Ising model: probability of a long path of minus under plus boundary conditions

Question

Consider for example the Ising model on a square lattice. Fix zero magnetic field and plus boundary conditions.

Low temperature, one minus spin. With a Peierls argument one can prove that, given a vertex $v$, the probability of having a minus spin on $v$, is bounded by $C\, e^{-c\beta}$ where $\beta$ is the inverse temperature and $C,c$ are suitable constants.

Low temperature, long path of minus spins. Now, given a simple path $P=(v_0,v_1,\dots,v_l)$ with $l$ large, is it true that the probability of having all minus spins on $P$ is bounded by something like $C^l\, e^{-c\beta\,l}$ (where the constants can be different from above)? The problem is that the spins are not independent, on the contrary the presence of a minus spin favours the presence of other neighbouring minus spins..

Answer 1

This doesn't look true as stated. What matters is something like the perimeter of the sites of the path (or of the smallest-perimeter box containing the sites of the path), not the length of the path itself. For example, if the path fills up an $m\times m$ box then the penalty for switching all the sites within the box from plus to minus is exponential in $m$ and not in $m^2$.

To prove such a bound (with the appropriate exponent) you can still use the Peierls argument; for a (connected) set to be all minus, there must be some contour separating the set from the boundary. If the set is large, the contour must be long, and you can bound the probability of seeing such a long contour (decaying exponentially in the length).