Open strings create endpoint defects

The operators $A_v,B_f$ are commuting self-adjoint involutions by [F1], so $\mathrm{\Pi }={\prod }_{v\in V}(I+A_v)\mathrm{/}2{\prod }_{f\in F}(I+B_f)\mathrm{/}2$ is the orthogonal projection onto their common $+1$ eigenspace. Expand this finite product. Each term is a tensor-product Pauli $X(a)Z(b)$ with coefficient $+1$: factors of the same type multiply without phases, and every vertex operator commutes with every face operator. For the ordinary matrix trace on ${\mathcal{H}}_K$, the trace of a tensor product is the product of the factor traces. A nonidentity Pauli therefore has trace zero because at least one edge factor is $X$, $Z$, or $XZ$, each of which has trace zero; an identity term has trace $2^{\mathrm{\mid }E\mathrm{\mid }}$. The empty subset contributes one identity term and no term contributes negative trace, so $\mathrm{Tr}\mathrm{\Pi }\ge 2^{\mathrm{\mid }E\mathrm{\mid }-\mathrm{\mid }V\mathrm{\mid }-\mathrm{\mid }F\mathrm{\mid }}>0$. Thus $\mathrm{\Pi }\mathrm{\ne }0$. Since each penalty is an orthogonal projection and the couplings are positive, for every $\psi$ the quadratic form is $⟨\psi ,H_K\psi ⟩={\sum }_{v\in V}{J}_v{\parallel \frac{I-A_v}{2}\psi \parallel }^2+{\sum }_{f\in F}{K}_f{\parallel \frac{I-B_f}{2}\psi \parallel }^2$. This is zero exactly when every penalty annihilates $\psi$, equivalently when every stabilizer has eigenvalue $+1$. Hence $\ker H_K$ is the common $+1$ eigenspace and is nonzero, so it contains a ground vector of energy zero.

Let $b\in {\mathbb{F}}_2^E$ be the edge chain of the primal path $\gamma$. Because the path visits no vertex twice and has distinct endpoints, each internal vertex has even path incidence and the endpoints have odd incidence, so ${\mathrm{\partial }}_{1}b=v_1+v_2$. For each vertex $v$, $Z(b)A_v=(-1{)}^{⟨b,{\mathrm{\partial }}_1^{T}v⟩}{A}_{v}Z(b)=(-1{)}^{({\mathrm{\partial }}_{1}b{)}_v}{A}_{v}Z(b)$; $Z(b)$ commutes with every $B_f$. Hence applying $Z(b)$ to any ground vector flips exactly $A_{v_1}$ and $A_{v_2}$ from $+1$ to $-1$ and leaves every other stabilizer eigenvalue unchanged.

Regard the dual path ${\gamma }^{\ast }$ as a path in the dual graph whose vertices are faces, and let $a\in {\mathbb{F}}_2^E$ indicate the primal edges crossed by its dual edges. By the definition of the dual incidence map as ${\mathrm{\partial }}_2^T$, the no-repeated-face hypothesis and distinct endpoint faces give ${\mathrm{\partial }}_2^{T}a=f_1+f_2$. For each face $f$, $X(a)B_f=(-1{)}^{⟨a,{\mathrm{\partial }}_{2}f⟩}{B}_{f}X(a)=(-1{)}^{({\mathrm{\partial }}_2^{T}a{)}_f}{B}_{f}X(a)$; $X(a)$ commutes with every $A_v$. Hence applying $X(a)$ to any ground vector flips exactly $B_{f_1}$ and $B_{f_2}$ and leaves every other stabilizer eigenvalue unchanged.

By step 2.1, the vector $Z(b)\psi$ has eigenvalue $-1$ precisely for the two vertex stabilizers at $v_1,v_2$ and eigenvalue $+1$ for every other stabilizer, when $\psi$ is a ground vector. Thus the only nonzero terms in $H_{K}Z(b)\psi$ are the penalties $J_{v_i}(I-A_{v_i})\mathrm{/}2$, each acting as multiplication by $J_{v_i}$; its $H_K$-eigenvalue, and hence its energy increase from the ground eigenvalue zero, is $J_{v_1}+J_{v_2}$.

By step 2.2, the vector $X(a)\psi$ has eigenvalue $-1$ precisely for $B_{f_1},B_{f_2}$ and eigenvalue $+1$ for every other stabilizer, when $\psi$ is a ground vector. Therefore only the two penalties $K_{f_i}(I-B_{f_i})\mathrm{/}2$ contribute, and its $H_K$-eigenvalue, hence energy increase from zero, is $K_{f_1}+K_{f_2}$.