# Is the self-dual point always a critical point?

I was studying duality maps in my Advanced Stat. Mech. class and it was told that all self-dual points need not correspond to critical point. I understand that critical points are points where analyticity of the partition function breaks down. If this point is unique then it can be given by the duality relationship between the constants which appear in the theory. But what happens if this point is not unique?

• I assume that you are referring to 2D Ising model on square lattice with $H = 0$. For square lattice this model is self dual because the dual of the square lattice is itself a square lattice. Yet that's not true for traiagular or kagome.
– Mass
Sep 28 at 10:21
• Could you please clarify what does it mean: "it can be given by the duality relationship between the constants which appear in the theory"? You ask what does happen if the are more than one critical point in theory? Sep 28 at 10:22
• @Mass , as I remember, classical 2D Ising model on the triangular lattice exhibits Krammers-Wannier duality at the critical point, too. The corresponding dual lattice is hexagonal. Sep 28 at 10:24
• Yes but that's not self dual. The traingle and hexagon is are related by an additional relation known as star-traingle transformation.
– Mass
Sep 28 at 10:25

Indeed, the self-dual point does not necessarily coincide with a critical point.

This happens, for instance, in the nearest-neighbor Ashkin-Teller model on $$\mathbb{Z}^2$$ (see, for instance, this answer for a definition). In this model, in some region of parameter space, the system undergoes two phase transitions and the self-duality transformation exchanges the two phase transition points; there is no transition at the self-dual point itself. Although this has been known for decades in the theoretical physics literature, a mathematically rigorous proof was only given recently; see this paper.

Let me give more details. The following picture is taken from the paper just mentioned:

Above (and up to) the diagonal, the bold line is a line of phase transition points and is composed of self-dual points. However, below the diagonal, the line of self-dual points is given by the dashed line and there is no phase transition at any point along this line; rather, the phase transitions occur along the two bold lines. These two lines are mapped onto each other by the duality transformation. (There is no contradiction with the usual argument used to locate the phase transition points when there is a unique point at which the free energy is singular, described for instance here, since the free energy is singular on both lines.)

It might be slightly simpler to interpret the results as follows. Rather than considering the full phase diagram as above, fix the coupling constants ($$J$$ and $$U$$ in the picture) and focus on a single line through the origin (the red line below). You can interpret the behavior along this red line as the behavior of the model at various values of the temperature (infinite temperature at the origin, decreasing towards zero as you get further along the line).

What happens is that there are two temperatures at which a phase transition occurs (located at the intersection points between the red line and the two bold lines). In contrast to the Ising model, one cannot use self-duality to locate these critical temperature, since the self dual point (the intersection between the red line and the dashed line) does not coincide with any of the two critical temperatures.

Referring to the question posed in the title, I would like to mention a further counterexample, which is moreover a lot simpler than the one Yvan has discussed.

Consider a toy model of integer-valued fields $$\varphi$$ on a lattice, governed by a Gaussian partition function of the form $$$$\tag{1} Z = \sum_{\{\varphi\}} \exp\left(- \frac{\beta J }{2} \sum_{\mathbf{r}} \varphi_{\mathbf{r}}^2 \right).$$$$ Using the Poisson summation formula one can elevate these integer-valued fields to real-valued ones, $$\phi$$, according to $$Z = \int \mathcal{D}\phi \sum_{\{\chi\}} \exp\left(- \frac{\beta J}{2} \sum_{\mathbf{r}} \phi_{\mathbf{r}}^2 + 2\pi \mathrm{i} \sum_{\mathbf{r}} \chi_{\mathbf{r}} \phi_{\mathbf{r}} \right).$$ By completing the square of the exponent $$-\frac{\beta J}{2} \phi^2 + 2\pi \mathrm{i} \phi \chi = -\frac{\beta J}{2} \left( \phi - 2\pi \mathrm{i} \frac{1}{\beta J} \chi \right)^2 - 2\pi^2 \frac{1}{\beta J} \chi^2,$$ we can do the functional integral over $$\phi$$ and be left with a new quadratic theory in terms of the integer-valued fields $$\chi$$ with partition function $$Z \sim \sum_{\{\chi\}} \exp\left( - \frac{2\pi^2}{\beta J} \sum_{\mathbf{r}} \chi_{\mathbf{r}}^2 \right).$$ Hence, this simple model is self-dual under the transformation $$\varphi \leftrightarrow \chi \quad \text{and} \quad \beta J \leftrightarrow (2\pi)^2 \frac{1}{\beta J}.$$ The unique self-dual point is determined by $$\beta = 2\pi J^{-1}$$.

On the other hand, doing the summation of equation (1) is even possible, using the Jacobi theta functions, and one finds $$Z = \left[\vartheta_{3}\left(0,\exp\left(- \frac{\beta J}{2}\right)\right)\right]^N,$$ which is an analytical function of its arguments, and hence an analytical function of $$T = \beta^{-1}$$. Hence, this model has a self-dual point which is not a critical point.