TITLE: Ising model on lattices with (vertical side length) $\neq$ (horizontal side length)
QUESTION [7 upvotes]: Consider the Ising model with nearest neighbours interactions on a rectangular lattice $L\times M$.
If $L=M$ ($2$-dimensional square lattice), it is known (e.g., by Peierls' argument or Onsager's explicit solution) that the model exhibits a phase transition when $L=M\to\infty$.
If instead we fix $L=1$ ($1$-dimensional line) and let $M\to\infty$, the model does not exhibit a phase transition.
My question is: Which relations among the side lengths $L,M$ guarantee the presence/absence of a phase transition? For example, what about the case $L=\log M$ ?

REPLY [5 votes]: Any increasing sequence $(\Lambda_n)_{n\geq 1}$ of finite subsets of $\mathbb{Z}^d$, $d\geq 2$, such that $\bigcup_{n\geq 1} \Lambda_n =\mathbb{Z}^d$ will do. All sequences $(\mu_{\Lambda_n}^+)_{n\geq 1}$ of finite-volume Gibbs measures in $\Lambda_n$ with $+$-boundary condition converge to the same infinite-volume Gibbs measure $\mu^+$, under which there is spontaneous magnetization as soon as the inverse temperature $\beta$ is large enough.
This can be proved easily using the FKG inequality (see, for example, the chapter on the Ising model here).

REPLY [2 votes]: Peierls's argument (in one of its modern forms, e.g. using chessboard estimates in the case of periodic boundary conditions) should work as long as $M$ and $L$ both go to $\infty$.