoid $B$ such that $B \otimes A \cong I$ in the monoidal bicategory of monoids, bimodules and bimodule homomorphisms.

2) a monoid $A$ that's separable (it's equipped with a section $s$ of the multiplication map $m: A \otimes A \to A$ in the category of $A,A$-bimodules) and central: the canonical map $p: A \to A$ projecting $A$ down to its center (defined using $s$) has image $I$.

John Baez (May 27 2023 at 17:25):

Item 2) could be somewhat wrong, and Vitale says it rather differently in that paper I linked to. After Proposition 1.3 he says a monoid $A$ is Azumaya if and only if it is faithfully projective in the category of $A,A$-bimodules and the canonical arrow $A \otimes A^{\text{op}} \to [A,A]$ is an isomorphism.

Here he's assuming $\mathsf{C}$ is a closed symmetric monoidal category, and $[A,A]$ is the internal hom. He uses a weird symbol for it... but not a lollipop.

Jean-Baptiste Vienney (May 27 2023 at 17:26):

Hmm I know how to draw string diagrams about monoids in a symmetric monoidal category, now I'm not sure how to draw the bimodules between them and the bimodule homorphisms.

John Baez (May 27 2023 at 17:28):

You can draw string diagrams in any bicategory, by shading the regions with objects. Believe it or not, this is actually explained on Wikipedia. (I keep telling people that whenever I learn anything I start by reading Wikipedia.)

Jean-Baptiste Vienney (May 27 2023 at 17:32):

Ok, nice if you want to work in a generic bicategory. But in fact here, it's not complicated to write string diagrams about bimodules between monoids in a symmetric monoidal category. You just have to write diagrams for the actions which are of the same form than the ones for the monoids.

Jean-Baptiste Vienney (May 27 2023 at 17:33):

So now, maybe you can add big arrows $\Rightarrow$ for the bimodules morphisms and success to write all your equations.

Jean-Baptiste Vienney (May 27 2023 at 17:35):

I'm not sure about this last sentence.

John Baez (May 27 2023 at 17:35):

You're right: the fact that we're dealing with a bicategory of bimodules of monoids in a symmetric monoidal category means we have a choice of how to draw things! The supposedly equivalent conditions 1) and 2) may be best drawn in two different styles, and then we'd need to show how those styles are related.

John Baez (May 27 2023 at 17:37):

It would be nice to reduce a bunch of this Azumaya-Brauer-Morita theory to pictures!

Jean-Baptiste Vienney (May 27 2023 at 17:38):

Sure!

Jean-Baptiste Vienney (May 27 2023 at 18:12):

To start, we can very well draw in string diagrams what are two monoids $A,B$, an $(A,B)$-bimodule and a morphism of $(A,B)$-bimodules, in a symmetric monoidal category.

Jean-Baptiste Vienney (May 27 2023 at 18:13):

I drew it here

Nathanael Arkor (May 27 2023 at 18:23):

John Baez said:

On the top of page 4 this lists four equivalent definitions of an Azumaya algebra in classic case of $R$-algebras, including the two I'm interested in, and threatens to generalize these to "Azumaya monoids" in a pretty general symmetric monoidal category. So that's what I want!

Note that the treatment in Azumaya Categories is more general: instead of working with monoids in a monoidal category $\mathcal V$, which are one-object $\mathcal V$-categories, they work with arbitrary $\mathcal V$-categories.

John Baez (May 27 2023 at 18:47):

Yes. It's very cool. I added a discussion Azumaya categories to the nLab article [[Azumaya algebra]], along with some other stuff. But I don't think I need this generalization for what I'm trying to do.

Jean-Baptiste Vienney (May 27 2023 at 18:48):

I think there is no problem to express in string diagrams that 1) There exists an equivalence $A^{op} \otimes A \sim I$ for a monoid $A$ and 2) there exists a section $A \rightarrow A \otimes A$ of $m:A\otimes A \rightarrow A$ such that $m$ followed by the symmetry $A \otimes A \rightarrow A \otimes A$ followed by the multiplication $A \otimes A \rightarrow A$ is equal to the counit $\epsilon:A \rightarrow I$ followed by the unit $\iota:I \rightarrow A$.

I have just one question. What is the definition of this counit?

Jean-Baptiste Vienney (May 27 2023 at 18:51):

Also, I'm not sure that a proof using the basic string diagrams of symmetric monoidal categories will be shorter. But it will be more visual for sure!

Jean-Baptiste Vienney (May 27 2023 at 19:07):

I think it must be easy to verify if "there exists an equivalence $A^{op} \otimes A \sim I$" and "there exists an equivalence $A \otimes A^{op} \sim I$" are equivalent using these string diagrams.

Jean-Baptiste Vienney (May 27 2023 at 19:08):

At least I can try to prove it. If it's false, it will be more difficult to prove that it is false.

Leopold Schlicht (May 27 2023 at 19:31):

Jean-Baptiste Vienney said:

If it's false, it will be more difficult to prove that it is false.

If it's true, it will be even more difficult to prove that it is false! :grinning_face_with_smiling_eyes:

Jean-Baptiste Vienney (May 27 2023 at 19:40):

A lemma that is super easily seen with string diagrams is that $(A \otimes B)^{op} = (A^{op} \otimes B^{op})$ and so $(A^{op} \otimes A)^{op} = A \otimes A^{op}$ (because ($A^{op})^{op} = A$).

Jean-Baptiste Vienney (May 27 2023 at 19:41):

($A^{op}$ is the monoid obtained by taking the object $A$, you keep the same unit, but you do a twist before the multiplication of $A$.)

Jean-Baptiste Vienney (May 27 2023 at 19:43):

And the multiplication of $A \otimes B$ is obtained by writing $A~B~A~B$ and using the two multiplication to arrive to $A~B$

Jean-Baptiste Vienney (May 27 2023 at 19:56):

Also $I^{op}=I$ and I believe that every $(A,B)$-bimodule $X$ is also a $(B^{op},A^{op})$-bimodule $X'$ (with same underlying object).

Jean-Baptiste Vienney (May 27 2023 at 20:09):

and every $(A,B)$-bimodule morphism $u:X \rightarrow Y$ induces a $(B^{op},A^{op})$-bimodule morphism $u':X' \rightarrow Y'$

John Baez (May 28 2023 at 00:11):

Jean-Baptiste Vienney said:

I think there is no problem to express in string diagrams that 1) There exists an equivalence $A^{op} \otimes A \sim I$ for a monoid $A$ and 2) there exists a section $A \rightarrow A \otimes A$ of $m:A\otimes A \rightarrow A$ such that $m$ followed by the symmetry $A \otimes A \rightarrow A \otimes A$ followed by the multiplication $A \otimes A \rightarrow A$ is equal to the counit $\epsilon:A \rightarrow I$ followed by the unit $\iota:I \rightarrow A$.

I have just one question. What is the definition of this counit?

That's a tricky and interesting question. I am most familiar with the case when our symmetric monoidal category $\mathsf{C}$ is the category of finite-dimensional vector spaces over a field $k$ so let me talk about that case first. Then $A$ is a finite-dimensional algebra over a field. In this case we can always define a counit $\epsilon: A \to k$ as follows:

$\epsilon(a) = \mathrm{tr}(L_a)$

where for any $a \in A$ the map $L_a: A \to A$ is given by left multiplication

$L_a(b) = a b$

You may enjoy showing

$\epsilon(a b) = \epsilon(b a)$

in this context.

John Baez (May 28 2023 at 00:13):

It's easy to generalize this to the case where $\mathsf{C}$ is any symmetric monoidal category and $A \in \mathsf{C}$ is any monoid whose underlying object has a dual $A^\ast$. That is, it's easy if you know about duals of objects in symmetric monoidal categories, and how they are used to define traces!

John Baez (May 28 2023 at 00:14):

As with any adjunction, the dual of an object in a symmetric monoidal category comes along with morphisms called the unit and counit, which are unfortunately completely different from the things I was calling the unit and counit here:

John Baez (May 28 2023 at 00:15):

there exists a section $A \rightarrow A \otimes A$ of $m:A\otimes A \rightarrow A$ such that $m$ followed by the symmetry $A \otimes A \rightarrow A \otimes A$ followed by the multiplication $A \otimes A \rightarrow A$ is equal to the counit $\epsilon:A \rightarrow I$ followed by the unit $\iota:I \rightarrow A$.

John Baez (May 28 2023 at 00:18):

So let me use some other names! Our object $A$ and its dual $A^\ast$ - which I'm assuming exists :warning: - come with morphisms

$\mu: I \to A \otimes A^\ast$
$\eta : A^\ast \otimes A \to I$

often called the unit and counit, obeying the [[triangle identities]].

John Baez (May 28 2023 at 00:24):

Because $A$ is a monoid with with multiplication $m: A \to A$ we can form this composite:

$A \xrightarrow{1_A \otimes \mu} A \otimes A \otimes A^\ast \xrightarrow{m \otimes 1_{A^\ast}} A \otimes A^\ast \xrightarrow{S_{A,A^\ast}} A^\ast \otimes A \xrightarrow{\eta} I$

John Baez (May 28 2023 at 00:24):

And this is the thing I was calling $\epsilon: A \to I$.

John Baez (May 28 2023 at 00:27):

In the long composite above I'm using the symmetry $S_{A,A^\ast}: A \otimes A^\ast \to A^\ast \otimes A$.

John Baez (May 28 2023 at 00:30):

It's a fun exercise to check that when $A$ is a monoid in the symmetric monoidal category of finite-dimensional vector spaces, this definition of $\epsilon: A \to I$ agrees with the one I mentioned before:

John Baez (May 28 2023 at 00:31):

$\epsilon(a) = \mathrm{tr}(L_a)$

where for any $a \in A$ the map $L_a: A \to A$ is given by left multiplication

$L_a(b) = a b$

John Baez (May 28 2023 at 00:32):

In short: we get $\epsilon: A \to I$ whenever $A$ has a dual. So the next question is: why, in our general abstract nonsense situation, should an Azumaya monoid have a dual?

John Baez (May 28 2023 at 00:33):

But that's a bit hard too, and I'm worn out for now.

John Baez (May 28 2023 at 00:44):

By the way, I find this thing:

$A \xrightarrow{1_A \otimes \mu} A \otimes A \otimes A^\ast \xrightarrow{m \otimes 1_{A^\ast}} A \otimes A^\ast \xrightarrow{S_{A,A^\ast}} A^\ast \otimes A \xrightarrow{\eta} I$

to be much clearer using string diagrams. I did a bunch of related string diagram calculations on pages 82-85 here.

Jean-Baptiste Vienney (May 28 2023 at 02:18):

Thanks for the explanations! I'm a bit tired right now me too, I'll try to understand all of this a bit later!

John Baez (May 28 2023 at 20:47):

A correction:

John Baez said:

Say $A$ is the algebra of $n \times n$ matrices with entries in $R$. Let $e_{i j} \in A$ be the elementary matrices. Then here's a $A,A$-bimodule homomorphism

$A \mapsto A \otimes_R A$

that's a right inverse for multiplication:

$e_{ij} \mapsto \frac{1}{n} \sum_{k=1}^n e_{ik} \otimes e_{kj}$

But notice, this only works if we can divide by $n$ in $R$.

John Baez (May 28 2023 at 20:49):

In fact there's an $A,A$-bimodule homomorphism that's a right inverse for multiplication even if we can't divide by $n$.

John Baez (May 28 2023 at 20:50):

The trick is to define it by

$1 \mapsto \sum_i e_{i j} \otimes e_{j i}$

where we don't sum over $j$.

John Baez (May 28 2023 at 20:53):

Then, for it to be a left $A$-module homomorphism it must do this:

$e_{k \ell} \mapsto \sum_i e_{k \ell} e_{ij} \otimes e_{ji} = e_{kj} \otimes e_{j \ell}$

John Baez (May 28 2023 at 20:54):

For it to be a right $A$-module homomorphism it must do this:

$e_{k\ell} \mapsto \sum_i e_{ij} \otimes e_{ji} e_{k \ell} = e_{kj} \otimes e_{j \ell}$

John Baez (May 28 2023 at 20:55):

Luckily these agree. So we use this as our rule for the map, and it's easy to see that multiplication sends $e_{k j} \otimes e_{j \ell}$ back to $e_{k\ell}$.

John Baez (May 28 2023 at 20:56):

This choice of splitting is scarily arbitrary since it depends on a choice of $j$.

John Baez (May 28 2023 at 20:57):

To avoid the arbitrariness we could sum over all $j$ and then divide by $n$... which is what I suggested before... but that only works if $n$ is invertible in our ring $R$.

Jean-Baptiste Vienney (May 28 2023 at 20:57):

Okay, I've been a bit silent because I was trying to make sense of traces and I can't figure out why if $f:A \rightarrow A$ and $g:A \rightarrow A$ are morphisms such that $A$ is dualizable, then $Tr(f;g) = Tr(g;f)$. I have seen graphical proofs in the context of compact closed categories but I don't succeed to understand how to translate these drawings in equations or commutative diagrams.

John Baez (May 28 2023 at 20:59):

Hi! It's very tiring to explain all the rules for translating string diagrams in compact closed categories into equations or commutative diagrams when you're in an environment where it's hard to draw pictures, like we are here. So I won't try.

John Baez (May 28 2023 at 21:01):

This particular proof is something that's incredibly beautiful with string diagrams: the "cyclic property of the trace", $\mathrm{tr}(fg) = \mathrm{tr}(gf)$ is proven just by cycling things around, using the fact that the trace is defined using a loop.

John Baez (May 28 2023 at 21:03):

But... okay, here's an explanation on Qiaochu Yuan's excellent blog:

• String diagrams, duality and trace.

John Baez (May 28 2023 at 21:04):

I don't know if it will help you - but by the way, I've been getting a lot of good information about separable algebras from his blog, and I think every mathematician who likes categories and algebra should read his blog.

Jean-Baptiste Vienney (May 28 2023 at 21:05):

Ah thanks! I've seen the pictures that make the morphisms turning around but I want to see how it comes from the equations, so I'm going to read that.

John Baez (May 28 2023 at 21:09):

He claims to explain it.

John Baez (May 28 2023 at 21:13):

Okay, I think my earlier remarks involving the center of a separable algebra were a bit wrong-headed. I think this paper has a better idea:

• Enrico M. Vitale, The Brauer and Brauer-Taylor groups of a symmetric monoidal category, 1996.

John Baez (May 28 2023 at 21:16):

I'm used to working with finite-dimensional separable algebras over a field, and there I can freely use the duals of those algebras - that is, their dual vector spaces - and the fact that finite-dimensional vector spaces form a [[compact closed category]]. But working over a general ring, or in a general symmetric monoidal category, it becomes questionable whether we're allowed to assume our separable algebras have duals.

John Baez (May 28 2023 at 21:16):

There's even the question of whether separable algebras over a field are automatically finite-dimensional or not!

John Baez (May 28 2023 at 21:17):

It seems they are - but the proof is rather hard:

• Is a "separable" algebra over a field finite-dimensional?

John Baez (May 28 2023 at 21:20):

This is different from [[Frobenius algebras]], where it's really easy to see they are finite-dimensional. All the [[separable algebras]] I really understand well are Frobenius algebras. These two concepts are deeply related, as you can see from the nLab articles (which I helped write back when I was using these algebras to study TQFT).

But when @Jean-Baptiste Vienney asked me to cough up the counit for an arbitrary separable algebra, I choked! Even working over a field, I couldn't do it without assuming the algebra is finite-dimensional!

John Baez (May 28 2023 at 21:23):

Luckily, Vitale has a better approach. Let's work in a [[Benabou cosmos]] $\mathsf{C}$, meaning a symmetric monoidal closed category $\mathsf{C}$ with small limits and colimits. A cosmos is a gold-plated sort of category that makes everything nice. We can reduce the assumptions later if we actually prove anything interesting!

In a cosmos $\mathsf{C}$, Vitale claims the following are equivalent:

John Baez (May 28 2023 at 21:25):

1. Azumaya monoids, defined as monoids $A \in \mathsf{C}$ that have a tensor inverse up to Morita equivalence, meaning a monoid $B$ such that $A \otimes B \cong I$ in the bicategory of monoids in $\mathsf{C}$, bimodules, and bimodule homomorphsms.

John Baez (May 28 2023 at 21:36):

2. Monoids $A \in \mathsf{C}$ that are faithfully projective in the category of $A,A$-bimodules and have the property that the obvious morphism $I \to [A,A]$ in the category of $A,A$-bimodules is an isomorphism.

John Baez (May 28 2023 at 21:37):

Now, the second one looks rather scary and technical to me, but I'm slowly learning to love it.

John Baez (May 28 2023 at 21:39):

For starters, the category of $A,A$-bimodules is monoidal closed, so it has an internal hom, which I'm denoting by $[-,-]$.

John Baez (May 28 2023 at 21:42):

(If you don't like bimodules remember that an $X,Y$-bimodule is the same as an $X \otimes Y^{\mathrm{op}}$-module, so bimodules can be seen as a special sort of modules. And in a symmetric monoidal category $\mathsf{C}$ that's a cosmos, I'm pretty sure any monoid $X$ has a closed monoidal category of modules!)

Jean-Baptiste Vienney (May 28 2023 at 21:47):

The property that $I \rightarrow [A,A]$ is an isomorphism is very strong, in the category of finite dimensional vector space, it means that $A$ is one-dimensional... if I'm not mistaken.

John Baez (May 28 2023 at 21:48):

You didn't yet read my explanation of what I meant by $[-,-]$.

John Baez (May 28 2023 at 21:48):

If I meant the internal hom in $\mathsf{C}$ then you'd be correct.

John Baez (May 28 2023 at 21:49):

Let me explain what $[A,A]$ is actually like when $\mathsf{C} = \mathsf{Vect}$. (By the way, finite-dimensional vector spaces don't form a cosmos. It's not a big deal, but it's worth noting.)

John Baez (May 28 2023 at 21:50):

So, when $\mathsf{C} = \mathsf{Vect}$, $A$ is an algebra, and $[A,A]$ is the algebra of all linear maps $f: A \to A$ that are actually $A,A$-bimodule endomorphisms:

$a f(b) c = f(abc)$

John Baez (May 28 2023 at 21:51):

Every left $A$-module endomorphism of $A$ is right multiplication by some element of $A$, so we know

$f (b) = b x$

for some fixed $x \in A$.

Then $a f(b) c = f(abc)$ implies $x$ in the center of $A$!

Jean-Baptiste Vienney (May 28 2023 at 21:52):

Oh, I see, ok, I was thinking to $I,I$-bimodule endomorphisms.

John Baez (May 28 2023 at 21:53):

Okay.

So: $[A,A]$ is just the center of $A$ in this case!

Jean-Baptiste Vienney (May 28 2023 at 21:54):

Ohhh, so this condition is the one that the center of $A$ is $I$ like before, it makes sense!

John Baez (May 28 2023 at 21:54):

Right: saying the "obvious" map $I \to [A,A]$ (I hope it's actually obvious) is an isomorphism, is saying that the center of $A$ consists just of scalars, i.e. elements of $I = k$, where $k$ is our field.

John Baez (May 28 2023 at 21:55):

But I don't think I used anything special about fields here; I believe everything I just said works when $\mathsf{C}$ is modules of any commutative ring $R$.

John Baez (May 28 2023 at 21:56):

So in this class of examples - the "classical" examples - we have a nice way of saying that the center of our $R$-algebra $A$ is just $R$.

John Baez (May 28 2023 at 21:56):

Such algebras are called central.

John Baez (May 28 2023 at 21:57):

So, in general, we'll say a monoid $A \in \mathsf{C}$ is central when the obvious morphism of $A,A$-bimodules $I \to [A,A]$ is an isomorphism!

John Baez (May 28 2023 at 21:58):

Here's a question I don't know how to answer, though the first few steps are clear:

Puzzle 1. - if $A,B$ are central monoids in a cosmos $\mathsf{C}$, is $A \otimes B$ central?

John Baez (May 28 2023 at 22:00):

In the case where $\mathsf{C} = \mathsf{Vect}$, this is true. And it's also easy to see that the tensor product of Azumaya monoids is Azumaya, so if 1. and 2. are really equivalent (like Benabou and Vitale claim), it's at least reasonable to hope that the tensor product of central monoids is central, even though it doesn't actually follow, since an Azumaya monoid is a central monoid with some extra property.

John Baez (May 28 2023 at 22:02):

If someone digs into Puzzle 1 they'll probably start wondering how $[A \otimes B, A \otimes B]$ is related to $[A,A]$ and $[B,B]$. Just remember that $[-,-]$ means something different in each expression!!!

Jean-Baptiste Vienney (May 28 2023 at 22:03):

John Baez said:

Right: saying the "obvious" map $I \to [A,A]$ (I hope it's actually obvious) is an isomorphism, is saying that the center of $A$ consists just of scalars, i.e. elements of $I = k$, where $k$ is our field.

Actually it's not so obvious. We have $A \otimes_{A} A \cong A$ in the category of $A,A$-bimodules and so you obtain $A \rightarrow [A,A]$ by monoidal closure and then you compose with the unit $I \rightarrow A$ to get your map $I \rightarrow [A,A]$.

Jean-Baptiste Vienney (May 28 2023 at 22:03):

If I'm not mistaken.

John Baez (May 28 2023 at 22:05):

Also, I don't see what, in the case of algebras over a field, this $A,A$-bimodule map $A \to [A,A]$ is supposed to be!

John Baez (May 28 2023 at 22:05):

We've seen $[A,A]$ is the center of $A$.

John Baez (May 28 2023 at 22:06):

But there's not a very good map from an algebra to its center, except in some special cases (like commutative algebras :upside_down:).

Jean-Baptiste Vienney (May 28 2023 at 22:07):

I'm currently thinking :upside_down:

John Baez (May 28 2023 at 22:08):

Your abstract argument getting a map from $A$ to $[A,A]$ looks convincing, but since I don't see what it gives in the "classical example" I'm afraid there's some confusion floating around here....

Jean-Baptiste Vienney (May 28 2023 at 22:09):

This thing of looking at bimodules over algebras over a commutative ring instead of just modules over the commutative ring makes my brain burning.

Jean-Baptiste Vienney (May 28 2023 at 22:10):

I always forget what are the "base rings".

John Baez (May 28 2023 at 22:11):

Right. I mentioned that bimodules over algebras are really just modules over (other) algebras:

John Baez said:

(If you don't like bimodules remember that an $X,Y$-bimodule is the same as an $X \otimes Y^{\mathrm{op}}$-module, so bimodules can be seen as a special sort of modules.

So, to think about bimodules over algebras over a commutative ring, you can just think about modules over an algebra over a commutative ring. And those behave a lot like modules over a not necessarily commutative ring.

John Baez (May 28 2023 at 22:14):

If we couldn't stack ideas on top of other ideas like this we wouldn't be category theorists. :upside_down:

John Baez (May 28 2023 at 22:15):

That burning sensation actually means you are growing new neurons.

Jean-Baptiste Vienney (May 28 2023 at 22:17):

Aaaah good, so I don't suffer for nothing.

David Michael Roberts (May 28 2023 at 22:19):

Is [A,-] a right adjoint here?

David Michael Roberts (May 28 2023 at 22:20):

Or does it not make sense to consider [A,-] as a functor here, with [A,A] as merely one value of it?

John Baez (May 29 2023 at 01:29):

Thanks for asking! It's supposed to be some sort of internal hom, right adjoint to some tensor product. It's described in item v on page 96 here:

• Enrico M. Vitale, The Brauer and Brauer-Taylor groups of a symmetric monoidal category, 1996.

using notation that's too baroque for me to reproduce here, and not very well explained.

John Baez (May 29 2023 at 01:32):

From the notation I assumed this [-,-] thing was an internal hom in the monoidal category of A,A-bimodules, or equivalently $A \otimes A^{\text{op}}$ modules.

John Baez (May 29 2023 at 01:33):

So I would guess that $[A,-]$ is right adjoint to the endofunctor $A \otimes_A -$, where $\otimes_A$ is the tensor product of $A,A$-bimodules.

John Baez (May 29 2023 at 01:35):

But now that you mention it, that can't be right! The endofunctor $A \otimes_A -$ is just the identity on the category of $A,A$-bimodules, so its right adjoint is just the identity!

John Baez (May 29 2023 at 01:35):

So, if this is what Vitale meant, we'd have $[A,A] \cong A$, and what he's claiming doesn't make sense.

John Baez (May 29 2023 at 01:37):

He's claiming that when the ambient symmetric monoidal category is $\mathsf{Vect}$, $[A,A]$ is the center of the algebra $A$.

John Baez (May 29 2023 at 01:39):

So I'm pretty sure that what he means is this: if $X$ and $Y$ are $A,A$-bimodules, $[X,Y]$ is the set of $A,A$-bimodule morphisms from $X$ to $Y$. This set has its own $A,A$-bimodule structure, unless I'm going insane.

John Baez (May 29 2023 at 01:40):

With this interpretation $[A,A]$ really is the center of $A$.

John Baez (May 29 2023 at 01:41):

But if this is what Vitale means, what is $[A,-]$ right adjoint to, if anything??? I'm left asking @David Michael Roberts's question.

John Baez (May 29 2023 at 01:46):

John Baez said:

So I'm pretty sure that what he means is this: if $X$ and $Y$ are $A,A$-bimodules, $[X,Y]$ is the set of $A,A$-bimodule morphisms from $X$ to $Y$. This set has its own $A,A$-bimodule structure, unless I'm going insane.

Okay, so I was going insane. The set of $A,A$-bimodule morphisms from $A$ to $A$ is only an $A,A$-bimodule when $A$ is commutative.

John Baez (May 29 2023 at 01:47):

For example suppose $f: A \to A$ is an $A,A$-bimodule morphism:

$f(abc) = a f(b) c$

John Baez (May 29 2023 at 01:48):

Then multiplying $f$ on the left (or right) by an element $x \in A$ ruins this property:

$xf (abc) \ne a \, xf(b) \,c$

John Baez (May 29 2023 at 01:53):

So, given a monoid $A$ in the cosmos $\mathsf{C}$ and $A,A$-bimodules $X$ and $Y$ in this cosmos, I believe Vitale wants $[X,Y]$ to be the $\mathsf{C}$-object of $A,A$-bimodule morphisms from $X$ to $Y$. It's not an $A,A$-bimodule itself, just an object in $\mathsf{C}$.

John Baez (May 29 2023 at 01:57):

However, $[X,X]$ is better: it's a monoid object in $\mathsf{C}$!

John Baez (May 29 2023 at 01:57):

And so it has a unit $I \to [X,X]$.

John Baez (May 29 2023 at 01:57):

And this fact, I believe, give the so-called "obvious" map $I \to [A,A]$ that I was mentioning:

John Baez (May 29 2023 at 01:58):

Jean-Baptiste Vienney said:

John Baez said:

Right: saying the "obvious" map $I \to [A,A]$ (I hope it's actually obvious) is an isomorphism, is saying that the center of $A$ consists just of scalars, i.e. elements of $I = k$, where $k$ is our field.

Actually it's not so obvious.

John Baez (May 29 2023 at 01:59):

It's like the old joke where a professor says something is obvious, the student asks him if it's really obvious, and the professor goes away, comes back three days later and says "yes, it's obvious".

David Kern (May 31 2023 at 11:20):

I just saw this discussion so I haven't had the time to read all the arguments (so I apologise if I missed something making this irrelevant), but I think you may be interested in this preprint that came out a few days ago, especially sections 5.5 and 6 addressing this issue in monoidal $(\infty,1)$-categories.

John Baez (Jun 01 2023 at 00:27):

Thanks! That looks helpful even though I'm not ready to plunge into an $(\infty,1)$-categorical treatment!

John Baez (Jun 01 2023 at 00:27):

I had not seen this before.

John Baez (Jun 01 2023 at 00:29):

I have been making a lot of progress this week thanks in part to this paper:

• Francis Borceux and Enrico Vitale, Azumaya categories, Applied Categorical Structures 10 (2002), 449-467.

and thanks in part to conversations with Todd Trimble.

John Baez (Jun 01 2023 at 00:30):

I'd like to outline some thoughts and state some theorems and conjectures that deserve nice simple proofs (if they're correct).

John Baez (Jun 01 2023 at 00:32):

Let me start by reviewing the basic ideas, which make sense at a high level of generality. Then I'll state some theorems and conjectures in the case of algebras over commutative rings. Maybe some of you can help find nice proofs and/or generalize them.

John Baez (Jun 01 2023 at 00:33):

Whoops, first I have to take a walk. :big_frown:

John Baez (Jun 01 2023 at 02:27):

Okay, let me start with the general context. Let's take V to be a very nice symmetric monoidal category: closed, complete and cocomplete. This is what Benabou called a cosmos.

John Baez (Jun 01 2023 at 02:46):

There is then a bicategory VMod where:

• objects are V-enriched categories (or V-categories for short)
• morphisms are V-enriched profunctors (or V-profunctors)
• 2-morphisms are V-enriched natural transformations between V-enriched profunctors (or simply V-natural transformations).

I will try to remember to write $\nrightarrow$ for V-profunctors and $\to$ for V-enriched functors (or V-functors, for short).

John Baez (Jun 01 2023 at 02:47):

I don't know good references for these facts:

Theorem 1 VMod is a symmetric monoidal bicategory.

Theorem 2. VMod is compact closed.

John Baez (Jun 01 2023 at 02:49):

Here's a good place to see the definitions spelled out:

• Mike Stay, Compact closed bicategories.

John Baez (Jun 01 2023 at 02:52):

The example that looms large in my mind is V = RMod, the category of modules of a commutative ring, with the usual tensor product $M \otimes_R N$ of modules giving the symmetric monoidal structure.

John Baez (Jun 01 2023 at 02:55):

In this case

• a one-object V-category is called an algebra over R,
• a V-enriched profunctor between these is called a bimodule,
• a V-natural transformation is called a bimodule homomorphism.

John Baez (Jun 01 2023 at 02:56):

When V = RMod we can more generally call a V-enriched category an algebroid over R, but there's vastly more literature on algebras.

John Baez (Jun 01 2023 at 03:01):

I'm trying to understand Azumaya algebras, but the general concept is this:

Definition. A V-category C is Azumaya if there is a V-category D such that C $\otimes$ D $\simeq$ I in VMod.

John Baez (Jun 01 2023 at 03:02):

Here $\otimes$ is the tensor product of V-enriched categories and I is the unit for this tensor product. So Azumaya V-categories are, morally speaking, the invertible ones.

John Baez (Jun 01 2023 at 03:03):

But notice that $\simeq$ stands for equivalence in the bicategory VMod, so Azumaya V-categories need only be invertible up to equivalence.

John Baez (Jun 01 2023 at 03:06):

My goal, ultimately, is to find nice proofs of a bunch of facts about Azumaya V-categories. Some of these facts are already known. One-object Azumaya V-categories where V = RMod for a commutative ring are called Azumaya algebras and there are some very interesting theorems about these.

Now I already said that VMod is compact closed. This means that every V-category $C$ has a dual $C^\ast$ which comes with a unit

$\iota: I \nrightarrow C \otimes C^\ast$

and counit

$\epsilon: C^\ast \otimes C \nrightarrow I$

John Baez (Jun 01 2023 at 03:07):

which obey the usual triangle identities up to isomorphism.

John Baez (Jun 01 2023 at 03:10):

In other words, they form a biadjunction. (The isomorphisms I alluded to can always be improved to obey some laws of their own called the 'swallowtail identities' - but this may not ever matter to us, though it would be fun for me if it did.)

John Baez (Jun 01 2023 at 03:13):

Now, there's another fundamental result that I don't know a reference for:

Theorem 3. Any two duals of an object in a symmetric monoidal bicategory are equivalent. Given a V-category $C$, the opposite V-category $C^{\text{op}}$, defined in a way analogous to the usual opposite of a category, serves as a dual for $C$ in VMod.

Jean-Baptiste Vienney (Jun 01 2023 at 03:17):

John Baez said:

Okay, let me start with the general context. Let's take V to be a very nice symmetric monoidal category: closed, complete and cocomplete. This is what Benabou called a cosmos.

Is it really necessary to suppose $V$ to be complete and cocomplete to prove your results? Is it sufficient to only suppose that $V$ is finitely complete and finitely cocomplete? In this case, you could also chose $V=FinVect_{k}$ for example.

John Baez (Jun 01 2023 at 03:21):

I am not going to try to find minimal assumptions for my results! It's important, but it would clutter the big picture I'm trying to paint. The proofs of the results will show us what the minimal assumptions are; these may vary from result to result.

John Baez (Jun 01 2023 at 03:27):

To repeat and expand on what I said:

Theorem 3. Given a V-category $C$, the opposite V-category $C^{\text{op}}$, defined in a way analogous to the usual opposite of a category, serves as a dual for $C$ in VMod. The counit

$\epsilon: C^{\text{op}} \otimes C \nrightarrow I$

can be taken to be the hom of $C$, which is the V-functor

hom: $C^{\text{op}} \otimes C \to V$

Furthermore, we can twist around this V-profunctor and get the unit

$\iota: I \nrightarrow C \otimes C^{\text{op}}$

Jean-Baptiste Vienney (Jun 01 2023 at 03:28):

John Baez said:

I am not going to try to find minimal assumptions for my results! It's important, but it would clutter the big picture I'm trying to paint. The proofs of the results will show us what the minimal assumptions are; these may vary from result to result.

(Ok, my intuition is that all of this will not use any "infinity" but I'm maybe mistaken. Mathematicians often suppose that some category is "complete" or "cocomplete" instead of simply supposing that all the limit and colimit they use exist which would be better.) But that's a bit out of subject, sorry, just a side note.

John Baez (Jun 01 2023 at 03:39):

Next, yet another result I don't know a reference for:

Theorem 4. If C, D are objects in a symmetric monoidal bicategory and C $\otimes$ D $\simeq$ I, then we can find a unit and counit making D into a dual of C.

Mike Shulman (Jun 01 2023 at 04:21):

John Baez said:

Theorem 1 VMod is a symmetric monoidal bicategory.

One way to prove this, which I unsurprisingly happen to think is nice and clean, is to combine Framed bicategories and monoidal fibrations with Constructing symmetric monoidal bicategories. In example 15.4 of FBMF I observed that if you make a symmetric monoidal category $V$ indexed/fibered over $\rm Set$ in the canonical way by taking families of objects, then it is frameable so you can make it into a double category, and then that double category has local coequalizers so you can take the double category of monoids and modules therein, which is your $V\rm Mod$. In the symmetric case all the monoidal structures carry through, so you can then apply CSMB to make the symmetric monoidal double category into a symmetric monoidal bicategory.

This fact was of course known to category theorists (particularly Australians) long before my papers, but I don't actually know of another place where it is proven carefully.

Mike Shulman (Jun 01 2023 at 04:24):

Jean-Baptiste Vienney said:

(Ok, my intuition is that all of this will not use any "infinity" but I'm maybe mistaken. Mathematicians often suppose that some category is "complete" or "cocomplete" instead of simply supposing that all the limit and colimit they use exist which would be better.)

If you want a bicategory $V\rm Mod$ whose objects are arbitrary small $V$-categories, then you need $V$ to have arbitrary small colimits in order to define composition of $V$-profunctors. You can get away with finite colimits if you restrict the objects of $V\rm Mod$ to be $V$-categories with finitely many objects. Or you can try to use something like a [[virtual double category]] instead, but that doesn't work so well for duality theory.

Mike Shulman (Jun 01 2023 at 04:32):

John Baez said:

Theorem 2. VMod is compact closed.
...
Theorem 3. Given a V-category $C$, the opposite V-category $C^{\text{op}}$, defined in a way analogous to the usual opposite of a category, serves as a dual for $C$ in VMod.

I'm not sure how one could prove theorem 2 without proving theorem 3 (which should have theorem 2 as an immediate consequence). One fairly early reference for this is section 7 of Day & Street's Monoidal bicategories and Hopf algebroids (1997), but they don't give any more details of the proof than you did. I don't offhand know of somewhere that gives more details.

Mike Shulman (Jun 01 2023 at 04:34):

John Baez said:

Theorem 4. If C, D are objects in a symmetric monoidal bicategory and C $\otimes$ D $\simeq$ I, then we can find a unit and counit making D into a dual of C.

I don't think I knew that (did I?). I find it very surprising. I guess it must use the symmetry fundamentally somehow? It's certainly not true that if $F$ and $G$ are functors and $F\circ G \cong \rm Id$ then they are adjoint.

John Baez (Jun 01 2023 at 05:22):

Thanks for all your comments, which will help me a lot if I ever write something up, e.g. on the nLab .

John Baez (Jun 01 2023 at 05:24):

You're right that Theorem 3 is how to prove Theorem 2; I was going for the gradual reveal.

John Baez (Jun 01 2023 at 05:35):

I hope Theorem 4 is right. The idea is to take an equivalence

$e: C \otimes D \nrightarrow I$

and take a weak inverse and compose it with the symmetry to get an equivalence

$i: I \nrightarrow D \otimes C$

and then finagle around until we get a dual of $C$ in the sense I defined. I'll try to scribble out a string diagram proof.

John Baez (Jun 01 2023 at 05:46):

Now, if Theorems 1-4 are true, we can conclude:

Corollary 5. A V-category $C$ is Azumaya iff its hom, made into a V-profunctor from $C^{\text{op}} \otimes C$ to $V$, is an equivalence in VMod.

John Baez (Jun 01 2023 at 05:53):

The $\Leftarrow$ direction is obvious but the $\Rightarrow$ direction uses Theorems 2-4 and (if it's true) it's a useful condition for Azumayaness, since it means we don't have to run around looking for a weak inverse to $C$: if it exists, it's the opposite $V$-category!

John Baez (Jun 01 2023 at 05:54):

I'll quit now... but so far I'm just warming up.

John Baez (Jun 01 2023 at 05:56):

Now is a great time for people to raise questions, raise objections, provide more references, or (dis)prove Theorem 4!

Mike Shulman (Jun 01 2023 at 13:16):

John Baez said:

I hope Theorem 4 is right. The idea is to take an equivalence $e: C \otimes D \to I$ and take a weak inverse and compose it with the symmetry to get an equivalence $i: I \to D \otimes C$...

Oh, of course! Once you have both $e$ and $i$, then you just need the fact that any equivalence can be improved to an adjoint equivalence.

Mike Shulman (Jun 01 2023 at 13:16):

(Applied in the delooping of your monoidal bicategory to a tricategory.)

Mike Shulman (Jun 01 2023 at 13:19):

In Corollary 5, do you mean to refer to the hom as a profunctor from $C^{\rm op}\otimes C$ to $I$? Writing it as a functor $C^{\rm op}\otimes C\to V$ doesn't indicate its domain and codomain qua profunctor, it could equally be a profunctor from $C$ to $C$.

Mike Shulman (Jun 01 2023 at 15:34):

By the way, have you looked at Niles Johnson's paper Azumaya Objects in Triangulated Bicategories? His characterization theorem 1.6 seems to be along the lines that you're talking about.

John Baez (Jun 01 2023 at 16:23):

Mike Shulman said:

In Corollary 5, do you mean to refer to the hom as a profunctor from $C^{\rm op}\otimes C$ to $I$? Writing it as a functor $C^{\rm op}\otimes C\to V$ doesn't indicate its domain and codomain qua profunctor, it could equally be a profunctor from $C$ to $C$.

I meant the hom is a profunctor from $C^{\rm op}\otimes C$ to $I$. I was forgetting the latex to write $\mathrm{hom} : C^{\rm op}\otimes C\nrightarrow V$ to indicate we've got a profunctor here, and sort of hoping I wouldn't need to introduce that notation.

**[EDIT: I later gave up, went back, and tried to consistently use $\nrightarrow$ for enriched profunctors and $\rightarrow$ for enriched functors.)

John Baez (Jun 01 2023 at 16:34):

Mike Shulman said:

John Baez said:

I hope Theorem 4 is right. The idea is to take an equivalence $e: C \otimes D \to I$ and take a weak inverse and compose it with the symmetry to get an equivalence $i: I \to D \otimes C$...

Oh, of course! Once you have both $e$ and $i$, then you just need the fact that any equivalence can be improved to an adjoint equivalence.

Right. For anyone who wants to see a string diagram proof of what we're talking about here, try the pictures in the proof of Theorem 14 on page 19 here. Here we've got a monoidal category where every object $x$ has an object $\overline{x}$ with isomorphisms $x \otimes \overline{x} \cong \overline{x} \otimes x \cong I$, and we improve one of those isomorphisms in order to make $\overline{x}$ into the dual of $x$ - i.e., in order to get isomorphisms obeying the [[triangle identities]]. The argument is visually quite pretty.

John Baez (Jun 01 2023 at 16:34):

So what I'm doing now is the same argument, but with all the equations in the string diagram calculation replaced by isomorphisms.

John Baez (Jun 01 2023 at 16:41):

Mike Shulman said:

By the way, have you looked at Niles Johnson's paper Azumaya Objects in Triangulated Bicategories? His characterization theorem 1.6 seems to be along the lines that you're talking about.

Oh yes, this looks like exactly what I was trying to do! My Corollary 5 proved the equivalence between Johnson's conditions iv and v. But what I'm really shooting for is the equivalence between ii and v: the connection between Azumaya algebras and central separable algebras, worked out at a high level of generality.

John Baez (Jun 01 2023 at 16:44):

I don't feel heartbroken that Johnson already did this, because this equivalence is already visible, at least in a special case, in this paper I was reading (thanks in part to @Nathanael Arkor):

• Francis Borceux and Enrico Vitale, Azumaya categories, Applied Categorical Structures 10 (2002), 449-467.

and I'm just trying to understand it enough that I can use it for what I'm really trying to do: some nefarious scheme I'll reveal in the fullness of time.

John Baez (Jun 01 2023 at 16:44):

So I'll just keep on here, now with a lot of help from Niles Johnson.

John Baez (Jun 02 2023 at 01:14):

So let's pick up here:

John Baez said:

Corollary 5. A V-category $C$ is Azumaya iff its hom, made into a V-profunctor from $C^{\text{op}} \otimes C$ to $V$, is an equivalence in VMod.

and see more concretely what happens when that hom, viewed as a V-profunctor, is an equivalence.

John Baez (Jun 02 2023 at 01:18):

I think now I'll break down and use $F: A \nrightarrow B$ to mean $F$ is a V-profunctor and $F: A \to B$ to mean $F$ is a V-functor, at least for a while.

John Baez (Jun 02 2023 at 01:37):

So, any V-profunctor $F : A \nrightarrow B$ is (by definition) a V-functor from $A \otimes B^{\mathrm{op}}$ to $V$, which amounts to the same thing as V-functors from $A$ to $\hat{B}$, the V-category of V-functors from $B^{\text{op}}$ to V.

$\hat{B}$ is called the category of V-enriched presheaves on $B$.

By a spinoff of the enriched Yoneda lemma, V-functors from $A$ to $\hat{B}$ in turn amount to the same thing as cocontinuous V-functors from $\hat{A}$ to $\hat{B}$.

John Baez (Jun 02 2023 at 01:47):

So, the symmetric monoidal bicategory I've been calling VMod, where

• objects are V-categories
• morphisms are V-profunctors
• 2-morphisms are V-natural transformations

is equivalent to one where

• objects are V-categories
• morphisms from $A$ to $B$ are cocontinuous V-functors from $\hat{A}$ to $\hat{B}$
• 2-morphisms are V-natural transformations

John Baez (Jun 02 2023 at 01:49):

So we get a new viewpoint on VMod. From this new viewpoint, what does it mean to say a V-category $C$ is Azumaya?

John Baez (Jun 02 2023 at 01:51):

In the old viewpoint $C$ is Azumaya if its hom, made into a V-profunctor from $C^{\text{op}} \otimes C$ to $V$, is an equivalence in VMod.

Mike Shulman (Jun 02 2023 at 01:55):

Don't you mean a $V$-profunctor from $C^{\rm op}\otimes C$ to $I$, the unit $V$-category?

John Baez (Jun 02 2023 at 01:56):

Yes. Thanks:

John Baez (Jun 02 2023 at 01:56):

In the old viewpoint $C$ is Azumaya if its hom, made into a V-profunctor from $C^{\text{op}} \otimes C$ to $I$, is an equivalence in VMod.