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On Negative Correlation of Arboreal Gas on Some Graphs

Xiangyu Huang***YMSC, Tsinghua University, Beijing 100084, China. [email protected]
Abstract

Key words: Negative correlation; Arboreal Gas; Finite connected graph.

1 Introduction to Arboreal Gas

Let G=(V,E)G=(V,E) be a finite connected graph. If there exists an edge eEe\in E connecting vertices u,vVu,v\in V, we say that u,vu,v are adjacent, denoted by uvu\sim v. We can also represent the edge ee as uvuv. A forest on GG is a subgraph that does not contain any cycle. Set \mathcal{F} to be the collection of all forests on GG. The Arboreal Gas with parameter βe>0\beta_{e}>0 for each eEe\in E is the measure on forests FF given by

β[F]:=1ZβeFβe,Zβ:=FeFβe.\mathbb{P}_{\beta}[F]:=\frac{1}{Z_{\beta}}\prod_{e\in F}\beta_{e},\hskip 17.07164ptZ_{\beta}:=\sum_{F\in\mathcal{F}}\prod_{e\in F}\beta_{e}.

Specifically, if βeβ\beta_{e}\equiv\beta for all edges ee, the measure will be given by

β[F]:=1Zββ|F|,Zβ:=Fβ|F|,\mathbb{P}_{\beta}[F]:=\frac{1}{Z_{\beta}}\beta^{|F|},\hskip 17.07164ptZ_{\beta}:=\sum_{F\in\mathcal{F}}\beta^{|F|},

where |F||F| stands for the number of edges in FF.

Let pperc\mathbb{P}^{\rm perc}_{p} denote the probablity of Bernoulli bond percolation with parameter pp. We can note that Arboreal Gas with a uniform parameter β\beta is equivalent to Bernoulli bond percolation with parameter pβ:=β/(1+β)p_{\beta}:=\beta/(1+\beta) conditioned to be acyclic:

pβperc[F|acyclic]=pβ|F|(1pβ)|E||F|Fpβ|F|(1pβ)|E||F|=β|F|Fβ|F|=β[F].\mathbb{P}^{\rm perc}_{p_{\beta}}[F|{\rm acyclic}]=\frac{p_{\beta}^{|F|}(1-p_{\beta})^{|E|-|F|}}{\sum_{F\in\mathcal{F}}p_{\beta}^{|F|}(1-p_{\beta})^{|E|-|F|}}=\frac{\beta^{|F|}}{\sum_{F\in\mathcal{F}}\beta^{|F|}}=\mathbb{P}_{\beta}[F].

Another notable observation is that Arboreal Gas emerges as the limit of the qq-states random cluster model as q0q\to 0, with p=βqp=\beta q, as mentioned in [13]. The uniform forest model, also mentioned in [13], can be seen as one particular instance of Arboreal Gas, where β=1\beta=1. It is important to note that this model is distinct from another known model referred to as uniform spanning forest in the literature. Furthermore, another specific case is the uniform spanning tree, which arises as a limit of Arboreal Gas as β\beta\to\infty.

An interesting phenomenon lies in the correlation between Arboreal Gas and hyperbolic spin systems. This kind of spin system is different from the classical spin systems with spherical symmetry, such as Ising model and classical Heisenberg model. Hyperbolic spin systems are also extensively studied in condensed matter physics due to their connection to the Anderson delocalisation-localisation transition of random Schödinger operators and related matrix models, as discussed in [7, 14, 15]. Researchers are interested in the existence of phase transitions in the hyperbolic spin systems. Considering Arboreal Gas, Bauerschmidt, Crawford, Helmuth and Swan presented a magic formula in [3] that elegantly describe its probability by the intergral in the hyperbolic spin system 0|2\mathbb{H}^{0|2}, which may also be seen in [1, 6].

The subcritical percolation and percolating phase transitions in Arboreal Gas have garnered significant attention. In the seminal works [4, 10, 12], it was established that Arboreal Gas, with a parameter β=α/N\beta=\alpha/N for fixed α\alpha, undergoes a phase transition on the complete graph KNK_{N} with NN vertices. This phase transition was also demonstrated by Bauerschmidt, Crawford, Helmuth, and Swan in their recent study [3]. Additionally, they proved the polynomial decay of connectivity probability from the origin to other vertices in all subgraphs of the lattice 2\mathbb{Z}^{2}. Based on this polynomial decay, and assuming for the existence of a weak limit of probabilities on asymptotic graphs of 2\mathbb{Z}^{2}, they demonstrated the absence of an infinite tree almost surely. On torus ΛN=d/LNd\Lambda_{N}=\mathbb{Z}^{d}/L^{N}\mathbb{Z}^{d} with LL fixed and NN\to\infty, Bauerschmidt, Crawford and Helmuth gave an asymptotic estimate of connectivity probability from the origin to other vertices for d3d\geq 3 in [2]. Recently, Halberstam and Hutchcroft verified the uniqueness of infinite tree in d\mathbb{Z}^{d} for d=3,4d=3,4 for any translation-invariant Arboreal Gas Gibbs measures in [9].

For any edge set SES\subset E, let β[S]\mathbb{P}_{\beta}[S] be the probability that all edges in SS belong to the forest. We simply write β[S]\mathbb{P}_{\beta}[S] as β[e1e2en]\mathbb{P}_{\beta}[e_{1}e_{2}\dots e_{n}] for S={e1,e2,,en}S=\{e_{1},e_{2},\dots,e_{n}\}. For two vertices u,vVu,v\in V, denote by β[uv]\mathbb{P}_{\beta}[u\leftrightarrow v] the probability that u,vu,v are in a same tree of the forest. The negative correlation of Arboreal Gas should be expressed as

β[e1e2]β[e1]β[e2],foranye1,e2E.\mathbb{P}_{\beta}[e_{1}e_{2}]\leq\mathbb{P}_{\beta}[e_{1}]\mathbb{P}_{\beta}[e_{2}],\hskip 17.07164pt{\rm for\ any\ }e_{1},e_{2}\in E. (1.1)

Until now, this question is still open. A weaker inequality has been proved recently in [5] by Brändén and Huh. They showed β[e1e2]2β[e1]β[e2]\mathbb{P}_{\beta}[e_{1}e_{2}]\leq 2\mathbb{P}_{\beta}[e_{1}]\mathbb{P}_{\beta}[e_{2}] by using the Lorentzian signature. Once (1.1) holds, the existence of the weak limit of Arboreal Gas Gibbs measures on increasing asymptotic graphs Gn2G_{n}\uparrow\mathbb{Z}^{2} as nn\to\infty was answered in [3]. Under this weak limit on 2\mathbb{Z}^{2}, all trees should be finite almost surely, see Corollary 1.4 in [3].

In this paper, we focus on the negative correlation of Arboreal Gas on finite connected graph GG. One way to consider this question is to realize the relevance between Arboreal Gas and the hyperbolic spin system 0|2\mathbb{H}^{0|2}. The negative correlation of Arboreal Gas can be implied by the monotonicity of a hyper-integral on 0|2\mathbb{H}^{0|2}, see section 5.2 in [1]. However, the monotonicity of hyper-integral is hard to be verified. In our paper, we consider Arboreal Gas directly. Here is the outline of this paper. In Section 2, we show our main results, and we give some preliminaries in Section 3. In Section 4, we show the negative correlation for adjacent edges and sufficiently large β\beta. In Section 5, we show the negative correlation of complete graphs KnK_{n} with nn vertices for sufficiently large nn and β\beta. In Section 6, we give the proof of the equivalence of the negative correlation on any graph and on its simplified version.

2 Negative Correlations and Main Results

Here are our main theorems.

Theorem 2.1.

Consider Arboreal Gas with parameter β\beta on finite connected graph GG. For sufficiently large β\beta and any adjacent distinct edges e1,e2Ee_{1},e_{2}\in E,

β[e1e2]β[e1]β[e2].\mathbb{P}_{\beta}[e_{1}e_{2}]\leq\mathbb{P}_{\beta}[e_{1}]\mathbb{P}_{\beta}[e_{2}].
Theorem 2.2.

Consider Arboreal Gas with parameter β\beta on complete graph KnK_{n} with nn vertices, where nn is large enough. For sufficiently large or sufficiently small β\beta and any distinct edges e1,e2Ee_{1},e_{2}\in E,

β[e1e2]β[e1]β[e2].\mathbb{P}_{\beta}[e_{1}e_{2}]\leq\mathbb{P}_{\beta}[e_{1}]\mathbb{P}_{\beta}[e_{2}].

We verify those two theorems by regarding the probability β[e1e2],β[e1],β[e2]\mathbb{P}_{\beta}[e_{1}e_{2}],\mathbb{P}_{\beta}[e_{1}],\mathbb{P}_{\beta}[e_{2}] as polynomials of β\beta. A fact is, the negative correlation holds for uniform spanning trees on finite connected graphs, see section 4.2 in [11]. An important observation is that the term with highest degree in the polynomials of β\beta mentioned before corresponds to the uniform spanning trees. This is our basic idea to show the negative correlation for sufficiently large β\beta.

Besides, we have another theorem to simplify the structure of graphs when we consider the negative correlation. Before we give this theorem, we first give some definitions.

Definition 2.3.

For eEe\in E, we call it pivotal for GG if extreme points of ee are not connected in Ge:=(V,E{e})G\setminus e:=(V,E\setminus\{e\}).

Definition 2.4.

For a graph GG with no pivotal edges, we give graph G~=(V~,E~)\tilde{G}=(\tilde{V},\tilde{E}) by induction:

  1. 1.

    Set G0=GG_{0}=G.

  2. 2.

    If there exists some vertex uu in VnV_{n} having degree 22, we assume v1,v2v_{1},v_{2} are adjacent to uu. Then we set Vn+1=Vn{u}V_{n+1}=V_{n}\setminus\{u\}. We give En+1E_{n+1} by adding a new edge e0e_{0} connecting v1,v2v_{1},v_{2} on En{uv1,uv2}E_{n}\setminus\{uv_{1},uv_{2}\}. Define fn+1:EnEn+1f_{n+1}:E_{n}\to E_{n+1} such that fn+1(uvi)=e0f_{n+1}(uv_{i})=e_{0} for i=1,2i=1,2 and fn+1(e)=ef_{n+1}(e)=e for euv1,uv2e\neq uv_{1},uv_{2}.

  3. 3.

    Stop if there is no vertex with degree 22.

Further, we give a map f:EE~f:E\to\tilde{E} by f=fnfn1f1f=f_{n}\circ f_{n-1}\circ\dots\circ f_{1}, where nn is the number of steps to get E~\tilde{E}.

Definition 2.5.

For a graph GG, we give a simple graph G=(V,E)G^{\prime}=(V^{\prime},E^{\prime}) by setting V=VV^{\prime}=V and letting u,vVu,v\in V^{\prime} adjacent in GG^{\prime} if they are adjacent in GG. Further, we give a map g:EEg:E\to E^{\prime} with g(e)=uvg(e)=uv if the end points of ee are u,vu,v.

Here we give our main theorem.

Theorem 2.6.

Consider Arboreal Gas with parameter βe\beta_{e} for each eEe\in E on finite connected graph GG. Give G^\hat{G} by deleting all pivotal edges in GG, then the negative correlation on GG is equivalent to that on each component of gf(G^)g\circ f(\hat{G}).

Corollary 2.7.

Consider Arboreal Gas with parameter βe\beta_{e} for each eEe\in E on finite connected graph GG. For any distinct edges e1,e2Ee_{1},e_{2}\in E, if e1e_{1} or e2e_{2} is pivotal, or gf(e1)=gf(e2)g\circ f(e_{1})=g\circ f(e_{2}), then we have

β[e1e2]β[e1]β[e2].\mathbb{P}_{\beta}[e_{1}e_{2}]\leq\mathbb{P}_{\beta}[e_{1}]\mathbb{P}_{\beta}[e_{2}].

This corollary immediately comes out by Theorem 2.6. We can also obtain the negative correlation on some specific graphs.

Corollary 2.8.

Consider Arboreal Gas with parameter βe\beta_{e} for each eEe\in E. . For any integer d0d\geq 0, the negative correlation holds on ladder 𝕃d:={1,2,,d}×{0,1}\mathbb{L}_{d}:=\{1,2,\dots,d\}\times\{0,1\}.

Proof. Note the fact that gf(𝕃d)=𝕃d2g\circ f(\mathbb{L}_{d})=\mathbb{L}_{d-2} for d2d\geq 2, where 𝕃0\mathbb{L}_{0} is a unique vertex, and 𝕃1\mathbb{L}_{1} is a line segment. By induction, the negative correlation on 𝕃d\mathbb{L}_{d} can be implied by that on 𝕃0\mathbb{L}_{0} and 𝕃1\mathbb{L}_{1}, which completes the proof.  

3 Preliminaries

First we show an equivalent condition of negative correlation.

Definition 3.1.

For disjoint S1,S2ES_{1},S_{2}\subset E, define

β[S1S2¯]=β[S1F,S2F=].\mathbb{P}_{\beta}[S_{1}\bar{S_{2}}]=\mathbb{P}_{\beta}[S_{1}\subset F,S_{2}\cap F=\emptyset].

Given a measure μ\mu defined on GG by

μ[S1S2¯]=β[S1S2¯]Zβ/eS1βe.\mu[S_{1}\bar{S_{2}}]=\mathbb{P}_{\beta}[S_{1}\bar{S_{2}}]\cdot Z_{\beta}/\prod_{e\in S_{1}}\beta_{e}.

Specifically, set μ[1]=Zβ\mu[1]=Z_{\beta}.

Proposition 3.2.

The following two conditions are equivalent:

  1. 1.

    β[e1e2]β[e1]β[e2]\mathbb{P}_{\beta}[e_{1}e_{2}]\leq\mathbb{P}_{\beta}[e_{1}]\mathbb{P}_{\beta}[e_{2}] for all distinct e1,e2Ee_{1},e_{2}\in E;

  2. 2.

    β[S1S2]β[S1]β[S2]\mathbb{P}_{\beta}[S_{1}S_{2}]\leq\mathbb{P}_{\beta}[S_{1}]\mathbb{P}_{\beta}[S_{2}] for all disjoint S1,S2ES_{1},S_{2}\subset E.

Proof.

121\Leftarrow 2: This result immediately comes out by setting Si=eiS_{i}=e_{i} for i=1,2i=1,2.

121\Rightarrow 2: First we prove the equivalence of condition 11 and decreasing of β[e0]\mathbb{P}_{\beta}[e_{0}] in each βe\beta_{e} for all e0Ee_{0}\in E and ee0e\neq e_{0}. Note that

1\displaystyle 1 β[e1e2]β[e1¯e2¯]β[e1e2¯]β[e1¯e2]\displaystyle\Leftrightarrow\mathbb{P}_{\beta}[e_{1}e_{2}]\mathbb{P}_{\beta}[\bar{e_{1}}\bar{e_{2}}]\leq\mathbb{P}_{\beta}[e_{1}\bar{e_{2}}]\mathbb{P}_{\beta}[\bar{e_{1}}e_{2}]
μ[e1e2]μ[e1¯e2¯]μ[e1e2¯]μ[e1¯e2]\displaystyle\Leftrightarrow\mu[e_{1}e_{2}]\mu[\bar{e_{1}}\bar{e_{2}}]\leq\mu[e_{1}\bar{e_{2}}]\mu[\bar{e_{1}}e_{2}]
βe1μ[e1e2¯]βe1μ[e1e2¯]+μ[e1¯e2¯]βe1μ[e1e2]βe1μ[e1e2]+μ[e1¯e2]\displaystyle\Leftrightarrow\frac{\beta_{e_{1}}\mu[e_{1}\bar{e_{2}}]}{\beta_{e_{1}}\mu[e_{1}\bar{e_{2}}]+\mu[\bar{e_{1}}\bar{e_{2}}]}\leq\frac{\beta_{e_{1}}\mu[e_{1}e_{2}]}{\beta_{e_{1}}\mu[e_{1}e_{2}]+\mu[\bar{e_{1}}e_{2}]}
βe2βe1μ[e1e2]+βe1μ[e1e2¯]βe2(βe1μ[e1e2]+μ[e1¯e2])+βe1μ[e1e2¯]+μ[e1¯e2¯]isdecreasinginβe2.\displaystyle\Leftrightarrow\frac{\beta_{e_{2}}\beta_{e_{1}}\mu[e_{1}e_{2}]+\beta_{e_{1}}\mu[e_{1}\bar{e_{2}}]}{\beta_{e_{2}}(\beta_{e_{1}}\mu[e_{1}e_{2}]+\mu[\bar{e_{1}}e_{2}])+\beta_{e_{1}}\mu[e_{1}\bar{e_{2}}]+\mu[\bar{e_{1}}\bar{e_{2}}]}{\rm\ is\ decreasing\ in\ }\beta_{e_{2}}. (3.2)

By βe2μ[e1¯e2]+μ[e1¯e2¯]=μ[e1¯]\beta_{e_{2}}\mu[\bar{e_{1}}e_{2}]+\mu[\bar{e_{1}}\bar{e_{2}}]=\mu[\bar{e_{1}}] and βe2μ[e1e2]+μ[e1e2¯]=μ[e1]\beta_{e_{2}}\mu[e_{1}e_{2}]+\mu[e_{1}\bar{e_{2}}]=\mu[e_{1}],

(3.2)=β[e1].(\ref{1 in ng equiv})=\mathbb{P}_{\beta}[e_{1}].

By the arbitrary of e1,e2e_{1},e_{2}, we obtain that (3.2) is equivalent to the decreasing of β[e0]\mathbb{P}_{\beta}[e_{0}] in each βe\beta_{e} for all e0Ee_{0}\in E and ee0e\neq e_{0}.

Secondly, we prove that the decreasing of β[e0]\mathbb{P}_{\beta}[e_{0}] in each βe\beta_{e} for all e0Ee_{0}\in E and ee0e\neq e_{0} deduces condition 22, which may complete the proof of 121\Rightarrow 2. To show this property, we claim that

β[|S]\mathbb{P}_{\beta}[\cdot|S] equals the limit of β[]\mathbb{P}_{\beta}[\cdot] as βe\beta_{e}\to\infty for all eSEe\in S\subset E.

By the decreasing of β[e0]\mathbb{P}_{\beta}[e_{0}] in each component of β\beta except βe0\beta_{e_{0}} for all e0Ee_{0}\in E, we know that for SES\subset E,

β[e|S]β[e]foralledgeeS.\mathbb{P}_{\beta}[e|S]\leq\mathbb{P}_{\beta}[e]{\rm\ for\ all\ edge\ }e\notin S.

For disjoint S1,S2ES_{1},S_{2}\subset E, Applying this to [|S1]\mathbb{P}[\cdot|S_{1}], we obtain

β[e|S1S2]β[e|S1]foralledgeeS1,S2.\mathbb{P}_{\beta}[e|S_{1}S_{2}]\leq\mathbb{P}_{\beta}[e|S_{1}]{\rm\ for\ all\ edge\ }e\notin S_{1},S_{2}.

This inequality is equivalent to

β[S2|eS1]β[S2|S1].\mathbb{P}_{\beta}[S_{2}|eS_{1}]\leq\mathbb{P}_{\beta}[S_{2}|S_{1}].

BY induction of S1S_{1}, we verify that

[S2|S1][S2],\mathbb{P}[S_{2}|S_{1}]\leq\mathbb{P}[S_{2}],

which implies condition 22.

Finally we prove our claim by setting

βex,S={βe,eS,xβe,eS.\beta_{e}^{x,S}=\left\{\begin{aligned} &\beta_{e},&e\notin S,\\ &x\cdot\beta_{e},&e\in S.\end{aligned}\right.

Then βx,S[F]=eFβex,S/FeFβex,S\mathbb{P}_{\beta^{x,S}}[F]=\prod_{e\in F}\beta_{e}^{x,S}/\sum_{F^{\prime}\in\mathcal{F}}\prod_{e\in F^{\prime}}\beta_{e}^{x,S}, which converges to (as xx\to\infty)

eFSβe1{FS}FSeFSβe1{FS}.\frac{\prod_{e\in F\setminus S}\beta_{e}\textbf{1}_{\{F\supset S\}}}{\sum_{F^{\prime}\supset S}\prod_{e\in F^{\prime}\setminus S}\beta_{e}\textbf{1}_{\{F^{\prime}\supset S\}}}.

By β[F|S]eFSβe1{FS}\mathbb{P}_{\beta}[F|S]\propto\prod_{e\in F\setminus S}\beta_{e}\textbf{1}_{\{F\supset S\}}, we show that βx,S[]\mathbb{P}_{\beta^{x,S}}[\cdot] converges to β[|S]\mathbb{P}_{\beta}[\cdot|S] in weak topology. Further, by the decreasing of βx,S[]\mathbb{P}_{\beta^{x,S}}[\cdot] in xx, we conclude that β[|S]β[]\mathbb{P}_{\beta}[\cdot|S]\leq\mathbb{P}_{\beta}[\cdot].  

Next we show the negative correlation of uniform spanning trees. Here we give the definition of the uniform spanning trees.

Definition 3.3.

A spanning tree TT on GG is a subgraph of GG containing all vertices in VV and being a tree. The uniform spanning tree is a uniform measure on all spanning trees on GG.

In our next part, denote by G\mathbb{P}_{\infty}^{G} the probability of uniform spanning trees on graph GG. In fact, uniform spanning trees can be generated by simple random walks. This method is called Wilson’s algorithm, see the details of which in Theorem 4.1 of [11]. To show the negative correlation of uniform spanning trees, we use this method and another tool called electrical network. This tool shows a nice relationship between the probability theory and potential theory. Here we define the electrical network.

For a finite connected graph GG with conductance cec_{e} on each edge, we build an electrical network. Given two disjoint vertex sets SS and TT to be source and sink. Define potential difference ϕ(e)\phi(\vec{e}) and current i(e)i(\vec{e}) on each direct edge e\vec{e}. For adjacent vertices u,vu,v,

ϕ(uv)=ϕ(vu),i(uv)=i(vu).\phi(\overrightarrow{uv})=-\phi(\overrightarrow{vu}),\hskip 8.53581pti(\overrightarrow{uv})=-i(\overrightarrow{vu}).

Here we normally take

uvi(uv){>0,vS,<0,vT.\sum_{u\sim v}i(\overrightarrow{uv})\left\{\begin{aligned} &>0,&v\in S,\\ &<0,&v\in T.\end{aligned}\right.

Further, the potential difference and current satisfies the following three laws, see [8].

Kirchhoff’s potential law. For any cycle v1v2vnvn+1v_{1}v_{2}\cdots v_{n}v_{n+1} with vn+1=v1v_{n+1}=v_{1},

i=1nϕ(vivi+1)=0.\sum_{i=1}^{n}\phi(\overrightarrow{v_{i}v_{i+1}})=0.

Kirchhoff’s current law. For any vSTv\notin S\cup T,

uvi(uv)=0.\sum_{u\sim v}i(\overrightarrow{uv})=0.

Ohm’s law. For any edge e=uve=uv,

i(e)c(e)=ϕ(e).i(\vec{e})c(e)=\phi(\vec{e}).

Kirchhoff’s potential law is equivalent to the existence of the function ϕ\phi on vertex set such that ϕ(uv)=ϕ(v)ϕ(u)\phi(\overrightarrow{uv})=\phi(v)-\phi(u) for adjacent vertices u,vu,v. Here ϕ\phi is also called potential function. Next we define the energy of current ii by

E(i)=12u,vVi2(uv)/c(uv).E(i)=\frac{1}{2}\sum_{u,v\in V}i^{2}(\overrightarrow{uv})/c(uv).

Define the unit currenct flow ii to be such a flow that vS,uvi(uv)=1\sum_{v\in S,u\sim v}i(\overrightarrow{uv})=1. Denote by

Reff(G,c)=E(i)foraunitcurrentiR_{\rm eff}(G,c)=E(i)\ {\rm for\ a\ unit\ current\ }i

the effective resistance of the electrical network GG with conductance cc. Here Reff(G,c)R_{\rm eff}(G,c) is equal to ϕ(v)\phi(v) for vSv\in S if we set ϕ(u)=0\phi(u)=0 for uTu\in T, where the current is unit.

For some finite connected graph GG, we have the following equation for its spanning trees.

Proposition 3.4 (Kirchhoff’s Effective Resistance Formula, [11]).

Let TT be the spanning tree of a finite connected graph GG, and ee be some edge of GG with endpoints e,e+e^{-},e^{+}, then

G[eT]=e[1sthitse+bytravellingalonge]=i(ee+)=c(e)Reff(G),\mathbb{P}_{\infty}^{G}[e\in T]=\mathbb{P}_{e^{-}}[1st\ hits\ e^{+}\ by\ travelling\ along\ e]=i(\overrightarrow{e^{-}e^{+}})=c(e)R_{\rm eff}(G),

where ii is unit current flow from ee^{-} to e+e^{+}.

By Kirchhoff’s Effective Resistance Formula, we deduce the negative correlation of uniform spanning trees. Before that, we give a lemma to show that the effective resistance decreases if the resistance on some edge decreases by the following lemma.

Lemma 3.5.

(Rayleigh principle, [8]) Consider an electrical network with unit current flow ii from ss to tt and conductance cec_{e} on each edge eEe\in E. For any edge e0Ee_{0}\in E, if i(e0)0i(\vec{e_{0}})\neq 0, then the effective resistance Reff(c)R_{\rm eff}(c) strictly decreases in c(e0)c(e_{0}). Otherwise, Reff(c)R_{\rm eff}(c) is a constant in c(e0)c(e_{0}).

Proof of Lemma 3.5. For the unit current flow ii and conductance cc, we consider the energy

Ec(i)=12u,vVi2(uv)/c(uv),E_{c}(i)=\frac{1}{2}\sum_{u,v\in V}i^{2}(\overrightarrow{uv})/c(uv),

which is equal to Reff(c)R_{\rm eff}(c). Set cc^{\prime} to be another conductance with

c(e){>c(e0),e=e0,=c(e),ee0.c^{\prime}(e)\left\{\begin{aligned} &>c(e_{0}),&e=e_{0},\\ &=c(e),&e\neq e_{0}.\end{aligned}\right.

If i(e0)0i(\vec{e_{0}})\neq 0, we may observe that Ec(i)<Ec(i)E_{c^{\prime}}(i)<E_{c}(i). Since the unit current flow ii^{\prime} on the electrical network with conductance cc^{\prime} has a smaller energy than flow ii, we obtain

Reff(c)=Ec(i)Ec(i)<Ec(i).R_{\rm eff}(c^{\prime})=E_{c^{\prime}}(i^{\prime})\leq E_{c^{\prime}}(i)<E_{c}(i).

If i(e0)=0i(\vec{e_{0}})=0, we have Ec(i)=Ec(i)E_{c^{\prime}}(i)=E_{c}(i). Here we prove that ii is the unit current flow on the electrical network with conductance cc^{\prime}. Consider the potential ϕ\phi on the electrical network with conductance cc, then for any edge eEe\in E,

i(e)c(e)=ϕ(e).i(\vec{e})c(e)=\phi(\vec{e}).

This implies that ϕ(e0)=0\phi(\vec{e_{0}})=0. On the electrical network with conductance cc^{\prime}, ϕ\phi and ii satisfy Kirchhoff’s potential law and Kirchhoff’s current law. As to Ohm’s law, for ee0e\neq e_{0},

i(e)c(e)=i(e)c(e)=ϕ(e).i(\vec{e})c^{\prime}(e)=i(\vec{e})c(e)=\phi(\vec{e}).

For e=e0e=e_{0},

i(e0)c(e0)=0=ϕ(e).i(\vec{e_{0}})c^{\prime}(e_{0})=0=\phi(\vec{e}).

Therefore, ii is a unit current flow on the electrical network with conductance cc^{\prime}, which implies

Reff(c)=Ec(i)=Ec(i).R_{\rm eff}(c^{\prime})=E_{c^{\prime}}(i)=E_{c}(i).

 

Proposition 3.6.

[11] The negative correlation holds for uniform spanning trees on finite connected graph GG.

Proof. We only need to verify that for any two distinct edges e1,e2e_{1},e_{2},

G[e1e2]G[e1]G[e2].\mathbb{P}_{\infty}^{G}[e_{1}e_{2}]\leq\mathbb{P}_{\infty}^{G}[e_{1}]\mathbb{P}_{\infty}^{G}[e_{2}].

This inequality is equivalent to

G/e2[e1]=G[e1e2]G[e2]G[e1],\mathbb{P}_{\infty}^{G/e_{2}}[e_{1}]=\frac{\mathbb{P}_{\infty}^{G}[e_{1}e_{2}]}{\mathbb{P}_{\infty}^{G}[e_{2}]}\leq\mathbb{P}_{\infty}^{G}[e_{1}],

where G/e2G/e_{2} stands for the contraction of GG by removing e2e_{2} and identifying its endpoints. By Proposition 3.4, this inequality is equivalent to

Reff(G/e2)Reff(G),R_{\rm eff}(G/e_{2})\leq R_{\rm eff}(G), (3.3)

where the unit current flow is from e1e_{1}^{-} to e1+e_{1}^{+}, and e1,e1+e_{1}^{-},e_{1}^{+} are endpoints of e1e_{1}. Actually, (3.3) holds since G/e2G/e_{2} can be regarded as graph GG with resistance 0 on e2e_{2}, leading to a smaller effective resistance.  

4 Proof of Theorem 2.1

For finite connected graph GG, set T[1]T[1] to be the number of spanning trees on GG. Then for each spanning tree, it has the probability 1/T[1]1/T[1]. For edge sets S1S_{1} and S2S_{2}, let T[S1S2¯]T[S_{1}\bar{S_{2}}] be the number of all spanning trees containing S1S_{1} and disjoint to S2S_{2}. Consider the Arboreal Gas with parameter β\beta. Observe that the highest term of μ[S1S2¯]\mu[S_{1}\bar{S_{2}}] in β\beta for some edge eFe\in F should be

T[S1S2¯]β|V|1.T[S_{1}\bar{S_{2}}]\beta^{|V|-1}.

If we consider the negative correlation of Arboreal Gas, i.e.,

μ[e1]μ[e2]μ[e1e2]μ[1]0,\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1]\geq 0,

we pay attention to the highest term of the left hand side of this inequality if β\beta is large enough. Here the highest term is

(T[e1]T[e2]T[e1e2]T[1])β2(|V|1).(T[e_{1}]T[e_{2}]-T[e_{1}e_{2}]T[1])\beta^{2(|V|-1)}.

By Proposition 3.6, this value should be non-negative. What we want is to show

T[e1]T[e2]T[e1e2]T[1]>0,T[e_{1}]T[e_{2}]-T[e_{1}e_{2}]T[1]>0,

by which we conclude the negative correlation for β\beta large enough. However, this does not always hold, such as that on trees. To prove Theorem 2.1, for adjacent edges e1e_{1} and e2e_{2}, we consider in two situations:

  • (a)

    there is a simple cycle crossing both e1e_{1} and e2e_{2};

  • (b)

    there is no simple cycle crossing both e1e_{1} and e2e_{2}.

In Case (a), if we start a unit current flow ii from e1e_{1}^{-} to e1+e_{1}^{+}, then i(e2)i(\vec{e_{2}}) is not vanishing.

Lemma 4.1.

For two adjacent edges e1,e2Ee_{1},e_{2}\in E, set ii to be the unit current flow from e1e_{1}^{-} to e1+e_{1}^{+}. If there is a simple cycle on GG crossing both e1e_{1} and e2e_{2}, then i(e2)i(\vec{e_{2}}) is not vanishing.

In Case (b), we prove that e1,e2e_{1},e_{2} can be separated into two parts of the graph GG, where the intersection of these two parts has only one vertex. By this property, the probability of Arboreal Gas on GG can be expressed as the product of the probabilities of Arboreal Gas on these two parts.

Lemma 4.2.

For two adjacent edges e1,e2Ee_{1},e_{2}\in E, assume that ei,ei+e_{i}^{-},e_{i}^{+} are two endpoints of edge eie_{i} for i=1,2i=1,2, where e1+=e2+e_{1}^{+}=e_{2}^{+}. If there is no simple cycle on G=(V,E)G=(V,E) crossing both e1e_{1} and e2e_{2}, then there exist subgraphs Gi=(Vi,Ei)G_{i}=(V_{i},E_{i}) for i=1,2i=1,2 such that

  • eiEie_{i}\in E_{i} for i=1,2i=1,2;

  • V1V2={e1+}V_{1}\cap V_{2}=\{e_{1}^{+}\}, E1E2=E_{1}\cap E_{2}=\emptyset;

  • V1V2=VV_{1}\cup V_{2}=V, E1E2=EE_{1}\cup E_{2}=E.

Now we prove Theorem 2.1.

Proof of Theorem 2.1.

(a). If there is a simple cycle crossing both e1e_{1} and e2e_{2}, by Lemmas 4.1 and 3.5, we obtain

Reff(G/e2)<Reff(G)R_{\rm eff}(G/e_{2})<R_{\rm eff}(G)

for the unit current flow from e1e_{1}^{-} to e1+e_{1}^{+}. Then the highest term in μ[e1]μ[e2]μ[e1e2]μ[1]\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1] is positive, i.e.,

(T[e1]T[e2]T[e1e2]T[1])β2|V|2=(Reff(G)Reff(G/e2))T[e2]T[1]β2|V|2>0,(T[e_{1}]T[e_{2}]-T[e_{1}e_{2}]T[1])\cdot\beta^{2|V|-2}=(R_{\rm eff}(G)-R_{\rm eff}(G/e_{2}))\cdot T[e_{2}]T[1]\beta^{2|V|-2}>0,

where the equality comes from Proposition 3.4. This implies the negative correlation for e1,e2e_{1},e_{2}, i.e., μ[e1]μ[e2]μ[e1e2]μ[1]0\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1]\geq 0, for sufficiently large β\beta.

(b). If there is no simple cycle crossing both e1e_{1} and e2e_{2} (assume e1+=e2+e_{1}^{+}=e_{2}^{+}), by Lemma 4.2, we separate GG into two parts G1=(V1,E1)G_{1}=(V_{1},E_{1}) and G2=(V2,E2)G_{2}=(V_{2},E_{2}) with

  • eiEie_{i}\in E_{i} for i=1,2i=1,2;

  • V1V2={e1+}V_{1}\cap V_{2}=\{e_{1}^{+}\}, E1E2=E_{1}\cap E_{2}=\emptyset;

  • V1V2=VV_{1}\cup V_{2}=V, E1E2=EE_{1}\cup E_{2}=E.

We claim that

there is no simple cycle crossing both edges in E1 and E2.\text{there is no simple cycle crossing both edges in }E_{1}\text{ and }E_{2}.

By this claim, we obtain that the subgraph FF in GG is a forest is equivalent to that F|GiF|_{G_{i}} are forests for i=1,2i=1,2, where F|Gi=(V(F)Vi,E(F)Ei)F|_{G_{i}}=(V(F)\cap V_{i},E(F)\cap E_{i}). Therefore, for i=1,2i=1,2, if we denote by μi\mu_{i} the measure for Arboreal Gas on GiG_{i}, i.e., the probability multiplying its partition function, then

μ[e1e2¯]=μ1[e1]μ2[e2¯],\displaystyle\mu[e_{1}\bar{e_{2}}]=\mu_{1}[e_{1}]\mu_{2}[\bar{e_{2}}],
μ[e1¯e2]=μ1[e1¯]μ2[e2],\displaystyle\mu[\bar{e_{1}}e_{2}]=\mu_{1}[\bar{e_{1}}]\mu_{2}[e_{2}],
μ[e1e2]=μ1[e1]μ2[e2],\displaystyle\mu[e_{1}e_{2}]=\mu_{1}[e_{1}]\mu_{2}[e_{2}],
μ[e1¯e2¯]=μ1[e1¯]μ2[e2¯].\displaystyle\mu[\bar{e_{1}}\bar{e_{2}}]=\mu_{1}[\bar{e_{1}}]\mu_{2}[\bar{e_{2}}].

By these properties, we deduce that

μ[e1e2¯]μ[e1¯e2]μ[e1e2]μ[e1¯e2¯]=0,\mu[e_{1}\bar{e_{2}}]\mu[\bar{e_{1}}e_{2}]-\mu[e_{1}e_{2}]\mu[\bar{e_{1}}\bar{e_{2}}]=0,

which implies the negative correlation for e1,e2e_{1},e_{2}.

Now we prove the claim by contradiction. Otherwise, there is a simple cycle C=u1u2unun+1C=u_{1}u_{2}\dots u_{n}u_{n+1} with un+1:=u1u_{n+1}:=u_{1} such that ukiuki+1Eiu_{k_{i}}u_{k_{i}+1}\in E_{i} for i=1,2i=1,2 and some k1k2k_{1}\neq k_{2}. Without loss of generality, assume 1k1<k2n1\leq k_{1}<k_{2}\leq n. Since E1E2=EE_{1}\cup E_{2}=E, there should be two distinct integers n1,n2n_{1},n_{2} with k1n1<k2k_{1}\leq n_{1}<k_{2} and n2<k1n_{2}<k_{1} or n2k2n_{2}\geq k_{2} such that un1un1+1,un2+1un2+2E1u_{n_{1}}u_{n_{1}+1},u_{n_{2}+1}u_{n_{2}+2}\in E_{1} and un1+1un2+2,un2un2+1E2u_{n_{1}+1}u_{n_{2}+2},u_{n_{2}}u_{n_{2}+1}\in E_{2}. This implies un1+1,un2+1V1V2u_{n_{1}+1},u_{n_{2}+1}\in V_{1}\cap V_{2}, i.e., un1+1=un2+1=e1+u_{n_{1}+1}=u_{n_{2}+1}=e_{1}^{+}, which is contradict to n1n2n_{1}\neq n_{2} and that CC is a simple cycle.  

Finally we prove Lemmas 4.1 and 4.2.

Proof of Lemma 4.1. Without loss of generality, assume e1e_{1}^{-} is a endpoint of e2e_{2}. We prove by contradiction. If i(e2)=0i(\vec{e_{2}})=0, we consider the simple cycle π\pi crossing both e1e_{1} and e2e_{2}, and set

π=e1u1u2une1+e1.\pi=e_{1}^{-}u_{1}u_{2}\dots u_{n}e_{1}^{+}e_{1}^{-}.

Note that the potentials ϕ\phi on e1,e1+e_{1}^{-},e_{1}^{+} are different. Precisely, ϕ(e1)>ϕ(e1+)=0\phi(e_{1}^{-})>\phi(e_{1}^{+})=0, which implies i(e1e1+)>0i(\overrightarrow{e_{1}^{-}e_{1}^{+}})>0. By Kirchhoff’s potential law and Ohm’s law, we know that there should be some edge ee1e\neq e_{1} with i(e)0i(\vec{e})\neq 0. Let m:=inf{k1:i(ukuk+1)0}m:=\inf\{k\geq 1:i(\overrightarrow{u_{k}u_{k+1}})\neq 0\}, where un+1:=e1+u_{n+1}:=e_{1}^{+}. By Kirchhoff’s current law, we deduce that there is a vertex v1v_{1} adjacent to v0:=umv_{0}:=u_{m} such that i(v0v1)<0i(\overrightarrow{v_{0}v_{1}})<0. By Kirchhoff’s potential law, v1e1+v_{1}\neq e_{1}^{+}. Otherwise, potentials on the cycle v0v1e1u1um1v0v_{0}v_{1}e_{1}^{-}u_{1}\dots u_{m-1}v_{0} do not obey Kirchhoff’s potential law. Now we construct a chain by induction:

For integer k1k\geq 1, assume there is a chain v0v1vkv_{0}v_{1}\dots v_{k} with i(vj1vj)<0i(\overrightarrow{v_{j-1}v_{j}})<0 for j=1,,kj=1,\dots,k and vj1vj2e1+v_{j_{1}}\neq v_{j_{2}}\neq e_{1}^{+} for distinct j1,j2=1,,kj_{1},j_{2}=1,\dots,k. By Kirchhoff’s current law, we can find a vertex vk+1v_{k+1} adjacent to vkv_{k} with i(vj1vj)<0i(\overrightarrow{v_{j-1}v_{j}})<0. Furthermore, vk+1vj,e1+v_{k+1}\neq v_{j},e_{1}^{+} for j=0,,kj=0,\dots,k. Otherwise, if vk+1=e1+v_{k+1}=e_{1}^{+}, potentials on the cycle v0v1vke1u1um1v0v_{0}v_{1}\dots v_{k}e_{1}^{-}u_{1}\dots u_{m-1}v_{0} do not obey Kirchhoff’s potential law. if vk+1=vjv_{k+1}=v_{j} for some j=0,1,,kj=0,1,\dots,k, potentials on the cycle vjvj+1vkvjv_{j}v_{j+1}\dots v_{k}v_{j} do not obey Kirchhoff’s potential law.

Thus we construct a chain v0v|V|v_{0}\dots v_{|V|} with vj1vj2v_{j_{1}}\neq v_{j_{2}} for distinct j1,j2=0,,|V|j_{1},j_{2}=0,\dots,|V|. That is, we find |V|+1|V|+1 different vertices, which is contradict to that the number of all vertices of GG is |V||V|.  

Proof of Lemma 4.2. Without loss of generality, assume GG is connected. For i=1,2i=1,2, set Gi=(Vi,Ei)G_{i}=(V_{i},E_{i}) to be the graph induced by the vertex set

Vi:={ei+,ei}{uV:thereisasimplepathfromeitounotvisitingei+}.V_{i}:=\{e_{i}^{+},e_{i}^{-}\}\cup\{u\in V:{\rm there\ is\ a\ simple\ path\ from\ }e_{i}^{-}\ {\rm to\ }u\ {\rm not\ visiting\ }e_{i}^{+}\}.

Set G3=(V3,E3)G_{3}=(V_{3},E_{3}) to be the graph induced by

V3:={e1+}[V(V1V2)].V_{3}:=\{e_{1}^{+}\}\cup[V\setminus(V_{1}\cup V_{2})].

Now we prove that G1,G2G3G_{1},G_{2}\cup G_{3} satisfy conditions in Lemma 4.2 by showing the following properties.

  • eiEie_{i}\in E_{i} for i=1,2i=1,2.

    Since ei,ei+Vie_{i}^{-},e_{i}^{+}\in V_{i} for i=1,2i=1,2, eiEie_{i}\in E_{i}.

  • ViVj={e1+}V_{i}\cap V_{j}=\{e_{1}^{+}\} and EiEj=E_{i}\cap E_{j}=\emptyset for distinct i,j=1,2,3i,j=1,2,3.

    If there is an edge eEiEje\in E_{i}\cap E_{j} for some distinct i,j=1,2,3i,j=1,2,3, then two endpoints of ee belong to ViV_{i} and VjV_{j}, contradict to ViVj={e1+}V_{i}\cap V_{j}=\{e_{1}^{+}\}. Thus we only need to show ViVj={e1+}V_{i}\cap V_{j}=\{e_{1}^{+}\} for distinct i,j=1,2,3i,j=1,2,3.

    For any i=1,2i=1,2, ViV3={e1+}V_{i}\cap V_{3}=\{e_{1}^{+}\} by the definition of V3V_{3}. We then show V1V2={e1+}V_{1}\cap V_{2}=\{e_{1}^{+}\}.

    For any vertex vV1v\in V_{1} with ve1+v\neq e_{1}^{+}, set π=e1u1unv\pi=e_{1}^{-}u_{1}\dots u_{n}v to be the simple path from e1e_{1}^{-} to vv not visiting e1+e_{1}^{+}. Then e2πe_{2}^{-}\notin\pi, otherwise there is some kk with uk=e2u_{k}=e_{2}^{-} and there will be a simple cycle e1+u1uke1+e_{1}^{+}u_{1}\dots u_{k}e_{1}^{+} crossing e1,e2e_{1},e_{2}, contradict to the condition in Lemma 4.2.

    We prove vV2v\notin V_{2} by contradiction. If vV2v\in V_{2}, vπv\in\pi implies ve2v\neq e_{2}^{-}. Therefore, there is a simple path π=v0v1vne2\pi^{\prime}=v_{0}v_{1}\dots v_{n^{\prime}}e_{2}^{-} from vv to e2e_{2}^{-} not visiting e1+e_{1}^{+}, where v0:=vv_{0}:=v. Set k:=sup{kn:vkπ}k^{\prime}:=\sup\{k\leq n^{\prime}:v_{k}\in\pi\}. Assume vk=ukv_{k^{\prime}}=u_{k} for some kk, where u0:=e1u_{0}:=e_{1}^{-} and un+1:=vu_{n+1}:=v. Then e1+u0ukvk+1vne2e1+e_{1}^{+}u_{0}\dots u_{k}v_{k^{\prime}+1}\dots v_{n^{\prime}}e_{2}^{-}e_{1}^{+} is a simple cycle crossing e1,e2e_{1},e_{2}, contradict to the condition in Lemma 4.2.

    Since we choose vV1v\in V_{1} arbitrarily, we obtain V1V2={e1+}V_{1}\cap V_{2}=\{e_{1}^{+}\}.

  • V1V2V3=VV_{1}\cup V_{2}\cup V_{3}=V, E1E2E3=EE_{1}\cup E_{2}\cup E_{3}=E.

    By definition of V3V_{3}, we immediately obtain V1V2V3=VV_{1}\cup V_{2}\cup V_{3}=V. Since ViVj={e1+}V_{i}\cap V_{j}=\{e_{1}^{+}\} for distinct i,j=1,2,3i,j=1,2,3, we show E1E2E3=EE_{1}\cup E_{2}\cup E_{3}=E by proving that there is no edge connecting Vi,VjV_{i},V_{j} except those edges with endpoint e1+e_{1}^{+}.

    For viVi{e1+}v_{i}\in V_{i}\setminus\{e_{1}^{+}\}, i=1,2i=1,2, and v3Vi{e1+}v_{3}\in V_{i}\setminus\{e_{1}^{+}\}, set πi\pi_{i} to be the simple path from eie_{i}^{-} to viv_{i} not visiting e1+e_{1}^{+}. Assume that there is an edge connecting vi,vjv_{i},v_{j}, i=1,2i=1,2, jij\neq i. If vjπiv_{j}\in\pi_{i}, then there is a simple path from eie_{i}^{-} to vjv_{j} not visiting e1+e_{1}^{+}, implying vjViv_{j}\in V_{i}. Otherwise vjπiv_{j}\notin\pi_{i}, then πvj\pi v_{j} is a simple path from eie_{i}^{-} to vjv_{j} not visiting ei+e_{i}^{+}, which also implies vjViv_{j}\in V_{i}. This is contradict to vjVj{e1+}v_{j}\in V_{j}\setminus\{e_{1}^{+}\} and ViVj={e1+}V_{i}\cap V_{j}=\{e_{1}^{+}\}.

By these three properties, we shows that

  • e1E1e_{1}\in E_{1}, e2E2E3e_{2}\in E_{2}\cup E_{3};

  • V1(V2V3)={e1+}V_{1}\cap(V_{2}\cup V_{3})=\{e_{1}^{+}\}, E1(E2E3)=E_{1}\cap(E_{2}\cup E_{3})=\emptyset;

  • V1(V2V3)=VV_{1}\cup(V_{2}\cup V_{3})=V, E1(E2E3)=EE_{1}\cup(E_{2}\cup E_{3})=E.

This completes the proof.  

5 Proof of Theorem 2.2

By Theorem 2.1, we only need to show the negative correlation for disjoint edges e1,e2e_{1},e_{2} in complete graph KnK_{n}. Consider the highest term of

μ[e1]μ[e2]μ[e1e2]μ[1].\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1].

We observe that the highest term (T[e1]T[e2]T[e1e2]T[1])β2|V|2(T[e_{1}]T[e_{2}]-T[e_{1}e_{2}]T[1])\beta^{2|V|-2} is vanishing. Because for any adjacent vertices u,vu,v, the unit current flow ii from e1e_{1}^{-} to e1+e_{1}^{+} is

i(uv)={2/n,u=e1,v=e1+,1/n,u=e1,ve1+, or ue1,v=e1+0,u,v{e1,e1+}.i(\overrightarrow{uv})=\left\{\begin{aligned} &2/n,&u=e_{1}^{-},v=e_{1}^{+},\\ &1/n,&u=e_{1}^{-},v\neq e_{1}^{+},\text{ or }u\neq e_{1}^{-},v=e_{1}^{+}\\ &0,&u,v\notin\{e_{1}^{-},e_{1}^{+}\}.\end{aligned}\right.

By Lemma 3.5 and Proposition 3.4,

T[e1]T[e2]T[e1e2]T[1]=T[e2]T[1](Reff(G)Reff[G/e2])=0.T[e_{1}]T[e_{2}]-T[e_{1}e_{2}]T[1]=T[e_{2}]T[1](R_{\rm eff}(G)-R_{\rm eff}[G/e_{2}])=0.

Therefore, in order to verify the negative correlation, we consider the second highest term of μ[e1]μ[e2]μ[e1e2]μ[1]\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1].

To give a more precise estimate, we use the following property.

Proposition 5.1 (Kirchhoff’s Matrix-tree Theory, [6]).

Let LL be the Laplacian matrix for the graph GG defined by

Lij={c(ij),ij,kic(ik),i=j.L_{ij}=\left\{\begin{aligned} &-c(ij),&i\neq j,\\ &\sum_{k\neq i}c(ik),&i=j.\end{aligned}\right.

Let L(i)L(i) be the matrix given by deleting ii-th row and column of LL. Then for any ii, the determinate of L(i)L(i) can be expressed as

detL(i)=Tis a spanning tree of G(eTc(e)).\det L(i)=\sum_{T\ \text{\rm is a spanning tree of }G}\left(\prod_{e\in T}c(e)\right).

By Kirchhoff’s Matrix-tree Theory, we give the following lemma to show the number of spanning trees on complete graph KnK_{n}.

Lemma 5.2.

Set Tn(A)T_{n}(A) to be the number of spanning trees on KnK_{n} satisfying event AA. Then for any disjoint edges e1,e2e_{1},e_{2} in KnK_{n},

Tn[1]=nn2,Tn[e1]=Tn[e2]=2nn3,Tn[e1e2]=4nn4.T_{n}[1]=n^{n-2},\ T_{n}[e_{1}]=T_{n}[e_{2}]=2n^{n-3},\ T_{n}[e_{1}e_{2}]=4n^{n-4}.

Proof. By Kirchhoff’s Matrix-tree Theory, we compute out Tn[1]T_{n}[1] directly. For Tn[e1]T_{n}[e_{1}] (this equals Tn[e2]T_{n}[e_{2}] by symmetry), there is a bijection from spanning trees on KnK_{n} containing e1e_{1} to spanning trees on Kn/e1K_{n}/e_{1}. Here for any positive integer kk, each multiple edge with multiplicity kk in Kn/e1K_{n}/e_{1} can be regarded as an edge with conductance kk. Precisely, when we consider the number of spanning trees, Kn/e1K_{n}/e_{1} can be regarded as Kn1K_{n-1} with conductance

c(uv)={2,u=v0,or v=v0,1,otherwise,c(uv)=\left\{\begin{aligned} &2,&u=v_{0},\ \text{or }v=v_{0},\\ &1,&\text{otherwise},\end{aligned}\right.

where v0v_{0} is the vertex contracted from edge e1e_{1}. Based on Kirchhoff’s Matrix-tree Theory, we compute out Tn[e1]=2nn3T_{n}[e_{1}]=2n^{n-3}.

For Tn[e1e2]T_{n}[e_{1}e_{2}], similarly, we consider Kn/e1,e2K_{n}/{e_{1},e_{2}}, which can be regarded as Kn2K_{n-2} with conductance

c(uv)={4,{u,v}={v1,v2},2,u{v1,v2},vv1,v2,or v{v1,v2},uv1,v2,1,otherwise,c(uv)=\left\{\begin{aligned} &4,&\{u,v\}=\{v_{1},v_{2}\},\\ &2,&u\in\{v_{1},v_{2}\},v\neq v_{1},v_{2},\ \text{or }v\in\{v_{1},v_{2}\},u\neq v_{1},v_{2},\\ &1,&\text{otherwise},\end{aligned}\right.

where viv_{i} is the vertex contracted from edge eie_{i} for i=1,2i=1,2. By Kirchhoff’s Matrix-tree Theory, we compute out Tn[e1e2]=4nn4T_{n}[e_{1}e_{2}]=4n^{n-4}.  

By Lemma 5.2, we verify that the second highest term of μ[e1]μ[e2]μ[e1e2]μ[1]\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1] is positive for disjoint edges e1,e2e_{1},e_{2} in KnK_{n}.

Lemma 5.3.

For disjoint edges e1,e2e_{1},e_{2} in KnK_{n}, if nn is large enough, then the coefficient of β2|V|3\beta^{2|V|-3} in μ[e1]μ[e2]μ[e1e2]μ[1]\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1] is positive.

By Lemma 5.3, we prove Theorem 2.2.

Proof of Theorem 2.2. For adjacent two edges e1,e2e_{1},e_{2}, by Theorem 2.1, we get the negative correlation for large enough β\beta. For disjoint two edges e1,e2e_{1},e_{2}, consider μ[e1]μ[e2]μ[e1e2]μ[1]\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1]. By Lemma 5.2, we deduce that the highest term should be

(Tn[e1]Tn[e2]Tn[e1e2]Tn[1])β2|V|2=(4n2n64n2n6)β2|V|2=0.(T_{n}[e_{1}]T_{n}[e_{2}]-T_{n}[e_{1}e_{2}]T_{n}[1])\beta^{2|V|-2}=(4n^{2n-6}-4n^{2n-6})\beta^{2|V|-2}=0.

By Lemma 5.3, the second highest term should be positive for sufficiently large nn, which implies the negative correlation for sufficiently large β\beta.  

Here we give the proof of Lemma 5.3.

Proof of Lemma 5.3. First we consider the second highest term of μ[S]\mu[S] for any edge set SS. This value should be

12VV,VV,FV[S]β|V|2.\frac{1}{2}\sum_{V^{\prime}\subset V,V^{\prime}\neq V,\emptyset}F_{V^{\prime}}[S]\cdot\beta^{|V|-2}.

Here FV[S]F_{V^{\prime}}[S] is the number of spanning forests containing SS such that there is exactly two trees T,TT,T^{\prime} in the forest, where TT is the spanning tree of VV^{\prime}, and TT^{\prime} is the spanning tree of VVV\setminus V^{\prime}.

We then estimate the coefficient of β2|V|3\beta^{2|V|-3} in μ[e1]μ[e2]μ[e1e2]μ[1]\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1], which equals

12VV,VV,Tn(e1)FV(e2)+Tn(e2)FV(e1)Tn(e1e2)FV(1)Tn(1)FV(e1e2)\displaystyle\frac{1}{2}\sum_{V^{\prime}\subset V,V^{\prime}\neq V,\emptyset}T_{n}(e_{1})F_{V^{\prime}}(e_{2})+T_{n}(e_{2})F_{V^{\prime}}(e_{1})-T_{n}(e_{1}e_{2})F_{V^{\prime}}(1)-T_{n}(1)F_{V^{\prime}}(e_{1}e_{2})
=k=1[n/2]VV,|V|=kTn(e1)FV(e2)+Tn(e2)FV(e1)Tn(e1e2)FV(1)Tn(1)FV(e1e2).\displaystyle=\sum_{k=1}^{[n/2]}\sum_{V^{\prime}\subset V,|V^{\prime}|=k}T_{n}(e_{1})F_{V^{\prime}}(e_{2})+T_{n}(e_{2})F_{V^{\prime}}(e_{1})-T_{n}(e_{1}e_{2})F_{V^{\prime}}(1)-T_{n}(1)F_{V^{\prime}}(e_{1}e_{2}).

Here we write Tn(e1)FV(e2)+Tn(e2)FV(e1)Tn(e1e2)FV(1)Tn(1)FV(e1e2)T_{n}(e_{1})F_{V^{\prime}}(e_{2})+T_{n}(e_{2})F_{V^{\prime}}(e_{1})-T_{n}(e_{1}e_{2})F_{V^{\prime}}(1)-T_{n}(1)F_{V^{\prime}}(e_{1}e_{2}) as aVa_{V^{\prime}} For the term with |V|=k|V^{\prime}|=k in this sum, denoted by ak:=VV,|V|=kaVa_{k}:=\sum_{V^{\prime}\subset V,|V^{\prime}|=k}a_{V^{\prime}}, we compute it in the following different cases:

  1. 1.

    the endpoints of e1,e2e_{1},e_{2} are in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals (n4k4)4nn4kk4(nk)nk-\binom{n-4}{k-4}4n^{n-4}k^{k-4}(n-k)^{n-k};

  2. 2.

    the endpoints of e1,e2e_{1},e_{2} are not in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals (n4k)4nn4kk(nk)nk4-\binom{n-4}{k}4n^{n-4}k^{k}(n-k)^{n-k-4};

  3. 3.

    all the endpoints of e1e_{1}, and one endpoint of e2e_{2} are in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals

    (n4k3)8nn4kk3(nk)nk1\binom{n-4}{k-3}8n^{n-4}k^{k-3}(n-k)^{n-k-1};

  4. 4.

    all the endpoints of e2e_{2}, and one endpoint of e1e_{1} are in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals

    (n4k3)8nn4kk3(nk)nk1\binom{n-4}{k-3}8n^{n-4}k^{k-3}(n-k)^{n-k-1};

  5. 5.

    only one endpoint of e1e_{1} is in VV^{\prime}, and the endpoints of e2e_{2} are not in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals (n4k1)8nn4kk1(nk)nk3\binom{n-4}{k-1}8n^{n-4}k^{k-1}(n-k)^{n-k-3};

  6. 6.

    only one endpoint of e2e_{2} is in VV^{\prime}, and the endpoints of e1e_{1} are not in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals (n4k1)8nn4kk1(nk)nk3\binom{n-4}{k-1}8n^{n-4}k^{k-1}(n-k)^{n-k-3};

  7. 7.

    the endpoints of e1e_{1} are in VV^{\prime}, and the endpoints of e2e_{2} are not in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals (n4k2)4nn4kk2(nk)nk2-\binom{n-4}{k-2}4n^{n-4}k^{k-2}(n-k)^{n-k-2};

  8. 8.

    the endpoints of e2e_{2} are in VV^{\prime}, and the endpoints of e1e_{1} are not in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals (n4k2)4nn4kk2(nk)nk2-\binom{n-4}{k-2}4n^{n-4}k^{k-2}(n-k)^{n-k-2};

  9. 9.

    one endpoint of e1e_{1}, and one endpoint of e2e_{2} are in VV^{\prime}, then the sum of aVa_{V^{\prime}} equals

    (n4k2)16nn4kk2(nk)nk2-\binom{n-4}{k-2}16n^{n-4}k^{k-2}(n-k)^{n-k-2}.

Based on these computations, we obtain aka_{k} by summing over all values:

ak=12nn3(n4k1)kk4(nk)nk4k(n+6)(nk)+2n2(nk1)(nk2).a_{k}=12n^{n-3}\binom{n-4}{k-1}k^{k-4}(n-k)^{n-k-4}\frac{-k(n+6)(n-k)+2n^{2}}{(n-k-1)(n-k-2)}.

For k=1k=1, a1=12nn3(n1)n5a_{1}=12n^{n-3}(n-1)^{n-5}. For k2k\geq 2, aka_{k} can be also written as

12nn3(n1)n5[(n4)!(k1)!(nk1)!kk4(nk)nk4k(n+6)(nk)2n2(n1)n5].-12n^{n-3}(n-1)^{n-5}\cdot\left[\frac{(n-4)!}{(k-1)!(n-k-1)!}k^{k-4}(n-k)^{n-k-4}\frac{k(n+6)(n-k)-2n^{2}}{(n-1)^{n-5}}\right].

For convenience, set

Ik=(n4)!(k1)!(nk1)!kk4(nk)nk4k(n+6)(nk)2n2(n1)n5.I_{k}=\frac{(n-4)!}{(k-1)!(n-k-1)!}k^{k-4}(n-k)^{n-k-4}\frac{k(n+6)(n-k)-2n^{2}}{(n-1)^{n-5}}.

We prove that for nn large enough, k=2[n/2]Ik<1\sum_{k=2}^{[n/2]}I_{k}<1.

By Stirling’s approximation, for sufficiently large nn and k=pnk=pn for p[2/n,12]p\in[2/n,\frac{1}{2}],

Ik\displaystyle I_{k} (n4)!(nk1)!(k1)!kk3(nk)nk3(n+6)1(n1)n5\displaystyle\leq\frac{(n-4)!}{(n-k-1)!(k-1)!}k^{k-3}(n-k)^{n-k-3}(n+6)\frac{1}{(n-1)^{n-5}}
2π(n4)2π(nk1)2π(k1)(n4)n4(nk1)nk1(k1)k1e2kk3(nk)nk3n+6(n1)(n5)\displaystyle\sim\frac{\sqrt{2\pi(n-4)}}{\sqrt{2\pi(n-k-1)}\sqrt{2\pi(k-1)}}\cdot\frac{(n-4)^{n-4}}{(n-k-1)^{n-k-1}(k-1)^{k-1}}\cdot e^{2}k^{k-3}(n-k)^{n-k-3}\frac{n+6}{(n-1)^{(}n-5)}
=e22π(n4)n72(n1)n72(nk)nk12(nk1)nk12kk12(k1)k12(n1)32(n+6)(nk)521k52\displaystyle=\frac{e^{2}}{\sqrt{2\pi}}\cdot\frac{(n-4)^{n-\frac{7}{2}}}{(n-1)^{n-\frac{7}{2}}}\cdot\frac{(n-k)^{n-k-\frac{1}{2}}}{(n-k-1)^{n-k-\frac{1}{2}}}\cdot\frac{k^{k-\frac{1}{2}}}{(k-1)^{k-\frac{1}{2}}}\cdot\frac{(n-1)^{\frac{3}{2}}(n+6)}{(n-k)^{\frac{5}{2}}}\cdot\frac{1}{k^{\frac{5}{2}}}
e2π(n1)32(n+6)(nk)521k52.\displaystyle\sim\frac{e}{\sqrt{2\pi}}\cdot\frac{(n-1)^{\frac{3}{2}}(n+6)}{(n-k)^{\frac{5}{2}}}\cdot\frac{1}{k^{\frac{5}{2}}}.

Note that

k=2[n/2]1[k(nk)]52\displaystyle\sum_{k=2}^{[n/2]}\frac{1}{[k(n-k)]^{\frac{5}{2}}} 1n21[x(nx)]52𝑑x\displaystyle\leq\int_{1}^{\frac{n}{2}}\frac{1}{[x(n-x)]^{\frac{5}{2}}}dx
1n121[n2p(1p)]52𝑑np\displaystyle\leq\int_{\frac{1}{n}}^{\frac{1}{2}}\frac{1}{[n^{2}p(1-p)]^{\frac{5}{2}}}dnp
=16n4[23n22(n1)12+13n2(n2)8(n1)32].\displaystyle=\frac{16}{n^{4}}\cdot\left[\frac{2}{3}\cdot\frac{n-2}{2(n-1)^{\frac{1}{2}}}+\frac{1}{3}\cdot\frac{n^{2}(n-2)}{8(n-1)^{\frac{3}{2}}}\right].

By this estimate, for nn sufficiently large,

k=2[n/2]Ike2π23+o(1)<0.8+o(1).\sum_{k=2}^{[n/2]}I_{k}\leq\frac{e}{\sqrt{2\pi}}\cdot\frac{2}{3}+o(1)<0.8+o(1).

Since the coefficient of β2|V|3\beta^{2|V|-3} is 12nn3(n1)n5(1k=2[n/2]Ik)12n^{n-3}(n-1)^{n-5}(1-\sum_{k=2}^{[n/2]}I_{k}), we complete our proof.  

6 Proof of Theorem 2.6

Before the proof of our main theorem, we post some lemmas.

Lemma 6.1.

If ee is pivotal for GG, then negative correlation on G{e}G\setminus\{e\} implies negative correlation on GG.

Proof. Assume that the negative correlation holds on G{e}G\setminus\{e\}, i.e.,

μ[e1e2e¯]μ[e¯]μ[e1e¯]μ[e2e¯]fordistincte1,e2e.\mu[e_{1}e_{2}\bar{e}]\mu[\bar{e}]\leq\mu[e_{1}\bar{e}]\mu[e_{2}\bar{e}]{\rm\ for\ distinct\ }e_{1},e_{2}\neq e. (6.4)

Set μ[S]=μ[Se¯]\mu^{\prime}[S]=\mu[S\bar{e}]. Since ee is pivotal for GG, whether ee exists or not does not influence the existence of cycle in GG. Hence, μ[Se¯]=μ[Se]\mu[S\bar{e}]=\mu[Se] for eSe\notin S, by which we have

μ[S]={μ[Se¯]+βeμ[Se]=(1+βe)μ[S],eS,μ[(S{e})e]=μ[S{e}],eS.\mu[S]=\left\{\begin{aligned} &\mu[S\bar{e}]+\beta_{e}\mu[Se]=(1+\beta_{e})\mu^{\prime}[S],&e\notin S,\\ &\mu[(S\setminus\{e\})e]=\mu^{\prime}[S\setminus\{e\}],&e\in S.\end{aligned}\right.

Combined with (6.4), we verify that

μ[e1e2]μ[1]μ[e1]μ[e2]foralle1,e2E.\mu[e_{1}e_{2}]\mu[1]\leq\mu[e_{1}]\mu[e_{2}]{\rm\ for\ all\ }e_{1},e_{2}\in E.

 

Lemma 6.2.

If the negative correlation holds on each component of GG, then it holds on GG.

Proof. We only need to prove negative correlation for e1,e2e_{1},e_{2} in different components of GG since the existence of cycle in a component cannot be influenced by other components.

Assume that eiEie_{i}\in E_{i} for i=1,2i=1,2, where {Gi=(Vi,Ei):i=1,,d}\{G_{i}=(V_{i},E_{i}):i=1,\dots,d\} are all components of GG. Set μi\mu_{i} to be the measure on GiG_{i} for i=1,,di=1,\dots,d, then we deduce that (j=1,2j=1,2)

μ[e1e2]\displaystyle\mu[e_{1}e_{2}] =μ1[e1]μ2[e2]i=3dμi[1],\displaystyle=\mu_{1}[e_{1}]\mu_{2}[e_{2}]\prod_{i=3}^{d}\mu_{i}[1],
μ[ej]\displaystyle\mu[e_{j}] =μj[ej]ijμi[1],\displaystyle=\mu_{j}[e_{j}]\prod_{i\neq j}\mu_{i}[1],
μ[1]\displaystyle\mu[1] =i=1dμi[1].\displaystyle=\prod_{i=1}^{d}\mu_{i}[1].

This implies that μ[e1e2]μ[1]μ[e1]μ[e2]\mu[e_{1}e_{2}]\mu[1]\leq\mu[e_{1}]\mu[e_{2}].  

Lemma 6.3.

The negative correlation holds on G=(V,E)G=(V,E) for any parameter {βe}eE\{\beta_{e}\}_{e\in E} if and only if it holds on G~=(V~,E~)\tilde{G}=(\tilde{V},\tilde{E}) for any parameter {β~e~}e~E~\{\tilde{\beta}_{\tilde{e}}\}_{\tilde{e}\in\tilde{E}}, where E~\tilde{E} is given by Definition 2.4.

Lemma 6.4.

The negative correlation holds on a complex graph GG for any parameter {βe}eE\{\beta_{e}\}_{e\in E} if and only if it holds on the simple graph G:=g(G)G^{\prime}:=g(G) for any {βe}eg(E)\{\beta_{e^{\prime}}^{\prime}\}_{e^{\prime}\in g(E)}, where gg is defined in Definition 2.5.

The proof of Lemmas 6.3 and 6.4 will be stated in the next subsection. Now we prove Theorem 2.6.

Proof of Theorem 2.6. By Lemmas 6.1 and 6.2, we know that the negative correlation on GG is equivalent to that on each component of a new graph G^\hat{G} given by deleting all pivotal edges. Thus if e1e_{1} or e2e_{2} is pivotal edges, or e1,e2e_{1},e_{2} belong to different component of G^\hat{G}, the negative correlation for e1,e2e_{1},e_{2} holds. Without loss of generality, assume G^\hat{G} is connected.

By Lemmas 6.3 and 6.4, we know that the negative correlation on G^\hat{G} is equivalent to that on gf(G^)g\circ f(\hat{G}), which immediately shows the negative correlation for e1,e2e_{1},e_{2} with gf(e1)=gf(e2)g\circ f(e_{1})=g\circ f(e_{2}).  

Finally we give our proof of Lemmas 6.3 and 6.4.

Proof of Lemma 6.3. Recall the definition of ff in Definition 2.4. For some subgraph FF with edge set E(F)E(F), set F~\tilde{F} to be a graph induced by the following edge set E~f(EE(F))\tilde{E}\setminus f(E\setminus E(F)). We write the edge set of F~\tilde{F} as E~(F~)\tilde{E}(\tilde{F}). Intuitively, for any edge eEe\in E, assume ef1(e~)e\in f^{-1}(\tilde{e}) for some e~E~\tilde{e}\in\tilde{E}, if eFe\notin F, then e~E~(F~)\tilde{e}\notin\tilde{E}(\tilde{F}). We show that FF\in\mathcal{F} is equivalent to that F~\tilde{F} is a forest on G~\tilde{G}.

If FF\in\mathcal{F}, set f1(F~)f^{-1}(\tilde{F}) to be a graph induced by edge set {eE:f(e)E~(F~)}\{e\in E:f(e)\in\tilde{E}(\tilde{F})\}. Note that f1(F~)Ff^{-1}(\tilde{F})\subset F. If there is a cycle in F~\tilde{F}, there should also exist a cycle in f1(F~)f^{-1}(\tilde{F}), contradict to f1(F~)Ff^{-1}(\tilde{F})\subset F\in\mathcal{F}.

If F~\tilde{F} is a forest, assume there is a simple cycle π\pi in FF. For any eπe\in\pi, we claim that

f1(f(e))π.f^{-1}(f(e))\subset\pi.

By this claim we know that f(π)F~f(\pi)\in\tilde{F}. Since f(π)f(\pi) is also a cycle, it is contradict to that F~\tilde{F} is a forest.

Here we prove the claim. Otherwise, there are two adjacent edges e1,e2f1(f(e))e_{1},e_{2}\in f^{-1}(f(e)) with e1πe_{1}\in\pi and e2πe_{2}\notin\pi. Setting vv to be the vertex adjacent to e1,e2e_{1},e_{2}, we find that there is no simple cycle crossing vv in FF because the degree of vv is 11, contradict to e1πe_{1}\in\pi.

Set μ~\tilde{\mu} to be the measure of Arboreal Gas on G~\tilde{G}, where the corresponding parameter

β~e~=ef1(e~)βeef1(e~)(1+βe)ef1(e~)βe.\tilde{\beta}_{\tilde{e}}=\frac{\prod_{e\in f^{-1}(\tilde{e})}\beta_{e}}{\prod_{e\in f^{-1}(\tilde{e})}(1+\beta_{e})-\prod_{e\in f^{-1}(\tilde{e})}\beta_{e}}.

Then for any forest F~\tilde{F}^{\prime} in G~\tilde{G},

μ({F:F~=F~})=μ~(F~)e~E~[ef1(e~)(1+βe)ef1(e~)βe].\mu(\{F\in\mathcal{F}:\tilde{F}=\tilde{F}^{\prime}\})=\tilde{\mu}(\tilde{F}^{\prime})\cdot\prod_{\tilde{e}\in\tilde{E}}\left[\prod_{e\in f^{-1}(\tilde{e})}(1+\beta_{e})-\prod_{e\in f^{-1}(\tilde{e})}\beta_{e}\right].

Here we write e~E~[ef1(e~)(1+βe)ef1(e~)βe]\prod_{\tilde{e}\in\tilde{E}}\left[\prod_{e\in f^{-1}(\tilde{e})}(1+\beta_{e})-\prod_{e\in f^{-1}(\tilde{e})}\beta_{e}\right] as C=C(βe,eE)C=C(\beta_{e},e\in E).

\Leftarrow: Consider two distinct edges e1,e2Ee_{1},e_{2}\in E.
(a). If f(e1)=f(e2)=e~0f(e_{1})=f(e_{2})=\tilde{e}_{0}, set

μ~[e~0]=β~e~0F1(β~e~,e~e~0),μ~[e~¯0]=F2(β~e~,e~e~0),\tilde{\mu}[\tilde{e}_{0}]=\tilde{\beta}_{\tilde{e}_{0}}\cdot F_{1}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{0}),\ \tilde{\mu}[\bar{\tilde{e}}_{0}]=F_{2}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{0}),

where F2(β~e~,e~e~0)F1(β~e~,e~e~0)>0F_{2}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{0})\geq F_{1}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{0})>0. To verify the negative correlation on GG, we need to prove μ[e1]μ[e2]μ[e1e2]μ[1]0.\mu[e_{1}]\mu[e_{2}]-\mu[e_{1}e_{2}]\mu[1]\geq 0. This is equivalent to

μ[e1e2¯]μ[e1¯e2]μ[e1e2]μ[e1¯e2¯]0.\mu[e_{1}\bar{e_{2}}]\mu[\bar{e_{1}}e_{2}]-\mu[e_{1}e_{2}]\mu[\bar{e_{1}}\bar{e_{2}}]\geq 0. (6.5)

The right hand side of (6.5) equals

C2βe1ef1(e~0),ee1,e2(1+βe)F2βe2ef1(e~0),ee1,e2(1+βe)F2\displaystyle C^{2}\cdot\beta_{e_{1}}\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}(1+\beta_{e})\ F_{2}\cdot\beta_{e_{2}}\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}(1+\beta_{e})\ F_{2}
C2βe1βe2[(ef1(e~0),ee1,e2(1+βe)ef1(e~0),ee1,e2βe)F2+ef1(e~0),ee1,e2βeF1]\displaystyle-C^{2}\cdot\beta_{e_{1}}\beta_{e_{2}}\left[\left(\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}(1+\beta_{e})-\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}\beta_{e}\right)\ F_{2}+\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}\beta_{e}\ F_{1}\right]
ef1(e~0),ee1,e2(1+βe)F2\displaystyle\hskip 56.9055pt\cdot\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}(1+\beta_{e})\ F_{2}
=C2β1β2ef1(e~0),ee1,e2(1+βe)ef1(e~0),ee1,e2βeF2(F2F1).\displaystyle=C^{2}\beta_{1}\beta_{2}\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}(1+\beta_{e})\prod_{e\in f^{-1}(\tilde{e}_{0}),e\neq e_{1},e_{2}}\beta_{e}\ F_{2}(F_{2}-F_{1}).

By F2F1>0F_{2}\geq F_{1}>0, we finish the proof of (6.5).

(b). If f(e1)=e~1e~2=f(e2)f(e_{1})=\tilde{e}_{1}\neq\tilde{e}_{2}=f(e_{2}), set

μ~[e~1e~2]=β~e~1β~e~2F12(β~e~,e~e~1,e~2),\displaystyle\tilde{\mu}[\tilde{e}_{1}\tilde{e}_{2}]=\tilde{\beta}_{\tilde{e}_{1}}\tilde{\beta}_{\tilde{e}_{2}}F_{12}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{1},\tilde{e}_{2}),\ μ~[e~1e~¯2]=β~e~1F12¯(β~e~,e~e~1,e~2),\displaystyle\tilde{\mu}[\tilde{e}_{1}\bar{\tilde{e}}_{2}]=\tilde{\beta}_{\tilde{e}_{1}}F_{1\bar{2}}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{1},\tilde{e}_{2}),
μ~[e~¯1e~2]=β~e~2F1¯2(β~e~,e~e~1,e~2),\displaystyle\tilde{\mu}[\bar{\tilde{e}}_{1}\tilde{e}_{2}]=\tilde{\beta}_{\tilde{e}_{2}}F_{\bar{1}2}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{1},\tilde{e}_{2}),\ μ~[e~¯1e~¯2]=F1¯2¯(β~e~,e~e~1,e~2).\displaystyle\tilde{\mu}[\bar{\tilde{e}}_{1}\bar{\tilde{e}}_{2}]=F_{\bar{1}\bar{2}}(\tilde{\beta}_{\tilde{e}},\tilde{e}\neq\tilde{e}_{1},\tilde{e}_{2}).

By the negative correlation on G~\tilde{G}, we have

F12¯F1¯2F12F1¯2¯0.F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}\geq 0.

To verify the negative correlation on GG, we also prove (6.5). For convenience, for i=1,2i=1,2, we rewrite ef1(e~i),eeiβe\prod_{e\in f^{-1}(\tilde{e}_{i}),e\neq e_{i}}\beta_{e}, ef1(e~i),eei(1+βe)\prod_{e\in f^{-1}(\tilde{e}_{i}),e\neq e_{i}}(1+\beta_{e}) as C1(ei)C_{1}(e_{i}) and C2(ei)C_{2}(e_{i}) respectively. Note the right hand side of (6.5) equals

C2βe1[[C2(e1)C1(e1)]C2(e2)F1¯2¯+C1(e1)C2(e2)F12¯]βe2[[C2(e2)C1(e2)]C2(e1)F1¯2¯+C1(e2)C2(e1)F1¯2]\displaystyle C^{2}\cdot\beta_{e_{1}}\Bigg{[}\left[C_{2}(e_{1})-C_{1}(e_{1})\right]C_{2}(e_{2})F_{\bar{1}\bar{2}}+C_{1}(e_{1})C_{2}(e_{2})F_{1\bar{2}}\Bigg{]}\cdot\beta_{e_{2}}\Bigg{[}\left[C_{2}(e_{2})-C_{1}(e_{2})\right]C_{2}(e_{1})F_{\bar{1}\bar{2}}+C_{1}(e_{2})C_{2}(e_{1})F_{\bar{1}2}\Bigg{]}
C2βe1βe2[[C2(e1)C1(e1)][C2(e2)C1(e2)]F1¯2¯+C1(e1)[C2(e2)C1(e2)]F12¯\displaystyle-C^{2}\cdot\beta_{e_{1}}\beta_{e_{2}}\Bigg{[}[C_{2}(e_{1})-C_{1}(e_{1})][C_{2}(e_{2})-C_{1}(e_{2})]F_{\bar{1}\bar{2}}+C_{1}(e_{1})[C_{2}(e_{2})-C_{1}(e_{2})]F_{1\bar{2}}
+C1(e2)[C2(e1)C1(e1)]F1¯2+C1(e1)C1(e2)F12]C2(e1)C2(e2)F1¯2¯\displaystyle+C_{1}(e_{2})[C_{2}(e_{1})-C_{1}(e_{1})]F_{\bar{1}2}+C_{1}(e_{1})C_{1}(e_{2})F_{12}\Bigg{]}\cdot C_{2}(e_{1})C_{2}(e_{2})F_{\bar{1}\bar{2}}
=C2i=1,2βeiC1(ei)C2(ei)(F12¯F1¯2F12F1¯2¯),\displaystyle=C^{2}\prod_{i=1,2}\beta_{e_{i}}C_{1}(e_{i})C_{2}(e_{i})(F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}),

which is non-negative because F12¯F1¯2F12F1¯2¯0F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}\geq 0. This completes the proof of (6.5).

\Rightarrow: For distinct e~1,e~2E~\tilde{e}_{1},\tilde{e}_{2}\in\tilde{E}, by (b) in “\Leftarrow”, the negative correlation on GG is equivalent to

F12¯F1¯2F12F1¯2¯0.F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}\geq 0.

This shows immediately that

μ~[e~1e~¯2]μ~[e~¯1e~2]μ~[e~1e~2]μ~[e~¯1e~¯2]0,\tilde{\mu}[\tilde{e}_{1}\bar{\tilde{e}}_{2}]\tilde{\mu}[\bar{\tilde{e}}_{1}\tilde{e}_{2}]-\tilde{\mu}[\tilde{e}_{1}\tilde{e}_{2}]\tilde{\mu}[\bar{\tilde{e}}_{1}\bar{\tilde{e}}_{2}]\geq 0,

which completes the proof of negative correlation on G~\tilde{G}.  

Proof of Lemma 6.4. Recall that gg maps edge sets to edge sets by deleting multiple edges. For some subgraph F=(V(F),E(F))F=(V(F),E(F)) in GG, denote by FF^{\prime} the graph induced by edge set E(F):=g(E(F))E^{\prime}(F^{\prime}):=g(E(F)). If FF is a forest, there should not be any multiple edges in FF. This implies that FF has the same structure as that of FF^{\prime}. On the other hand, if FF^{\prime} is a forest, set g1(F)g^{-1}(F^{\prime}) to be a graph induced by edge set {eE:g(e)E(F)}\{e\in E:g(e)\in E^{\prime}(F^{\prime})\}. Then all the subgraph of g1(F)g^{-1}(F^{\prime}) without multiple edges is a forest.

Set μ\mu^{\prime} to be the measure of Arboreal Gas on GG^{\prime} with the parameter

βe=eg1(e)βe.\beta_{e^{\prime}}^{\prime}=\sum_{e\in g^{-1}(e^{\prime})}\beta_{e}.

Then for any forest F′′F^{\prime\prime} in GG^{\prime},

μ({F:F=F′′})=μ(F′′).\mu(\{F\in\mathcal{F}:F^{\prime}=F^{\prime\prime}\})=\mu^{\prime}(F^{\prime\prime}).

\Leftarrow: Consider two distinct edges e1,e2Ee_{1},e_{2}\in E.
(a). If e1,e2g1(e0)e_{1},e_{2}\in g^{-1}(e_{0}^{\prime}) for some e0Ee_{0}^{\prime}\in E^{\prime}, set

μ[e0]=βe0F1(βe,ee0),μ[e0¯]=F2(βe,ee0),\mu^{\prime}[e_{0}^{\prime}]=\beta_{e_{0}^{\prime}}^{\prime}F_{1}(\beta_{e^{\prime}}^{\prime},e^{\prime}\neq e_{0}^{\prime}),\ \mu^{\prime}[\bar{e_{0}^{\prime}}]=F_{2}(\beta_{e^{\prime}}^{\prime},e^{\prime}\neq e_{0}^{\prime}),

where F2F1>0F_{2}\geq F_{1}>0. The negative correlation on GG is equivalent to

μ[e1e2¯]μ[e1¯e2]μ[e1e2]μ[e1¯e2¯]0.\mu[e_{1}\bar{e_{2}}]\mu[\bar{e_{1}}e_{2}]-\mu[e_{1}e_{2}]\mu[\bar{e_{1}}\bar{e_{2}}]\geq 0. (6.6)

Note that μ[e1e2]=0\mu[e_{1}e_{2}]=0. The right hand side of (6.6) is equal to μ[e1e2¯]μ[e1¯e2]0\mu[e_{1}\bar{e_{2}}]\mu[\bar{e_{1}}e_{2}]\geq 0.

(b). If g(e1)=e1e2=g(e2)g(e_{1})=e_{1}^{\prime}\neq e_{2}^{\prime}=g(e_{2}), set

μ(e1e2)=βe1βe2F12(βe,ee1,e2),\displaystyle\mu^{\prime}(e_{1}^{\prime}e_{2}^{\prime})=\beta_{e_{1}^{\prime}}^{\prime}\beta_{e_{2}^{\prime}}^{\prime}F_{12}(\beta_{e^{\prime}}^{\prime},e^{\prime}\neq e_{1}^{\prime},e_{2}^{\prime}),\ μ(e1e2¯)=βe1F12¯(βe,ee1,e2),\displaystyle\mu^{\prime}(e_{1}^{\prime}\bar{e_{2}^{\prime}})=\beta_{e_{1}^{\prime}}^{\prime}F_{1\bar{2}}(\beta_{e^{\prime}}^{\prime},e^{\prime}\neq e_{1}^{\prime},e_{2}^{\prime}),
μ(e1¯e2)=βe2F1¯2(βe,ee1,e2),\displaystyle\mu^{\prime}(\bar{e_{1}^{\prime}}e_{2}^{\prime})=\beta_{e_{2}^{\prime}}^{\prime}F_{\bar{1}2}(\beta_{e^{\prime}}^{\prime},e^{\prime}\neq e_{1}^{\prime},e_{2}^{\prime}),\ μ(e1¯e2¯)=F1¯2¯(βe,ee1,e2).\displaystyle\mu^{\prime}(\bar{e_{1}^{\prime}}\bar{e_{2}^{\prime}})=F_{\bar{1}\bar{2}}(\beta_{e^{\prime}}^{\prime},e^{\prime}\neq e_{1}^{\prime},e_{2}^{\prime}).

By the negative correlation on GG^{\prime}, we obtain

F12¯F1¯2F12F1¯2¯0.F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}\geq 0.

Now we verify (6.6) to prove the negative correlation on GG. Rewrite eg1(ei),eeiβe\prod_{e\in g^{-1}(e_{i}),e\neq e_{i}}\beta_{e} as C(ei)C(e_{i}) for i=1,2i=1,2. Then the right hand side of (6.6) should be

βe1(C(e2)F12+F12¯)βe2(C(e1)F12+F1¯2)\displaystyle\beta_{e_{1}}(C(e_{2})F_{12}+F_{1\bar{2}})\cdot\beta_{e_{2}}(C(e_{1})F_{12}+F_{\bar{1}2})
βe1βe2F12(C(e1)C(e2)F12+C(e1)F12¯+C(e2)F1¯2+F1¯2¯)\displaystyle-\beta_{e_{1}}\beta_{e_{2}}F_{12}\cdot(C(e_{1})C(e_{2})F_{12}+C(e_{1})F_{1\bar{2}}+C(e_{2})F_{\bar{1}2}+F_{\bar{1}\bar{2}})
=βe1βe2(F12¯F1¯2F12F1¯2¯).\displaystyle=\beta_{e_{1}}\beta_{e_{2}}(F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}).

By F12¯F1¯2F12F1¯2¯0F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}\geq 0, we deduce (6.6).

\Rightarrow: For distinct e1,e2Ee_{1}^{\prime},e_{2}^{\prime}\in E^{\prime}, by (b) in “\Leftarrow”, the negative correlation on GG is equivalent to

F12¯F1¯2F12F1¯2¯0,F_{1\bar{2}}F_{\bar{1}2}-F_{12}F_{\bar{1}\bar{2}}\geq 0,

which implies that

μ[e1e2¯]μ[e1¯e2]μ[e1e2]μ[e1¯e2¯]0.\mu^{\prime}[e_{1}^{\prime}\bar{e_{2}^{\prime}}]\mu^{\prime}[\bar{e_{1}^{\prime}}e_{2}^{\prime}]-\mu^{\prime}[e_{1}^{\prime}e_{2}^{\prime}]\mu[\bar{e_{1}^{\prime}}\bar{e_{2}^{\prime}}]\geq 0.

This shows the negative correlation on GG^{\prime}.  

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