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Topological Orders in (4+1)-Dimensions

by Theo Johnson-Freyd, Matthew Yu

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Authors (as registered SciPost users): Matthew Yu
Submission information
Preprint Link:  (pdf)
Date submitted: 2022-07-21 03:25
Submitted by: Yu, Matthew
Submitted to: SciPost Physics
Ontological classification
Academic field: Physics
  • Condensed Matter Physics - Theory
  • High-Energy Physics - Theory
  • Mathematical Physics
Approach: Theoretical


We investigate the Morita equivalences of (4+1)-dimensional topological orders. We show that any (4+1)-dimensional super (fermionic) topological order admits a gapped boundary condition -- in other words, all (4+1)-dimensional super topological orders are Morita trivial. As a result, there are no inherently gapless super (3+1)-dimensional theories. On the other hand, we show that there are infinitely many algebraically Morita-inequivalent bosonic (4+1)-dimensional topological orders.

Current status:
Has been resubmitted

Reports on this Submission

Anonymous Report 1 on 2022-7-21 (Invited Report)

  • Cite as: Anonymous, Report on arXiv:2104.04534v2, delivered 2022-07-21, doi: 10.21468/SciPost.Report.5433


The authors seem to have overlooked my following comments

"My main concern is about the physical interpretation on the "super" case. Being a trivial n-supercategory nSVec does not physically guarantee gappability. Let's consider the example in 2+1D : if the operators are described by SVec, the corresponding classification (of invertible fermionic topological order) is given by $\mathbb Z$, or $\mathbb Z_{16}$ (modulo fermionic E8 state), and their boundaries are all gapless except the really trivial one with zero chiral central charge. Given such possibility, it is not justified to claim that "All 4d fermionic boundaries can be gapped". Further physical arguments should be provided to support the claim. However, I tend to believe that there are indeed obstructions to gap out the boundary: In analogy to the 16-fold way in 2+1D, the $\mathbb Z_{16}$ could be obtaind as the kernel of $\mathcal W\to \mathcal SW$; the results in this work basically say that in 4+1D $\mathcal SW$ is trivial while $\mathcal W$ is of infinite rank."

Let me add a few more sentences to clarity my point: if the analogy between 2+1D and 4+1D indeed works, the kenerl of $\mathcal W\to \mathcal SW$ still classify fermionic invertible phases in 4+1D. Then, by the mathematical results in this paper, such kernel is necessarily of infinite rank. It is hard to believe that a nontivial invertible phase can have a gapped boundary condition.

I'm also worried if the authors put enough care in the revision. I found, for example, in eq.(7)(8), there are still errors: in (7) $b_{y|x}$ should be $(b_{x|y})^{-1}$; in (7)(8) the direction of $R$ $S$ does not agree with the convention in (6). The typos and errors are not limited to the ones listed in my previous report and the authors should perform a careful enough proof reading.

  • validity: -
  • significance: -
  • originality: -
  • clarity: -
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Author:  Matthew Yu  on 2022-07-21  [id 2675]

(in reply to Report 1 on 2022-07-21)
answer to question

We thank the referee for their patience and providing feedback to our previous report.
What the paper proves is that every 4+1D topological order admits a topological boundary condition. The referee's comparison is completely different: by a well-known folklore theorem (a version of which is carefully proved by Freed-Teleman, but it is really 20 or 30 years older than their work), the Witt class of a 2+1D topological order is an obstruction to that order admitting topological boundary. We agree that perhaps if one is given some arbitrary 3+1D boundary condition for your 4+1D topological order, it might be interesting to ask what you have to do to make it gapped.  This is no easier or harder than asking: given some random 3+1D theory living on its own, i.e. with no gravitational anomaly, what do you need to do to make it gapped? One reasonable answer is: pick any boundary condition at all (presumably some very rough boundaries exist, and in any case since the theory is already gapless it is meaningless to ask the boundary condition to be gapped or anything), and then drive a phase transition as follows: create some tiny bubbles of vacuum with the boundary condition around the bubble, all in a dense network, then expand the bubbles, and merge them, and do this all at a mesoscopic scale; now you've just driven your theory to the vacuum. 

We will fix the typos and give the draft a more careful reading before the next submission.

Anonymous on 2022-07-26  [id 2689]

(in reply to Matthew Yu on 2022-07-21 [id 2675])

Thanks for the authors' quick reply. I'm not asking about the details to gap a particular boundary.
I agree that this paper proves that there is a gapped domain wall between every 4+1D topological order and the "trivial order" described by 4SVec.
The concern is that 4SVec is a description up to invertible phases of matter. Surely there is a physically truely trivial order for 4SVec, and topological orders connected to this one do have gapped boundary. However, there may be an invertible phase (say W) whose fusion category is 4SVec, but W does not have a gapped boundary. The obstruction will be inherited to the topological orders connected to W. Such situation indeed exists in 2+1D (p+ip state); it is the authors' responsability to rule it out in 4+1D if they want to make a stronger physical claim than Morita trivial.

That being said, every 4+1D topological order has a gapped boundary is not a physically acceptable claim. I suggest to use the conservative statement that every 4+1D topological order is Morita trivial instead of claiming that their boundaries are physically gappable.

Anonymous on 2022-07-28  [id 2695]

(in reply to Anonymous Comment on 2022-07-26 [id 2689])
answer to question

We thank the referee again for his comment, and would like to make a short reply.

We would like to clarity that the definition of "topological order" is in the sense of [Kong--Wen, Johnson-Freyd], where any invertible phase is considered as the trivial topological order. When we conclude that every sVec-enriched 4+1D topological order has a gapped boundary, we are using the fact that "topological order" means "topological phase up to invertible phases", and as such, for a topological order to have a gapped boundary, this means that the corresponding phase has a gapped interphase to an invertible phase.

We will add a short remark to clarify this definition in a new version of the paper, and then submit again for publication.

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