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  String-net models on non-trivalent lattices

+ 4 like - 0 dislike

I have just started reading about string net models. The following aspect wasn't entirely clear to me:

String net models are most naturally defined on trivalent networks, that is to say networks where we attach 3 "legs" to each "point" of the lattice. The model is then fully specified by defining the string types, branching rules (the "$N_{ij}^{k}$") and the orientation of the strings.

While this approach appears natural (especially based on what I have read on Tensor categories) most of the "simplest" String net models (such as the $Z_2$ Kitaevs Toric Code or more generically the $Z_N$ Wen Plaquette Model) can also be defined on square lattices where we seem to have 4 "legs" attached to each "point".

I was wondering how and whether one can always "reduce" a String net model on an arbitrary N-valent lattice (N "legs" attached to each "point") to a String Net Model on a trivalent lattice. Similarly: Can one construct arbitrary String Net Models on N-Valent lattices by understanding the model on some trivalent lattice?

But in order not to complicate things: What is the "recipe" to reduce the Z2 Kitaev Toric Code on a planar square lattice to some "trivalent lattice"?

I am looking forward to your responses!

This post imported from StackExchange Physics at 2014-04-11 15:46 (UCT), posted by SE-user MrLee
asked Mar 12, 2014 in Theoretical Physics by MrLee (45 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

All the non-trivalent graph can be obtain from the trivalent graph by combining some vertices together. For example, the Z2 string-net model is defined on a honeycomb lattice, which is a trivalent lattice. But if you combine the two sites in each unit cell together, and consider the whole unit cell as your "site", then the honeycomb lattice simply becomes the square lattice, and the resulting model is exactly the Kitaev toric code on square lattice. In fact Kitaev deduced his square lattice toric code model from a honeycomb model in his original paper following the same way.

In general, for any site that is non-trivalent, you can consider it as a composite site and split it into several trivalent sites connected together. Such a reduction is just like the triangulation of 2d manifold in topology (the center of the triangle is dual to the site in the trivalent graph). Since there are actually more than one ways to split the sites, you may worry that if all different splitting will give you the same physics. The pentagon relation of the fusion category actually guarantee that different splitting will be equivalent to each other. So we can always reduce the non-trivalent lattice to the trivalent one. So once we know the physics of string-net on the trivalent lattice, we know it on all lattices, that is why we only need to study the trivalent lattice. Also there is no ambiguity of how the strings are fused at the vertex on the trivalent lattice, so it is much more convenient to work with trivalent lattice.

This post imported from StackExchange Physics at 2014-04-11 15:46 (UCT), posted by SE-user Everett You
answered Mar 15, 2014 by Everett You (785 points) [ no revision ]
Thanks for this very nice answer! One more question based on curiosity: There do exist String net models for fermionic topological order (understood in the sense that the K-Matrix of the corresponding TQFT has a very specific form, see for example: arxiv.org/abs/1309.7032) In these models it there is the additional requirement that one defines a direction on each trivalent graph. Does this effect the argument that you have been given in your response?

This post imported from StackExchange Physics at 2014-04-11 15:46 (UCT), posted by SE-user MrLee
@MrLee Even if the graph is directive, you can still combine sites to make non-trivalent graphs, just remember to keep the link direction unchanged. I can see your concern, the directive graph is indeed more complicated, for example you can not rotate the graph, and you also need to take care of the bending. On a non-orientable manifold you may not even be able to assign the directions. But if you already start from a well-defined directive trivalent graph, there is no ambiguity.

This post imported from StackExchange Physics at 2014-04-11 15:46 (UCT), posted by SE-user Everett You

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