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Stream: learning: questions

Topic: Semisimple commutative rings

Jean-Baptiste Vienney (Dec 21 2024 at 22:40):

Maybe irrelevant to the discussion, but to me semisimple commutative rings (product of fields) are useful since the category of (left) modules over such a ring resembles a lot the one of vector spaces over a field: it is a symmetric monoidal category enriched over abelian groups and every module is flat, meaning that the tensor product preserves finite limits.

In fact, I like them only because they are Von Neumann regular commutative rings. The category of modules over such a ring is exactly as I say above, but other rings that products of fields are Von Neumann regular commutative rings, for example boolean rings.

Todd Trimble (Dec 21 2024 at 23:47):

It's true that Boolean rings are von Neumann regular, but not all Boolean rings are products of fields. Those that are are the power set Boolean algebras (complete atomic Boolean algebras).

Jean-Baptiste Vienney (Dec 22 2024 at 00:40):

I've never said that all Boolean rings are products of fields, I've said that other rings that products of fields are von Neumann regular commutative rings, for example boolean rings. What I didn't say is that some Boolean rings are product of fields, so I should have written "most boolean rings".

Jean-Baptiste Vienney (Dec 22 2024 at 00:48):

Something that I would like to see on an example is how a commutative von Neumann regular ring is exactly a subring of a product of fields closed under taking weak inverses as they say on Wikipedia.

Jean-Baptiste Vienney (Dec 22 2024 at 00:52):

For instance, how is the boolean ring $\mathbb{B}=\{\mathrm{true},\mathrm{false}\}$ a subring of a product of field closed under taking weak inverses?

John Baez (Dec 22 2024 at 00:53):

When you treat $\mathbb{B}$ as a Boolean ring, it is a field!

Jean-Baptiste Vienney (Dec 22 2024 at 00:58):

Oh sorry, I was confused. $\mathbb{B}$ is a multiplicative idempotent semiring but not even a ring. It is a boolean algebra and the associated boolean ring is $\mathbb{Z}_2$ which is a field.

John Baez (Dec 22 2024 at 00:59):

I used to know something about von Neumann regular rings, because they were invented in an approach to the foundations of quantum mechanics, and I used to be interested in all those approaches. But I seem to have forgotten everything about them.

Jean-Baptiste Vienney (Dec 22 2024 at 01:11):

I've just understood a little bit more about them: on Wikipedia, they write:
Screenshot 2024-12-21 at 8.10.23 PM.png
But in a boolean ring, the weak inverse "y" is just the unity "1". So it seems to mean that the boolean rings are just the subrings of products of fields.

Jean-Baptiste Vienney (Dec 22 2024 at 01:12):

Oh no sorry

Jean-Baptiste Vienney (Dec 22 2024 at 01:14):

I don't understand what is this canonical weak inverse in a boolean ring. It can't be $1$ all the time as I wrote since you also want $y^2x=x$ and not only $x^2y=x$.

John Baez (Dec 22 2024 at 01:45):

So it seems to mean that the boolean rings are just the subrings of products of fields.

In my version of mathematical English you'd only say this if every subring of a product of fields is a boolean ring, which of course is false. If you say

So it seems to mean that boolean rings are subrings of products of fields.

this problem goes away. (I was going to study all 8 versions of leaving out words "just" and "the" from your sentence, but I decided this was a bit obsessive.)

Talking about "fields" here is unnecessary. I believe every boolean ring is a subring of a product of copies of $\mathbb{F}_2$. I don't think considering other fields helps.

Todd Trimble (Dec 22 2024 at 02:28):

While we're on the issue of mathematical English, the phrase

other rings that [are] products of fields are Von Neumann regular commutative rings, for example boolean rings

strongly suggests to me that the writer means "other rings that are products of fields are Von Neumann regular commutative rings, for example boolean rings [seeing that these are products of fields]". And since this seems like an easy trap to fall into, I thought I should say something. Just in case.

But while we're on the topic of von Neumann regular rings: this from the Wikipedia article seems pretty interesting:

because these rings R are characterized by the fact that every left R-module is flat

Jean-Baptiste Vienney (Dec 22 2024 at 02:47):

Todd Trimble said:

because these rings R are characterized by the fact that every left R-module is flat

I don't know about you but that's the main reason why I know that there is something called von Neumann regular ring.

Todd Trimble (Dec 22 2024 at 03:26):

I confess that I've never studied von Neumann regular rings; the only definition I had ever seen was this to-me goofy-looking "for all $a$ there exists $x$ such that $axa = a$". Which I don't find terribly illuminating. The flat module characterization, which I did not know, has to me a more sensible shape to it.

Now I'm a little curious...

Notification Bot (Dec 22 2024 at 03:31):

15 messages were moved here from #learning: questions > Field with one element by Madeleine Birchfield.

Todd Trimble (Dec 22 2024 at 03:32):

Jean-Baptiste Vienney said:

I don't understand what is this canonical weak inverse in a boolean ring. It can't be $1$ all the time as I wrote since you also want $y^2x=x$ and not only $x^2y=x$.

Looks like $y = x$ works.

Jean-Baptiste Vienney (Dec 22 2024 at 03:33):

How yes, you're right

Valery Isaev (Dec 22 2024 at 10:59):

Another interesting property of Von Neumann regular commutative rings is that a finitely generated algebra over such a ring is integral if and only if it is 0-dimensional.

John Baez (Dec 22 2024 at 18:34):

Todd Trimble said:

I confess that I've never studied von Neumann regular rings; the only definition I had ever seen was this to-me goofy-looking "for all $a$ there exists $x$ such that $axa = a$". Which I don't find terribly illuminating.

"Young man, in mathematics you don't understand things. You just get used to them." - John von Neumann. :smiling_devil:

I thoroughly disagree with this quote but it's funny that his name is attached to this rather cryptically defined kind of ring. So I decided to look at his 1936 paper Regular rings and see how he introduced them.

By page 2 he has shown the following are equivalent for any element $a$ of any ring $A$:

• There exists $x$ with $a x a = a$.
• There exists an idempotent $e$ such that the left ideal $A e$ equals the left ideal $A x$.
• There exists an idempotent $e$ such that the right ideal $e A$ equals the right ideal $x A$.

The latter two conditions are pretty nice, and it's pretty surprising that they're equivalent.

John Baez (Dec 22 2024 at 18:38):

This equivalence is Lemma 6, and he ramps up to it at high speed, like a aircraft that takes off vertically at high gees.

Lemma 1 is that if $e$ is an idempotent, $1 - e$ is an idempotent: I guess back then not everyone reading an algebra paper would be expected to know that!

John Baez (Dec 22 2024 at 18:40):

Lemma 2 is also one I know, and it's cute: if $e$ is idempotent, $x \in e A$ iff $e x = x$.

Everyone who doesn't know this one should try to prove it - it's very easy.

Todd Trimble (Dec 22 2024 at 19:00):

Well, this looks very nice so far, and these are the kinds of things where it's nice just to see the statements and then prove them by one's self for as long as one can.

I had looked at the Wikipedia article, especially at the remarks on equivalent ways to view commutative von Neumann rings $R$, and can figure out some of it, but I don't see yet how to prove existence of the "weak inverse" of an element $r \in R$. (Uniqueness of the weak inverse is fairly straightforward, and there are other things I can do like no non-zero nilpotents and Krull dimension zero and triviality of the Jacobson radical and embedding into a product of fields, but existence of the weak inverse so far has me stumped.)

I like the idea of using unique existence, because it shows that commutative von Neumann rings form a variety of algebras [i.e. are models of a Lawvere theory], whereas von Neumann rings generally don't seem to be that way.

John Baez (Dec 22 2024 at 19:01):

I was just getting to the good part....

John Baez (Dec 22 2024 at 19:14):

I left off here:

Lemma 2: if $e$ is idempotent, $x \in e A$ iff $e x = x$.

Things get interesting at Lemma 5. It says $a A = e A$ for some idempotent $e$ iff for some $x \in A$ we have $a x a = a$. When these hold, we can take $e = a x$.

Let's see his proof of the forwards direction! Assume $a A = e A$ for some idempotent $e$. This is equivalent to $e \in a A$, $a \in e A$, and $e^2 = e$.

The first condition says $e = ax$ for some $x \in A$. By Lemma 2, the second condition is equivalent to $a = e a$, but since $e = a x$ this says $a = a x a$. Done!

Okay, now let's do the backwards condition. Suppose for some $x \in A$ we have $a x a = a$. Thus $a x a x = a x$ so $e = a x$ is idempotent. We just need to show $a A = e A$. It's clear that $e A \subseteq a A$ because $e = a x$. Why is $a A \subseteq e A$? That's because $a = a x a = e a$.

Whew!

John Baez (Dec 22 2024 at 19:22):

I'd say this argument is the key to why this mysterious condition "for every $a$ there exists $x$ with $a = a x a$" shows up in the theory of von Neumann regular rings.

Jean-Baptiste Vienney (Dec 22 2024 at 19:24):

I've learned about semisimple rings in an algebra course this semester and I always liked when idempotents were used in propositions or proofs. So these first lemmas look very exciting to me! Still stuff with idempotents but to study more general rings than semisimple ones.

John Baez (Dec 22 2024 at 21:40):

I have uses von Neumann's lemmas above to improve the nLab article [[von Neumann regular ring]], so now that article says something about what the equation $a = a x a$ is good for.

Jean-Baptiste Vienney (Dec 22 2024 at 21:47):

I think I will try to add the proofs of the implications between the equivalent definitions once I understand them if no one do it before me.

Jean-Baptiste Vienney (Dec 22 2024 at 21:47):

For now I'm looking at von Neumann's paper

John Baez (Dec 22 2024 at 22:01):

In the current list of definitions on nLab, I gave the proof that 1 $\iff$ 2, and the proof that 1 $\iff$ 3 is the same (that is, symmetrical). The proof that $2 \implies 4$ is easy, as is $3 \implies 5$.

(I am using the correct numbers rather than the screwed-up numbers I now see, where there are two definitions called "1". I could fix the numbering, but you're editing the page now!)

Jean-Baptiste Vienney (Dec 22 2024 at 22:03):

That's funny that von Neumann doesn't require an ideal to be nonempty. It doesn't help that the symbol $\emptyset$ for the empty set started to be used only in 1939 according to this webpage while von Neumann's paper is from 1936.

Jean-Baptiste Vienney (Dec 22 2024 at 22:07):

I don't think I changed any numbering.

Jean-Baptiste Vienney (Dec 22 2024 at 22:30):

I think for von Neumann, in this paper, a set is a nonempty set since he says that $(0)$ is the minimal left ideal which would be false if he considered $\emptyset$ as a set.

Jean-Baptiste Vienney (Dec 22 2024 at 23:16):

I was very confused but I've finally understood that what he means by "inverse ideal of an ideal J" is an ideal $I$ s.t.$I \oplus J=R$ as $R$-modules, ok.

Todd Trimble (Dec 23 2024 at 16:31):

Todd Trimble said:

I had looked at the Wikipedia article, especially at the remarks on equivalent ways to view commutative von Neumann rings $R$, and can figure out some of it, but I don't see yet how to prove existence of the "weak inverse" of an element $r \in R$. (Uniqueness of the weak inverse is fairly straightforward, and there are other things I can do like no non-zero nilpotents and Krull dimension zero and triviality of the Jacobson radical and embedding into a product of fields, but existence of the weak inverse so far has me stumped.)

I like the idea of using unique existence, because it shows that commutative von Neumann rings form a variety of algebras [i.e. are models of a Lawvere theory], whereas von Neumann rings generally don't seem to be that way.

Oh for heaven's sake: existence of a weak inverse for any element $a$ (in a von Neumann regular ring) turns out to be really elementary. I found this in the Wikipedia article on the closely related article on regular semigroups: suppose we have $x$ such that $axa = a$. The claim is that $y = xax$ is a weak inverse of $a$:

• $aya = axaxa = (axa)xa = axa = a$

• $yay = (xax)a(xax) = x(axa)(xax) = xa(xax) = x(axa)x = xax = y$

For uniqueness of a weak inverse in a commutative semigroup, suppose we have the four equations

$axa = a \qquad xax = x \qquad aya = a \qquad yay = y$

We want to prove $x = y$. First prove $ax = ay$:

$ax = xa = xaya = (xay)a = a(xay) = (axa)y = ay$.

Then prove $x = y$:

$x = x(ax) = x(ay) = y(ax) = y(ay) = y$

where only the middle equality uses commutativity.

Todd Trimble (Dec 23 2024 at 17:05):

Jean-Baptiste Vienney said:

Something that I would like to see on an example is how a commutative von Neumann regular ring is exactly a subring of a product of fields closed under taking weak inverses as they say on Wikipedia.

For whatever interest this may have at this point, the following is not an example, but a proof. Actually, instead of spitting out the proof, let me just outline the proof by splitting it up into a series of "puzzles", a ploy I'm copying from John. They all concern commutative von Neumann regular rings $R$. My previous comment proved that weak inverses of elements uniquely exist, so that commutative von Neumann regular rings can be defined equationally: they are commutative rings $R$ equipped with a unary operation $a \mapsto \overline{a}$ such that $a \overline{a} a = a$ and $\overline{a} a \overline{a} = \overline{a}$.

• Puzzle 1: show there are no nonzero nilpotents in $R$. For this, it is enough to show Claim: $r^2 = 0 \Longrightarrow r = 0$ (why is this claim enough?). Hint for claim: consider a weak inverse of $r$; use commutativity.

• Puzzle 2: show that an [[integral domain]] $R$ that is von Neumann regular must be a field.

• Puzzle 3: show that all prime ideals in $R$ are maximal.

• Puzzle 4: show that the Jacobson radical of $R$, i.e., the intersection of all maximal ideals of $R$, is zero. Hint: it helps to know something about the intersection of all prime ideals, and to use Puzzles 1 and 3.

• Puzzle 5: show that $R$ embeds in a product of fields. Hint: show that the obvious ring map $R \to \prod_{\mathfrak{m}} R/\mathfrak{m}$, where the product is over all maximal ideals, is injective.

Now the statement from the quotation of Jean-Baptiste follows from Puzzle 5, since any ring homomorphism preserves weak inverses.

Jean-Baptiste Vienney (Dec 23 2024 at 17:50):

I'm not sure I will solve all the puzzles now but thank you!!

Todd Trimble (Dec 23 2024 at 17:58):

John Baez said:

In the current list of definitions on nLab, I gave the proof that 1 $\iff$ 2, and the proof that 1 $\iff$ 3 is the same (that is, symmetrical). The proof that $2 \implies 4$ is easy, as is $3 \implies 5$.

(I am using the correct numbers rather than the screwed-up numbers I now see, where there are two definitions called "1". I could fix the numbering, but you're editing the page now!)

Just a quick observation, that all the arguments given in John's recent edition of the nLab article hold in the context of regular semigroups: addition wasn't really used anywhere.

I'd still like to wend my way to the characterization via flat modules, though.

And somehow I'd like to see still more examples of von Neumann regular rings!

David Michael Roberts (Dec 24 2024 at 00:10):

I added a couple of examples to [[von Neumann regular ring]], but certainly more can be explained, or in more detail

John Baez (Dec 24 2024 at 01:24):

That's cool how you were able to define commutative von Neumann rings as a variety of rings by treating 'weak inverse' as an operation, @Todd Trimble. This instantly gives us a lot of theorems.

It's also cool that none of this uses the additive structure, so that we get a concept of 'von Neumann regular commutative monoid'. Does this concept have a name? I imagine the hard-core universal algebraists would have studied it already.

John Baez (Dec 24 2024 at 01:32):

@Todd Trimble says that in a ring,

$r^2 = 0 \Longrightarrow r = 0$

is enough to imply zero is the only nilpotent. I didn't know that!

Let me do one case. Suppose $r^3 = 0$. Then $r^4 = 0$, so using the implication above we get $r^2 = 0$, and using it again we get $r = 0$.

Let me do another case. Suppose $r^4 = 0$. Oh, that's too easy, obviously we get $r = 0$.

So let me do another case. Suppose $r^5 = 0$. Then $r^8 = 0$, so using the above implication three times we get $r^4 = 0$, and then $r^2 = 0$, and then $r = 0$.

So yeah, the pattern is clear. Nice!

I used to know physicists who said some element $x$ was nilpotent iff it obeyed $x^2 = 0$. I made fun of them, saying physicists were really impatient, so if something didn't square to zero then to hell with it - they didn't have time to raise it to higher powers! But it turns out they weren't completely wrong: a ring with no nonzero elements that square to zero has no nonzero nilpotents.

Todd Trimble (Dec 24 2024 at 03:33):

John Baez said:

That's cool how you were able to define commutative von Neumann rings as a variety of rings by treating 'weak inverse' as an operation, Todd Trimble. This instantly gives us a lot of theorems.

It's also cool that none of this uses the additive structure, so that we get a concept of 'von Neumann regular commutative monoid'. Does this concept have a name? I imagine the hard-core universal algebraists would have studied it already.

Yeah, I'm pretty sure regular monoids and regular semigroups have been studied a lot. Benjamin Steinburg could tell you all about them.

John Baez (Dec 24 2024 at 04:54):

Steinberg. Yes, he's the expert on monoids and their representations!

Todd Trimble (Dec 24 2024 at 17:11):

Thanks for the spelling correction! Yes, something looked funny to me about "Steinburg" even as I wrote it, and if only I bothered to consult the side of me that is trying to learn German these days, "Steinberg" would have seemed rather more plausible.

Anyhow, I'm having fun learning more about von Neumann regular rings. I still want to hear more examples, for example any infinite-dimensional examples that emerged from von Neumann's mathematical study of quantum mechanics. Apparently von Neumann rings are closely connected with "continuous geometries" (certain complemented modular lattices) which were very much a hot topic in a different era, but I don't hear much about them any more.

Here is one delightful fact that I learned a few minutes ago, from a MathStackexchange post by @Ryan Schwiebert: "von Neumann regularity is a Morita invariant property, and hence it will be preserved by matrix rings". In other words, if $R$ is von Neumann regular, then so is the matrix algebra $M_n(R)$.

I think this fact could be waggled into a proof of one of the facts mentioned in the Wikipedia article, that a finitely generated submodule $N$ of a finitely generated free module $R^n$ is a direct summand of $R^n$. In other words, that the inclusion $N \hookrightarrow R^n$ splits. And from here, one can argue that every (let's say left) module of a von Neumann regular ring $R$ is a flat $R$-module, as follows:

Any module is a filtered colimit of finitely presented modules -- this is something basic and general, falling under the basic framework of Gabriel-Ulmer duality or locally finitely presentable categories or what have you. And a filtered colimit of flat modules is again flat. So it would suffice to prove that a finitely presented $R$-module $M$ is flat. So suppose we have an exact sequence

$0 \to N \to R^n \to M \to 0$

where $N$ is finitely generated. This exact sequence splits, by the previous paragraph. So then $M$ is a direct summand of $R^n$, hence is projective. But projective modules are flat. Done!

Somewhere deep in the von Neumann paper, around lemmas 15 and 16, it is shown that principal left (or right) ideals of von Neumann regular rings are closed under meets and joins in the lattice of all left ideals. (And principal ideals are generated by idempotents; it's not too hard to see from there that the lattice of principal ideals is a complemented [[modular lattice]].) This lore is part of the chain of results in the Wikipedia article that lead up to the flat module result, and anyhow it's pretty cool to see von Neumann argue his way up to lemmas 15, 16.

John Baez (Dec 24 2024 at 18:29):

Great stuff! I couldn't see why all modules of a von Neumann regular ring should be flat, but now I roughly get the idea. It seems intuitively plausible to me that von Neumann regularity is Morita invariant, because I think of von Neumann regular rings as an algebraist's distillation of von Neumann algebras, and any matrix algebra with entries in a von Neumann algebra is a von Neumann algebra. But that's an argument of the form "it's true for the examples I understand, so maybe it's true".

John Baez (Dec 24 2024 at 18:29):

Any module is a filtered colimit of finitely presented modules -- this is something basic and general, falling under the basic framework of Gabriel-Ulmer duality or locally finitely presentable categories or what have you. And a filtered colimit of flat modules is again flat.

This is the kind of black magic that makes me glad you're my friend, not my enemy. I'm sure it's all quite obvious once you understand it! But from the outside it seems quite intimidating.

John Baez (Dec 24 2024 at 18:39):

Apparently von Neumann rings are closely connected with "continuous geometries" (certain complemented modular lattices) which were very much a hot topic in a different era, but I don't hear much about them any more.

These days people seem interested in things like type $\text{II}$ von Neumann algebras for other reasons, like their connection to quantum groups, conformal field theories and other related things.

For anyone who has never thought about this stuff:

Type $\text{II}$ [[von Neumann algebras]] are certain algebras of bounded operators on a Hilbert space, that have a "trace" obeying

$\mathrm{tr}(AB) = \mathrm{tr}(BA)$

for which the trace of an idempotent can take a continuous range of values, either $[0,1]$ for the type $\text{II}_1$ guys or $[0,\infty]$ for the scarier $\text{II}_\infty$ guys. Since the trace of a projection onto a subspace of a Hilbert space is its dimension, we think of the trace of an idempotent as a "dimension", so these type $\text{II}$ von Neumann algebras are describing geometries where instead of points, lines, planes, etc. with natural number dimensions, we have figures with continuously variable dimension.

Todd Trimble (Dec 24 2024 at 20:14):

So it may be a silly question. but are von Neumann algebras von Neumann regular as rings? (This is the sort of thing I was hoping to hear, as a source of more examples.)

John Baez said:

Any module is a filtered colimit of finitely presented modules -- this is something basic and general, falling under the basic framework of Gabriel-Ulmer duality or locally finitely presentable categories or what have you. And a filtered colimit of flat modules is again flat.

This is the kind of black magic that makes me glad you're my friend, not my enemy. I'm sure it's all quite obvious once you understand it! But from the outside it seems quite intimidating.

I'll try to whiten it up. A left $R$-module $M$ is flat if the functor $- \otimes_R M: \mathsf{Ab}^{R^{op}} \to \mathsf{Ab}$, from right modules to abelian groups, is left exact: preserves finite limits. Meanwhile, filtered colimits of modules or of abelian groups commute with finite limits of modules or of abelian groups, because this is true at the underlying set level, and the forgetful functor from modules/abelian groups to sets preserves and reflects filtered colimits and finite limits. (The same is true replacing modules by algebras of any finitary = filtered colimit-preserving monad.)

So if $M_i$ is a filtered diagram of flat modules, we can check that $M = \mathrm{colim}_i M_i$ is flat, i.e., that $- \otimes_R M$ preserves finite limits:

$- \otimes_R M \cong - \otimes_R (\mathrm{colim}_i M_i) \cong \mathrm{colim}_i (- \otimes_R M_i)$

where the last functor is a filtered colimit of finite limit-preserving functors. This filtered colimit also preserves finite limits, by the commutation of filtered colimits and finite limits.

John Baez (Dec 24 2024 at 21:15):

Nice! I think I just need to grok and thus remember these facts about finitary monads, which seems doable:

The forgetful functor from modules/abelian groups to sets preserves and reflects filtered colimits and finite limits. (The same is true replacing modules by algebras of any finitary = filtered colimit-preserving monad.)

John Baez (Dec 24 2024 at 21:19):

I think any von Neumann algebra is von Neumann regular, and when I get to my laptop I can sketch an argument. (It gets really tiring to type math on a cell phone.)

Ryan Schwiebert (Dec 24 2024 at 22:08):

Got a ping here when my name got mentioned. It'd be nice to be able to help.

Ryan Schwiebert (Dec 24 2024 at 22:09):

A lot of the comments above go into "show more" and I'm reluctant to chase them all down for outstanding questions :) If there are any important ones still in the air I would be happy to be brought up to speed...

Todd Trimble (Dec 24 2024 at 22:21):

That's funny that von Neumann doesn't require an ideal to be nonempty.

He doesn't have to. Any ideal of $R$ is an $R$-module, and modules always have a zero element.

Todd Trimble (Dec 24 2024 at 22:26):

Ryan Schwiebert said:

A lot of the comments above go into "show more" and I'm reluctant to chase them all down for outstanding questions :) If there are any important ones still in the air I would be happy to be brought up to speed...

Thanks for responding!

Sorry, it's Christmas I know, and maybe it wasn't a good time to ping you. But whenever there's available time, I'd enjoy hearing more about von Neumann regularity being Morita invariant. This sounds very believable, but I haven't tried to put together a proof.

(I was happy to discover that you maintain a database for ring theory and are something of an expert here.)

John Baez (Dec 24 2024 at 22:29):

So @Todd Trimble, maybe it'd amuse you to watch me think about loud about why von Neumann algebras should be von Neumann regular rings.

A von Neumann algebra is a subalgebra $X$ of $B(H)$, the bounded operators on a Hilbert space $H$, which is closed under taking adjoints and closed in a weak enough topology that all the stuff we usually do with bounded operators on a Hilbert space stays in $X$ if we start with operators in $X$. So for example we can apply any bounded Borel-measurable function $f: \mathbb{R} \to \mathbb{C}$ to a self-adjoint operator $A \in X$ and get an operator $f(A) \in X$, and the usual rules hold - this is the Borel functional calculus, which is one way of thinking about the spectral theorem.

Also for any $T \in X$ you get self-adjoint operators $T T^\ast$ and $T^\ast T$ in $X$ which are self-adjoint, so you can apply the Borel functional calculus even when you've got an operator $T$ that is not itself self-adjoint. You can use this to get a polar decomposition of any $T \in X$. That is, you can write $T = U A$ where $A \in X$ is nonnegative and thus self-adjoint, while $U \in X$ is - alas, not unitary, but close: it's a partial isometry, meaning it's an isometry when restricted to some closed subspace of $H$.

So these are some tricks we have up our sleeve when dealing with von Neumann algebras.

Now I don't instantly see how to use these to get, for any $T \in X$, an operator $S$ such that $T S T = T$. If I could find such an $S \in X$, we'd be done: that's one criterion for a ring to be von Neumann regular. There could easily be some clever formula for this $S$, but I don't see it.

Instead, I'll try to show the right ideal generated by any $T \in X$ is also generated by some idempotent $P \in X$. That's another criterion for a ring to be von Neumann regular, and this one feels less artificial to me.

My intuition goes like this: $T$ is an operator on $H$, it has some range, for any $S \in X$ we have $\text{ran} T S \subseteq \text{ran} T$, so every operator in the right ideal generated by $T$ has a range contained in $\text{ran} T$.

So, maybe the right ideal generated by $T$ consists of exactly the $S \in X$ whose range is contained in $\text{ran} T$.

Thus, to get an idempotent that generates the same right idea, we should try taking $P$ to be the self-adjoint projection onto $\text{ran} T$. Of course we need $P \in X$.

John Baez (Dec 24 2024 at 23:01):

But here our bag of tricks might help us. Use the polar decomposition to write $T = U A$ where $U \in X$ is a partial isometry and $A \in X$ is non-negative. By the way, this is not magic: the idea is to take

$A = |T| := \sqrt{T^\ast T}$

and then there's a unique partial isometry $U$ that makes $T = U |T|$, where the range of $U$ is the range of $T$.

(Or more precisely their ranges have the same closure - I should be saying 'closure of the range' all along, but I won't bother.)

Remember, I wanted a self-adjoint idempotent $P$ whose range is the same as that of $T$. But it's not hard to get from our partial isometry $U$ to a self-adjoint idempotent with the same range.

In fact, for any partial isometry $U$, both $U U^\ast$ and $U^\ast U$ are self-adjoint idempotents! The first is the projection onto the range of $U$, while the second is the projection onto the 'corange' of $U$, which is the range of $U^\ast$, and also the orthogonal complement of the kernel of $U$.

John Baez (Dec 24 2024 at 23:03):

So let's take

$P = U U^\ast$

It's in our von Neumann algebra $X$. And with any luck, this idempotent will generate the same right ideal as $T$.

John Baez (Dec 24 2024 at 23:14):

Take a guy in the right ideal generated by $T$, say

$T S$

This equals

$U |T| S$

but I bet $U |T| = U U^\ast U |T|$ because $U^\ast U$, being a projection, acts like the identity on some subspace, and I bet the range of $|T|$ is exactly that subspace.

I'm counting on the gods to smile on me here. If they do, we have

$T S = U |T| S = U U^\ast U |T| S = P U |T| S$

so the right ideal generated by $T$ is contained in the right ideal generated by our idempotent $P$, and we're halfway home! Or more than halfway, since the hard part was coming up with a plan.

John Baez (Dec 24 2024 at 23:38):

Conversely, let's see whether the right ideal generated by $P$ is contained in that generated by $T$. So take a guy in the right ideal generated by $P$:

$P S$

and start fiddling around. $P = U U^\ast$ so

$P S = U U^\ast S$

Thus, the right idea generated by $P$ is contained in the right ideal generated by $U$. But why is it in the right ideal generated by $T$?

Well, what's $U$? I never really said! Now it's time to think about that. Since the polar decomposition says

$T = U|T|$

you might optimistically hope

$U = T |T|^{-1}$

This would be great, because it would immediately imply that the right ideal generated by $U$ is contained in that generated by $T$, and we'd be done - except for the place I needed the gods to smile on me.

Indeed it's basically true that $U = T|T|^{-1}$, except for one little problem: there's no reason in the world for $|T|$ to be invertible.

Here's where the Borel functional calculus comes in handy - and problem 20 in Chapter VII of Reed and Simon's Functional Analysis, which says this:

If we take an increasing sequence of bounded functions $f_n: \mathbb{R} \to \mathbb{R}$ that converge pointwise to the function $1/x$ for $x > 0$ and are zero for $x < 0$, then

$T f_n(|T|) \psi \to U \psi$

for all $\psi$ in our Hilbert space $H$.

This is a sneaky way of saying that while $|T|^{-1}$ doesn't make sense by itself, $T |T|^{-1}$ still does: very roughly, $|T|^{-1}$ only blows up in ways that $T$ is gonna come along and annihilate. And we can say it like this:

$T f_n(|T|) \to U$

in the strong operator topology. von Neumann algebras are closed in the strong operator topology, and now I will suddenly claim that every time I said "ideal" I meant strongly closed ideal. In operator algebra theory, we'd be dead in the water without some topology to help us like this: there's a vast and woolly wilderness of non-closed ideals.

So, in the strong operator topology

$P S = U U^\ast S = \lim_{n \to \infty} T f_n(|T|) U^\ast S$

And this says that the right ideal generated by $P$ is contained in the closure of the ideal generated by $T$.

John Baez (Dec 24 2024 at 23:40):

So I seem to be getting a weaker result than I originally claimed: for any element $T$ in a von Neumann algebra, the closure in the strong operator topology of the right ideal it generates equals the right ideal generated by some self-adjoint idempotent $P$.

John Baez (Dec 24 2024 at 23:45):

It's possible that with more work I could do better and eliminate the italicized phrase. For example, I never used the $U^\ast$ in those formulas above. It could be that $f_n(|T|) U^\ast$ converges to some operator $Q$, and then $PS = T Q S$, so I could cross out the italicized phrase and triumph.

But anyway, it was good trying to remember some analysis.

John Baez (Dec 25 2024 at 01:00):

Okay, @Todd Trimble, I now think I see that von Neumann algebras aren't von Neumann regular rings. They're only some sort of operator algebra analogue of von Neumann regular - as witnessed by the sudden intrusion of "ideals closed in the strong operator topology" in my attempted proof above.

John Baez (Dec 25 2024 at 01:02):

Here's a counterexample. Take the von Neumann algebra $X = L^\infty(\mathbb{R})$, where elements are equivalence classes of of bounded Lebesgue-measurable functions $f: \mathbb{R} \to \mathbb{C}$, where two are equivalent if they're equal a.e.

John Baez (Dec 25 2024 at 01:04):

Let $f \in L^\infty(\mathbb{R})$ be (the equivalence class of) a continuous function that's supported in $[0,1]$ and positive on $(0,1)$, so it goes to zero as $x \to 0$ or $x \to 1$.

John Baez (Dec 25 2024 at 01:08):

Look at the ideal $f L^\infty(\mathbb{R})$. Any function in here goes to zero as $x \to 0$ or $x \to 1$ (at least if we exclude points in a set of measure zero).

This idea is not generated by an idempotent, because an idempotent in $L^\infty(\mathbb{R})$ must be the characteristic function of a subset of $\mathbb{R}$, say $\chi_S$, and there's no way that can work.

For example $\chi_{[0,1]} L^\infty(\mathbb{R})$ will consist of all $L^\infty$ functions supported on $[0,1]$, but they won't obey that condition about going to zero as $x \to 0$ or $x \to 1$.

Todd Trimble (Dec 25 2024 at 04:57):

Thanks for looking into it! I would need to dust off my functional analysis, particularly the subtleties with various operator algebra topologies, to fully appreciate what you wrote. But the counterexample is easy enough to understand.

So it seems von Neumann algebras are too large a class to expect them all to be von Neumann regular, but perhaps there's a reasonable subclass of them that are, or particularly noteworthy examples among them that are? I'm beginning to warm a little to von Neumann regular rings, but I still would like more juicy examples.

John Baez (Dec 25 2024 at 06:21):

The strong operator topology is just the one where we have convergence $T_\alpha \to T$ iff $T_\alpha \psi \to T \psi$ in the norm topology for any vector $\psi$ in the Hilbert space on which the operators act. If analysts weren't trying to show off it would be called the topology of pointwise convergence!

von Neumann algebras are closed in all our favorite topologies: norm topology (like C*-algebras of operators on a Hilbert space are) but also the strong topology, the weak topology and even some others.

John Baez (Dec 25 2024 at 06:26):

Now you've got me wondering what von Neumann did with von Neumann regular rings and "continuous geometries" - this suggests that at least a few infinite-dimensional von Neumann algebras are von Neumann regular rings.

John Baez (Dec 25 2024 at 06:47):

Hmm, it's very interesting to read the foreword, written by Israel Halperin, of von Neumann's lecture note book, Continuous Geometries: it sketches what von Neumann did on this subject, and then how he got stuck and quit in 1936. His energy was remarkable for what seems like a bit of a niche topic. He generalizes the usual axioms for finite-dimensional projective geometry in terms of a modular lattices, and the usual theorems relating these to matrix algebras over division rings (the lattice of left ideals corresponding to the lattice of points, lines, planes and higher-dimensional figures in in the geometry) - but now the "dimension" of the figures can take arbitrary values in $[0,1]$, and instead of a matrix algebra over a division algebra we have something more general: a von Neumann regular ring!

The foreword says some words that convince me type $\text{II}_1$ von Neumann algebras must be von Neumann regular rings, since that class of von Neumann algebras were one of von Neumann's great discoveries when he was trying to generalize matrix algebras, and the foreword says these were the main example of continuous geometries.

John Baez (Dec 25 2024 at 06:49):

I don't know if anyone here has read Varadarajan's Geometry of Quantum Theory, which proves the "usual" theorems. Von Neumann's book feels like the logical continuation of that - though it was written much earlier.

Todd Trimble (Dec 25 2024 at 16:30):

I'll have to admit that I don't know much about this stuff, in particular I don't have mathematical feeling for type $\Pi_1$ von Neumann algebras (I spent a few minutes looking at the Wikipedia article, but I'm not grasping the significance). I think I've heard that they figure in important work of Vaughan Jones, maybe in connection with the Jones polynomial, but that was an avenue I never studied.

But this

convince me type $\Pi_1$​ von Neumann algebras must be von Neumann regular rings

sounds like it should go into any introductory article on von Neumann rings. It sounds too important to omit!

As I said earlier, I think continuous geometries used to be hot stuff. Here's a reminiscence of Irving Kaplansky from the collection More Mathematical People (which by the way is a great book -- it would be or would've been a great Christmas gift for a mathematician):

MP -- One last question. Of all the things you've done in mathematics is there any one thing -- you can mention even a couple of things -- about which you have a special sense of pride?

IK -- I actually already have an answer to that in writing. I'm not sure where it appears, but I remember giving it once. I am proudest of the paper "Any orthocomplemented complete modular lattice is a continuous geometry". It's not that it's especially important, but nevertheless I am proudest of it. It required a sustained effort of about six months, an effort of a kind that I don't think I could ever put out again [...] I think there is a level of unexpectedness about it -- I don't want to use a word like ingenuity -- and a high level of infinite algebra -- the kind that I'm especially fond of, as I mentioned in the earlier part of this conversation.

Todd Trimble (Dec 25 2024 at 17:06):

So one definition of von Neumann algebra is that's a $C^\ast$-algebra that admits a predual, i.e., if we consider the $C^\ast$-algebra as a Banach space $B$, then there is a Banach space whose dual (= Banach space of bounded linear functionals to the ground field $\mathbb{C}$) gives back $B$. It turns out a predual, if one exists, is unique up to isomorphism. That seems like a memorable definition.

Is the trace operation you mention obtained by the pairing of the $C^\ast$-algebra with its predual, according to the usual way of defining trace in monoidal categories with duals?

John Baez (Dec 25 2024 at 19:00):

Todd Trimble said:

So one definition of von Neumann algebra is that's a $C^\ast$-algebra that admits a predual, i.e., if we consider the $C^\ast$-algebra as a Banach space $B$, then there is a Banach space whose dual (= Banach space of bounded linear functionals to the ground field $\mathbb{C}$) gives back $B$. It turns out a predual, if one exists, is unique up to isomorphism. That seems like a memorable definition.

That's memorable and conceptually nice because it's "intrinsic" - it doesn't require treating the von Neumann algebra as a subalgebra of the algebra of bounded linear operators on a Hilbert space. But if you want to quickly understand a bunch of stuff about von Neumann algebras, I claim you want to use the "extrinsic" definition: a von Neumann algebra is a $\ast$-algebra of bounded linear operators on a Hilbert space that's closed in the strong topology (the topology where $T_\alpha \to T$ iff $\|(T_\alpha - T)\psi\| \to 0$ for all $\psi$ in the Hilbert space).

John Baez (Dec 25 2024 at 19:02):

This mimics how we can define a C*-algebra to be a $\ast$-algebra of operators on a Hilbert space that's closed in the norm topology (the topology where $T_\alpha \to T$ iff $\|T_\alpha - T\| \to 0$). Despite their names, this is clearly stronger than the strong topology!

John Baez (Dec 25 2024 at 19:07):

As you probably know, C*-algebras also have a beautiful 'intrinsic' definition, probably invented by my advisor Irving Segal, and the equivalence of the intrinsic and extrinsic definitions is the Gelfand-Naimark-Segal theorem. You can pretty efficiently use the intrinsic definition to get a lot done.

The intrinsic definition of von Neumann algebras is much less useful, in my opinion, and it came along a lot later. People used to call a $C^\ast$-algebra with a predual a $W^\ast$-algebra, and I believe some Japanese mathematician studied them and proved they are equivalent to von Neumann algebras. But it's so damn easy to prove stuff using the extrinsic definition that I never worked with the intrinsic one. (There are some clever things you can do with the predual, but I think of those as more sophisticated than anything I ever messed with.)

John Baez (Dec 25 2024 at 19:16):

It's too bad the Wikipedia article doesn't give an intuition for type $\text{II}_1$ von Neumann algebras, because there's one, the type $\text{II}_1$ hyperfinite factor, which is quite easy to understand and a really fundamental object in math.

Is the trace operation you mention obtained by the pairing of the C∗-algebra with its predual, according to the usual way of defining trace in monoidal categories with duals?

I don't think so! At least I've never heard it defined that way.

Here's the basic idea - I'll telegraph it, and what I say may or may not make sense. There's an obvious inclusion of $2^n \times 2^n$ matrices into $2^{n+1} \times 2^{n+1}$ matrices that sends a matrix $T$ into the block diagonal matrix

$T$ $0$
$0$ $T$

Take the colimit of these algebras. You can define a trace on the nth of these algebras to be the usual matrix trace divided by $2^n$, and then the inclusions are all trace-preserving! So, the colimit gets a well-defined "trace" on it. Note the trace of the identity is 1. And this algebra has there lots of projections (= self-adjoint idempotents) with traces between 0 and 1. I bet you can find infinitely many.

This algebra, defined as a colimit, is not yet the hyperfinite type $\text{II}_1$ factor: that's the completion of this algebra in a pretty obvious metric. (Cauchy completion in the good old metric space sense!) That algebra has projections with traces taking arbitrary values in $[0,1]$, which is the reason for the subscript "1" in the name $\text{II}_1$.

John Baez (Dec 25 2024 at 19:18):

The sci-fi sounding term "hyperfinite" just means that we can construct this algebra as a completion of a colimit of matrix algebras.

John Baez (Dec 25 2024 at 19:22):

Thanks very much for that snippet of an interview with Kaplansky. The history of this stuff is interesting! There was a kind of decline of popularity in von Neumann algebras, and then a revival due to the work of Connes, and an explosion when Vaughan Jones connected them to knot invariants (by having lunch with Joan Birman at a place where two conferences were being held, one on knot theory and one on operator algebras - I guess you know that story)... but I feel that some of the old themes, like "continuous geometries", got lost in the shuffle.

Who is connecting continuous geometries to quantum groups? They're both connected to the hyperfinite type $\text{II}_1$ factor. So somehow incidence geometries where the geometrical figures can have dimensions taking arbitrary values in $[0,1]$ are related to quantum groups and $q$-deformation. Perhaps only indirectly. And that's something I've never heard about.

John Baez (Dec 25 2024 at 22:17):

Merry Christmas, everyone! :evergreen_tree:

The introduction here gives a nice taste of the Japanese line of research on continuous geometries and regular rings:

• Fumitomo Maeda, Continuous Geometry, English translation by Tristan Bice.

Unfortunately only the introduction and first chapter of the book has been translated here (so far). The chapter titles look intriguing, e.g. there's a whole chapter on regular rings.

Here's how the introduction starts:

Traditionally, projective geometry was based on fundamental concepts like points and lines, and it was thought that this was unavoidable. However, around 1935, it was shown by G. Birkhoff and K. Menger that, from a lattice theoretic standpoint, projective geometries are irreducible finite dimensional complemented modular lattices. Here, just as points are ‘included’ in lines, ‘order’ becomes the fundamental concept and, due to the finite dimensionality restriction, objects like points and lines naturally arise. Thus if we eliminate the finite dimensionality condition, we would expect to get a new geometry that, in a lattice theoretic sense, has the same structure as a projective geometry but without points and lines. But constructing such a geometry is no easy matter. Still, J. von Neumann accomplished this difficult task in 1936-1937. If we change the way we express dimension slightly, a linear subspace like a point or line in an n − 1 dimensional projective geometry can take dimension values 0, 1/n, 2/n,. . . , n−1/n and 1. Specifically, the dimension of the ‘empty’ subspace is 0, and the dimension of points and lines is 1/n nd 2/n respectively. And the whole space has dimension 1. Von Neumann showed that sometimes elements of an irreducible continuous complemented modular lattice can take all real numbers from 0 to 1 as dimension values. In this case, there are elements having dimension arbitrarily close to 0 and so there is no such thing as a ‘point’. In other words, this is a continuous geometry (in the strict sense). On the other hand, it is known that in a projective geometry of dimension at least 3, coordinates can be introduced by a certain skew field which makes the linear subspace lattice isomorphic to the lattice of right ideals of the matrix ring over this skew field. Von Neumann generalized this, proving that a complemented modular lattice of order at least 4 is isomorphic the the principal right ideals of a certain matrix ring. This is the essence of von Neumann’s continuous geometry (in the broad sense). In other words, continuous geometry can be roughly divided into two parts, dimension theory and representation theory.

Todd Trimble (Dec 26 2024 at 18:14):

Thanks, John -- this is all very helpful. I'll certainly take your word for concrete von Neumann algebras being a much quicker port of entry into the subject than abstract von Neumann algebras.

by having lunch with Joan Birman at a place where two conferences were being held, one on knot theory and one on operator algebras - I guess you know that story

I'm afraid I don't! I looked a little at google results, and read a little about their mutual excitement in Birman's office when the two of them were trying out Jones's newly hatched polynomials on various knots and determined (1) no, this was not just the Alexander polynomial in a different guise, and (2) it was different from the Alexander polynomial in a pretty deep way. I guess the fact that the Jones polynomial could distinguish a right trefoil from a left trefoil, which the Alexander polynomial could not, is pretty well known lore.

The only lunch I read about is recounted here, where Jones proposes buying a bottle of champagne for Birman and that they be co-authors, and Birman said something like, "thanks, but I don't need the champagne, and I'll decline to be a co-author [she felt she didn't have the time then], but the only thing I ask is to be given appropriate credit". Which of course Jones gladly gave.

John Baez (Dec 26 2024 at 18:27):

Hmm, I'm sure I've heard that Jones and Birman met at a conference center where there were simultaneously occuring separate conferences on knot theory and operator algebras, and they talked over lunch, and Jones wrote down equations like $xyx = yxy$ and Birman said "hey, those are the braid group relations!" But this seems to be wrong. Jones first met a friend of Joan Birman. From an interview:

Rob: Joan, can you tell us about the origins of your connection to Vaughan Jones and his knot polynomial?

Joan: Sure, I’m happy to do that. In the spring of 1984 I had been working with Caroline Series, who was on sabbatical and had spent some time at the Institute for Advanced Study (IAS) in Princeton, New Jersey. She met Vaughan Jones there, He told her about his Hecke Algebra representations of Artin’s braid group that he had discovered. He knew they gave representations of Artin’s braid group, and he had read Artin’s paper, and he was searching for someone who could help him to understand their meaning better. Caroline said, “Well you must go and talk to Joan about this”, and that’s why he contacted me.

Rob: Did he know about your book?

Joan Yes, because he included it as a reference in a paper he had presented at a conference in Kyoto Japan in July 1983. He proves, in that paper, that his Hecke algebra representation of the braid groups $B_n, n \ge 2$ was reducible, and included as a summand the Burau representation of $B_n$. He had discovered a trace function on the Hecke algebra. He knew (in a vague way, I think) that the Alexander polynomial of a knot or link was determined by the Burau representation of a braid that determined it.

Following Caroline’s suggestion, he contacted me and we arranged to get together in my office at Columbia on Monday, May 14, 1984. At the end of that meeting we arranged a second meeting on Tuesday, the 22nd, just eight days later. (I know both dates precisely because I had made a note of both meetings in my 1984 “daily reminder”.) I happen to have kept all my old little books, which date back to the mid 1970’s.

At our first meeting I told Vaughan about knots and links being formed by closed braids, and I told him about Markov’s Theorem. He told me about his Hecke algebra representation of $B_n$. He was very aware of the curious fact that a representation of the braid group should have appeared in connection with type $\text{II}_1$ factors, and eager to understand more. I don’t know whether he knew about Markov’s Theorem. After our first meeting, he got to work. In between those two meetings we may have talked on the telephone once or twice, but the main point is that by the time our second meeting began Vaughan knew he had a knot polynomial.

Rob: Vaughan wrote a letter to you on May 31st, which we include in this volume. The letter starts: “Dear Joan, First of all, my deepest thanks for putting me onto this. None of it would have begun had it not been for our seemingly unproductive first meeting. Let me begin by summarizing what we need from operator algebras,” and then he goes into the first theorem “For every $t$ greater than zero, …”

Todd Trimble (Dec 26 2024 at 18:30):

IIRC, Markov's theorem says that every link occurs as the trace of some braid? (Of course I could look this up myself.)

John Baez (Dec 26 2024 at 18:31):

Yes, and more importantly how to get all the braids that give the same knot, so you can tell when a braid invariant gives a knot invariant.

John Baez (Dec 26 2024 at 18:32):

We now understand that type $\text{II}_1$ von Neumann algebras show up as algebras of observables connected to a quantum field theory, Chern-Simons theory, for which the expected value of the holonomy of a connection around a knot gives the Jones polynomial. It was mainly Witten who saw how the Jones polynomial is connected to Chern-Simons theory. This famous paper of his is what eventually got me interested in braided monoidal categories and later $n$-categories.

John Baez (Dec 26 2024 at 18:32):

Ever since then I've been working on simpler and simpler things. :upside_down:

John Baez (Dec 26 2024 at 18:34):

But the whole mess of ideas I just described falls out of something fairly simple: a certain braided monoidal category, roughly the category of representations of the Hopf algebra "quantum SU(2)", a deformation of the category of representations of SU(2).

John Baez (Dec 26 2024 at 18:36):

It turns out there's a 1-parameter way to deform the category of representations of any compact simple group, like SU(2), so that it becomes a braided but not symmetric monoidal category. And this is the magic that leads to all the above stuff!

Todd Trimble (Dec 26 2024 at 18:36):

I'm going to have to pester Jim into reminding me of his thoughts on quantum SU(2)... the earlier conversation was in days when groupoidification and Hecke algebras was very much on his mind.

John Baez (Dec 26 2024 at 18:39):

Yes, I wrote some papers on that stuff, like my paper on groupoidification....

John Baez (Dec 26 2024 at 18:40):

So if we ever get tired of thinking about all the 2-rigs in our big diagram of 2-rigs, we could study how a bunch of them, probably all of them, have a "q-deformation" where they become braided 2-rigs.

John Baez (Dec 26 2024 at 18:40):

I think people have studied q-deformed symmetric functions.... or am I just imagining that?

Todd Trimble (Dec 26 2024 at 18:41):

Your paper with Hoffnung and Walker, HDA VII?

John Baez (Dec 26 2024 at 19:46):

Yes. But I didn't mean to act like I understand everything you or Jim did! I get the impression you guys came up with a topos that unifies the symmetric group algebra and the Hecke algebras for q a prime power. I don't understand that, but any sort of unification like this might help q-deform that whole diagram of 2-rigs we've been studying.

For example people already know how to q-deform the nxn matrix algebra, or more precisely the corresponding bialgebra, or its category of comodules. But this should be part of a bigger story where we q-deform 'everything'.

Todd Trimble (Dec 26 2024 at 20:12):

And I didn't mean to suggest that Jim, not you, should be my main source! I can pester you too, of course, to tell me more (maybe tonight?). It's just that he might remember the conversation he and I had on this.

I did want to discuss some of my takes on groupoidification tonight, as I foreshadowed last week.

John Baez (Dec 26 2024 at 20:55):

Yes!

Graham Manuell (Dec 27 2024 at 09:32):

John Baez said:

That's cool how you were able to define commutative von Neumann rings as a variety of rings by treating 'weak inverse' as an operation, Todd Trimble. This instantly gives us a lot of theorems.

It's also cool that none of this uses the additive structure, so that we get a concept of 'von Neumann regular commutative monoid'. Does this concept have a name? I imagine the hard-core universal algebraists would have studied it already.

These are called commutative inverse monoids. Inverse semigroups are very well studied.

Ryan Schwiebert (Dec 27 2024 at 15:52):

John Baez said:

Now you've got me wondering what von Neumann did with von Neumann regular rings and "continuous geometries" - this suggests that at least a few infinite-dimensional von Neumann algebras are von Neumann regular rings.

My go-to reference for von Neumann regular rings is Goodearl's book. I'm attaching photos of the introduction for all of you to peruse.

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Ryan Schwiebert (Dec 27 2024 at 16:33):

Todd Trimble said:

I'd enjoy hearing more about von Neumann regularity being Morita invariant.

My intuition says that flatness has a categorical formulation, and that it should be preserved by an equivalence. However, my inexperience with categorical arguments, as well as lack of understanding of the situation with exact sequences in category theory hampers me.

So what would this be... a right module amounts to a presheaf $F\in [R^{op}, Ab]$. The definition of flat module says "the tensor functor with left modules has to be exact." I'm guessing we can show that an equivalence between this category of presheaves and the category $[S^{op}, Ab]$ will preserve flatness of $F$.

At present when I try to connect the dots between the presheaf and exactness of a functor, I'm still bewildered. It's a huge and stupid gap I know, and all the more frustrating because I can state barely more than the definition. It would be educational for me to see the way, though.

Ryan Schwiebert (Dec 27 2024 at 16:44):

Jean-Baptiste Vienney said:

Something that I would like to see on an example is how a commutative von Neumann regular ring is exactly a subring of a product of fields closed under taking weak inverses as they say on Wikipedia.

I'm not familiar with that characterization, but it does sound related to these two standard lemmas:

For a commutative ring $R$ (with nonzero identity)

if $R$ has no nonzero nilpotent elements then it is a subdirect product of integral domains (and in particular it is a subring of a product of integral domains.)

if $R$ is von Neumann regular it is a subdirect product of fields (and in particular is a subring of a product of fields.)

In both cases, the product ring can be taken to be the product formed of all the $R/P$ for all the minimal prime ideals $P$. In the case of VNR rings the minimal primes are maximal ideals, so that makes $R/P$ of course a field...

Jean-Baptiste Vienney (Dec 27 2024 at 17:10):

Oh thanks @Ryan Schwiebert, it is very helpful!

Jean-Baptiste Vienney (Dec 27 2024 at 17:28):

To be completely clear, when you write “minimal prime ideal” does it mean “prime ideal minimal among the nonzero prime ideals”?. I’m asking this because the usual definition of “maximal ideal” is “ideal maximal among the proper ideals” and I guess we don’t the zero ideal to be the unique minimal prime ideal in every integral domain.

Todd Trimble (Dec 27 2024 at 18:18):

Ryan Schwiebert said:

Todd Trimble said:

I'd enjoy hearing more about von Neumann regularity being Morita invariant.

My intuition says that flatness has a categorical formulation, and that it should be preserved by an equivalence. However, my inexperience with categorical arguments, as well as lack of understanding of the situation with exact sequences in category theory hampers me.

So what would this be... a right module amounts to a presheaf $F\in [R^{op}, Ab]$. The definition of flat module says "the tensor functor with left modules has to be exact." I'm guessing we can show that an equivalence between this category of presheaves and the category $[S^{op}, Ab]$ will preserve flatness of $F$.

At present when I try to connect the dots between the presheaf and exactness of a functor, I'm still bewildered. It's a huge and stupid gap I know, and all the more frustrating because I can state barely more than the definition. It would be educational for me to see the way, though.

Thank you, Ryan! What you are saying makes sense to and it adds to the general landscape.

For the narrow purpose I was pursuing, I wanted to wend my way from the regular logic definition of von Neumann ring (every element has a weak inverse) to derive the second order logic definition (every module is flat), and I also wanted to do that in reverse. Going forward, assuming the regular logic definition (but not assuming the flatness condition), I was hoping there would be an easy proof of Morita invariance, so I couldn't use your idea directly without being circular. Anyway, I'll keep thinking about this. Something like your idea will probably work for me.

Ryan Schwiebert (Dec 27 2024 at 18:32):

Jean-Baptiste Vienney said:

“prime ideal minimal among the nonzero prime ideals”

No: "minimal prime" means "minimal among all prime ideals." Of course, the ones in which zero is minimal are the prime rings (integral domains, in commutative rings.)

Ryan Schwiebert (Dec 27 2024 at 18:35):

Todd Trimble said:

Going forward, assuming the regular logic definition (but not assuming the flatness condition), I was hoping there would be an easy proof of Morita invariance

Understood! I will think about it that way for a while. My brain is occupied with the mindset I described, and I hope I eventually understand that path too, but if I think of anything more elementary I will let you know.

John Baez (Dec 27 2024 at 20:56):

Ryan Schwiebert said:

Todd Trimble said:

I'd enjoy hearing more about von Neumann regularity being Morita invariant.

My intuition says that flatness has a categorical formulation, and that it should be preserved by an equivalence.

My first guess was the same, but then I realize: a ring is absolutely flat if for every $R$-module $M$, the functor

$M \otimes -: R\mathsf{Mod} \to R\mathsf{Mod}$

preserves finite limits (which is equivalent to it preserving equalizers, and equivalent to it preserving kernels). This property is defined using not just the structure of $R\mathsf{Mod}$ as an $\mathsf{Ab}$-enriched category (which is what we need to instantly see that this property is Morita invariant, from the definition of Morita equivalence) but using the monoidal structure as well.

John Baez (Dec 27 2024 at 21:02):

Even worse, for a noncommutative ring we can't tensor modules, so this characterization of von Neumann regularity only works for commutative rings - and commutative rings are Morita equivalent iff they are isomorphic!!!

So, I broke down and looked at my beloved book Rings and Categories of Modules by Anderson and Fuller, and they give this problem as an exercise, with a hint to look at Exercise 20.14, which says:

Prove that a ring $R$ is von Neumann regular iff every finitely generated left $R$-module is projective.

This does the job. And this connects nicely with something Todd may have taught me: flat modules are precisely the directed colimits of finitely generated projective modules.

Ryan Schwiebert (Dec 27 2024 at 21:10):

John Baez said:

Even worse, for a noncommutative ring we can't tensor modules

The flatness criterion is the same for noncommutative rings/modules/von neumann regularity though. Is there something wrong with the same argument using the functor from $R-mod\to Ab$?

John Baez (Dec 27 2024 at 21:14):

Oh, okay. Yes, I'm being silly, we say an $R$-module is flat iff

$M \otimes _R - : R\mathsf{Mod} \to \mathsf{Ab}$

is left exact (i.e. it preserves finite limits, or equivalently equalizers, or equivalently kernels). We don't need $R$ to be commutative for this to make sense.

John Baez (Dec 27 2024 at 21:18):

It's still not obvious (to me) why this characterization is Morita invariant, since it's using not just the $\mathsf{Ab}$-enriched category $R\mathsf{Mod}$ itself but this functor. Maybe it's not hard to see. But I'm glad I gave up, because Anderson and Fuller's characterization is obviously Morita invariant, depending only on the $\mathsf{Ab}$-enriched category $R\mathsf{Mod}$ itself, up to equivalence.

(I keep saying "$\mathsf{Ab}$-enriched" because that's in the definition of Morita equivalence, and I think that matters, though I don't know examples of rings that aren't Morita equivalent whose categories of modules are equivalent as mere categories.)

Ryan Schwiebert (Dec 27 2024 at 21:24):

Anderson and Fuller is one of those books I really value but did not read through entirely... The same goes for Wisbauer's notes. I know they lean more heavily with category theoretic aspects.

John Baez (Dec 27 2024 at 21:26):

I don't know Wisbauer's notes. I got ahold of Anderson and Fuller long before I became really interested in category theory, and back then it was nice to have a book that clearly explained a lot of stuff about modules, using categories but not talking like category theorists do.

Ryan Schwiebert (Dec 27 2024 at 21:27):

John Baez said:

though I don't know examples of rings that aren't Morita equivalent whose categories of modules are equivalent as mere categories.

If such an example existed, I think I'd want to know. So the nuance is that the equivalence needs to respect the homs as abelian groups, and not merely as sets?

Ryan Schwiebert (Dec 27 2024 at 21:28):

@John Baez To be clear, this is what I was thinking of : https://www.math.uni-duesseldorf.de/~wisbauer/book.pdf

Stuff like $\sigma(M)$ is something he makes extensive use of, and I never got a chance to sit down to learn it...

John Baez (Dec 27 2024 at 21:38):

Thanks, I will probably learn a lot from this book! I had no idea what $\sigma(M)$ is, but it's the full subcategory of modules that are subgenerated by $M$, meaning submodules of modules that are generated by $M$, which means what I'd called separated by $M$: a module $N$ is separated by $M$ iff for any two distinct morphisms $f, g: N \to N'$ there's a morphism $a: M \to N$ such that $f a \ne g a$.

John Baez (Dec 27 2024 at 21:39):

I would love to say that $\sigma(M)$ consists of all the modules that are colimits of diagrams built using $M$, or maybe limits of such diagrams, but I don't see that either of these are true, so the concept of $\sigma(M)$ seems a bit slippery to me.

Ryan Schwiebert (Dec 27 2024 at 22:46):

@John Baez Don't let this message get buried either. I have a feeling you'd like that book too.

Ryan Schwiebert (Dec 27 2024 at 22:53):

@Todd Trimble There's a nice characterization that sometimes simplifies checking Morita invariant properties. It turns out that if a property passes to all matrix rings, and also corner rings of the form $eRe$ where $e$ is an idempotent such that $R=ReR$, then the property is Morita invariant.

Actually it's very simple to show that $eRe$ is von Neumann regular if $R$ is, without any further conditions on the idempotent. If you figure out how to show $M_n(R)$ is von Neumann regular, you're done.

John Baez (Dec 28 2024 at 00:26):

Ryan Schwiebert said:

John Baez Don't let this message get buried either. I have a feeling you'd like that book too.

I find it hard to read the jpegs with my bad vision, so I'll probably just get ahold of the book! Thanks for taking those pictures, though. Back in my misspent youth, when I was trying to understand all known algebraic approaches to quantum theory, I used to know something about how von Neumann regular rings are related to Baer $\ast$-rings. Now I completely forget! This book seems to say what's going on. So at least I got that much out the pictures.

Also Goodearl seems to say von Neumann algebras are only von Neumann regular rings when they are finite dimensional! This kills off my hope that at least the type $\text{II}$ von Neumann algebras were von Neumann regular rings. Instead he says there's a way to massage a von Neumann algebra and get a von Neumann regular ring.

John Baez (Dec 28 2024 at 00:27):

Ryan Schwiebert said:

Todd Trimble There's a nice characterization that sometimes simplifies checking Morita invariant properties. It turns out that if a property passes to all matrix rings, and also corner rings of the form $eRe$ where $e$ is an idempotent such that $R=ReR$, then the property is Morita invariant.

Oh, that's nice.

Todd Trimble (Dec 28 2024 at 01:09):

John Baez said:

It's still not obvious (to me) why this characterization is Morita invariant, since it's using not just the $\mathsf{Ab}$-enriched category $R\mathsf{Mod}$ itself but this functor. Maybe it's not hard to see. But I'm glad I gave up, because Anderson and Fuller's characterization is obviously Morita invariant, depending only on the $\mathsf{Ab}$-enriched category $R\mathsf{Mod}$ itself, up to equivalence.

(I keep saying "$\mathsf{Ab}$-enriched" because that's in the definition of Morita equivalence, and I think that matters, though I don't know examples of rings that aren't Morita equivalent whose categories of modules are equivalent as mere categories.)

Equivalence functors are product-preserving (also they are coproduct-preserving), and if an ordinary functor $F$ between categories that are enriched in commutative monoids (for example, categories enriched in abelian groups) is product-preserving, then the maps on hom-sets $\hom(a, b) \to \hom(F(a), F(b))$ are automatically maps of commutative monoids. And since the forgetful functor $\mathsf{Ab} \to \mathsf{CMon}$ is fully faithful, a product-preserving functor between $\mathsf{Ab}$-enriched categories is automatically an $\mathsf{Ab}$-enriched functor.

John Baez (Dec 28 2024 at 01:13):

Wow! Nice! That could be used to simplify a bunch of textbook definitions of Morita equivalence (though the argument might have to be translated into language ordinary ring theorists can understand).

Todd Trimble (Dec 28 2024 at 01:18):

So the basic point is that equivalences preserve [[biproducts]] in categories that have them, and that biproduct-preserving functors between additive categories are $\mathsf{CMon}$-enriched (and conversely).

John Baez (Dec 28 2024 at 01:18):

The power of category theory!

Todd Trimble (Dec 28 2024 at 01:20):

Yes, it's a very pleasant business to be in. :big_smile:

Todd Trimble (Dec 28 2024 at 01:24):

Incidentally -- veering away slightly from the current thread but following up on a tidbit from the chat we had last night -- I saw this intriguing-looking MO post where there may be some weird connection between $\mathbb{F}_1$-geometry and continuous geometry, due to Connes. But I have no idea what to make of this (if we can't get some sort of firm account of Connes's talk, then I guess we'd have to treat it as gossip).

John Baez (Dec 28 2024 at 01:36):

I'd noticed that! Connes has written several papers about $\mathbb{F}_1$, and they look interesting, but I haven't seen anything in those papers about continuous geometry!

Todd Trimble (Dec 28 2024 at 01:51):

Also Goodearl seems to say von Neumann algebras are only von Neumann regular rings when they are finite dimensional! This kills off my hope that at least the type II von Neumann algebras were von Neumann regular rings. Instead he says there's a way to massage a von Neumann algebra and get a von Neumann regular ring.

Kills off my hope too :crying_cat: for getting more interesting examples of von Neumann regular rings. The list given on Wikipedia was at first glance underwhelming to me, except that just now when I look again, I see some muttering about "The ring of affiliated operators of a finite von Neumann algebra is von Neumann regular."

Ryan Schwiebert (Dec 28 2024 at 01:52):

@Todd Trimble Actually, maybe it is not so hard. Suppose $T$ is a f.g. right ideal of $M_n(R)$. Choose a finite generating set $X$. The f.g. right ideal of $R$ generated by the entries of matrices in $X$ is principal, say of the form $xR$. Now, each generator is of the form $(xI_n)B$ for some matrix $B\in M_n(R)$, so every linear combination of them lies in $(xI_n)M_n(R)$. Hmm, I didn't quite complete the circle there. Not sure if $(xI_n)M_n(R)\subseteq T$. Still needs work...

Todd Trimble (Dec 28 2024 at 01:53):

Well, I'm rooting you on, Ryan!

Todd Trimble (Dec 28 2024 at 02:03):

There are a couple of scattered references to continuous geometry and the old work of von Neumann and Murray in Connes's massive Noncommutative geometry, which lends some sort of support to John's speculations about type $\Pi_1$ von Neumann algebras, but this stuff gets waaaaay over my head.

Ryan Schwiebert (Dec 28 2024 at 02:19):

Todd Trimble said:

"The ring of affiliated operators

I think that's what John's reference to "massaging" was about :)

Todd Trimble (Dec 28 2024 at 02:23):

I assumed that as well :-)

Oh, also @Ryan Schwiebert when you have a chance: the accepted MSE answer which appeared below yours said

A ring $R$ is Von Neumann regular if $\forall a\in R$ there exists $x\in R$ such that $a=axa$.
We can show that the ring of all $n\times n$ matrices over a regular ring is regular. You can find the proof in: K.R. Goodearl , Von Neumann regular Rings, pag. 4.

and since you seem to own that book, I wonder, if it's not too much trouble, if you could share that page? (I was just trying to look this up for myself, but didn't grab hold of his book yet.)

Ryan Schwiebert (Dec 28 2024 at 02:26):

I know Goodearl's book discusses types I, II, III of von Neumann regular rings, but I do not know if they are correspondent to the types of von Neumann algebras mentioned above. These rings are probably not finite dimensional as algebras (because then they'd just be plain old boring semisimple rings) but they might have other types of dimensions that I'm not versed in.

Todd Trimble (Dec 28 2024 at 02:32):

Thanks!

I have tunnel vision at the moment, and I'm interested in drilling down specifically on Emilio Novati's MSE answer that I just quoted, which seems to tell me where I can find a proof of the result I'm after, that if $R$ is von Neumann then so is $M_n(R)$, allegedly on page 4 of Goodearl's book. I hate to pester people and wish I were better at getting hold of books myself, but if it's not too inconvenient, could you post an image of the relevant page, please?

Ryan Schwiebert (Dec 28 2024 at 02:32):

Todd Trimble said:

ould share that page?

Oh! Wow, lol. Looks like I've been down this path before.

IMG_0845.jpeg
IMG_0846.jpeg

Ryan Schwiebert (Dec 28 2024 at 02:33):

He makes it look easy...

Todd Trimble (Dec 28 2024 at 02:34):

Aw, hell yeah! Thanks so much, Ryan! I will study this...

John Baez (Dec 28 2024 at 07:24):

Todd Trimble said:

the result I'm after, that if $R$ is von Neumann then so is $M_n(R)$, allegedly on page 4 of Goodearl's book.

This follows from the fact that von Neumann regularity is Morita invariant together with the fact that for every ring, the category of modules of $M_n(R)$ is equivalent to that of $R$. But maybe you don't like this because we here haven't proved the former fact, just quoted various Morita-invariant characterizations of von Neumann regularity?

Anyway, there should be a 'direct' proof of what you want.

Todd Trimble (Dec 28 2024 at 13:36):

The direct proof is there on pages 3 and 4 (Lemma 1.6 and Theorem 1.7).

And yes, this result was a stepping stone I was using to show that von Neumann regularity implies that all modules are flat. Back here.

Todd Trimble (Dec 28 2024 at 16:33):

Finally, let's prove the converse, that if all $R$-modules are flat, then $R$ is von Neumann regular.

Proof: As indicated in [[flat module]], a module is flat iff it is a filtered colimit of finitely generated free modules. In particular, given an element $a \in R$, the cyclic module $M = R/aR$, being flat by hypothesis, is a filtered colimit of finitely generated free modules, say $M \cong \mathrm{colim}_i F_i$. At the same time, $M$ is finitely presented by a short exact sequence

$0 \to aR \to R \to M \to 0$

and by the usual Gabriel-Ulmer theory, finite presentability of $M$ is equivalent to $\hom(M, -)$ preserving filtered colimits. Hence $\hom(M, -)$ preserves the colimit $M \cong \mathrm{colim}_i F_i$: the canonical map

$\mathrm{colim}_i \hom(M, F_i) \to \hom(M, M)$

is an isomorphism. In particular, it is surjective, and therefore there is some element $[f: M \to F_i] \in \mathrm{colim}_i \hom(M, F_i)$ that maps to $1_M$, i.e., there is a map $g: F_i \to M$ appearing in the colimit cocone such that $g \circ f = 1_M$. Therefore $M$ is a retract of the free module $F_i$, hence $M$ is projective. It follows that the exact sequence above that we used to present $M$ splits. Therefore $aR$ itself is a retract of $R$: there is an idempotent $e \in R$ such that $aR = eR$, and you can take it from here (follow the arguments given in [[von Neumann regular ring]]).

Todd Trimble (Dec 29 2024 at 18:40):

I found another proof of this Theorem that $R$ von Neumann regular => $M_n(R)$ von Neumann regular, in Kaplansky's fields and rings, starting at page 110. He makes frightfully clever use of a delightfully kooky little lemma, called McCoy's lemma, which I'll write out here. Call an element $a$ in a ring regular if there is an $x$ with $axa = a$.

McCoy's Lemma: if $axa - a$ is regular, then so is $a$.

Proof: This is super-easy. Write $(axa - a)y(axa - a) = axa - a$ for some $y$. Then rewrite this with $a$ on one side of the equation:

$a = axa - (axa - a)y(axa - a)$

and just notice that the right-hand side is of the form $a$ times something times $a$. QED

On to the proof of the theorem. First do the case of a $2 \times 2$ matrix

$\left( \begin{array}{ccc} a & & b\\ c & & d \end{array}\right)$

which we want to prove is regular. Suppose $crc = c$ for some $r$, and calculate that

$\left( \begin{array}{ccc} a & & b\\ c & & d \end{array}\right)\left( \begin{array}{ccc} 0 & & r\\ 0 & & 0 \end{array}\right)\left( \begin{array}{ccc} a & & b\\ c & & d \end{array}\right) - \left( \begin{array}{ccc} a & & b\\ c & & d \end{array}\right) = \left( \begin{array}{ccc} \ast & & \ast\\ crc - c & & \ast \end{array}\right) = \left( \begin{array}{ccc} \ast & & \ast\\ 0 & & \ast \end{array}\right)$

so by McCoy's lemma, we are reduced to proving that upper triangular matrices are regular. So now do the case of a $2 \times 2$ matrix

$\left( \begin{array}{ccc} a & & b\\ 0 & & d \end{array}\right)$

which we want to prove is regular. Write $axa = a$ and $dyd = d$ for some $x, y$, and calculate that

$\left( \begin{array}{ccc} a & & b\\ 0 & & d \end{array}\right)\left( \begin{array}{ccc} x & & 0\\ 0 & & y \end{array}\right)\left( \begin{array}{ccc} a & & b\\ 0 & & d \end{array}\right) - \left( \begin{array}{ccc} a & & b\\ 0 & & d \end{array}\right) = \left( \begin{array}{ccc} axa - a & & \ast\\ 0 & & dyd - d \end{array}\right) = \left( \begin{array}{ccc} 0 & & \ast\\ 0 & & 0 \end{array}\right)$

so by McCoy's lemma, we are reduced to proving that strictly upper triangular matrices are regular. So now do the case of a $2 \times 2$ matrix

$\left( \begin{array}{ccc} 0 & & b\\ 0 & & 0 \end{array}\right)$

which we want to prove is regular. Write $bzb = b$ for some $z$, and calculate that

$\left( \begin{array}{ccc} 0 & & b\\ 0 & & 0 \end{array}\right)\left( \begin{array}{ccc} 0 & & 0\\ z & & 0 \end{array}\right)\left( \begin{array}{ccc} 0 & & b\\ 0 & & 0 \end{array}\right) = \left( \begin{array}{ccc} 0 & & bzb\\ 0 & & 0 \end{array}\right) = \left( \begin{array}{ccc} 0 & & b\\ 0 & & 0 \end{array}\right)$

and now we are done with the $2 \times 2$ case.

For the $4 \times 4$ case, write this in block form as

$\left( \begin{array}{ccc} A & & B\\ C & & D \end{array}\right)$

where $A, B, C, D$ are $2 \times 2$ matrices, which reduces us to the $2 \times 2$ case just done. By induction, we can get to the $2^k \times 2^k$ case.

Finally, for any $n$, find $k$ such that $2^k \geq n$. The ring of $n \times n$ matrices sits in the upper left corner of the ring of $2^k \times 2^k$ matrices. Kaplansky concludes (paraphrasing), "The desired result then follows from a remark which we leave as an exercise for the reader: if $R$ is von Neumann regular and $e$ is an idempotent in $R$, then $eRe$ is von Neumann regular."

John Baez (Dec 29 2024 at 22:16):

Zounds! Whoever proved that lemma was the real McCoy.

This is the kind of algebra that sends many category theorists running for exit door: clever tricks and nary a functor visible. But it's got its own charm, and many a gleaming tower of abstraction is built on foundations such as these.

I'm hoping you... or, in the worst case, we... put some of these arguments on the nLab, to preserve them.

Todd Trimble (Dec 29 2024 at 22:22):

Well, part of me then thinks, "so much the worse for those category theorists". I consider the proof both efficient and elegant, and certainly creative. Another part of me sort of understands, so then the challenge for category theorists who don't like it is to try to shine more conceptual light on it, and make it more memorable. This could be possible!

Certainly, I can add to the nLab!

Todd Trimble (Dec 29 2024 at 22:23):

Oh, you edited in the McCoy joke while I was typing! Good one!

John Baez (Dec 29 2024 at 22:24):

Great! I've been improving the page on [[von Neumann regular rings]], but the characterization in terms of flatness, and the Morita invariance of the concept, have no proof nor even a proof sketch.

Sometimes a proof sketch is more enlightening than a full proof, especially since we can point to details in the references.

Todd Trimble (Dec 30 2024 at 02:50):

By the way, I caught a mistake in the set of equivalences on the nLab: it is not generally true that finitely generated modules over a von Neumann ring are projective. A counterexample is the von Neumann ring consisting of an infinite product of fields, $R = \prod_i k_i$. The infinite direct sum $I = \bigoplus_i k_i$ is an ideal, and the $R$-module $R/I$ is finitely generated by the quotient $R \to R/I$. But this quotient isn't projective: we don't have a splitting.

I'm making an adjustment.

Todd Trimble (Dec 30 2024 at 03:15):

Okay, I think I'm done making edits to the nLab article [[von Neumann regular ring]]. I included a lot of detail.

John Baez (Dec 30 2024 at 05:43):

Todd Trimble said:

By the way, I caught a mistake in the set of equivalences on the nLab: it is not generally true that finitely generated modules over a von Neumann ring are projective. A counterexample is the von Neumann ring consisting of an infinite product of fields, $R = \prod_i k_i$. The infinite direct sum $I = \bigoplus_i k_i$ is an ideal, and the $R$-module $R/I$ is finitely generated by the quotient $R \to R/I$. But this quotient isn't projective: we don't have a splitting.

I'm making an adjustment.

So the correct version is that a ring is von Neumann regular iff every finitely presented module is projective?

I was the one who put in the incorrect version. I was just trying to copy some nice characterizations from various sources, but I can't find what I thought I read. I may have been reading this on Wikipedia:

• every finitely generated submodule of a projective left R-module P is a direct summand of P.

This is pretty clearly equivalent to a characterization that's already on the list, so it's not really worth adding.

Todd Trimble (Dec 30 2024 at 05:47):

So the correct version is that a ring is von Neumann regular iff every finitely presented module is projective?

Yes. I wasn't trying to ding you, just heading off at the pass a potential question, "hey, what happened to that condition I thought I wrote down?"

In all honesty, I tried to get that condition in there, but didn't see how, and then chanced by luck on another place in the nLab where the counterexample was given. Then I could relax.

Todd Trimble (Dec 30 2024 at 05:52):

Most of the long list of lemmas that establish the equivalence of conditions have fairly short and snappy proofs. The one exception is the "McCoy proof", which could be eliminated according to taste (referring instead to the page in Kaplansky's book, which incidentally I forgot to list as a reference), but I the McCoy proof put a smile on my face and so I left it in. :-)

John Baez (Dec 30 2024 at 05:52):

No problem, I'm not hurt, just a bit confused about why I thought that was true. (Not from thinking hard about it!) It's nice if "presented" saves the day.

John Baez (Dec 30 2024 at 05:53):

I'll add the Kaplansky book as a reference, in penance. :upside_down:

Todd Trimble (Dec 30 2024 at 06:14):

Ryan Schwiebert said:

Todd Trimble There's a nice characterization that sometimes simplifies checking Morita invariant properties. It turns out that if a property passes to all matrix rings, and also corner rings of the form $eRe$ where $e$ is an idempotent such that $R=ReR$, then the property is Morita invariant.

Actually it's very simple to show that $eRe$ is von Neumann regular if $R$ is, without any further conditions on the idempotent. If you figure out how to show $M_n(R)$ is von Neumann regular, you're done.

Could you supply a source for that nice comment, Ryan? It would round out nicely the list of results now in the nLab article [[von Neumann regular ring]], where I've been recording main points of the discussion here.

John Baez (Dec 30 2024 at 06:16):

Yes, right now the Morita invariance of von Neumann regularity is relegated to an unreferenced "Example" near the end, which is a pity since Todd put in a proof that

$R$ von Neumann $\implies$ $M_n(R)$ von Neumann.

(This article is infinitely better than a week ago, and it now surpasses the Wikipedia article and the Encyclopedia of Mathematics article, though the latter has a lot of weird and interesting comments about related concepts.)

Todd Trimble (Dec 30 2024 at 06:34):

Incidentally, although I may be about to move on from this little fling with von Neumann regular rings, a thought occurred to me that the McCoy proof, although it might seem too clever by half, could perhaps be seen as an adaptation of the easier to think about example of $M_n(k)$, where $k$ is a field. In the field context, we can take advantage of usual linear algebra and Gaussian elimination: according to the Wikipedia article, every operator $P \in M_n(k)$ can be written as

$P = U\left(\begin{array}{ccc} I_r & & 0 \\ 0 & & 0 \end{array}\right) V$

where $U, V$ are invertible and $I_r$ is an $r \times r$ identity matrix nestled in the top left corner. Wikipedia then observes that $X = V^{-1}U$ satisfies $PXP = P$, establishing von Neumann regularity. It might be the case that the sequence of reductions used in the "McCoy proof" can be seen as a series of things like row reductions, except instead of taking advantage of inversion of nonzero elements $a$ in a field $k$, we use the next best thing which is this near-inverse in the form of an element $x$ such that $a = axa$.

I'm not sure this pans out, but I'm recording the idea for what it's worth.

Ryan Schwiebert (Dec 30 2024 at 12:02):

Todd Trimble said:

Well, part of me then thinks, "so much the worse for those category theorists". I consider the proof both efficient and elegant, and certainly creative.

I guess I can't speak for all algebraists but speaking for myself as a ring theorist, I also sometimes despair at mystifying elementary arguments. It's very much like a chess engine that can make great -but sometimes inscrutable- moves.

I've also seen once or twice someone very keen on categorical arguments overestimate the power of arguments-without-elements (as he unfortunately tried to use an element-free argument to prove something that had a counterexample). So, apparently both approaches complement each other and have a lot to offer...

Ryan Schwiebert (Dec 30 2024 at 12:14):

One (meta)property of VNR rings which I can't seem to find reflected above or in the nlab article is that the class of VNR rings is equational. Since it is a universal-algebraic property I think it's of interest to category theorists..

Ryan Schwiebert (Dec 30 2024 at 12:16):

Another "missing" characterization, IMO, is that this is also an equivalent to VNR:

• All right modules are principally injective

meaning that every homomorphism from a principal right ideal of $R$ into a module extends to a homomorphism of $R$ into the module. It is close to a dual notion of flatness, and therefore again possibly of interest to category theorists...

Ryan Schwiebert (Dec 30 2024 at 12:36):

Todd Trimble said:

Could you supply a source for that nice comment, Ryan?

That characterization of Morita equivalent rings is in Proposition 18.33 p 491 of Lam's Lectures on Modules and Rings. I really have to believe it appears in Anderson and Fuller as well but I didn't catch it on my first scan.

Todd Trimble (Dec 30 2024 at 13:06):

@Ryan Schwiebert wrote

One (meta)property of VNR rings which I can't seem to find reflected above or in the nlab article is that the class of VNR rings is equational. Since it is a universal-algebraic property I think it's of interest to category theorists..

I didn't know it! I know that commutative VNRs are equational, and an argument towards that appears in the nLab. Yes, that fact should go into the nLab. I don't know how to prove it (perhaps for not having tried it yet).

Ryan Schwiebert said:

Another "missing" characterization, IMO, is that this is also an equivalent to VNR:

• All right modules are principally injective

meaning that every homomorphism from a principal right ideal of $R$ into a module extends to a homomorphism of $R$ into the module. It is close to a dual notion of flatness, and therefore again possibly of interest to category theorists...

I saw something that looks related on the Wikipedia page, again for commutative VNRs, about "V-rings".

That characterization of Morita equivalent rings is in Proposition 18.33 p 491 of Lam's Lectures on Modules and Rings. I really have to believe it appears in Anderson and Fuller as well but I didn't catch it on my first scan.

Thank you!

Ryan Schwiebert (Dec 30 2024 at 13:25):

At risk of distracting you further, there's another interesting generalization of semisimplicity: quasi-Frobenius rings. After semisimple rings, QF and VNR rings could be considered the "next nicest" classes. QF rings don't form an equational class, but they do have some properties which seem like they verge on categorical interest (namely, they are characterized by their projective modules coinciding with their injective modules.) The first time I really saw someone leveraging functors in a coding theory paper was a preprint by Jay Wood about algebraic codes over QF rings.

Ryan Schwiebert (Dec 30 2024 at 13:27):

Todd Trimble said:

VNRs are equational,

I've always thought of VNR rings as a sort of generalization of division rings, what with the pseudoinverses for everything and all. Don't VNR rings form an algebraic variety? If they are a variety containing division rings I wonder if it is not the smallest...

Todd Trimble (Dec 30 2024 at 13:31):

Don't VNR rings form an algebraic variety?

Well, I don't know. I was just trying to find where anyone proves this, and was hoping you'd know!

It would be fine if pseudo-inverses in a noncommutative VNR were uniquely determined. Are they? (i.e., given $a$, an element $x$ such that $axa = a$ and $xax = x$.)

Todd Trimble (Dec 30 2024 at 13:50):

In this MO question, Thomas Klimpel asserts that vN regular rings do not form a variety of algebras, saying that pseudo-inverses in the sense I just mentioned are not uniquely determined. That falls short of a proof that they absolutely, positively cannot be a variety of algebras, but that's my current spidey-sense as well. (Mind you, I've only been learning this stuff in the past few days, so what do I know?)

Ryan Schwiebert (Dec 30 2024 at 15:01):

Todd Trimble said:

It would be fine if pseudo-inverses in a noncommutative VNR were uniquely determined. Are they?

I had the same question! I do not know.

Todd Trimble said:

Thomas Klimpel asserts that vN regular rings do not form a variety of algebras, saying that pseudo-inverses in the sense I just mentioned are not uniquely determined

Is that an obstruction to equationality? To say that pseudoinverses exist is to say thereis a function $P:R\to R$ such that $x=x*p(x)*x$ for all $x$, and to my understanding that constituted an equational definition for the class. At least, this is how it seemed to me and the more expert users on math.se explaining it to me.

I get the feeling there might be a nuance here between varieties and what I'm calling equational classes... The answer there seems to hint there is something like that.

Chris Grossack (they/them) (Dec 30 2024 at 15:14):

VNR rings (in the language of rings) can't possibly be equationally defined since they aren't closed under substructure (and all varieties are closed under quotients, substructure, and products). Indeed, $\mathbb{R}$ is VNR, since it's a field, but $\mathbb{Z}$ isn't, since there's no solution to $2y2 = 2$ in $\mathbb{Z}$.

Todd Trimble (Dec 30 2024 at 15:22):

Chris, I'm confused by your reasoning here. Commutative vN regular rings do form a variety. Why wouldn't your example apply there?

Chris Grossack (they/them) (Dec 30 2024 at 15:22):

This isn't enough to buy us nonuniqueness of pseudoinverses, though... (I do think the pseudoinverses are nonunique, but I don't have an example). Indeed, it turns out that in commutative VNR rings the pseudoinverse is unique! See Qiaochu's answer to his own question here, so we can view commutative VNR rings as a retract of a theory with a symbol for the pseudoinverse (this theory is often called the theory of meadows in (parts of) the literature). The reason $\mathbb{Z}$ is not a meadow, then, is because it's not a substructure in this larger language, since it isn't closed taking pseudoinverses!

Todd Trimble (Dec 30 2024 at 15:24):

Indeed, it turns out that in commutative VNR rings the pseudoinverse is unique!

Yes, this observation is in the nLab (no mention of meadows, though) and also appears earlier in this very long thread.

Todd Trimble (Dec 30 2024 at 15:31):

In any event, what I'm not seeing clearly is why there couldn't be some other crazy operations and equations, possibly quite other than pseudo-inverse, such that the class of rings with those operations and equations winds up being the same class as vN regular rings.

Maybe some idea similar to HSP, or some anomaly about the category of vN regular rings (such as, oh I don't know, maybe not being Barr exact for example), could be used to rule out the possibility, but I'm not seeing it at the moment.

Ryan Schwiebert (Dec 30 2024 at 15:32):

@Chris Grossack (they/them) My understanding is that the only substuctures that need to satisfy that would have to also fit the equation defining the parent structure. It' not simply all subrings.

Chris Grossack (they/them) (Dec 30 2024 at 15:33):

Yes, that's why I added a parenthetical that VNR rings can't be equationally defined in the language of rings, you would need to extend the language like we do in the commutative case.

Chris Grossack (they/them) (Dec 30 2024 at 15:34):

I've just changed my language from "a variety" to "equationally defined in this language", which is hopefully clearer, since "variety" is probably best saved as a presentation-invariant word anyways

Ryan Schwiebert (Dec 30 2024 at 15:36):

@Chris Grossack (they/them) I guess I'm still dissonant in definitions. The definition I was given regarding equational classes is here. Does that differ from "equationally defined in this language"?

I'm still a bit unclear on why anyone cares at all if the pseudoinverses are unique or not. That inverses in a division ring are unique is one thing, but it's not really an issue for VNR rings definitionally. No doubt you'll point out to me what I'm missing...

Chris Grossack (they/them) (Dec 30 2024 at 15:46):

Well regardless, this should get the job done: Goodearl's book on VNR Rings says (example 1.10, on pg 5):

There exists a regular ring $R$ with isomorphic regular subrings $R_1 \supseteq R_2 \supseteq \cdots$ such that $\bigcap R_n$ is not regular. Thus, inverse limits of regular rings need not be regular.

I haven't read the example too closely (I'm about to leave to see a friend), but it's crucially noncommutative, which gives some weight to the idea that if you look at it more carefully this is really a statement about the category of noncommutative VNR Rings, which thus can't form a variety since, as Todd pointed out, the category of any variety will have limits.

Ryan Schwiebert (Dec 30 2024 at 15:50):

Ok, so it's sounding more and more like I have a version of equational class that does not amount to "variety."

Todd Trimble (Dec 30 2024 at 15:50):

Just to get something nomenclatural out of the way: Variety of algebras. Generally, I just think of a variety of algebras as a category of models for a Lawvere theory, although maybe other mathematicians don't speak of the category per se.

I'm still a bit unclear on why anyone cares at all if the pseudoinverses are unique or not. That inverses in a division ring are unique is one thing, but it's not really an issue for VNR rings definitionally. No doubt you'll point out to me what I'm missing...

Just because there exists a pseudo-inverse for every element doesn't mean that there's a coherent global inversion operation that would give you that. But even if there were, having two different inversion structures would mean that the identity map on the underlying set is not an isomorphism within this variety, while it is an isomorphism on the underlying vN rings. So the categories would be different.

Todd Trimble (Dec 30 2024 at 15:52):

It looks like Chris may have spotted a crucial counterexample, but I'd like to have a look.

Ryan Schwiebert (Dec 30 2024 at 15:56):

Todd Trimble said:

ust because there exists a pseudo-inverse for every element doesn't mean that there's a coherent global inversion operation that would give you that.

How does selecting a $y_x$ for every $x$ to satisfy $x=xy_xx$ differ from a 'global inversion operation'? Are there some constraints beyond what I think there are?

Todd Trimble (Dec 30 2024 at 15:58):

Well, let me not try to belabor that particular point. Perhaps the more important one was about the categories, i.e., the morphisms involved.

Ryan Schwiebert (Dec 30 2024 at 15:59):

Yeah, I'm not following the angles being suggested to me. All I can tell is that the definition of equational class I'm working with may be nonstandard. I do recall that in the discussion the selection of the function was a piece of data for the class, and maybe what we're discussing here is a version that encompasses more than that.

Todd Trimble (Dec 30 2024 at 16:06):

The basic point is that you could have two genuinely different structures (within the same signature) that are the same as vN regular rings, so vN rings are not genuinely equational in nature. But maybe you got that already.

Ryan Schwiebert (Dec 30 2024 at 16:12):

@Todd Trimble Ok, meaning that including the function as part of the signature is not standard. What's a good reference for the version you're talking about? I had an exceedingly hard time finding them, and had to rely on math.se friends.

For reference, I first got onto it in nlab on the Bezout ring page. I could swear it used to comment somewhere that the theory was equational, and that would be witnessed by similar functions and identities. But there again, the inputs to the identities are by no means uniquely determined.

All I can see there now is "algebraic theory." It looks like algebraic theories are not quite the same as varieties? "Variety" is only mentioned once on the algebraic theory page.

Todd Trimble (Dec 30 2024 at 16:29):

Yes, you could put it that way. It comes down to how you define the morphisms between the objects of interest. The usual notion of morphism between vN regular rings is (correct me if I'm wrong) just a ring homomorphism. But the notion of morphism between rings equipped with pseudo-inversion would preserve that operation.

Without meaning to antagonize, let me quote you again:

One (meta)property of VNR rings which I can't seem to find reflected above or in the nlab article is that the class of VNR rings is equational. Since it is a universal-algebraic property I think it's of interest to category theorists..

This would be fabulous to know, but it would be of interest to category theorists because varieties as categories are quite wonderful. But here, paying attention to what the morphisms are is critical.

The notion of "variety" as category I'm using is the one used by category theorists such as Adamek and Rosicky and their various co-authors: it's a category of models of a [[Lawvere theory]]. I'm having trouble linking to it directly, but see for example

• Adámek, V. Koubek and J. Velebil, A duality between infinitary varieties and algebraic theories, Comment. Math. Univ. Carolin. 41 (2000)

near the top of page 280.

Todd Trimble (Dec 30 2024 at 16:39):

Ooh, that article on [[Bezout ring]] looks flawed, for exactly the reasons we're discussing. There's a world of difference between saying "has functions such that..." and "is equipped with functions such that". The former is merely an existential condition. The latter posits a specific choice as part of the data.

Looking into the history, the article was created by some anonymous users, and we may have had problems in the past with one of them.

John Baez (Dec 30 2024 at 18:00):

Ryan Schwiebert said:

All I can see there now is "algebraic theory." It looks like algebraic theories are not quite the same as varieties? "Variety" is only mentioned once on the algebraic theory page.

They're closely linked: a variety is the same as the category of 'models' of an algebraic theory of the kind called a Lawvere theory.

Simply put, a Lawvere theory is a category $T$ with products whose objects are $x^0, x^1, x^2, \dots$ for some $x$, and a model of this theory is a product-preserving functor $f: T \to \mathsf{Set}$. A morphism of models is a natural transformation between such functors. The category of models of a Lawvere theory is called a variety.

This is not a definition of 'variety' that you'd ever usually see people give in the field of 'universal algebra', but it winds up being equivalent.

For example: there's a Lawvere theory $T$ of rings, where the morphisms $x^n \to x$ are all the $n$-ary operations you can define in the language of rings, with all the usual equations between them. A model of $T$ is a ring, and morphism of models is a ring homomorphism. The category of models is the category of rings, and this is a variety.

John Baez (Dec 30 2024 at 18:12):

I seem to recall reading these claims earlier in this discussion:

1. A ring $R$ is von Neumann regular if it can be equipped with an operation $\overline{\phantom{x}}$ obeying the equational laws $x \overline{y} x = x$ and $\overline{x} y \overline{x} = y$.

2. Such an operation on a ring, obeying these equations, is unique if it exists.

(Someone correct me if these are wrong.)

If so we can make up a Lawvere theory of vN regular rings by taking the theory of rings and throwing in this extra operation obeying these equations. Its category of models is a variety. Objects are vN regular rings, and morphisms are ring homomorphisms preserving the $\overline{\phantom{a}}$ operation.

This category will thus have a forgetful functor to the category of rings. This forgetful functor is faithful, and 2. claims that it's essentially injective, but is it full? I.e., does every ring homomorphism between vN regular rings preserve the $\overline{\phantom{a}}$ operations?

Todd Trimble (Dec 30 2024 at 18:21):

John Baez said:

I seem to recall reading these claims earlier in this discussion:

1. A ring $R$ is von Neumann regular if it can be equipped with an operation $\overline{\phantom{x}}$ obeying the equational laws $x \overline{y} x = x$ and $\overline{x} y \overline{x} = y$.

2. Such an operation on a ring, obeying these equations, is unique if it exists.

(Someone correct me if these are wrong.)

If so we can make up a Lawvere theory of vN regular rings by taking the theory of rings and throwing in this extra operation obeying these equations. Its category of models is a variety. Objects are vN regular rings, and morphisms are ring homomorphisms preserving the $\overline{\phantom{a}}$ operation.

This category will thus have a forgetful functor to the category of rings. This forgetful functor is faithful, and 2. claims that it's essentially injective, but is it full? I.e., does every ring homomorphism between vN regular rings preserve the $\overline{\phantom{a}}$ operations?