Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Frobenius algebras and orientability

Jules Hedges (Nov 28 2021 at 13:01):

I've been trying to understand in more detail how the monoidal theory of Frobenius algebras relates to 2d surfaces. Let $\mathbf{Frob}$ be the free symmetric monoidal category generated by a commutative (not special) Frobenius algebra. Here's a thing I believed until yesterday: $\mathbf{Frob}$ is isomorphic to the category of 2-cobordisms. The essence of this equivalence is contained in pictures like this: Screenshot-2021-11-28-at-12.51.32.png

Consequently, the monoid of scalars $\mathbf{Frob}(0, 0)$ is (homeomorphism? isotopy? diffeomorphism?) equivalence classes of not-necessarily-connected compact surfaces, made into a monoid by disjoint union. And since compact surfaces are classified by their genus, $\mathbf{Frob}(0, 0)$ is isomorphic as a monoid to finite multi-subsets of $\mathbb N$ with multiset union

Anyway, yesterday I realised that orientability messes this up slightly. I used to believe these 2 facts: (1) every compact surface admits a pants decomposition, and (2) string diagrams in $\mathbf{Frob}$ are the same thing as pants decompositions. I then spent several hours trying to describe a Klein bottle, which is a compact surface, as a string diagram in $\mathbf{Frob}$ before convincing myself that it's impossible.

Here's some things I currently believe, can anyone confirm or deny these? (1) $\mathbf{Frob}$ is equivalent to the category of oriented 2-cobordisms. (2) $\mathbf{Frob}(0, 0)$ is the monoid of oriented (not necessarily connected) compact surfaces with disjoint union, which is isomorphic to the monoid of finite multi-subsets of natural numbers. (3) Pants decompositions are not the same thing as string diagrams in $\mathbf{Frob}$ , for example this is a pair of pants (up to homeomorphism) that doesn't correspond to anything in $\mathbf{Frob}$ : Screenshot-2021-11-28-at-12.54.21.png

(4) Every (not necessarily oriented) compact surface admits a pants decomposition, it's just that the Klein bottle uses that "twisted" pants

Heading out of "questions that somebody here could probably answer in their sleep" and into the realm of wild speculation, I wondered what happens if you add an extra generator to the monoidal theory of commutative Frobenius algebras, corresponding to the "Klein twist": Screenshot-2021-11-28-at-12.57.42.png

I wonder what equations ought this generator to satisfy in order to classify not-necessarily-orientable cobordisms? I came up with a completely out-there hypothesis that this thing ought to behave like a Hadamard gate in phase-free ZX, and that the theory of "commutative Frobenius algebras + Klein twist" is equivalent to the theory of interacting Hopf algebras. Then the idea would be that the 2 coloured spiders of ZX correspond to keeping track of the 2 possible "local orientations". But I don't have much evidence for this beyond wishful thinking

John Baez (Nov 28 2021 at 13:17):

(1) $\mathbf{Frob}$ is equivalent to the category of oriented 2-cobordisms.

Yes, I think a good reference is

Joachim Kock, Frobenius Algebras and 2D Topological Quantum Field Theories.

John Baez (Nov 28 2021 at 13:19):

The link goes to the PDF of a short version of Joachim's book. In the introduction, he says:

In other words, 2Cob is the free symmetric monoidal category on a commutative Frobenius object (Theorem 3.4.14).

John Baez (Nov 28 2021 at 13:22):

The theorem is older than this book. Joachim has a page including errata for the book and a pleasant history of the origin of the Frobenius equation.

John Baez (Nov 28 2021 at 13:22):

This is the Frobenius equation:

John Baez (Nov 28 2021 at 13:25):

(2) $\mathbf{Frob}(0, 0)$ is the monoid of oriented (not necessarily connected) compact surfaces with disjoint union, which is isomorphic to the monoid of finite multi-subsets of natural numbers.

Yes, that follows. Over on Twitter I noticed that this is the free commutative monoid on (the underlying set of) the free commutative monoid on one generator!

John Baez (Nov 28 2021 at 13:26):

In other words, it's the free commutative monoid on $\mathbb{N}$ .

John Baez (Nov 28 2021 at 13:27):

That's interesting to me because the set of Young diagrams is the free commutative monoid on $\mathbb{N} - 0$ .

John Baez (Nov 28 2021 at 13:29):

A Young diagram is a thing like this:

John Baez (Nov 28 2021 at 13:30):

It has no rows with zero boxes, so Young diagrams are elements of the free commutative monoid on $\mathbb{N} - 0$ .

John Baez (Nov 28 2021 at 13:32):

. (3) Pants decompositions are not the same thing as string diagrams in $\mathbf{Frob}$ for example this is a pair of pants (up to homeomorphism) that doesn't correspond to anything in $\mathbf{Frob}$ .

I don't know your technical definition of 'pants decomposition', but that thing you showed is an unorientable surface with boundary.

John Baez (Nov 28 2021 at 13:37):

(4) Every (not necessarily oriented) compact surface admits a pants decomposition, it's just that the Klein bottle uses that "twisted" pants.

Again, I don't know what your definition of 'pants decomposition' is so I can't confirm this.

Jules Hedges (Nov 28 2021 at 13:39):

I don't have a definition (it's a phrase I picked up from wikipedia), I'm just eyeballing it. It ought to mean something like "bunch of cuts that partition a surface into finitely many pieces, each of which is (homeomorphic? isotopic? diffeomorphic? shrug) to either a pair of pants or a hemisphere

John Baez (Nov 28 2021 at 13:40):

Heading out of "questions that somebody here could probably answer in their sleep" and into the realm of wild speculation, I wondered what happens if you add an extra generator to the monoidal theory of commutative Frobenius algebras, corresponding to the "Klein twist".

That looks right. I believe the category of unoriented 2d cobordisms is equivalent to the free symmetric monoidal category on one object $x$ equipped with a morphism $t: x \to x$ obeying $t^2 = 1$ . This $t$ is your 'Klein twist'.

Jules Hedges (Nov 28 2021 at 13:40):

Oh, neat! Do you know a reference for that?

John Baez (Nov 28 2021 at 13:48):

Actually my guess was wrong because you can't get the Moebius strip from just that! You need to include the Moebius strip (a cobordism from the circle to the empty set) and the Klein twist, and some relations. This has a reference:

Generators and relations for the 2-dimensional unoriented cobordism category, MathOverflow.

John van de Wetering (Nov 29 2021 at 11:24):

Jules Hedges said:

Heading out of "questions that somebody here could probably answer in their sleep" and into the realm of wild speculation, I wondered what happens if you add an extra generator to the monoidal theory of commutative Frobenius algebras, corresponding to the "Klein twist": Screenshot-2021-11-28-at-12.57.42.png

I wonder what equations ought this generator to satisfy in order to classify not-necessarily-orientable cobordisms? I came up with a completely out-there hypothesis that this thing ought to behave like a Hadamard gate in phase-free ZX, and that the theory of "commutative Frobenius algebras + Klein twist" is equivalent to the theory of interacting Hopf algebras. Then the idea would be that the 2 coloured spiders of ZX correspond to keeping track of the 2 possible "local orientations". But I don't have much evidence for this beyond wishful thinking

So @Alexander Cowtan and I have been thinking about this kinda stuff a bit, based on the work done by David Reuter on understanding bialgebras as 2d surfaces whose cross-section is a square that is two-coloured (not sure if I'm describing this very well). In that setting a Hadamard corresponds to a 90-degree rotation. Then you don't get that Hadamard*Hadamard = identity, rather using Pauli's belt-trick you get that Hadamard^8 = identity, which corresponds to the identity that is actually provable in general Hopf algebras (this isn't a formal proof yet, but more of an intuition). Interestingly, using just topological deformations you can prove the axioms of a weak bialgebra.

John van de Wetering (Nov 29 2021 at 11:25):

And Cowtan told me that if you make some kind of quotient on the surfaces that you can prove all the Hopf algebra equations

Cole Comfort (Nov 29 2021 at 11:48):

John van de Wetering said:

Jules Hedges said:

Heading out of "questions that somebody here could probably answer in their sleep" and into the realm of wild speculation, I wondered what happens if you add an extra generator to the monoidal theory of commutative Frobenius algebras, corresponding to the "Klein twist": Screenshot-2021-11-28-at-12.57.42.png

I wonder what equations ought this generator to satisfy in order to classify not-necessarily-orientable cobordisms? I came up with a completely out-there hypothesis that this thing ought to behave like a Hadamard gate in phase-free ZX, and that the theory of "commutative Frobenius algebras + Klein twist" is equivalent to the theory of interacting Hopf algebras. Then the idea would be that the 2 coloured spiders of ZX correspond to keeping track of the 2 possible "local orientations". But I don't have much evidence for this beyond wishful thinking

So Alexander Cowtan and I have been thinking about this kinda stuff a bit, based on the work done by David Reuter on understanding bialgebras as 2d surfaces whose cross-section is a square that is two-coloured (not sure if I'm describing this very well). In that setting a Hadamard corresponds to a 90-degree rotation. Then you don't get that Hadamard*Hadamard = identity, rather using Pauli's belt-trick you get that Hadamard^8 = identity, which corresponds to the identity that is actually provable in general Hopf algebras (this isn't a formal proof yet, but more of an intuition). Interestingly, using just topological deformations you can prove the axioms of a weak bialgebra.

This is getting way off-topic, but I have thought about this quite a lot. Partly crossposting from twitter, this is almost surely related to the Weyl representation of stabilizer circuits.
In this setting, the Fourier transform corresponds to a twist with an antipode on one of the wires.
If we look at the phase free fragment and double the generators and swap the colour on one side, we can see the colour change law follows topologically (the grey box is the antipode)

Selection_018.png

This compositional account for the Weyl representation is given in our (shameless self-promotion) paper:
https://arxiv.org/pdf/2105.06244.pdf

An advantage of this doubling approach is that by mixing singled and doubled wires, you can talk about measurement and state preparation (paper forthcoming)

A similar thing is going on in this language quon, where qubits look like squares, the identity is a rectangle and the Fourier tranform is a twist. I am a bit skeptical of this paper because they don't even prove soundness and claim that this representation owrks for a universal class of quantum circuits, not just the stabilizer ones.
https://arxiv.org/pdf/1612.02630.pdf

Cole Comfort (Nov 29 2021 at 11:52):

John van de Wetering said:

And Cowtan told me that if you make some kind of quotient on the surfaces that you can prove all the Hopf algebra equations

In his talk, David quotients by adjoints to get the equations of a bialgebra:
http://www.cs.ox.ac.uk/people/david.reutter/Perimeter2017.pdf

If I recall correctly, you have to additionally quotient by some identity to cut a Möbius strip to get the equations of a Hopf algebra.

Jules Hedges (Nov 29 2021 at 11:53):

Hah, I came across those slides yesterday, I was literally about to ask if the square-ish diagrams in this is what you were talking about

Cole Comfort (Nov 29 2021 at 11:57):

If you grant that the Frobenius algebras are commutative, then the Fourier transform actually has order 4 instead 8. So if you see the Fourier transform from the point of view of the Weyl representation, this is evident. Because the Fourier transform has order 4, somehow the doubling in the symplectic picture feels kind of like a quadrupling. I would be really interested to find out the connection between the 3-categorical picture and the symplectic picture. There is pretty clearly some sort of functor going in the one direction

Cole Comfort (Nov 29 2021 at 12:01):

Like if double all of the wires in the symplectic picture once again, sending the identity to two wires, the antipode to a swap and the frobenius algebras to doubling them (but not changing the colours again) I think this analogy could actually be made precise.

Cole Comfort (Nov 29 2021 at 12:02):

Where we just imagine now that there is a surface between all of the wires that are bundled up together.

John van de Wetering (Nov 29 2021 at 12:15):

Cole Comfort said:

John van de Wetering said:

And Cowtan told me that if you make some kind of quotient on the surfaces that you can prove all the Hopf algebra equations

In his talk, David quotients by adjoints to get the equations of a bialgebra:
http://www.cs.ox.ac.uk/people/david.reutter/Perimeter2017.pdf

If I recall correctly, you have to additionally quotient by some identity to cut a Möbius strip to get the equations of a Hopf algebra.

Yes, those are the slides Cowtan and I have been staring at