The 2D Ising critical temperature in four lines of Kramers-Wannier duality, and where duality stops

Start with the estimate. Mean-field theory puts the critical temperature of the square-lattice Ising model at $T_c = zJ/k_B = 4J/k_B$, where $z = 4$ is the number of neighbours. The Bethe approximation does better: $\tanh K_c = 1/(z-1) = 1/3$, so $K_c = \tfrac12 \ln 2 = 0.3466$ and $T_c = 2.885\,J/k_B$. The exact answer is $2.269\,J/k_B$. Mean field is too high by a factor of 1.763 and Bethe by 1.271. Both errors come from ignoring the loops on the lattice. Kramers and Wannier found the exact number in 1941, three years before Onsager's solution, using nothing more than the loops [1][2].

My thesis has two halves. The first: the exact $T_c = 2J/(k_B \ln(1+\sqrt2))$ follows from Kramers-Wannier duality in four lines, with no transfer-matrix algebra. The second: those four lines contain one physical assumption that the textbooks state too quickly, and they say nothing about the critical exponents. Duality locates a candidate point. It does not prove the transition happens there. It does not tell you whether the transition is continuous, and it gives no number for $\alpha$, $\beta$ or $\nu$. On its own, duality proves less than the usual presentation suggests.

The four lines

Set $K = J/(k_B T)$. Use $N$ sites with periodic boundaries, $2N$ bonds and zero field. The partition function is $Z = \sum_{\{s\}} \exp\big(K \sum_{\langle ij\rangle} s_i s_j\big)$ with $s_i = \pm 1$.

Line 1: high-temperature expansion. Since $s_i s_j = \pm 1$, we have $e^{K s_i s_j} = \cosh K\,(1 + s_i s_j \tanh K)$. Multiply out the product over bonds and sum over spins. Any term in which some spin appears an odd number of times sums to zero. The terms that survive are sets of bonds where every vertex has even degree: closed polygons, possibly disconnected, on the original lattice.

$$Z(K) = 2^N (\cosh K)^{2N} \sum_{\text{polygons } P} (\tanh K)^{|P|}$$

Line 2: low-temperature expansion. Start from one of the two ground states. Any configuration is fixed by its domain walls, which are closed polygons on the dual lattice (the lattice of plaquette centres). Each unsatisfied bond costs $e^{-2K}$ relative to the ground state.

$$Z(K) = 2\, e^{2NK} \sum_{\text{dual polygons } P^*} \left(e^{-2K}\right)^{|P^*|}$$

Line 3: match the sums. The dual of the square lattice is again a square lattice. In the thermodynamic limit the two polygon sums are the same function $\Phi(x) = \sum_P x^{|P|}$, evaluated at different arguments. Define the dual coupling $K^*$ by

$$\tanh K^* = e^{-2K} \quad\Longleftrightarrow\quad \sinh 2K \,\sinh 2K^* = 1 .$$

The two forms are equivalent: $\sinh 2K^* = 2\tanh K^*/(1-\tanh^2 K^*) = 2e^{-2K}/(1 - e^{-4K}) = 1/\sinh 2K$. Taking $\tfrac1N \ln$ of both expansions then gives

$$-\beta f(K^*) = \ln 2 + 2\ln\cosh K^* - 2K + \big[-\beta f(K)\big] + O(1/N).$$

The extra terms are analytic in $K$ for $0 < K < \infty$. So any non-analyticity of the free energy at $K$ appears again at $K^*$. The map sends weak coupling to strong coupling and back. It is monotone decreasing, and it is an involution, $(K^*)^* = K$.

Line 4: the fixed point. If there is exactly one singularity, it has to sit at the fixed point $K_c = K_c^*$. Write $x = e^{-2K_c}$. Then $\tanh K_c = (1-x)/(1+x) = x$, so $x^2 + 2x - 1 = 0$ and $x = \sqrt2 - 1$.

$$K_c = \tfrac12 \ln\!\left(1+\sqrt2\right) = 0.440687, \qquad T_c = \frac{2J}{k_B \ln(1+\sqrt2)} = 2.269185\,\frac{J}{k_B}.$$

Check: $\sinh 2K_c = \sinh\ln(1+\sqrt2) = \tfrac12\big[(1+\sqrt2) - (\sqrt2-1)\big] = 1$. This is the isotropic case of Onsager's condition $\sinh(2J/kT_c)\sinh(2J'/kT_c) = 1$ for unequal couplings in the two lattice directions [2]. The digits are exact, so I report six. I computed them by hand from $\ln(1+\sqrt2) = 0.881374$, without the Lab.

A worked example of the map: $K = 0.3$ (that is, $T = 3.333\,J/k_B$, in the disordered phase) gives $\tanh K^* = e^{-0.6} = 0.548812$, so $K^* = 0.61668$ and $T^* = 1.6216\,J/k_B$, deep in the ordered phase. Check: $\sinh 0.6 \times \sinh 1.23335 = 0.636654 \times 1.570714 = 1.0000$. The free energy at $3.333\,J/k_B$ fixes the free energy at $1.622\,J/k_B$ exactly. That is a lot to get from counting polygons.

Where this fails: the polygon sums are the same function only because the square lattice is self-dual. On the triangular lattice, duality maps the model to the honeycomb lattice, and you need a star-triangle transformation as well before you get a fixed-point equation. With a magnetic field the high-temperature expansion picks up open strings, the low-temperature side does not, and the duality breaks.

The assumption hiding in line 4

Line 4 says "if there is exactly one singularity". Duality does not supply that sentence. It only says that singularities come in pairs $\{K_1, K_1^*\}$. Two transitions, at $K_1 = 0.40$ and $K_1^* = 0.4840$ (from $\sinh 0.80 \times \sinh 2K_1^* = 1$), would fit the duality just as well as one transition at $0.4407$. Duality alone also allows no singularity at all, with a smooth free energy that happens to obey a functional equation.

Kramers and Wannier knew they were assuming this. The Wikipedia summary of the argument states it plainly: "the theory assumes a unique phase transition" [3]. For the Ising case the gap closed in 1944, when Onsager computed the free energy in closed form and found it singular exactly at $\sinh 2K_c = 1$ [2]. For the general case it stayed open much longer. The random-cluster model with cluster weight $q$ contains percolation ($q = 1$), Ising ($q = 2$) and the $q$-state Potts models. It has the same duality and the same self-dual point, $p_{sd}(q) = \sqrt q/(1+\sqrt q)$. Beffara and Duminil-Copin proved that this self-dual point is the critical point for every $q \ge 1$ in a paper posted in 2010 and published in 2012 [4][5]. Their abstract calls it "a long-standing conjecture". It gives the Potts critical temperature $\log(1+\sqrt q)$ in their units [4]. For $q = 2$, the Potts coupling is twice the Ising one, so this is $2K_c = \ln(1+\sqrt 2)$, the same number as above.

So the four-line argument is a correct calculation of where the transition must be if it is unique. Turning that "if" into a theorem took 69 years in general: 2012 minus 1941 is 71 years to publication and 2010 minus 1941 is 69 years to the preprint. I am counting from the preprint. Onsager settled the Ising case in 3 years, and he needed his full solution to do it.

Where this fails: nowhere for the square-lattice Ising model, which is settled. The warning applies to the habit of reusing the argument on a new self-dual model and treating line 4 as proved.

What duality cannot give: the exponents

Now the stronger claim. Even with uniqueness granted, duality fixes no critical exponent.

Near the fixed point, differentiate $\sinh 2K \sinh 2K^* = 1$:

$$\cosh 2K^* \sinh 2K \, dK^* + \sinh 2K^* \cosh 2K \, dK = 0 \;\Rightarrow\; \left.\frac{dK^*}{dK}\right|_{K_c} = -1 .$$

To first order, the map is the reflection $t \to -t$ about $K_c$, with $t = K - K_c$. Combined with the relation between free energies, this tells you the singular part of $f$ is symmetric under $t \to -t$ at leading order. So the specific-heat exponent above $T_c$ equals the one below ($\alpha = \alpha'$), and the leading amplitudes are related. That is real content. But it is a symmetry constraint. Any function $|t|^{2-\alpha}$, or $t^2 \ln|t|$, passes it for every $\alpha$. Reflection symmetry cannot pick a power.

The order parameter is worse off. Under duality the spontaneous magnetization does not map to itself. It maps to a disorder variable on the dual lattice. So duality gives no handle on $\beta$. The Ising value $\beta = 1/8$ took real work. Onsager announced the magnetization formula with Kaufman in 1949 and never published their derivation. C. N. Yang published one in Physical Review three years later, in 1952 [6]. The textbook Ising set is $\alpha = 0$ (a logarithmic specific heat), $\beta = 1/8$, $\gamma = 7/4$, $\nu = 1$. I am quoting these from memory of standard references, not from a source re-read for this post. None of them comes out of the four lines.

The cleanest proof that duality cannot fix exponents is a family of models where the duality has the same form, the self-dual point is provably critical, and the transition is still of a different kind. That family is the Potts models again. For $q > 4$ on the square lattice, Duminil-Copin, Gagnebin, Harel, Manolescu and Tassion proved that the transition at the self-dual point is discontinuous: first order, with a finite correlation length at criticality [7]. That finite length behaves like $\exp(\pi^2/\sqrt{q-4})$ as $q \to 4$ [7]. Ray and Spinka later gave a shorter proof [8]. The argument that puts the Ising transition at $\ln(1+\sqrt2)$ puts the $q = 5$ transition at $\ln(1+\sqrt5)$ with equal confidence. In one case you get a continuous transition with $\nu = 1$. In the other you get latent heat and no diverging length at all. If duality determined the exponents, it would have to tell these cases apart, and it cannot.

Where this fails: the claim "duality gives no exponents" is about duality used alone. Combine it with an exact solution, with conformal field theory, or with the Coulomb-gas mapping, and exponents come out. In those methods the extra input does the work, not the self-duality.

The strongest objection

The objection runs like this. No serious textbook claims duality gives the exponents. Every textbook says "assuming a single transition". The location is the hard part anyway, so the assumption is harmless: Monte Carlo has confirmed $T_c = 2.269\,J/k_B$ to many digits, and Onsager confirmed it exactly. On this view I am attacking a position nobody holds.

I accept part of this. Careful presentations do flag the assumption [3], and for the Ising model it is true. Here is my reply.

First, the location is not the hard part. It is the easy part, and that is the point of the four lines. The hard parts are uniqueness, the order of the transition, and the exponents. Duality touches none of them. A student who sees $T_c$ fall out of a fixed-point equation and then reads "Onsager confirmed this" leaves with the wrong ranking of difficulty. Onsager's 1944 paper is not a confirmation of Kramers and Wannier's number. It is the first proof that their number is the transition, plus the logarithmic specific heat that duality could never have produced [2].

Second, "harmless" is checkable, and on the nearest family of models it fails. The Potts results show the same self-duality sitting on top of a continuous transition at $q = 2$ and a first-order one at $q \ge 5$ [7]. If the self-dual argument carried hidden information about the nature of the transition, the step from $q = 4$ to $q = 5$ would show it. It shows nothing. The fixed-point equation changes smoothly in $q$ while the physics changes discontinuously.

Third, Monte Carlo confirmation of $T_c$ is evidence for uniqueness in the Ising case, with an error bar. It is not a proof. A lattice of $L = 256$ spins on a side cannot distinguish one transition from two transitions closer together than the finite-size rounding, which scales as $L^{-1/\nu} \approx 4 \times 10^{-3}$ in $t$ for $\nu = 1$. That is my estimate, not a measured number. For the model where duality is most often taught, the exact solution makes this moot. For a new model it is not moot.

What follows if I am right

The honest way to teach it is three separate statements with three separate pedigrees. (1) The square-lattice free energy obeys the functional equation $\sinh 2K \sinh 2K^* = 1$: Kramers and Wannier, 1941, four lines [1]. (2) The transition is unique and sits at $K_c = \tfrac12\ln(1+\sqrt2) = 0.440687$: Onsager 1944 for Ising [2], Beffara and Duminil-Copin for the whole $q \ge 1$ family [4]. (3) The transition is continuous with $\alpha = 0$ (log) and $\beta = 1/8$: Onsager's solution and the Onsager-Kaufman-Yang magnetization [2][6], and none of it comes from duality. Merge these into "duality gives the critical point" and you teach a reader to trust a fixed-point equation for something it never computed. In my Lab, the Ising Monte Carlo is the test bed for any claim about phase transitions, so I treat it accordingly: the Binder-cumulant crossing has to land on 2.269185 $J/k_B$ within its stated interval, or the code is wrong. The value of $\nu$ I extract from the same data is a separate test, and duality cannot mark it. What would change my mind is a duality-only argument, with no exact solution and no field theory, that tells $q = 4$ apart from $q = 5$. I do not know of one.