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Stream: theory: mathematics

Topic: Categorical Description of the Brauer Group as a functor.

Jack Jia (Nov 20 2024 at 21:18):

I saw that nlab has some very high-up explanation on this that is out of my reach. I am more interested of just Brauer groups over fields, and would like to know juat some basic properties.

Recall that given a field k, we can construct its Brauer group Br(k):
Elements are central simple algebras over k, quotient by the equivalence relation: 2 CSAs A,B over k are equivalnet if there exist positive integers m,n such that the matrix algebras M_m(A) is isomorphic to M_n(B).
Group operation is given by tensoring any two representatives of any equivalence classes.

We can view this as a functor Br: Fields -> Ab, where each k is sent to Br(k), and for each extension i:E->F, we send the class [A] (A is a CSA/E) to [A (x)_E F] (after extension of scalars, this is now a CSA/F).

My question this: What properties does this functor have? Does it admit left/right adjoints? Does it preserve/reflect certain kinds of limits/colimits?

I am also wondering if there is a categorical classification of this functor. After struggling a while, I find myself unable to say anything about a functor in [Fields,Ab]: Fields has too few arrows and it just feels very restirctive.

John Baez (Nov 20 2024 at 21:41):

As you may know, the category of fields has very limits or products: for example there's no product nor coproduct of fields. One useful colimit that exists is the "increasing union" of fields, or (slightly more generally) [[directed colimit]].

John Baez (Nov 20 2024 at 21:45):

It may be possible to use this to show the functor Br: Fields $\to$ Ab doesn't have a left adjoint, since Ab has all colimits, and the putative left adjoint would need to preserve these, but Fields is unlikely to have them. A similar argument using limits might rule out the possibility that Br has a right adjoint. You might try this. Or, you might try directly to ask what field the putative left or right adjoint would assign to the group $\mathbb{Z}$, and prove a contradiction.

John Baez (Nov 20 2024 at 21:47):

However, if you want to apply category theory to Brauer theory, it's really making your life much harder to restrict yourself to Brauer groups of fields. The Brauer group of a commutative ring is a perfectly well-defined concept, and the category of commutative rings is much better behaved.

John Baez (Nov 20 2024 at 21:48):

If you do this you need to work with [[Azumaya algebras]], which are the appropriate generalization of central simple algebras.

John Baez (Nov 20 2024 at 21:49):

So yes, I am like the guy who answers the question "how do I do this in C++?" by saying "you should use Python instead, it's better."

However, unlike that guy I would be happy to improve the nLab article on Brauer groups so it's easier to understand. It's incredibly beautiful stuff, which I want to blog about more someday.

Mike Shulman (Nov 20 2024 at 22:12):

I don't know if that's completely fair: Python isn't a generalization of C++.

Damiano Mazza (Nov 20 2024 at 22:23):

Hopefully there aren't too many programming language theorists around here, a sentence like "Python is better than C++" could start a war! :wink:

Kevin Carlson (Nov 20 2024 at 22:24):

I'm sure any real PL theorists would hardly look up from their ML for that kind of thing.

Damiano Mazza (Nov 20 2024 at 22:27):

Of course, both would be considered garbage :wink: But it doesn't mean you can compare one kind of garbage to another, and have all sorts of very strong opinions about it... :big_smile:

John Baez (Nov 20 2024 at 23:39):

See, I know how to get everyone to join this conversation. Only problem: we're not talking about Brauer groups anymore. :smirk:

John Baez (Nov 20 2024 at 23:46):

To be more useful to @Jack Jia, I should think about whether the functor from CommRing to AbGp sending any commutative ring $R$ to its Brauer group $Br(R)$ has good categorical properties. For those not in the know, what we do is this: first we form the monoidal bicategory $Alg(R)$ of

• algebras over $R$
• bimodules between algebras
• bimodule homomorphisms

The monoidal structure is just tensoring algebras over $R$.

Then we look at the invertible objects in this monoidal bicategory, which are called Azumaya algebras: these are the algebras $A$ over $R$ for which there's an algebra $B$ over $R$ such that $A \otimes_R B$ and $B \otimes_R A$ are equivalent, as objects in $Alg(R)$, to the unit object (which is $R$ as an algebra over itself).

(In fact if one of them is, then so is the other, because our monoidal bicategory is symmetric monoidal.)

John Baez (Nov 20 2024 at 23:48):

By the way, equivalence of objects in the bicategory $Alg(R)$ is called Morita equivalence.

Then we look at Morita equivalence classes of Azumaya algebras, and these form the Brauer group $Br(R)$.

John Baez (Nov 20 2024 at 23:50):

So to see if $Br: \mathsf{CommRing} \to \mathsf{AbGp}$ has any good categorical propertiers (like preserving limits or colimits), it probably pays to start by seeing if there's a functor $Alg : \mathsf{CommRing} \to \mathbf{MonBicat}$ and seeing how nice it is.

But I can certainly see why someone who is not comfortable with bicategories would want to avoid this approach!

John Baez (Nov 20 2024 at 23:53):

A less highbrow approach would begin by asking: if we have a homomorphism $f \colon R \to S$ of commutative rings, and $A$ is an Azumaya algebra over $S$, is $f^\ast A = R \otimes_S A$ an Azumaya algebra over $R$? And I bet the answer is yes.

John Baez (Nov 20 2024 at 23:56):

So, with further work along those lines, we should be able to see if there's a functor $Br : \mathsf{CommRing}^{\text{op}} \to \mathsf{AbGp}$ and what its properties are. (It seems to be coming out contravariant, as that notation $f^\ast$ hints.)

John Baez (Nov 20 2024 at 23:58):

My hunch is that $Br$ should send binary products of rings to binary coproducts of abelian groups (which are confusingly the same as binary products).

Jack Jia (Nov 21 2024 at 03:03):

Thank you @John Baez so much for such detailed reply! I will have to parse this more but this is exactly what I was looking for! I just have some questions regarding the $Alg$ functor: If I have a map f:R->S and an R-algebra R' and an S-algebra S', I think there are in general many maps from R' to S', right? Where should I send f? Also I am unfamiliar with higher categories but would like to learn more: Can a monomidal bicategory be viewed as a one object tricategory? Should the morphisms in the catgory MonBiCat be the lax-monoidal 2-functors?

Jack Jia (Nov 21 2024 at 03:07):

Also I am not sure how to use LaTeX here, I would aprreciate it if someone can tell me how.

John Baez (Nov 21 2024 at 03:13):

Unfortunately you need to use double dollar signs here for LaTeX to work, e.g.

a map $$f \colon R \to S$$

produces

a map $f \colon R \to S$

John Baez (Nov 21 2024 at 03:26):

Jack Jia said:

I just have some questions regarding the $Alg$ functor: If I have a map $f:R \to S$ and an $R$-algebra $R'$ and an $S$-algebra $S'$, I think there are in general many maps from $R'$ to $S'$, right? Where should I send $f$?

I'm a bit confused by your question (since it's probably not quite the right question), but I think this is how $Alg$ works:

1) it sends any commutative ring $R$ to the monoidal bicategory $Alg(R)$ of

• algebras over $R$
• bimodules between algebras
• bimodule homomorphisms

2) it sends any map of commutative rings $f: R \to S$ to the monoidal functor $Alg(f) \colon Alg(S) \to Alg(R)$ that sends

• any algebra $A$ over $R$ to the algebra $f^\ast A := S \otimes_R A$ over $S$, where the tensor product is defined by making $S$ into an $R$-module using $f$.
• any $(A,B)$-bimodule $M$ to an $(f^\ast A, f^\ast B)$-bimodule which I'll call $f^\ast M$. I can figure out how it works if you want, or maybe you can do it: there should be an obvious choice.
• any $(A,B)$-bimodule homomorphism $\phi \colon M \to N$ to an $(f^\ast A, f^\ast B)$-bimodule homomorphism which I'll call $f^\ast \phi$. Again, I can figure this out if you want.

John Baez (Nov 21 2024 at 03:34):

Jack Jia said:

Can a monoidal bicategory be viewed as a one object tricategory?

Yes, exactly.

Should the morphisms in the category MonBiCat be the lax-monoidal 2-functors?

MonBiCat can be made into something richer than a category - actually if you want to get very fancy you can probably think of it as a tetracategory since each monoidal bicategory can be thought of as a tricategory! However, you can 'truncate' it in various ways to make it less fancy, and this is fine if you're ultimately interested in the Brauer group $Br(R)$ rather than the whole monoidal bicategory $Alg(R)$. Since groups form a category, you can probably use a mere category where the morphisms are certain equivalence classes of monoidal 2-functors.

You could use lax monoidal 2-functors or you use strong monoidal 2-functors; I'd prefer to use strong ones here, but we'd have to see what the above 2-functor $Alg(f)$ is actually like. I'm hoping it's strong, which is nicer than lax.

I would use 'monoidal natural equivalence classes of strong monoidal 2-functors' to get a mere category. 'Monoidal natural equivalence' is the next thing up from 'monoidal natural isomorphism'.

Again, a lot of this complexity will eventually be hidden if you focus on Brauer groups. Someone has probably already worked out the functor $Br$ from $\mathsf{CommRing}$ to $\mathsf{AbGp}$.

Jack Jia (Nov 21 2024 at 03:57):

John Baez said:

Jack Jia said:

I just have some questions regarding the $Alg$ functor: If I have a map $f:R \to S$ and an $R$-algebra $R'$ and an $S$-algebra $S'$, I think there are in general many maps from $R'$ to $S'$, right? Where should I send $f$?

I'm a bit confused by your question (since it's probably not quite the right question), but I think this is how $Alg$ works:

1) it sends any commutative ring $R$ to the monoidal bicategory $Alg(R)$ of

• algebras over $R$
• bimodules between algebras
• bimodule homomorphisms

2) it sends any map of commutative rings $f: R \to S$ to the monoidal functor $Alg(f) \colon Alg(S) \to Alg(R)$ that sends

• any algebra $A$ over $R$ to the algebra $f^\ast A := S \otimes_R A$ over $S$, where the tensor product is defined by making $S$ into an $R$-module using $f$.
• any $(A,B)$-bimodule $M$ to an $(f^\ast A, f^\ast B)$-bimodule which I'll call $f^\ast M$. I can figure out how it works if you want, or maybe you can do it: there should be an obvious choice.
• any $(A,B)$-bimodule homomorphism $\phi \colon M \to N$ to an $(f^\ast A, f^\ast B)$-bimodule homomorphism which I'll call $f^\ast \phi$. Again, I can figure this out if you want.

Thanks! I see that indeed I was not parsing this coreectly. I will try to work out the details myself.

Jack Jia (Nov 21 2024 at 03:59):

Also thank you for the amazing explanation! I think I have obtained enough hints and will try to work out the details.

John Baez (Nov 21 2024 at 05:52):

I'm glad it helped. But in fact, let me get straight to the point and describe what functor

$Br: \mathsf{CommRing}^{\text{op}} \to \mathsf{AbGp}$

we actually get if we follow the above strategy. We can describe the final result more easily than the process of getting it!

• $Br$ sends any commutative ring to the group $Br(R)$ whose elements are Morita equivalence classes of Azumaya algebras over $R$. The multiplication in this group is given by
  $[A] [B] = [A \otimes_R B ]$
  where the brackets mean 'Morita equivalence class'.

• $Br$ sends any homomorphism of commutative rings $f: R \to S$ to the group homomorphism $Br(f) : Br(S) \to Br(R)$ which sends any Morita equivalence class $[A]$ of Azumaya algebras over $S$ to the Morita equivalence class $[f^\ast A]$ where
  $f^\ast A = R \otimes_S A$
  as described earlier.