On the Derivation of Mean-Field Percolation Critical Exponents from the Triangle Condition

Abstract

We give a new derivation of mean-field percolation critical behaviour from the triangle condition that is quantitatively much better than previous proofs when the triangle diagram $\nabla _{p_c}$ is large. In contrast to earlier methods, our approach continues to yield bounds of reasonable order when the triangle diagram $\nabla _p$ is unbounded but diverges slowly as $p \uparrow p_c$, as is expected to occur in percolation on ${\mathbb {Z}}^d$ at the upper-critical dimension $d=6$. Indeed, we show in particular that if the triangle diagram diverges polylogarithmically as $p \uparrow p_c$ then mean-field critical behaviour holds to within a polylogarithmic factor. We apply the methods we develop to deduce that for long-range percolation on the hierarchical lattice, mean-field critical behaviour holds to within polylogarithmic factors at the upper-critical dimension. As part of the proof, we introduce a new method for comparing diagrammatic sums on general transitive graphs that may be of independent interest.

Comparing Diagrams: A Sojourn into Linear Algebra

In this section we prove Proposition 1.8 in full generality, removing all the Fourier analysis from the proof given for the Euclidean and hierarchical cases in Sect. 2. We will use the alternative expressions for $A_\beta$ and $B_\beta$ derived from the mass-transport principle in (1.21) and (1.22) throughout our calculations. While we have made this section into an appendix due to its tangential relationship to the rest of the paper, we stress that the tools introduced here are in our view highly elegant and, in some cases, do not appear to have been used in probability theory or classical statistical mechanics before.

The general proof of Proposition 1.8 will be based on the theory of infinite positive (semi)definite matrices and the Schur product theorem, with a particularly important role being played by a theorem of Ando [6] concerning the interplay between the Hadamard product (i.e., the entrywise product) and the usual matrix product. Since the results we use are spread around several disparate sources and since we expect the content of this section to be highly unfamiliar to most of our audience, we give a thorough and self-contained treatment of most of the results that we use. A secondary purpose of this self-contained account is to prove infinite-dimensional cases of various results that are usually stated only in the finite-dimensional case. In addition to the paper on Ando [6], the primary sources we used to prepare this section were [9] and [14, Appendix A], both of which are written in a very accessible manner.

Let V be a countable (finite or infinite) set, and denote by ${\mathcal {M}}(V) = {\mathbb {R}}^{V \times V}$ the set of all matrices indexed by V. Let $L^0(V)$ be the set of finitely supported functions $L^0(V) := \{ f \in {\mathbb {R}}^V : f(v) =0$ for all but finitely many $v\}$, so that every $T \in {\mathcal {M}}(V)$ defines a linear map

$$\begin{aligned} \begin{array}{rcl} L^0(V) &{}\longrightarrow &{} {\mathbb {R}}^V\\ f &{} \longmapsto &{} Tf \end{array} \qquad \text { where } \qquad Tf(u)=\sum _{v\in V} T(u,v) f(v) \quad \text { for each }u\in V \end{aligned}$$

in the natural way, where the assumption that $f \in L^0(V)$ means that convergence is not an issue.

A matrix $T \in {\mathcal {M}}(V)$ is said to be symmetric if $T(u,v) = T(v,u)$ for every $u,v \in V$. A matrix $T \in {\mathcal {M}}(V)$ is said to be positive semidefinite if it is symmetric and satisfies $\langle T f, f \rangle := \sum _{u,v \in V} f(u) T(u,v) f(v) \ge 0$ for every $f \in L^0(V)$ and is said to be positive definite if there exists $\lambda >0$ such that $\langle T f, f \rangle \ge \lambda \langle f,f \rangle$ for every $f\in L^0(V)$. (Again, restricting to $f \in L^0(V)$ means that both inner products are trivially well-defined.) Given $S,T \in {\mathcal {M}}(V)$, we write $S \ge T$ to mean that $S-T$ is positive semidefinite and write $S>T$ to mean that $S-T$ is positive definite.

The relevance of these notions to percolation theory is established by the following lemma of Aizenman and Newman [4, Lemma 3.3]. We include a proof for completeness.

Lemma A.1

(Aizenman and Newman) Let G be a countable weighted graph. Then $T_\beta$ is positive semidefinite for every $\beta \ge 0$. Moreover, if G has $\sup _{v\in V} \sum _{e\in E^\rightarrow _v} J_e < \infty$ then $T_\beta$ is positive definite for every $\beta \ge 0$.

Proof of Lemma A.1

The matrix $T_\beta$ is trivially symmetric. Let ${\mathscr {C}}$ be the set of clusters of $\omega$, each of which is a set of vertices. Then for each finitely supported $f: V \rightarrow {\mathbb {R}}$ we have that

$$\begin{aligned} \langle T_\beta f, f \rangle&= \sum _{u,v \in V} f(u)f(v)\mathbb {P}_\beta (u \leftrightarrow v) = {\mathbb {E}}_\beta \left[ \sum _{u,v \in V} f(u)f(v)\mathbb {1}(u \leftrightarrow v) \right] \nonumber \\&= {\mathbb {E}}_\beta \left[ \sum _{C \in {\mathscr {C}}} \Bigl (\sum _{u\in C} f(u) \Bigr )^2\right] \ge 0, \end{aligned}$$

(Appendix A.1)

so that $T_\beta$ is positive semidefinite as required. Similarly, letting ${\mathscr {C}}_1 \subseteq {\mathscr {C}}$ be the set of singleton clusters of $\omega$, that is, clusters containing exactly one vertex, we have by (Appendix A.1) that

$$\begin{aligned}&\langle T_\beta f, f \rangle \ge {\mathbb {E}}_\beta \left[ \sum _{C \in {\mathscr {C}}_1} \Bigl (\sum _{u\in C} f(u) \Bigr )^2\right] = \sum _{u \in V} f(u)^2 \mathbb {P}_\beta (\{u\} \text { is a singleton cluster}\}) \nonumber \\&\quad = \sum _{u \in V} f(u)^2 \prod _{e\in E^\rightarrow _v} e^{-\beta J_e} \end{aligned}$$

(Appendix A.2)

which is easily seen to establish the second claim. $\square$

Remark A.2

The proof that the two-point matrix is positive semidefinite works for any percolation process, that is, any random subgraph of a given graph, and is not specific to Bernoulli percolation. Similarly, the proof of positive definiteness applies to any percolation process in which the probability that a vertex lies in a singleton cluster is bounded away from zero.

Since we will want to consider powers of positive semidefinite matrices, it will be useful to work within an algebra where such powers are well-defined. Define ${\mathcal {M}}^2(V) \subseteq {\mathcal {M}}(V)$ to be the algebra of V-indexed matrices T such that

$$\begin{aligned} \Vert T\Vert _{2\rightarrow 2} := \sup \left\{ \frac{\langle Tf,Tf\rangle ^{1/2}}{\langle f,f\rangle ^{1/2}} : f\in L^0(V), \langle f ,\, f \rangle > 0 \right\} < \infty . \end{aligned}$$

Since $L^0(V)$ is dense in $L^2(V)$, each matrix $T\in {\mathcal {M}}^2(V)$ defines a unique bounded linear operator $f\mapsto Tf$ on $L^2(V)$. Conversely, we can always recover a matrix from its associated operator using the formula $T(u,v) = \langle T \mathbb {1}_v,\mathbb {1}_u\rangle$; for $S,T \in {\mathcal {M}}^2$ the product of S and T as operators coincides with the usual matrix product $ST(u,v)=\sum _w S(u,w) T(w,v)$. A symmetric matrix $T \in {\mathcal {M}}^2(V)$ is positive semidefinite if and only if $\langle T f, f \rangle \ge 0$ for every $f\in L^2(V)$ and is positive definite if and only if there exists $\lambda >0$ such that $\langle Tf, f \rangle \ge \lambda \langle f,f\rangle$ for every $f\in L^2(V)$.

It is a consequence of the Riesz-Thorin theorem (see e.g. [20, Theorem 9.3.3]) that

$$\begin{aligned} \Vert T\Vert _{2\rightarrow 2} \le \Vert T\Vert _{1\rightarrow 1}^{1/2} \Vert T\Vert _{\infty \rightarrow \infty }^{1/2} := \sqrt{\sup _{u \in V}\sum _{v\in V} |T(u,v)|}\sqrt{\sup _{u \in V}\sum _{v\in V} |T(v,u)|} \end{aligned}$$

for every $T \in {\mathcal {M}}(V)$ (see e.g. [47, Theorem 19 and Exercise 20] for the fact that the $1\rightarrow 1$ and $\infty \rightarrow \infty$ norms can be expressed as the maximum $\ell ^1$ norm of a column and of a row respectively) and when T is symmetric this simplifies to

$$\begin{aligned} \Vert T\Vert _{2\rightarrow 2} \le \Vert T\Vert _{1\rightarrow 1} := \sup _{u \in V}\sum _{v\in V} |T(u,v)|. \end{aligned}$$

In particular, it follows that if G is a transitive weighted graph then the associated two-point matrix $T_\beta$ satisfies $\Vert T_\beta \Vert _{2\rightarrow 2} \le \Vert T_\beta \Vert _{1\rightarrow 1} = \mathcal {X}_\beta$ for every $\beta \ge 0$ and hence by the sharpness of the phase transition that $T_\beta$ defines a bounded operator on $L^2(V)$ for every $0\le \beta <\beta _c$. (In fact it is possible in some contexts for this boundedness to continue hold for $\beta$ strictly larger than $\beta _c$, see [26] for a thorough discussion.) Moreover, it follows by spectral considerations that if $T \in {\mathcal {M}}^2(V)$ is positive (semi)definite then so is $T^k$ for every $k\ge 1$, so that if G is an infinite, connected, locally finite, quasi-transitive graph then $T_\beta ^k$ is positive definite for every $0 \le \beta < \beta _c$ and $k\ge 1$.

Let us now prove the easy part of Proposition 1.8.

Lemma A.3

Let G be a connected, unimodular transitive weighted graph. Then $A_\beta \le \nabla _\beta ^{2}$ for every $0 \le \beta \le \beta _c$.

Proof of Lemma A.3

By left continuity of $T_\beta$, it suffices to prove the claim for every $0 \le \beta < \beta _c$, in which case $T_\beta$ is both positive semidefinite and belongs to ${\mathcal {M}}^2(V)$, so that $T_\beta ^3$ is well-defined as an element of ${\mathcal {M}}^2(V)$ and positive semidefinite also. Fix one such $\beta$. Since G is unimodular, we have by (1.21) that $A_\beta = \sum _{v\in V} T_\beta (o,v) T_\beta ^2(o,v) T_\beta ^3(o,v).$ Since $T_\beta ^3$ is positive definite, it has the property that

$$\begin{aligned} \langle T_\beta ^3(\mathbb {1}_u-\mathbb {1}_v) , (\mathbb {1}_u-\mathbb {1}_v) \rangle = T_\beta ^3(u,u) + T_\beta ^3(v,v) - 2 T_\beta ^3(u,v) \ge 0 \end{aligned}$$

for every $u,v \in V$ and hence by transitivity that

$$\begin{aligned} T_\beta ^3(u,v) \le \frac{1}{2}\left[ T_\beta ^3(u,u) + T_\beta ^3(v,v) \right] = T_\beta ^3(o,o)=\nabla _\beta \end{aligned}$$

for every $u,v \in V$. It follows that

$$\begin{aligned} A_\beta&\le \nabla _\beta \sum _{v\in V} T_\beta (o,v) T_\beta ^2(v,o) = \nabla _\beta T_\beta ^3(o,o)=\nabla _\beta ^2 \end{aligned}$$

as claimed. $\square$

It remains to prove that $B_\beta \le A_\beta$. This proof will rely on various more sophisticated pieces of technology, which we now begin to introduce. Recall that if $S,T \in {\mathcal {M}}(V)$ are V-indexed matrices, the Hadamard product $S \circ T \in {\mathcal {M}}(V)$ is defined by entrywise multiplication

$$\begin{aligned}{}[S \circ T](u,v) : = S(u,v)T(u,v) \qquad \text { for every }u,v \in V. \end{aligned}$$

Note that if $S,T \in {\mathcal {M}}^2(V)$ then $S\circ T \in {\mathcal {M}}^2(V)$ also and indeed satisfies $\Vert S\circ T\Vert _{2\rightarrow 2}\le \Vert S\Vert _{2\rightarrow 2}\Vert T\Vert _{2\rightarrow 2}$ [9, Eq. 2.2]. The Hadamard product may be used to give various alternative expressions for the diagrams $A_\beta$ and $B_\beta$. The most obvious way to do this, which is analogous to what we did in (2.8) and (2.9), is to use (1.21) and (1.22) to write

$$\begin{aligned} A_\beta =\sum _{v\in V} \Bigl [T_\beta \circ T_\beta ^2 \circ T_\beta ^3\Bigr ](o,v) \qquad and \qquad B_\beta = \sum _{v\in V} \Bigl [T_\beta ^{2} \circ T_\beta ^2 \circ T_\beta ^2 \Bigr ](o,v). \end{aligned}$$

However, we will see that the equivalent expressions

$$\begin{aligned} A_\beta = \Bigl [T_\beta \circ T_\beta ^3\Bigr ]T_\beta ^2(o,o) \qquad and \qquad B_\beta = \Bigl [T_\beta ^{2} \circ T_\beta ^2\Bigr ]T_\beta ^2(o,o) \end{aligned}$$

are better suited to our needs. We stress that these expressions contain both Hadamard products and ordinary matrix products: for example, $T_\beta ^2$ denotes the ordinary square of $T_\beta$ while $[T_\beta \circ T_\beta ^3]T_\beta ^2$ denotes the ordinary matrix product of $T_\beta \circ T_\beta ^3$ with $T_\beta ^2$.

The key estimate needed to complete the proof of Proposition 1.8 is the following proposition, which can be thought of as a generalisation of the inequality (2.10).

Proposition A.4

Let V be a countable set and let $T \in {\mathcal {M}}^2(V)$ be positive semidefinite. Then

$$\begin{aligned} T^2 \circ T^2 \le T \circ T^3. \end{aligned}$$

We will deduce this proposition as a special case of a theorem of Ando [6], which in turn is a (non-obvious) consequence of the Schur product theorem. While all these theorems are usually stated for finite matrices, they are also valid for infinite matrices; indeed, we will see that the infinite cases of these theorems can be immediately reduced to the finite cases.

Theorem A.5

(Schur product theorem) Let V be countable, and let $S,T \in {\mathcal {M}}(V)$. If S and T are both positive semidefinite, then $S \circ T$ is positive semidefinite.

Proof of Theorem A.5

Recall that if V is a countable set and $T \in {\mathcal {M}}(V)$ the matrices $T|_W \in {\mathcal {M}}(W)$ where $W \subseteq V$ and $T|_W(u,v)= T(u,v)$ for every $u,v \in W$ are referred to as the principal submatrices of T. By the definitions, a matrix $T \in {\mathcal {M}}(V)$ is positive definite if and only if all its finite dimensional principal submatrices are. Thus, since the Hadamard product commutes with taking principal submatrices in the sense that $S|_W \circ T|_W = (S\circ T)|_W$ for every $S,T \in {\mathcal {M}}(V)$ and $W \subseteq V$, it suffices to consider the case that V is finite. That is, the countably infinite case of the Schur product theorem immediately reduces to the finite case, which is classical; see https://en.wikipedia.org/wiki/Schur_product_theorem for three distinct proofs. (A further proof works by observing that $S \circ T$ is a principal submatrix of the tensor product $S \otimes T$, which in turn is easily seen to be positive definite as a direct consequence of the relevant definitions.) $\square$

Before stating Ando’s theorem on Hadamard products, we recall that for every positive semidefinite matrix $T \in {\mathcal {M}}^2(V)$ there exists a unique positive semidefinite matrix $T^{1/2} \in {\mathcal {M}}^2(V)$, known as the principal square root of T, such that $(T^{1/2})^2 =T$. (Be careful to note that there may exist other matrices S for which $S^2=T$ or $S S^\perp =T$, but any such matrix other than $T^{1/2}$ will not be positive semidefinite.) Moreover, the principal square root $T^{1/2}$ of T commutes with T under multiplication in the sense that $T^{1/2}T=TT^{1/2}$. Indeed, the matrix $T^{1/2}T$ is also equal to the principal square root of $T^3$ and is denoted simply by $T^{3/2}$. Similarly, one may define for each $\alpha \ge 0$ a symmetric, positive semidefinite matrix $T^\alpha$ such that $(T^\alpha )_{\alpha \ge 0}$ is a semigroup in the sense that $T^\alpha$ depends continuously on $\alpha$ and $T^\alpha T^\beta =T^\beta T^\alpha = T^{\alpha +\beta }$ for every $\alpha ,\beta \ge 0$. All of these claims are immediate consequences of the spectral theorem for bounded self-adjoint operators, see e.g. [43, Chapter VII] for further background. Similar spectral considerations imply that every positive definite matrix $T \in {\mathcal {M}}^2(V)$ has a well-defined inverse $T^{-1} \in {\mathcal {M}}^2(V)$ with $\Vert T^{-1}\Vert _{2\rightarrow 2}^{-1}=\sup \{ \lambda : \langle Tf,f\rangle \ge \lambda \langle f,f\rangle$ for every $f\in L^0(V)\}$.

We are now ready to state the aforementioned theorem of Ando, which is a special case of [6, Theorem 12]. While Ando stated his theorem for finite-dimensional matrices, the proof extends easily to the infinite-dimensional case. We remark that the paper of Ando contains a huge number of further related inequalities.

Theorem A.6

(Ando) Let V be a countable set, and let $S,T \in {\mathcal {M}}^2(V)$ be commuting, positive semidefinite matrices. Then

$$\begin{aligned} (ST)^{1/2} \circ (ST)^{1/2} \le S \circ T. \end{aligned}$$

As discussed above, we include a full proof of this theorem with the dual aims of making it accessible to probabilists and verifying the infinite-dimensional case. Before starting the proof, let us note that the theorem immediately implies Proposition A.4.

Proof of Proposition A.4given Theorem A.6 Let $T \in {\mathcal {M}}^2(V)$ be positive semidefinite. Applying Theorem A.6 to the commuting, positive semidefinite matrices T and $T^3$ yields that $T^2 \circ T^2 \le T \circ T^3$ as claimed. $\square$

The proof of Theorem A.6 will proceed by applying the Schur product theorem to various judiciously defined block matrices; see [9] for further applications of this technique. Let V be a countable set and consider the disjoint union $V \sqcup V= V \times \{1,2\}$. Given $A,B,C,D \in {\mathcal {M}}(V)$, we define the block matrix

$$\begin{aligned} \left[ \begin{matrix} A &{} B \\ C &{} D \end{matrix}\right] \in {\mathcal {M}}(V \sqcup V) \qquad \text {by} \qquad \left[ \begin{matrix} A &{} B \\ C &{} D \end{matrix}\right] \bigl ((u,i),(v,j)\bigr ) = {\left\{ \begin{array}{ll} A(u,v) &{} i=1,\, j=1\\ C(u,v) &{} i=2,\, j=1\\ B(u,v) &{} i=1,\, j=2\\ D(u,v) &{} i=2,\, j=2 \end{array}\right. } \end{aligned}$$

so that if $f,g \in L^0(V)$ and we define

$$\begin{aligned} \left[ \begin{matrix} f \\ g \end{matrix}\right] \!\in \! L^0(V \sqcup V) \quad \text {by} \quad \left[ \begin{matrix} f \\ g \end{matrix}\right] (v,i) \!=\! {\left\{ \begin{array}{ll} f(v) &{} i=1 \\ g(v) &{} i=2 \end{array}\right. } \quad \text {then} \quad \left[ \begin{matrix} A &{} B \\ C &{} D \end{matrix}\right] \left[ \begin{matrix} f \\ g \end{matrix}\right] \!=\! \left[ \begin{matrix} A f + Bg \\ Cf + Dg \end{matrix}\right] . \end{aligned}$$

Similarly, if $A,B,C,D,E,F,G,H \in {\mathcal {M}}^2(V)$ then we can compute the products of the associated block matrices in the usual way

$$\begin{aligned} \left[ \begin{matrix} A &{} B \\ C &{} D \end{matrix}\right] \left[ \begin{matrix} E &{} F \\ G &{} H \end{matrix}\right] = \left[ \begin{matrix} A E + BG &{} AF + BH \\ C E + D G &{} CG + DH \end{matrix}\right] \in {\mathcal {M}}^2(V \sqcup V). \end{aligned}$$

For each $T \in {\mathcal {M}}(V)$, we define $T^\perp$ to be the transpose matrix $T^\perp (u,v)=T(v,u)$. The following lemma is an infinite-dimensional version of [9, Theorem 1.3.3].

Lemma A.7

Let V be a countable set, and let $S,T,X \in {\mathcal {M}}^2(V)$ be such that S is positive semidefinite and T is positive definite. Then

$$\begin{aligned} \left[ \begin{matrix} S &{} X \\ X^\perp &{} T \end{matrix}\right] \ge 0 \qquad \text { if and only if } \qquad S \ge X T^{-1} X^\perp . \end{aligned}$$

Before proving this lemma, let us note that if $S,T \in {\mathcal {M}}^2(V)$ are congruent in the sense that there exists a matrix $C \in {\mathcal {M}}^2(V)$ with inverse $C^{-1} \in {\mathcal {M}}^2(V)$ such that $S = CTC^\perp$, then S is positive semidefinite if and only if T is. Indeed, first note that congruence is an equivalence relation since if $S = CTC^\perp$ then $T = C^{-1} S (C^{-1})^\perp$. Congruence is also trivially seen to preserve symmetry. To conclude, note that if T is positive semidefinite then we have that $\langle Sf,f \rangle = \langle CTC^\perp f,f \rangle = \langle T (C^\perp f), (C^\perp f)\rangle \ge 0$ for every $f\in L^2(V)$ so that S is positive semidefinite as claimed.

Proof of Lemma A.7

Since T is positive definite, it is invertible with inverse $T^{-1}\in {\mathcal {M}}^2(V)$. Write I and O for the identity and zero matrices indexed by V respectively. The block matrix we are interested in is congruent to

$$\begin{aligned} \left[ \begin{matrix} I &{} -XT^{-1} \\ O &{} I \end{matrix}\right] \left[ \begin{matrix} S &{} X \\ X^\perp &{} T \end{matrix}\right] \left[ \begin{matrix} I &{} -XT^{-1} \\ O &{} I \end{matrix}\right] ^\perp&= \left[ \begin{matrix} I &{} -XT^{-1} \\ O &{} I \end{matrix}\right] \left[ \begin{matrix} S &{} X \nonumber \\ X^\perp &{} T \end{matrix}\right] \left[ \begin{matrix} I &{} O \\ -T^{-1}X^\perp &{} I \end{matrix}\right] \\&= \left[ \begin{matrix} S - X T^{-1} X^\perp &{} O \\ O &{} T \end{matrix}\right] , \end{aligned}$$

(Appendix A.3)

where we note that this is indeed a congruence since

$$\begin{aligned} \left[ \begin{matrix} I &{} -XT^{-1} \\ O &{} I \end{matrix}\right] \qquad \text { is invertible with inverse } \qquad \left[ \begin{matrix} I &{} -XT^{-1} \\ O &{} I \end{matrix}\right] ^{-1} = \left[ \begin{matrix} I &{} XT^{-1} \\ O &{} I \end{matrix}\right] . \end{aligned}$$

The matrix on the right hand side of (Appendix A.3) is positive semidefinite if and only if

$$\begin{aligned} \left\langle \left[ \begin{matrix} S - X T^{-1} X^\perp &{} O \\ O &{} T \end{matrix}\right] \left[ \begin{matrix} f \\ g \end{matrix}\right] , \left[ \begin{matrix} f \\ g \end{matrix}\right] \right\rangle = \bigl \langle (S -X T^{-1}X^\perp ) f,f\bigr \rangle + \langle T g, g\rangle \ge 0, \end{aligned}$$

for every $f,g \in L^0(V)$, and since T is positive definite this is the case if and only if $S -X T^{-1}X^\perp$ is positive semidefinite as claimed. $\square$

We will also use the following easy fact. Note that the implication here is only in one direction.

Lemma A.8

Let V be countable and let $S,T \in {\mathcal {M}}(V)$ be symmetric matrices.

• 1. If the block matrix $\left[ \begin{matrix} S &{} T \\ T &{} S \end{matrix}\right]$ is positive semidefinite then $S - T$ is positive semidefinite.

• 2. If the block matrix $\left[ \begin{matrix} S &{} T \\ T &{} S \end{matrix}\right]$ is positive definite then $S - T$ is positive definite.

Proof of Lemma A.8

We can compute that

$$\begin{aligned} \left\langle \left[ \begin{matrix} S &{} T \\ T &{} S \end{matrix}\right] \left[ \begin{matrix} f \\ -f \end{matrix}\right] , \left[ \begin{matrix} f \\ -f \end{matrix}\right] \right\rangle = 2\bigl \langle (S-T) f, f \bigr \rangle \end{aligned}$$

for every $f \in L^0(V)$, and the claim follows from the definitions. $\square$

We are now ready to prove Theorem A.6.

Proof of Theorem A.6

The assumption that S and T commute implies that $(ST)^{1/2} = (TS)^{1/2} = ((ST)^{1/2})^\perp$. In particular, we have that $S = (ST)^{1/2} T^{-1} (ST)^{1/2}$ and $T = (ST)^{1/2} S^{-1} (ST)^{1/2}$, and we deduce from Lemma A.7 that the block matrices

$$\begin{aligned} \left[ \begin{matrix} S &{} (ST)^{1/2} \\ (ST)^{1/2} &{} T \end{matrix}\right] \qquad \text {and} \qquad \left[ \begin{matrix} T &{} (ST)^{1/2} \\ (ST)^{1/2} &{} S \end{matrix}\right] \end{aligned}$$

are both positive semidefinite. Applying the Schur product theorem we deduce that

$$\begin{aligned} \left[ \begin{matrix} S \circ T &{} (ST)^{1/2} \circ (ST)^{1/2} \\ (ST)^{1/2} \circ (ST)^{1/2} &{} S \circ T \end{matrix}\right] \end{aligned}$$

is positive semidefinite also, so that the claim follows from Lemma A.8. $\square$

The trace on the group von Neumann algebra We now wish to apply Ando’s theorem to prove Proposition 1.8. To do this, we will employ the notion of the trace on the von Neumann algebra associated to the action of ${\text {Aut}}(G)$ on G. We assure the unfamiliar reader that, despite the intimidating name, this is in fact a very simple notion.

Given a connected, transitive weighted graph $G=(V,E,J)$ with automorphism group ${\text {Aut}}(G)$, we define the von Neumann algebra ${\mathcal {A}}(G)$ to be the set of matrices $T \in {\mathcal {M}}^2(V)$ that are diagonally invariant under the action of ${\text {Aut}}(G)$ on V in the sense that $T(\gamma u,\gamma v)=T(u,v)$ for every $u,v \in V$ and $\gamma \in {\text {Aut}}(G)$. Note that ${\mathcal {A}}(G)$ is indeed a von Neumann algebra in the sense that it is a weak$^*$-closed subspace of ${\mathcal {M}}^2(V) \cong {\mathcal {B}}(L^2(V))$ that is closed under multiplication and adjunction. It follows trivially from the definitions that ${\mathcal {A}}(G)$ is also closed under Hadamard products. We will also need the slightly less obvious fact that ${\mathcal {A}}(G)$ is closed under taking principal square roots of its positive semidefinite elements, which we now prove.

Lemma A.9

Let $G=(V,E,J)$ be a connected, transitive weighted graph. If $T \in {\mathcal {A}}(G)$ is positive semidefinite then its principal square root $T^{1/2}$ also belongs to ${\mathcal {A}}(G)$.

Proof of Lemma A.9

It suffices to prove that $T^{1/2}$ is diagonally invariant, i.e., that $T^{1/2}(\gamma u, \gamma v) = T^{1/2}(u,v)$ for every $\gamma \in {\text {Aut}}(G)$ and $u,v\in V$. Fix $\gamma \in {\text {Aut}}(G)$ and let $S \in {\mathcal {M}}(V)$ be defined by $S(u,v)=T^{1/2}(\gamma u,\gamma v)$ for each $u,v\in V$. Since $\gamma \in {\text {Aut}}(G)$ was arbitrary, it suffices to prove that S is equal to $T^{1/2}$. Since the principal square root of T is unique, it suffices to prove that S is positive semidefinite and satisfies $S^2=T$. We begin by proving that S is bounded and positive semidefinite, noting that is is clearly symmetric. For each $f \in L^0(V)$ let $\gamma f \in L^0(V)$ be given by $\gamma f (u)=f(\gamma ^{-1}u)$. Since $\gamma$ is a bijection we have that

$$\begin{aligned}&Sf(u)=\sum _{v\in V} S(u,v)f(v) = \sum _{v\in V} T^{1/2}(\gamma u, \gamma v)f( v) \\&\quad = \sum _{v\in V} T^{1/2}(\gamma u, v)f( \gamma ^{-1} v) = [T^{1/2}(\gamma f)](\gamma u) \end{aligned}$$

for every $f\in L^0(V)$ and $u\in V$ and hence that

$$\begin{aligned} \Vert Sf\Vert =\Vert T^{1/2}(\gamma f)\Vert \le \Vert T\Vert _{2\rightarrow 2}^2 \Vert \gamma f\Vert =\Vert T\Vert _{2\rightarrow 2}^2 \Vert f\Vert \end{aligned}$$

and

$$\begin{aligned} \langle Sf,f\rangle&= \sum _{u\in V} \sum _{v\in V} T^{1/2}(\gamma u, v)f(u)f( \gamma ^{-1} v) = \sum _{u\in V} \sum _{v\in V} T^{1/2}( u, v)f( \gamma ^{-1} u)f( \gamma ^{-1} v) \\&\quad = \langle T^{1/2}(\gamma f),\gamma f\rangle \end{aligned}$$

for every $f\in L^0(V)$. Since $T^{1/2}$ is bounded and positive semidefinite it follows that S is bounded and positive semidefinite also. We now verify that $S^2=T$ by computing that

$$\begin{aligned} S^2(u,v)&= \sum _{w\in V} S(u,w)S(w,v) = \sum _{w\in V} T^{1/2}(\gamma u, \gamma w)T^{1/2}(\gamma w, \gamma v) \\&= \sum _{w\in V} T^{1/2}(\gamma u, w)T^{1/2}(w, \gamma v) =T(\gamma u,\gamma v)=T(u,v) \end{aligned}$$

for every $u,v\in V$. We deduce that $S=T^{1/2}$ by uniqueness of the principal square root of T. $\square$

We now define the trace on ${\mathcal {A}}(G)$ and establish its basic properties.

Proposition A.10

(The trace on the von Neumann algebra ${\mathcal {A}}(G)$) Let $G=(V,E,J)$ be a unimodular, connected, transitive weighted graph. Define ${\text {Tr}}:{\mathcal {A}}(G) \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} {\text {Tr}}(T) = T(o,o) \qquad \text { for every }T\in {\mathcal {A}}(G). \end{aligned}$$

Then ${\text {Tr}}$ is a tracial state on ${\mathcal {A}}(G)$. This means that the following conditions hold:

• 1. ${\text {Tr}}:{\mathcal {A}}(G) \rightarrow {\mathbb {R}}$ is linear.

• 2. ${\text {Tr}}(I)=1$, where $I(u,v)=\mathbb {1}(u=v)$ is the identity matrix.

• 3. ${\text {Tr}}(T T^\perp ) \ge 0$ for every $T \in {\mathcal {A}}(G)$, with equality if and only if $T=0$.

• 4. ${\text {Tr}}(ST) = {\text {Tr}}(TS)$ for every $S,T \in {\mathcal {A}}(G)$.

See e.g. [5, §5] and [35, Chapter 10.8] for previous uses of similar notions in probabilistic contexts.

The proof of this proposition will use the signed mass-transport principle, which states that if $G=(V,E,J)$ is a unimodular transitive weighted graph and $F:V\times V\rightarrow {\mathbb {R}}$ is diagonally invariant in the sense that $F(\gamma u,\gamma v)=F(u,v)$ for every $u,v\in V$ and $\gamma \in {\text {Aut}}(G)$ then $\sum _{v\in V}F(o,v) = \sum _{v\in V}F(v,o)$ provided that F satisfies the integrability condition

$$\begin{aligned} \sum _{v\in V}|F(o,v)| = \sum _{v\in V}|F(v,o)| <\infty . \end{aligned}$$

(Appendix A.4)

This can be verified by applying the usual mass-transport principle (1.8) to the positive and negative parts of F separately.

Proof of Proposition A.10

The first two items are trivial. For the third, simply note that

$$\begin{aligned} {\text {Tr}}(TT^\perp ) = \sum _{v\in V} T(o,v)T^\perp (v,o) = \sum _{v\in V} T(o,v)^2 \end{aligned}$$

for every $T \in {\mathcal {A}}(G)$, from which the claim follows trivially. Finally, for the fourth item, we apply the mass-transport principle to the diagonally invariant function $F(u,v) = S(u,v)T(v,u)$, which satisfies the integrability hypothesis (Appendix A.4) since $S,T \in {\mathcal {M}}^2(V)$, to deduce that

$$\begin{aligned} {\text {Tr}}(ST)&= \sum _{v\in V} S(o,v)T(v,o) = \sum _{v\in V} S(v,o)T(o,v) = {\text {Tr}}(TS). \end{aligned}$$

as claimed. $\square$

The following is an adaptation to our setting of an inequality due to Fejér [14, Theorem A.8].

Lemma A.11

Let $G=(V,E,J)$ be a unimodular, connected, transitive weighted graph. If $S,T \in {\mathcal {A}}(G)$ are positive semidefinite then ${\text {Tr}}(ST) \ge 0$.

It follows in particular from this lemma that if two matrices $S,T\in {\mathcal {A}}(G)$ satisfy $S\le T$ then ${\text {Tr}}(SX)\le {\text {Tr}}(TX)$ for every positive semidefinite matrix $X\in {\mathcal {A}}(G)$.

Proof of Lemma A.11

We have by Item 4. of Proposition A.10 that

$$\begin{aligned} {\text {Tr}}(ST)={\text {Tr}}(S^{1/2}S^{1/2}T^{1/2}T^{1/2}) = {\text {Tr}}(S^{1/2}T^{1/2} T^{1/2}S^{1/2}). \end{aligned}$$

Since $(S^{1/2} T^{1/2})^\perp = (T^{1/2})^\perp (S^{1/2})^\perp = T^{1/2}S^{1/2}$, it follows from item 3 of Proposition A.10 that ${\text {Tr}}(ST) \ge 0$ as claimed. $\square$

We are finally ready to prove that $B_\beta \le A_\beta$ and conclude the proof of Proposition 1.8.

Proof of Proposition 1.8

Let $G=(V,E,J)$ be a unimodular, connected, transitive weighted graph and let $0\le \beta <\beta _c$. We have by Lemma A.3 that $A_\beta \le \nabla _\beta ^{2}$, so that it suffices to prove that $B_\beta \le A_\beta$. The two-point matrix $T_\beta$ is positive definite by Lemma A.1 and belongs to ${\mathcal {M}}^2(V)$ and hence to ${\mathcal {A}}(G)$ by sharpness of the phase transition as discussed above. It follows from Proposition A.4 that $T_\beta ^2 \circ T_\beta ^2 \le T_\beta \circ T_\beta ^3$ and hence by Lemma A.11 that

$$\begin{aligned} B_\beta= & {} \bigl [T_\beta ^2 \circ T_\beta ^2\bigr ]T_\beta ^2(o,o) = {\text {Tr}}\Bigl (\bigl [T_\beta ^2 \circ T_\beta ^2\bigr ]T_\beta ^2\Bigr ) \le {\text {Tr}}\Bigl (\bigl [T_\beta \circ T_\beta ^3\bigr ]T_\beta ^2\Bigr ) \\= & {} \bigl [T_\beta \circ T_\beta ^3\bigr ]T_\beta ^2(o,o) = A_\beta \end{aligned}$$

as claimed. Together with Proposition 1.7 this also concludes the proof of Theorem 1.1. $\square$