1 Introduction to Arboreal Gas
Let be a finite connected graph. If there exists an edge connecting vertices , we say that are adjacent, denoted by . We can also represent the edge as . A forest on is a subgraph that does not contain any cycle. Set to be the collection of all forests on . The Arboreal Gas with parameter for each is the measure on forests given by
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Specifically, if for all edges , the measure will be given by
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where stands for the number of edges in .
Let denote the probablity of Bernoulli bond percolation with parameter . We can note that Arboreal Gas with a uniform parameter is equivalent to Bernoulli bond percolation with parameter conditioned to be acyclic:
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Another notable observation is that Arboreal Gas emerges as the limit of the -states random cluster model as , with , as mentioned in [13]. The uniform forest model, also mentioned in [13], can be seen as one particular instance of Arboreal Gas, where . 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 .
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 , 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 for fixed , undergoes a phase transition on the complete graph with 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 . Based on this polynomial decay, and assuming for the existence of a weak limit of probabilities on asymptotic graphs of , they demonstrated the absence of an infinite tree almost surely. On torus with fixed and , Bauerschmidt, Crawford and Helmuth gave an asymptotic estimate of connectivity probability from the origin to other vertices for in [2]. Recently, Halberstam and Hutchcroft verified the uniqueness of infinite tree in for for any translation-invariant Arboreal Gas Gibbs measures in [9].
For any edge set , let be the probability that all edges in belong to the forest. We simply write as for . For two vertices , denote by the probability that are in a same tree of the forest. The negative correlation of Arboreal Gas should be expressed as
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(1.1) |
Until now, this question is still open. A weaker inequality has been proved recently in [5] by Brändén and Huh. They showed by using the Lorentzian signature. Once (1.1) holds, the existence of the weak limit of Arboreal Gas Gibbs measures on increasing asymptotic graphs as was answered in [3]. Under this weak limit on , 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 . One way to consider this question is to realize the relevance between Arboreal Gas and the hyperbolic spin system . The negative correlation of Arboreal Gas can be implied by the monotonicity of a hyper-integral on , 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 . In Section 5, we show the negative correlation of complete graphs with vertices for sufficiently large and . In Section 6, we give the proof of the equivalence of the negative correlation on any graph and on its simplified version.
3 Preliminaries
First we show an equivalent condition of negative correlation.
Definition 3.1.
For disjoint , define
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Given a measure defined on by
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Specifically, set .
Proposition 3.2.
The following two conditions are equivalent:
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for all distinct ;
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for all disjoint .
: This result immediately comes out by setting for .
: First we prove the equivalence of condition and decreasing of in each for all and . Note that
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(3.2) |
By and ,
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By the arbitrary of , we obtain that (3.2) is equivalent to the decreasing of in each for all and .
Secondly, we prove that the decreasing of in each for all and deduces condition , which may complete the proof of . To show this property, we claim that
equals the limit of as for all .
By the decreasing of in each component of except for all , we know that for ,
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For disjoint , Applying this to , we obtain
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This inequality is equivalent to
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BY induction of , we verify that
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which implies condition .
Finally we prove our claim by setting
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Then , which converges to (as )
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By ,
we show that converges to in weak topology. Further, by the decreasing of in , we conclude that .
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 on is a subgraph of containing all vertices in and being a tree. The uniform spanning tree is a uniform measure on all spanning trees on .
In our next part, denote by the probability of uniform spanning trees on graph . 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 with conductance on each edge, we build an electrical network. Given two disjoint vertex sets and to be source and sink. Define potential difference and current on each direct edge .
For adjacent vertices ,
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Here we normally take
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Further, the potential difference and current satisfies the following three laws, see [8].
Kirchhoff’s potential law. For any cycle with ,
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Kirchhoff’s current law. For any ,
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Ohm’s law. For any edge ,
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Kirchhoff’s potential law is equivalent to the existence of the function on vertex set such that for adjacent vertices . Here is also called potential function. Next we define the energy of current by
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Define the unit currenct flow to be such a flow that . Denote by
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the effective resistance of the electrical network with conductance .
Here is equal to for if we set for , where the current is unit.
For some finite connected graph , we have the following equation for its spanning trees.
Proposition 3.4 (Kirchhoff’s Effective Resistance Formula, [11]).
Let be the spanning tree of a finite connected graph , and be some edge of with endpoints , then
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where is unit current flow from to .
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 from to and conductance on each edge . For any edge , if , then the effective resistance strictly decreases in . Otherwise, is a constant in .
Proof of Lemma 3.5. For the unit current flow and conductance , we consider the energy
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which is equal to .
Set to be another conductance with
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If , we may observe that . Since the unit current flow on the electrical network with conductance has a smaller energy than flow , we obtain
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If , we have . Here we prove that is the unit current flow on the electrical network with conductance . Consider the potential on the electrical network with conductance , then for any edge ,
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This implies that . On the electrical network with conductance , and satisfy Kirchhoff’s potential law and Kirchhoff’s current law. As to Ohm’s law, for ,
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For ,
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Therefore, is a unit current flow on the electrical network with conductance , which implies
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Proposition 3.6.
[11]
The negative correlation holds for uniform spanning trees on finite connected graph .
Proof. We only need to verify that for any two distinct edges ,
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This inequality is equivalent to
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where stands for the contraction of by removing and identifying its endpoints. By Proposition 3.4, this inequality is equivalent to
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(3.3) |
where the unit current flow is from to , and are endpoints of . Actually, (3.3) holds since can be regarded as graph with resistance on , leading to a smaller effective resistance.
4 Proof of Theorem 2.1
For finite connected graph , set to be the number of spanning trees on . Then for each spanning tree, it has the probability . For edge sets and , let be the number of all spanning trees containing and disjoint to . Consider the Arboreal Gas with parameter . Observe that the highest term of in for some edge should be
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If we consider the negative correlation of Arboreal Gas, i.e.,
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we pay attention to the highest term of the left hand side of this inequality if is large enough.
Here the highest term is
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By Proposition 3.6, this value should be non-negative. What we want is to show
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by which we conclude the negative correlation for large enough. However, this does not always hold, such as that on trees. To prove Theorem 2.1, for adjacent edges and , we consider in two situations:
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(a)
there is a simple cycle crossing both and ;
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(b)
there is no simple cycle crossing both and .
In Case (a), if we start a unit current flow from to , then is not vanishing.
Lemma 4.1.
For two adjacent edges , set to be the unit current flow from to . If there is a simple cycle on crossing both and , then is not vanishing.
In Case (b), we prove that can be separated into two parts of the graph , where the intersection of these two parts has only one vertex. By this property, the probability of Arboreal Gas on can be expressed as the product of the probabilities of Arboreal Gas on these two parts.
Lemma 4.2.
For two adjacent edges , assume that are two endpoints of edge for , where . If there is no simple cycle on crossing both and , then there exist subgraphs for such that
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for ;
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, ;
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, .
Now we prove Theorem 2.1.
(a). If there is a simple cycle crossing both and , by Lemmas 4.1 and 3.5, we obtain
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for the unit current flow from to .
Then the highest term in is positive, i.e.,
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where the equality comes from Proposition 3.4.
This implies the negative correlation for , i.e., , for sufficiently large .
(b). If there is no simple cycle crossing both and (assume ), by Lemma 4.2, we separate into two parts and with
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for ;
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, ;
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, .
We claim that
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By this claim, we obtain that the subgraph in is a forest is equivalent to that are forests for , where . Therefore, for , if we denote by the measure for Arboreal Gas on , i.e., the probability multiplying its partition function, then
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By these properties, we deduce that
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which implies the negative correlation for .
Now we prove the claim by contradiction. Otherwise, there is a simple cycle with such that for and some . Without loss of generality, assume . Since , there should be two distinct integers with and or such that and . This implies , i.e., , which is contradict to and that is a simple cycle.
Finally we prove Lemmas 4.1 and 4.2.
Proof of Lemma 4.1. Without loss of generality, assume is a endpoint of .
We prove by contradiction. If , we consider the simple cycle crossing both and , and set
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Note that the potentials on are different. Precisely, , which implies . By Kirchhoff’s potential law and Ohm’s law, we know that there should be some edge with . Let , where . By Kirchhoff’s current law, we deduce that there is a vertex adjacent to such that . By Kirchhoff’s potential law, . Otherwise, potentials on the cycle do not obey Kirchhoff’s potential law.
Now we construct a chain by induction:
For integer , assume there is a chain with for and for distinct . By Kirchhoff’s current law, we can find a vertex adjacent to with . Furthermore, for . Otherwise, if , potentials on the cycle do not obey Kirchhoff’s potential law. if for some , potentials on the cycle do not obey Kirchhoff’s potential law.
Thus we construct a chain with for distinct . That is, we find different vertices, which is contradict to that the number of all vertices of is .
Proof of Lemma 4.2.
Without loss of generality, assume is connected. For , set to be the graph induced by the vertex set
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Set to be the graph induced by
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Now we prove that satisfy conditions in Lemma 4.2 by showing the following properties.
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for .
Since for , .
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and for distinct .
If there is an edge for some distinct , then two endpoints of belong to and , contradict to . Thus we only need to show for distinct .
For any , by the definition of . We then show .
For any vertex with , set to be the simple path from to not visiting . Then , otherwise there is some with and there will be a simple cycle crossing , contradict to the condition in Lemma 4.2.
We prove by contradiction. If , implies . Therefore, there is a simple path from to not visiting , where . Set . Assume for some , where and . Then is a simple cycle crossing , contradict to the condition in Lemma 4.2.
Since we choose arbitrarily, we obtain .
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, .
By definition of , we immediately obtain . Since for distinct , we show by proving that there is no edge connecting except those edges with endpoint .
For , , and , set to be the simple path from to not visiting . Assume that there is an edge connecting , , . If , then there is a simple path from to not visiting , implying . Otherwise , then is a simple path from to not visiting , which also implies . This is contradict to and .
By these three properties, we shows that
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, ;
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, ;
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, .
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 in complete graph . Consider the highest term of
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We observe that the highest term is vanishing. Because for any adjacent vertices , the unit current flow from to is
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By Lemma 3.5 and Proposition 3.4,
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Therefore, in order to verify the negative correlation, we consider the second highest term of .
To give a more precise estimate, we use the following property.
Proposition 5.1 (Kirchhoff’s Matrix-tree Theory, [6]).
Let be the Laplacian matrix for the graph defined by
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Let be the matrix given by deleting -th row and column of . Then for any , the determinate of can be expressed as
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By Kirchhoff’s Matrix-tree Theory, we give the following lemma to show the number of spanning trees on complete graph .
Lemma 5.2.
Set to be the number of spanning trees on satisfying event . Then for any disjoint edges in ,
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Proof. By Kirchhoff’s Matrix-tree Theory, we compute out directly. For (this equals by symmetry), there is a bijection from spanning trees on containing to spanning trees on . Here for any positive integer , each multiple edge with multiplicity in can be regarded as an edge with conductance . Precisely, when we consider the number of spanning trees, can be regarded as with conductance
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where is the vertex contracted from edge .
Based on Kirchhoff’s Matrix-tree Theory, we compute out .
For , similarly, we consider , which can be regarded as with conductance
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where is the vertex contracted from edge for . By Kirchhoff’s Matrix-tree Theory,
we compute out .
By Lemma 5.2, we verify that the second highest term of is positive for disjoint edges in .
Lemma 5.3.
For disjoint edges in , if is large enough, then the coefficient of in is positive.
By Lemma 5.3, we prove Theorem 2.2.
Proof of Theorem 2.2.
For adjacent two edges , by Theorem 2.1, we get the negative correlation for large enough .
For disjoint two edges , consider . By Lemma 5.2, we deduce that the highest term should be
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By Lemma 5.3, the second highest term should be positive for sufficiently large , which implies the negative correlation for sufficiently large .
Here we give the proof of Lemma 5.3.
Proof of Lemma 5.3.
First we consider the second highest term of for any edge set . This value should be
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Here is the number of spanning forests containing such that there is exactly two trees in the forest, where is the spanning tree of , and is the spanning tree of .
We then estimate the coefficient of in , which equals
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Here we write as
For the term with in this sum, denoted by , we compute it in the following different cases:
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the endpoints of are in , then the sum of equals ;
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2.
the endpoints of are not in , then the sum of equals ;
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all the endpoints of , and one endpoint of are in , then the sum of equals
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all the endpoints of , and one endpoint of are in , then the sum of equals
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only one endpoint of is in , and the endpoints of are not in , then the sum of equals ;
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only one endpoint of is in , and the endpoints of are not in , then the sum of equals ;
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the endpoints of are in , and the endpoints of are not in , then the sum of equals ;
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the endpoints of are in , and the endpoints of are not in , then the sum of equals ;
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one endpoint of , and one endpoint of are in , then the sum of equals
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Based on these computations, we obtain by summing over all values:
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For , . For , can be also written as
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For convenience, set
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We prove that for large enough, .
By Stirling’s approximation, for sufficiently large and for ,
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Note that
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By this estimate, for sufficiently large,
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Since the coefficient of is , 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 is pivotal for , then negative correlation on implies negative correlation on .
Proof. Assume that the negative correlation holds on , i.e.,
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(6.4) |
Set . Since is pivotal for , whether exists or not does not influence the existence of cycle in .
Hence, for , by which we have
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Combined with (6.4), we verify that
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Lemma 6.2.
If the negative correlation holds on each component of , then it holds on .
Proof. We only need to prove negative correlation for in different components of since the existence of cycle in a component cannot be influenced by other components.
Assume that for , where are all components of . Set to be the measure on for , then we deduce that ()
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This implies that .
Lemma 6.3.
The negative correlation holds on for any parameter if and only if it holds on for any parameter , where is given by Definition 2.4.
Lemma 6.4.
The negative correlation holds on a complex graph for any parameter if and only if it holds on the simple graph for any , where 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 is equivalent to that on each component of a new graph given by deleting all pivotal edges. Thus if or is pivotal edges, or belong to different component of , the negative correlation for holds. Without loss of generality, assume is connected.
By Lemmas 6.3 and 6.4, we know that the negative correlation on is equivalent to that on , which immediately shows the negative correlation for with .
Finally we give our proof of Lemmas 6.3 and 6.4.
Proof of Lemma 6.3. Recall the definition of in Definition 2.4. For some subgraph with edge set , set to be a graph induced by the following edge set
. We write the edge set of as .
Intuitively, for any edge , assume for some , if , then . We show that is equivalent to that is a forest on .
If , set to be a graph induced by edge set . Note that . If there is a cycle in , there should also exist a cycle in , contradict to .
If is a forest, assume there is a simple cycle in . For any , we claim that
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By this claim we know that . Since is also a cycle, it is contradict to that is a forest.
Here we prove the claim. Otherwise, there are two adjacent edges with and . Setting to be the vertex adjacent to , we find that there is no simple cycle crossing in because the degree of is , contradict to .
Set to be the measure of Arboreal Gas on , where the corresponding parameter
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Then for any forest in ,
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Here we write as .
: Consider two distinct edges .
(a).
If , set
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where . To verify the negative correlation on , we need to prove
This is equivalent to
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(6.5) |
The right hand side of (6.5) equals
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By , we finish the proof of (6.5).
(b). If , set
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By the negative correlation on , we have
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To verify the negative correlation on , we also prove (6.5).
For convenience, for , we rewrite , as and respectively. Note the right hand side of (6.5) equals
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which is non-negative because . This completes the proof of (6.5).
: For distinct , by (b) in “”, the negative correlation on is equivalent to
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This shows immediately that
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which completes the proof of negative correlation on .
Proof of Lemma 6.4. Recall that maps edge sets to edge sets by deleting multiple edges. For some subgraph in , denote by the graph induced by edge set . If is a forest, there should not be any multiple edges in . This implies that has the same structure as that of . On the other hand, if is a forest, set to be a graph induced by edge set . Then all the subgraph of without multiple edges is a forest.
Set to be the measure of Arboreal Gas on with the parameter
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Then for any forest in ,
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: Consider two distinct edges .
(a). If for some , set
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where . The negative correlation on is equivalent to
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(6.6) |
Note that . The right hand side of (6.6) is equal to .
(b). If , set
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By the negative correlation on , we obtain
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Now we verify (6.6) to prove the negative correlation on . Rewrite as for . Then the right hand side of (6.6) should be
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By , we deduce (6.6).
: For distinct , by (b) in “”, the negative correlation on is equivalent to
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which implies that
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This shows the negative correlation on .