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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?

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$N_\Lambda^w(E)$ 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 $y$.

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    $\begingroup$ If you believe that, as currently written, this is unclear in the book, then I'll add this to the errata (available on my homepage). $\endgroup$ Sep 19, 2020 at 17:03

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