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Topological order is defined to be a phase that has ground state degeneracy (GSD) not described by the Landau SSB paradigm but exhibits some Long Range Entanglement property. Mathematically, it is defined to be a tensor category.

My question is: can we define topological order in QFT, especially SUSY field theories and string theory? It interests me whether the tensor category has an elegant interplay with algebraic geometry used in superstring theory.

When we study non-invertible symmetry in-depth, we will encounter the so-called categorical Landau paradigm (see recent works of Sakura Schäfer-Nameki et al, e.g. 2310.03786). I wonder, is it the same thing as topological order? Generally, can we rewrite all "phase transition beyond the Landau paradigm" into a generalized Landau paradigm in terms of generalized symmetry? Concretely, can we define a "Long Range Entangled" vacuum in QFT? What will the IR effective field theory of it be like? Can we compute its entanglement entropy? Can it be strictly computed in SUSY field theories? Does it have something to do with String landscape and quantum gravity?

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Many questions, I will tackle them one by one.

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In hep-th language, topological order simply means a topological field theory, also known as a TQFT. It is just a special type of QFT: one that does not depend on the metric of spacetime. The canonical examples are BF theories $$ \mathcal L=A\wedge\mathrm dB $$ and Chern-Simons theories, $$ \mathcal L=A\wedge\mathrm dA $$ (neglecting various coefficients for simplicity)

These are standard QFTs, but also manifestly topological, since I can write them without ever introducing a metric. They describe topological order. The latter models, for example, the fractional Hall effect, which is the most important example of topological order.

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TQFTs have no interplay with SUSY. The former only describe vacua but the latter has no implications on vacua: SUSY pairs up excited states but has no effect on ground states. Equivalently, every TQFT is formally supersymmetric but in a very trivial way: TQFTs have vanishing Hamiltonian and vanishing supercharge, so they satisfy $H=\{Q,Q\}$ in a boring, $0=0$ way. This is precisely the reason the Witten index is an interesting observable.

This of course does not mean TQFTs do not appear in SUSY studies. They very much do: many SUSY theories flow to TQFTs at large distances. The usual example is super-Yang-Mills, which flows to BF theory in the bulk and Chern-Simons on domain walls, see the classic paper of Acharya and Vafa.

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Finally, yes, categorical Landau is basically the extension of the usual Landau paradigm that includes non-invertible symmetries. If you spontaneously break an invertible discrete symmetry, you get degenerate vacua without topological order, and if you spontaneously break a non-invertible discrete symmetry, you get degenerate vacua with topological order.

That being said, it is not known if all the "beyond Landau" phases of matter can be described by "categorical Landau": there might be some phases that do not arise via breaking of any symmetries, whether invertible or not. This is an open question although as far as I know, the expectation is that "categorical Landau", suitably defined, will capture all phases of matter.

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The IR effective description of a TQFT is the TQFT itself: TQFTs do not flow, they are already in the IR. Regular QFTs might flow to TQFTs in the IR. In fact, any QFT that is gapped will.

And yes, you can compute the entanglement entropy of such states. Just google for example "entanglement entropy Chern-Simons".

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  • $\begingroup$ I know that gapped phase is the symptom of TQFT, but I haven't heard that TO is equivalent to TQFT. How can we distinguish SRE and LRE in TQFT? Besides, there's still a question left: can we define long-range-entangled QFT vacuum? $\endgroup$
    – Yankun Ma
    Commented Jan 21 at 14:38
  • $\begingroup$ @YankunMa TO is just a special case of a gapped phase, so it is indeed a TQFT. SRE means non-degenerate vacua (so $|Z[M]|=1$ for any manifold $M$) and LRE means degenerate vacua (so $|Z[M]|>1$ for generic $M$). This is of course under the assumption of stability (so $Z(S^d):=1$), which we can always make WLOG. $\endgroup$ Commented Jan 23 at 15:27

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