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Introduction

Axiomatic TQFTs

An axiomatic $n$-dimensional TQFT is a symmetric monoidal functor $\mathcal{Z}\colon \operatorname{Bord}_n \to \operatorname{Hilb}$ from $n$-dimensional oriented bordisms to Hilbert spaces (other targets are certainly possible, but for the purposes of this question, we will stay with Hilbert spaces).

It is unitary if $\mathcal{Z}$ is a unitary, or $\dagger$-functor. Written out, this means: Let $\Sigma\colon M_1 \to M_2$ be an (oriented) bordism. The orientation reversed bordism goes in the other direction, i.e. $\overline{\Sigma}\colon M_2 \to M_1$. The TQFT must then satisfy $\mathcal{Z}(\overline{\Sigma}) = \mathcal{Z}(\Sigma)^\dagger$, where $\dagger$ is the adjoint of maps of Hilbert spaces.

Hamiltonian TQFTs

An $(n+1)$-dimensional Hamiltonian TQFT is a local Hamiltonian lattice quantum system on an $n$-dimensional manifold $M$, that has a gapped ground state which is invariant under certain local perturbations. It is unitary in the sense that the Hamiltonian $H$ is self-adjoint, and thus the time evolution is unitary.

The correspondence to axiomatic TQFTs lies in the lowest eigenspace. Any $n$-manifold has a (possibly degenerate) ground state, and it corresponds to the Hilbert space assigned to that manifold by an $(n+1)$-dimensional axiomatic TQFT.

Question

My question is now, how these two notions of unitarity fit together. Are they equivalent, or is one stronger?

Observations

Observation 1: Cylinders

As a first observation, it's clear that the two notions of unitarity agree when we just consider cylinders, i.e. bordisms of the form $I \times M$. This is the typical situation for Hamiltonian TQFTs.

(I'm not sure how the mapping class group is typically represented on Hamiltonian TQFTs, though.)

Observation 2: Topology changes

Most of the interesting information about axiomatic TQFTs is found in the way they behave on non-cylinder bordisms. If the boundaries to such a bordism are assigned Hilbert spaces of different dimensions, the bordism can never give rise to a unitary map of Hilbert spaces, though!

From the Hamiltonian perspective, it's obvious that it can't. When topology changes, the Hamiltonian changes (even if the lattice stays the same), and thus the lowest eigenspace. Take the orthogonal decomposition of the physical Hilbert space $\mathcal{H}$ into eigenspaces:

$$\mathcal{H} = \mathcal{H}_0 \oplus \mathcal{H}_1 \oplus \cdots$$ $$1_{\mathcal{H}} = \pi_0 + \pi_1 + \cdots$$ $$\pi_k\pi_{k'} = 1_{\mathcal{H}_k} \delta_{kk'}$$

The projection $\pi_0$ onto the lowest eigenspace of the new Hamiltonian should correspond to $\mathcal{Z}(\Sigma)$, where $\Sigma$ is the non-cylinder bordism. But why should this map be unitary in the axiomatic-TQFT-sense?

Question

Assume we are given an axiomatic TQFT that can be modelled by a local Hamiltonian lattice system. (The Hamiltonian is of course assumed to be self-adjoint.) Is the axiomatic TQFT then automatically unitary in the sense of a unitary functor? What are the topology changing maps?

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    $\begingroup$ It is my understanding that understanding the relationship between these two notions is wide open. It's even unclear how to precisely describe the correspondence you mentioned. There are simple examples of nonisomorphic TQFTs whose Hilbert spaces of states are isomorphic for all $n$-manifolds. The correspondence must somehow see the difference between these TQFTs with a lattice and Hamiltonian, but everything in dimension $n$ looks the same. $\endgroup$
    – Arun Debray
    Commented Apr 26, 2018 at 18:32
  • $\begingroup$ @ArunDebray, do you think the correspondence is hard to describe precisely because it's unclear how exactly to define a Hamiltonian TQFT? Can you give examples for such TQFTs you mentioned? (Maybe Dijkgraaf-Witten with the same group but different cocycles?) $\endgroup$
    – Manuel Bärenz
    Commented Apr 27, 2018 at 9:58
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    $\begingroup$ As for your first question, I think the answer is yes, that a complete definition would help address this correspondence, but it still won't be immediate. For example, in dimension 2+1, it took several years after the introduction of the Levin-Wen model and TVBW TQFT before a proof of their equivalence on codimension 1 manifolds appeared, and if I understand correctly it's still not clear how to see this on partition functions. $\endgroup$
    – Arun Debray
    Commented May 5, 2018 at 19:49
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    $\begingroup$ You can compute some partition functions, but in general not enough to solve that problem. For example, if you can describe a mapping class group action on your lattice model, you can compute the partition functions of mapping tori: if $[f]\in\mathrm{MCG}(M)$, the partition function of the mapping torus $[0,1]\times M/((0, x) \sim (1, f(x)))$ is the trace of the action of $f$ on the state space of $M$. However, these mapping tori cannot distinguish the two TQFTs I mentioned above. (continued...) $\endgroup$
    – Arun Debray
    Commented May 9, 2018 at 2:24
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    $\begingroup$ However, some ideas for getting the partition functions of $\mathbb{RP}^n$ and lens spaces in a Hamiltonian TQFT can be found in Shiozaki-Shapourian-Ryu. $\endgroup$
    – Arun Debray
    Commented May 9, 2018 at 2:26

2 Answers 2

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The answer is no: There are unitary Hamiltonian TQFTs (ie there are gapped lattice Hamiltonian systems in physics) which are not a unitary axiomatic TQFTs.

An example can be given in 3+1-dimension space-time. A unitary Hamiltonian TQFT can be a stacking of layers of $\nu=1/2$ bosonic fractional quantum Hall states. Such a unitary Hamiltonian TQFT is not a unitary axiomatic TQFT. This was discussed by us in arXiv:1406.5090. Only topological orders (ie gapped quantum liquids) may correspond to a unitary axiomatic TQFT. For a more detailed statement, see Bosonic topological orders and unitary fully dualizable fully extended TQFT (In some sense, we require the lattice Hamiltonian can be defined on any random lattice, and still give rise to ground states in the same phase)

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On the Hamiltonian level, the (axiomatic) TQFT tensors correspond to the imaginary time evolution of a microscopic system, not the real-time evolution (which would be a unitary operator). So there's no contradiction to the fact that the TQFT tensors are not unitary maps.

More precisely, if you put the imaginary time evolution of a gapped local microscopic Hamiltonian onto $M\times I$ and scale the whole thing up in time (and space) it will exponentially tend to the ground state projector. In general if you put your imaginary time evolution onto some cobordism and increase the lattice size, for each vector space corresponding to a connected component of the boundary, it will only be supported inside the ground state subspace of this vector space (up to an arrow exponential in the system size). The restriction of the imaginary time evolution operator to this support on every connected component is the (axiomatic) TQFT tensor associated to the corbordism.

Of course, in order to put your imaginary time evolution on an arbitrary cobordism, you have to be able to put it on any microscopic lattice (not just the square lattice). It is very unclear how to do this for a given microscopic Hamiltonian. Models where this can be done in a nice way in 2+1 dimensions are the Turaev-Viro models which however only yield non-chiral TQFTs (those arising from Drinfeld centers via Reshethikin-Turaev).

So if you accept that the imaginary time evolution operator of a local Hamiltonian system on a cobordism, no matter how you'll define that, should be the adjoint from the imaginary time evolution operator on the inversed cobordism then: Yes, the axiomatic TQFT arising from a microscopic Hamiltonian model should always be unitary.

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