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The 2d Ising model is extremely well studied, nevertheless I have encountered two facts which seem to contradict one another, and I have not been able to find the resolution in the literature. The puzzle is the following.

The critical Ising model is well known to be described by a CFT, and in particular a minimal model. This is described in many places, for example Ginsparg's CFT notes https://arxiv.org/abs/hep-th/9108028. To find the critical temperature, for which the CFT description is valid, perhaps the easiest way is to exploit the Kramers-Wannier duality, which relates the high-temperature/weak-coupling theory to the low-temperature/strong-coupling theory. The critical temperature is then given by the self-dual temperature. This makes it clear that the critical theory is just the usual 2d Ising Hamiltonian, but with the critical value of hte coupling constant $\beta J \equiv K = K_{*}$.

The defining property of the theory at the critical point is that it is invariant under RG flows. In general if $\mathcal{R}$ denotes the RG operation (in any given scheme, for example block-spin RG) and if the Hamiltonian $H$ depends on the coupling constants $\lambda_{1} \lambda_{2}, ..$, then this may be written schematically as

$$ \mathcal{R} H[\lambda_1^*, \lambda_2^* , ... ]= H[\lambda_1^*, \lambda_2^* , ... ],$$

where $*$ denotes fixed point quantities. Here is where the puzzle arises. Applied to the 2d Ising model with nearest-neighbor (NN) interactions only, the standard block-spin RG generates next-to-nearest-neighbor (NNN) interactions, and even NNNN interactions. See this for a demonstration of this fact. By examining the RG recursion relations, one finds that for no finite $K$ do these new interactions vanish. Therefore, with this RG scheme at least, the 2d Ising model with NN interactions can never be a fixed point of the RG transformation. Any critical theory will necessarily involve additional higher-spin couplings.

So is the critical Ising model $H = - J_* \sum_{<i,j>} s_i s_j$, with only NN interactions, or are there an infinite number of additional higher-spin interactions (which may become negligible in the continuum limit)?

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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.

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    $\begingroup$ Similarly, in the field-theory framework, even the RG fixed-point couplings can depend on your choice of RG scheme: physicsoverflow.org/32483/…. A lot more things are scheme-dependent than you might think; pretty much only the critical exponents are scheme-independent and therefore directly measurable. $\endgroup$ – tparker Mar 14 '17 at 1:41
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    $\begingroup$ If for some reason you ever happen to find yourself near Southampton University, you could ask this guy about his paper discussing what doesn't and doesn't depend on your choice of RG scheme: arxiv.org/abs/hep-th/0511107 $\endgroup$ – tparker Mar 14 '17 at 1:48
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    $\begingroup$ Maybe it is worth mentioning that the renormalization group seen as acting on the "space of all Hamiltonians" is in general ill-defined mathematically. See this paper for a nice review of this problem. $\endgroup$ – Yvan Velenik Mar 14 '17 at 11:21
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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:

  1. "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.
  2. " 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 $W^s$.
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  • $\begingroup$ This SE answer is the single most illuminating paragraph on renormalization I have ever read. Thank you. I wish I had the mathematical skills to read your article. $\endgroup$ – user2723984 Feb 3 at 13:51
  • $\begingroup$ Do you happen to have an accessible reference that expands on a somewhat rigorous definition of renormalization on a lattice, like the one you provided here? I.e. that defines these stable and unstable manifolds, and draws a connection between the critical surface and scale invariance? $\endgroup$ – user2723984 Feb 3 at 14:19
  • $\begingroup$ If you want, I would be glad if you could take a look at my lastest question (sorry if it's bad site etiquette to do this, feel free to ignore this comment) $\endgroup$ – user2723984 Feb 3 at 16:00
  • $\begingroup$ @user2723984: thank you. I tried to answer your new question. I don't have much time at the moment, so it is just a quick answer with some details missing. $\endgroup$ – Abdelmalek Abdesselam Feb 3 at 22:36
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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.

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