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For a 2D Ising model with zero applied field, it seems logical to me that the phases above and below T_c will have different percolation behaviour. I would expect that percolation occurs (and hence there is a spanning cluster) below T_c and that it does not occur (no spanning clusters) above T_c. I'm looking for conformation that this is true, but haven't found anything so far. Is my conjecture correct?

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Yes your conjecture is right. The Ising model is connected to the bond-correlated percolation model. See this paper. – Vijay Murthy May 4 '12 at 9:33
Thanks, though my suspicions were that it is related to site percolation. Specifically, I would expect that the spin configuration of the high temperature Ising model can be mapped to the configuration of sites for p \approx 1/2 bond percolation. I suppose that the bond-correlated percolation mapping also works at finite temperature. – Matthew Matic May 4 '12 at 9:42
@VijayMurthy it would be great if you post that as an answer (and just add a few words about the argument given in the paper). – David Z May 4 '12 at 15:50
up vote 3 down vote accepted

You conjecture is correct. One can relate the 2d Ising model with the bond correlated percolation model. The details are in the paper Percolation, clusters, and phase transitions in spin models.

The basic idea is to consider interacting (nearest neighbor) spins as forming a bond with a certain probability. One can then show that the partition function of the Ising model is related to the generating function of the bond-correlated percolation model.

The above paper demonstrates that the bond-correlated percolation model has the same critical temperature and critical exponents as the 2d Ising model. However, the values of $T_c$ and the critical exponents seem to be dependent on exactly how one defines a bond. See section III.A.1 in Universality classes in nonequilibrium lattice systems (or arxiv version).

Nonetheless your intuitive picture that there would be spanning clusters below $T_c$ and no such clusters above $T_c$ remains valid.

EDIT 21 May 2012

I found a pedagogical paper that discusses this issue.

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