A Liouville hyperbolic souvlaki

We construct a transient bounded-degree graph no transient subgraph of which embeds in any surface of finite genus. Moreover, we construct a transient, Liouville, bounded-degree, Gromov--hyperbolic graph with trivial hyperbolic boundary that has no transient subtree. This answers a question of Benjamini. This graph also yields a (further) counterexample to a conjecture of Benjamini and Schramm.


Introduction
A well-known result of Benjamini & Schramm [3] states that every nonamenable graph contains a non-ambenable tree. This naturally motivates seeking for other properties that imply a subtree with the same property. However, there is a simple example of a transient graph that does not contain a transient tree [3] (such a graph had previously also been obtained by McGuinness [13]). We improve this by constructing -in Section 7-a transient bounded-degree graph no transient subgraph of which embeds in any surface of finite genus (even worse, every transient subgraph has the complete graph K r as a minor for every r). This answers a question of I. Benjamini (private communication Given these examples, it is natural to ask for conditions on a transient graph that would imply a transient subtree. In this spirit, Benjamini [4, Open Problem 1.62] asks whether hyperbolicity is such a condition. We answer this in the negative by constructing -in Section 6-a transient hyperbolic (bounded-degree) graph that has no transient subtree. While preparing this manuscript, T. Hutchcroft and A. Nachmias (private communication) provided a simpler example with these properties, which we sketch in Section 6.1.
A related result of Thomassen states that if a graph satisfies a certain isoperimetric inequality, then it must have a transient subtree [15].
The starting point for this paper was the following problem of Benjamini and Schramm We remark that in order to disprove -the second sentence of-this, it suffices to find a transient, Gromov-hyperbolic bounded valence (aka. degree) graph with the Liouville property; for given such a graph G, one can attach a disjoint 1-way infinite path to each vertex of G, to obtain a graph having 1-dense bi-infinite geodesics while preserving all other properties. As pointed out by I. Benjamini (private communication), it is not hard to prove that any 'lattice' in a horoball in 4-dimensional hyperbolic space has these properties. We prove that our example also has these properties, thus providing a further counterexample to Conjecture 1.1. A perhaps surpising aspect of our example is that all of its geodesics eventually coincide despite its transience; see Section 2.
Although we do not formally provide a counterexample to the first sentence of Conjecture 1.1, we believe it is easy to obtain one by blowing up the edges of our graph into tubes.
In Section 2.1 we provide a sketch of this construction, from which the expert reader might be able to deduce the details.

The hyperbolic Souvlaki
In this section we construct a bounded-degree graph Ψ with the following properties 1. it is hyperbolic, and its hyperbolic boundary consists of a single point; 2. for every vertex x of Ψ, there is a unique infinite geodesic starting at x, and any two 1-way infinite geodesics of Ψ eventually coincide;

Sketch of construction
Let us sketch the construction of this graph Ψ, and outline the reasons why it has the above properties. It consists of an 1-way infinite path S = s 0 s 1 . . . , on which we glue a sequence M i of finite increasing subgraphs of an infinite '3-dimensional' hyperbolic graph H 3 . For example, H 3 could be the 1-skeleton of a regular tiling of 3-dimensional hyperbolic space, and the M i could be taken to be copies of balls of increasing radii around some origin in H 3 , although it was more convenient for our proofs to construct different H 3 and M i .
In order to glue M i on S, we identify the subpath s 2 i . . . s 2 i+2 −1 with a geodesic of the same length in M i . Thus M i intersects M i−1 and M i+1 but no other M j , and this intersection is a subpath of S; see Figure 4. (Our graph can be quasi-isometrically embedded in H 5 , but probably not in H 4 .) We call this graph a hyperbolic souvlaki , with skewer S and meatballs M i . We detail its construction in Section 2.
To prove that this graph is transient, we construct a flow of finite energy from s 0 to infinity (Section 4). This flow carries a current of strength 2 −i inside M i out of each vertex in s 2 i . . . s 2 i+1 −1 , and distributes it evenly to the vertices in s 2 i+1 . . . s 2 i+2 for every i. These currents can be thought of as flowing on spheres of varying radii inside M i , avoiding each other, and it was important to have at least three dimensions for this to be possible while keeping the energy dissipated under control.
To prove that our graph has the Liouville property, we observe that random walk has to visit S infinitely often, and has enough time to 'mix' inside the M i between subsequent visits to S (Section 5).

Formal construction
We now explain our precise construction, which is similar but not identical to the above sketch. We start by constructing a hyperbolic graph H 3 which we will use as a model for the 'meatballs' M i ; more precisely, the M i will be chosen to be increasing subgraphs of H 3 .
Let T 3 denote the infinite tree with one vertex r, which we call the root, of degree 3 and all other vertices of degree 4. For n = 1, 2, . . . , we put a cycle -of length 3 n -on the vertices of T 3 that are at distance n from r in such a way that the resulting graph is planar 1 ; see Figure 1. We denote this graph by H 2 . It is not hard to see that H 2 is hyperbolic. Recall that a ray is a 1-way infinite path. We will now turn H 2 into a '3-dimensional' hyperbolic graph H 3 , in such a way that each ray inside T 3 (or H 2 ) starting at r gives rise to a subgraph of H 3 isomorphic to the graph W of Figure 2, which is a subgraph of the Cayley graph of the Baumslag-Solitar group BS(1, 2). Formally, we construct W from infinitely many 1 Formally, we pick a cyclic ordering on the neighbours of r and a linear ordering on the outer neighbours of every other vertex of T3. Given a cyclic ordering on the vertices at level n of T3, we get a cyclic ordering at level n + 1 by replacing each vertex by the linear ordering on its outer neighbours. Now we add edges between any two vertices that are adjacent in any of these cyclic orderings.  To define H 3 , we let the height h(t) of a vertex t ∈ V (H 2 ) be its distance d(r, t) from the root r. For a vertex w of W , we say that its height h(w) is n if w lies in D n , the nth horizontal double ray in Figure 2.
We define the vertex set of H 3 to consist of all ordered pairs (t, w) where t is a vertex of H 2 and w is a vertex of W and h(w) = h(t). The edge set of H 3 consists of all pairs of pairs (t, w)(t , w ) such that either • tt ∈ E(H 2 ) and ww ∈ E(W ), or • tt ∈ E(H 2 ) and w = w , or • t = t and ww ∈ E(W ).
Thus every vertex t of H 2 gives rise to a double ray in H 3 , which consists of those vertices of H 3 that have t as their first coordinate. Similarly, every vertex w of W gives rise to a cycle in H 3 , the length of which depends on h(w). We call the vertices on any such cycle cocircular . Every ray of T 2 starting at r gives rise to a copy of W , and if two such paths share their first k vertices, then the corresponding copies of W share their first k levels of h. It is not hard to prove that H 3 is a hyperbolic graph, but we will omit the proof as we will not use this fact. We next construct Ψ by glueing a sequence of finite subgraphs M n of H 3 along a ray S. We could choose the subgraph M n to be a ball in H 3 , but we found it more convenient to work with somewhat different subgraphs of H 3 : we let M n be the finite subgraph of H 3 spanned by those vertices (t, w) such that w lies in a certain box B n ⊆ W of W defined as follows. Consider a subpath P n of the bottom double-ray of W of length 3 · 2 n , and let B n consist of those vertices w that lie in or above P n (as drawn in Figure 2) and satisfy h(w) ≤ n.
This completes the definition of M n . We let S n denote the vertices of M n corresponding to P n , and we index the vertices of S n as {r(x), 0 ≤ x ≤ 3 · 2 n }. Note that S n is a geodesic of M n . We subdivide S n into three parts: We define the ceiling F n of M n to be its vertices of maximum height, i.e. the vertices (t, w) ∈ V (M n ) with h(w) = n.
Finally, it remains to describe how to glue the M n together to form Ψ. We start with a ray S, the first vertex of which we denote by o and call the root of Ψ. We glue M 1 on S by identifying S 1 with the initial subpath of S of length |S 1 |. Then, for n = 2, 3, . . ., we glue M n on S in such a way that L n is identified with R n−1 (where we used the fact that |L n | = |R n−1 | = 2 n by construction), m n is identified with the following vertex of S, and R n is identified with the subpath of S following that vertex and having length |R n | = 2 n+1 . Of course, we perform this identification in such a way that the linear orderings of L n and R n are given by the induced linear ordering of S. We let Ψ denote the resulting graph. We think of M n as a subgraph of Ψ.

Properties of Ψ
The following fact is easy to see.
For every n, R n separates L n (and o) from infinity. (1) The following property will be important for the proof of Liouvilleness.
There is a uniform lower bound p > 0 for the probability P v [τ Fn < τ Sn ] that random walk from any vertex of L n will visit the ceiling F n before returning to S n . (2) Indeed, we can let p be the probability for random walk on H 2 starting at the root o to never visit o again; this is positive because H 2 is transient. Then (2) holds because in a random walk from S n on M n , any steps inside the copies of H 2 behave like random walk on H 2 until hitting F n , and the steps 'parallel' to S n do not have any influence.

Hyperbolicity
In this section we prove that Ψ is hyperbolic in the sense of Gromov [8].
Lemma 3.1. The graph Ψ is hyperbolic, and has a one-point hyperbolic boundary.
Proof. We claim that for every vertex x ∈ V (Ψ), there is a unique 1-way infinite geodesic starting at x. Indeed, this geodesic x 0 x 1 . . ., takes a step from x i towards the root of T 3 inside the copy of H 2 corresponding x i whenever such an edge exists in Ψ, and it takes a horizontal step in the direction of infinity whenever such an edge does not exist.
The hyperbolicity of Ψ now follows from a well-known fact saying that a space is hyperbolic if and only if any two geodesics with a common starting point are either at bounded distance or diverge exponentially in a certain sense; see [14] for details. We skip the details here as in our case the condition is trivially satisfied do to the above claim.
As all infinite geodesics eventually coincide with S, we also immediately have that the hyperbolic boundary of G consists of just one point.

Transience
In this section we prove that Ψ is transient. We do so by displaying a flow from o to infinity having finite Dirichlet energy; transience then follows from Lyons' criterion: Theorem 4.1 (T. Lyons' criterion (see [11] or [10])). A graph G is transient, if and only if G admits a flow of finite energy from a vertex to infinity.
We refer the reader to [10] or [7] for the basics of electrical networks needed to understand this theorem.
To construct this flow f , we start with the flow t on the tree T 3 ⊂ H 2 which sends the amount 3 −n through each directed edge of T 3 from a vertex of distance n − 1 from the root to a vertex of distance n from the root. Note that t has finite Dirichlet energy.
Our flow f will be as described in the introduction, that is, it is composed of flows g(n) in M n . These flows flow from L n to R n . The flow g(n) in turn is composed of 'atomic' flows, one for each v ∈ L n . Roughly, these atomic flows imitate t from above for some levels, then use the edges parallel to S n to bring it 'above' R n , and then collect it back to (two vertices of) S n imitating t in the inverse direction. A key idea here is that although the energy dissipated along the long paths parallel to S n is proportional to their length, by going up enough levels with the t-part of these flows, we can ensure that the flow i carried by each such path is very small compared to its length . Thus its contribution i 2 to the Dirichlet energy can be controlled: although going up one level doubles , and triples the number of long paths we have, each of them now carries 1/3 of the flow, and so its contribution to the energy is multiplied by a factor of 1/9. Thus all in all, we save a factor of 6/9 by going up one more level -and we have made the M i high enough that we can go up enough levels.
We now describe g(n) precisely. For every n ∈ N, let us first enumerate the vertices of L n as l j = l j n , with j ranging from 1 to |L n | = 2 n , in the order they appear on S n as we move from the midpoint m n towards the root o. Likewise, we enumerate the vertices of R n as r j = r j n , with j ranging from 1 to |R n | = 2|L n |, in the order they appear on S n as we move from the midpoint m n towards infinity. Thus r 1 , l 1 are the two neighbours of m n on S. We will let g(n) be the union of |L n | subflows g j = g j n , where g j flows from l j into r 2j and r 2j−1 . More precisely, g j sends 1/|L n | = 2 −n units of current out of l j , and half as many units of current into each of r 2j and r 2j−1 .
We define g j as follows. In the copy of H 2 containing the source l j of g j , we multiply the flow t from above by the factor 2 −n , and truncate it after j layers; we call this the out-part of g j . Then, from each endpoint x of that flow, we send the amount of flow that x receives from l j , which equals 2 −n 3 −j , along the horizontal path P x joining x to the copy C 1 of H 2 containing r 2j−1 . We let half of that flow continue horizontally to reach the copy C 2 of H 2 containing r 2j ; call this the middle-part of g j . Finally, inside each of C 1 , C 2 , we put a copy of the out-part of g j multiplied by 1/2 and with directions inverted; this is called the in-part of g j . Note that the union of these three parts is a flow of intensity 2 −n from l j to r 2j and r 2j−1 , each of the latter receiving 2 −n−1 units of current.
Let us calculate the energy E(g j ). The contribution to E(g j ) by its outpart is bounded above by 2 −2n E(t) because that part is contained in the flow 2 −n t. Similarly, the contribution of the in-part is half of the contribution of the out-part. The contribution of the middle-part is 3 j ·(2j +1)2 j ·(2 −n 3 −j ) 2 : the factor 3 j counts the number of horizontal paths used by the flow, each of which has length (2j + 1)2 j , and carries 2 −n 3 −j units of current (except for its last 2 j edges, from C 1 to C 2 , which carry half as much, but we can afford to be generous). Note that this expression equals 2 −2n (2j + 1)(6/9) j , which is upper bounded by k2 −2n for some constant k.
Adding up these contributions, we see that E(g j ) ≤ K2 −2n for some constant K (which depends on neither n nor j). Now let g(n) be the union of the 2 n flows g j . Note that g j , g i are disjoint for i = j, and therefore the energy E(g(n)) of g is just the sum j<2 n E(g j ).
By the above bound, this yields E(g(n)) ≤ K2 −n . Now let f = n∈N g(n) be the union of all the flows g(n). Then g(n), g(m) are disjoint for n = m, because they are in different M i s. Thus E(f ) = n E(g(n)) ≤ K is finite. Since g(n) removes as much current from each vertex of L n as g(n − 1) inputs, f is a flow from o to infinity. Hence Ψ is transient by Lyons' criterion (Theorem 4.1).

Liouville property
In this section we prove that Ψ is Liouville, i.e. it admits no bounded nonconstant harmonic functions.
We remark that a well-known theorem of Ancona [1] states that in any non-amenable hyperbolic graph the hyperbolic boundary coincides with the Martin boundary. We cannot apply this fact to our case in order to deduce the Liouville property from the fact that our hyperbolic boundary is trivial, because our graph turns out to be amenable.
We will use some elementary facts about harmonic functions that can be found e.g. in [6].
Let h be a bounded non-constant harmonic functions on a graph G. We may assume that the range of h is contained in [0, 1]. Recall that, by the bounded martingale convergence theorem, if (X n ) n∈N is a simple random walk on G, then h(X n ) converges almost surely. We call such a function h sharp, if this limit lim n h(X n ) is either 0 or 1 almost surely. It is well-known that if a graph admits a bounded non-constant harmonic function, then it admits a sharp harmonic function, see [6,Section 4].
So let us assume from now on that h : V (Ψ) → [0, 1] is a sharp bounded harmonic function on Ψ.
We first recall some basic facts from [6, Section 7]; we repeat some of the proofs for the convenience of the reader.   Let A be a shift-invariant event of our random walk, i.e. an event not depending on the first n steps for every n. (The only kind of event we will later consider is the event 1 s that s(Z n ) converges to 1, where s is our fixed sharp harmonic function.) For r ∈ (0, 1/2], let Note that A r ∩ Z r = ∅ for every such r. By Lemma 5.1, if we let A := 1 s then we have A r = {v ∈ V | s(v) > 1−r} and Z r = {v ∈ V | s(v) < r}.
Proof. Start a random walk (Z n ) at v, and consider a stopping time τ at the first visit to V \ A δ . If τ is finite, let z = Z τ be the first vertex of random walk outside A δ . Since z ∈ A δ , the probability that s(X n ) converges to 1 for a random walk (X n ) starting from z is at least δ by the definition of A δ . Thus, subject to visiting V \ A δ , the event A fails with probability at least δ since it is a shift-invariant event. But A fails with probability less than because v ∈ A , and so P v [visit V \ A δ ] < /δ as claimed.
The second assertion follows by the same arguments applied to the complement of A.
Corollary 5.4. If a random walk from v ∈ A (respectively, v ∈ Z ) visits a set W ⊂ V with probability at least κ, then there is a v-W path all vertices of which lie in A /κ (resp. Z /κ ).
Proof. Apply Lemma 5.3 with δ = /κ. Then the probability that random walk always stays within A /κ is larger than 1 − κ. Hence there is a nonzero probability that random walk meets W and along its trace only has vertices from A /κ . Easily, h is uniquely determined by its values on the skewer S. Indeed, for every other vertex v, note that random walk X n from v visits S almost surely, and so h(v) = Eh(X τ (R) ), where τ (S) denotes the first hitting time of S by X n . The same argument implies that h is radially symmetric, i.e. for every two cocircular vertices v, w, we have h(v) = h(w). (3) Indeed, this follows from the fact that cocircular vertices have the same hitting distribution to S, which is easy to see (for any vertex on a circle, random walk has the same probability to move to some other circle).
We claim that, given an arbitrarily small > 0, all but finitely many of the L n contain a vertex in Z .
Indeed if not, then since random walk from o has to visit all L n by transience and (1) (where we use the fact that L n = R n−1 ), we would have P [lim h(X n ) = 1] = 0 for random walk X n from o. But that probability is equal to h(o) by Lemma 5.1, and if it is zero, then using Lemma 5.1 again easily implies that h is identically zero, contrary to our assumption that it is not constant.
Similarly, all but finitely many of the L n contain a vertex in A . Thus we can find a late enough M n such that L n contains a vertex a ∈ A as well as a vertex z ∈ Z . We assume that a and z are the last vertices of L n (in the ordering of L n induced by the well-ordering of S) that are in A and Z , respectively. Assume without loss of generality that a appears before z in the ordering of L n .
Note that, since R n separates a from infinity (1), random walk from a visits R n almost surely. Thus we can apply Corollary 5.4 with W := R n and κ = 1 to obtain an a-R n path P a with all its vertices in A . We may assume that P a ⊂ M n by taking a subpath contained in M n if needed. Indeed, P a can meet L n only in vertices that are not past a in the linear ordering of L n .
Let O a denote the set of vertices {x = (t, w) ∈ M n | there is (t , w ) ∈ V (P a ) with w = w} obtained by 'rotating' P a around S. By (3), we have O a ⊂ A since P a ⊂ A . Note that O a separates z from the ceiling F n of M n . But as random walk from z ∈ Z visits F n before returning to S with probability uniformly bounded below by (2), we obtain a contradiction to Lemma 5.3 with δ = 1/2 for small enough compared to that bound.

A transient hyperbolic graph with no transient subtree
In this section we explain how our souvlaki construction can be slightly modified so that it does not contain any transient subtrees but remains transient and hyperbolic (and Liouville). This answers a question of I. Benjamini (private communication). The question is motivated by the fact that it is not too easy to come up with transient graphs that do not have transient subtrees [3]. We start with a very fast growing function f : N → N, whose precise definition we reveal at the end of the proof. Roughly speaking, we will attach a sequence of finite graphs (M f (n) ) n∈N similar to the 'meatballs' from above to a ray S (the 'skewer') in such a way that most of the intersection of S with a fixed meatball is not contained in any other meatball. Formally, we let P m be the 'bottom path' of M m as defined in Section 2, and we tripartition P f (n) as follows: Let L n consist of the first 2 n vertices on P f (n) , and R n consist of the last 2 n+1 . The set of the remaining vertices of P f (n) we denote by Z n , which by our choice of f will be much larger than R n . As before, we glue the M f (n) on S by identifying P n with a subpath of S. We start by glueing M f (1) on the initial segment of S of the appropriate length. Then we recursively glue the other M f (n) in such a way that L n is identified with R n−1 . We call the resulting graphΨ.
Theorem 6.1.Ψ is a bounded degree transient gromov-hyperbolic graph that does not contain a transient subtree.
Proof. The hyperbolicity and the transienceΨ can be proved by the arguments we used for the original souvlaki Ψ. So it remains to show thatΨ does not have a transient subtree.
Let T be any subtree ofΨ. We want to prove that T is not transient. Easily, we may assume that T does not have any degree 1 vertices. We will show that the following quotient Q of T is not transient: for each n, we identify all vertices in L n to a new vertex v n .
Note that the vertices v n and v n+1 are cut-vertices of Q; let Q n be the block of Q incident with both v n and v n+1 containing these two vertices (a block is a maximally 2-connected subgraph). We will show that in Q n the effective resistence from v n to v n+1 is bounded away from 0, from which the recurrence of T will follow using Lyons' criterion.
Let d = |L n+1 |. We claim that there is some constant c = c(d) only depending on d such that there are at most c vertices of Q n with a degree greater than 2: indeed, Q n \{v n , v n+1 } is a forest with d(v n )+d(v n+1 ) leaves.
Next, we observe that Q n has maximum degree at most d. Furthermore, the distance between v n and v n+1 in Q n is at least Z n , which -by the choice of f -is huge compared to d and so also compared to c. Hence it remains to prove the following: Lemma 6.2. For every constant C and every m there is some s = s(m, C), such that for every finite graph K with maximum degree at most C and at most C vertices of degree greater than 2, and any two vertices x, y of K at distance x and y at least s, the effective resistence between x and y in K is at least m.
Proof. We start with a large R ∈ N the value of which we reveal later, and set s = R · C.
Let K be the graph obtained from K by contracting all vertices of degree 2. We colour an edge of K black if it is subdivided at least R times in K. Note that K has at most C vertices. Thus every x-y-path in K has length at most C, but in K any such path has length at least s. Therefore each x-ypath in K contains a black edge. Hence in K there is an x-y-cut consisting of black edges only. This cut has at most C 2 edges. Thus the effective resistence in K between x and y is at least the one of that cut considered as a set of paths in K, which is as large as we want: indeed, we can pick R so large that the latter resistance exceeds m. Now we reveal how large we have picked f (n): recall that d = 2 n+1 and that |Z n | = f (n) − 3 · 2 n . We pick f (n) large enough that |Z n | ≥ s(1, max(c(d), d)), where s is as given by the last lemma. With these choices the effective resistence between v n and v n+1 in Q n is at least 1. So Q cannot be transient by Lyons' criterion (Theorem 4.1). By Rayleigh's monotonicity law [10], T is recurrent too.

Another transient hyperbolic graph with no transient subtree
We • y ⊃ x, y ∈ D i and x ∈ D i+1 for some i ∈ {0, . . . n − 1}, or • x, y ∈ D i for some i ∈ {0, . . . , n} and x ∩ y is a square.
Then the graphs G n are uniformly Gromov hyperbolic and, since the subgraph of G n induced by D n is a cube in Z 3 (of size 4 n ), the effective resistance between two corners this cube are bounded above uniformly in n. Moreover, the distance between these two points in G n is at least n. Let T be a binary tree, and let G be the graph formed by replacing each edge of T at height k from the root with a copy of G 3 k , so that the endpoints of each edge of T are identified with opposite corners in the corresponding copy of D 3 k . Since the graphs G n are uniformly hyperbolic and T is a tree, it is easily verified that G is also hyperbolic. The effective resistance from the root to infinity in G is at most a constant multiple of the effective resistance to infinity of the root in T , so that G is transient. However, G does not contain a transient tree, since every tree contained in G is isomorphic to a binary tree in which each edge at height k from the root has been stretched by at least 3 k , plus some finite bushes.

A transient graph with no embeddable transient subgraph
We say that a graph H has a graph K as a minor , if K can be obtained from H by deleting vertices and edges and by contracting edges. Let K r denote the complete graph on r vertices.
Proposition 7.1. There is a transient bounded degree graph G such that every transient subgraph of G has a K r minor for every r ∈ N.
In particular, G has no transient subgraph that embeds in any surface of finite genus.
We now construct this graph G. We will start with the infinite binary tree with root o, and replace each edge at distance r from o with a gadget D 2 r which we now define. Given n (= 2 r ), the vertices of D n are organized in 2n+1 levels numbered −n, . . . , −1, 0, 1, . . . , n. Each level i has 2 n−|i| vertices, and two levels i, j form a complete bipartite graph whenever |i − j| = 1; otherwise there is no edge between levels i, j. Any edge of D n from level i ≥ 0 to level i + 1 or from level −i to level −(i + 1) is given a resistance equal to 2 n−|i| (we will later subdivide such edges into paths of that many edges each having resistance 1). With this choice, the effective resistance R i between levels i and i + 1 of D n is 2 n−|i| divided by the number of edges between those two levels, that is, R i = 2 n−|i| 2 n−|i| 2 n−|i|−1 = 2 −n+|i|+1 , and so the effective resistance in D n between its two vertices at levels n and −n is O(1) Let G be the graph obtained from the infinite binary tree with root o by replacing each edge e at distance n from o with a disjoint copy of D n , attaching the two vertices at levels n and −n of D n to the two end-vertices of e. We will later modify G to obtain a bounded degree G with similar properties satisfying Proposition 7.1.
Note that as D n has effective resistance O(1), the graph G is transient by Lyons' criterion. We are claiming that if H is a transient subgraph of G , then H has a K r minor for every r ∈ N.
This will follow from the following basic fact of finite extremal graph theory [12,9,5] Theorem 7.2. For every r ∈ N there is a constant c r such that every graph of average degree at least c r has a K r minor. Proof. Suppose that H has no K r minor for some r, and fix any m ∈ N. Consider for every copy C of the gadget D n in G where n > m, the bipartite subgraph G m = G m (C) of H spanned by levels m and m + 1 of C ∩ H. By Theorem 7.2, the average degree of G m is at most c r . Thus, if we identify each of the partition classes of G m into one vertex, we obtain a graph with 2 vertices and at most 3 2 2 n−m c r parallel edges, each of resistance 2 n−m , so that the effective resistance of the contracted graph is greater than 1/c r =: C r . Now repeating this argument for m+1, m+2, . . ., we see that the effective resistance between the two partition classes of G m+k (which is edge-disjoint to G m ) is also at least the same constant C r . This easily implies that the effective resistance between the two endvertices of C ∩ H for any copy C of D n is Ω(n). Since G has 2 r copies of D 2 r at each 'level' r, we obtain that the effective resistance from o (which we may assume without loss of generality to be contained in H) to infinity in H is Ω( r 2 r /2 r ) = ∞.
Thus H can have no electrical flow from a vertex to infinity, and by Lyons' criterion (Theorem 4.1) it is not transient.
Recall that the edges of G had resistances greater than 1. By replacing each edge of resistance k by a path of length k with edges having resistance 1, we do not affect the transience of G . We now modify G further into a graph G of bounded degree, which will retain the desired property.
Let x be a vertex of some copy C of D n , at some level j = n, −n of C. Then x sends edges to the two neighbouring levels j ± 1. Each of those levels L, L , sends 2 k±1 edges to x for some k. Now disconnect all the edges from L to x, attach a binary tree T L of depth k ± 1 to x, and then reconnect those edges, one at each leaf of T L . Do the same for the other level L , attaching a new tree T L of appropriate depth to x. Note that after doing this for every such x, the graph G obtained has maximum degree 6 (we do not need to modify the vertices at levels n, −n in C, as they already had degree 6. Now let's check that G is still transient, by considering the obvious flow to infinity. Each new tree of the form T L we attached has effective resistance from its root to the union of its leaves less that 1. Moreover, between any two levels of size about 2 k in some D n we introduced as many such trees as there are vertices in the levels. An easy calculation yields that the extra effective resistance we introduced between two levels is about 2/2 k ; hence the total resistance we introduced to each D n taking into account all its levels is bounded by a constant. Thus the effective resistance of each D n remains bounded by a constant (independent of n), and so G is still transient.
Note that G can be obtained from G by contracting edges.

Problems
It is not hard to see that our hyperbolic souvlaki Ψ is amenable, that is, we Similarly, one can ask Problem 2. Is there a non-amenable, hyperbolic graph with bounded-degrees, C-dense infinite geodesics, and the Liouville property, the hyperbolic boundary of which consists of a single point?
Here we did not ask for transience as it is implied by non-amenability [3].
We conclude with further questions asked by I. Benjamini (private communication) Problem 3. Is there a uniformly transient counterexample to Conjecture 1.1? Is there an 1-ended counterexample?
Here uniformly transient means that there is an upper bound on the effective resistance between any vertex of the graph and infinity.