Uniqueness and Non-Uniqueness for Spin-Glass Ground States on Trees

We consider a Spin Glass at temperature $T = 0$ where the underlying graph is a locally finite tree. We prove for a wide range of coupling distributions that uniqueness of ground states is equivalent to the maximal flow from any vertex to $\infty$ (where each edge $e$ has capacity $|J_{e}|$) being equal to zero which is equivalent to recurrence of the simple random walk on the tree.

where ∂B denotes the set of edges with exactly one end in B. We denote the set of ground states of G with couplings J by G(J). The main goal of this paper is to determine the cardinality of G(J), when the underlying graph G = (V, E) is a tree and J is distributed according to some distribution satisfying ν((−ǫ, ǫ)) = Θ(ǫ) for ǫ → 0; that means there exist 0 < c < C < ∞ such that c · ǫ ≤ ν((−ǫ, ǫ)) ≤ C · ǫ for all ǫ small enough. A distribution ν satisfying this will also be called a distribution of linear growth. The existence of ground states is a consequence of compactness of the space {−1, +1} V . Clearly σ is a ground state if and only if −σ is a ground state, hence |G(J)| is even (or infinity) and greater or equal than 2. For trees there are two natural ground states, namely the ones satisfying J xy σ x σ y ≥ 0 for every (x, y) ∈ E. In Section 2 we will see a necessary and sufficient condition which ensures that the two natural ground states are the only ground states.
For a tree T = (V, E) we choose one vertex and call it the root of the tree or simply 0. All concepts presented in the following are independent of the specific choice of the root. For x ∈ V let |x| be the length of the shortest (and hence only non intersecting) path starting from zero and ending in x. For e = (x, y) ∈ E |e| := min {|x|, |y|}. By x y we mean that x is part of every path connecting 0 and y. This directly implies that |x| ≤ |y|. Note that 0 y ∀y ∈ V . We say that x → y if x y and |x| = |y| − 1, i.e. (x, y) ∈ E and x is the vertex located closer to the root. For x ∈ V we define the subgraph T x by the tree containing all vertices y, s.t. x y and all edges of the form (u, v), s.t. (u, v) ∈ E, x u and x v. For n ∈ N we define T ≤n = (V ≤n , E ≤n ) by the tree containing all vertices x s.t. |x| ≤ n and and edges e s.t. |e| ≤ n − 1. By T ≥n = (V ≥n , E ≥n ) we mean the forest containing all vertices x s.t. |x| ≥ n and and edges e s.t. |e| ≥ n. Throughout we will assume that all edges are oriented towards infinity, i.e. that (x, y) ∈ E implies x y. For some edge e = (x, y), we denote the shortest path connecting the root 0 to y by P e .
A subset Π ⊂ E is called a cutset separating x and infinity if every infinite non self-intersecting path starting at x contains at least one edge in Π. If we do not mention a specific vertex x, we always mean that the cutset separates 0 from ∞.

The main theorem
The main goal of this section is to prove the following theorem: The equivalence of i) and ii) will be proven in section 2.1, the equivalence of ii) and iii) is just a well known extension of the Max-Flow Min-Cut -Theorem of L.R. Ford and D.R. Fulkerson [2]. In the sections 2.2 and 2.3 we will show the equivalence of iii) and iv). In sections 2.1-2.3 we will for technical reasons assume that T is a tree without finite branches, that means that T x is infinite for every x ∈ V . In section 2.4 we will prove that this assumption is in fact not necessary.  For a given set of nonnegative values (κ e ) e∈E we define the maximal flow from 0 to ∞ as MaxFlow(0 → ∞, κ e ) := max{strength(θ) : θ is a flow w.r.t. κ e } .
Remember that due to the enhanced version of the Max-Flow Min-Cut Theorem, see for example ( [6], Chapter 3) With this, we are ready to prove the equivalence of i) and ii) of Theorem 2.1.
Take a cutset Π such that f ∈Π |J f | < |J e | and e lies on the same side as the root, a picture of such a situation is given in Figure 1. Define Π e as all elements f ∈ Π satisfying f e. Now let B ⊂ V be the set of all vertices which are seperated from infinity by {e} ∪ Π e . Then So σ is not a ground state. i) ⇒ ii): For a tree T let T 1 = (V 1 , E 1 ), T 2 = (V 2 , E 2 ) be two subtrees such that where MaxFlow T1 is the maximal flow in T 1 , respectively T 2 . We choose the subtrees T 1 and T 2 before we choose the root here. Then we can orient all edges pointing away from 0. Let h = (u, v) ∈ E 2 , such that for every ǫ > 0 That means that increasing the coupling value at h increases the Maximal Flow from the root to ∞ in T and therefore also in T 2 . In particular |J h | ≤ M axF low T2 (0 → ∞, |J e | ); The almost sure existence of such an edge and such subtrees will be discussed in Lemma 2.2.
for (x, y) ∈ E. σ is unique up to a global spin flip and σ is a ground state. To see this we will show (2) for three different cases of finite sets B ⊂ V . The three cases correspond to the three trees (from left to right) in Figure 2. Case 1. (u, v) / ∈ ∂B : Here (2) is clearly true, as (u, v) / ∈ ∂B and J xy σ x σ y ≥ 0 for every (x, y) ∈ E, (x, y) = (u, v). Case 2. u ∈ B, v / ∈ B : First note that |J h | = min {κ f : f ∈ P h } due to the construction. If ∂B contains an edge f ∈ P h (2) is true, as |J h | ≤ |J f |. Otherwise ∂B contains a cutset Π separating 0 and ∞ in T 1 . Hence and σ is a ground state.
As vanishing of the maximal flow from the root to ∞ does not depend on the values of finitely many couplings (we assumed ν({0}) = 0) we get that ν E (M axF low(0 → 0 φ(ñ) ∞, |J e | ) = 0) ∈ {0, 1} by Kolmogorov's 0-1-law. Hence uniqueness of ground states is an a.s. property for trees and coupling distributions of linear growth; Later we will see that |G(J)| = ∞ in the case of non-uniqueness. ii) and iii) are clearly equivalent for every graph. For the implication ii) ⇒ i) we did not use the linear growth assumption in the proof, so this implication holds, whenever ν({0}) = 0.
Lemma 2.2. Let T = (V, E) be a tree and J e be distributed according to some distribution of linear growth ν such that MaxFlow(0 → ∞, |J e | ) > 0 a.s., then there exists some vertex0 and subtrees T 1 = (V 1 , E 1 ) and and for every ǫ > 0 where we orient all edges in the trees T 1 and T 2 such that they are pointing away from 0. Moreover, there exist even infinitely many such divisions into two subtrees and respective edges h satisfying (4) and (5).
Proof. Let φ : N → E 2 be a bijective enumeration of E 2 which starts with all edges adjacent to 0, then all edges e ∈ E 2 s.t. |e| = 1 and so on. So in particular n → |φ(n)| is increasing. Define This means, we set the capacities at the edges {φ (1) for some k large enough. Hence we obtain M F (0) < L by a Borel-Cantelli-argument and independence of the J f ′ s. Letñ be the smallest integer such that M F (ñ) > M F (0). Asñ is the smallest integer, |J g | ≥ |J φ(ñ) | ∀g ∈ P φ(ñ) , so actually we can choose0 as the vertex adjacent to φ(ñ) and nearer to 0. Then (4) and (5) hold true when one considers the subtrees T φ(ñ) and T \ T φ(ñ) with appropriate edge and vertex sets. A picture of our construction is given in Figure 3. We can apply the construction above also to the tree T 1 instead of T and get subtrees T 1,1 and T 1,2 of T 1 and an edge h 1 ∈ E 1,2 , such that T 1,1 is the tree connected to the root 0 and (4) and (5) hold true in T 1 . By iterating this idea, we get that there exist infinitely many partitions of T and edges h satisfying (4) and (5). But as every such edge corresponds to a uniquely defined ground state pair (the one where h is the only unsatisfied edge), we directly get |G(J)| = ∞.
Actually it suffices to require ν ({0}) = 0 and the lower bound of the linear growth condition. It seems plausible that Lemma 2.2, and hence the equivalence of i) and ii) of Theorem 2.1 even hold true, as soon as ν((−ǫ, ǫ)) > 0 ∀ǫ > 0 and ν({0}) = 0, but there is no proof known to us.
Proof. As p c < 1 there exists some ǫ > 0 and infinitely many subtreesT = (Ṽ ,Ẽ) satisfying |J e | > ǫ ∀e ∈Ẽ, hence L = ∞ for such a tree. By the same arguments as above Lemma 2.2 holds true in this case, too.

Random Walks and Maximal Flows for Exponential Couplings
In this section we prove a one-to-one correspondence between the maximal flow and recurrence/transience of random walks. The proof is based on the following theorem by R. Lyons, R. Pemantle and Y. Peres, see [5].
Theorem 2.4. Let G be a finite graph and κ e be independent exponentially-distributed random variables with mean c e and Z ⊂ V, 0 ∈ V. Then Furthermore, if G is a tree, 0 its root and Z its leaves, then Before going to the proof of Theorem 2.4 we need to prove the following Lemma, see also [5]. For self-containedness we will repeat the proofs by Lyons, Pemantle and Peres of Theorem 2.4 and Lemma 2.5, but just for trees, where notation is a bit simpler.
Lemma 2.5. Let θ be a flow from 0 to Z. Then there exists a measure µ on selfavoiding paths from 0 to Z so that Proof. We use induction on the number of edges satisfying θ(e) = 0. For n = 0 the statement is clearly true. Now let n + 1 be the number of edges satisfying θ(e) = 0 and let P be a self-avoiding path from 0 to Z satisfying α := min e∈P θ(e) > 0. Let θ 1 be the unit flow along P. Then θ 2 = θ − α · θ 1 is also a flow from 0 to Z with number of edges satisfying θ 2 (e) = 0 less or equal than n. So we can find a measure µ 2 satisfying (8) for θ 2 instead of θ. But now the measure µ := µ 2 + α · δ P has the desired property (8) for θ.
With this we are now ready to prove Theorem 2.4.
Proof. Let θ be the current flow of strength 1 from 0 to Z; we denote the voltage function by V . Let µ be a measure on paths from 0 to Z such that (8) holds. Since θ is a unit flow µ is a probability measure. Define a new flow ψ by ψ is also a flow with respect to κ e since As the κ e are exponentially distributed and independent we have for all s > 0 For (7) we distinguish two cases. If deg(0) ≥ 2 (7) is true by linearity of expectation, as we can split up the tree into two or more subtrees. If deg(0) = 1 and f = (0, a) ∈ E and Z are the leaves of the tree, we can w.l.o.g. assume that c f = 1, as Conductance, Expectation and MaxFlow are all linear under positive scalings. Furthermore X := MaxFlow (a → Z, κ e ) and Y := κ f are independent random variables, say on probability spaces (Ω 1 , µ 1 ) and (Ω 2 , µ 2 ). Let C := Conduct (a → Z, c e ). Then The first inequality follows by Jensen's inequality, the second inequality by the induction assumption. The last inequality is equivalent to (1 − C)e 2C ≤ 1 + C .
For C ≥ 1 this is clearly true, for 0 ≤ C < 1 the result is obtained by dividing by 1 − C on both sides and developing the functions as power series. This concludes the proof.
Having the theorem above at hand, we can prove the following: Corollary 2.6. Let T=(V,E) be a locally finite infinite tree and (κ e ) e∈E be independent and exponentially distributed with mean 1. Let ν be the associated probability measure. Then the following are equivalent: ii) The simple random walk on T = (V,E) is recurrent a.s.
Proof. Take Z = V n := {x ∈ V : |x| = n}. As we assume as always that T is a tree without finite branches we get the inequalities and Now take n → ∞ on both sides. Note that which has finite expectation. So by dominated convergence we can interchange limit and expectation on the left side of the inequalities. Hence we get

Maximal Flows for more general Couplings
In the sections above we saw that there is a 1-to-1 correspondence between the maximal flow and the number of ground states for any distribution of linear growth and there is a connection between the maximal flow and recurrence/transience properties of the simple random walk on the tree, if ν is the law of an exponential distribution. The goal of this section is to prove Corollary 2.6 for all distributions of linear growth. With this theorem we can prove the following corollary: For the proof of Theorem 2.7 we use the quantile function. Let X be a nonnegative random variable with distribution function F (x) = P(X ≤ x). Then the function Q : (0, 1) → R defined by is called the quantile function of the random variable X. Now assume that U is uniformly distributed on the interval (0,1). Then Q(U ) has the same distribution as X.
With this we are now ready to prove Theorem 2.7.
So we have seen that for distributions of linear growth there is a connection between the maximal flow and the conductance. It is a natural question to ask, whether this holds true for all absolutely continuous distributions with support at 0. In fact it does not hold true. In chapter 3 we will give an example of a tree T and an absolutely continuous distribution ν, such that the simple random walk on T is transient, but the maximal flow from 0 to ∞ with respect to some capacities |J e |, which are i.i.d. with law ν, equals 0 almost surely. Up to now we did all proofs assuming that T has no finite branches. We only assumed this for technical reasons, Theorem 2.1 holds true for every infinite tree. To see this, note that for any tree T = (V, E) we can define a new treeT = (Ṽ ,Ẽ) bỹ T is another locally finite tree with |Ṽ | = ∞. It is the tree obtained by removing all finite branches from T , we will callT also the backbone of the tree.

Randomness of |G(J)|
So far we saw that the number of ground states is an a.s. constant for any tree T and a coupling distribution of linear growth and either 2 or ∞. For the case of the half plane and for many coupling distributions it has been proven by L.-P. Arguin and M. Damron in [1] that |G(J)| is either 2 or ∞ a.s.. For graphs which have some translational symmetry it can be shown, see [1], that the number of ground states is also an almost sure constant. In this section we give an example of a tree T = (V, E) and a distribution ν, such that P (|G(J)| = 2) = P (|G(J)| = 4) = 1 2 . To achieve this we have to drop the condition ν((−ǫ, ǫ)) > 0 for every ǫ > 0 .
Let ν be the uniform distribution on the interval (1, 3) and let T = (V, E) be the tree of Figure 5. We have one edge f in the middle and both vertices adjacent to f are starting points of two halflines going to ∞. Here the number of ground states depends on the coupling value J f . If σ is a ground state, then all edges e ∈ E \ {f } have to be satisfied, as for every J e > 1 there almost surely exists an edge h in the same halfline such that J h < J e . Therefore f is the only edge which can be satisfied or unsatisfied in a ground state. As all couplings are positive the natural ground states are the spin configurations satisfying either σ x = +1 ∀x ∈ V or σ x = −1 ∀x ∈ V . f Figure 5 If J f > 2 the natural ground states are the only ground states. When f is not satisfied we can almost surely find edges h 1 in the upper right and h 2 in the lower right halfline such that J h1 + J h2 < J f , which contradicts (2). So the natural ground states are the only ground states. If J f ≤ 2 the spin configurations where f is the only unsatisfied edge are ground states. Note that this are precisely the spin configurations which are +1 on the righthand side of the graph and −1 on the left-hand side, or vice versa. The reason for this is that for B ⊂ V and f ∈ ∂B there are at least two other edges in ∂B. Additional to those ground states the natural ones still exist, so |G(J)| = 4 in this case.

Dependence of |G(J)| on the coupling distribution
The main goal of this chapter is to give a tree T and two equivalent distributions, such that |G(J)| is 2 or ∞ almost surely, depending from which of the two distributions the couplings are drawn.
Let T be a tree such that the simple random walk on T is recurrent. Let (ω n ) n∈N be a summable sequence of positive real numbers. Then, as already noted in ( where P e is the path connecting e and 0. The idea is to find a condition of a similar form as (9) which ensures that M axF low(0 → ∞, κ e ) = 0 a.s., where the (κ e ) e∈E are i.i.d. and exponentially distributed with parameter 1.
One can extend Theorem 3.2 assuming the slightly weaker condition lim inf Π→∞ e∈Π 1 |e| < ∞, but to prove this one first needs to deduce several other lemmas. The proof is given in the Appendix.
In section 2 we saw that |G(J)| is either 2 or ∞ and this is independent of the specific choice of the coupling distribution ν, as long as ν is a distribution of linear growth. Below we show that |G(J)| still depends on the coupling distribution ν. We do this by giving an example of a tree T = (V, E) and i.i.d. random variables (J e ) e∈E with absolutely continuous distribution ν with support at 0, such that the simple random walk on T is transient but M axF low(0 → ∞, |J e | ) = 0 almost surely. Let (I e ) e∈E be i.i.d. uniformly distributed on (0, 1). Then |G(I)| = ∞ a.s., but |G(J)| = 2 a.s.. For the rest of this chapter we assume that T is a spherical symmetric tree such that the simple random walk on T is transient and n 2 ≤ |E n | ≤ 2n 2 ∀n. To show existence of such a tree, note that if θ is the flow satisfying θ(e) = 1 |E |e| | for every e ∈ E (1 − s) n · 3s 2 ds .
Now use the following formula, see for example [3]: where Γ is the Gamma function and both α and β are positive. This implies E min 1≤i≤n X i = 6Γ(n + 1) Γ(n + 4) ≤ 6 n 3 .
Using the same steps as in the proof of Theorem 3.2 and using that the (J e ) e∈E are i.i.d. with distribution ν we get that where the last equality follows by considering the cutsets E n and using that |E n | ≤ 2n 2 . Hence, by Theorem 2.1 and Theorem 2.7, we have constructed a tree, such that the natural ground states are the only ones, when the couplings have law ν. However, when the couplings are uniformly distributed on the interval (0, 1), there are infinitely many.

Open Problems
We conclude the paper with three open problems: In section 2 we saw that the number of ground states is an a.s. constant for every tree T and every distribution of linear growth and either 2 or ∞. However, there still are trees and distributions, where |G(J)| is a non degenerate random variable, see section 2.5. It remains unsolved, whether there exists a graph G and a distribution ν satisfying ν((−ǫ, ǫ)) > 0 ∀ǫ > 0, where |G(J)| is not an a.s. constant or where |G(J)| ∈ N \ {2} has positive probability.
In section 2 we also saw that there is a connection between the number of ground states, percolation and random walks for various distributions. Is there also and connection between those for more general graphs, for example lattices?
For any tree T inf Π cutset e∈Π is a necessary and sufficient condition for ensuring p c = 1. Furthermore we saw that inf Π cutset e∈Π is necessary and that lim inf is sufficient to ensure recurrence of the simple random walk on the tree. It is still unknown, whether one can find a condition of the same form as (10), which is equivalent to recurrence of the simple random walk. Proof. In the same way as we did in the proof of Corollary 2.6 we get that where the J e are independent exponentially distributed with mean c e . Now note that we have by Thomson's Principle, see for example ( As F δ is finite the laws of (κ e ) e∈E and (I e ) e∈E are equivalent. Therefore P(MaxFlow(0 → ∞, κ e ) > δ −1 ) > 0 .
As δ was arbitrary we have that M axF low(0 → ∞, κ e ) is unbounded as soon as the simple random walk is transient. If the simple random walk is recurrent, we know from Corollary 2.6 that M axF low(0 → ∞, κ e ) = 0 a.s.. Note that we do not have equality in general, as Ξ does not have to be injective. So we have to show that Π min → ∞, a soon as Π → ∞. Therefore assume that we have a sequence of cutsets Π n → ∞, but Π min n ∞. This is equivalent to the existence of some e ∈ E such that |Ξ −1 (e)| = ∞. Consider the tree G = (V e , Ξ −1 (e)), where V e is the set of vertices which are adjacent to at least one edge in Ξ −1 (e). This is an infinite subtree of T satisfying κ f < κ e ∀f ∈ Ξ −1 (e). Such a tree occurs with probability 0, as p c = 1, hence we get that there exists almost surely no e ∈ E such that |Ξ −1 (e)| = ∞. Hence Π min n → ∞ a.s. and lim inf Π→∞ e∈Π κ min e ≤ M almost surely. Using this, we directly get (12).
The left-hand side of (12) is an a.s. constant by Kolmogorov's 0-1-law. Lemma 6.2 directly implies that the right-hand side is also an almost sure constant. In the proof of the desired theorem we use the concept of branching numbers: Let T be a tree with root 0. Define the branching number brT of T by brT := inf λ > 0 : inf Π cutset e∈Π There is a connection between bond percolation and the branching number of a tree, which is due to R. Lyons, see [4] for details. Let p c be the critical probability for bond percolation on the tree T . Then p c = 1 brT .

Theorem 6.3. Let T be a tree such that
Then the simple random walk on T is recurrent.
Proof. First note that (13) implies p c = brT −1 = 1. Let (κ e ) e∈E be i.i.d. exponentially distributed with mean 1. Note that for every positive random variable X and any constant M ≥ E[X] P(X ≤ 2M ) ≥ 1 2 hence we have that for C ∈ R and a sequence of cutsets Π n satisfying e∈Πn 1 |e| ≤ C ∀n ∈ N and Π n → ∞ for n → ∞. Then But this implies that M axF low(0 → ∞, κ e ) is bounded by 2C. By Lemma 6.1 we obtain M axF low(0 → ∞, κ e ) = 0 a.s.. Therefore the simple random walk on T is recurrent, by Corollary 2.6.