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Stream: deprecated: mathematics

Topic: generalizing Wedderburn-Artin

John Baez (May 24 2023 at 14:21):

In the category of vector spaces over a field $k$, Wedderburn's theorem says a semisimple algebra over $k$ is a finite product of matrix algebras with entries in some [[division algebra]] over $k$.

John Baez (May 24 2023 at 14:22):

For example a finite-dimensional simple algebra over $\mathbb{R}$ is an $n \times n$ matrix algebra with entries in $\mathbb{R}, \mathbb{C}$ or $\mathbb{H}$, and a semisimple algebra over $\mathbb{R}$ is a finite product of such matrix algebras.

John Baez (May 24 2023 at 14:27):

In the category of abelian groups, the Wedderburn-Artin theorem says a semisimple ring is finite product of rings with entries in division rings.

John Baez (May 24 2023 at 14:35):

I think I've seen a version of Wedderburn-Artin for the category of $\mathbb{Z}/n$-graded vector spaces over a field. There's a notion of $\mathbb{Z}/n$-graded division algebra which is actually a bit subtle, etc.

John Baez (May 24 2023 at 14:36):

So, I'm wondering if someone has tried to generalize Wedderburn-Artin to, say, a sufficiently nice symmetric monoidal abelian category $\mathsf{A}$. I think we can define 'ideals' for a monoid object in $C$. And I think we can define a 'division algebra' in $\mathsf{A}$ to be a monoid object $R$ in $\mathsf{A}$ that has no left or right ideals except $0$ and $R$ itself.

John Baez (May 24 2023 at 14:37):

I think we can define a 'semisimple' monoid in $\mathsf{A}$ to be a monoid whose category of actions is a [[semisimple category]].

John Baez (May 24 2023 at 14:38):

Then we can try to prove any semisimple monoid in $\mathsf{A}$ is a finite product of matrix algebras over division algebras in $\mathsf{A}$.

Morgan Rogers (he/him) (May 24 2023 at 14:41):

That would be a monoid for which the only non-trivial congruence is the maximal one, I think.

John Baez (May 24 2023 at 14:43):

"That" meaning what I was calling a "division algebra", I guess?

Morgan Rogers (he/him) (May 24 2023 at 14:43):

No, the semisimple monoid

John Baez (May 24 2023 at 14:44):

Oh, better.

John Baez (May 24 2023 at 14:45):

But isn't that more appropriate for simple monoids? Isn't there a nontrivial congruence on the semisimple real algebra $\mathbb{R} \oplus \mathbb{R}$, for example? There's certainly a nontrivial two-sided ideal.

Morgan Rogers (he/him) (May 24 2023 at 14:50):

Hmmm... I missed the "abelian" condition and was drawing intuition from the set-monoid case, where the action of a monoid on itself is not decomposable in its category of actions. Is $\R \oplus \R$ decomposable in the way suggested by the notation within its category of actions?

Morgan Rogers (he/him) (May 24 2023 at 14:51):

(I think the answer is yes; I had also ignored the abelianness condition on semisimple categories)

John Baez (May 24 2023 at 14:54):

Yes, $\mathbb{R} \oplus \mathbb{R}$ is the common notation for the commutative ring that deserves to be called $\mathbb{R} \times \mathbb{R}$ in the category of commutative rings, and $\mathbb{R} \oplus \mathbb{R}$ is the sum of two $\mathbb{R} \oplus \mathbb{R}$-modules.

John Baez (May 24 2023 at 14:55):

(For nonexperts: it's just the ring of ordered pairs of real numbers, with componentwise $+$ and $\times$. A module of this thing is just a pair of real vector spaces.)

Morgan Rogers (he/him) (May 24 2023 at 14:56):

When you say "category of actions", you need some finite generation condition, right? $\R^{\N}$ is not a finite direct sum of simple $\R$-vector spaces, for example.

John Baez (May 24 2023 at 14:57):

You're right... let me think about how that usually gets handled.

John Baez (May 24 2023 at 15:00):

Oh, I screwed up. This is wrong:

John Baez said:

I think we can define a 'semisimple' monoid in $\mathsf{A}$ to be a monoid whose category of actions is a [[semisimple category]].

In the case where $\mathsf{A} = \mathsf{Ab}$, for example, I want a semisimple monoid to be a semisimple ring. And this is not a defined to be a ring whose category of modules is semisimple; it's a ring that's semisimple as a left module over itself!

John Baez (May 24 2023 at 15:01):

The word "semisimple" is massively overloaded in algebra - I just used in 3 different standard senses.

John Baez (May 24 2023 at 15:02):

So let me define the key one here: a module of a ring is semisimple if it's a direct sum of simple modules, which in turn are defined to be nonzero modules with no proper submodules.

John Baez (May 24 2023 at 15:04):

Okay, here's the theorem Morgan probably wanted me to say: "Moreover, a ring R is semi-simple if and only if the category of finitely generated R-modules is semisimple. "

Morgan Rogers (he/him) (May 24 2023 at 15:09):

A crux of the usual proof is that if $I$ is a minimal left ideal of $R$ then $\mathrm{End}(I)$ is a division ring. To make sense of endomorphism monoids/rings we might need a cartesian-closedness condition on our abelian category?

Morgan Rogers (he/him) (May 24 2023 at 15:17):

If we have that, I think the argument should proceed as in the classical case - $I^I$ will be a division ring in the sense you described, since if it had a non-trivial ideal we could take its annihilator to contradict minimality (minimality meaning simple-ness, I expect), and we can also deduce that the endomorphism ring is a direct sum of these division rings.

Morgan Rogers (he/him) (May 24 2023 at 15:18):

The converse might be trickier? It's hard for me to be sure.

John Baez (May 24 2023 at 15:37):

Thanks! I was lazily hoping someone had already done this generalization, but you're diving in and starting to do it!

Morgan Rogers (he/him) (May 24 2023 at 15:54):

Heh Wedderburn-Artin is the first result I encountered in a "grown-up" representation theory course in my Masters, and it's remained a little mysterious to me so I was keen to look at it more closely. I'm not on top of the literature for this, though.

John Baez (May 24 2023 at 17:11):

I'm far from an expert on Wedderburn-Artin, but I seem to need to generalize it for my work on the tenfold way and its generalizations.

By the way, did you really mean cartesian closedness? I've often seen symmetric monoidal closed categories of modules of a commutative ring, with an internal hom right adjoint to the tensor product, but never cartesian closed ones. (In fact I think there should be some obstruction to interesting cartesian closed categories when the product is a biproduct, as it is in abelian categories.)

John Baez (May 24 2023 at 17:24):

Hmm, here's a generalization that goes in a different direction:

• Artin-Wedderburn in fusion categories.

John Baez (May 24 2023 at 17:25):

So [[fusion categories]] are merely monoidal, not symmetric monoidal, which is more general than I need, but also they have other properties that I may not want to assume. Most notably, they force us to work over a field, so Ostrik's theorem doesn't include the classical Artin-Wedderburn theorem:

"A fusion category is a rigid semisimple linear (Vect-enriched) monoidal category (“tensor category”), with only finitely many isomorphism classes of simple objects, such that the endomorphisms of the unit object form just the ground field k."

John Baez (May 24 2023 at 17:26):

The author claims Victor Ostrik showed something like this:

Any semisimple algebra object in a fusion category C is isomorphic (as an algebra object) to the internal endomorphisms End(X) for some object X in a semisimple module category M over C.

Morgan Rogers (he/him) (May 24 2023 at 17:56):

John Baez said:

By the way, did you really mean cartesian closedness?

I suppose not, it seems like monoidal closed is what I need, since I probably want $\mathrm{Hom}(R,I^I) \cong \mathrm{Hom}(I,I) = \mathrm{Hom}(R \otimes I, I)$? Or at least that's the relationship that holds when we're really working in a category of (left) $R$-modules, although the tensor product isn't globally defined...

Morgan Rogers (he/him) (May 24 2023 at 17:58):

The main thing is that I want to be able to talk about the internal monoid of endomorphisms of my object.

John Baez (May 24 2023 at 20:28):

Yes, I think you get all those things in the category of modules of a commutative ring, which is a paradigmatic example that I want to be able to handle.

John Baez (May 24 2023 at 20:32):

I don't know what you mean by saying the tensor product isn't "globally defined". For (left) modules of a commutative ring, which automatically become bimodules, the tensor product is well-defined.

John Baez (May 24 2023 at 20:35):

I'd be delighted if I could prove Wedderburn-Artin in any symmetric monoidal closed abelian category.

John Baez (May 24 2023 at 20:36):

Or if anyone could.

Jean-Baptiste Vienney (May 24 2023 at 21:00):

A very first step would consist into defining what is a matrix monoid over a monoid $D$ in a symmetric monoidal closed category with biproducts. I would say that the underlying object of the object of matrices $n \times p$ is $D^{\oplus n} \multimap D^{\oplus p}$ and then you should be able to define multiplication of matrices as some morphism $\nabla_{n,p,q}:(D^{\oplus n} \multimap D^{\oplus p}) \otimes (D^{\oplus p} \multimap D^{\oplus q}) \rightarrow (D^{\oplus n} \multimap D^{\oplus q})$. I thought to that the other time but I didn't how to define this multiplication.

Jean-Baptiste Vienney (May 24 2023 at 21:02):

(In our case, I guess that the case $n=p=q$ is sufficient, in general it would give a kind of graded monoid which should make commute the appropriate associativity diagram and with a unit $\eta:D \rightarrow D \multimap D$)

Jean-Baptiste Vienney (May 24 2023 at 21:06):

That's stuff which seems obvious to generalize to a pure categorical setting but is not, to me at least. I would be glad if you could explain to me. I'm not very experienced with monoidal closed categories.

Jean-Baptiste Vienney (May 24 2023 at 21:13):

So yeah, let's define matrices first :upside_down: , that's a tool I would be glad to be able to use

Jean-Baptiste Vienney (May 24 2023 at 21:17):

If you had matrices, you could also try to define their determinants which would be fun

Morgan Rogers (he/him) (May 24 2023 at 21:32):

John Baez said:

I don't know what you mean by saying the tensor product isn't "globally defined". For (left) modules of a commutative ring, which automatically become bimodules, the tensor product is well-defined.

I mean in the non-commutative case; the commutative case of Wedderburn-Artin is not quite so exciting.

Jean-Baptiste Vienney (May 24 2023 at 21:58):

My advancement in defining matrices is currently at this point:

• (1) from the multiplication $D \otimes D \rightarrow D$ I apply the monoidal closed adjunction to obtain a morphism $D \rightarrow D \multimap D$ which send an object to the $1 \times 1$ corresponding matrix.

Jean-Baptiste Vienney (May 24 2023 at 21:59):

I would now be glad to succeed to define

• (2) the multiplication of $1 \times 1$ matrices. So I must find a morphism $(D \multimap D) \otimes (D \multimap D) \rightarrow D \multimap D$ from the multiplication $D \otimes D \rightarrow D$

Jean-Baptiste Vienney (May 24 2023 at 22:01):

One other useful step would be to define

• (3) a multiplication $D^{\oplus n} \otimes D^{\oplus n} \rightarrow D^{\oplus n}$. That doesn't seem too difficult

Jean-Baptiste Vienney (May 24 2023 at 22:05):

Hopefully, you will be able to define the multiplication $\nabla_{n,p,q}$ of matrices from $(2)$ and $(3)$

Jean-Baptiste Vienney (May 24 2023 at 22:06):

• Another useful step is to define a morphism $D \rightarrow (D^{\oplus n} \multimap D^{\oplus n})$ which kinda gives the identity matrix $n \times n$ (more precisely, in the case of a monoid in $Mod_{R}$ it would send $x$ to $diag(x,...,x)$)

John Baez (May 25 2023 at 04:36):

Jean-Baptiste Vienney said:

A very first step would consist into defining what is a matrix monoid over a monoid $D$ in a symmetric monoidal closed category with biproducts. I would say that the underlying object of the object of matrices $n \times p$ is $D^{\oplus n} \multimap D^{\oplus p}$ and then you should be able to define multiplication of matrices as some morphism $\nabla_{n,p,q}:(D^{\oplus n} \multimap D^{\oplus p}) \otimes (D^{\oplus p} \multimap D^{\oplus q}) \rightarrow (D^{\oplus n} \multimap D^{\oplus q})$. I thought to that the other time but I didn't how to define this multiplication.

There should be a way to do it. Here is something simpler, maybe helpful, that doesn't use the monoid structure on $D$ at all.

There's a morphism called internal composition:

$\circ_{x,y,z} :(x \multimap y) \otimes (y \multimap z) \rightarrow (x \multimap z)$

in any symmetric monoidal closed category. (In fact a monoidal closed category is sufficient if we tensor things in the correct order, but I'm too lazy to remember the correct order so I'll use a symmetric monoidal closed category.) Using the hom-tensor adjunction we can get the above morphism if we have a morphism

$x \otimes (x \multimap y) \otimes (y \multimap z) \rightarrow z \qquad (\star)$

But notice that in any symmetric monoidal closed category there are evaluation morphisms

$x \otimes (x \multimap y) \to y$

$y \otimes (y \multimap z) \to z$

By suitably composing/tensoring these we get the desired morphism $(\star)$.

John Baez (May 25 2023 at 05:07):

By the way, it's also true that internal composition is "associative" and "unital", so for any object $x$ in a symmetric monoidal closed category, the object $x \multimap x$ becomes a monoid object with internal composition as its multiplication.

There's a special case when we take $x = I^{\oplus n}$, the $n$-fold direct sum of the unit object. If our symmetric monoidal category is the category of modules of some commutative ring $k$, $I^{\oplus n} \multimap I^{\oplus n}$ consists of $n \times n$ matrices with entries in $k$, and the procedure I'm sketching makes it into a matrix algebra with the usual matrix multiplication.

If we instead want to generalize the algebra of $n \times n$ matrices with entries in some algebra $A$ over $k$, we can do this:

Let $A$ be any monoid in our symmetric monoidal closed category. Then $A \otimes ( I^{\oplus n} \multimap I^{\oplus n})$ becomes a monoid since it's a tensor product of monoids.

John Baez (May 25 2023 at 05:12):

So, I guess the statement I'd like to prove goes roughly like this:

Suppose $\mathsf{A}$ is a symmetric monoidal closed category. We can define left, right, and two-sided ideals in any monoid object $R \in \mathsf{A}$. We say $R$ is simple if it has no two-sided ideals except $R$ and $0$, and division if it has no left or right ideals except $R$ and $0$.

John Baez (May 25 2023 at 05:25):

Wedderburn for $\mathsf{A}$: every simple monoid $R \in \mathsf{A}$ is isomorphic to one of the form $D \otimes (I^{\oplus n} \multimap I^{\oplus n})$ where $D$ is some division monoid.

John Baez (May 25 2023 at 05:34):

I don't expect this is true without some extra conditions on $\mathsf{A}$, like it being an abelian category. I know it's not true unless we add some sort of "finiteness" condition on $R$.

For example, if $\mathsf{A}$ is the category of vector spaces, we have:

Wedderburn for $\mathsf{Vect}$: every finite-dimensional simple monoid $R \in \mathsf{Vect}$ is isomorphic to one of the form $D \otimes (I^{\oplus n} \multimap I^{\oplus n})$ where $D$ is some division monoid in $\mathsf{Vect}$.

John Baez (May 25 2023 at 05:40):

Perhaps the only finiteness condition we need is that $R$ has some minimal left ideal. For example Nicholson gives a short proof of this version of Wedderburn's theorem when $\mathsf{A}$ is the category of abelian groups.

Wedderburn for $\mathsf{Ab}$: if $R \in \mathsf{Ab}$ is a simple monoid with a minimal left ideal, then it's isomorphic to one of the form $D \otimes (I^{\oplus n} \multimap I^{\oplus n})$ where $D$ is some division monoid in $\mathsf{Ab}$.

John Baez (May 25 2023 at 05:55):

Morgan pointed out the importance of this "minimal left ideal" business. Paraphrasing him:

A crux of the usual proof is that if $J$ is a minimal left ideal of $R$ then $\mathrm{End}(J)$ is a division ring.

John Baez (May 25 2023 at 05:58):

Nicholson mentions something called "Brauer's Lemma", which also involves a minimal left ideal and a division ring.

David Egolf (May 25 2023 at 15:04):

I haven't seen the notation $x \multimap y$ before. Is this another notation for $[x, y]$ where $[-,-]$ is an internal hom functor?

Jean-Baptiste Vienney (May 25 2023 at 15:18):

Yes, this is the notation used in linear logic for the linear implication, ie. if $x,y$ are proposition, $x \multimap y$ is the proposition which means "from one unit of $x$, I get one unit of $y$" which is not the same that $x \otimes x \multimap y$ which means "from two units of "x", I get one unit of $y$"

Jean-Baptiste Vienney (May 25 2023 at 15:19):

Categorically, the connectors $\otimes$ and $\multimap$ of linear logic correspond to symmetric monoidal closed categories.

Jean-Baptiste Vienney (May 25 2023 at 15:20):

Whereas, the connectors $\wedge$ and $\Rightarrow$ of intuitionist logic correspond to cartesian closed categories.

John Baez (May 25 2023 at 15:39):

David Egolf said:

I haven't seen the notation $x \multimap y$ before. Is this another notation for $[x, y]$ where $[-,-]$ is an internal hom functor?

Yes, it's an internal hom. This notation emphasizes that we've got an internal hom related to a tensor product $\otimes$ that's not necessarily cartesian. For example in Set or any other topos we've got

$\mathrm{hom}(X \times Y, Z) \cong \mathrm{hom}(X, Z^Y)$

but in Vect we've got

$\mathrm{hom}(X \otimes Y, Z) \cong \mathrm{hom}(X, Y \multimap Z)$

where $Y \multimap Z$ is the vector space of linear maps from $Y$ to $Z$. But as Jean-Baptiste noted, it's mainly linear logicians who use the symbol $\multimap$. Also category theorists. $[Y , Z]$ is perhaps more common.

Josselin Poiret (May 25 2023 at 15:41):

(that symbol is called a lollipop by the way)

David Egolf (May 25 2023 at 15:42):

Awesome, thanks for clarifying!

John Baez (Jun 10 2023 at 23:27):

By the way, one reason I gave up on this thread is that I realized that this version of Wedderburn's theorem is a bit limited:

Wedderburn for the symmetric monoidal closed category $\mathsf{A}$: every simple monoid $R \in \mathsf{A}$ is isomorphic to one of the form $D \otimes (I^{\oplus n} \multimap I^{\oplus n})$ where $D$ is some division monoid.

For example, if $\mathsf{A}$ is the category of complex vector bundles over a topological space, a monoid in $\mathsf{A}$ is an algebra bundle, and I believe there are simple monoids not of this form, coming from algebra bundles that are locally isomorphic to trivial bundles of matrix algebras, but not globally.

So, I think we should expect "Wedderburn for $\mathsf{A}$" to hold in the above form only under some conditions, e.g. when all projective objects are free. It would still be interesting (to me) to find a bunch of symmetric monoidal closed categories for which this limited form of Wedderburn holds.... or to state Wedderburn in a more general way. But I decided to go in another direction, which took me to separable algebras and Azumaya algebras.

John Baez (Jun 12 2023 at 18:12):

Okay, I finally understand the Wedderburn-Artin theorem well enough to have a chance of generalizing it. As part of my struggles I rewrote these articles:

• Wikipedia, Wedderburn-Artin theorem.
• nLab, [[Wedderburn-Artin theorem]].

The Wikipedia article Simple ring had a completely bogus proof of the Wedderburn-Artin theorem which one editor had been complaining about for a year! I deleted it.

John Baez (Jun 12 2023 at 19:34):

So let me try to sketch how Wedderburn-Artin might be generalized. Let $\mathsf{A}$ be a closed symmetric monoidal Ab-enriched category with small limits and colimits, though I would like to reduce these hypotheses once I've actually proved what I want to prove!

John Baez (Jun 12 2023 at 19:35):

Every monoid $R \in \mathsf{A}$ has a category of actions, which I'll call modules and denote as $R \mathsf{Mod}$.

If you want to understand what I'm doing, it's good to think about the example $\mathsf{A} = \mathsf{Ab}$, where $R$ will be a ring. This is what I'm trying to generalize. Another example might be a category of sheaves of abelian groups, or sheaves of modules of some fixed commutative ring.

John Baez (Jun 12 2023 at 19:39):

Conjecture A. $R \mathsf{Mod}$ is Ab-enriched with small limits and colimits, and symmetric monoidal if $R$ is commutative.

John Baez (Jun 12 2023 at 19:42):

Conjecture B. $R \mathsf{Mod}$ is $\mathsf{A}$-enriched.

John Baez (Jun 12 2023 at 19:45):

The idea here is that given $M, N \in R \mathsf{Mod}$ we can first form $[M,N]$ using the internal hom in $\mathsf{A}$, but then pick out a subobject of this which serves as the object of $R$-module homomorphisms. Maybe we can call that ${}_R \mathrm{hom}(M,N) \in \mathsf{A}$ if we're using left $R$-modules.

John Baez (Jun 12 2023 at 19:45):

This is different than the abelian group of $R$-module homomorphisms from $M$ to $N$, which also should exist.

John Baez (Jun 12 2023 at 19:49):

We say $R$ is semisimple if $R \mathsf{Mod}$ is a [[semisimple category]] (viewed externally, I guess, as an Ab-enriched category).

John Baez (Jun 12 2023 at 19:50):

$R$ is always a left module over itself in a standard way, giving an object I'll call ${}_R R \in R \mathsf{Mod}$.

John Baez (Jun 12 2023 at 19:50):

When $R$ is semisimple this object ${}_R R$ will be a coproduct of finitely many simple objects.

John Baez (Jun 12 2023 at 19:53):

Conjecture (External Wedderburn-Artin Theorem). When $R$ is semisimple, the ring of left $R$-module endomorphisms of $R$ is a finite direct sum of matrix algebras over division rings.

John Baez (Jun 12 2023 at 19:55):

This should actually be pretty easy, at least if I have enough of the right assumptions on $\mathsf{A}$: for example, I believe the endomorphism ring of any simple object in any abelian category is a division ring by Schur's lemma, and the rest of the argument should mimic this.

John Baez (Jun 12 2023 at 19:57):

(But I didn't actually say $\mathsf{A}$ is abelian - does it follow? - and also we may not need the full force of abelianness to do this argument.)

John Baez (Jun 12 2023 at 20:01):

But what I want is an internal version of Wedderburn-Artin which describes $R$ as a monoid in $\mathsf{A}$, not the ring of left $R$-module endomorphisms of $R$.

Note that when $\mathsf{A} = \mathsf{AbGp}$ these are the same! That is, we have an isomorphism of rings between $R$ and the ring of $R$-module endomorphisms of $R$ as a left $R$-module. (Hmm, I guess I need to treat it as a right $R$-module instead, for this to be true.)

John Baez (Jun 12 2023 at 20:01):

But in the case of general $\mathsf{A}$ I want to "internalize" the external Wedderburn-Artin conjecture to get a description of $R$ itself.

Mike Shulman (Jun 12 2023 at 20:18):

I'd be surprised if your assumptions imply that $\mathsf{A}$ is abelian.

Mike Shulman (Jun 12 2023 at 20:23):

But conjectures A and B should be pretty straightforrward enriched category theory, even with Ab replaced by some other Benabou cosmos. E.g. for conjecture B, you can think of $R\mathsf{Mod}$ as an $\mathsf{A}$-enriched presheaf category on $R$ regarded as a one-object $\mathsf{A}$-enriched category. Then to get the Ab-enrichment, you can use the fact that a monoidal Ab-enriched category is the same as an ordinary monoidal category $\mathcal{A}$ with a monoidal adjunction $\mathrm{Ab} \rightleftarrows \mathcal{A}$, and apply change-of-enrichment along the right adjoint. That right adjoint sends an object $X\in \mathsf{A}$ to the Ab-enriched hom $\hom_{\mathsf{A}}(I,X)$, where $I$ is the monoidal unit of $\mathsf{A}$.

John Baez (Jun 12 2023 at 21:34):

Nice, thanks! That's slick.

Mike Shulman (Jun 12 2023 at 21:40):

I suppose you could get the monoidal structure on $R\mathsf{Mod}$ by noting that when $R$ is commutative, its corresponding one-object $\mathsf{A}$-category is a monoidal $\mathsf{A}$-category, and so its presheaf category has a Day convolution monoidal structure. But it might be easier to just write down the tensor product explicitly.

John Baez (Jun 12 2023 at 21:48):

The point at which I'm tempted to use the abelianness of $\mathsf{A}$ comes later, in proving Schur's lemma. People define an object to be simple if it has no subobjects other than 0 and itself, but in an abelian category this is equivalent to saying it has no quotients other than 0 or itself. So, in an abelian category you can see that an endomorphism of a simple object is either zero or invertible.

John Baez (Jun 12 2023 at 21:48):

This makes the endomorphism ring a division ring.

John Baez (Jun 12 2023 at 21:50):

Maybe we need less than abelianness for this, or else I could just decree a simple object has no subobjects or quotients other than 0 and itself.

Mike Shulman (Jun 12 2023 at 21:51):

Yeah, that was the first thing that occurred to me too. But why don't you want to assume that $\mathsf{A}$ is abelian?

John Baez (Jun 12 2023 at 21:56):

Maybe I'm being silly, but I might want $\mathsf{A}$ to be the category of real vector bundles on a topological space, which is not abelian.

Mike Shulman (Jun 12 2023 at 21:59):

Ok. So first of all, that example shows that your assumptions don't imply $\mathsf{A}$ is abelian, right? (-:

John Baez (Jun 12 2023 at 21:59):

Right.

Mike Shulman (Jun 12 2023 at 22:00):

Do you have some intuition for what a "simple vector bundle" should be?

John Baez (Jun 12 2023 at 22:22):

A line bundle I guess. Those are the ones whose endomorphism algebra bundles are division algebra bundles.
But I see now that's a kind of "internal" version of being simple: their rings of endomorphisms are huge and not at all division rings.

But btw, also, the place I tend to want abelianness is not $\mathsf{A}$ but for $R \mathsf{Mod}$ where $R$ is a monoid in $\mathsf{A}$.

John Baez (Jun 12 2023 at 22:23):

Anyway, I should think more about this example and see if I even want to handle it.

Mike Shulman (Jun 12 2023 at 22:30):

Note that $\mathsf{A}$ is a special case of $R\mathsf{Mod}$ for $R$ a monoid in $\mathsf{A}$, namely $R=I$ the monoidal unit.

Mike Shulman (Jun 12 2023 at 22:31):

Anyway, that makes me wonder whether the category of vector bundles is "internally abelian" in some sense.

John Baez (Jun 13 2023 at 22:53):

Luckily I have other motivating examples that don't force me to stretch so far: interesting examples of symmetric monoidal abelian $\mathsf{A}$ where it seems the 'semisimple algebras' in $\mathsf{A}$ are very interesting.

John Baez (Jun 13 2023 at 22:55):

My favorite is the symmetric monoidal category of super-vector spaces over a field, i.e. $\mathbb{Z}/2$-graded vector spaces where we use the symmetry that puts in a minus sign when we switch two objects. This is connected to the tenfold way. But people have also generalized a lot of the results to $\mathbb{Z}/2$-graded vector spaces!

John Baez (Jun 13 2023 at 22:56):

Okay, here's something a bit different an attempt to set up a theory of something like semisimple categories and division algebras with very low prerequisites:

John Baez (Jun 13 2023 at 22:56):

Let $\mathsf{A}$ be a semiadditive category, i.e. a category with binary biproducts and a zero object (a both initial and terminal object).

John Baez (Jun 13 2023 at 22:58):

Such a category is automatically $\mathsf{CommMon}$-enriched.

John Baez (Jun 13 2023 at 23:47):

Let's say an object is atomic if all its endomorphisms are zero or invertible.

Let's say two objects are separate if the only morphism between them is the zero morphism.

John Baez (Jun 13 2023 at 23:48):

Let's say our category $\mathsf{A}$ is molecular if there exists a set of mutually separate atomic objects such that every object is a finite biproduct of copies of these objects.

I don't intend these terms to be permanent terminology, so please don't anyone hassle me about them - I just need some words for now to state a few easy results. But if you know more standard terms please let me know - or even better, useful theorems related to these ideas! (They are much more familiar for additive categories.)

John Baez (Jun 13 2023 at 23:50):

I believe for every atomic object $x$, $\mathrm{End}(x)$ is a division rig, i.e. a [[rig]] such that every nonzero element is invertible.

John Baez (Jun 13 2023 at 23:52):

Thanks to how biproducts work, if $x^n$ is the n-fold biproduct $x \oplus \cdots \oplus x$ then

$\mathrm{End}(x^n) \cong M_n(\mathrm{End}(x))$

that is, is the rig of $n \times n$ matrices with entries in $\mathrm{End}(x)$.

John Baez (Jun 13 2023 at 23:56):

In a molecular category we then get a kind of Wedderburn-Artin-like result. Every object is of the form

$y \cong x_1^{n_1} \oplus \cdots \oplus x_k^{n_k}$

John Baez (Jun 13 2023 at 23:59):

and so we get

$\displaystyle{ \mathrm{End}(y) \cong \prod_{i = 1}^k M_{n_i}(\mathrm{End}(x_i)) }$

John Baez (Jun 14 2023 at 00:00):

This is like how every semisimple algebra is a finite product of matrix algebras over division rings!

Mike Shulman (Jun 14 2023 at 05:27):

Interesting. How much of that has moved the theorem into the definitions? That is, if you specialize that to the classical Wedderburn-Artin theorem, is the statement "such-and-such category is molecular" closer to the hypotheses of the theorem or to its conclusion?

John Baez (Jun 14 2023 at 05:48):

A few things to say here:

• "Schur's lemma", or some abstraction of the usual version, says that in an abelian category an atomic object is the same as a simple object. This has a very short and conceptual proof.

• As an immediate consequence, an abelian category is semisimple iff it's molecular.

• A semisimple ring can be defined as one whose category of right modules is semisimple, but by the above that's equivalent to saying its category is molecular.

• Then by my above generalization of Wedderburn-Artin we see that if $R$ is a semisimple ring, $\mathrm{End}(R_R)$ is a finite product of matrix algebras over division rings.

• Then we need an extra fact for rings, $\mathrm{End}(R_R) \cong R$, to see that if $R$ is semisimple it's a finite product of matrix algebras over division rings. This is the "traditional" Wedderburn-Artin theorem.

John Baez (Jun 14 2023 at 05:54):

So, I think the proof of my generalization of Wedderburn-Artin is pretty much the same as the good proof of the "traditional" version, except that the "traditional" one also needs Schur's lemma to check that simple objects in an abelian category are atomic, i.e. have division rings of endomorphisms.

John Baez (Jun 14 2023 at 05:54):

It's all pretty easy stuff (now that I understand it).

John Baez (Jun 14 2023 at 05:58):

One thing I want to generalize away from the traditional case of rings is $\mathrm{End}(R_R) \cong R$, so that my generalized Wedderburn-Artin, which describes endomorphism rigs of objects, can actually describe the objects themselves in some cases.

John Baez (Jun 14 2023 at 05:59):

This is some sort of closed category / enriched presheaf stuff.

John Baez (Jun 14 2023 at 21:26):

Okay, I wrote up a blog post on generalizing the Wedderburn-Artin theorem, and I think I'll declare victory for now, though there still are some issues, like how generally we can pull a trick like $\mathrm{End}(R_R) \cong R$:

• The Wedderburn-Artin theorem, n-Category Cafe, June 6, 2023.

@Jean-Baptiste Vienney will be happy to see that I've reduced the assumptions to a bare minimum - so for example, my result applies to rigs.

Jean-Baptiste Vienney (Jun 14 2023 at 21:30):

Thank you, I wanted to protest when you were talking about abelian categories but I didn't want to be annoying. I'm glad that it applies to rigs!

John Baez (Jun 14 2023 at 21:34):

Yes, me too! I try to get the basic ideas working before removing assumptions. It was only recently that I understand Wedderburn-Artin well enough to generalize it.