One-ended spanning trees in amenable unimodular graphs

We prove that every amenable one-ended Cayley graph has an invariant spanning tree of one end. More generally, for any 1-ended amenable unimodular random graph we construct a factor of iid percolation (jointly unimodular subgraph) that is almost surely a spanning tree of one end. In [2] and [1] similar claims were proved, but the resulting spanning tree had 1 or 2 ends, and one had no control of which of these two options would be the case.

As usual, let o be the root of our unimodular graph. Denote by x a uniformly chosen neighbor of o. If H is a subgraph of G, v and w two vertices, we let v ↔ H w stand for the event that v and w are in the same component of H. For an arbitrary forest F and vertex v, let F(v) be the component of v in F.
We will define spanning forests F n of G that all have only finite components, (G, F n ) is jointly unimodular, and their limit will be the tree in the claim. Let H 0 := G and F 0 := ∅.
Let k(n) be a strictly increasing sequence of positive integers, to be defined later, with k(0) = 0. Suppose recursively that F n has been defined, all its edges are in G \ H k(n) , and every component of it is adjacent to H k(n+1) . Suppose further that The recursive assumptions trivially hold for n = 0. Figure 1 illustrates the steps of the construction that are explained next.
For every component C of F n , let v(C) be a randomly chosen vertex of H k(n+1) that is adjacent to C.
Define F + n as the union of F n and all the edges of the form {v(C), u}, where u ∈ C and C is a component . Grow a forest within H k(n) \ H k(n+1) starting from ∂ up H k(n) iteratively as follows. As i = 0, 1, . . ., consider the set U i of vertices at distance i from ∂ up H k(n) (so U 0 = ∂ up H k(n) ), and for i ≥ 1 pick a randomly chosen edge between each vertex in U i and some vertex in U i−1 . As i → ∞, we end up with a forest F − n+1 in H k(n) \ H k(n+1) , which has the property that each of its components contains a unique point of ∂ up H k(n) (by the connectedness of H k(n) ), and consequently, each component if finite (by the MTP).
Let A(n, m) be the event that there is a path with consecutive vertices p 1 , . . . , p , between v n (x) and v n (o) (p 1 = v n (x), p = v n (o)), with the F − n+1 (p i ) all fully contained in the same class of P m . By definition Choose m(n) large enough so that Such a choice is possible by the recursive assumption (1). For each class K of P m(n) consider the set of components of F − n+1 that lie entirely in K, and add a maximal number of edges to them (following some otherwise arbitrary rule) so that the result is still cycle-free. Call the resulting forest F n+1 (so F n+1 is F − n+1 with all these added edges). Then, by (2),  define F n+1 as F n+1 ∪ F + n . By construction, the recursive assumptions are satisfied by F n+1 .
Let F be the limit of the increasing sequence F n . It is clearly a forest, and by (1), F is a spanning tree.
To see that F has one end, pick an arbitrary vertex v, and let n ∈ {0, 1, . . .} be such that v ∈ H k(n) \H k(n+1) .
infinity by one point, as we wanted.
In what follows we are going to construct a sequence of fiid connected subgraphs of H n with marginals tending to 0, as in Lemma 3. This will then establish Theorem 1.
From now on, intervals always mean discrete intervals, e.g. [a, b] with a, b ∈ Z is the set {a, a + 1, . . . , b}.
An interval may only consist of 1 point. Given a set of intervals, it will automatically define an interval graph, as the graph whose vertices are the given intervals, and two are adjacent if they intersect. By a slight sloppiness, we will refer to the graph induced by a set I of intervals by the same notation I. 1. |x − ι(x)| ≤ 2δ for every x ∈ V(I).
2. If x ≤ y, x, y ∈ V(I), then ι(x) ≤ ι(y). In particular, the interval graph defined by I : [a, b] ∈ I } is such that ι maps adjacent (intersecting) intervals in I to adjacent intervals in I .

Every point of [a, b]
is contained in at most two elements of I .
Proof. Choose a path I 0 , . . . , I m (with I i ∩ I i+1 = ∅) in the interval graph I with the property that a ∈ I 0 , b ∈ I m (this latter we refer to by saying that the path bridges a and b), and make the choice so that m is there exist three distinct intervals that contain k. It is easy to check that then one can choose two of these three such that their union contains the third one. But then this third one could be dropped from I 0 , . . . , I b , and one would still be left with a connected graph (and a path that bridges a and b in it), contradicting the minimality of m. Hence I := {I 0 , . . . , I m } satisfies the first assertion. See Figure 2 for the pattern of the intervals and the naming introduced in the next paragraph.
Still using that the I we defined is a minimal path, one can check the following. Denote the endpoints of I 0 by x 0 and x 2 , with x 0 < x 2 . Denote the endpoints of I m by x 2m−1 and x 2m+1 , where x 2m−1 < x 2m+1 .
Finally, for 0 < k < m, let the endpoints of I k be x 2k−1 and x 2k+2 , where x 2k−1 < x 2k+2 . Then x 2k−1 ≤ x 2k , because I k−1 intersects I k , and the latter is closer to b than the former. Similarly, for k ≥ 1 we have x 2k < x 2k+1 , because I k−1 ∩ I k+1 = ∅ (by the assumption that every point of [a, b] is in at most two of the intervals).
It is easy to check that ι satisfies the requirements.
Lemma 5. Let G and B be random graphs, B being a biinfinite path, and suppose that (G, B) is jointly unimodular, V (G) = V (B) and E(B) ⊂ E(G). Suppose further that G has only one end. Let x n and x −n be the two vertices whose distance from the root is n in B. Then

Proof.
There are two graph isomorphisms from B to Z that take the root to 0, pick one of the two randomly with probability 1/2 and fix is, for simpler reference. Through this isomorphism, we can refer to the points of B as integers; we will use B and Z interchangeably. This way, to every edge e = {k, } ∈ E(G), we can assign the interval I(e) = [k, ], which can be thought of as the unique path in B between the endpoints of e. We refer to − k as the length of the edge e. Note that by subadditivity the limit exists, ). We need to prove that this number is 0.
Because of the one-endedness of G, for any point x ∈ Z, there are infinitely many intervals I(e), e ∈ E(G), that contain x. In other words, there are infinitely many edges whose endpoints belong to because every interval in I ⊂ I has length at least D.
Let P 2 ⊂ I be the set of those points that are contained in exactly two elements of I , and P 1 = I \ P 2 be the set of those that are contained in one. Now, let S 2 be the set of maximal connected subintervals induced by P 2 . Consider also the set of maximal connected subintervals induced by P 1 , and partition it into two subsets, using the natural ordering on these intervals from left to right: let S 1 be the subset of these intervals that are at odd positions at this ordering, and S 3 be the set of those that are at even positions.
where the last inequality follows by unimodularity (via E(dist G (iδ, (i + 1)δ)) = E(dist G (0, δ))) and the definition of c , and the inequality before it uses Lemma 4. We conclude that This holds for every large enough N , contradicting c > 15c /16.
Proof of Theorem 1. Let T 0 be a fiid spanning tree of G with one or two ends. Such a tree exists, as a straightforward generalization of Theorem 8.9 of [1] to the fiid setting.
If T 0 has one end, then the claim is proved, so let us assume that it has 2 ends. Let B be the biinfinite path in T 0 . To every vertex x in B, define B x as the subgraph induced in T 0 by x and all vertices that are in We define a new unimodular graph B + on the vertex set of B, as a deteministic function of (G, T 0 ). For an edge e = {v, w} in G, define e + = {b(v), b(w)}, and let E(B + ) := {e + : e ∈ E(G)}. We will define an fiid sequence (K n ) of subgraphs of B + that satisfy the following: 1. K n is connected; 2. lim n→∞ P(o ∈ K n ) = 0.
Once we have (K n ), we will define a sequence (H n ) of subgraphs of G, where H n : It is easy to check that if (K n ) satisfies conditions (1) and (2), then so does (H n ), and thus the theorem follows from Lemma 3. It remains to construct the K n .
Fix n and consider Bernoulli(2 −n ) percolation on V (B), independently from all other randomness that we have (the unimodular graph and the iid labels). For every pair of open vertices x and y such that every vertex of B on the path between x and y is closed, choose a connected finite subgraph C x,y of minimal size of B + that contains both x and y. Let K n be the union of all these C x,y . Then the K n are connected. We will show that they also satisfy item 2.
As in the proof of Lemma 5, choose a random uniform isomorphism between B and Z that maps o to the origin, for simpler reference. When convenient, we will refer to the vertices of B as elements of Z. For an arbitrary x ∈ V (B), let x + be the smallest x + > x that is open, and let x − be the largest ∞ j=1 P(o + = j)E(dist B + (o, j)/j), using the independence of the percolation process. The first M terms of this sum are less than /2, while the sum ∞ j=M +1 P(o + = j)E(dist B + (o, j)/j) is also bounded by /2. We obtain that P(o ∈ K n ) < , as we wanted.