1D Ising Renormalization-Group Flow

Decimate every other spin of the 1D Ising chain and the remaining spins obey the same model with renormalized couplings. The transformation is exact: $K'=\tfrac14\ln\frac{\cosh(2K+h)\cosh(2K-h)}{\cosh^2 h}$, $h'=h+\tfrac12\ln\frac{\cosh(2K+h)}{\cosh(2K-h)}$, with $K=\beta J$, $h=\beta H$. Iterating this map is the renormalization group. The plane is $(u=\tanh K,\;h)$, so $K=0$ (infinite temperature) is the left edge and $K\to\infty$ (zero temperature) the right. Every trajectory flows left into the disordered sink at the origin: there is no finite-temperature fixed point, the exact statement that the 1D Ising chain has no phase transition. The only critical point sits at $T=0$ ($u=1$), and it is unstable. Summing the per-step free-energy constants reconstructs the exact transfer-matrix free energy.