# Critical 2d Ising Model

The critical point is not the same thing as the RG fixed point. Let $\mathcal{T}$ denote "theory space" meaning the set of all possible probability measures on real valued fields on the fixed unit lattice $\mathbb{Z}^2$. Block-spin or decimation etc. give you a map $R:\mathcal{T}\rightarrow \mathcal{T}$, namely, a renormalization group transformation. The picture to have in mind is that there is a special point $V_{\ast}\in \mathcal{T}$ such that $R(V_{\ast})=V_{\ast}$. That's the RG fixed point. It is hyperbolic and has a stable manifold $W^s$ (the critical surface) as well as an unstable manifold $W^u$. Now $\beta\mapsto {\rm Ising}(\beta)$ is a special one parameter curve in the big space $\mathcal{T}$. It is the curve made of clean NN Ising Hamiltonians with no extra terms. The critical point is the intersection of this curve with $W^s$ and it corresponds to a special value $\beta_c$ of the inverse temperature. For more about the various ways of seeing a QFT: a set of correlation functions, a probability measure on the space of distributions, a point in $\mathcal{T}$ or a complete orbit of $R$ in $\mathcal{T}$ see the end of Section 4 of my short article QFT, RG, and all that, for mathematicians, in eleven pages.

Summary:

- "The defining property of the theory at the critical point is that it is invariant under RG flows" is wrong because $R({\rm Ising}(\beta_c))\neq {\rm Ising}(\beta_c)$. In fact, by definition of stable manifold, $\lim_{n\rightarrow \infty} R^n({\rm Ising}(\beta_c))=V_{\ast}$, the true fixed point.
- " Any critical theory will necessarily involve additional higher-spin couplings." is wrong too because ${\rm Ising}(\beta_c)$ is NN and is also a critical theory. The set of critical theories is all of the stable manifold $W^s$.

This is a great question. Pages 15 and 16 of these notes argues that no nontrivial spin Hamiltonian can ever be a fixed point under spin decimation, but I don't understand why their argument doesn't hold in the 1D case. The notes end with the cryptic comment

there are many RG’s. The goal is not to see how many don’t work, but rather to find one that does.

I suspect that under repeated spin decimation, the 2D Ising model Hamiltonian will eventually converge to an extremely complicated fixed-point Hamiltonian with $n$-spin couplings for all even $n$ (odd-$n$ couplings would break the Hamiltonian's $\mathbb{Z}_2$ symmetry $\sigma_i \to -\sigma_i$), with coupling coefficients that decay exponentially with $n$. While obviously much more complicated than the original Hamiltonian, this fixed-point Hamiltonian would lie in the same Ising universality class and would therefore have the same long-distance physics. Since the particular choice of spin renormalization scheme is UV-sensitive, the resulting fixed-point Hamiltonian will depend on your choice of RG scheme - but every possible RG scheme should yield a fixed-point Hamiltonian in the same universality class, which is all that matters for the low-energy physics.

While there are a lot of technical details involved in this topic, what is going on here can be explained in very simple terms. Suppose you have found an exact RG scheme. Then starting at the critical point of the Ising model you'll end up at the critical fixed point in the scaling limit. But we know that this fixed point theory can also be described as a field theory (a free fermion model in the 2d case). But how can you end up with a field theory that has symmetries such as invariance under translations and rotations if all you are doing is applying block spin transforms?

The answer is that these symmetries will end up getting implemented by all those higher spin coupling terms. As you approach the critical point, the vast number of couplings on your lattice will start to emulate the terms of the field theory better and better. You can describe the approach to the critical fixed point also from the field theory perspective by adding so-called irrelevant operators to the fixed point model that flow to zero. The fact that the RG flow started out from a lattice model means that before you reach the infinite scaling limit, there will be remnants of translational symmetry breaking that will yet have to flow to zero. The presence of such irrelevant operators affects the critical behavior of the model, the exponents for the correction to scaling behavior depends on the scaling dimension of the irrelevant operators.