The Ising Model on the Random Planar Causal Triangulation: Bounds on the Critical Line and Magnetization Properties

Abstract

We investigate a Gibbs (annealed) probability measure defined on Ising spin configurations on causal triangulations of the plane. We study the region where such measure can be defined and provide bounds on the boundary of this region (critical line). We prove that for any finite random triangulation the magnetization of the central spin is sensitive to the boundary conditions. Furthermore, we show that in the infinite volume limit, the magnetization of the central spin vanishes for values of the temperature high enough.

Appendices

Appendix 1: Proof of Lemma 2

Consider the multiple series

$$\begin{aligned} W_{n+1,l}(x) = \sum _{k_1 = 1}^\infty \cdots \sum _{k_n = 1}^\infty \prod _{i=1}^{n} \left( {\begin{array}{c}k_{i+1}+k_i-1\\ k_i-1\end{array}}\right) x^{k_i}, \end{aligned}$$

(A.1)

with $k_{n+1}=l$. Summing over $k_1$ we obtain

$$\begin{aligned} \sum _{k_1 = 1}^\infty \left( {\begin{array}{c}k_1+k_2-1\\ k_1 -1\end{array}}\right) x^{k_1} = \frac{x}{1-x} \left( \frac{1}{1-x} \right) ^{k_2} = x B_1^{k_2+1}, \end{aligned}$$

(A.2)

where we denoted $B_1 = (1- x)^{-1}$. Inserting it in the equation and summing over $k_2$ we obtain

$$\begin{aligned} \sum _{k_2 = 1}^\infty \left( {\begin{array}{c}k_2+k_3-1\\ k_2 -1\end{array}}\right) x B_1 (x B_1)^{k_2} = \frac{x^2 B_1^2}{1 - x B1} \left( \frac{1}{1 - x B_1}\right) ^{k_3} = x^2 B_1^2 B_2^{k_3+1}, \end{aligned}$$

(A.3)

where $B_2 = (1- x B_1)^{-1}$. Summing over the remaining $k_i$’s, we obtain

$$\begin{aligned} W_{n+1,l}(x) = x^n B_n(x)^{l+1}\prod _{i=1}^{n-1} B_i(x)^2, \end{aligned}$$

(A.4)

where $B_i(x)$ is the solution to the recursion relation

$$\begin{aligned} B_i&= \frac{1}{1-x B_{i-1}}, \nonumber \\ B_1&= \frac{1}{1-x}, \end{aligned}$$

(A.5)

This reads

$$\begin{aligned} B_i(x) = 2 \frac{c_+(x)^{i+1} - c_-(x)^{i+1}}{c_+(x)^{i+2} - c_-(x)^{i+2}}, \end{aligned}$$

(A.6)

with

$$\begin{aligned} c_\pm (x) = 1 \pm \sqrt{1-4x}. \end{aligned}$$

(A.7)

For $0<x<1/4$, substituting (A.6) into (A.4) we get

$$\begin{aligned} W_{n+1,l}(x) = 2^{l+3} (1-4x) (4x)^n \frac{(c_+(x)^{n+1} - c_-(x)^{n+1})^{l-1}}{(c_+(x)^{n+2} - c_-(x)^{n+2})^{l+1}}, \end{aligned}$$

(A.8)

which gives

$$\begin{aligned} \lim _{n \rightarrow \infty } W_{n,l}(x) = 0. \end{aligned}$$

(A.9)

In particular we have

$$\begin{aligned} W_{n+1,l}(x) \sim f_{l}(x) \left( \frac{4x}{c_+(x)^2} \right) ^{n}, \quad \text {for } n \rightarrow \infty , \end{aligned}$$

(A.10)

where

$$\begin{aligned} f_l(x) = (1-4x) \left( \frac{2}{c_+(x)} \right) ^{l+3} . \end{aligned}$$

(A.11)

This yields the first statement of Lemma 2. The rest follows directly by the results of [20].

Appendix 2: Proof of Corollary 4

We have that, for any $x \in (0,1/4)$, $B_i(x)$ is monotonically increasing

$$\begin{aligned} \frac{B_{i-1}(x)}{B_{i}(x)}&= \frac{(c_+(x)^{n+2} - c_-(x)^{n+2})(c_+(x)^{n} - c_-(x)^{n})}{(c_+(x)^{n+1} - c_-(x)^{n+1})^2} \nonumber \\&= \frac{c_+(x)^{n+1} + c_-(x)^{n+1} - (c_-(x) c_+(x))^{n} (c_+^2 + c_-^2)}{(c_+(x)^{n+1} - c_-(x)^{n+1})^2} \nonumber \\&= 1 - \frac{(c_-(x) c_+(x))^{n} (c_+ - c_-)^2}{(c_+(x)^{n+1} - c_-(x)^{n+1})^2} \nonumber \\&= 1 - \frac{4 (4x)^n (1-4x)}{(c_+(x)^{n+1} - c_-(x)^{n+1})^2} < 1. \end{aligned}$$

(B.1)

Therefore, using that

$$\begin{aligned} \lim _{i \rightarrow \infty } B_i(x) = \frac{2}{c_+(x)}, \end{aligned}$$

(B.2)

we obtain that, for any $i \in \mathbb {N}$ and $x \in (0,1/4)$,

$$\begin{aligned} B_i(x) < \frac{2}{1 + \sqrt{1-4x}}. \end{aligned}$$

(B.3)

Now, consider the series

$$\begin{aligned} W_{n+1}(x,y) = \sum _{l \ge 1} y^l W_{n+1,l}(x). \end{aligned}$$

(B.4)

By Eq. (A.4), we have that the series is convergent if and only if

$$\begin{aligned} y B_{n}(x) < 1. \end{aligned}$$

(B.5)

Formula (B.3) together with Lemma 2 yield that the inequality (B.5) is satisfied for any $n \in \mathbb {N}$ and if and only if $x \in (0,1/4]$ and

$$\begin{aligned} \frac{2 y}{1 + \sqrt{1-4x}} <1, \end{aligned}$$

(B.6)

that is

$$\begin{aligned} y^2 - y + x < 0. \end{aligned}$$

(B.7)

This proves the corollary.