 Solution

The joint law is the Edwards-Sokal coupling. For a fixed spin configuration, the sum over each edge state contributes $(1-p)+p{\delta }_e(\sigma )$. Since ${\delta }_e$ is either zero or one, setting $\beta =-\log (1-p)$ givesConsequently the spin marginal distribution is the Potts measureThis follows from a finite sum that factorizes over edges, without any independence assumption on the spin marginal.

For a fixed edge configuration, the joint weight vanishes unless the spins agree across every open edge, equivalently unless they are constant on each open percolation cluster. Each of the $k(\omega )$ clusters, including isolated vertices, may independently be assigned any of $q$ colors. Thus exactly $q^{k(\omega )}$ spin configurations have nonzero weight, and their common weight is the edge-product factor divided by $Z$. The edge marginal distribution is therefore the random-cluster measureIn this convention $Z^{\prime \prime }=Z$. Dividing the joint mass by this positive marginal gives the conditional distributionEquivalently the clusters receive independent uniformly chosen colors. Conversely, conditionally on the spins, edges are independent: an unequal-spin edge must be closed, while an equal-spin edge is open with probability $p$. These two conditional descriptions explain the Edwards-Sokal coupling directly.

If $x$ and $y$ lie in the same open cluster, their conditional spin-equality probability is one. If they lie in different clusters, their two independent uniform colors agree with probability $1/q$. Averaging over the random-cluster measure yieldsand hence the spin-connectivity identity for the Potts modelThe identity also includes $x=y$, when connectivity and spin equality are both certain.