I'm considering the Ising model in a field on a square lattice in $d$ dimensions: $$H = -\sum_{\langle i j \rangle} \sigma_i \sigma_j - h \sum_{i} \sigma_i$$ As usual, $\langle i j\rangle$ refers to nearest neighbor couplings.
When $h>0$ and $0< T < \infty$, it follows that there is a finite magnetization density of $\mathbb{E}[\sigma_i] = m(h, T)>0$ in the thermodynamic limit.
One can imagine this might occur because there are more or larger finite domains of up spins than down spins, and that either of the following situations could arise:
- The up spins percolate in the system, so that there is a nonzero probability that a given spin is part of an infinite domain of up spins.
- The up spins do not percolate in the system, so that there is no infinite cluster of up spins containing a finite density of spins, but the fact that there are more or larger finite domains of up spins than down spins is sufficient to give a nonzero magnetization.
In the $1$d model, the latter mechanism without an infinite cluster occurs. In the $2$d model, I believe that the former mechanism with an infinite cluster occurs, at least for low temperatures. However, it's not clear to me that this holds at high enough temperatures. I go a little bit back and forth - I'm tempted that high temperatures (say large relative to $T_c$) should have only finite domains of either size, but I feel this would necessitate a phase transition on changing $T$ at nonzero $h$ which should not occur.
In $d > 1$, $h>0$, do the up spins percolate at all $0<T<\infty$? I believe they do at low temperature, but am less sure about high temperature (high relative to $T_c$).