Consider a 2D ferromagnetic Ising model below the critical temperature, on a rectangular strip with infinite height but finite width N. However, there is a twist: consider changing all of the horizontally-oriented bonds on one of the (finitely-many) columns to be antiferromagnetic. This special column is, say, 1/3 of the way across the strip (for our purposes, it could really be anywhere less than half-way across, but let's just say 1/3 for simplicity).

The partition function, using the column transfer matrix, is

$$Z=\sum_{i,j=1}^2\sum_{s_1,s_N}\langle s_1|GS_i\rangle\langle GS_i| T_{AF}|GS_j\rangle \langle GS_j| s_N\rangle$$

where $T_{AF}$ is the special antiferromagnetic transfer matrix for column $I=N/3$, and we are implicitly assuming that the width of the rectangle is much larger than the correlation length, and $|GS_{i=1,2}\rangle$ are the maximal eigenvectors of the bulk column transfer matrix $T_{FM}$. Now the hard question comes: *How does the kink in the spontaneous magnetization manifest itself in this particular representation of the partition function?*

In particular, consider applying a uniform external field $\epsilon\to 0^+$ to the entire lattice. Now, if the antiferromagnetic column were 1/2, instead of 1/3 of the way across the lattice, then (now thinking using the row transfer matrix) this external field will induce no splitting among degenerate transfer matrix eigenvectors, and the induced spontaneous magnetization will be zero. However, here, there is an asymmetry with the antiferromagnetic column 1/3 of the way across the lattice, and there will be a unique energy-favorable configuration: 1/3 of the spins misaligned with the external field, then 2/3 of the spins aligned. **However, in our column representation of the transfer matrix, there is no information that distinguishes these (very) different scenarios!**