## 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?

3more comments