Oriented finite-group lattice gauge operators

Let $G$ be a finite group with identity $e$. A closed connected surface means a compact, connected topological $2$-manifold without boundary. A finite cellulation here is a finite decomposition of such a surface into vertices, edges, and faces, where each face interior is an open disk and its attaching boundary is a finite cyclic walk along edges. An oriented cellulation means that each edge has a chosen direction, recorded by its source and target; no orientation of the surface is assumed. Let $\mathrm{\Sigma }$ be such a cellulation and write $V,E,F$ for its vertex, edge, and face sets. Each edge $a\in E$ has source $s(a)$ and target $t(a)$. For a configuration $\mathbf{g}=(g_a{)}_{a\in E}\in G^E$, the edge Hilbert space is ${\mathcal{H}}_{\mathrm{\Sigma }}={⨂}_{a\in E}\mathbb{C}[G]$, with orthonormal configuration basis $\mathrm{\mid }\mathbf{g}⟩$.

For $v\in V$ and $x\in G$, put $k_u=x$ if $u=v$ and $k_u=e$ otherwise. Define the vertex gauge permutation $U_v(x)$ on basis vectors by

$U_v(x)\mathrm{\mid }\mathbf{g}⟩=\mathrm{\mid }{\mathbf{g}}^{\mathrm{\prime }}⟩,g_a^{\mathrm{\prime }}=k_{s(a)}{g}_a{k}_{t(a)}^{-1}(a\in E)\mathrm{.}$

For a face $f\in F$, choose its oriented boundary walk $a_1^{{ϵ}_1}\cdots a_m^{{ϵ}_m}$, with ${ϵ}_j=1$ when the walk follows the selected orientation of edge $a_j$ and ${ϵ}_j=-1$ otherwise. Its based holonomy on $\mathbf{g}$ is

$H_f(\mathbf{g}):=g_{a_1}^{{ϵ}_1}\cdots g_{a_m}^{{ϵ}_m}\mathrm{.}$

Define the vertex average, face flatness projector, and finite lattice Hamiltonian by

$A_v:=\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{x\in G}{U}_v(x),B_f\mathrm{\mid }\mathbf{g}⟩:={\mathbf{1}}_{H_f(\mathbf{g})=e}\mathrm{\mid }\mathbf{g}⟩,H_{\mathrm{\Sigma }}:={\sum }_{v\in V}(I-A_v)+{\sum }_{f\in F}(I-B_f)\mathrm{.}$

Explicit finite-group quantum-double data

A. Yu. Kitaev, “Fault-tolerant quantum computation by anyons,” §4

Alexander Cowtan and Shahn Majid, “Quantum Double Aspects of Surface Code Models,” §3

In the diagram, arrows specify edge orientations and $g_a$ labels the corresponding group element. The counterclockwise boundary walk from $v_0$ is $a_1{a}_2{a}_3^{-1}{a}_4^{-1}$, so its holonomy is $g_1{g}_2{g}_3^{-1}{g}_4^{-1}$. A gauge action $x$ at $v_0$ changes the two outgoing edge labels by left multiplication and therefore changes this based holonomy by conjugation, $H_f\mapsto x{H}_f{x}^{-1}$. The diagram illustrates the convention; it supplies no proof step.