A Cluster Expansion Approach to the Heilmann–Lieb Liquid Crystal Model

Abstract

A monomer-dimer model with a short-range attractive interaction favoring collinear dimers is considered on the lattice $\mathbb Z^2$. Although our choice of the chemical potentials results in more horizontal than vertical dimers, the horizontal dimers have no long-range translational order—in agreement with the Heilmann–Lieb conjecture (Heilmann and Lieb in J Stat Phys 20(6):679–693, 1979).

Appendices

Appendix A: 1D Systems

Consider a finite line L, that is a finite connected sub-lattice of $\mathbb Z$. Consider a monomer-dimer model on L given by the following partition function:

$$\begin{aligned} Z_L \,=\, \sum _{\alpha \in \mathscr {D}_L} e^{-\beta H_L(\alpha )}\ e^{I_\mathrm{l}(\alpha _{x_\mathrm{l}})}\ e^{I_\mathrm{r}(\alpha _{x_\mathrm{r}})} . \end{aligned}$$

$\mathscr {D}_L$ denotes the set of monomer-dimer configurations on L (allowing also external dimers at the endpoints of L); the Hamiltonian is defined as

$x_\mathrm{l},x_\mathrm{r}$ denote the left and the right endpoint of L respectively; $I_\mathrm{l},I_\mathrm{r}$ represent the interaction among the configuration on L and the boundary condition outside its endpoints.

This one-dimensional system can be described by a transfer matrix T over the three possible states of a site, $l\equiv$“left-dimer”, $r\equiv$“right-dimer”, $m\equiv$“monomer”:

$$\begin{aligned} T \,\equiv \, \begin{pmatrix} T(l,l) &{} T(l,r) &{} T(l,m) \\ T(r,l) &{} T(r,r) &{} T(r,m) \\ T(m,l) &{} T(m,r) &{} T(m,m) \end{pmatrix} \,:=\, \begin{pmatrix} 0 &{} 1 &{} \sqrt{ab} \\ 1 &{} 0 &{} 0 \\ 0 &{} \sqrt{ab} &{} a \end{pmatrix} , \end{aligned}$$

(5.7)

where to shorten the notation we set $\sqrt{a}:=e^{-\beta \frac{\mu _{\mathrm{h}}+J}{4}}$ the transfer contribution of a monomer²⁰, $\sqrt{b}:=e^{-\beta \frac{J}{2}}$ the transfer contribution of a site with a dimer but neighbor to a monomer. Two vectors are also needed to encode the boundary conditions:

$$\begin{aligned} \begin{array}{ll} B_\mathrm{l}\,\equiv \, \begin{pmatrix} B_\mathrm{l}(l)&B_\mathrm{l}(r)&B_\mathrm{l}(m) \end{pmatrix} \,:=\, \begin{pmatrix} e^{I_\mathrm{l}(l)} &{} e^{I_\mathrm{l}(r)} &{} \sqrt{a}\,e^{I_\mathrm{l}(m)} \end{pmatrix} ,\\ B_\mathrm{r}\,\equiv \, \begin{pmatrix} B_\mathrm{r}(l) \\ B_\mathrm{r}(r) \\ B_\mathrm{r}(m) \end{pmatrix} \,:=\, \begin{pmatrix} e^{I_\mathrm{r}(l)} \\ e^{I_\mathrm{r}(r)} \\ \sqrt{a}\,e^{I_\mathrm{r}(m)} \end{pmatrix} . \qquad \qquad \quad \end{array} \end{aligned}$$

(5.8)

Proposition 5.1

The partition function of the system rewrites as a bilinear form:

$$\begin{aligned} Z_L \,=\, B_\mathrm{l}\;T^{|L|-1}\,B_\mathrm{r}. \end{aligned}$$

(5.9)

Proof

According to the previous definitions it is clear that for every configuration $\alpha \in \{l,r,m\}^{|L|}$

$$\begin{aligned}&\mathbbm {1}(\alpha \in \mathscr {D}_L)\ e^{-\beta H_L(\alpha )} \\&\qquad \qquad \qquad =\, \sqrt{a}^{\,\mathbbm {1}(\alpha _1=m)}\; T(\alpha _1,\alpha _2)\;T(\alpha _2,\alpha _3)\,\dots \,T(\alpha _{|L|-1},\alpha _{|L|})\; \sqrt{a}^{\,\mathbbm {1}(\alpha _{|L|}=m)}\ . \end{aligned}$$

Therefore

$$\begin{aligned} \begin{array}{ll} Z_L \,&{}= \sum _{\,\alpha \in \{l,r,m\}^{|L|}} B_\mathrm{l}(\alpha _1)\; T(\alpha _1,\alpha _2)\;T(\alpha _2,\alpha _3)\,\dots \,T(\alpha _{|L|-1},\alpha _{|L|})\; B_\mathrm{r}(\alpha _{|L|}) \\ &{}=\, B_\mathrm{l}\;T^{|L|-1}\,B_\mathrm{r}. \end{array} \end{aligned}$$

$\square$

Assume for the moment that the transfer matrix T is diagonalizable. Denote by $\lambda _1,\lambda _2,\lambda _3$ its eigenvalues and by $E_\mathrm{r}^{(1)},E_\mathrm{r}^{(2)},E_\mathrm{r}^{(3)}$, $E_\mathrm{l}^{(1)},E_\mathrm{l}^{(2)},E_\mathrm{l}^{(3)}$ the corresponding right (column) eigenvectors and left (row) eigenvectors, normalized so that $E_\mathrm{l}^{(i)} E_\mathrm{r}^{(i)} = 1$ for $i=1,2,3$.

Corollary 5.2

$$\begin{aligned} Z_L \,=\, \sum _{i=1,2,3} \lambda _i^{|L|-1}\, B_\mathrm{l}\,E_\mathrm{r}^{(i)}\,E_\mathrm{l}^{(i)}\,B_\mathrm{r}. \end{aligned}$$

(5.10)

Proof

Since we are assuming that T is diagonalizable, it holds $T=P\,D\,P^{-1}$ where D is the diagonal matrix of eigenvalues, P is the matrix with the right eigenvectors on the columns, $P^{-1}$ has the left eigenvectors on the rows. Then $T^{|L|-1}=P\,D^{|L|-1}\,P^{-1}$ and

$$\begin{aligned} B_\mathrm{l}\;T^{|L|-1}\,B_\mathrm{r}\,=\, (B_\mathrm{l}\,P)\,D^{|L|-1}\,(P^{-1}\,B_\mathrm{r}) \,=\, \sum _{i=1}^3 \,(B_\mathrm{l}\,E_\mathrm{r}^{(i)})\,\lambda _i^{|L|-1}\,(E_\mathrm{l}^{(i)}B_\mathrm{r}) . \end{aligned}$$

$\square$

Now our purpose is to diagonalise the transfer matrix T when $\beta$ is large.

Proposition 5.3

For all $\beta >0$ the transfer matrix T is diagonalizable over $\mathbb R$. Its eigenvalues are

$$\begin{aligned} \begin{array}{ll} &{} \lambda _1 \,=\, 1 + \frac{ab}{2}\, (1+o(1)) \\ &{} \lambda _2 \,=\, -1 + \frac{ab}{2}\, (1+o(1)) \\ &{} \lambda _3 \,=\, a - ab- a^3b\, (1+o(1)) \end{array} \end{aligned}$$

(5.11)

as $\beta \rightarrow \infty$.

Proof

The eigenvalues $\lambda _1,\lambda _2,\lambda _3$ are the (complex) roots of the characteristic polynomial of T, that is

$$\begin{aligned} p(\lambda ) := \det (\lambda I-T) = -ab + (\lambda -a)(\lambda ^2-1) . \end{aligned}$$

For all $\beta >0$ it turns out that p has 3 distinct real roots²¹ , hence T is diagonalizable over the reals.

As $\beta \rightarrow \infty$, $p(\lambda )\rightarrow \lambda (\lambda ^2-1)$ hence $\lambda _1\rightarrow 1$, $\lambda _2\rightarrow -1$, $\lambda _3\rightarrow 0$. Thus it is convenient to write $\lambda _1=1+\varepsilon _1$, $\lambda _2=-1+\varepsilon _2$, $\lambda _3=a+\varepsilon _3$ with $\varepsilon _i\rightarrow 0$ as $\beta \rightarrow \infty$ for $i=1,2,3$. Now expand the polynomial p in powers of $\varepsilon _i$ and truncate it at the first order:

$$\begin{aligned} \begin{array}{ll} &{} 0 \,=\, p(\lambda _1) \,=\, -ab+(1-a+\varepsilon _1)\,(2\varepsilon _1+\varepsilon _1^2) \,=\, -ab+2\varepsilon _1\,(1+o(1)) \\ &{}\quad \Rightarrow \, \varepsilon _1=\frac{ab}{2}\,(1+o(1)) ;\\ &{} 0 \,=\, p(\lambda _2) \,=\, -ab+(-1-a+\varepsilon )\,(-2\varepsilon _2+\varepsilon _2^2) \,=\, -ab+2\varepsilon _2\,(1+o(1)) \\ &{}\quad \Rightarrow \, \varepsilon _2=\frac{ab}{2}\,(1+o(1)) ;\\ &{} 0 \,=\, p(\lambda _3) \,=\, -ab+\varepsilon _3\,\big ((a+\varepsilon _3)^2-1\big ) \,=\, -ab-\varepsilon _3\,(1+o(1)) \\ &{}\quad \Rightarrow \, \varepsilon _3=-ab\,(1+o(1)) . \end{array} \end{aligned}$$

In order to find the following order of $\lambda _3$, now one can write $\lambda _3=a-ab\,(1+\varepsilon _3')$ with $\varepsilon _3'\rightarrow 0$ as $\beta \rightarrow \infty$ and repeat the procedure:

$$\begin{aligned} \begin{array}{ll} &{} 0 \,=\, \frac{p(\lambda _3)}{-ab} \,=\, 1 + (1+\varepsilon _3')\left( a^2\,(1+o(1))-1\right) \,=\, a^2\,(1+o(1)) - \varepsilon _3'\,(1+o(1)) \\ &{}\quad \Rightarrow \, \varepsilon _3'=a^2\,(1+o(1)) . \end{array} \end{aligned}$$

$\square$

Proposition 5.4

The right eigenvectors of the transfer matrix T are

$$\begin{aligned} \begin{array}{ll} &{} E_\mathrm{r}^{(1)} \,=\, \frac{1}{\sqrt{2}}\, \begin{pmatrix} 1-\frac{a}{2}\,(1+o(1)) \\ 1-\frac{a}{2}\,(1+o(1)) \\ \sqrt{ab}\,(1+o(1)) \end{pmatrix} \\ &{} E_\mathrm{r}^{(2)} \,=\, \frac{1}{\sqrt{2}}\, \begin{pmatrix} 1+\frac{a}{2}\,(1+o(1)) \\ -1-\frac{a}{2}\,(1+o(1)) \\ \sqrt{ab}\,(1+o(1)) \end{pmatrix} \\ &{} E_\mathrm{r}^{(3)} \,=\, \begin{pmatrix} -a\sqrt{ab}\,(1+o(1)) \\ -\sqrt{ab}\,(1+o(1)) \\ 1+a\,(1+o(1)) \end{pmatrix} \end{array} \end{aligned}$$

(5.12)

and moreover

$$\begin{aligned} \begin{array}{ll} &{} E_\mathrm{r}^{(2)}(1)+E_\mathrm{r}^{(2)}(2)+\sqrt{ab}\,E_\mathrm{r}^{(2)}(3) \,=\, \frac{ab}{2\sqrt{2}}\,(1+o(1)) \\ &{} E_\mathrm{r}^{(3)}(2)+\sqrt{ab}\,E_\mathrm{r}^{(3)}(3) \,=\, -a^2\sqrt{ab}\,(1+o(1)) \end{array} \end{aligned}$$

as $\beta \rightarrow \infty$. The left eigenvectors are obtained by a simple transformation: $E_\mathrm{l}^{(i)} = \sigma \big (E_\mathrm{r}^{(i)}\big )$ for $i=1,2,3$, where

$$\begin{aligned} \sigma \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} \,:=\, \begin{pmatrix} v_2&v_1&v_3 \end{pmatrix} . \end{aligned}$$

Proof

The right eigenvectors $E_\mathrm{r}$ associated to the eigenvalue $\lambda$ are the non-zero solutions of the linear system

$$\begin{aligned} (\lambda I - T)\,E_\mathrm{r}= 0 \ \Leftrightarrow \ E_\mathrm{r}= \begin{pmatrix} \lambda (\lambda -a) \\ \lambda -a \\ \sqrt{ab} \end{pmatrix} t \ ,\ t\in \mathbb R. \end{aligned}$$

And the left eigenvectors $E_\mathrm{l}$ associated to the same eigenvalue $\lambda$ are the non-zero solutions of the linear system

$$\begin{aligned} E_\mathrm{l}\,(\lambda I - T) = 0 \ \Leftrightarrow \ E_\mathrm{l}= \begin{pmatrix} \lambda -a&\lambda (\lambda -a)&\sqrt{ab}\, \end{pmatrix} t \ ,\ t\in \mathbb R. \end{aligned}$$

The desired normalization $E_\mathrm{l}\,E_\mathrm{r}= 1$ can be obtained by choosing

$$\begin{aligned} t=\sqrt{2\lambda (\lambda -a)+ab} \end{aligned}$$

in both cases. Now to conclude the proof it is sufficient to exploit the estimates of the eigenvalues given by the proposition 5.3. $\square$

The formula (5.10) together with the estimates of propositions 5.3, 5.4 give us a complete control on the one-dimensional system on L at low temperature, for every choice of the boundary conditions.

We concentrate on providing an estimation of the quantity $R_L$ defined by (3.9), since it is needed in the Sect. 3. We have to distinguish three cases, according to where the endpoints of L lie.

Lemma 5.5

The ratios of the eigenvalues of the transfer matrix T are

$$\begin{aligned} \frac{\lambda _2}{\lambda _1} \,=\, -1+ab\,(1+o(1)) \quad ,\quad \frac{\lambda _3}{\lambda _2} \,=\, -a+ab\,(1+o(1)) \end{aligned}$$

as $\beta \rightarrow \infty$. In particular setting $m:=-\log |\lambda _2/\lambda _1|$ it holds

$$\begin{aligned} e^{-m} \,=\, 1-e^{-\beta \frac{\mu _{\mathrm{h}}+3J}{2}}\,(1+o(1)) \quad \text {as }\beta \rightarrow \infty . \end{aligned}$$

(5.13)

Proof

It follows immediately from the proposition 5.3. $\square$

Lemma 5.6

If $x_\mathrm{l}\in \partial ^{\mathrm{ext}} _\mathrm{r}S_j$, then as $\beta \rightarrow \infty$

$$\begin{aligned} \begin{array}{ll} &{} B_\mathrm{l}\,E_\mathrm{r}^{(1)} = \frac{\sqrt{b}}{\sqrt{2}}\,(1+o(1)) \;\\ &{} B_\mathrm{l}\,E_\mathrm{r}^{(2)} = -\frac{\sqrt{b}}{\sqrt{2}}\,(1+o(1)) \;\\ &{} B_\mathrm{l}\,E_\mathrm{r}^{(3)} = \sqrt{a}\,(1+o(1)) . \end{array} \end{aligned}$$

If $x_\mathrm{r}\in \partial ^{\mathrm{ext}} _\mathrm{l}S_j$, then the same estimates hold for $E_\mathrm{l}^{(1)}B_\mathrm{r}$, $E_\mathrm{l}^{(2)}B_\mathrm{r}$, $E_\mathrm{l}^{(3)}B_\mathrm{r}$ respectively.

Proof

If $x_\mathrm{l}\in \partial ^{\mathrm{ext}} _\mathrm{r}S_j$ then by (3.7) and (5.8) the vector describing the boundary condition on the left side of the line L is $B_\mathrm{l}= \begin{pmatrix} 0&\sqrt{b}&\sqrt{a} \end{pmatrix}$. Then the estimates for $B_\mathrm{l}\,E_\mathrm{r}^{(i)}$, $i=1,2,3$, are computed using the proposition 5.4. $\square$

Lemma 5.7

If $x_\mathrm{l}\in \partial _\mathrm{l}\Lambda$, then as $\beta \rightarrow \infty$

$$\begin{aligned} B_\mathrm{l}\,E_\mathrm{r}^{(1)} = {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2}}\,\big (1-\frac{a}{2}\,(1+o(1))\big ) &{} {\textit{if the h-dimer on}}\,x_\mathrm{l}-(1,0)\,{\textit{has~fixed~position}} \\ \sqrt{2}\,\big (1-\frac{a}{2}\,(1+o(1))\big ) &{} {\textit{if the h-dimer on}}\,x_\mathrm{l}-(1,0)\,{\textit{has~free~position}} \end{array}\right. } \end{aligned}$$

$$\begin{aligned} B_\mathrm {l}\,E_\mathrm{r}^{(2)} = {\left\{ \begin{array}{ll} -\frac{1}{\sqrt{2}}\,\big (1+\frac{a}{2}\,(1+o(1))\big ) &{} {\textit{if the h-dimeron}}\,x_\mathrm{l}-(1,0)\,{\textit{is~fixed~to~the~left}} \\ \frac{1}{\sqrt{2}}\,\big (1+\frac{a}{2}\,(1+o(1))\big ) &{} {\textit{if the h-dimer on}} x_\mathrm{l}-(1,0)\,{\textit{is~fixed~to~the~right}} \\ \frac{ab}{2\sqrt{2}}\,(1+o(1)) &{} {\textit{if~the~h-dimer~on}} x_\mathrm{l}\!-\!(1,0)\,{\textit{has~free~position}} \end{array}\right. } \end{aligned}$$

If $x_\mathrm{r}\in \partial _\mathrm{r}\Lambda$, then the same estimates hold respectively for $E_\mathrm{l}^{(1)}B_\mathrm{r}$, $E_\mathrm{l}^{(2)}B_\mathrm{r}$, $E_\mathrm{l}^{(3)}B_\mathrm{r}$ after substituting: $x_\mathrm{l}-(1,0)$ by $x_\mathrm{r}+(1,0)$, “left” by “right” and “right” by “left”.

Proof

If $x_\mathrm{l}\in \partial _\mathrm{l}\Lambda$ then by (3.7) and (5.8) the vector describing the boundary condition on the left side of the line L is: $B_\mathrm{l}= \begin{pmatrix} 0&1&\sqrt{ab} \end{pmatrix}$ if a left-dimer is fixed on $x_\mathrm{l}\!-\!(1,0)$; $B_\mathrm{l}= \begin{pmatrix} 1&0&0 \end{pmatrix}$ if a right-dimer is fixed on $x_\mathrm{l}\!-\!(1,0)$; $B_\mathrm{l}= \begin{pmatrix} 1&1&\sqrt{ab} \end{pmatrix}$ if on $x_\mathrm{l}\!-\!(1,0)$ there is a h-dimer with free position. Then the estimates for $B_\mathrm{l}\,E_\mathrm{r}^{(i)}$, $i=1,2,3$, are computed using the proposition 5.4. $\square$

Proposition 5.8

Denote by o(1) any function $\omega (\beta ,\mu _{\mathrm{h}},J)$ that goes to zero as $\beta \rightarrow \infty$ and does not depend on the choice of the line L nor on $\Lambda$. Then for every line $L\in {\mathscr {L}}_{\Lambda }(\cup _jS_j)$, $S_j\in \mathscr {S}_\Lambda$ pairwise disconnected, $\Lambda \subset \mathbb Z^2$ finite, it holds

$$\begin{aligned} |R_L| \,\le \, e^{-m|L|}\,\gamma _L \end{aligned}$$

(5.14)

where the quantity $\gamma _L$ can be chosen as follows:

$$\begin{aligned} \gamma _{L} := {\left\{ \begin{array}{ll} \left( \frac{e^{-\beta J}}{2} + e^{-\beta \frac{\mu _{\mathrm{h}}+J}{2}|L|} \right) (1+o(1)) &{} \text {if }x_\mathrm{l}\in \cup _i\partial _\mathrm{r}^{\mathrm{ext}} S_i,\ x_\mathrm{r}\in \cup _i\partial _\mathrm{l}^{\mathrm{ext}} S_i \\ \frac{e^{-\beta \frac{J}{2}}}{\sqrt{2}}\,(1+o(1)) &{} \text {if }x_\mathrm{l}\in \cup _i\partial _\mathrm{r}^{\mathrm{ext}} S_i,\ x_\mathrm{r}\in \cup _i\partial _\mathrm{r}\Lambda \\ &{} \text {or vice versa }x_\mathrm{l}\in \cup _i\partial _\mathrm{l}\Lambda ,\ x_\mathrm{r}\in \cup _i\partial _\mathrm{l}^{\mathrm{ext}} S_i \\ 1+o(1) &{} \text {if }x_\mathrm{l}\in \partial _\mathrm{l}\Lambda ,\ x_\mathrm{r}\in \partial _\mathrm{r}\Lambda \end{array}\right. } \end{aligned}$$

(5.15)

Proof

• Suppose $x_\mathrm{l}\in \partial ^{\mathrm{ext}} _\mathrm{r}S_i$ and $x_\mathrm{r}\in \partial ^{\mathrm{ext}} _\mathrm{l}S_j$. The definition (3.9) and the corollary 5.2 give

  $$\begin{aligned} \begin{array}{ll} \lambda _1\,R_L \,&{}=\, \frac{Z_L}{\lambda _1^{|L|-1}} \,-\, B_\mathrm{l}E_\mathrm{r}^{(1)}E_\mathrm{l}^{(1)}B_\mathrm{r}\\ &{}=\, \left( \frac{\lambda _2}{\lambda _1}\right) ^{\!|L|-1}\!B_\mathrm{l}E_\mathrm{r}^{(2)}E_\mathrm{l}^{(2)}B_\mathrm{r}\,+\, \left( \frac{\lambda _3}{\lambda _1}\right) ^{\!|L|-1}\!B_\mathrm{l}E_\mathrm{r}^{(3)}E_\mathrm{l}^{(3)}B_\mathrm{r}. \end{array} \end{aligned}$$

  By the lemma 5.5 $|\lambda _3/\lambda _1|\le a\,|\lambda _2/\lambda _1|$ when $\beta$ is sufficiently large. Therefore, using also the estimates of lemma 5.6, one finds

  $$\begin{aligned} |R_L| \,\le \, \left| \frac{\lambda _2}{\lambda _1}\right| ^{|L|-1} \left( \frac{b}{2} + a^{|L|} \right) (1+o(1)) . \end{aligned}$$

• Suppose now $x_\mathrm{l}\in \partial ^{\mathrm{ext}} _\mathrm{r}S_j$ and $x_\mathrm{r}\in \partial _\mathrm{r}\Lambda$. The definition (3.9) and the corollary 5.2 give

  $$\begin{aligned} \begin{array}{ll} \lambda _1^{1/2}\,R_L \,&{}=\, \frac{Z_L}{\lambda _1^{|L|-1}E_\mathrm{l}^{(1)}B_\mathrm{r}} \,-\, B_\mathrm{l}E_\mathrm{r}^{(1)}\\ &{}=\, \left( \frac{\lambda _2}{\lambda _1}\right) ^{\!|L|-1} \frac{B_\mathrm{l}E_\mathrm{r}^{(2)}E_\mathrm{l}^{(2)}B_\mathrm{r}}{E_\mathrm{l}^{(1)}B_\mathrm{r}} \,+\, \left( \frac{\lambda _3}{\lambda _1}\right) ^{\!|L|-1} \frac{B_\mathrm{l}E_\mathrm{r}^{(3)}E_\mathrm{l}^{(3)}B_\mathrm{r}}{E_\mathrm{l}^{(1)}B_\mathrm{r}} . \end{array} \end{aligned}$$

  By the lemma 5.5 $|\lambda _3/\lambda _1|\le a\,|\lambda _2/\lambda _1|$ when $\beta$ is sufficiently large. Therefore, using also the estimates of lemmas 5.6, 5.7, one finds

  $$\begin{aligned} |R_L| \,\le \, \left| \frac{\lambda _2}{\lambda _1}\right| ^{|L|-1}\, \frac{\sqrt{b}}{\sqrt{2}}\,(1+o(1)) . \end{aligned}$$

• Suppose now $x_\mathrm{l}\in \partial _\mathrm{l}\Lambda$ and $x_\mathrm{r}\in \partial _\mathrm{r}\Lambda$. The definition (3.9) and the corollary 5.2 give

  $$\begin{aligned} \begin{array}{ll} R_L \,&{}=\, \frac{Z_L}{\lambda _1^{|L|-1}B_\mathrm{l}E_\mathrm{r}^{(1)}E_\mathrm{l}^{(1)}B_\mathrm{r}} \,-\, 1 \\ &{}=\, \left( \frac{\lambda _2}{\lambda _1}\right) ^{\!|L|-1} \frac{B_\mathrm{l}E_\mathrm{r}^{(2)}E_\mathrm{l}^{(2)}B_\mathrm{r}}{B_\mathrm{l}E_\mathrm{r}^{(1)}E_\mathrm{l}^{(1)}B_\mathrm{r}} \,+\, \left( \frac{\lambda _3}{\lambda _1}\right) ^{\!|L|-1} \frac{B_\mathrm{l}E_\mathrm{r}^{(3)}E_\mathrm{l}^{(3)}B_\mathrm{r}}{B_\mathrm{l}E_\mathrm{r}^{(1)}E_\mathrm{l}^{(1)}B_\mathrm{r}} . \end{array} \end{aligned}$$

  By the lemma 5.5 $|\lambda _3/\lambda _1|\le a\,|\lambda _2/\lambda _1|$ when $\beta$ is sufficiently large. Therefore, using also the estimates of lemma 5.7, one finds

  $$\begin{aligned} |R_L| \,\le \, \left| \frac{\lambda _2}{\lambda _1}\right| ^{|L|-1}\,(1+o(1)) . \end{aligned}$$

$\square$

Appendix B: Cluster Expansion

In this Appendix we state the main results about the general theory of cluster expansion used in this paper. The condition that we adopt to guarantee the convergence of the expansion is due to Kotecky–Preiss [11]. For a modern proof we refer to [17].

Let $\mathscr {P}$ be a finite set, called the set of polymers. Let $\varrho :\mathscr {P}\rightarrow \mathbb C$, called the polymer activity, and $\delta :\mathscr {P}\times \mathscr {P}\rightarrow \{0,1\}$, called the polymer hard-core interaction, such that $\delta (P,P)=0$ and $\delta (P,P')=\delta (P',P)$ for all $P,P'\in \mathscr {P}$. Consider the polymer partition function:

$$\begin{aligned} \begin{array}{ll} \mathcal {Z}\,&{}:=\, \sum _{\mathscr {P}'\subseteq \mathscr {P}}\; \prod _{P\in \mathscr {P}'}\varrho (P) \prod _{\begin{array}{c} P,P'\in \mathscr {P}'\\ P\ne P' \end{array}}\delta (P,P') \\ &{}=\, \sum _{q\ge 0}\,\frac{1}{q!}\sum _{P_1,\dots ,P_q\in \mathscr {P}}\, \prod _{t=1}^q\varrho (P_t)\; \prod _{t<s}\delta (P_t,P_s) . \end{array} \end{aligned}$$

(5.16)

A family of polymers $(P_1,\dots ,P_q)$ is called compatible if $\delta (P_t,P_s)=1$ for all $t\ne s$; otherwise it is called incompatible. Observe that in the partition function $\mathcal {Z}$ only the compatible families of polymers give non-zero contributions.

A family of polymers $(P_1,\dots ,P_q)$ is called a cluster if the graph with vertex set $\{1,\dots ,q\}$ and edge set $\{(t,s)\,|\,\delta (P_t,P_s)=0\}$ is connected.

Theorem 5.9

Suppose that there exists $a\!:\mathscr {P}\rightarrow [0,\infty [$, called size function, such that the Kotecky–Preiss condition is satisfied, namely:

$$\begin{aligned} \sum _{\begin{array}{c} P\in \mathscr {P}\\ \delta (P,P^*)=0 \end{array}} |\varrho (P)|\,e^{a(P)} \,\le \, a(P^*) \quad \forall \,P^*\!\in \mathscr {P}. \end{aligned}$$

(5.17)

Then:

$$\begin{aligned} \log \mathcal {Z}\,=\, \sum _{q\ge 0}\,\frac{1}{q!}\sum _{P_1,\dots ,P_q\in \mathscr {P}} \left( \prod _{t=1}^q\varrho (P_t)\right) u(P_1,\dots ,P_q) \end{aligned}$$

(5.18)

where the series on the r.h.s. is absolutely convergent and

$$\begin{aligned} u(P_1,\dots ,P_q) \,:= \sum _{\begin{array}{c} G=(V,E)\!\text { connected graph}\\ V=\{1,\dots ,q\}\\ E\subseteq \{(t,s)\,|\,\delta (P_t,P_s)=0\} \end{array}} (-1)^{|E|} . \end{aligned}$$

(5.19)

Moreover, for all $\mathscr {E}\subseteq \mathscr {P}$

$$\begin{aligned} \sum _{q\ge 0}\,\frac{1}{q!}\sum _{\begin{array}{c} P_1,\dots ,P_q\in \mathscr {P}\\ \exists \,t:\,P_t\in \mathscr {E} \end{array}} \bigg |\prod _{t=1}^q\varrho (P_t)\bigg | \left| u(P_1,\dots ,P_q)\right| \,\le \, \sum _{\begin{array}{c} P\in \mathscr {P}\\ P\in \mathscr {E} \end{array}} |\varrho (P)|\,e^{a(P)} . \end{aligned}$$

(5.20)

It is worth to observe that if $(P_1,\dots ,P_q)$ is not a cluster then $u(P_1,\dots ,P_q)=0$. Therefore only the clusters of polymers (that are infinitely many) give non-zero contributions to the expansion (5.18) of $\log \mathcal {Z}$.