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For an edge $e=\{a,b\}$ of conductance $c_e$ in a finite electrical network, its inclusion probability in the conductance-weighted uniform spanning tree is $c_e{R}_{\mathrm{eff}}(a,b)$. The same formula holds for the wired limit using the limiting wired effective resistance.

In a finite unweighted connected graph, take distinct graph vertices $a\ne z$ and orient the unique graph path from $a$ to $z$ in a uniform spanning tree and assign signed unit current to its edges. Its mean over trees is the electrical unit flow from $a$ to $z$. The Kirchhoff node law follows by averaging graph path divergences. For the graph cycle law, let ${\mathcal{F}}_{az}$ be the two-component spanning forests separating the terminals and $A_F$ the component containing $a$. Deleting a tree-path edge and conversely adjoining an edge across a forest cut are inverse operations. Hence the mean signed current on an edge $u\to v$ isThis is the gradient of $h(u)=|\mathcal{T}(G){|}^{-1}{\sum }_F{1}_{A_F}(u)$, proving the graph cycle law. In particular $R_{\mathrm{eff}}(a,z)=|{\mathcal{F}}_{az}|/|\mathcal{T}(G)|$. For an oriented existing edge $e=(a,z)$, that edge is used by the tree graph path exactly when it belongs to the tree, and then has positive orientation. Thus its mean current is $\mathbb{P}(e\in T)$, giving the edge-inclusion formula for a uniform spanning tree by Ohm's law.

Orient the unique path from $s$ to $t$ in a uniform spanning tree $T$, and put $j_{xy}^T=1$ when it uses the edge from $x$ to $y$, $-1$ when it uses the reverse direction, and zero otherwise. Then the proposed current is $i_{xy}=\mathbb{E}{j}_{xy}^T$; set it to zero on nonedges. It is antisymmetric.

For each tree path, every intermediate graph vertex has one incoming and one outgoing path edge, while the source has one outgoing edge and the sink one incoming edge. HenceTaking expectations givesthe Kirchhoff node law for a unit flow.

To establish the Kirchhoff cycle law, it is not enough to check it separately for each tree path: those path currents need not satisfy the voltage law on the original graph. Instead, let $\mathcal{T}$ be all spanning trees and let ${\mathcal{F}}_{st}$ be all spanning forests with two components, one containing $s$ and the other $t$. For $F\in {\mathcal{F}}_{st}$ denote the source component by $C_s(F)$.

For an existing edge $xy$, deleting it from a tree whose $s$-$t$ path uses $x\to y$ produces precisely a forest $F\in {\mathcal{F}}_{st}$ with $x\in C_s(F)$ and $y\notin C_s(F)$. Conversely, adding $xy$ to any such forest produces exactly that tree and path orientation. This is a bijection. Applying it to both orientations givesDefine $h(x)=|\mathcal{T}{|}^{-1}{\sum }_F{1}_{x\in C_s(F)}$. Then $i_{xy}=h(x)-h(y)$ on every edge. Since the resistances are one, this is Ohm's law, and the voltage drops telescope around every oriented cycle:Together with the node law, this proves that the mean tree-path current is the electrical unit current. This argument is the mean spanning-tree path current identity.

For clarity, these laws determine the current uniquely. The difference of two solutions is a potential gradient $k(x)-k(y)$ with zero divergence. Multiplying its divergence by $k(x)$ and summing over graph vertices gives ${\sum }_{\{x,y\}}(k(x)-k(y){)}^2=0$. Thus every difference current is zero.

Let $\mathcal{T}(G)$ be the finite set of spanning trees of the connected graph. A spanning tree has all the graph vertices, is connected, and contains no graph cycle. Connectedness ensures that at least one exists, for example by repeatedly deleting an edge of a graph cycle while retaining connectedness. The uniform spanning tree is the random edge set $T$ withEvery tree has $|V|-1$ edges, so uniformity is over complete trees, not over arbitrary subsets of that size. Such subsets can be disconnected or cyclic and do not belong to $\mathcal{T}(G)$. Nor are indicators of the tree edges generally independent. The matrix-tree theorem gives $|\mathcal{T}(G)|$ as any cofactor of the unweighted Graph Laplacian if a numerical normalizing constant is needed.

This definition uses the ordinary undirected unweighted graph convention. With positive conductances the analogous law for spanning trees weights $t$ proportionally to ${\prod }_{e\in t}{c}_e$ and is generally not uniform. When the graph has one graph vertex the empty edge set is the unique spanning tree, so its probability is one.

For finite $G$, choose a root $r$ and an ordering of the other graph vertices. Begin with a tree consisting only of $r$. At each stage take the first graph vertex not already in the tree, run an independent simple random walk from it until its first hit of the existing tree, erase the loops chronologically, and add the resulting graph path. In chronological loop erasure, revisiting a graph vertex already in the current list deletes all list entries after that graph vertex; otherwise append the new graph vertex. The retained final graph path is simple, with new internal graph vertices and exactly one endpoint in the old tree. Adding it therefore preserves connectedness and creates no graph cycle. Graph vertices already included are skipped. On a finite connected graph each walk hits the existing tree almost surely; after finitely many attachments all graph vertices are present.

The Wilson algorithm theorem states that this tree is a uniform spanning tree, independently of the chosen root and order. The required hypotheses here are an undirected connected finite graph and transition probabilities proportional to conductance; unit conductances give simple random walk and the uniform law. This is the sampling result being described, not a claim that arbitrary walk transition probabilities would produce a uniform tree.

For an infinite connected locally finite recurrent graph, choose a fixed finite root $r$ and enumerate all graph vertices. A connected locally finite graph is countable, since its finite-radius balls are finite. Run the same algorithm successively. Recurrence and irreducibility imply that a walk from any graph vertex hits $r$ almost surely, hence hits the current tree. Every attached graph path is therefore finite. The countable intersection of these probability-one events has probability one, and the increasing union is an acyclic connected subgraph containing all graph vertices: each graph vertex receives a finite route to $r$.

This gives the uniform spanning tree of a recurrent infinite graph. It is not a uniform counting measure on all infinite spanning trees. Equivalently, its joint probability distributions on finite collections of edges are the common limits of uniform spanning trees along any graph exhaustion, using the approximations defining the free uniform spanning forest and the wired uniform spanning forest. To check that equivalence, fix a finite edge set $F$ and run the chosen enumeration through a finite prefix containing all its endpoints. Once those graph vertices are in the tree, later attachments cannot add an edge with both endpoints already included, so membership of $F$ is settled permanently. Only finitely many random walks have been used; each has a finite trajectory almost surely. All their visited graph vertices and all their graph neighbours lie in a sufficiently large exhaustion set. Couple finite-volume and infinite-volume walks with the same steps there. Whether the exterior is deleted or wired to one boundary graph vertex, these initial trajectories and their loop erasures then agree. This proves convergence of the probabilities of cylinder sets determined by those edges.

The limiting law is unchanged by the root and ordering because the laws on finite graphs have this invariance. The same coupling shows that changing the exhaustion also leaves the law unchanged. ThusThe result is a single spanning tree because the graph is recurrent. On a transient graph, infinite-volume spanning-tree limits may instead be uniform spanning forests; recurrence cannot be omitted from this conclusion.

First take $x\ne y$. Sample a uniform spanning tree using Wilson's algorithm rooted at $y$, with $x$ processed first. The first walk runs from $x$ until its first hit of $y$, so its loop erasure is exactly the unique $x$-to-$y$ graph path of the resulting tree. Later attachments cannot alter that graph path. This is the uniform-spanning-tree path representation of loop-erased random walk.

Instead root Wilson's algorithm at $x$ and process $y$ first. The resulting tree still has the same uniform law, while its first attachment has the law of loop-erased random walk from $y$ to $x$. A tree's graph path between two fixed graph vertices has the same unoriented edge set in both directions. Pushing the common tree law through this graph path map therefore givesHere $\mathcal{E}$ means the set of unoriented edges. In fact, reversing the complete graph vertex sequence from the first loop-erased walk has the law of the second, giving reversibility of loop-erased random walk. If $x=y$ and first hitting allows time zero, both graph paths are empty and the equality is immediate.

This is a distributional assertion, not a pathwise identity between chronological loop erasure and reversal. For example, the valid walk trajectory $x,a,b,a,c,b,y$ in a graph containing the triangle on $a,b,c$ and the edges $xa,by$ erases forward to $x,a,c,b,y$. Erasing the reversed trajectory and then reversing the result gives $x,a,b,y$, a different graph path. Wilson's root-independent tree law is what proves the stated equality of probability distributions.

Use the usual implicit convention that the infinite connected graph is a locally finite graph. This is needed for the simple random walk and the finite wired constructions to be defined. Choose finite vertex sets $V_1\subset V_2\subset \cdots$ exhausting $V$, with each induced subgraph $G[V_k]$ connected.

For the free uniform spanning forest, sample a uniform spanning tree of $G[V_k]$, using only the internal edges. Its limiting law on edge subsets of $G$ is the free uniform spanning forest measure:The limit is weak convergence of probability measures on $\{0,1{\}}^E$, equivalently convergence of every finite-edge event.

For the wired uniform spanning forest, form $G_k^{\mathrm{w}}$ by identifying every vertex outside $V_k$ to one boundary vertex ${\partial }_k$. Retain all edges with at least one endpoint in $V_k$, preserve parallel edges, and discard loops at ${\partial }_k$. Sample a uniform spanning tree of this finite multigraph and restrict it to the original internal edges. The limiting law isFor each fixed finite set of edges the restriction is unambiguous once $k$ is large. The standard exhaustion theorem guarantees that both limits exist and do not depend on the chosen connected exhaustion. Both are random spanning forests; despite their names, they need not be connected.

Fix a root $r$. On a recurrent graph, a simple random walk started at any vertex hits $r$ almost surely. Run Wilson's algorithm with root $r$: successively attach the loop-erased random walk from each new starting vertex, stopped when it hits the existing tree. Every walk terminates almost surely, because that tree already contains $r$.

To compare the two finite approximations, fix a finite set of edges $K$ and put all their endpoints first in the ordering of starting vertices. In the infinite graph, these finitely many Wilson's algorithm walks have finite lengths almost surely. Their visited vertices, together with their neighbors, are therefore contained in $V_k$ for all sufficiently large $k$ on each realization.

Couple the walks in $G[V_k]$, in $G_k^{\mathrm{w}}$, and in $G$ using the same neighbor choices while they are in the interior of $V_k$. On the event just described, they encounter neither boundary, so the walks, their chronological loop erasures, and the resulting partial trees agree. Once every endpoint of $K$ has entered the partial tree, later walks cannot add any edge of $K$: they stop on their first hit of the existing tree. Thus the indicators of membership for all edges of $K$ agree in the free and wired uniform spanning trees, with probability tending to one.

The two limiting measures agree on every finite-edge event, and thereforeThis argument also shows that the common uniform spanning forest is a single spanning tree on a recurrent graph. The equality refers to probability measures, not to independently sampled forests being identical.

A spanning tree of a finite connected graph $G$ is a connected acyclic subgraph containing every vertex. A uniform spanning tree is a random spanning tree chosen uniformly from the finite set of all spanning trees of $G$.

For an infinite locally finite recurrent connected graph, take an increasing exhaustion by finite connected subgraphs and sample a uniform spanning tree in each. The restrictions to any fixed finite edge set converge in distribution; on a recurrent graph the free and wired limits coincide and form one tree. This infinite-volume law is called the uniform spanning tree of the recurrent graph.

It can be generated by Wilson's algorithm. Fix a root and an enumeration of the remaining vertices. Starting with the root, run a random walk from the first vertex not yet in the tree until it hits the existing tree, erase its loops chronologically, and add the resulting path. Recurrence ensures every walk hits the finite tree almost surely. Repeating this operation produces the infinite uniform spanning tree, independently of the enumeration.

Exhaust ${\mathbb{Z}}^2$ by finite boxes $G_n$ and let $T_n$ be a uniform spanning tree of $G_n$. Every finite tree satisfies the handshaking lemma, soChoose the root uniformly from $V(G_n)$. The proportion of roots within any fixed distance of the boundary tends to zero, and the rooted trees converge locally to the uniform spanning tree $T$ of ${\mathbb{Z}}^2$. Since every degree is at most four, expectations also converge. Translation invariance therefore givesThe four edges incident to $0$ have equal inclusion probability by the rotations and reflections of the square lattice. If that common probability is $r$, then $4r=\mathbb{E}{\mathrm{deg}}_T(0)=2$. Consequentlyfor every edge $e$ by translation invariance.

Let ${\mu }_n$ be the uniform spanning tree measure on the finite connected induced graph $G_n$. Uniform spanning-tree edge indicators are negatively associated: increasing events depending on disjoint edge sets have nonpositive covariance. Together with the spatial Markov property, this implies the free-boundary monotonicity under graph enlargement: for every increasing cylinder event $A$, ${\mu }_n(A)$ is eventually nonincreasing once $G_n$ contains all edges on which $A$ depends.

Define the free uniform spanning forest measure byfor increasing cylinder events. These limits determine a unique probability measure, independent of the exhaustion; equivalently, ${\mu }_n$ converges weakly to ${\mu }_F$ on the product space.

Run Wilson algorithm with root $a$, beginning with a simple random walk from $x_1$. Its chronological loop erasure is added as the first branch, so the resulting tree contains a path from $x_1$ to $a$ with exactly the law of the stated loop-erased random walk.

For completeness, the algorithm produces a uniform spanning tree. For any prescribed ordered collection of loop-erased branches, the random-walk decomposition gives a product of local transition factors and diagonal killed Green functions. The transition factors depend only on the resulting oriented tree, while the product of Green functions is independent of the vertex order by Question 3. Thus every spanning tree receives the same probability. Since a tree has a unique simple path between two vertices, its $x_1$-to-$a$ path has the claimed law.

Sample a uniform spanning tree $T$. Rooting Wilson's algorithm at $a$ shows that the path in $T$ oriented from $x_1$ to $a$ has the law of $\mathrm{LE}(x_1\to a)$. Rooting the same uniform law at $x_1$ shows that the same undirected path with its orientation reversed has the law of $\mathrm{LE}(a\to x_1)$. ConsequentlyThis is reversibility of loop-erased random walk.

Take an increasing exhaustion $G_n$ of $G$ by finite connected vertex sets, identify every vertex outside $G_n$ to one boundary vertex, and choose a uniform spanning tree of the resulting finite wired graph. The weak limit as $n\to \infty$ is the wired uniform spanning forest of $G$.

For Wilson algorithm rooted at infinity, enumerate $V$ as $v_1,v_2,\dots$. Run a simple random walk from $v_1$ forever and add its chronological loop erasure. Transience makes the infinite loop erasure well-defined. Next, from the first vertex not already in the forest, run an independent random walk until it hits the existing forest; if it never hits, run it forever. Add its loop erasure and continue through the enumeration. Wilson's theorem rooted at infinity says that the resulting forest has the wired uniform spanning forest law.

For $n\ge 2$, choose $n-2$ independent random variables, each with the uniform distribution on a finite set $\{1,\dots ,n\}$, and decode their values as a Prüfer sequence. Every sequence has probability $n^{-(n-2)}$, and the bijection gives each labelled tree the same probability. This produces a uniform spanning tree of the complete graph without repeatedly proposing sets of edges and rejecting those that fail to be trees. It supplies random inputs for testing implementations of breadth-first search and a tractable model of random connections. ^[ref]

This distribution is uniform over fixed-label trees, rather than over unlabelled trees. With four labels, the star graph shape has probability $4/16=1/4$ and the path graph shape has probability $12/16=3/4$. Also, for a fixed label $v$, the Prüfer degree formula givesHere the binomial distribution counts the independent code positions equal to $v$. The expected value agrees with the average vertex degree from the degree-sum formula. Individual labels have the same distribution by symmetry, though their vertex degrees are constrained together by their fixed sum. When $n=1$, simply return the single isolated vertex.

A uniform spanning forest is an infinite-volume limit of uniform spanning trees on finite graph approximations. Free and wired boundary identifications give the free uniform spanning forest and wired uniform spanning forest, respectively.

If $p,q\downarrow 0$ with $q/p\to 0$, the random-cluster model on a finite connected graph converges to the uniform spanning tree. Relative to a tree, an acyclic configuration with $k>1$ components pays a factor $(q/p{)}^{k-1}$, while every additional cycle pays a factor asymptotic to $p$.