On the Singularities in the Susceptibility Expansion for the Two-Dimensional Ising Model

Abstract

For temperatures below the critical temperature, the magnetic susceptibility for the two-dimensional isotropic Ising model can be expressed in terms of an infinite series of multiple integrals. With respect to a parameter related to temperature and the interaction constant, the integrals may be extended to functions analytic outside the unit circle. In a groundbreaking paper, Nickel (J Phys A 32:3889–3906, 1999) identified a class of singularities of these integrals on the unit circle. In this note we show that there are no other singularities on the unit circle.

Appendices

Appendix 1

For $T<T_c$ and $N\ge 0$ we have the following Fredholm determinant representation of the spin-spin correlation function (see [13, p. 375] or [12, p. 142]):

$$\begin{aligned} \langle \sigma _{00}\,\sigma _{MN}\rangle =\mathcal M^2 \det (I+g_{MN}). \end{aligned}$$

The operator has kernel

$$\begin{aligned} g_{MN}(\theta _1,\theta _2) = e^{i M\theta _1-N\gamma (e^{i\theta _1})}\,h(\theta _1,\theta _2), \end{aligned}$$

where

$$\begin{aligned} h(\theta _1,\theta _2)={\sinh {1\over 2}(\gamma (e^{i\theta _1})-\gamma (e^{i\theta _2}))\over \sin {1\over 2}(\theta _1+\theta _2)}, \end{aligned}$$

and $\gamma (z)$ is defined by

$$\begin{aligned} \cosh \gamma (z)= s+s^{-1}-(z+z^{-1})/2, \end{aligned}$$

with the condition that $\gamma (z)$ is real and positive for $|z|=1$. The operator acts on $L^2(-\pi ,\pi )$ with weight function

$$\begin{aligned} {1\over 2\pi \,\sinh \gamma (e^{i\theta })}. \end{aligned}$$

Using the identity (see [13, (5.5)] or [12, (2.69)])

$$\begin{aligned} \det \left( h(\theta _j,\theta _k)\right) =\prod _{j<k} [h(\theta _j,\theta _k)]^2, \end{aligned}$$

and the Fredholm expansion we obtain that $\langle \sigma _{00}\,\sigma _{MN}\rangle$ equals

$$\begin{aligned} \mathcal M^2\,\sum _{n=0}^\infty {1\over (2n)!} {1\over (2\pi )^{2n}}\int \limits _{-\pi }^\pi \ldots \int \limits _{-\pi }^\pi \prod _{j<k} [h(\theta _j,\theta _k)]^2\, \prod _j e^{i M\theta _j-N\gamma (e^{i\theta _j})} {d\theta _j\over \sinh \gamma (e^{i\theta _j})}. \end{aligned}$$

(7)

Here all indices run from 1 to $2n$. We used the fact that since the matrix $(h(\theta _j,\theta _k))$ is antisymmetric its odd-order determinants vanish.

We have the identity, observed in [11],

$$\begin{aligned} {\sinh ({1\over 2}(\gamma (e^{i\theta _1})-\gamma (e^{i\theta _2}))\over \sin ({1\over 2}(\theta _1+\theta _2))}= {\sin ({1\over 2}(\theta _1-\theta _2))\over \sinh ({1\over 2}(\gamma (e^{i\theta _1})+\gamma (e^{i\theta _2}))} \end{aligned}$$

Therefore, with $x_j=e^{i\theta _j}$,

$$\begin{aligned}{}[h(\theta _1,\theta _2)]^2= {e^{-\gamma (x_1)}-e^{-\gamma (x_2)}\over 1-e^{-\gamma (x_1)-\gamma (x_2)}} \, {x_1-x_2\over 1-x_1 x_2}. \end{aligned}$$

With $D(x,y;s)$ defined by (2) a short calculation shows that

$$\begin{aligned} y\,D(x,y;s)=-{1\over 2}(y-e^{-\gamma (x)})\,(y-e^{\gamma (x)}). \end{aligned}$$

Thus inside the unit circle $1/(y\,D(x,y;s))$ has a pole at $y=e^{-\gamma (x)}$ with residue $1/\sinh \gamma (x)$. It follows that for $r$ sufficiently close to 1,

$$\begin{aligned} {1\over (2\pi i)^{2n}}\,\int \limits _{\mathcal C_r}\ldots \int \limits _{\mathcal C_r}\prod _{j<k}{y_j-y_k\over 1-y_jy_k}\, \prod _j{y_j^{N-1}\,dy_j\over D(x_j,y_j;s)}= \prod _j{e^{-N\gamma (x_j)}\over \sinh \gamma (x_j)}\,\prod _{j<k}{e^{-\gamma (x_j)}-e^{-\gamma (x_k)}\over 1-e^{-\gamma (x_j)-\gamma (x_k)}}. \end{aligned}$$

We deduce that the integral in (7) equals

$$\begin{aligned} {1\over (2\pi )^{2n}}\int \limits _{\mathcal C_r}\ldots \int \limits _{\mathcal C_r}\prod _{j<k}{y_j-y_k\over 1-y_jy_k}\,{x_j-x_k\over 1-x_jx_k}\,\prod _j{x_j^{M}\,y_j^{N}\over D(x_j,y_j;s)}\,\prod _j{dx_j\over x_j}\,{dy_j\over y_j}. \end{aligned}$$

(8)

It remains to compute

$$\begin{aligned} \sum _{M,N\in \mathbb Z}\left\{ \langle \sigma _{0,0}\,\sigma _{M,N} \rangle -\mathcal M^2 \right\} . \end{aligned}$$

Subtracting $\mathcal M^2$ in the summand is the same as taking the sum in (7) only over $n>0$.

To compute the sum over $M,N\in \mathbb Z$ we use the fact that $\langle \sigma _{0,0}\,\sigma _{M,N} \rangle$ is even in $M$ and in $N$, so

$$\begin{aligned} \sum _{M,N}=4\,\sum _{M,N\ge 0}-2\,\sum _{M=0,\,N\ge 0}-2\,\sum _{N=0,\,M\ge 0} +\ \mathrm the (0,0) term , \end{aligned}$$

and find that after summing, the factor $\prod _j x_j^{M}\,y_j^{N}$ in the integrand in (8) gets replaced by

$$\begin{aligned} {(1+\prod _j x_j)\,(1+\prod _j y_j)\over (1-\prod _j x_j)\,(1-\prod _j y_j)}. \end{aligned}$$

This gives (3).

Appendix 2

Suppose $f$ and $g$ are $C^\infty$ functions on $\mathbb R^d$, with $f$ having compact support, and we have an integral

$$\begin{aligned} \int f(\theta )\,e^{g(\theta )}\,d\theta . \end{aligned}$$

We write it as

$$\begin{aligned} \int f(\theta ){\nabla g(\theta )\over |\nabla g(\theta )|^2}\,\cdot \, \nabla e^{g(\theta )}\,d\theta . \end{aligned}$$

If define the operator $L$ by

$$\begin{aligned} (Lf)(\theta )=-\nabla \,\cdot \,f(\theta ){\nabla g(\theta )\over |\nabla g(\theta )|^2}, \end{aligned}$$

then $q$ applications of the divergence theorem show that the integral equals

$$\begin{aligned} \int (L^qf)(\theta )\;e^{g(\theta )}\;d\theta . \end{aligned}$$

Now we have

• (a) $L^q f$ is a linear combination of (partial) derivatives of $f$ with coefficients that are homogeneous polynomials of degree $q$ in derivatives of the components of $\nabla g/|\nabla g|^2$;

• (b) each $p$th derivative of each component of $\nabla g/|\nabla g|^2$ equals $1/|\nabla g|^{2p+2}$ times a homogeneous polynomial of degree $2p+1$ in derivatives of $g$. Assume that we also have

• (c) $|\nabla g(\theta )|\ge \mu$ and each derivative of $g(\theta )$ is $O(\mu )$;

• (d) each derivative of $f(\theta )$ is $O(1)$.

Then assuming that $\mathcal Re\,g$ is uniformly bounded above, we can conclude that

$$\begin{aligned} \int \limits _{\mathbb R^d} f(\theta )\,e^{g(\theta )}\,d\theta =O(\mu ^{-q})\ \ \mathrm for all \ q. \end{aligned}$$

In the application in Lemma 3 we have $d=2n$, $g$ is the sum of the exponents in the integrals, $f$ is the product of other integrands, and $\mu$ can be taken to be a small constant times the sum of the coefficients in the subsum of (5).

Appendix 3

Suppose $\mathcal U$ is an open set in $\mathbb T$, that $f$ is analytic in the region

$$\begin{aligned} \Omega =\{Rs:s\in \mathcal U,\ \ 1<R<1+\delta \}, \end{aligned}$$

and that $f$ and each of its derivatives is bounded in $\Omega$. We show that $f$ extends to a $C^\infty$ function on $\Omega \cup \mathcal U$.

Pick any $s_0\in \Omega$. We have for each $k\ge 0$ and $s'\in \Omega$,

$$\begin{aligned} f^{(k)}(s')=f^{(k)}(s_0)+\int \limits _{s_0}^{s'} f^{(k+1)}(t)\,dt, \end{aligned}$$

with the path of integration in $\Omega$. Since $f^{(k+1)}$ is bounded, this shows that that $f^{(k)}$ extends continuously to $\Omega \cup \mathcal U$. Denote by $f_k(s)$ this extension. In paticular $f_0$ is the continuous extension of $f$. We show that it belongs to $C^\infty$.

We show by induction that $f_0\in C^k$. We know this for $k=0$. Assuming this for $k$, we see that for $s\in \mathcal U$,

$$\begin{aligned} {d^k\over ds^k}f_0(s)=\lim _{s'\rightarrow s}{d^k\over ds'{^k}} f(s')= f^{(k)}(s_0)+\int \limits _{s_0}^s f_{k+1}(t)\,dt. \end{aligned}$$

It follows that $f_0$ is $k+1$ times differentiable and

$$\begin{aligned} {d^{k+1}\over ds^{k+1}}f_0(s)=f_{k+1}(s)= \lim _{s'\rightarrow s} {d^{k+1}\over ds'{^{k+1}}}f(s'). \end{aligned}$$

This gives the assertion.