The Uniform Spanning Tree and related

models

Antal A. Jarai

December 18, 2009

1 Introduction

The following type of question is characteristic of discrete probability: given alarge finite set of objects, defined by some combinatorial property, what doesa typical element ”look like”, that is, can we describe some of its statisticalfeatures. Simple as it is to state such questions, they can lead to surprisinglydeep and difficult mathematical problems.

In this course we are going to look at a very special example: randomspanning trees of graphs (see precise definitions later in this introduction).Spanning trees have been well-known in combinatorics for a long time; see[6]. However, the study of random spanning trees is relatively recent. Thetopic proves to be very interesting from several points of view. There isa surprising connection between spanning trees, random walk and electriccircuits. The connection with random walk yields efficient algorithms togenerate a spanning tree of a graph uniformly at random. We will see twosuch beautiful algorithms. Uniform spanning trees are also interesting fromthe point of view of statistical physics, as they are a special case of so-called random cluster measures; see Exercise 1.1. The notion of amenability,that comes from group theory, will enter some of our discussions. Anotherinteresting aspect of random spanning trees is that they are an example of adeterminantal process (see its definition later in this course).

We now introduce some basic notions.

1.1 Spanning Trees

A graph is called a tree, if it is connected, and contains no cycles. A graphthat contains no cycles, and is not necessarily connected is called a forest(its connected components are trees). Let G = (V,E) be a connected graph.

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Figure 1: A subgraph of the square grid and one of its spanning trees

A spanning subgraph of G is one that contains every vertex of V (some ofthese may be isolated vertices not incident on any edge of the subgraph). Aspanning tree of G is a spanning subgraph of G that is a tree. Similarly wedefine spanning forest. See Figure 1. We can identify a spanning subgraphof G with an element of the space Ω = 0, 1E, by writing a 1 if an edge ispresent, and 0 if it is not. That is, ω = (ωe)e∈E ∈ Ω represents the spanningsubgraph which contains precisely those edges e ∈ E for which ωe = 1.Assume now that G is a finite graph. By the Uniform Spanning Tree on G,we mean the probability distribution UST on Ω that assigns equal mass toeach spanning tree of G, and no mass to subgraphs that are not spanningtrees:

UST[ω] =

1

N(G)if ω is a spanning tree of G;

0 otherwise,,

where N(G) is the number of spanning trees of G. By drawing an exampleor looking at Figure 1, you can convince yourself that a general connectedgraph usually has many spanning trees, so this definition meaningful.

Having introduced the space Ω, we can view the Uniform Spanning Tree asa 0-1 valued random process (ωe) indexed by the edges ofG. It should be clearthat this is very far from an i.i.d. process, since cycles are forbidden, and thesubgraph is constrained to be connected. We will write T = e ∈ E : ωe = 1.Some questions we will consider: what is the probability that a given edgee ∈ E belongs to T ? More generally, given K ⊂ E, what is UST[K ⊂ T ]?A different type of question that proves to be very interesting is as follows.Suppose now that G is the d-dimensional integer lattice: G = (Zd,Ed), d ≥ 2,with x, y ∈ Zd connected by an edge, if |x − y| = 1, that is, Ed = x, y :x, y ∈ Z

d, |x−y| = 1. This G has infinitely many spanning trees, hence it isnot clear what should be meant by picking one uniformly at random. It turns

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out, as will be discussed later in the course, that there is a very natural waythis can be done. Consider a sequence V1 ⊂ V2 ⊂ . . . Vn ⊂ · · · ⊂ Zd, of finiteconnected subsets of Z

d, such that ∪nVn = Zd. Let Gn = (Vn, En) be the

subgraph induced by Vn, that is the graph with vertex set Vn and edge set theset of all edges of G that connect vertices in Vn (Figure 1 shows an exampleV when d = 2.) On each Gn, we can construct the measure USTn. Do theseapproximate, in a suitable sense, a measure on Ω = 0, 1E

d? Making this

precise: is it true that for a fixed e ∈ Ed, the probabilities USTn[e ∈ T ]

converge to some limit (noting that the probability is indeed well definedas soon as e ∈ En)? More generally, for K ⊂ Ed finite, does USTn[K ⊂ T ]converge to some limit? The reader is invited to give some thought to what itwould involve to prove such convergence, before moving on. Once we provethat the limit exists, we will examine in detail what type of graph is thelimiting T .

Exercise 1.1 (Random Cluster Model). Let G = (V,E) be a finite graph,and consider the following probability measure on Ω = 0, 1E.

µp,q(ω) =1

Zp,qpn(ω)(1 − p)|E|−n(ω)qK(ω), ω ∈ Ω,

where 0 < p < 1, and q > 0 are parameters, n(ω) =∑

e∈E ωe, K(ω) =number of connected components of the graph ω, and Zp,q is a constant nor-malizing µp,q to be a probability distribution. This is called the RandomCluster Measure on G.(a) Check that when q = 1, the edges are i.i.d. This is called the PercolationModel on G with edge density p.(b) Prove that as p→ 0 and q/p→ 0, µp,q[ω] → UST[ω].

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2 The Aldous-Broder algorithm

A large graph usually has many spanning trees; often exponentially manyin the number of edges of the graph. (The famous Matrix-Tree Theorem incombinatorics [6, Corollary II.13] gives the exact number of them.) Hencefor a general graph, it is not obvious how to simulate a uniformly randomspanning tree in reasonable time; that is, polynomial in the number of edges.The algorithm below that achieves this, was discovered independently byAldous [1] and Broder [7].

Let G = (V,E) be a connected finite graph. We will write deg(u) for thenumber of edges incident on u. Let (Xn)n≥0 be the simple random walk onG, that is, the Markov chain with state space V and transition probability

puv =

1

deg(u)if u, v ∈ E;

0 otherwise.

Since G is connected, (Xn) is irreducible. Let Tv be the first hitting time ofv:

Tv := infn ≥ 0 : Xn = v.Let C be the cover time of the random walk:

C := maxvTv = infn ≥ 0 : X0, X1, . . . , Xn = V .

By irreducibility, C is finite with probability 1.

Algorithm 2.1 (Aldous-Broder Algorithm). Choose X0 arbitrarily. Runsimple random walk on G up to the cover time C. Consider the set of edges:

T := XTw−1, XTw : w 6= X0.

That is, every time you visit a vertex that you have not seen before, recordwhich edge was used to enter that vertex. Then T is a spanning tree, and itis uniformly distributed over all spanning trees of G.

Let us note that it is easy to see that T is a spanning tree: we alwaysdraw edges to vertices that have not been visited before, so no cycles areformed during the process. By induction, T is connected, since every edgehas an endvertex that has been visited already.

The starting point for the proof that the algorithm works is a beautifuland surprising theorem that shows a connection between spanning trees andMarkov chains.

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2.1 The Markov Chain-Tree Theorem

Let us put aside simple random walk for a moment, and let (Xn)n≥0 bean arbitrary irreducible finite state Markov chain on a state space V , withtransition matrix P = (pxy). We associate to it a directed graph G = (V,E),where [x, y] ∈ E, if pxy > 0. That is, we draw an edge from x to y, if it ispossible to move from x to y in one step. By a rooted tree (t, r), we mean atree t with a distinguished vertex r, called the root. From every vertex of thetree, there is a unique path to the root that involves no backtracking. Thusfor every vertex v of a rooted tree, there is a unique edge v, w of the treethat leads one step closer to the root, and we orient this edge towards theroot, that is as [v, w]. It is easy to see that this way each edge of the rootedtree gets an orientation.

Consider a rooted spanning tree (t, r) of G. We assign to this the weight

q(t, r) =∏

[v,w]∈E(t,r)

pvw.

That is, the weight of a rooted spanning tree is the product of the transitionprobabilities along the edges of the tree. Let

p(v) :=∑

(t,v)

q(t, v), v ∈ V,

where the sum is over all rooted spanning trees of G with root v.

Theorem 2.1 (Markov Chain-Tree Theorem). The stationary distributionof X is proportional to p(v).

Proof. The key to the proof is that one can build a Markov chain on rootedtrees on top of (Xn). Consider the following operation. Given a rootedspanning tree (t′, v′) of G, and an edge [v′, v] ∈ E, we define a new rootedspanning tree (t, v). We first add the edge [v′, v] to t′. Since t′ alreadycontains a directed path from v to v′, this together with the edge [v′, v] formsa directed cycle. Note that due to the orientation of edges in t′, all otheredges are oriented towards this cycle. Hence if we now remove the outgoingedge from v, [v, w] say, then we obtain a rooted spanning tree t rooted atthe new vertex v. See Figure 2. Note that it may happen that w = v′, inwhich case the cycle consists of the edges [v′, v], [v, v′]. It is also no problemif pv′v′ > 0, and hence the loop-edge [v′, v′] is present. In this case we add[v′, v′] and then remove it. Let us note here what the backwards operationis. If (t, v) and w are specified, then adding the edge [v, w] we get back thedirected cycle, and then v′ is uniquely determined as the vertex preceeding

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Figure 2: The operation that changes a spanning tree rooted at v′ into aspanning tree rooted at v. The dashed lines indicate the unique orientedcycle that is created by adding the edge [v′, v].

v in this cycle, and hence also t′ is uniquely determined. Also note that forfixed (t, v), to any w with [v, w] ∈ E there corresponds a ”precursor tree”(t′, v′). Again, since adding [v, w] creates a unique directed cycle, etc. When(t, v) is fixed, let us write (t′, v′) = prec((t, v), w), to indictae the relationshipbetween (t, v), w and (t′, v′).

We turn the above operation into a Markov chain (Yn), by letting theroot perform the underlying Markov chain (Xn). That is,

P[Yn+1 = (t, v) | Yn = (t′, v′)] := P[Xn+1 = v |Xn = v′] = pv′v.

We will see in Exercise 2.1, that (Yn) is irreducible.We claim that the stationary distribution of (Yn) is proportional to q(t, v).

We can write

q(t, v) =∑

w:[v,w]∈E

pvwq(t, v). (1)

Let (t′, v′) denote the rooted tree uniquely determined by (t, v), w. Then

pvwq(t, v) = pv′vq(t′, v′).

Hence the right-hand side of (1) can be rewritten

∑

(t′,v′)

pv′vq(t′, v′). (2)

Since pv′v is the transition probability for moving from (t′, v′) to (t, v), thisproves the claim (assuming Exercise 2.1).

The Theorem now follows, since p(v) is proportional to the long runfraction of time that the root equals v, which is the long run fraction of timeXn = v.

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Exercise 2.1. Show that given a rooted tree (t, v), and for any X0 andY0, there exists N and a sequence X1, . . . , XN of allowed steps such thatYN = (t, v). Conclude that the chain (Yn) is irreducible. (Hint: traverse theedges of t in a suitable order.)

2.2 The backwards tree construction

Let . . . , Y−1, Y0, Y1, . . . be the stationary tree-chain run from time −∞. Wewill write P for the probability measure governing this chain. Then Xn, theroot of Yn is the stationary X chain.

We define the backward tree at time n as follows. For w 6= Xn, let

Lnw = supk : k < n, Xk = w.

Due to irreducibility, Lnw > −∞ with probability 1. Then we have

Yn = [XLnw, XLn

w+1] : w 6= Xn.That is, if for every w 6= Xn, we mark the edge that was used on the last exitfrom w prior to time n, we get Yn. This is easily verified from the definitionof the Y -chain.

2.3 Verification of the Aldous-Broder algorithm

We now return to the task of generating a uniform spanning tree of a con-nected graph G = (V,E). This is done by reversing the time in the backwardstree construction.

Let Xn be the simple random walk on G. Then the directed graph con-structed from X is the same as G, with each edge being present with bothorientations. It is easy to check that the stationary distribution is propor-tional to deg(u). We still write P for the probability measure correspondingto the stationary chain.

The simple random walk is reversible, that is, P[X0 = x0, . . . , Xk = xk] =P[Xk = x0, . . . , X0 = xk]. In particular,

P[X1 = x1, . . . , Xk = xk |X0 = v] = P[X−k = xk, . . . , X−1 = x1 |X0 = v].(3)

Proof. [Aldous-Broder Algorithm works] Orient each edge of T towards theroot. Let us prove that T is uniformly distributed. For simple random walk,the weight of a rooted tree (t, v) is

q(t, v) =∏

w 6=v

1

deg(w)= const · deg(v). (4)

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It is clear from the definitions of the backwards tree and T , that the forwardtree constructed from the sequence v, x1, x2, . . . will be (t, v) if and only ifthe backwards tree constructed from . . . , x2, x1, v is (t, v). Hence (3) gives

P[T = (t, v) |X0 = v] = P[Y0 = (t, v) |X0 = v] =P[Y0 = (t, v)]

P[X0 = v]

=const · q(t, v)const · deg(v)

.

Due to (4), the right hand side is a constant independent of (t, v). Henceeach tree rooted at v is equally likely to result from the algorithm. Forgettingthe root then yields a uniformly distributed spanning tree.

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3 Wilson’s algorithm

In this section, we present a second algorithm to generate random spanningtrees, due to Wilson [34]. It turns out to be extremely useful not only as asimulation tool, but also for theoretical analysis. Before presenting the algo-rithm, we generalize the notion of random spanning trees from the uniformcase to a weighted case.

3.1 Some terminology

Let G = (V,E) be an unoriented graph. We allow multiple edges or loops inthe graph. For many purposes we will be able to disregard loops, as they cannever be contained in a tree. Often we will use oriented edges, and in thiscase, each edge will be present in E with both orientation. For an orientededge e we write e = [e, e], and call e the tail of e, and e the head of e. Wedefine e = [e, e] the reversal of e.

By a network, we mean a pair (G,C), where C : E → (0,∞), defined onunoriented edges (or, equivalently, C is symmetric: C(e) = C(e)). We callC(e) the conductance of e. For any vertex v ∈ V , we call Cv :=

∑e:e=v C(e),

the conductance of v. For infinite networks, we assume that Cv < ∞ for allv. The resistance of e is defined as R(e) := 1/C(e). Every graph carries thedefault network (G, 1), where every edge has conductance 1.

For every v ∈ V , the network random walk started at v is the Markovchain with Pv[X(0) = v] = 1 and transition probabilities Pv[X(n + 1) =w |X(n) = u] = C(u, w)/Cu. Here, C(u, w) =

∑e:e=u,e=w C(e), allowing

multiple edges.On a finite connected network, we define the weight of a spanning tree by

weight(t) =∏

e∈tC(e). In the sequel, by a random spanning tree, we meanone chosen with probability proportional to weight. The distribution UST isobtained as a special case when the conductances are 1.

3.2 Loop-erased random walk

Wilson’s method is based on Loop-Erased Random Walk, which we nowdefine. If P = [u0, u1, . . . , un] is a path in G, we define its loop-erasure bychronologically removing loops from P. Formally, this is defined as follows.We first set γ0 := u0. Let s1 := maxm : um = γ0. If s1 = n, thenLE(P) = [γ0]. Otherwise, let γ1 := u(s1 + 1). Assuming γ0, . . . , γk havebeen defined, we let sk+1 := maxm : um = γk. If sk+1 = n, we stop andLE(P) = [γ0, . . . , γk]. Otherwise, we let γk+1 := u(sk+1 + 1). Note that theorder in which we remove loops does matter, and in the above definition, we

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remove loops, as they are created, following the path. When loop-erasure isapplied to a random walk path, we talk about Loop-Erased Random Walk(LERW).

3.3 Wilson’s method

Let (G,C) be a finite connected network. Pick any r ∈ V , called the ”root”.We define a growing sequence of subtrees T (i), i ≥ 0 of G. We let T (0) :=r. Let 〈v1, . . . , vn−1〉 be an enumeration of V \ r. Suppose T (i) has beengenerated. Start a network random walk at vi+1, and stop when it hits T (i).Let

T (i+ 1) := T (i) ∪ LE(path from vi+1 to T (i)).

The output of the algorithm is T = T (n− 1).

Theorem 3.1. T is distributed proportional to weight.

Proof. To each v ∈ V , v 6= r, we assign an infinite stack of random orientededges (arrows) ev

j , j = 1, 2, . . . , such that evj = v, and P[ev

j = e] = C(e)/Cv.

All elements in all the stacks are independent. We colour the i-th element ofeach stack with colour i.

The arrows on top of the stacks define a random oriented graph withvertex set V . This may contain oriented cycles. If there are no orientedcycles, then, since every vertex v 6= r has a unique outgoing edge, the graphis a tree directed towards the root r.

Consider the notion of cycle popping. If C = 〈e1, . . . , ek〉 is an orientedcycle on top of the stacks, by popping C, we mean removing all the edgesej , j = 1, . . . , k from their stacks, and lifting those stacks up by one level, sothat now new edges sit of top of those stacks.

The collection of stacks provides a probability space on which the al-gorithm can be defined. Since the arrows at each vertex were chosen withthe random walk transition probabilities, the random walk started at v1 canfollow the oriented edges. When a vertex is revisited for the first time, anoriented cycle has been found. If we pop this cycle from the stacks, then therevisited vertex receives a fresh, independent arrow, which can be followed tocontinue the random walk. It is easy to see that if every completed cycle ispopped, then following the arrows on top of the current stacks is precisely thenetwork random walk. Moreover, when r is hit, on top of the stacks we seethe LERW from v1 to r, as well as ”fresh” arrows (independent of previouslyvisited ones) on top of the stacks of v 6∈ T (1). Similarly, the LERW fromv2 correpsonds to following the arrows from v2 and popping cycles as they

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are encountered. When the algorithm is complete, the algorithm outputs thetree on top of the stacks.

The above shows that the algorithm defines a particular way of poppingcycles until a tree is uncovered. Note that each popped cycle is coloured(each of its oriented edges has a colour), and a particular coloured cycle canbe popped at most once. We now prove that we may in fact pop cycles inany order we wish, and regardless of what we do, the same coloured cycleswill be popped, and the same tree uncovered.

Let C1, . . . , Cm be a sequence of oriented coloured cycles that can bepopped (in this order), with a tree t resulting. We know from the algorithm,that such a finite sequence exists with probability 1. Let D1, D2, . . . be anyother possible sequence of coloured oriented cycles that can be popped. Weshow that the D-sequence consists of the same coloured cycles as the C-sequence. We prove this by induction on m. If m = 0, then there are nocycles to be popped, hence the D-sequence is also empty. Assume now thatm ≥ 1, and that the statement is true for C-sequences of length less than m.Let Di be the first cycle in the D-sequence that is not disjoint from C1. HenceDi and C1 share a vertex w1. Since D1, . . . , Di−1 are disjoint from C1, they donot contain w1, and hence the colour of w1 is the same in C1 andDi. It followsthat w1 has the same successor w2 in C1 andDi. Using again that w2 does notoccur in D1, . . . , Di−1, we see that w2 has the same colour in C1 and Di, andso on. We get that C1 and Di are the same coloured cycle. Popping Di = C1

commutes with popping D2, . . . , Di−1, hence popping the D-sequence hasthe same effect as popping Di = C1, D2, D3, . . . , Di−1, Di+1, . . . . Now pop C1

from the top of the stacks. By the induction hypothesis, C2, . . . , Cm consistsof the same coloured cycles as D2, . . . , Di−1, Di+1, . . . , and it uncoveres thesame tree t. The claim follows.

We now show that the tree uncovered by cycle popping has probabilityproportional to weight. Let C1, . . . , Cm be any sequence of coloured orientedcycles that can be popped in this order, and t any tree rooted at r. Theprobability that C1 can be popped is the probability that the arrows in C1

all point ”the right way”, that is∏

e∈C1C(e)/Ce. Using independence of the

elements in the stacks, the conditional probability that Ci can be popped,given that C1, . . . , Ci−1 can be popped is

∏e∈Ci

C(e)/Ce. Finally, using againindependence of the elements in the stacks, conditioned on all the cyclespopped, the probability that the tree t is uncovered, is

∏e∈tC(e)/Ce. Hence

we have

P[C1, . . . , Cm can be popped and as a result t is uncovered]

=m∏

i=1

∏

e∈Ci

C(e)

Ce×∏

e∈t

C(e)

Ce.

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The last factor is const · weight(t). Note that t varies independently of thecycles: given any sequence of cycles popped, the tree uncovered may be anytree rooted at r. This shows more than we need: even if we condition on thesequence of cycles popped, the tree has the claimed distribution.

Remark. The proof showed not only that the result of the algorithm is inde-pendent of the sequence v1, . . . , vn, but also that we may, if we wish, pick vi+1

depending on what happened in the algorithm up to that point. The proofalso showed that the distribution of the number of random walk steps neededis independent of how v1, . . . , vn was chosen. The proof in fact establishes aprobability space on which any choice results in the same number of steps.This is called a coupling of the different instances of the algorithm.

Remark. The same proof works for any Markov chain in place of a networkrandom walk. That is, if G is the directed graph constructed from a Markovchain as in Section 2, the algorithm outputs a rooted tree (t, r) proportionalto q(t, r) =

∏[v,w]∈t pvw.

Remark. The expected number of Markov chain steps used by the algorithmcan be found as follows. To find the expected number of times that a stepout of u is used, we may assume, by the proof, that v1 = u. Then we need tofind E[# visits to u before τr], where τr = infn ≥ 0 : X(n) = r. For anyirreducible Markov chain, this expectation equals π(u)[Euτr + Erτu], where πis the stationary distribution. [2, Chapter 2]. Hence the expected runningtime of the algorithm (measured by the number of Markov chain steps) is

∑

u 6=r

π(u)[Euτr + Erτu].

Note that this is always at most 2EC, where C is the cover time, and maybe smaller.

Remark. Later we will see variants of Wilson’s algorithm on infinite graphsgive useful information. As a first example, consider the square lattice G =(Z2,E2). Simple random walk is recurrent, that is, for all v, w ∈ Z

2,

Pv[X(n) = w for some n ≥ 1] = 1.

Pick r ∈ Z2, and let v1, v2, . . . list all the vertices of Z

2. Since T (i) is hitwith probability 1, the algorithm can be run, and results in a spanning treeof (Z2,E2).

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4 Electric networks and spanning trees

In this section we describe the relation between electrical networks, spanningtrees and random walk. To a great extent, we follow the exposition in [5].The goal is to define the notion of current, and use it to describe the jointprobability that edges e1, . . . , ek belong to a random spanning tree. Althoughthe physical interpretations of ”resistance”, ”current”, etc. will not be needed,we note that a nice introduction to these and their connection with randomwalk can be found in [11].

4.1 The gradient and divergence operators

Let (G,C) be a finite network. We denote by ℓ2(V ) the space of real-valuedfunctions on V , with the inner product:

(f, g)C :=∑

v∈V

Cvf(v)g(v), (5)

and norm ‖f‖C . We denote by ℓ2−(E) the space of antisymmetric functionson oriented edges, that is, functions θ : E → R satisfying θ(e) = −θ(e) forall e ∈ E. We equip this space with the inner product:

(θ, θ′)R :=1

2

∑

e∈E

R(e)θ(e)θ′(e) =∑

e∈E1/2

R(e)θ(e)θ′(e), (6)

where E1/2 contains each edge of G with exactly one orientation. The energyof θ is

E(θ) := (θ, θ)R = ‖θ‖2R.

The gradient operator is ∇ : ℓ2(V ) → ell2−(E), defined by

(∇F )(e) := C(e)(F (e) − F (e)). (7)

The divergence operator is div : ℓ2−(E) → ℓ2(V ), defined by

(div θ)(v) :=1

Cv

∑

e:e=v

θ(e). (8)

−∇ and div are adjoints of each other: (θ,−∇F )R = (div θ, F )C. To seethis, write

(θ,−∇F )R =1

2

∑

e∈E

R(e)θ(e)C(e)(F (e) − F (e)) =1

2

∑

e∈E

θ(e)(F (e) − F (e)).

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Since θ(e)(−F (e)) = θ(e)F (e), this can be written as

∑

e∈E

θ(e)F (e) =∑

e∈E

∑

v∈V

I[e = v]F (v)θ(e) =∑

v∈V

CvF (v)1

Cv

∑

e∈E

I[e = v]θ(e)

=∑

v∈V

CvF (v)div θ(v).

To motivate what follows, imagine that the network (G,C) is an electricnetwork, where edge e is a resitor with resistance R(e). Suppose that wehook up a battery between the two endpoints of the edge e, and supposethat a unit of current flows through the battery. Let Ie(f) be the amount ofcurrent that flows along the edge f . How to determine Ie(f)? We know thatcurrent is conserved at each vertex v 6= e, e. Hence, div Ie(v) = 0. A unit ofcurrent comes in at e, and a unit of current is taken out at e, so

div Ie =1

Ce1e −

1

Ce1e. (9)

This does not uniquely specify Ie, however, there is a physical principle calledThompson’s principle that states that Ie has minimal energy among all flowsthat have divergence equal to (9). Here we will adopt this characterizationas the mathematical defintion of the current in (10) below.

4.2 The gradient and cycle spaces

We define the unit flow along the edge e: χe ∈ ℓ2−(E) defined by χe := 1e−1e,where 1f denotes the function that is 1 on the edge f and zero elsewhere.Note that this has the required divergence (9). Let Gr denote the space ofgradients:

Gr := ∇ℓ2(V ).

The flow −∇1v =∑

e;e=v χe is called the star at v. The space Gr is spanned

by the stars. If e1, . . . , ek is an oriented cycle in G, then the flow∑k

i=1 χei ∈

ℓ2−(E) is called a cycle. We call

Cyc := linear span of all cycles ⊂ ℓ2−(E).

Lemma 4.1. We have ℓ2−(E) = Gr ⊕ Cyc.

Proof. Stars are orthogonal to cycles. For this note that if v is not on thecycle e1, . . . , ek, then the star at v and this cycle are supported on disjointedges, and hence are orthogonal. When v = ei = ei+1, then the inner productis R(ei)(−C(ei)) + R(ei+1)C(ei+1) = 0 (note that if v occurs multiple times

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in the cycle, a similar calculation remains valid). Therefore, what we needto show is that if θ is orthogonal to Cyc, then it is a gradient. This goesby a well-known argument. Fix o ∈ V , and for any v ∈ V let e1, . . . , ek beany path in G from o to v. Define F (v) :=

∑j R(ej)θ(ej). This definition is

independent of the path chosen, since if e′1, . . . , e′l is another path, then θ is

orthogonal to the cycle e1, . . . , ek, e′l, . . . , e

′1. It is easy to see that ∇F = θ,

and this completes the proof.

For a subspace Z of ℓ2−(E), let PZ be the orthogonal projection onto Z,and let P⊥

Z be the orthogonal projection onto the orthogonal complement ofZ.

We are ready to defineIe := PGrχ

e. (10)

Since Ie − χe ⊥ Gr, we have, for any F 0 = (Ie − χe,∇F )R = −(div Ie −div χe, F )C , which is equivalent to div Ie = div χe. Hence Ie indeed hassmallest energy among flows with divergence (9).

4.3 Connection with spanning trees

Let e and f be oriented edges of G. We define

β(e, f) = P[path from e to e in random spanning tree uses f ].

Start a network random walk at e, and stop when it hits e. Let Je(f) be thenet number of times edge f is used, that is

Je(f) := E[# times f is used − # times f is used].

Theorem 4.1. We have

β(e, f) − β(e, f) = Je(f) = Ie(f).

In particular,

P[e ∈ T ] = Pe[first hit e via the edge e] = Ie(e). (11)

Remark. The equality of the spanning tree quantity and the current is due toKirchhoff [19], and the equality of the random walk quantity and the currentis due to Doyle and Snell [11].

Proof. By Wilson’s algorithm, that path in the random spanning tree is aLERW from e to e. Hence

β(e, f)−β(e, f) = E[# times LERW uses f−# times LERW uses f ]. (12)

15

A cycle is traversed an equal number of times in expectation in both direc-tions, since if e1, . . . , ek is a cycle with v = e1, then

P[e1, . . . , ek are traversed |X(n) = v]

=C(e1)

Ce1

C(e2)

Ce2

. . .C(ek)

Cek

= P[ek, . . . , e1 are traversed |X(n) = v].

(13)

This imples that adding the loops of the random walk in the right hand sideof (12), does not change the expected net number of times f is used, hencethe β(e, f) − β(e, f) = Je(f).

To prove the second equality, let

F (v) := Ee[# visits to v up to τe].

Since for every v 6= e, e, any incoming step to v is balanced by an outgoingstep from v, we have div Je(v) = 0. At v = e there is one more outgoing stepthan incoming step, and at v = e, there is one incoming step and no outgoingstep. Hence div Je = (1/Ce)1e − (1/Ce)1e = div Ie. Hence, Je − Ie ⊥ Gr,and it is enough to show that Je ∈ Gr. Let

θv = Pv[first step of r.w. uses f ] − Pv[first step of r.w. uses f ]

= − 1

Cv

1v ∈ Gr.

Then it is easy to check that Je =∑

v F (v)θv ∈ Gr. It follows that Je = Ie,and the proof is complete.

Exercise 4.1. Complete the argument regarding loops being traversed anequal number of times on average. Condition on the loop-erasure of therandom walk path from e to e, and decompose the random walk path into itsloop-erasure v0 = e, v1, . . . , vN = e, and the oriented cycles C0, C1, . . . , CN−1,where Ci is a cycle based at vi. Further condition on the cycles withouttheir orientation, and deduce from (13) that the two orientations are equallylikely. Deduce that the expectation of the quantity defining Je(f) is equalto β(e, f) − β(e, f).

The matrix Y (e, f) = Ie(f) = 1R(e)

(Ie, χf)R = C(e)(PGrχe, χf)R is called

the transfer current matrix.

4.4 Contracting edges in a network

Definition 4.1. Let G = (V,E) be a graph, F ⊂ E. We denote by G/Fthe graph obtained by identifying, for every f ∈ F , the endpoints of f . We

16

identify the edges of G with those of G/F , where some edges in G havebecome loops in G/F . Hence we also have an identification of the space ℓ2(E)for the two graphs.

Let us write TG for the random spanning tree of G.

Proposition 4.1. Assume that G is a finite connected network, and that nocycle can be formed of the edges in F . Then TG conditioned on F ⊂ TG hasthe same distribution as TG/F ∪ F .

Proof. For any C ⊂ E, C ∩ F = ∅, we have that C ∪ F contains a cycleof G if and only if C contains a cycle of G/F . Hence C ∪ F is a spanningtree of G if and only if C is a spanning tree of G/F . Since weight(C ∪ F ) =∏

e∈C C(e)∏

f∈F C(f), we get the statement.

The above proposition shows that if we condition on a certain set ofedges F to be present in TG, the conditional probabilities are given by thecontracted network G/F . Hence we would like to see how the current Ie

changes when we contract edges in a network. This will turn out to be givenby applying an orthogonal projection to Ie.

Recall that we identify edges in G and edges in G/F (where the edges inF became loops in G/F ). This gives an identification of the spaces ℓ2−(E)for the two graphs.

Let Gr denote the gradients in G/F , and let Cyc denote the linear spanof cycles in G/F .

Lemma 4.2.Cyc = Cyc + 〈χF 〉,

where 〈χF 〉 is short for the linear span of χf : f ∈ F.

Proof. It is clear that cycles in G remain cycles in G/F , so Cyc is a subspace

of Cyc. Since edges in F become loops in G/F , it is also clear that 〈χF 〉is a subspace of Cyc. This shows that the right hand side is contained inthe left hand side. Suppose now that e1, . . . , ek form a cycle in G/F . Weshow that we can insert edges from F into this sequence to get a cycle in G,which will show that Cyc is contained in the right hand side. If ei−1 = ei

in the graph G, then there is no need to insert an edge between ei−1 and ei.If ei−1 = v1 6= v2 = ei in G, then v1 and v2 belong to the same connectedcomponent of F (since they are contracted to the same vertex in G/F ).Hence we can insert edges f1, . . . fk(i) ∈ F to get a path from v1 to v2.

17

Now writeℓ2−(E) = Gr ⊕ Cyc = Gr ⊕ Cyc,

where Cyc ⊃ Cyc, and consequently Gr ⊂ Gr. We can write Gr = Gr ⊕(Gr ∩ Cyc) and Cyc = (Gr ∩ Cyc) ⊕ Cyc. Hence

ℓ2−(E) = Gr ⊕ (Gr ∩ Cyc) ⊕ Cyc.

We show that the middle subspace equals Z := PGr〈χF 〉. For this note that

Gr ∩ Cyc = PGrCyc = PGrCyc + PGr〈χF 〉 = Z.

Henceℓ2−(E) = Gr ⊕ Z ⊕ Cyc.

Let now e be an edge that does not form a cycle together with edges from F(so that it is not a loop in G/F ). Then writing Ie for the current in G/F ,we have

Ie = PcGrχe = P⊥

Z PGrχe = P⊥

Z Ie. (14)

4.5 The transfer current theorem

The expression (14) allows us to prove the following beautiful theorem on thejoint probability that edges e1, . . . , ek belong to the random spanning tree.The theorem is due to [8], the proof we present is from [5].

Theorem 4.2 (Transfer current theorem). [Burton, Pemantle; 1993] Let Gbe a finite connected network. For distinct edges e1, . . . , ek ∈ E, we have

P[e1, . . . , ek ∈ TG] = det[Y (ei, ej)]1≤i,j≤k,

where Y (e, f) = Ie(f).

Proof. The left hand side is 0, if a cycle can be formed from the edgese1, . . . , ek, and we need to show that in this case the determinant is also0. Write the cycle as

∑j αjχ

ej ∈ Cyc (where αj ∈ −1, 0, 1). Then thefollowing linear combination of columns of the matrix Y vanishes:

∑

j

αjR(ej)Y (ei, ej) =∑

j

αjR(ej)Iei(ej)χ

ej(ej) =∑

j

αj(Iei, χej)R

= (Iei,∑

j

αjχej )R = (PGrχ

ei ,∑

j

αjχej)R = 0.

The last equality follows, since gradients are orthogonal to cycles. Therefore,the determinant is also 0.

18

Assume now that no cycle can be formed from the edges e1, . . . , ek. Wewrite, using that PGr = P 2

Gr, and that PGr is self-adjoint:

Y (e, f) = Ie(f) =1

R(f)(Ie, χf)R = C(f)(Ie, χf)R

= C(f)(PGrχe, χf )R = C(f)(PGrχ

e, PGrχf )R

= C(f)(Ie, If)R.

(15)

Hence

det[Y (ei, ej)] =k∏

i=1

C(ei) detYk, (16)

where Yk(ei, ej) = (Iei, Iej)R. The matrix Yk has the following structure: wehave k vectors, Ie1, . . . , Iek, and the (i, j) element is the inner product of the i-th and j-th vectors. Such a matrix is called a Gram matrix. The determinantof a gram matrix equals the squared volume of the paralelepiped spannedby its determining vectors. This is easy to see if the determining vectorsare pairwise orthogonal, as then the matrix is diagonal with diagonal entriesequal to the squared length of the determining vectors. When the vectors arenot orthogonal, we can applied the Gram-Schmidt orthogonalization processto them, and it is not hard to check that the determinant does not change,and neither the squared volume of the paralelepiped spanned by the vectors.Therefore, we can write (16) as

k∏

i=1

C(ei)‖P⊥ZiIei‖2

R, (17)

whereZi = spanIe1, . . . , Iei−1 = PGr(spanχe1, . . . , χei−1).

Now we have

P[e1, . . . , ek ∈ TG] =

k∏

i=1

P[ei ∈ TG | e1, . . . , ei−1 ∈ TG]

Prop. 4.1=

k∏

i=1

P[ei ∈ TG/e1,...,ei−1](11)=

k∏

i=1

Iei(ei)

(15)=

k∏

i=1

C(ei)(Iei, Iei)R

(14)=

k∏

i=1

C(ei)‖P⊥ZiIei‖2

R.

The last expression is the same as (17), which completes the proof.

19

Remark. A 0–1 valued process Xαα∈I is called determinantal if

P[Xα1, . . . , Xαk

= 1] = det[K(αi, αj)]1≤i,j,≤k

for some kernel K.

4.6 Monotonicity properties of currents

We need some monotnicity properties, that will be very important when welook at limits of spanning trees on infinite graphs.

Let (G,C) be a finite network and e ∈ E. When we need to indicate thata current is computed in a network H , we write Ie

H .

Proposition 4.2. (a) If G′ is a subgraph of G containing e, then IeG′(e) ≥

IeG(e).

(b) If F ⊂ E and F ∪ e has no cycles containing e, then IeG/F (e) ≤ Ie

G(e).

Proof. In both cases, the inequality results from the fact that one of thecurrents can be written as an orthogonal projection of the other, that reducesthe norm.

(a) Write G′ = (V ′, E ′). Since E ′ ⊂ E, there is a natural embeddingof ℓ2−(E ′) into ℓ2−(E), by setting θ(e) = 0 for all e ∈ E \ E ′. We haveℓ2−(E ′) = Gr′⊕Cyc′, where Cyc′ ⊂ Cyc (any cycle in G′ is also a cycle in G).We have χe ∈ ℓ2−(E ′) ⊂ ℓ2−(E). Since Ie

G′ is the projection of χe onto Gr′, wecan write χe = Ie

G′ + f ′, with f ′ ∈ Cyc′. Then

IeG = PGrχ

e = PGrIeG′ + PGrf

′ = PGrIeG′.

Hence

IeG(e)

(15)= C(e)(Ie

G, IeG)R = C(e)‖Ie

G‖2R = C(e)‖PGrI

eG′‖2

R ≤ C(e)‖IeG′‖2

R

= IeG′(e).

(b) We have IeG/F = P⊥

Z IeG. Hence, similarly to (a):

IeG/F (e) = C(e)‖Ie

G/F‖2R = C(e)‖P⊥

Z IeG‖2

R ≤ C(e)‖IeG‖2

R = IeG(e).

Proposition 4.2 has the following consequence for random spanning trees.

20

Proposition 4.3. Let G = (V,E) be a finite connected network, F ⊂ E.(a) If G′ is a subgraph of G containing F , then

P[F ⊂ TG′] ≥ P[F ⊂ TG].

(b) If e 6= f thenP[f ∈ TG | e ∈ TG] ≤ P[f ∈ TG].

More generally, if F ∩ F ′ = ∅, then

P[F ⊂ TG/F ′] ≤ P[F ⊂ TG].

Proof. (a) We may assume that F contains no cycles. We induct on |F |. IfF = e, then by (11) and Proposition 4.2(a),

P[e ∈ TG′ ] = IeG′(e) ≥ Ie

G(e) = P[e ∈ TG].

Induction step: F = F1∪e. Then G′/F1 is a subgraph of G/F1. Therefore,

P[F ⊂ TG′ ] = P[e ∈ TG′/F1]P[F1 ⊂ TG′ ] ≥ P[e ∈ TG/F1

]P[F1 ⊂ TG]

= P[F ⊂ TG].

(b) We have

P[f ∈ TG|e ∈ TG] = P[f ∈ TG/e] = IfG/e(f) ≤ If

G(f) = P[f ∈ TG].

General case follows by induction on |F |. If F = e, then

P[e ∈ TG/F ′ ] = IeG/F ′(e) ≤ Ie

G(e) = P[e ∈ TG].

For the induction step, write F = F1 ∪e. Note that G/F ′/F1 is a contrac-tion of G/F1. Therefore,

P[F ⊂ TG/F ′ ] = P[e ∈ TG/F ′/F1]P[F1 ⊂ TG/F ′]

≤ P[e ∈ TG/F1]P[F1 ⊂ TG] = P[F ⊂ TG].

21

5 Random spanning forests of infinite graphs

Some properties of random spanning trees become especially transparentwhen we consider the limit of an infinite graph. As an analogy, think aboutsimple random walk, and how the phenomenon of recurrence/transience ismost naturally expressed as a property of an infinite walk (although it couldbe formulated as a limiting property of finite walks).

It is not obvious, how to define a ”uniform” spanning tree on an infinitegraph, and we will encounter some surprises. The first is that there willbe more than one natural way to define the limit, and the results can bedifferent. The second is that in the context of infinite graphs, the naturalobjects will be spanning forests, rather than spanning trees.

To construct the limiting objects on infinite graphs, we exploit the monot-nicity properties proved in the previous section. For this section, G = (V,E)is an infinite connected network.

5.1 Measurable space

We will work on the space Ω = 0, 1E. This is a compact metric space inthe product topology. The Borel σ-algebra is generated by the elementarycylinders, that is, sets of the form

AB,K = F ⊂ E : F ∩K = B,

where B ⊂ K are finite sets of edges. The event AB,K expresses that theedges in B are present and the edges in K \B are not.

5.2 Exhaustion

Consider V1 ⊂ V2 ⊂ · · · ⊂ V , such that ∪nVn = V , Vn is finite. Let Gn =(Vn, En) be the subgraph induced by Vn. The sequence 〈Gn〉 is called anexhaustion of G by finite subgraphs.

5.3 Free spanning forest

Let µFn be the random spanning tree measure on Gn (this measure can be

realized on Ω).Since Gn is a subgraph of Gn+1, Proposition 4.3 implies that for any fixed

B ⊂ E finite we have

µFn [B ⊂ T ] ≥ µF

n+1[B ⊂ T ].

22

Note that this makes sense for large enough n, as then B ⊂ En. Hence wecan define

µF [B ⊂ T ] := limn→∞

µFn [B ⊂ T ], (18)

where the limit exists by monotonicity.

Exercise 5.1. Show that if B ⊂ K are fixed finite sets of edges, then

µF [T ∩K = B] := limn→∞

µFn [T ∩K = B],

exists. Hint: use the inclusion-exclusion principle to reduce the statement to(18).

By the result of the exercise, µF is defined on all elementary cylinders.By Kolmogorov’s extension theorem, µF has a unique extension to a measureon Ω.

Exercise 5.2. Show that µFn ⇒ FSF, in the sense of weak convergence of

probability measures on Ω.

The limit does not depend on the exhaustion: if 〈G′n〉 is another exhaus-

tion, we can findVn1

⊂ V ′n′

1⊂ Vn2

⊂ V ′n′

2⊂ . . .

and the limit along this third exhaustion has to coincide with both the limitalong 〈Gn〉 and the limit along 〈G′

n〉.The measure µF is called the (weighted) free spanning forest measure on

G, we will denote it FSF. We will write F for the set of edges present inan element of Ω. Then FSF[F contains a cylce] = 0, since any specific cyclebeing present is a cylinder event that has 0 probability under any µF

n , andhence also under FSF. It is obvious that FSF[F is spanning] = 1. HenceFSF[F is a spanning forest] = 1.

5.4 Wired spanning forest

The word ”free” refers to the fact how in defining µFn , we disconnected Gn

from the rest of the graph G. There is another natural way of taking thelimit, that is to force all connections outside of Gn to occur.

Define the graph GWn as the result of identifying all vertices in V \Vn to a

single vertex zn. This will result in infitely many loop edges at zn, which weomit, as they cannot be part of a spanning tree. Hence another descriptionof the graph GW

n is that we add to the graph Gn a new vertex zn, and foreach edge of G connecting a vertex v ∈ Vn with a vertex in V \ Vn, we placean edge between v and zn.

23

Let µWn be the random spanning tree measure on GW

n (also realized onΩ). Now GW

n is obtained by contracting some edges in GWn+1. Therefore, by

Proposition 4.3, we have

µWn [B ⊂ T ] ≤ µW

n+1[B ⊂ T ],

where this makes sense as soon as B ⊂ En. We again conclude that

µW [B ⊂ T ] := limn→∞

[B ⊂ T ]

exists, and similalry for µW [T ∩K = B]. The limit again does not depend onthe exhaustion. The measure defined this way is called the (weighted) wiredspanning forest measure, and we call it WSF. It is again concentrated onspanning forests of G.

The name ”wired” refers to how the complement of Gn has been short-circuited in the graph GW

n .

5.5 Examples

Let us pause a little to consider some simple examples of the constructionsabove.

Consider the d-dimensional integer lattice, that is the graph with vertexset Zd, where x, y is an edge if and only if |x − y| = 1. We set C(e) = 1for all edges. Let Vn = [−n, n]d ∩Zd, and let Gn = (Vn, En) be the subgraphinduced by Vn. We considered the random spanning tree measure (in this casethis is the uniform spanning tree measure) µF

n , and showed that it convergesweakly to a measure FSF. This was first proved by Pemantle [30]. The word”free” refers to boundary condition we used by disconnecting Gn from therest of the lattice. We also considered a different boundary condition, called”wired”, where we obtained the graph GW

n by adding a new vertex zn to Gn

and for every pair v ∈ Vn, w ∈ V \ Vn, v ∼ w, we place an edge between vand zn. The random spanning tree measure on GW

n was µWn , and we showed

it has a weak limit WSF. We will see later that on Zd, and on many othergraphs, we have FSF = WSF. That is, in the case of Zd, the difference inboundary conditions washes away as n→ ∞.

When are the two measures different? Here is a simple example wherethey are. Let G = (V,E) be a 3-regular tree, o a fixed vertex in V , andVn = x ∈ V : dist(o, x) ≤ n. Since Gn is a tree, its only spanning tree isitself, and hence µF

n concentrates on the single point Gn. It follows thatFSF concentrates on the single point G. The limit is more interesting forthe wired boundary condition. Let a be a neighbour of o, and use Wilson’smethod to generate the random spanning tree on GW

n , starting with the

24

vertices o and a. Since a random walk started at o has probability 2/3 tostep further from o and probability 1/3 to step closer to o, whenever not ato, we have

Po[Xk hits zn before a] ≥ α > 0

uniformly in n, and similarly,

Pa[Xk hits zn before o] ≥ α > 0.

It follows that

WSF[oa 6∈ F ] = limnµW

n [oa 6∈ F ] ≥ α2 > 0.

Therefore, WSF 6= FSF. Later we will see that under WSF, there are in-finitely many trees a.s.

We will also see later that independent random walks on Zd with d ≥5 have a positive probability of never intersecting, and we will be able toconclude that in this case there are also infinitely many trees a.s.

5.6 Wilson’s method on transient networks

Assume that (G,C) is an infinite network on which the network random walkis transient. The following variant of Wilson’s method was introduced in [5].

We construct a random spanning forest of G. Let F0 := ∅, and letv1, v2, . . . be an enumeration of V . We inductively define Fn as follows. Starta network random walk at vn, and stop it if it hits Fn−1, otherwise run itindefinitely. Let Pn be the path of this walk. Due to transience, Pn visits anyvertex finitely often. Hence LE(Pn) is well-defined. Set Fn := Fn−1∪LE(Pn).Finally, let F := ∪nFn. It is clear that F contains no cycles, and it spans G,so F is a random spanning forest of G. We call this process Wilson’s methodrooted at infinity.

Lemma 5.1. [5] The distribution of F does not depend on the chosen enu-meration of V .

This lemma can be proved similarly to the finite case. Its statement alsofollows from Theorem 5.1 below, however, the proof gives more: it showsthat one can realize the method with any choice of enumeration on the sameprobability space in such a way that the same random forest is generated.

Proof. Consider stacks and cycle popping as in the finite case. Fix all thestacks. Let us call a sequence D1, D2, . . . of coloured cycles legal, if (i) thecycles can be popped in this sequence; (ii) if at any stage an (uncoloured)

25

cycle is present then it is popped at some later stage. We say that a legalsequence terminates, if any finite F ⊂ E, becomes void of cycles eventually.For a legal, terminating sequence, the end result of popping all the cyclesD1, D2, . . . is well-defined, and is a spanning forest.

A fixed choice of v1, v2, . . . yields, almost surely, a legal, terminating se-quence C1, C2, . . . of coloured cycles. Fix the stacks so that this holds. Thenwe show that any other other legal sequence D1, D2, . . . is necessarily alsoterminating, and results in the same spanning forest. Let Di1 be the firstcycle in the D-sequence that is not disjoint from C1. As in the finite case, wesee that C1 = Di1 as coloured cycles. Popping Di1 commutes with poppingD1, . . . , Di1−1, hence D′ = 〈Di1 , D1, D2, . . . , Di1−1, Di1+1, . . . 〉 is also legal.Pop now C1 = Di1 from the stacks. Repeating the argument, we see thatthere exists i2 ∈ 1, 2, . . . \ i1, such that C2 = Di2 as coloured cycles, andthat moving Di2 to the beginning, we get a legal sequence. Inductively, wefind ik ∈ 1, 2, . . . \ i1, . . . , ik−1, such that Ck = Dik as coloured cycles.We show that i1, i2, . . . = 1, 2, . . .. Consider Dk. The C-sequence guar-antees that the uncoloured cycle Dk is not present after popping C1, . . . , CN

for some N . Hence, k ∈ i1, . . . , iN. We have thus proved that the C- andD-sequences consist of the same coloured cycles. It follows that D1, D2, . . . isterminating, and results in the same spanning forest. The result follows.

We now identify the distribution of the resulting spanning forest.

Theorem 5.1. [5] On any transient network, Wilson’s method rooted atinfinity yields a spanning forest with distribution WSF.

Proof. If P = 〈xk : k ≥ 0〉 is a deterministic path that visits every vertexfinitely often, then

LE(〈xk : k ≤ K〉) K→∞−−−→ LE(P).

The meaning of convergence here is that for any i, the i-th step of the pathon the left hand side is the same for all large enough K, and coincides withthe i-th step on the right hand side. If P is a random walk path, almostsurely,

LE(〈Xk : k ≤ K〉) K→∞−−−→ LE(P),

by transience.Recall that GW

n is obtained from G by identifying V \Vn to a single vertexzn. Consider the network random walk on G up to the hitting time of V \Vn.This is identical in law to the network random walk on GW

n up to the hittingtime of zn, hence we can use the network random walk on G for Wilson’salgorithm on GW

n with root zn.

26

We write T (n) for the random spanning tree on GWn , and we write G for

its limit, the wired spanning forest on G. Fix e1, . . . , eM ∈ E. Let u1, . . . , uL

contain all vertices that are endpoints of some ej . Let 〈Xk(uj)〉 be the randomwalk started at uj. Wilson’s method on GW

n , rooted at zn, uses the stoppingtimes τn

j that are the first time when the earlier part of the spanning treeis hit. Wilson’s method on G, rooted at infinity, uses the stopping times τj .Let

Pnj := 〈Xk(uj) : k ≤ τn

j 〉P := 〈Xk(uj) : k ≤ τj〉γn

j := LE(Pnj )

γj := LE(Pj)

Fnj := Fn

j−1 ∪ γnj .

We haveP[e1, . . . , eM ∈ T (n)] = P

[e1, . . . , eM ∈ ∪L

j=1γnj

]. (19)

We claim that almost surely, τnj → τj , γ

nj → γj, j = 1, . . . , L. This can be

proved by induction on j.When j = 1, τ1 = ∞. Here τn

1 = hitting time of V \ Vn, that goes toinfinity as n→ ∞.

Assume now j ≥ 2. Assume first the event τj = ∞. Fix a large integerN . By the induction hypothesis, we have γn

i ∩ GN = γi ∩ GN for all largeenough n, i < j. (The statement is clear if γi is a finite path. When γi

is infinite, the statement holds if Pi does not return to GN after time τni .)

Therefore, we have

Fnj−1 ∩GN = ∪i<jγ

ni ∩GN = ∩i<jγi ∩GN = Fj−1 ∩GN .

Since Pj does not hit Fj−1, it follows that τnj is greater than the exit time

from GN . Letting N → ∞, we get τnj → ∞.

Assume now the event τj <∞. For large N , we have Pj ⊂ GN . Again,from the induction hypothesis, Fn

j−1 ∩GN = Fj−1 ∩GN for all large enoughn. For such n, we have τn

j = τj , Pnj = Pj and γn

j = γj. The left hand side of(19) approaches P[e1, . . . , eM ∈ G]. Due to the convergence of the τn

j ’s andγn

j ’s, the right hand side of (19) converges to

P[e1, . . . , eM ∈ ∪Lj=1γj] = P[e1, . . . , eM ∈ F ].

This completes the proof.

27

5.7 Automorphisms

By a graph automorphism of G we mean a pair of bijections φ : V → V ,φ : E → E (for convenience, we denote them by the same letter), suchthat e is an edge between v1 and v2, if and only if φ(e) is an edge betweenφ(v1) and φ(v2). By a network automorphism of (G,C), we mean a graphautomorphism of G that in addition satisfies C(φ(e)) = C(e) for all e ∈ E.

A graph automorphism induces a map φ : Ω → Ω by letting e ∈ φ(F) ifand only if φ−1(e) ∈ F .

Proposition 5.1. FSF and WSF are invariant under any network automor-phisms.

Proof. If 〈Gn〉 is any exhaustion, then 〈φ−1(Gn)〉 is also an exhaustion. Forany finite B ⊂ E, we have

WSF[B ⊂ φ(F)] = WSF[φ−1(B) ⊂ F ] = limnµW

φ−1(Gn)[φ−1(B) ⊂ T ]

= limnµW

Gn[B ⊂ T ] = WSF[B ⊂ F ].

5.8 Trees are infinite

Proposition 5.2. On any infinite network (G,C), all components are infi-nite WSF-a.s. and FSF-a.s.

Proof. For any finite tree t ⊂ G, the event t is a component is a cylinderevent, that has probability 0 with respect to µF

n for all large n (since underµF

n , T is connected). Hence this event has probability 0 in the limit, andthere are countably many such events. The proof for WSF is the same.

5.9 Wilson’s method on recurrent networks / equality

Let (G,C) be an infinite recurrent network. We can make sense of Wilson’smethod onG, if we place the root at a fixed vertex r ∈ V . Indeed, if v1, v2, . . .is any enumeration of V \ r, then let F := r, let Pn be the path of thenetwork random walk started at vn, stopped when it hits Fn−1, which is finitealmost surely, by recurrence. Then let Fn := Fn−1∪LE(Pn), and F := ∪nFn.It is clear that F is a spanning tree of G.

Proposition 5.3. [5] On any recurrent network, and for any enumeration,Wilson’s method rooted at r yields a tree with distribution FSF = WSF.

28

Proof. Let B be a cylinder event, 〈Gn〉 an exhaustion. Let K0 be the end-points of edges on which B depends, and K := vj : ∃i ≥ j, vi ∈ K0.Write

∂intGn = x ∈ Vn : ∃y ∈ V \ Vn, x ∼ y.Run Wilson’s method on GW

n , rooted at r to generate TGWn

(note here thatwe use r, rather than zn, as the root, which is the same, due to Theorem3.1). Run Wilson’s method, rooted at r, to generate F . The random walks inthe two constructions are indistinguishable until one of ∂intGn is hit. Hencewe can put the two constructions on the same probability space in sucha way that they use the same random walks until ∂intGn is first hit. LetC1 := F ∈ B, C2 := TGW

n∈ B. then

∣∣P[F ∈ B] − µWn (B)

∣∣ = |P[C1] − P[C2]|≤ P[C1∆C2]

≤ P[some random walk started in K hits ∂intGn]

≤∑

v∈K

Pv[τ∂intGn < τr]n→∞−−−→ 0,

by recurrence. This shows that F has the distribution of WSF. In exactlythe same way, we obtain that F has the distribution of FSF.

It follows from the above theorem that on Z2, FSF = WSF.

5.10 Stochastic domination

In this section, we prove that WSF is always stochastically smaller thanFSF; see (20) below for what this means precisely. This is a very powerfulcomparison of the two measures, that sometimes allows us to conclude thatthe two measures are in fact the same.

Let 〈Gn〉 be an exhaustion of G by finite subnetworks. Since Gn is asubgraph of GW

n , we have, by Proposition 4.3(a),

µFn [e ∈ T ] ≥ µW

n [e ∈ T ]

for any edge e ∈ En. Letting n→ ∞, this implies that

FSF[e ∈ F ] ≥ WSF[e ∈ F ], e ∈ E.

We now state a more general inequality.

Definition 5.1. An event A ⊂ Ω is called increasing (or upwardly closed) ifF1 ∈ A, F2 ⊃ F1 imply that F2 ∈ A.

29

That is, an event is increasing, if whenever it occurs, adding more edgesto the configuration preserves the event. A simple example is:

at least one of e1, e2, e3 is present.

Proposition 5.4. [5, Section 5] For any increasing event A depending onthe edges in En, we have µF

n [A] ≥ µWn [A].

Corollary 5.1. For any increasing cylinder event A,

FSF[A] ≥ WSF[A]. (20)

It can be deduced from a theorem of Strassen [32] that (20) implies thatthere exists a measure ν on Ω × Ω with marginals WSF and FSF, such thatthe first coordinate is ν-almost surely contained in the second coordinate.That is,

ν[(F1,F2) : F1 ⊂ F2] = 1

ν[(F1,F2) : F1 ∈ B1] = WSF[B1]

ν[(F1,F2) : F2 ∈ B2] = FSF[B2].

(21)

Such a ν is called a monotone coupling of the measures WSF and FSF, andwe say that WSF stochastically dominates FSF. For a proof of Strassen’stheorem and the existence of the coupling ν, see [26, Section 10.2].

It is easy to see that (20) is necessary for (21). Indeed, if ν with properties(21) exists, then for any increasing event A

WSF[A] = ν[(F1,F2) : F1 ∈ A] = ν[(F1,F2) : F1 ∈ A, F1 ⊂ F2]

≤ ν[(F1,F2) : F2 ∈ A] = FSF[A].

Hence (20) and the existence of ν satisfying (21) are equivalent. The pointof Strassen’s theorem is to prove the non-trivial statement that (20) implies(21).

5.11 Sufficient conditions for FSF = WSF

We can use the monotone coupling to give sufficient conditions for the equal-ity FSF = WSF. These will be used later to show that the measures agreeon Zd as well as on some other graphs.

We saw that WSF[e ∈ F ] ≤ FSF[e ∈ F ] for all edges e ∈ E.

Proposition 5.5. If WSF[e ∈ F ] = FSF[e ∈ FSF] for all edges e ∈ E, thenFSF = WSF.

30

Proof. We have e ∈ F1 ⊂ e ∈ F2 ν-a.s. By assumption, we also have,

ν[e ∈ F1] = WSF[e ∈ F ] = FSF [e ∈ F ] = ν[e ∈ F2].

Hence e ∈ F1 = e ∈ F2 ν-a.s., and F1 = F2 ν-a.s. Note that withoutthe monotone coupling, we could could gain little from the equality of theone-dimensional marginals.

The following proposition is based very much on the same idea. It saysthat if the expected degree of vertex v is the same in the free spanning forestas in the wired spanning forest, for each v ∈ V , then the free and wiredspanning forests coincide.

Proposition 5.6. [5] If E[degF (v)] is the same under the measures FSF andWSF for all v ∈ V , then FSF = WSF.

Proof. Under the measure ν, the set of edges incident on v in F1 is a subsetof the set of edges incident on v in F2. Due to the assumption, the two setsof edges coincide ν-a.s., and hence F1 = F2 ν-a.s.

31

6 The number of trees on Zd

In this section we look at the question: is the random spanning forest on Zd

connected (that is, a single tree) or not? We first state a theorem that wewill prove in greater generality in the next section. The theorem is implicitin [30], and is proved explicitly in [15].

Theorem 6.1. On Zd, we have FSF = WSF for all d ≥ 1.

In view of this theorem, we can speak simply of the ”Uniform SpanningForest” on Zd: there is no need to specify the boundary condition, and wecall it uniform, since we use C(e) ≡ 1.

We will frequently use the notation: f ≍ g to denote that the positivequantities f and g (depending on some argument), are of the same order,that, is, there exist constants c1, c2 > 0, such that c1g ≤ f ≤ c2g.

The main result of this section is the following theorem of Pemantle [30].

Theorem 6.2. The uniform spanning forest on Zd is a.s. a single tree, ifd ≤ 4, and it has infinitely many components a.s., if d ≥ 5. When d ≥ 5,u 6= v ∈ Zd, then

P[u, v in the same component of F ] ≍ |u− v|4−d. (22)

In what follows, we will also use the notation ‖|u‖| := 1 + |u|, so thatwe do not have to make explicit exceptions for a negative power of 0, whenu = 0. Note that the right hand side of (22) is comparable to ‖|u− v‖|4−d.

Proof for d = 1, 2. The case d = 1 is easy to see. Since Gn = [−n, n] ∩ Zd isa tree, the FSF concentrates on the single point Z.

For d = 2, we saw in Proposition 5.3 that the forest is connected.

Henceforth we assume d ≥ 3, so the random walk is transient, and dueto Theorem 5.1 we can use Wilson’s method rooted at infinity to generatethe spanning forest. The proof we present is adapted from [27], [4] and [26],where substantially more general theorems are proved.

Fix u ∈ Zd, and let P0 := 〈Xk : k ≥ 0〉 and Pu := 〈Yk : k ≥ 0〉 be thepaths of independent simple random walks started at X0 = 0 and Y0 = u,respectively. Using Wilson’s method with an enumeration starting with 0, u,we immediately obtain the following proposition.

Proposition 6.1. Vertices 0 and u belong to the same component a.s., ifand only if P[LE(P0) ∩ Pu 6= ∅] = 1.

32

We will first focus on the case d ≥ 5.Preliminaries. The Green function of the walk is defined by

G(x, y) :=

∞∑

n=0

pn(x, y) = E

[ ∞∑

n=0

I[Xn = y]∣∣∣X0 = x

]

= E[# visits of Xn to y|X0 = x] = G(y, x).

Here pn(x, y) is the probability for simple random walk started at x to be at yafter n steps. The last equality follows from symmetry of pn(x, y) (reversibil-ity). Good estimates on pn(x, y) yield important information for G(x, y).Such estimates are provided by the Local Central Limit Theorem. We statethis below (although we will not need such a detailed estimate in what fol-lows). We say that x ∈ Zd is even if the sum of its coordinates are.

Theorem (Local CLT; [22, Section 1.2]). If x has the same parity as n ≥ 1,then

pn(0, x) = 2

(d

2πn

)d/2

e−d|x|2

n + E(n, x),

where

E(n, x) =

O(n− d

2−1)

O(

n−d/2

|x|2

).

Summing over n, one can deduce (see [22, Theorem 1.5.4]) that thereexists a constant ad > 0, such that

G(x, y) ∼ ad

|x− y|d−2, as |x− y| → ∞.

In particular,G(x, y) ≍ ‖|x− y‖|2−d. (23)

We first consider how likely is it that P0 and Pu intersect. Consider

V :=

∞∑

n=0

∞∑

m=0

I[Xn = Ym] = # intersections of P0 and Pu.

We find

EV =∑

z

∑

n

∑

m

P[Xn = z = Ym] =∑

z

(∑

n

pn(0, z)

)(∑

m

pm(u, z)

)

=∑

z

G(0, z)G(u, z) ≍∑

z

‖|z‖|2−d‖|u− z‖|2−d.

33

Lemma 6.1. If d ≥ 5, we have∑

z

‖|z‖|2−d‖|u− z‖|2−d ≍ ‖|u‖|4−d. (24)

Proof. Suppose that 2N ≤ ‖|u‖| < 2N+1. By symmetry, it is enough toconsider the contribution from ‖|z‖| ≤ ‖|u− z‖|. Then, separating the termswith n ≤ N + 1 and n > N +1, and using that there are order (2n)d verticeswith 2n ≤ ‖|z‖| < 2n+1, we get

LHS of (24) ≍∞∑

n=0

∑

z:2n≤‖|z‖|<2n+1

‖|z‖|≤‖|u−z‖|

‖|z‖|2−d‖|u− z‖|2−d

≍N+1∑

n=0

(2n)2−d (2N)2−d

(2n)d +

∞∑

n=N+2

(2n)d−2 (2n)d−2 (2n)d

≍(2N)4−d ≍ ‖|u‖|4−d.

This estimate is enough to yield the quantitative upper bound:

P[0 and u in the same component] = P[LE(P0) ∩ Pu 6= ∅]≤ P[P0 ∩ Pu 6= ∅] = P[V ≥ 1] ≤ E[V ] ≤ c‖|u‖|4−d.

We can also conclude that there are infinitely many trees a.s. Fix K ≥ 1, andpick vertices u1, . . . , uK far away from each other, so that ‖|ui − uj‖| ≥ ε−1,i 6= j. Then

P[∃ at least K components] ≥ P[u1, . . . , uK are in different components]

≥ 1 −∑

i6=j

c‖|ui − uj‖|4−d

≥ 1 − C(K)εd−4.

Letting ε ↓ 0, we get that there exists at least K components with probability1. Since this holds for any K ≥ 1, we get that there are infinitely manycomponents a.s.

Proving the quantitative lower bound is a bit more involved. We computethe second moment of V .

E[V 2] =∑

z

∑

n,i

∑

m,j

P[Xn = z = Ym, Xi = w = Yj]

=∑

z,w

∑

n,i

P[Xn = z, Xi = w]∑

m,j

P[Ym = z, Yj = w].(25)

34

By separating the cases n ≤ i and i < n, we can write

∑

n,i

P[Xn = z, Xi = w] =∞∑

n=0

∞∑

i=n

P[Xn = z]P[Xi = w|Xn = z]

+

∞∑

i=0

∞∑

n=i+1

P[Xi = w]P[Xn = z|Xi = w]

≤ G(0, z)G(z, w) +G(0, w)G(w, z).

We similarly get

∑

m,j

P[Ym = z, Yj = w] ≤ G(u, z)G(z, w) +G(u, w)G(w, z).

The product of the two sums gives four terms. Two and two of these areidentical, when summed over z, w, by symmetry, so we get

E[V 2] ≤ 2∑

z,w

G(0, z)G(u, z)G(z, w)2 + 2∑

z,w

G(0, z)G(z, w)2G(w, u)

≍∑

z

‖|z‖|2−d‖|u− z‖|2−d∑

w

‖|z − w‖|4−2d

+∑

z

‖|z‖|2−d∑

w

‖|z − w‖|4−2d‖|w − u‖|2−d.

(26)

Since 4−2d = −d+(4−d) < −d, the sum over w in the first term is boundedby a finite constant. The remaining sum over z, by Lemma 6.1, is bounded by‖|u‖|4−d. In the second term, the sum over w gives ‖|z − u‖|2−d; this can beseen by a decomposition similar to that used in the proof of Lemma 6.1, andwe do not give the details. The remaing sum over z can then be estimatedusing Lemma 6.1. Putting things together, we get E[V 2] ≤ C‖|u‖|4−d.

By the Cauchy-Schwarz inequality,

E[V 2]P[V ≥ 1] ≥ (E[V I[V ≥ 1]])2 = (EV )2.

HenceP[P0 ∩ Pu 6= ∅] = P[V ≥ 1] ≥ c‖|u‖|4−d.

We need more, to estimate P[LE(P0) ∩ Pu 6= ∅]. Here is the basic idea, howwe handle the loop-erasure. Suppose that the paths P0 and Pu intersectat Xn = z = Ym. Loop-erase P0 up to the intersection at time n, andcall the loop-erasure γ. If the earliest intersection of γ (as measured alongγ) with 〈Ym+k : k ≥ 0〉 comes no later than the earliest intersection with〈Xn+k : k ≥ 0〉, then the intersection of γ with 〈Ym+k : k ≥ 0〉 stays after

35

the rest of X is loop-erased. Given the intersection Xn = z = Ym, this eventwill happen with conditional probability at least 1/2, which will yield thestatement.

To write the argument precisely, we denote

LE(〈Xk : k ≤ n〉) =: 〈γn(i) : i ≤ K(n)〉σ(n) := min0 ≤ i ≤ K(n) : γn(i) = Xk for some k ≥ n

τ(m,n) := min0 ≤ i ≤ K(n) : γn(i) = Yk for some k ≥ m.

Let Im,n := I[Xn = Ym]I[τ(m,n) ≤ σ(n)]. Given the event Xn = z = Ym,the paths 〈Xn+k : k ≥ 0〉 and 〈Ym+k : k ≥ 0〉 are exchangeable. Therefore,

P[τ(m,n) ≤ σ(n)|Xn = z = Ym] ≥ 1/2. (27)

PutW :=

∑

n

∑

m

Im,n,

and note that W ≥ 1 implies that LE(P0) and Pu intersect. Then

EW =∑

n

∑

m

EIm,n =∑

z

∑

n

∑

m

E[Im,n|Xn = z = Ym]P[Xn = z = Ym]

≥ 1

2EV.

Since W ≤ V , we also have EW 2 ≤ EV 2. Hence

P[LE(P0) ∩ Pu 6= ∅] ≥ P[W ≥ 1] ≥ (EW )2

E[W 2]≥ (EV )2

4E[V 2]≥ c‖|u‖|4−d.

This concludes the proof of Theorem 6.2 in the case d ≥ 5.The quantitative estimate shows that the trees in the uniform spanning

forest are 4-dimensional, when d ≥ 5: consider

Vn := |B(n) ∩ component of 0|=∑

z:|z|≤n

I[z and 0 are in the same component].

Then E[Vn] ≍∑z:|z|≤n ‖|z‖|4−d ≍ n4. It is remarkable, that the dimension of

the trees is stable, and does not depend on d for large d. See [4] for furtherinteresting geometric properties of the uniform spanning forest.

We now deal with the case d = 3, 4. It is easy to see, using (23) thatEV =

∑z G(0, z)G(u, z) = ∞, when d = 3, 4. The difficulty is in handling

36

the loop-erasure, and show that an intersection occurs with probability 1.It turns out to be helpful to prove more, and show that LE(P0) and Pu

intersect infinitely often, with probability 1. To work with something finite,we define

VN :=N∑

n=0

N∑

m=0

I[Xn = Ym]

GN(x, y) =

N∑

n=0

pn(x, y) = GN(y, x).

Lemma 6.2. Assume that X0 = 0 = Y0. Then

(EVN)2

E[V 2N ]

≥ 1

4. (28)

Proof. We have

EVN =∑

z

N∑

n=0

N∑

m=0

pn(0, z)pm(u, z) =∑

z

GN(0, z)2 =: bN .

Similarly to the computation in (26), and using the Cauchy-Schwarz inequal-ity for the second term arising, we get EV 2

N ≤ 4b2N .

We will need the fact that the lower bound (28) still holds asymptotically,when the walks start at arbitrary vertices u, v ∈ Zd. Write Pu,v, Eu,v to denotethat X0 = u, Y0 = v.

Lemma 6.3. Assume now X0 = u and Y0 = v. Then we have

lim infn→∞

(Eu,vVN)2

Eu,v[V2N ]

≥ 1

4. (29)

Proof. Similarly to the computations in (26), and using Cauchy-Schwarz, weget Eu,vV

2N ≤ 4b2N . For the first moment, we write:

Eu,vVN =∑

z

N∑

n=0

N∑

m=0

pn(u, z)pm(v, z) =N∑

n,m=0

pn+m(u, v).

Similarly,

E0,0VN =N∑

n,m=0

pn+m(0, 0). (30)

By the Local CLT, as n + m → ∞ we have, pn+m(u, v) ∼ pn+m(0, 0). Sincethe sum in (30) diverges as N → ∞, we conclude that Eu,vVN ∼ E0,0VN , asN → ∞. The claim follows.

37

The key to the proof is the following lemma, which says that infinitelymany intersections will occur with a probability that is uniform over thestarting points u and v, and an arbitrary initial segment of the X-path.

Lemma 6.4. Fix a path 〈xj〉−1j=−ℓ, and set Xj := xj for −ℓ ≤ j ≤ −1. Let

X0 = u, Y0 = v. Then

P [|LE(〈Xn : n ≥ −ℓ〉 ∩ 〈Ym〉| = ∞] ≥ 1

16.

Proof. Denote

〈γn(i) : 0 ≤ i ≤ K(n)〉 := LE(〈Xk : −ℓ ≤ k ≤ n〉)σ(n) := min0 ≤ i ≤ K(n) : γn(i) = Xk for some k ≥ n

τ(m,n) := min0 ≤ i ≤ K(n) : γn(i) = Yk for some k ≥ m.

Let Im,n = I[Xn = Ym]I[τ(m,n) ≤ σ(n)], and

WN :=

N∑

n=0

N∑

m=0

Im,n.

As in (27), we see that EWN ≥ 12Eu,vVN → ∞, as N → ∞. Also, EW 2

N ≤Eu,vV

2N . Cauchy-Schwarz gives

E[W 2N ]P[WN ≥ εEWN ] ≥ (E[WNI[WN ≥ εEWN ]])2 ≥ (EWN − εEWN)2

= (1 − ε)2(EWN )2.

Hence, for large N ,

P[WN ≥ εEWN ] ≥ (1 − ε)2 (EWN )2

E[W 2N ]

≥ (1 − ε)2

4

(Eu,vVN)2

Eu,v[V2N ]

≥ (1 − ε)2

4

(1

4− ε

).

Since WN is monotone, this implies that WN → ∞ with probability at least14(1 − ε)2(1

4− ε). Letting ε ↓ 0, we get that P[WN → ∞] ≥ 1/16.

Finally, notice that on the event WN → ∞, the intersection LE(〈Xn :n ≥ −ℓ〉)∩〈Ym : m ≥ 0〉 is infinite. This is because every intersection countedin WN is counted at most a finite number of times, by transience.

Proof of Theorem 6.2 when d = 3, 4. Let E = |LE(P0) ∩ Pu| = ∞. ByLevy’s 0–1 law [12, Section 4.5], we have

I[E]a.s.= lim

nP0,u[E|X1, . . . , Xn, Y1, . . . , Yn]. (31)

38

By the Markov property, the conditional law of the continuation 〈Yn+k : k ≥0〉 is a random walk started at v = Yn, and the law of the continuation〈Xn+k : k ≥ 0〉 is a random walk started at u = Xn. Noting that for anyn ≥ 0, E is the same event as |LE(P0) ∩ 〈Yn+k : k ≥ 0〉| = ∞, we are inthe setting of Lemma 6.4, and can conclude that the right hand side of (31)equals

limn

PXn,Yn[E|X1, . . . , Xn] ≥ 1

16.

Therefore, I[E] ≥ 1/16 a.s., and hence P[E] = 1.

39

7 Average degrees and amenability

We start this section by proving Theorem 6.1, following [5]. It will turn outthat the proof is applicable in much greater generality. The generalization isrelated to the notion of amenability, and we explore this concept in the restof this section.

Proof of Theorem 6.1. Let Vn = [−n, n]d ∩ Zd, Gn = (Vn, En) the inducedsubgraph. First consider a fixed (deterministic) spanning forest F of Zd, suchthat all components of F are infinite. For K ⊂ V , we define the externaledge boundary of K as the set of edges between K and V \K:

∂EK := e ∈ E : e ∈ K, e ∈ V \K.

Let kn be the number of trees in F ∩En. Then kn ≤ |∂EVn|, since an infinitetree intersecting En contains a boundary edge, and different tree componentsof F ∩En cannot use the same boundary edge. We claim that (uniformly inF)

limn→∞

1

|Vn|∑

x∈Vn

degF(x) = 2. (32)

To see this, we note that the sum of the degrees counts each edge in F ∩En

twice, and it counts each edge in F ∩ ∂EVn once. Therefore,∑

x∈Vn

degF(x) ≤ 2|F ∩En| + |∂EVn| = 2(|Vn| − kn) + |∂EVn|

≤ 2|Vn| + |∂EVn|.

At the equality sign, we used that a tree component of F∩En with n verticeshas n− 1 edges, and that F is spanning. On the other hand, we have

∑

x∈Vn

degF(x) ≥ 2|F ∩ En| = 2(|Vn| − kn) ≥ 2|Vn| − 2|∂EVn|.

Hence

2 − 2|∂EVn||Vn|

≤∑

x∈Vn

degF(x) ≤ 2 +|∂EVn||Vn|

.

Since |∂EVn| = O(nd−1), we get (32).Now take expectation in (32) with respect to either FSF or WSF. By the

dominated convergence theorem, we have

limn→∞

1

|Vn|∑

x∈Vn

E[degF(x)] = 2.

40

By translation invariance, Proposition 5.1, we see that E[degF (x)] does notdepend on x, and in fact E[degF(x)] = 2 for all x ∈ Zd. This shows that theexpected degree of each vertex is the same under both FSF and WSF. ByProposition 5.6, we conclude that FSF = WSF.

The above proof used properties of Zd only in two places: we used trans-lation invariance, and we used that the sets Vn have small boundary relativeto their size. We now generalize these properties.

Definition 7.1. A graph or network is called transitive, if given x, y ∈ V ,there is an automorphism φ such that φ(x) = y.

By Proposition 5.1, for a transitive network, E[degF(x)] does not dependon x for either FSF or WSF.

A large class of examples of transitive graphs is provided by Cayley graphs.Let Γ be a countable group. We say that a set S ⊂ Γ generates Γ, if thesmallest subgroup containing S is Γ. In what follows we will assume thatΓ is finitely generated, that is, a finite generating set S exists. We will alsoassume that S is symmetric, that is, whenever s ∈ S, we also have s−1 ∈ S.Then any x ∈ Γ can be written as x = s1s2 . . . sn, sj ∈ S.

Definition 7.2. The (right-) Cayley graph of (Γ, S) is the graph G = (V,E)with V = Γ, and E = [x, y] : x = ys for some s ∈ S.

Left multiplication by any γ ∈ Γ is an automorphism: if φγ(x) = γx, wehave x = ys if and only if γx = γys. It follows that any Cayley graph istransitive, since given x, y ∈ V , φyx−1(x) = yx−1x = y.

Examples of Cayley graphs:

1. Zd with S = ±ei : i = 1, . . . , d, where ei are the unit coordinatevectors.

2. The free group on two letters, with S = a, b, a−1, b−1. This groupcan be described as the set of all finite words formed of elements of Swith no occurrence of the strings aa−1, a−1a, bb−1, b−1b. Composition isby concatenation, and removal of any forbidden strings. Some thoughtreveals that the Cayley graph is a 4-regular tree.

3. The free product of three copies of Z2. This can be described as the setof all finite words in the letters a, b, c such that no aa, bb, cc occur. Mul-tiplication is again by concatenation, and removal of forbidden strings.With S = a, b, c, the Cayley graph is a 3-regular tree.

41

4. The free product of Z2 and Z3. This can be described as the set ofall words in the letters a, b, such that no aa or bbb occurs. With S =a, b, bb, the Cayley graph consists of triangles joined together withline segments in a tree-like fashion.

Now we come to the generalization of the property of small boundary tovolume ratio.

Definition 7.3. The edge-isoperimetric constant of a graph G = (V,E) isdefined by

ιE(G) := inf

|∂EK||K| : K ⊂ V finite

.

We call the graph amenable, if ιE(G) = 0. We call the graph non-amenable,if ιE(G) > 0.

Hence amenability of a graph is equivalent to the existence of finite subsetsKn such that limn |∂EKn|/|Kn| = 0. In a non-amenable graph, each finitesubset has a ”large” edge boundary, in the sense that the size of the boundaryis at least a constant fraction of the volume. In an amenable graph, thereexist sets with small boundary.

Remark. A more general notion of amenability is introduced in [26, Chapter6].

Exercise 7.1. Show that a d-regular tree is non-amenable if d ≥ 3.

With the above notions, the same proof we had for Zd, yields the followingtheorem.

Theorem 7.1. On a transitive, amenable network, FSF = WSF.

The notion of amenability was originally introduced in the context ofgroups, and in the rest of this section we will illustrate it from the pointof view of Cayley graphs of groups. This will also explain the origin of thename.

We define the external vertex boundary of K as

∂VK := x ∈ V \K : ∃ x ∈ K such that x ∼ y.

Definition 7.4. The vertex-isoperimetric constant of a graph G = (V,E) isdefined by

ιV (G) := inf

|∂VK||K| : K ⊂ V finite

.

42

Proposition 7.1. For any transitive graph, ιE(G) > 0 if and only if ιV (G) >0.

Proof. Clear from the inequalities |∂VK| ≤ |∂EK| ≤ D|∂VK|, where D isthe degree of a vertex.

In particular, for Cayley graphs, amenability is equivalent to

∃ Kn ⊂ V finite such that limn

|∂VKn|/|Kn| = 0. (33)

The origin of the name ”amenable” is explained by the concept of aninvariant mean on a group. Let ℓ∞(Γ) be the space of all bounded realfunctions on Γ. A linear map µ : ℓ∞(Γ) → R is called a mean if µ(1) =1, and µ(f) ≥ 0 for f ≥ 0. For γ ∈ Γ and f ∈ ℓ∞(Γ), we define thefunction Rγf ∈ ℓ∞(Γ) by Rγf(x) := f(xγ). The mean µ is called invariant,if µ(Rγf) = µ(f) for all γ ∈ Γ and f ∈ ℓ∞(Γ). Hence an invariant meanis a way of averaging functions on Γ in such a way that the average valueis invariant under the transformations Rγ . As a play on words, we call Γamenable, if an invariant mean exists on Γ.

Let us see the connection of invariant means to the notion introducedfor graphs. We remind that we restrict the discussion to finitely generatedgroups, although amenability of groups applies in greater generality. If theCayley graph G is amenable, as witnessed by the sequence Kn, then we canconsider the means

µn(f) :=1

|Kn|∑

x∈Kn

f(x).

We now show that these means are ”almost invariant”. For γ ∈ S, we have

|µn(f) − µn(Rγf)| =1

|Kn|

∣∣∣∣∣∑

x∈Kn

(f(x) − f(xγ))

∣∣∣∣∣ .

If x1 = x2γ, x1, x2 ∈ Kn, then the terms f(x1) and f(x2γ) cancel in the sum.The terms that do not cancel are: x1 ∈ Kn, such that x2 = x1γ

−1 6∈ Kn, andx2 ∈ Kn such that x1 = x2γ 6∈ Kn. Hence the number of terms that do notcancel is bounded by 2|∂VKn|, and therefore,

|µn(f) − µn(Rγf)| ≤ 2‖f‖∞|∂VKn||Kn|

n→∞−−−→ 0. (34)

Since any γ ∈ Γ can be written as a product of elements of S, we concludethat (34) holds for all γ ∈ Γ.

43

In order to use (34) to extract an invariant mean, we need a few factsfrom functional analysis. The space X = ℓ∞(Γ) is a Banach space with thesupremum norm, and means belong to the unit ball of the dual space X∗.The weak* topology on X∗ is given by specifying the following base: withµ ∈ X∗, f1, . . . , fk ∈ X, and ε > 0, the sets

U(µ, f1, . . . , fk, ε) := ν ∈ X∗ : |ν(fj) − µ(fj)| < ε, j = 1, . . . , k

form a base. Since in our case X is not separable, care should be taken asthis topology is not metrizable. Let B denote the unit ball in X∗. By atheorem of Alaoglu [25, Theorem 12.3], B is compact in the weak* topology.The sets AN := µn : n ≥ N have the finite intersection property, and hencethe intersection ∩NAN is non-empty. Let µ ∈ ∩NAN . Then µ is a weak*cluster point of the sequence µn, that is, given any neighbourhood U of µ,and any N , there exists n ≥ N such that µn ∈ U . Now apply this to theneighbourhoods U(µ, f, Rγf, ε) to conclude that that µ(f)−µ(Rγf) = 0. Wehave shown that the condition (33) implies that an invariant mean µ existson Γ.

A remarkable theorem of Følner [13] implies that the converse is also true:

Theorem 7.2. If an invariant mean exists on Γ, then for any Cayley graphof Γ, (33) holds.

We do not prove this here.We conclude this section with a few results on the connection between

algebraic properties of groups and their amenability. Write

B(n) := x ∈ Γ : x = s1 . . . sk, k ≤ n, sj ∈ S;

this is the ball of radius n centred at the identity in the Cayley graph.

Proposition 7.2. If |B(n)| does not grow exponentially, then Γ is amenable.

Proof. We have

|B(n+ 1)| = |B(n) ∪ ∂VB(n)| ≥ (1 + ιV (G))|B(n)|.

Hence, if Γ is non-amenable, then, ιV (G) > 0, and |B(n)| grows exponen-tially.

Proposition 7.3. An Abelian group is amenable.

Proof. We have |B(n)| ≤ (2n + 1)|S|, hence by Proposition 7.2, the group isamenable.

44

Proposition 7.4. If Γ is amenable, H is a subgroup of Γ, then H is alsoamenable.

Proof. Let µ be an invariant mean on Γ. We construct an invariant mean onH . Let f ∈ ℓ∞(H). The idea is to ”lift” f to a function on Γ, and use theinavriant mean on Γ. In each left coset of H , we fix an element x0, so thatthe coset takes the form x0H . We define f(x0h) = f(h), h ∈ H . We putµ(f) := µ(f). The requirments µ(1) = 1 and positivity are immediate. Toprove invariance, let g ∈ H . Chasing the defintions, we have

Rgf(x0h) = (Rgf)(h) = f(hg) = f(x0hg) = (Rg f)(x0h),

hence Rgf = Rgf . It follows that

µ(Rgf) = µ(Rgf) = µ(Rgf) = µ(f) = µ(f).

A consequence of Proposition 7.4 is that if Γ contains a non-amenablesubgroup, then it has to be non-amenable. For example, if Γ contains a freegroup, then, by Exercise 7.1 and Theorem 7.2, Γ is non-amenable.

Proposition 7.5. If H is a normal subgroup of Γ, and H and Γ/H areamenable, then Γ is amenable.

Corollary 7.1. Solvable groups are amenable.

The proof of Proposition 7.5 is based on somewhat similar ideas as theproof of Proposition 7.4, and we do not give it here. (We can average overcosets using the invariant mean on H , then average the averages using theinvariant mean on Γ/H .)

45

8 Currents on infinite networks

In this section we extend the definition of current to infinite networks. Therewill be two natural ways to do this, that will correspond to the free and wiredboundary conditions.

Let (G,C) be an infinite network. The space ℓ2−(E) consists of antisym-metric functions θ on the edges such that E(θ) = ‖θ‖2

R = 12

∑e∈E R(e)θ(e)2 <

∞. This is a real Hilbert space with inner product defined by (6). The Hilbertspace ℓ2(V ) has inner product (5). We define the operators ∇ and div bythe same formulas as in (7) and (8).

Exercise 8.1. Show that ‖∇F‖R ≤√

2‖F‖C, and ‖divθ‖C ≤√

2‖θ‖R, andthat −∇ and div are adjoints of each other.

Just as in the finite case, we see that each function ∇1v is orthogonal toeach

∑e∈C χ

e, C a cycle in G. It follows that if we put

Cyc0 := span〈∑

e∈C

χe : C a cycle in G〉,

Gr0 := span〈∇1v : v ∈ V 〉,

then Gr0 ⊥ Cyc0. We do not know if Gr0 and Cyc0 together span ℓ2−(E) ornot, and as we will see, this is equivalent to the question whether FSF =WSF.

We define the free current in G as IeF := P⊥

Cyc0χe. We define the wired

current in G as IeW := PGr0χ

e.

8.1 Free boundary condition

Let 〈Gn〉 be an exhaustion of G by finite subnetworks, Gn = (Vn, En). ForGn, Lemma 4.1 gives the decomposition:

ℓ2−(En) = Grn ⊕ Cycn ⊂ ℓ2−(E),

where ⊂ denotes the natural inclusion of ℓ2−(En) in ℓ2−(E). We earlier defined

IeGn

= PGrnχe = P⊥

Cycn.

Here P⊥Cycn

is a projection taking place in the space ℓ2−(En). By the natural

inclusion, we may view it as taking place in ℓ2−(E).

Proposition 8.1. We have ‖IeGn

− IeF‖R → 0 as n→ ∞, and

E(IeF ) = Ie

F (e)R(e). (35)

46

Proof. As a cycle in Gn is also a cycle in Gn+1, we have

· · · ⊂ Cycn ⊂ Cycn+1 ⊂ · · · ⊂ Cyc0.

Since any cycle in G is contained in some Gn, we have Cyc0 = ∪nCycn. Thefirst claim now follows from the following exercise.

Exercise 8.2. For any θ ∈ ℓ2−(E), we have limn ‖P⊥Cycn

θ − P⊥Cyc0θ‖R = 0.

Proposition 8.1 indeed follows, since

IeGn

= P⊥Cycn

χe → P⊥Cyc0χ

e = IeF .

For the second claim, write

R(e)IeF (e) = (P⊥

Cyc0χe, χe)R = (P⊥

Cyc0χe, P⊥

Cyc0χe)R = E(Ie

F ). (36)

8.2 Wired boundary condition

We have the ”wired graph” GWn = (Vn ∪ zn, EW

n ). Again we have, byLemma 4.1, the decomposition

ℓ2−(EWn ) = GrW

n ⊕ CycWn ⊂ ℓ2−(E).

Letting ∇(n) denote taking the gradient in the graph GWn , we have

GrWn = span〈∇(n)1v : v ∈ Vn ∪ zn〉.

Lemma 8.1. We have GrWn ⊂ Gr0.

Proof. For v ∈ V , we have ∇(n)1v = ∇1v ∈ Gr0, as functions in ℓ2−(E). Forzn, we can write 1zn = 1 −∑v∈Vn

1v, and hence

∇(n)1zn = −∑

v∈Vn

∇(n)1v =∑

v∈Vn

∇1v ∈ Gr0.

Proposition 8.2. We have ‖IeGW

n− Ie

W‖R → 0 as n→ ∞, and

E(IeW ) = Ie

W (e)R(e). (37)

Moreover, IeW has minimal energy among all θ ∈ ℓ2−(E) satisfying divθ =

divχe.

47

Proof. We have, using Lemma 8.1,

· · · ⊂ GrWn ⊂ GrW

n+1 ⊂ · · · ⊂ Gr0.

For any v ∈ V , ∇1v is in some GrWn , so we have Gr0 = ∪nGrW

n . By a similarargument as in Exercise 8.2, we get

IeGW

n= PGrW

nχe → PGr0χ

e = IeW .

The statement on E(IeW ) follows as in (36). For the second part of the

proposition, observe that

divθ = divχe ⇔ ∀v ∈ V 0 = (div(θ − χe), 1v)C = (θ − χe,−∇1v)R

⇔ θ − χe ⊥ Gr0.

Hence the projection PGr0χe has minimal norm.

Remark. From the definition of the free and wired currents, it is clear thatE(Ie

W ) ≤ E(IeF ), and that equality holds if and only if Ie

W = IeF .

We now describe when the free and wired currents are the same. We callF : V → R harmonic at v, if div∇F (v) = 0. From the definitions we have

div∇F (v) =[

1Cv

∑w C(v, w)F (w)

]− F (v), so this is a discrete version of

∆F = 0. We say that F : V → R is a Dirichlet function, if E(∇F ) < ∞.Let HD(G) denote the space of harmonic Dirichlet functions.

Proposition 8.3. ℓ2−(E) = Gr0 ⊕ Cyc0 ⊕∇HD.

Proof. Suppose that θ ∈ (Gr0)⊥∩(Cyc0)⊥. Due to orthogonality to cycles, wesee, just as in the finite case, that θ = ∇F for some F : V → R. In particular,F is a Dirichlet function. Due to the first orthogonality assumption, we havefor all v ∈ V :

(div∇F, 1v)C = (∇F,−∇1v)R = 0.

Hence F is harmonic, and we have shown that (Gr0)⊥ ∩ (Cyc0)⊥ ⊂ ∇HD.For the reverse inclusion, assume that ∇F ∈ ∇HD. Due to harmonicity,

for all v ∈ V we have

(∇F,∇1v)R = (div∇F, 1v)C = 0,

so ∇F ⊥ Gr0. Also, being a gradient, ∇F is orthogonal to cycles, hence∇F ⊥ Cyc0. The proposition follows.

48

Constant functions are in HD(G), but their gradients do not contributeto the subspace ∇HD. Sometimes constants are the only harmonic Dirichletfunctions. The following theorem was proved by Benjamini, Lyons, Peresand Schramm [5, Theorem 7.3].

Theorem 8.1. The following are equivalent.(i) FSF = WSF;(ii) Ie

W = IeF for all e ∈ E;

(iii) ℓ2−(E) = Gr0 ⊕ Cyc0;(iv) HD(G) ∼= R.

Proof. We first show the equivalence of (i) and (ii). Due to Proposition 5.5,equality of FSF and WSF is equivalent to WSF[e ∈ F ] = FSF[e ∈ F ] for alle ∈ E. This is the same as Ie

W (e) = IeF (e) for all e ∈ E, which by (35) and

(37) is equivalent to E(IeW ) = E(Ie

F ) for all e ∈ E. Due to Remark 8.2, thisis equivalent to (ii).

To see the equivalence of (ii) and (iii), note that (iii) is equivalent toP⊥

Cyc0 = PGr0. Since χe forms a basis, this is equivalent to P⊥Cyc0χe = PGr0χ

e

for all e ∈ E, which is (ii).The equivalence of (iii) and (iv) is immediate from Proposition 8.3.

49

9 Scaling limits

An important question is what can we say about our models when the graphZd is replaced by a very fine grid δZd, and δ → 0. Does the model convergein a suitable sense to a continuum model, often called scaling limit? We willonly look at this question for the basic building block of spanning trees, theloop-erased random walk.

9.1 LERW in dimensions d ≥ 5

We give the proof of the result of Lawler [22] that suitably scaled loop-erasedwalk in dimensions d ≥ 5 converges to Brownian motion. The idea of theproof is to show that a positive fraction of points are not erased from therandom walk path generating the loop-erased walk, and hence the resultwill follow from the convergence of scaled simple random walk to Brownianmotion.

Let X and Y be independent simple random walks, X0 = 0 = Y0. LetV denote the number of intersections of the two paths. We have shown inSection 6, that EV =

∑n

∑m P[Xn = Ym] <∞, if d ≥ 5. We show that this

implies that there is positive probability that the intersection of the two pathsonly consists of the intersection at n = 0 = m. Write (n,m) (n1, m1), ifn ≤ n1 and m ≤ m1, and write ≺, when equality does not hold. Define

h := P[X(n) 6= Y (m), (0, 0) ≺ (n,m)].

Lemma 9.1. If d ≥ 5, then h > 0.

Proof. Call (n,m) a *-last intersection, ifX(n) = Y (m), andX(n1) 6= Y (m1)for (n,m) ≺ (n1, m1). Note that in general, *-last intersections are notunique. Since P[V <∞] = 1, there exists at least one *-last intersection a.s.This implies that

1 ≤∑

n

∑

m

P[(n,m) is a *-last intersection]

=∑

n

∑

m

P[X(n) = Y (m), X(n1) 6= Y (m1) for (n,m) ≺ (n1, m1)]

=∑

n

∑

m

P[X(n) = Y (m)]h = hEV.

This shows that h ≥ (EV )−1 > 0.

We introduce the notation 〈X(i) : i ≥ 0〉 := LE(〈X(n) : n ≥ 0〉) Let

σ(i) = time of last visit to X(i) = supn : X(n) = X(i).

50

Letρ(j) = i for σi ≤ j < σi+1.

Then it is clear that ρ(σ(i)) = i, and X(i) = X(σ(i)). We will use theshorthand X[k, l] = 〈X(n) : k ≤ n ≤ l〉, and similalry X(k, l], etc. Let

Zn :=

1 if σ(i) = n for some i ≥ 0;

0 otherwise.

Then we have

ρ(n) =n∑

j=1

Zj

= # points remaining of X[0, n] after loop-erasure.

It will be useful to extend X to a two-sided walk:

Xn :=

Xn 0 ≤ n <∞;

Y−n −∞ < n ≤ 0.

Note that the increments X(n+ 1) −X(n), −∞ < n <∞ are i.i.d. We callj loop-free if X(−∞, j] ∩X(j,∞) = ∅. Then

b := P[j loop-free] = P[X(−∞, j] ∩X(j,∞] = ∅] ≥ h > 0.

Lemma 9.2. If d ≥ 5, with probability 1, there are infinitely many positiveand negative loop-free points.

Proof. Let U := # positive loop-free points. We call j n-loop-free, if X[j −n, j] ∩X(j, j + n] = ∅. Then

P[j n-loop-free] =: bn → b as n→ ∞.

Let

Vi,n := (2i− 1)n is loop-freeWi,n := (2i− 1)n is n-loop-free.

Then Wi,n, i = 1, 2, . . . are independent. We have

P[U ≥ k] ≥ P

[m∑

i=1

I[Vi,n] ≥ k

]

≥ P

[m∑

i=1

I[Wi,n] ≥ k

]− P

[m∑

i=1

I[Wi,n \ Vi,n] ≥ 1

]

51

Given ε > 0, the first term can be made at least 1 − ε, by choosing m large(uniformly in n). The absolute value of the second term is at most m(bn−b),which can be made less than ε by choosing n large. Hence P[U ≥ k] ≥ 1−2ε.Letting ε→ 0 and k → ∞, we get P[U = ∞] = 1.

Theorem 9.1. If d ≥ 5, there exists a = a(d) > 0, such that

limn→∞

ρ(n)

n= a a.s.

Proof. Let j0 := infj ≥ 0 : j loop-free. Let the sequence of loop-free pointsbe

· · · < j−2 < j−1 < j0 < j1 < j2 < . . .

Erase loops on each piece X[ji, ji+1] separately. Let

Zn := I[n-th point is not erased]

= I[LE(X[ji, n]) ∩X(n, ji+1] = ∅],

where ji ≤ n ≤ ji+1. The following observation is crucial: if n ≥ j0, thenZn = Zn. This is due to the following. Since j0 is loop-free, LE(X[0, j0])does not influence loop-erasure of the continuation X(j0,∞]. Similarly, sincej1 is loop-free, LE[j0, j1] does not influence loop-erasure of the continuationX(j1,∞), etc. This implies the claim.

Shifting the path X so that X(n) becomes the origin, we see that thesequence 〈Zn : n ≥ 0〉 is stationary. It is also ergodic, as a function of thei.i.d. process X(n + 1) − X(n), −∞ < n < ∞. Hence by the observationabove, and the ergodic theorem [12, Section 6.2], we have

limn→∞

ρ(n)

n= lim

n→∞

1

n

n∑

j=1

Zj = limn→∞

1

n

n∑

j=1

Zj = E[Z0] =: a.

We have a ≥ P[0 loop-free] > 0.

Since σ(n) → ∞ a.s., it follows from Theorem 9.1 that almost surely,

a = limn→∞

ρ(σ(n))

σ(n)= lim

n→∞

n

σ(n). (38)

This gives the time-rescaling necessary so that we can compare scaled loop-erased walk to scaled simple random walk. Let us write ⇒ for weak conver-gence in the space C[0, 1] of continuous functions on [0, 1] with the supremummetric. Write B(t)0≤t≤1 for a standard d-dimensional Brownian motion.

52

Theorem 9.2. Let d ≥ 5. Put

Wn(t) :=d√aX(nt)√n

,

(with linear interpolation in place). Then Wn(t) ⇒ B(t).

Proof. Put

Wn(t) :=d√aX(nt

a)√

n.

Then Wn(t) ⇒ B(t). From (38), we get

sup0≤t≤1

∣∣∣∣aσ(nt)

n− t

∣∣∣∣→ 0 in probability, as n→ ∞.

We can rewrite Wn(t) = d√aX(σ(nt))/

√n. Given ε > 0, choose K ⊂ C[0, 1]

compact so that P[Wn(t) 6∈ K] ≤ ε, n = 1, 2, . . . This can be done due totightness of the sequence Wn(t). By Arzela’s Theorem [20], compactnessof K implies that we can find δ > 0, such that for all f ∈ K, |f(t)−f(s)| ≤ ε,if |t− s| ≤ δ. Then

P[ sup0≤t≤1

|Wn(t) −Wn(t)| ≥ ε]

≤ ε+ P

[sup

0≤t≤1

∣∣∣∣d√aX(σ(nt))√

n− d

√aX(nt

a)√

n

∣∣∣∣ ≥ ε, Wn(t) ∈ K

]

≤ ε+ P

[sup

0≤t≤1

∣∣∣∣aσ(nt)

n− t

∣∣∣∣ ≥ δ

]

≤ 2ε

for n large. Hence Wn(t) −Wn(t) → 0 in probability in the space C[0, 1],and therefore they have the same weak limit.

In d = 4, Lawler [22] proves the following. Let

an = P[n-th point is not erased] = P[LE(X[0, n]) ∩X(n,∞) = ∅].

Lawler shows that this sequence grows only logarithmically, and

ρ(n)(nan)−1 → 1 in probability.

Then he proves that Wn(t) = d√anX(nt)/

√n⇒ B(t).

Regarding d = 3, Kozma has shown that a scaling limit exists [21] and isinvariant under dilations and rotations.

53

9.2 LERW in dimension d = 2

In two dimensions, LERW has a deep connection to complex analytic func-tions. We informally describe this connection in this section. We identifyR2 ∼= C. Let U, V ⊂ C be domains. Recall that f : U → V is called confor-mal, if it is analytic and one-to-one. The Riemann mapping Theorem statesthat given any two simply connected domains U, V ⊂ C, U, V 6= C, thereexists a conformal mapping between them.

Locally around a point z, an analytic function f has the form f(w) =f(z) + f ′(z)(w − z) + o(|w − z|). Hence f is approximately the compositionof a translation, a rotation and a dilation. The traslation and rotation ofa Brownian path is again a Brownian path, and the dilation of a Brownianpath is again Brownian path (with time rescaled). It turns out that for ananalytic function f and Brownian motion B(t), f(B(t)) is a time-changedBrownian motion. The proof of this relies on stochastic integrals and Ito’sformula. Recall that Ito’s formula in one dimension says that for a sufficientlysmooth function f : R → R

f(B(t)) − f(B(0)) =

∫ t

0

f ′(B(t))dB(t) +1

2

∫ t

0

f ′′(B(s))ds,

where the first term is a stochastic integral. For a function f : R2 → R, theformula takes the form

f(B(t)) − f(B(0)) =

∫ t

0

∇f(B(s)) · dB(t) +1

2

∫ t

0

∆f(B(s))ds. (39)

Suppose now that f : C → C is analytic, and write f = u+ iv. The Cauchy-Riemann equations ∂1u = ∂2v, ∂2u = −∂1v imply that ∆u = 0 and ∆v = 0,hence the terms with ∆u and ∆v vanish in (39). Write B(t) = B1(t)+ iB2(t)for a complex Brownian motion. Then formally applying Ito’s formula weget:

f(B(t)) − f(B(0)) =

∫ t

0

f ′(B(s)) · dB(t), (40)

where · represents complex multiplication, and dB(s) = dB1(s) + idB2(s).The right hand side is continuous a.s., has independent increments, that are”infinitesimally” centred Gaussians. By properties of stochastic integrals:

E

∣∣∣∣∫ t

0

|f ′(B(s))dB(s)

∣∣∣∣2

=

∫ t

0

|f ′(B(s))|2ds =: ζ(t).

This suggests, that f(B(t)) is a Brownian motion looked at at time ζ(t).For a proof of the following theorem, see [29].

54

Theorem 9.3. Let f : U → C be analytic, z ∈ U , and let B(t) be a planarBrownian motion started at z. Let τU := inft ≥ 0 : B(t) 6∈ U. Thenf(B(t)) = B(ζ(t)), 0 ≤ t ≤ τU , for some planar Brownian motion B startedat f(z).

Heuristically, the above theorem suggests that the scaling limit of LERWin d = 2 should also be conformally invariant, if it exists, since a one-to-onetransformation ”does not affect the loop-structure”. In fact, it is natural toask, if one can erase loops from Brownian motion to define the scaling limit.This does not quite work: there are loops on all scales, so one cannot definechronological loop-erasure.

Let us see a bit more precisely what conformal invariance of LERW shouldmean. Suppose that 0 ∈ U ⊂ C is a domain, and consider LERW on thelattice U ∩ δZ2, from 0 to ∂(U ∩ δZ2). Saying that a scaling limit exists,means that the random path of the LERW converges in suitable sense tosome random path in U from 0 to ∂U . If now f : U → V is a conformalmap, where for simlicity f(0) = 0, then conformal invariance means that thelimit we get from LERW on V ∩ δZ2 from 0 to ∂(V ∩ δZ2) has the samedistribution as the image of the random curve in U under f .

The scaling limit process was discovered by Schramm [24]. It is most con-veniently described in the setting U = z ∈ C : |z| < 1, as a curve growingfrom ∂U towards 0 (rather than from 0 to ∂U). Hence we consider the time re-versal of the LERW. By a result of Lawler [23], the time reversal of LERW hasthe same distribution as the reverse loop-erasure of the random walk path.That is, for a random walk pathX[0, τUc] from 0 to ∂U , LE(X[0, τUc ]) has thesame distribution as the loop-erasure of X(τUc), X(τUc − 1), . . . , X(1), X(0).(Here τUc is the exit time from U .) The following simple Markovian propertyis fundamental.

Lemma 9.3. (Markovian property) Let γ = γ[0, ℓ] denote the reverse loop-erasure of X[0, τUc ]. Conditioned on γ[0, k] = [ω0, . . . , ωk], γ[k, ℓ] has thedistribution of LERW from γ(k) to 0 in the domain U \ ω0, . . . , ωk.

The scaling limit curve has the following description, see [33].Let γ : [0,∞) → U be a continuous, simple curve, with γ(0,∞) ⊂ U .

Assume that γ(0) = 1. Let gt : U \ γ[0, t] → U be the unique conformalmap that satisfies the normalization gt(0) = 0 and g′t(0) > 0. One can showthat g′t(0) is strictly increasing and continuous. Hence one can choose theparametrization of γ in such a way that g′t(0) = et. Let

W (t) = limz→γ(t)

z∈U\γ[0,t]

gt(z),

55

which lies on the unit circle. Then W : [0,∞) → ∂U is continuous, andsatisfies the Loewner differential equation:

d

dtgt(z) = −gt(z)

gt(z) +W (t)

gt(z) −W (t)

g0(z) = z.

The path γ can be uniquely recovered from W , using the Loewner equation.Crucially, the Markovian property Lemma 9.3, implies that if the scaling

limit exists and is conformally invariant, then W (t) has stationary and inde-pendent increments. The scaling limit of LERW is obtained when W (t) =exp(i

√κB(t)), with κ = 2, for a standard one-dimensional Brownian motion

B(t), and is called radial SLE2. Other values of κ > 0 give radial SLEκ, andthey arise in the context of the random cluster measures of Exercise 1.1.

A precise theorem about the LERW is the following. Let D be a domainin Z2 whose boundary consists of edges of Z2. Assume that 0 ∈ D. The innerradius of D is defined as inf|z| : z ∈ Dc. Let γ be the path of the reverseloop-erasure of a random walk in D from 0 to ∂D. View γ as a continuospath. For t ≥ 0, let ft : D \ γ[0, t] → U be the unique conformal mappingsuch that ft(0) = 0 and f ′

t(0) > 0. Assume that γ has been parametrizedso that f ′

t(0)/f ′0(0) = et. Let W (t) = ft(γ(t)), and write W (t) = exp(iϑ(t))

with ϑ continuous.

Theorem 9.4 (Lawler, Schramm, Werner; 2004 [24]). For every T > 0and ε > 0, there exists r1 = r1(ε, T ), such that for all domains D as abovewith inner radius at least r1, there is a coupling of γ with a standard one-dimensional Brownian motion B(t) starting at a uniform point in [0, 2π),such that

P

[sup

0≤t≤T|ϑ(t) − B(2t)| > ε

]< ε.

Thus the theorem says that the ”driving function” of the LERW γ con-verges to a B(2t) (which has the same distribution as

√2B(t)).

56

10 The Abelian sandpile / Chip-firing game

In this last section we look at the Abelian sandpile model, also called thechip-firing game. It has a surpising connection with spanning trees, that isnot at all apparent from its definition. See [16] for some very interestingresults that are not discussed here. See [9, 10] for the relevance of the modelto the concept of self-organized criticality, and [31] for an overview.

Let G = (V,E) be a finite, connected graph, with a distinguished vertexδ, called the sink. Write V0 = V \ δ, and n = |V |. We will use the matrixindexed by V0 × V0:

∆xy :=

degG(x) if x = y;

−1 if x ∼ y;

0 otherwise,

x, y ∈ V0.

By a chip-configuration we mean an assignment of finitely many chips to thenon-sink vertices, where each vertex can hold any non-negative number ofchips. Chip-configurations are collected in the set

XG :=∏

x∈V0

0, 1, 2, . . ..

We say that η ∈ XG is stable, if ηx < degG(x), x ∈ V0. If η has an unstablevertex, that is ηx ≥ degG(x), that vertex can fire, which means that one chipis sent along each edge from x, and any chips arriving at the sink are removed.This can be written as follows. When x ∈ V0 fires, we replace η by Txη, where(Txη)y = ηy − ∆xy, y ∈ V0. More concisely, Txη = η − ∆x, where ∆x is therow of ∆ corresponding to x. Note that if the vertex fired is not a neighbourof the sink, then the number of chips is conserved, while if a neighbour ofthe sink is fired, then the number of chips is reduced by the number of edgesto the sink. The following lemma shows that any chip-configuration can bestabilized, in a unique way.

Lemma 10.1. (i) For any η ∈ XG there exists k ≥ 0 and x1, . . . , xk ∈ V0,such that Txk

. . . Tx1η is defined and stable.

(ii) If Tyℓ. . . Ty1

η is another stabilization, then ℓ = k and y1, . . . , yℓ is apermutation of x1, . . . , xk.

Proof. (i) This property is ensured by the existence of the sink. Fix x ∈ V0.Since G is connected, there exists a path x = y0, . . . , yr = δ. Since thereare finitely many chips, yr−1 can fire only finitely many times, because eachtime it fires, it sends a chip to the sink. It follows that yr−2 can also fire

57

only finitely many times, since each time it fires, it sends a chip to yr−1. Byinduction, x = y0 can only fire finitely many times.

(ii) Suppose there is a counterexample to the statement, and assume kis minimal. Since x1 is unstable in η, there exists a smallest j such thatyj = x1. Now note that if z, w are both unstable in a configuration ξ, thenTzTwξ = TwTzξ (both are equal to ξ − ∆z − ∆w). It follows that Tyj

can becommuted through Tyj−1

, . . . , Ty1, so that

Tyℓ. . . Ty1

η = Tyℓ. . . Tyj+1

Tyj−1. . . Ty1

Tx1η.

This shows that η′ = Tx1η provides a shorter counterexample, a contradic-

tion.

Let us denote by

ΩG :=∏

x∈V0

0, 1, . . . , degG(x) − 1

the set of stable configurations. Lemma 10.1 shows that there is a well-defined stabilization map S : XG → ΩG, that is the result of carrying out allpossible firings (in any order). Let δx denote the configuration with one chipat x and no other chips. We define the addition operators

ax : XG → ΩG, axη = S(η + δx), x ∈ V0,

that is, we add one chip at x and stabilize.

Lemma 10.2 (Abelian porperty). axay = ayax for all x, y ∈ V0.

Proof. Consider a sequence of firings that stabilizes η+δx. The same sequencecan be legally applied to η+ δx + δy, since the extra chip at y only makes theconfiguration larger, and the same vertices can be fired. This sequence takesη + δx + δy to S(η + δx) + δy. Stabilizaing further we obtain, using Lemma10.1(ii), that S(η + δx + δy) = S(S(η + δx) + δy) = ayaxη. Exchanging theroles of x and y, we have S(η + δx + δy) = axayη.

Corollary 10.1. Applying axk. . . ax1

gives the same result as adding all thechips at the beginning, and stabilizing.

Stabilization involves subtracting rows of the the matrix ∆. Hence it isplausible that the process of stabilization is related to equivalence modulothe row span of ∆, and this what we will show soon. It is not hard to seethat if we keep adding chips and stabilize, not all stable configurations willbe seen in the long run. For example, if we start with a configuration that

58

has two zeroes next to each other, then after adding chips at one of thosevertices, we never see the two zeroes again: if one of the vettices has justfired, the neighbour cannot be zero at the same time. This motivates thedefinition of recurrent configuration below.

Definition 10.1. The sandpile group of G is defined as

KG := Zn−1/Zn−1∆,

where n− 1 = |V0|, and Zn−1∆ is the integer row span of ∆.

We adopt the definition of recurrent configuration and the proof of Propo-sition 10.1 below from [14]. See [16] for several other equivalent definitions.

Definition 10.2. A stable configuration η is called recurrent, if there existarbitrarily large ξ ∈ XG such that S(ξ) = η. We write RG for the set ofrecurrent configurations.

Proposition 10.1. Every equivalence class mod ∆ contains exactly one el-ement of RG.

Proof. (i) We first show that every equivalence class contains a recurrentconfiguration. Let F be all elements of a fixed equivalence class that arelarger componentwise than degG(x). This set is infinite, and since ΩG isfinite, some element of S(F ) has infinitely many preimages, and that elementis then necessarily recurrent.

(ii) We now show that if η1, η2 ∈ RG and η1∆∼ η2, then η1 = η2. Take ξ1, ξ2

such that S(ξ1) = η1 and S(ξ2) = η2. We may assume that (ξ2)x ≥ degG(x),

x ∈ V0. In particular, (ξ2)x > (η1)x, x ∈ V0. Since ξ2∆∼ η1, we can write

ξ2 − η1 =∑

x∈V0

nx∆x, (41)

for some integers nx. We claim that nx ≥ 0, x ∈ V0. Put W := x ∈V0 : nx < 0, and assume that W is non-empty. The relation (41) says thatif we allow the possibility of negative number of chips, and we ”unfire” thevertices in W and then fire the vertices in V0 \W , according to nx, thenthis takes ξ2 to η1. So first ”unfire” the each vertex x ∈ W , −nx times, anddenote the result ξ′2. The total number of chips in W in ξ′2 is at least what itwas in ξ2. Hence there exists y ∈ W such that (ξ′2)y ≥ (ξ2)y. Now fire eachx ∈ V0 \W nx times to get η1. This can only increase the number of chipsat y, and we get:

(η1)y ≥ (ξ′2)y ≥ (ξ2)y > (η1)y.

59

This is a contradiction, soW = ∅, and all coefficients in (41) are non-negative.Consider now the configuration ξ3 = ξ1+ξ2−η1. We want to start with ξ3

and fire each vertex x ∈ V0, nx times, to take the configuration ξ3 to ξ1. Thismay be non-legal, as it may create negative number of chips, somewhere.But if ξ1 is large enough this cannot happen, and we are allowed to assumethis. Hence for ξ1 large enough, we have S(ξ3) = S(ξ1) = η1.

Now consider the sequence of firings that takes ξ1 to η1. Adding extrachips accoridng to ξ2 − η1 > 0 does not affect the legality of these firigns,and then this sequence of firings takes ξ3 to η1 + ξ2 − η1 = ξ2. It follows thatS(ξ3) = S(ξ2) = η2. We have proved that η1 = S(ξ3) = η2.

Lemma 10.3. ax leaves RG invariant for all x ∈ V0.

Proof. Take η ∈ RG, with η = S(ξ). Then axη = S(η + δx) = S(ξ + δx).Since ξ can be arbitrarily large, it follows that axη is recurrent.

Corollary 10.2. For η ∈ RG, axη is the unique element of RG equivalentto η + δx mod ∆.

Corollary 10.3. ax : RG → RG is one-to-one.

Proof. The map η 7→ η + δx respects equivalence classes.

Proposition 10.2. With the operation

η ⊕ ζ := S(η + ζ), η, ζ ∈ RG,

RG is an Abelian group isomorphic to KG.

Proof. Commutativity is clear from the definition. Associativity is clear from

(η ⊕ ζ) ⊕ ξ = S(η + ζ + ξ) = η ⊕ (ζ ⊕ ξ).

Let I ∈ RG be equivalent to 0 mod ∆. Then η + I∆∼ η, and hence by

Corollary 10.2, S(η+ I) = η. Hence I is an identity element. Given η ∈ RG,

let η ∈ RG be equivalent to −η mod ∆. Then η+ η∆∼ 0, hence S(η+ η) = I.

Hence inverses exist.The map φ : RG → Zn−1/Zn−1∆ given by η 7→ [η]∆ is one-to-one by

Proposition 10.1. It is clearly a homomorphism, and hence an isomorphism.

Remark. It is an intriguing question what the configuration I looks like. See[16] for some pictures.

60

We now define a natural Markov chain with statespace ΩG. At eachtimestep, pick x ∈ V0 uniformly at random, add a chip at x to the currentconfiguration, and stabilize (equivalently, apply ax).

Proposition 10.3. (i) The set of states that are recurrent in the Markovchain sense is precisely RG.(ii) The stationary distribution is uniform on RG.

Proof. (i) Suppose η ∈ ΩG is recurrent in the Markov chain sense. Givenany ξ, there is positive probability for the Markov chain to go from S(η + ξ)to η, which implies that there exists α such that η = S(η + ξ + α). Sinceη + ξ + α can be arbitrarily large, we get η ∈ RG. We have proved that allstates in ΩG \ RG are transient, hence the Markov chain, starting from anystate will enter RG with probability 1 and stay there. We prove that all ofRG is a single recurrent class, by showing that the restriction of the Markovchain to RG is irreducible. Indeed, given any η, ζ ∈ RG, there exists ξ (forexample ζ ⊖ η ∈ RG), such that S(η + ξ) = ζ . This implies irreducibility.

(ii) The transition matrix on RG is doubly stochastic:

∑

η∈RG

p(η, ξ) =∑

η∈RG∃x∈V0:axη=ξ

1

|V0|=∑

x∈V0

1

|V0|= 1.

Remark. Statement (ii) above is an example of the general fact that a (invari-ant) random walk on a group always has the uniform measure as stationarydistribution.

We have |RG| = |KG| = det(∆). The matrix-tree theorem in combina-torics [6, Corollary II.13] states that det(∆) is also the number of spanningtrees of G. Our goal in the remainder of this section is to give a mappingbetween RG and TG = set of spanning trees of G.

Let

βx := degG(x) − degV0(x) = # edges from x to δ, x ∈ V0.

The following proof of the so called buring algorithm is from [16].

Lemma 10.4 (Burning algorithm). η ∈ RG if and only if S(η + β) = η. Ifη ∈ RG, each vertex x ∈ V0 fires exactly once in stabilizing η + β.

Proof. (i) Suppose S(η+β) = η. Then S(η+Mβ) = η for allM ≥ 1. Startingfrom η+Mβ, we can selectively fire vertices and obtain a configuration thatcan be made arbitrarily large by choosing M large. Hence η is recurrent.

61

(ii) Suppose now that η ∈ RG. Observe that β =∑

x∈V0∆x. Hence we

have η + β∆∼ η, therefore, S(η + β) = η.

(iii) Let cx = # times x fires in stabilizing η + β. Then

η = S(η + β) = η + β −∑

x∈V0

cx∆x.

Since the rows of ∆ are linearly independent, cx = 1, x ∈ V0.

We can view the statement of the lemma like this: start from a config-uration η. First ”fire the sink”, meaning that βx chips are placed on eachneighbour x of the sink. Then fire as many vertices as possible. If all verticesfired, we get back η and η was recurrent.

Following the sequence of firings allows us to define a map from RG toTG. Fix η ∈ RG. When stabilizing η + β, at each time step t = 1, 2, . . . fireall vertices, simultaneously, that can be fired. Let

W0 := δWt := vertices that fire at time t, t = 1, 2, . . .

We of course have eventually Wt = ∅, and V = ∪t≥0Wt a disjoint union. Aspanning tree ψ(η) will be defined by connecting a vertex x ∈ Wt (t ≥ 1)with a vertex x ∈Wt−1, as follows. Fix x ∈Wt, t ≥ 1. Let

Nx := y ∼ x : y ∈ ∪s<tWsMx := y ∼ x : y ∈Wt−1 .

Up to time t − 1, x has received Nx chips, and it fires at time t. Therefore,ηx +Nx ≥ degG(x). Before time t− 1, x has received Nx −Mx chips, and itdoes not fire at time t− 1, therefore, ηx +Nx −Mx < degG(x). Hence

degG(x) −Nx ≤ ηx < degG(x) −Nx +Mx.

The number of possible values of ηx equals Mx, which equals the numberof neighbours of x in Wt−1. Associate each possible value with a neighbourx ∈ Wt−1, and draw an edge xx. Then xx : x ∈ V0 is a spanning tree of G.

Theorem 10.1 (Majumdar-Dhar correspondence [28]). The map ψ definedabove is one-to-one.

Note that under the correspondence, the stationary distribution of theMarkov chain is mapped to the uniform spanning tree measure. This opensup the possibility of studying the sandpile model using the uniform spanningtree. See [3, 17, 18] for result in this direction.

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