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John Baez (Jun 27 2023 at 18:39):I decided to start writing a bunch of blog posts about the ideas that have been obsessing me lately. Here's one of many possible entry points.
Sets and vector spaces are two of the most useful structures in mathematics. But in a certain sense we can embed set theory into linear algebra! For any set there's a vector space with that set as basis. And any function between sets gives a linear map between their vector spaces, sending basis vectors to basis vectors.
But can we find set theory lurking in linear algebra? That is, starting from the category of vector spaces, can we recognize the category of sets embedded in it?
Yes we can! I explain how here:
The free vector space on a set acquires extra structure from having come from a set (that is, having a chosen basis): it's a 'cocommutative coseparable algebra', which is a gadget I define. And I have a conjecture:
Conjecture 3. The category of cocommutative coseparable coalgebras in , and coalgebra homomorphisms between these, is equivalent to the category of sets and functions.
But so far the only proof I know relies on a bunch of results including an offhand remark without proof in the first paper on coseparable coalgebras. There should be a better proof, with all the details filled in.
Patrick Nicodemus (Jun 27 2023 at 19:10):I would be interested to see whether you can relate this to the theorem that sets are equivalent to coalgebras with respect to the forgetful-free comonad on vector spaces.
John Baez (Jun 27 2023 at 19:14):Hmm, that sounds like it could be a more systematic way to approach this problem. In "my" solution, we get cocommutative coseparable coalgebras, which are defined by a binary and a nullary co-operation obeying some equational laws and, alas, a property that involves an existential quantifier.
Fabrizio Genovese (Jun 28 2023 at 11:03):Does this have to do with the fact that every vector space is free? If that's so, how much can you generalize this, e.g. to 'embedding set into any category of free objects'?
John Baez (Jun 28 2023 at 14:07):Strangely the fact that every vector space is free doesn't seem to be crucial here: we are finding all the algebraic structure that a vector space gets from being free on a set, i.e. being equipped with a basis. This structure is called "being a commutative special Frobenius algebra". But the fact that every vector space has the property that you can give it a basis doesn't seem so important.
John Baez (Jun 28 2023 at 14:09):In other words, the fact that the free vector space on a set functor is faithful means there's a copy of sitting inside . The fact that you pointed out says that is essentially surjective, and that just means this copy contains all the objects (up to iso).
John Baez (Jun 28 2023 at 14:12):The free group on a set functor is also faithful, and in a comment on my blog article Oscar Cunningham discusses how to recover the category of sets from the category of groups. But this functor is not essentially surjective: not every group is free.
John Baez (Jun 28 2023 at 14:13):So it's definitely interesting to try to generalize this business of finding set theory hiding inside linear algebra, or group theory. At least it's interesting to me.
Fabrizio Genovese (Jun 28 2023 at 14:19):John Baez said:
Strangely the fact that every vector space is free doesn't seem to be crucial here: we are finding all the algebraic structure that a vector space gets from being free on a set, i.e. being equipped with a basis. This structure is called "being a commutative special Frobenius algebra". But the fact that every vector space has the property that you can give it a basis doesn't seem so important.
Well then my question just becomes: Can you generalize this to every category whoe objects are equipped with a commutative special Frobenius algebra?
Cole Comfort (Jun 28 2023 at 14:36):Fabrizio Genovese said:
John Baez said:
Strangely the fact that every vector space is free doesn't seem to be crucial here: we are finding all the algebraic structure that a vector space gets from being free on a set, i.e. being equipped with a basis. This structure is called "being a commutative special Frobenius algebra". But the fact that every vector space has the property that you can give it a basis doesn't seem so important.
Well then my question just becomes: Can you generalize this to every category whoe objects are equipped with a commutative special Frobenius algebra?
I was under the impression that this correspondence between orthonormal bases and frobenius algebras only works in for finite dimensional Hilbert spaces (where the monoid/comonoid are additionally required to be adjoint to each other). If you want to talk about the "closest thing to a free Hilbert space on a set", then which takes the special commutative natural -semi-frobenius algebras induced by the diagonal and codiagonal maps and in to -algebras in .
So I would expect that the functor picks out the special semi-frobenius algebras which are induced by the composite . But this is not a dagger functor, so there is something fishy going on.
John Baez (Jun 29 2023 at 15:25):Fabrizio Genovese said:
John Baez said:
Strangely the fact that every vector space is free doesn't seem to be crucial here: we are finding all the algebraic structure that a vector space gets from being free on a set, i.e. being equipped with a basis. This structure is called "being a commutative special Frobenius algebra". But the fact that every vector space has the property that you can give it a basis doesn't seem so important.
Well then my question just becomes: Can you generalize this to every category whose objects are equipped with a commutative special Frobenius algebra?
John Baez (Jun 29 2023 at 15:25):I think it's not quite that general: some special facts about vector spaces are being used, like duals. Also, the case of vector spaces over an algebraically closed field, like works a lot more simply than for other fields. Aurelio Carboni has written a nice paper trying to generalize the idea a lot, and I'll be talking about this in future blog articles.
By the way I must have been half-asleep when I wrote the above, because it's not quite right. A finite-dimensional vector space with a basis gives a commutative special Frobenius algebra. I found out how to generalize that to infinite sets yesterday, when Theo Johnson-Freyd helped me prove Conjecture 3 in my blog article.
Conjecture 3. The category of cocommutative coseparable coalgebras in , and coalgebra homomorphisms between these, is equivalent to the category of sets and functions.
John Baez (Jun 29 2023 at 15:36):Cole Comfort said:
I was under the impression that this correspondence between orthonormal bases and frobenius algebras only works in for finite dimensional Hilbert spaces (where the monoid/comonoid are additionally required to be adjoint to each other).
John Baez (Jun 29 2023 at 15:37):Finite-dimensionality is indeed necessary to get a Frobenius algebra (all Frobenius algebras are finite-dimensional), but a Hilbert space structure is not. It sounds like you're talking about @Bob Coecke, @Jamie Vicary and @dusko Pavlovic's 2008 paper A new description of orthogonal bases. This shows that equipping a finite-dimensional complex Hilbert space with an orthonormal basis is equivalent to making it into a commutative special dagger-Frobenius algebra.
But my blog article explains a line of work starting with Grothendieck, whose results from 1960-1961 easily imply that equipping a finite-dimensional vector space over any algebraically closed field is equivalent to making it into a commutative special Frobenius algebra!
I don't know if the newer authors were familiar with this older line of work, which often goes under the name "The fundamental theorem of Grothendieck's Galois theory".
John Baez (Jun 29 2023 at 15:37):The Fundamental Theorem of Grothendieck Galois Theory. The opposite of the category of commutative separable algebras over a field is equivalent to the category continuous actions on finite sets of the absolute Galois group of .
There's a nuance here: a separable algebra is an algebra with the property that there exists some way of making it into a commutative special Frobenius algebra.
John Baez (Jun 29 2023 at 15:44):My blog article only covers the case where the Galois group is trivial, which happens when is 'separably closed'. To make it less intimdating, for most of the article I only talk about the example of . But near the end I state and outline the proof of this:
Theorem. If a field is separably closed, the opposite of the category of commutative separable algebras over is equivalent to .
Fabrizio Genovese (Jun 29 2023 at 15:47):Another way to get Frobenius structures from bases in the context of separable hilbert spaces can be found in this old paper by me and Stefano Gogioso: https://arxiv.org/pdf/1605.04305.pdf
Fabrizio Genovese (Jun 29 2023 at 15:50):The trick is more or less this: You switch to non-standard Hilbert spaces and truncate them up to to some fixed infinite natural number (a finite natural equal to the dimension of the space if you are starting from a finite dimensional one). Then you can show that for each orthonormal basis in a separable space H, you get a weakly unital, weakly special commutative dagger Frobenius algebra in the non-standard counterpart.
Fabrizio Genovese (Jun 29 2023 at 15:50):I don't think this approach is particularly relevant here, I just wanted to point out that you can do this as well if you are trying to connect frobenius algebras and orthonormal bases in the infinite-dimensional case.
John Baez (Jun 29 2023 at 15:51):Neat! I see you tackle infinite-dimensional Hilbert spaces. When I tackled infinite-dimensional vector spaces in my (now-proved) Conjecture 3, I needed some gadgets called 'coseparable coalgebras', which turn out to have been invented by Richard Gustavus Larsson in 1972.
Fabrizio Genovese (Jun 29 2023 at 15:52):Bluntly put, our idea here was immensely simple: In defining the frobenius algebra from a basis, at some point you need to take a sum over all the basis vectors. If you have infinitely many, this sum is not part of your hilbert space, because it diverges. So you are missing some pieces (such as the counit), which should nevertheless 'morally exist'.
Fabrizio Genovese (Jun 29 2023 at 15:52):The idea is that, modulo some details, if you go non-standard, then infinites become just numbers, that sum exists, and you are happy. Lol.
Fabrizio Genovese (Jun 29 2023 at 15:56):(And yes one then has to prove that everything works as expected etc, which we did). Our main aim here was recovering the pictorial formalism in terms of spiders and string diagrams (à la 'Picturing quantum processes') also for the infinite-dimensional case.
John Baez (Jun 29 2023 at 15:57):Nice. The deal with coalgebras is that they allow you to avoid these infinite sums, yet still get the job done.
Fabrizio Genovese (Jun 29 2023 at 15:57):We also had two follow up papers on this, but as time went by they got more and more closer to quantum field theory and I started understanding things less and less since I'm not a field theorists. So in the end I dropped this line of research
Jamie Vicary (Jun 29 2023 at 16:01):Thanks for this, yes we were aware when writing the paper that it was
already known that bases correspond to special commutative Frobenius
algebras -- because you told us so, and we thank you for that in the paper!
However this does not trivialise our results on orthogonal and orthonormal
bases. The overall situation can be summarised as follows:
Arbitrary basis --- Commutative special Frobenius algebra
Orthogonal basis --- Commutative †-Frobenius algebra
Orthonormal basis --- Commutative special †-Frobenius algebra
Note that in the orthogonal case, the Frobenius algebra is not special! So
the known result does not immediately apply.
This is a head-scratcher. An orthogonal basis is certainly a basis. So how
is it possible that the associated Frobenius algebra is not special?
The answer is that the basis only fixes the comultiplication, as the unique
map cloning the basis elements. The choice of the multiplication remains
free up to a choice of scalar factor. Given an orthogonal basis which is
not orthonormal, different choices of that scalar let you choose if you
want the structure to be special Frobenius (and not dagger), or dagger
Frobenius (and not special.)
In the case of an orthonormal basis this freedom of choice disappears.
Fabrizio Genovese (Jun 29 2023 at 16:01):John Baez said:
Nice. The deal with coalgebras is that they allow you to avoid these infinite sums, yet still get the job done.
Yup, I'm reading about it now. In a sense it's the dual approach. You replace counits with an existential requirement as in 'this multiplication must exist', whereas we force them in the picture by making the infinite sums converge
John Baez (Jun 29 2023 at 16:48):Jamie Vicary said:
Thanks for this, yes we were aware when writing the paper that it was
already known that bases correspond to special commutative Frobenius
algebras -- because you told us so, and we thank you for that in the paper!
Bwahaha - I'm getting so old and forgetful I forgot I even knew about that back then, much less told you, much less that you thanked me for it.
John Baez (Jun 29 2023 at 16:56):However this does not trivialise our results on orthogonal and orthonormal bases.
No, indeed it does not. I wasn't trying to say that; sorry if it came across that way. I'm just finding there to be a strange disconnect between Grothendieck's old work on separable algebras, and the work on topos theory it led to, and the newer thoughts on Frobenius algebras in quantum foundations. They're both about 'logic', in a general sense, and how topos logic fits inside quantum logic. But the old work never mentions the word 'quantum', and the newer work doesn't seem to pay much attention to the older work (in part because it's focused on rather than more general fields or commutative rings).
I could be wrong, of course - I apparently don't even know what's in your paper, much less all the papers that followed. :rolling_eyes: But I feel there's some realm to be explored, that seems mainly to have gotten incursions from two separate sides. So I want to write some blog articles about how they fit together.
John Baez (Jun 29 2023 at 21:38):Now that I think about it, @Jamie Vicary, I probably got interested in special Frobenius algebras from Fukuma, Hosono and Kawai's 1992 paper on 2d lattice TQFT, which was later polished to mathematical perfection in two papers by Lauda and Pfeiffer. The latter discussed strongly separable algebras in 2006, so I think by then I knew pretty well that every commutative special Frobenius algebra over is isomorphic to the algebra of functions on a finite set. I also remember Aaron telling me that over non-algebraically-complete fields of characteristic there are semisimple algebras that are not separable. But at the time I regarded that as a curiosity. Only recently have I gotten excited about the Fundamental Theorem of Grothendieck Galois Theory, which says that for every field the opposite of the category of commutative separable algebras forms the topos of continuous actions of its Galois group on finite sets! Back when I told you something about commutative special Frobenius algebras, I had no inkling of this. So, my attitude towards this subject has completely changed, perhaps blotting out some earlier memories.