Gauge averaging and flatness commute

For a fixed vertex $v$, applying $U_v(y)$ and then $U_v(x)$ sends each edge label to $(xy)g_a$ at an outgoing incidence, to $g_a(xy{)}^{-1}$ at an incoming incidence, and to $(xy)g_a(xy{)}^{-1}$ on a loop based at $v$. All other labels remain fixed. Thus $U_v(x)U_v(y)=U_v(xy)$; each $U_v(x)$ is a permutation of an orthonormal basis and is unitary.

In the configuration basis, each $B_f$ is diagonal with entries in $\{0,1\}$. Therefore $B_f^{\ast }=B_f$ and $B_f^2=B_f$, so every $B_f$ is an orthogonal projection.

For distinct vertices $v,w$, their gauge permutations commute: on a shared edge they act at opposite ends, as left multiplication by one group element and right multiplication by the inverse of the other, and these operations commute; on all other edges at least one action is the identity. Thus $A_v{A}_w=A_w{A}_v$. Since all $B_f$ are diagonal in the same configuration basis, they commute pairwise.

Under a gauge change $k_u=x$ at $u=v$ and $k_u=e$ elsewhere, a face-boundary holonomy based at a vertex $u_0$ changes to $k_{u_0}{H}_f{k}_{u_0}^{-1}$: along the closed boundary walk the intermediate factors $k_u^{-1}{k}_u$ cancel. The condition $H_f=e$ is invariant under conjugation, so $U_v(x)$ preserves every eigenspace of $B_f$ and commutes with $B_f$. Averaging over $x$ gives $A_v{B}_f=B_f{A}_v$.

Taking adjoints in the finite average gives $A_v^{\ast }=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_x{U}_v(x^{-1})=A_v$. Also $A_v^2=\mathrm{\mid }G{\mathrm{\mid }}^{-2}{\sum }_{x,y}{U}_v(xy)=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_z{U}_v(z)=A_v$, since each $z\in G$ has exactly $\mathrm{\mid }G\mathrm{\mid }$ factorizations $xy=z$. Hence $A_v$ is an orthogonal projection.

Each complement $I-A_v$ and $I-B_f$ is an orthogonal projection, hence positive. Their finite sum $H_{\mathrm{\Sigma }}$ is positive.

For a vertex assignment $\mathbf{y}=(y_v{)}_{v\in V}\in G^V$, define the pure-gauge edge configuration $g_a=y_{s(a)}{y}_{t(a)}^{-1}$. The holonomy around every face telescopes to $e$, so each such basis vector lies in the $+1$ eigenspace of every $B_f$. The set $\mathcal{P}$ of distinct pure-gauge configurations is finite because $G^V$ is finite, and nonempty by taking every $y_v=e$. Each $U_v(x)$ permutes $\mathcal{P}$, because it replaces $y_v$ by $x{y}_v$ and leaves other vertex labels fixed. Therefore $\psi :={\sum }_{\mathbf{g}\in \mathcal{P}}\mathrm{\mid }\mathbf{g}⟩$ is nonzero, $U_v(x)\psi =\psi$ for every $v,x$, and consequently $A_v\psi =B_f\psi =\psi$ for every $v,f$. This gives a common zero vector of all positive terms in $H_{\mathrm{\Sigma }}$, proving frustration-freeness and ground energy zero.