Sharp Asymptotics for the Truncated Two-Point Function of the Ising Model with a Positive Field

Abstract

We prove that the correction to exponential decay of the truncated two points function in the homogeneous positive field Ising model is $c\Vert x\Vert ^{-(d-1)/2}$. The proof is based on the development in the random current representation of a “modern” Ornstein–Zernike theory, as developed by Campanino et al. (Ann Probab 36(4):1287–1321, 2008).

Appendices

Appendix A. A Few Random Current Properties

We collect here a few properties of the random current together with proofs.

1.1 Insertion tolerance

Lemma A.1

For any graph $G=(V_G,{\mathbf {J}})$ and any $e\in E_G$, uniformly over the values of $n_f, f\ne e$, one has:

$$\begin{aligned} {\mathbf {P}}^A_{G}({\mathbf {n}}_e>0 ,{\mathbf {n}}_f=n_f\forall f\ne e) \ge c({\tilde{J}}_e){\mathbf {P}}^A_{G}({\mathbf {n}}_f=n_f\forall f\ne e) \end{aligned}$$

where $c({\tilde{J}}_e)=\frac{\cosh ({\tilde{J}}_e)-1}{\cosh ({\tilde{J}}_e)}$.

Proof

If the values $n_f$ implies ${\mathbf {n}}_e=1\mod 2$, then ${\mathbf {P}}^A_{G}({\mathbf {n}}_e>0 \,|\,{\mathbf {n}}_f=n_f\forall f\ne e)=1$ and it is over. Otherwise, ${\mathbf {P}}^A_{G}({\mathbf {n}}_e>0 \,|\,{\mathbf {n}}_f=n_f\forall f\ne e)$ is the probability for a Poisson random variable of parameter ${\tilde{J}}_e$ to be positive conditionally on being even. $\quad \square$

1.2 Exponential ratio mixing when $h>0$

We describe here an adaptation of an argument due to Duminil-Copin [9] to obtain exponential mixing under the random current measure with a field (a version of this idea is used in [10]).

We describe the results for a finite weighted graph $\Lambda$. As we consider random current measures, the support of a local event is a set of edges; to handle distances between supports define, for $E\subset E_{\Lambda _g}$,

$$\begin{aligned} V=V(E) = \bigcup _{e\in E}e\cap V_{\Lambda } \end{aligned}$$

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the set of non-ghost endpoints of edges in $E$.

Theorem A.2

There exist $R\ge 0$ and $C\ge 0$ such that, for any $E_1,E_2$ sets of edges, $V_i=V(E_i)$, any $A\subset V_{\Lambda }$ and any events $D$$E_1$-measurable and $D'$$E_2$-measurable, if $d_{\Lambda }(A\cup V_1,V_2) >R$, then

$$\begin{aligned} \Big |\log \Big ( \frac{{\mathbf {P}}_{\Lambda _g}^{A} (D,D')}{{\mathbf {P}}_{\Lambda _g}^{A} (D){\mathbf {P}}_{\Lambda _g}^{A} (D')}\Big )\Big |\le C\cosh (h)^{-d_{\Lambda }(A\cup V_1,V_2)}. \end{aligned}$$

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Proof

Fix two disjoint sets of edges $E_1$ and $E_2$. Denote ${\tilde{\Lambda }}_g$ the graph obtained from $\Lambda _g$ by removing the edges of $E_1\cup E_2$. Then the key observation is that for any configurations $n_1, m_1\in {\mathbb {Z}}_{+}^{E_1}$ and $n_2, m_2\in {\mathbb {Z}}_{+}^{E_2}$,

$$\begin{aligned} \frac{{\mathbf {P}}_{\Lambda _g}^{A} ( n_1, n_2){\mathbf {P}}_{\Lambda _g}^{A} ( m_1, m_2)}{{\mathbf {P}}_{\Lambda _g}^{A} ( n_1, m_2){\mathbf {P}}_{\Lambda _g}^{A} ( m_1, n_2)} = \frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1\Delta \partial n_2)Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m_1\Delta \partial m_2)}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1\Delta \partial m_2)Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m_1\Delta \partial n_2)}. \end{aligned}$$

Then, the RHS can be written

$$\begin{aligned}&\frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1\Delta \partial n_2)Z_{{\tilde{\Lambda }}_g}(\varnothing )}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1)Z_{{\tilde{\Lambda }}_g}(\partial n_2)} \frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m_1\Delta \partial m_2)Z_{{\tilde{\Lambda }}_g}(\varnothing )}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m_1)Z_{{\tilde{\Lambda }}_g}(\partial m_2)} \\&\quad \times \frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1)Z_{{\tilde{\Lambda }}_g}(\partial m_2)}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1\Delta \partial m_2)Z_{{\tilde{\Lambda }}_g}(\varnothing )} \frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m_1)Z_{{\tilde{\Lambda }}_g}(\partial n_2)}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m_1\Delta \partial n_2)Z_{{\tilde{\Lambda }}_g}(\varnothing )}. \end{aligned}$$

Now, using the switching lemma,

$$\begin{aligned} 1-\frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1)Z_{{\tilde{\Lambda }}_g}(\partial n_2)}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n_1\Delta \partial n_2)Z_{{\tilde{\Lambda }}_g}(\varnothing )}= {\mathbf {P}}_{{\tilde{\Lambda }}_g}^{A\Delta \partial n_1\Delta \partial n_2,\varnothing }\big ( {\mathcal {E}}_{\partial n_2}^c \big ). \end{aligned}$$

Now, if $(A\Delta \partial n_1)\cap \partial n_2\cap \Lambda \ne \varnothing$, then simply bound the probability by $1$. Otherwise, $(A\Delta \partial n_1)\cap \partial n_2=\{g\}$ or $\varnothing$. In both cases, the combination of the sources constraint and ${\mathcal {E}}_{\partial n_2}^c$ implies the existence of an edge-self-avoiding path $\gamma$ going from $(A\Delta \partial n_1)$ to $\partial n_2$ in the first current and such that $\gamma \nleftrightarrow g$ in the sum of the two currents. One thus gets,

$$\begin{aligned} {\mathbf {P}}_{{\tilde{\Lambda }}_g}^{A\Delta \partial n_1\Delta \partial n_2,\varnothing }\big ( {\mathcal {E}}_{\partial n_2}^c \big )&\le \cosh (h)^{-d_{\Lambda }\big ((A\Delta \partial n_1)\cap \Lambda ,\partial n_2\cap \Lambda \big )}\\&\le \cosh (h)^{-d_{\Lambda }(A\cup V_1,V_2)}, \end{aligned}$$

as $d_{\Lambda }\big ((A\Delta \partial n_1)\cap \Lambda ,\partial n_2\cap \Lambda \big )\ge d_{\Lambda }(A\cup V_1,V_2)$, where $V_i=\bigcup _{e\in E_1}e\cap \Lambda$. Using this, there exist $C\ge 0$ and $R\ge 0$ such that

$$\begin{aligned} \Big |\log \Big (\frac{{\mathbf {P}}_{\Lambda _g}^{A} ( n_1, n_2){\mathbf {P}}_{\Lambda _g}^{A} ( m_1, m_2)}{{\mathbf {P}}_{\Lambda _g}^{A} ( n_1, m_2){\mathbf {P}}_{\Lambda _g}^{A} ( m_1, n_2)}\Big )\Big |\le C\cosh (h)^{-d_{\Lambda }(A\cup V_1,V_2)} \end{aligned}$$

whenever $d_{\Lambda }(A\cup V_1,V_2)\ge R$. Now, fix $A\subset \Lambda$, take $E_1,E_2$ two sets of edges, let $V_1,V_2$ be defined as before. Suppose $d_{\Lambda }(A\cup V_1,V_2)= L>R$, then for any two events $D,D'$ supported on $E_1,E_2$ respectively,

$$\begin{aligned} {\mathbf {P}}_{\Lambda _g}^{A} (D,D')&= \sum _{\begin{array}{c} n_1\in {\mathbb {Z}}_{+}^{E_1}\\ n_1\in D \end{array}}\sum _{\begin{array}{c} n_2\in {\mathbb {Z}}_{+}^{E_2}\\ n_2\in D' \end{array}}{\mathbf {P}}_{\Lambda _g}^{A} ( n_1, n_2)\sum _{ m_1\in {\mathbb {Z}}_{+}^{E_1}}\sum _{ m_2\in {\mathbb {Z}}_{+}^{E_2}}{\mathbf {P}}_{\Lambda _g}^{A} ( m_1, m_2)\\&\le \exp {C\cosh (h)^{-L}}\sum _{\begin{array}{c} n_1\in D\\ n_2\in D' \end{array}}\sum _{ m_1, m_2 }{\mathbf {P}}_{\Lambda _g}^{A} ( n_1, m_2){\mathbf {P}}_{\Lambda _g}^{A} ( m_1, n_2)\\&= \exp {C\cosh (h)^{-L}}{\mathbf {P}}_{\Lambda _g}^{A} (D){\mathbf {P}}_{\Lambda _g}^{A} (D'). \end{aligned}$$

In the same fashion,

$$\begin{aligned} {\mathbf {P}}_{\Lambda _g}^{A} (D,D') \ge \exp {-C\cosh (h)^{-L}}{\mathbf {P}}_{\Lambda _g}^{A} (D){\mathbf {P}}_{\Lambda _g}^{A} (D'). \end{aligned}$$

So,

$$\begin{aligned} \Big |\log \Big ( \frac{{\mathbf {P}}_{\Lambda _g}^{A} (D,D')}{{\mathbf {P}}_{\Lambda _g}^{A} (D){\mathbf {P}}_{\Lambda _g}^{A} (D')}\Big )\Big |\le C\cosh (h)^{-L}. \end{aligned}$$

$\square$

Using the same technique, one can obtain:

Lemma A.3

There exist $R\ge 0$ and $C\ge 0$ such that, for any $E$ set of edges, $V=V(E)$, any $A\subset V_{\Lambda }$ and any event $D$$E$-measurable, if $d_{\Lambda }(A,V) >R$, then

$$\begin{aligned} \Big |\log \Big ( \frac{{\mathbf {P}}_{\Lambda _g}^{A} (D)}{{\mathbf {P}}_{\Lambda _g}^{\varnothing } (D)}\Big )\Big |\le C\cosh (h)^{-d_{\Lambda }(A,V)}. \end{aligned}$$

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Proof

As before, let $n\in {\mathbb {Z}}_{+}^{E}$, and let ${\tilde{\Lambda }}_g$ be the graph obtained by removing edges in $E$ from $\Lambda _g$. Then

$$\begin{aligned} {\mathbf {P}}_{\Lambda _g}^{A}( n) = \sum _{ m\in {\mathbb {Z}}_{+}^{E} } \frac{{\mathbf {P}}_{\Lambda _g}^{A}( n){\mathbf {P}}_{\Lambda _g}^{\varnothing }( m)}{{\mathbf {P}}_{\Lambda _g}^{\varnothing }( n){\mathbf {P}}_{\Lambda _g}^{A}( m)}{\mathbf {P}}_{\Lambda _g}^{\varnothing }( n){\mathbf {P}}_{\Lambda _g}^{A}( m). \end{aligned}$$

So, one just need to control the fraction term:

$$\begin{aligned} \frac{{\mathbf {P}}_{\Lambda _g}^{A}( n){\mathbf {P}}_{\Lambda _g}^{\varnothing }( m)}{{\mathbf {P}}_{\Lambda _g}^{A}( m){\mathbf {P}}_{\Lambda _g}^{\varnothing }( n)} = \frac{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial n)Z_{{\tilde{\Lambda }}_g}(\partial m)}{Z_{{\tilde{\Lambda }}_g}(A\Delta \partial m)Z_{{\tilde{\Lambda }}_g}(\partial n)}. \end{aligned}$$

Proceeding as in the proof of Theorem A.2, one get the wanted estimate. $\quad \square$

Appendix B. Toolbox

1.1 A combinatorial lemma

Lemma B.1

Let $G=(V_G,E_G)$ be a finite connected graph. For any $A\subset V_G$ of even cardinality, there exists $\omega \subset E_G$ with $\partial \omega = A$.

Proof

As $G$ is connected, it admits a spanning tree. So it is sufficient to prove the result for trees. Assume $G$ is a tree. Let $A=\{a_1,\ldots ,a_n\}$. We proceed by induction over $n=|A|$. For $n=2$, set $\omega$ to be the (unique) path going from $a_1$ to $a_2$. For $n$ even, suppose one has constructed $\omega '$ with $\partial \omega ' = \{a_1,\ldots ,a_{n-2}\}$. Let $\gamma$ be the unique path going from $a_{n-1}$ to $a_n$ in $G$. Set $\omega = \omega '\Delta \gamma$. As the sources of the symmetric difference is the symmetric difference of the sources, we have $\partial \omega = \partial \omega '\Delta \partial \gamma =\{a_1,\ldots ,a_{n-2}\}\Delta \{a_{n-1},a_n\}= A$. $\quad \square$

1.2 A geometrical lemma

For $\xi$ a norm on ${\mathbb {R}}^d$, $t\in \partial {\mathbf {K}}_{\xi }$ (see Subsection 2.5) and $\delta \in (0,1)$, define cones

$$\begin{aligned} {\mathcal {Y}}^\blacktriangleleft _{\delta }=\big \{ x\in {\mathbb {R}}^d: (t,x)_d>(1-\delta )\xi (x) \big \},\qquad {\mathcal {Y}}^\blacktriangleright _{\delta } = -{\mathcal {Y}}^\blacktriangleleft _{\delta }(t). \end{aligned}$$

Remark that ${\mathcal {Y}}^\blacktriangleleft _{\delta }$ are increasing sets in $\delta$. Let $A$ be a compact subset of ${\mathbb {R}}^d$ and $x\in {\mathbb {R}}^d$. We say that $A$$\delta$-sees$x$ if there exists $y\in A$ with $x\in y+{\mathcal {Y}}^\blacktriangleleft _{\delta }$; we say that $A$$\delta$-blocks$x$ if $A\not \subset x+{\mathcal {Y}}^\blacktriangleright _{\delta }$ (in other words, $A$$\delta$-blocks $x$ if $x$ does not $\delta$-backward-see $A$).

Lemma B.2

Let $A$ be a bounded subset of ${\mathbb {R}}^d$. Let $\delta ,\delta '>0$ such that $\delta +\delta '<1$. Then, the diameter of

$$\begin{aligned} V=\{x\in {\mathbb {R}}^d: A\ \delta \text {-sees}\ x, A\ (\delta +\delta ')\text {-blocks}\ x \} \end{aligned}$$

is upper bounded by a constant depending only on $\delta ,\delta ',A,d$.

Proof

As the only parameters of our problem are $\delta ,\delta ',A,d$, one only need to show that $V$ is bounded. The first observation is that if $x$$\delta$-sees $A$, then $x$$\delta$-sees $y$ for any $y$$\delta$-seen by $A$; indeed, if $y\in x+{\mathcal {Y}}^\blacktriangleleft _{\delta }$ then $(y+{\mathcal {Y}}^\blacktriangleleft _{\delta } )\subset (x+{\mathcal {Y}}^\blacktriangleleft _{\delta })$. The second observation is that if $x$ is not $\delta$-blocked by $A$, then so are all $y\in x+{\mathcal {Y}}^\blacktriangleleft _{\delta }$. Now, as $A$ is bounded, there exist $a,b\in {\mathbb {R}}^d$ such that

• $a$$\delta$-sees $A$,

• $b$ is not $(\delta +\delta ')$-blocked by $A$.

The two observations made before imply that $V$ is a subset of $(a+{\mathcal {Y}}^\blacktriangleleft _{\delta }) \cap (b+{\mathcal {Y}}^\blacktriangleleft _{\delta +\delta '})^c$. The final observation is that, as $\delta '>0$, the previous set is bounded. This implies the lemma. $\quad \square$