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: theory: mathematics

Topic: Braided categorical torus

Alonso Perez-Lona (Aug 06 2025 at 20:35):

I believe this is a simple question but I am not sure what is the explicit form. What exactly is the braiding of a categorical torus of rank 1, the one presented as a crossed module (Z \times U(1) \to R) with boundary map (m,z) \mapsto z, and action R \times Z \times U(1) \to Z \times U(1) as (x,m,z) \mapsto (m,z exp(J(m,x))) for J the bilinear form that defines the extension.
I am having trouble seeing the braiding as a bracket { , }: R \times R \to U(1) as one could do per Joyal and Street.

Or is the categorical torus of rank 1 actually not braided?

Kevin Carlson (Aug 06 2025 at 21:38):

Hi Alonso,it'd be nice if you'd typeset your mathematical notation. Just use double dollar signs and type otherwise ordinary $\LaTeX.$

Morgan Rogers (he/him) (Aug 07 2025 at 06:52):

@Alonso Perez-Lona like this:

I believe this is a simple question but I am not sure what is the explicit form. What exactly is the braiding of a categorical torus of rank 1, the one presented as a crossed module $(\Z \times U(1) \to \R)$ with boundary map $(m,z) \mapsto z$, and action $\R \times \Z \times U(1) \to \Z \times U(1)$ as $(x,m,z) \mapsto (m,z \mathrm{exp}(J(m,x)))$ for $J$ the bilinear form that defines the extension.
I am having trouble seeing the braiding as a bracket $\{ , \}: \R \times \R \to U(1)$ as one could do per Joyal and Street.

Or is the categorical torus of rank 1 actually not braided?

For anyone that was wondering, searching for "categorical torus" turns up this paper: https://arxiv.org/abs/1406.7046 and related work by Nora Ganter but not much else; I don't think this is yet a widespread enough term for you to use it without context and expect people to instantly know what you're talking about, Alonso!

David Michael Roberts (Aug 07 2025 at 08:43):

Well, I knew what it was, and I put a little effort to look up the definition of braided crossed module, and it's slightly complex, but amounts to encoding the second component of the natural isomorphism data $\mathbb{R}\times \mathbb{R} \to \mathbb{R}\times (\mathbb{Z}\times U(1))$ for a (putative) braiding for the monoidal groupoid corresponding to the crossed module (this is the action groupoid of $\mathbb{Z}\times U(1)$ on $\mathbb{R}$, via $(n,z)\cdot x = x+n$). The first component is fixed by the requirement that the source of the braiding map is one order of the monoidal product.

David Michael Roberts (Aug 07 2025 at 08:48):

Since the monoidal structure on objects is addition in $\mathbb{R}$, which is commutative, then a braiding isomorphism for $(x,y)$ is $b_{x,y} = (n_{x,y},z_{x,y}) \colon x+y \to y+z$, but this forces $n_{x,y}=0$, so that the braiding would be determined by a map $\mathbb{R}^2\to U(1)$. This has to satisfy a bunch of axioms: https://xabier.garcia.martinez.webs.uvigo.gal/CT2024/files/Slides/Fern%C3%A1ndezFari%C3%B1a_Slides.pdf (the slide "Braiding for crossed modules of groups" - you can also see the reasoning behind this definition in the slides)

John Baez (Aug 07 2025 at 08:48):

I think it's gonna be less complicated than it sounds when you put it into the right context. Braided crossed modules are really braided 2-groups, which are infinity-groupoids with vanishing homotopy groups except for... let's see... the 3rd and 4th? So they're classified by two different abelian groups and some Postnikov data.

David Michael Roberts (Aug 07 2025 at 08:50):

They are the same as 2-crossed modules $H\to G \to 1$, which are pointed, connected, simply-connected homotopy 3-types. So it's vanishing homotopy groups except the 2nd and 3rd.

John Baez (Aug 07 2025 at 08:51):

Whoops! The hardest part of n-category theory is remembering what n equals.

But to understand that "bunch of data", we category theorists would prefer to see this Postnikov data as something like an associator and braiding obeying pentagon and hexagon identities up to pentagonators and Yang-Baxterators, which in turn obey some identities.

David Michael Roberts (Aug 07 2025 at 08:53):

One has to unwind the definition here for the special case where $G=\mathbb{Z}\times U(1)$ and the first component of $\{-,-\}$ is trivial:

Screenshot 2025-08-07 at 6.22.25 PM.png

I suspect it is not going to be horrible to apply those identities and see what happens, and make a guess...

John Baez (Aug 07 2025 at 08:57):

This looks very mild compared to the full-fledged case I was talking about - i.e., the associator and pentagonator look trivial here, so no pentagonator identity (= 3d Stasheff polytope).

Alonso Perez-Lona (Aug 07 2025 at 08:58):

I should have provided that slide along with my question. The first identity says that the image is only a $U(1)$ phase. The last two say that it is bilinear in $\mathbb{R}\times \mathbb{R}$, since the action is trivial whenever the integer component is zero. I thought the braiding would be something like $(x,x') \mapsto (0, \exp( J(x',x))$ but somehow the second and third identities are sensitive to whether the integer is on the left or on the right, as the rhs are $(0, \exp( -J(m',x))$ and $(0, \exp( J(m',x))$, respectively, for $m$ the image of $g:=(m,z)$. So I am not sure.

David Michael Roberts (Aug 07 2025 at 09:14):

@John Baez yes, it's a strict 2-group!

David Michael Roberts (Aug 07 2025 at 09:15):

@Alonso Perez-Lona also, please note that $J$ is initially only defined on the integers, if one takes an abstract approach and considers it as a cocycle defining an extension by $U(1)$....

David Michael Roberts (Aug 07 2025 at 09:16):

Is it $\mathbb{R}$-bilinear, or $\mathbb{Z}$-bilinear? I can't check right now. If the former, you only need define it at the basis vectors (0,1) and (1,0), no?

David Michael Roberts (Aug 07 2025 at 09:35):

Hmm, ok I guess it's only Z-bilinear, i.e. additive in each slot. So knowing it on R-generators isn't enough,

Alonso Perez-Lona (Aug 07 2025 at 09:35):

David Michael Roberts said:

Alonso Perez-Lona also, please note that $J$ is initially only defined on the integers, if one takes an abstract approach and considers it as a cocycle defining an extension by $U(1)$....

So one starts with the $\mathbb{Z}$-bilinear form $J: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, and then tensors with $\mathbb{R}$ to obtain the form $J: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ that is used to define the action.

David Michael Roberts (Aug 07 2025 at 09:36):

But, yes, it might have no braiding. Just like you can get non-abelian groups by central extensions of abelian groups....

Alonso Perez-Lona (Aug 07 2025 at 09:41):

David Michael Roberts said:

But, yes, it might have no braiding. Just like you can get non-abelian groups by central extensions of abelian groups....

I had been looking at the computation of the Drinfeld center of categorical tori presented in Proposition 4.1 here: https://arxiv.org/abs/2202.01271 where the objects are endowed with an additional character that precisely makes that distinction between the integer being on the left or on the right that I mentioned above. But it was suggested to me that the categorical torus itself was already braided, though I had never seen the braiding explicitly anywhere...

Alonso Perez-Lona (Aug 07 2025 at 09:42):

Which is why I was under the impression that the categorical torus being braided was a folk result of sorts, yet it sounds like it isn't?

David Michael Roberts (Aug 07 2025 at 11:04):

I was there at the birth of Nora's paper, or at least visited her and contributed some advice, and I don't recall it being discussed. And it didn't come up when her student visited me recently and we talked a lot about geometry that involved it. I couldn't say I know everything around this construction, but I think I would have noticed someone mentioning a fact like the braiding at some point, because I do like it as an example.

Alonso Perez-Lona (Aug 07 2025 at 14:24):

Alonso Perez-Lona said:

David Michael Roberts said:

But, yes, it might have no braiding. Just like you can get non-abelian groups by central extensions of abelian groups....

I had been looking at the computation of the Drinfeld center of categorical tori presented in Proposition 4.1 here: https://arxiv.org/abs/2202.01271 where the objects are endowed with an additional character that precisely makes that distinction between the integer being on the left or on the right that I mentioned above. But it was suggested to me that the categorical torus itself was already braided, though I had never seen the braiding explicitly anywhere...

Looking at the computation of the Drinfeld center here, it would seem to me that the categorical torus of rank 1 $T_J$ is braided only when the extension class is trivial. The reason being that for a nontrivial class, the objects of the Drinfeld center are $\mathbb{R}$, which project to those of $T_J$ via the exponential map $\mathbb{R} \to U(1)$. But if $T_J$ admitted a braiding then it would imply a group homomorphism section of the exponential map, which does not exist. On the other hand, if the extension class is trivial, then the objects of the Drinfeld center of $T_{J=0}$ are $U(1) \times \mathbb{Z}$, which map to the objects of $T_0$ by projecting out the $\mathbb{Z}$. Equipping $T_0$ with a braiding is the section of this homomorphism, which just amounts to a choice of character $\mathbb{Z} = \text{Hom}(U(1),U(1))$.

Is this argument correct, or is it overlooking something?

David Michael Roberts (Aug 07 2025 at 21:41):

The object group of the categorical torus of rank 1 is R, not U(1). So I don't understand the argument.

Alonso Perez-Lona (Aug 07 2025 at 21:47):

David Michael Roberts said:

The object group of the categorical torus of rank 1 is R, not U(1). So I don't understand the argument.

I'm talking about the cokernel, the $\pi_0$

David Michael Roberts (Aug 07 2025 at 22:48):

Then I don't understand this sentence:

the objects of the Drinfeld center are $\mathbb{R}$, which project to those of $T_J$ via the exponential map $\mathbb{R} \to U(1)$.

David Michael Roberts (Aug 08 2025 at 02:34):

@Alonso Perez-Lona ok, there is an obstruction to being that we can check without knowing anything other than the crossed module itself: Lemma 4 in https://arxiv.org/abs/2109.00981 says that for a braided crossed module the action of the abelian group pi_0 on pi_1 is trivial.

Sadly, now I check this, the action is trivial, but that doesn't tell us anything about if the categorical circle is actually braided!

Alonso Perez-Lona (Aug 08 2025 at 09:06):

David Michael Roberts said:

Then I don't understand this sentence:

the objects of the Drinfeld center are $\mathbb{R}$, which project to those of $T_J$ via the exponential map $\mathbb{R} \to U(1)$.

Prop 4.1 in https://arxiv.org/abs/2202.01271 computes the Drinfeld center $\mathcal{Z}$ of $T_J$. While $\pi_0 T_J= U(1)$, at the top of p.17 in said article one sees that $\pi_0 \mathcal{Z} = \mathbb{R}$ if the class is nontrivial, and $\pi_0 \mathcal{Z} = U(1) \times \mathbb{Z}$ if the class is trivial. The forgetful functor $\mathcal{Z} \to T_J$ in particular has a group homomorphism $\pi_0\mathcal{Z} \to \pi_0 T_J$. If $T_J$ could be equipped with a braiding, then one could construct a functor in the opposite direction $T_J \to \mathcal{Z}$, but this in particular needs an injective group homomorphism $\pi_0 T_J \to \pi_0 \mathcal{Z}$, which for nontrivial extension is $U(1) \to \mathbb{R}$ but for trivial extension is $U(1) \to U(1) \times \mathbb{Z}$.

David Michael Roberts (Aug 08 2025 at 10:37):

@Alonso Perez-Lona ah, I see. I was missing the "if the categorical circle could be equipped with a braiding", one would get a (pseudo)retraction or similar to the Drinfeld centre. Thanks!