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Given a spherical fusion category $\mathcal C$, the Levin-Wen model constructs a lattice field theory: to each oriented surface with a triangulation, it assigns a state space $\mathcal H$ and a Hamiltonian $H$, whose space of ground states is independent of the choice of triangulation; and the Turaev-Viro model uses $\mathcal C$ to construct a (2+1)-dimensional oriented TQFT $Z_{\mathcal C}$.

Kirrilov (2011) showed that for any spherical fusion category $\mathcal C$ and oriented surface $\Sigma$, the space of ground states of the Levin-Wen model for $\mathcal C$ on $\Sigma$ is canonically isomorphic to the state space $Z_{\mathcal C}(\Sigma)$ in the Turaev-Viro TQFT.

Is there an analogous result in one dimension lower?

Presumably this construction would, given a semisimple commutative Frobenius algebra $A$, define a lattice model on the circle whose space of ground states can be canonically identified with $A$. Someone's probably worked on this, but I haven't found any references.

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Yes, there is an analogous 1d result, but it's not very interesting (which probably accounts for you not finding it in the literature).

In order for there to be a corresponding lattice model, you want the TQFT to be fully extended. This means you should start with a noncommutative Frobenius algebra (the Hilbert space of an interval), not a commutative Frobenius algebra (the Hilbert space of a circle). We can drop the "Frobenius" part, since we are not planning on constructing a path integral for 2-manifolds. Any semisimple algebra is Morita equivalent to a sum of trivial algebras. A trivial algebra corresponds to the trivial TQFT, and the associated lattice Hamiltonian is the trivial Hamiltonian.

If we consider a direct sum of trivial TQFTs, then the resulting lattice Hamiltonian is a ferromagnet (I think that's the right name). At each site on the 1d lattice we put a label (spin) indicating a summand of the algebra. The Hamiltonian projects to the subspace where adjacent labels are equal.

In conclusion, (1+ε)-dimensional TQFTs and 1d commuting projection Hamiltonians are both sort of boring, and these two facts are related.

(But if you add group actions to the picture, then things become more interesting. The lattice Hamiltonian corresponding to a trivial (1+ε)-dimensional TQFT equipped with a non-trivial finite group action is a 1d SPT phase. Both of these are indexed by group 2-cocycles.)

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  • $\begingroup$ Great, thanks so much! When you say "the resulting lattice Hamiltonian is a ferromagnet," does that mean it's an Ising model? If not, what do you mean by a ferromagnet? $\endgroup$ – Arun Debray Jun 1 '17 at 12:35
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    $\begingroup$ Sorry, I don't know the physics terminology that well, and "ferromagnet" might not have been the right name. I meant the local degrees of freedom are $\mathbb C^d$, and the local term in the Hamiltonian projects to the subspace spanned by $e_i\otimes e_i$. $\endgroup$ – Kevin Walker Jun 1 '17 at 12:51
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The 1D result is discussed in detail here: 1607.06766. (See also 1607.06504.)

The analog of the TVBW invariant is the FHK state sum, which takes as input a separable algebra $A$. As Kevin points out, this algebra has an interpretation as the space of states on the interval with some boundary conditions. The algebra has a canonical special symmetric Frobenius form, which is used to build the invariant. Not all 2d TQFTs, which are classified by commutative Frobenius algebras, have an FHK state sum; but up to deformation (appropriate for gapped phases), they all do. The construction is also redundant, as the TQFT only senses the Morita class of $A$. In particular, the space of states on the circle (the vector space underlying the commutative Frobenius algebra) is the center of $A$.

The analog of the LW model is the fixed-point Matrix Product State (MPS) system, which also takes as input a separable algebra. It consists of a parent Hamiltonian, whose ground states are have a simple wavefunction representation called MPS. For each module $V$ over $A$, there is a ground state $\vert\psi_V\rangle$ on the circle; the space of ground states is spanned by the indecomposable modules. Therefore, as was the case for the TQFT, the space of ground states is the center of the separable algebra $A$.

The correspondence is made more explicit by extending the state sum to incorporate physical (brane) boundaries. These are associated to modules $V$ over $A$. If one evaluates the state sum for the anulus with boundary condition $V$ on one boundary circle and outgoing cut boundary on the other, one obtains the MPS state $\vert\psi_V\rangle\in A^{\otimes N}$ before projecting to the TQFT state space (the center of $A$). The parent Hamiltonian arises as the projector assigned to the cylinder.

I agree with Kevin's opinion that the 1D result only becomes interesting when new structure is added. This is due to the trivial algebra $\mathbb{C}$ being the only indecomposable separable algebra in $\text{Vect}$. See the references above for systems with finite group actions. There, one works with separable algebras in $\text{Rep}(G)$ and has a Morita class of algebras for each pair $(H,\omega)$, an unbroken symmetry group and a cocycle characterizing the SPT order. For the correspondence between 1D fermionic systems and field theories that are sensitive to spin structure, see 1610.10075.

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  • $\begingroup$ This is also really helpful. Thank you! $\endgroup$ – Arun Debray Jun 27 '17 at 13:52

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