# 2D Ising model and FK-percolation

Consider the 2D Ising model on the finite lattice $$\Lambda$$ with $$+$$ boundary conditions, i.e., all spins outside of $$\Lambda$$ are $$=+1$$. Let $$\mathscr{E}_\Lambda^b$$ denote the edges in $$\Lambda$$ and the edges connecting $$\Lambda,\Lambda^c$$ so that the Hamiltonian is given by $$H = H_{\Lambda;\beta,0}^+ (\sigma) = -\beta \sum_{kl\in \mathscr{E}_\Lambda^b}\sigma_k\sigma_l$$ By writing $$e^{\beta \sigma_k \sigma_l}=e^\beta((1-p)+p1_{\sigma_k=\sigma_l}), \quad p=1-e^{-2\beta}$$ We can deduce that the partition function (as done in Velenik's Stat Mech of Lattice Systems, Chap 3.10.6) is given by $$Z_{\Lambda}^+ = e^{\beta |\mathscr{E}_\Lambda^b} \sum_{E\subset \mathscr{E}_\Lambda^b} p^{|E|}(1-p)^{|\mathscr{E}_\Lambda^b \backslash E|} \sum_{\omega\in \Omega_\Lambda^+} \prod_{kl\in E} 1_{\sigma_k(\omega)=\sigma_l(\omega)}$$ where $$\Omega_\Lambda^+$$ denotes the possible spin configurations on $$\Lambda$$ with all spins outside of $$\Lambda$$ fixed to be $$=+1$$. In the next step, Velenik claims that $$Z_{\Lambda}^+ = e^{\beta |\mathscr{E}_\Lambda^b} \sum_{E\subset \mathscr{E}_\Lambda^b} p^{|E|}(1-p)^{|\mathscr{E}_\Lambda^b \backslash E|} 2^{N_\Lambda^w(E)-1}$$ where $$N_\Lambda^w(E)$$ is the number of connected components of the graph $$(\mathbb{Z}^d, E\cup \mathscr{E}_{\Lambda^c}$$).

Question. Shouldn't it be $$\sum_{\omega \in \Omega_\Lambda^+} \prod_{kl\in E} 1_{\sigma_k(\omega)=\sigma_l(\omega)} = 2^{N_\Lambda^w(E)-1} 2^{|\Lambda \backslash V_E|}$$ where $$V_E$$ is the set of vertices of $$E$$, since the spins on $$\Lambda \backslash V_E$$ are free to change? If so, why would the 2D Ising model correspond to the FK-percolation process now that we have an extra $$2^{|\Lambda \backslash V_E|}$$ term?

$$N_\Lambda^w(E)$$ counts all connected components of the graph including the isolated vertices. That is, we are interested in clusters of vertices and two vertices $$x$$ and $$y$$ belong to the same cluster of $$E\cup\mathcal{E}_{\Lambda^c}$$ if either $$x=y$$, or $$E\cup\mathcal{E}_{\Lambda^c}$$ contains a path connecting $$x$$ to $$y$$.