Cutoff on Trees is Rare

We study the simple random walk on trees and give estimates on the mixing and relaxationtimes.RelyingonaseminalresultbyBasu,HermonandPerescharacterizing cutoffontrees,wegivegeometriccriteriathatareeasytoverifyandallowtodetermine whetherthecutoffphenomenonoccurs.Weprovideageneralcharacterizationof familiesoftreeswithcutoff,andshowhowourcriteriacanbeusedtoprovethe absenceofcutoffforseveralclassesoftrees,includingsphericallysymmetrictrees, Galton–Watsontreesofaﬁxedheight,andsequencesofrandomtreesconvergingto theBrowniancontinuumrandomtree


Introduction and Main Results
In recent breakthroughs, random walks on various families of random graphs were shown to exhibit the cutoff phenomenon, which is a sharp transition in the convergence to equilibrium [8,9,24].Motivated by these results, we address the question, 'For which families of (random) trees does the simple random walk have/exhibit cutoff?' First natural candidates to study are the simple random walk on Galton-Watson trees, when the offspring distribution μ has mean m ≥ 1 and finite variance σ 2 ∈ (0, ∞), and the simple random walk on spherically symmetric trees.Our results on cutoff for these families of trees are summarized as follows.
Theorem 1. 1 The simple random walk does (almost surely) not exhibit cutoff if the underlying trees (G n ) n≥1 come from one of the following three constructions: • A supercritical Galton-Watson tree (m > 1) conditioned on survival and truncated at generation n. • A family of Galton-Watson trees conditioned on having n sites.
• A spherically symmetric tree of bounded degree truncated at generation n.
We refer to Sect. 5 for precise formal definitions and statements.Intuitively, Theorem 1.1 is due to the fact that Galton-Watson trees are typically short and fat, as shown by Addario-Berry [1], while spherically symmetric trees are balanced in the number of sites by construction.Our main contribution is to show that trees are typically long, thin and imbalanced when the simple random walk exhibits cutoff; see Theorems 1.3-1.6.Our arguments rely on a characterization of the cutoff phenomenon on trees that was recently proven by Basu, Hermon and Peres [7].With their characterization at hand, we provide simple geometric criteria for the occurrence of cutoff, which are easy to verify for specific examples of trees.The criteria in Theorems 1.3 and 1.4 are sharp in the sense that we can either exclude the occurrence of cutoff or slightly modify the trees via a retraction such that the simple random walk exhibits cutoff.This allows us to construct a very simple example for a family of trees on which the simple random walk has cutoff; see Corollary 1.5.Previously, a first example of trees with cutoff was obtained in [31].
We now give a brief introduction to the theory of mixing times and refer to Levin et al. [22] for a more comprehensive treatment.Let G = (V , E, o) be a finite, rooted graph.Let (η t ) t≥0 denote the (continuous-time) simple random walk on G in the variable speed model, i.e., the walker moves along every edge of the tree at rate 1.For two probability measures μ, ν on V , let their total-variation distance be μ − ν TV := max A⊆V |μ(A) − ν(A)| . ( The ε-mixing time of (η t ) t≥0 is now given, for ε ∈ (0, 1), by where π is the uniform distribution on V .For a sequence of graphs (G n ) n≥1 , let (t n mix (ε)) n≥1 denote the collection of mixing times of the random walks on (G n ).We say that the family of random walks on (G n ) exhibits cutoff if for any ε ∈ (0, 1) The cutoff phenomenon was first verified in [15], and obtained its name in the seminal paper of Aldous and Diaconis [6].Ever since there has been a lot of activity toward showing that specific examples of Markov chains exhibit cutoff; see, for example, [8,9,24].The ultimate goal is to produce a necessary and sufficient condition for cutoff.
In turns out that cutoff has deep connections to the spectral properties of the underlying random walk; see the results [17,23,27,28] of Miclo and others for a discussion of the spectrum of random walks on trees, and the work of Jonasson [20] for a collection of results on the spectrum of random walks on random graphs.It was conjectured by Peres that a necessary and sufficient condition for cutoff is that for some ε ∈ (0, 1) (or, equivalently, for all ε ∈ (0, 1)) where λ n denotes the spectral gap of the random walk on G n ; see also Sect. 2 for a formal definition.However, while this product criterion is indeed necessary, Aldous showed that it is not sufficient; see Chapter 18 of Levin et al. [22].In a recent breakthrough, Salez provides a sufficient entropy criterion for cutoff which is for example satisfied by nonnegatively curved Markov chains [32].Nevertheless, it is believed that the product criterion in ( 4) is sufficient for cutoff for a wide range of families of Markov chains.
For birth-and-death chains, Ding et al. showed that the product criterion is sharp [16].Moreover, Chen and Saloff-Coste provided simple conditions such that (4) holds [12,13].Using their results, we will be able to exclude cutoff on spherically symmetric trees in Sect.5.1, following a projection argument due to Nestoridi and Nguyen for d-regular trees [29].It was verified for the simple random walk on trees by Basu et al. [7] that the product criterion is sufficient to see cutoff; see also Theorem 3.3 for a formal statement of their result.A first example of a family of trees with cutoff was found by Peres and Sousi [31].Note that although their results are stated for a discretetime model of the simple random walk, they can be converted to the continuous-time setup using Chen and Saloff-Coste [11].We will re-obtain the result of Peres and Sousi using Theorem 1.6; see Corollaries 4.5 and 1.5 for a simplified example of a family of trees with cutoff.
In the following, we will focus on random walks on finite trees and assume that the underlying graphs will always be a sequence of rooted trees.Before we come to the main theorems, we introduce some notations.We let The existence of a δ-center of mass is guaranteed for all δ ∈ [0, 1  3 ] by the following result.For its proof, we refer to the appendix.
Fig. 1 Example of a tree G on the left-hand side and its corresponding v-retraction G on the right-hand side.The edges in the segment (v) connecting the root and the site v are depicted in dashed blue Proposition 1.2 For all trees G, there must be at least one vertex which is a 1 3 -center of mass.
From now on, for a given family of trees (G n ), we fix some ε > 0 and assume without loss of generality that the root o n is chosen to be a δ-center of mass for some δ > 0 which does not depend on n.For a pair of real-valued, nonnegative sequences ( f (n)) n≥0 and (g(n)) n≥0 , we write f g if f and g have the same order of magnitude, and we write f g if f (n)/g(n) converges to 0 for n going to infinity.We start with a criterion on trees which allows us to determine when cutoff does not occur for the simple random walk.

Theorem 1.3 For a sequence of trees
Then, the simple random walk on (G n ) does not exhibit cutoff.
Using Theorem 1.3, one can directly check that, for example, the simple random walk on the segment of size n or on the regular tree truncated at level n does not exhibit cutoff.In order to state a converse theorem, which guarantees the existence of cutoff, we require the following definition.Fix a tree G and a vertex For each v i , let Ti be the largest subtree which is attached to v i and does not intersect with (v) in any site other than v i .A v-retraction of G is a tree G rooted at some site õ, which consists of a segment of size | (v)| with õ at one of its endpoints.In addition, we let a binary tree of size | Ti | emerge from each site in the segment at distance i from the root õ (if | Ti | is not a power of two, consider the binary tree with leaves missing at the last level).Intuitively, the v-retraction replaces 'long and thin' subtrees by 'short and fat' binary trees (Fig. 1).

Theorem 1.4 For every G n in (G
If we have that then for any sequence ( Gn ) of v * n -retractions of (G n ), we have that the simple random walk on ( Gn ) exhibits cutoff.
Note that taking retractions is necessary for cutoff in Theorem 1.4 as we will see in Sect.5.1 where we study random walks on spherically symmetric trees; see Remark 5.3.Further, note that instead of attaching binary trees, we may allow also for attaching other classes of graphs with sufficiently fast growth; see Remark 4.3 for precise conditions on the attached trees.We give now a simple example of a family of trees on which the simple random walk exhibits cutoff.This follows as a consequence of Theorem 1.4.

Corollary 1.5 Consider the family of trees (G n ) n≥1 , where G n consists of a segment of size
√ n , rooted at one of its endpoints, and binary trees of size n/(i + 1) 2 attached at distance i from the root for all i ∈ {0, 1, . . ., √ n }.Then, the root is a δ-center of mass for δ = 1  6 and the simple random walk on (G n ) exhibits cutoff.Our last main result is a criterion which is particularly suited to verify cutoff for thin and long trees; see also the trees constructed in [31].For all k ≥ 0, we set Theorem 1.6 Recall (9), i.e., the set of all vertices at distance at least k from the root.Suppose that a family of trees (G n ) with maximum degrees ( n ) satisfies Then, the simple random walk on (G n ) exhibits cutoff.
This paper will be organized as follows.In Sect.2, we discuss preliminary estimates on the spectral gap of the simple random walk on trees.We present two bounds on the spectral gap, a first bound using a characterization of the spectral gap via a discrete Hardy inequality, and a second bound using weighted paths which follows directly from the first characterization.In Sect.3, we present preliminary facts on upper and lower bounds on the mixing time using a representation as hitting times of large sets.Building on these results, we prove our main criteria for the occurrence of cutoff in Sect. 4. Section 5 is dedicated to applying these criteria to the families of trees in Theorem 1.1, and showing that the simple random walk does not exhibit cutoff.In particular, we verify the absence of cutoff for the simple random walk on spherically symmetric trees, supercritical Galton-Watson trees conditioned on survival, and families of trees which converge to the Brownian continuum random tree.
The latter includes Galton-Watson trees conditioned to contain a certain number of sites.We conclude with an outlook on open problems.

Some Facts About the Spectral Gap on Trees
In order to study cutoff for the simple random walk on trees, our key tool will be to give bounds on the spectral gap of the random walk (η t ) t≥0 .Let L denote the generator of the random walk on a tree G = (V , E).We study pairs of eigenvalues and eigenfunctions for the random walk, i.e., we want to find (κ, f ) for κ ∈ C and f : V → C such that holds for all x ∈ V .Note that since (η t ) t≥0 is reversible, all eigenvalues of L are realvalued and non-positive.Moreover, the function f ≡ 1 is always an eigenfunction with respect to the eigenvalue λ = 0. Our goal is to investigate the spectral gap λ of the process, i.e., the absolute value of the second largest eigenvalue of L, respectively, the relaxation time Recall the following variational characterization of λ; see Definition 2.1.3 in [33].
Lemma 2.1 Let λ be the spectral gap of simple random walk (η t ) t≥0 on the tree G.Then, we have that where we set The quantities E( f ) and Var ( f ) are the Dirichlet form and the variance of the function f ; see Chapter 13 of Levin et al. [22] for a more general introduction.

A Discrete Hardy Inequality on Trees
Using Lemma 2.1, we obtain the following characterization of the spectral gap in terms of a (discrete) Hardy inequality on trees; see also [17].

Lemma 2.2
Recall that o is a δ-center of mass for some δ > 0, and let T and T be two trees which satisfy (5).Let A be the smallest constant such that we have for all functions g : E(T ) → R with g ≡ 0 as well as for all functions g : E( T ) → R with g ≡ 0.Then, we have that the spectral gap λ of the simple random walk on G satisfies Proof For any function f : for all v ∈ V (G).Using the definition of Var( f ) and that for positive a, b, c, d holds, we see that where we set We set g T (e) := f T (e + ) − f T (e − ) for e ∈ E(T ) and g T (e) := f T (e + ) − f T (e − ) for e ∈ E( T ).The inequalities ( 14) and ( 15) applied to g T and g T yield A f T .Now, we use Lemma 2.1 and ( 19) to conclude the lower bound on λ claimed in (16).
For the corresponding upper bound on λ, we note that the minimizer g * for A is given by ( 14) with equality and satisfies g * (e) ≥ 0 for all e ∈ E(T ) (consider |g * | otherwise to see that this has to be the case).Define a function h : Recall that π is the uniform measure on V (G) and use the Paley-Zygmund inequality to see that and hence, we must have Using the characterization of λ in Lemma 2.1, this concludes the proof.
We will use Lemma 2.2 in Sects.4.1 and 4.3 in order to obtain upper and lower bounds on the spectral gap, respectively.

A Bound on the Relaxation Time by Weighted Paths
Next, we give an estimate which allows us to achieve upper bounds on the relaxation time for the simple random walk on trees.Note that this bound was obtained already in [23] in a more general setup, including a corresponding lower bound of the same form within a factor of 2. For the convenience of the reader, we provide a short proof for our special case of the simple random walk.Recall (12).

Proposition 2.3 Let (a e ) e∈E(G) be any family of positive edge weights for the tree G.
For any choice of the (a e )'s, we have that Proof Recall the trees T , T for G from (5).By Lemmas 2.1 and 2.2, if we have for some constant C that for all functions g : E(T ) → R, and a similar statement with respect to T , we conclude t rel ≤ C. In the following, we only show (23) with respect to the tree T .Fix positive edge weights a e and note that by the Cauchy-Schwarz inequality, Rearranging the sum according to the edges and assuming without loss of generality that g is nonnegative, we get Hence, for any choice of (a e ), we can take the right-hand side of (22) as C in (23).
Note that we can choose positive weights (a e ) in any particular way to obtain an upper bound.In the following, we present three special cases of (a e )'s and the respective bounds on the relaxation time.We start with the case where we set a e = 1 |e| in Proposition 2.3 to obtain the following bound.
Corollary 2. 4 We have that where diam(G) denotes the diameter of the tree G.
In Proposition 4.1, we give a corresponding lower bound on the relaxation time, but without the additional factor of log(diam(G)).Next, consider a e = 1 f (|e|) for some function f : N → R, whose reciprocals are summable.We obtain the following bound as a consequence of Proposition 2.3.|T e |.

Corollary 2.5 Suppose that we have some function f on the integers with
We will see in the upcoming section that the right-hand side in (26) gives also an upper bound on the mixing time for the simple random walk.

Some Facts About the Mixing Time on Trees
In this section, we discuss mixing time estimates for the simple random walk (η t ) t≥0 on a tree G = (V , E, o).The main result presented in this section is that the ε-mixing time can be bounded in terms of the hitting time of the root o.For sites x, y ∈ V , let be the hitting time of x and let E y [τ hit (x)] denote the expected hitting time of x when starting the random walk from y.The following proposition gives an upper bound on the ε-mixing time.
Proposition 3.1 Let t mix (ε) be the ε-mixing time of (η t ) t≥0 .There is a universal constant C > 0 such that we have for all ε ∈ (0, 1 2 ) In particular, Proof Note that the first inequality in ( 28) is immediate from Theorem 5 in [2] together with Theorem 20.3 in Levin et al. [22] and Lemma 5.2 of Peres and Sousi [30].To see that the second inequality holds, we claim that for two adjacent sites v, w ∈ V with |v| < |w|, we have To see this, consider the embedded discrete-time simple random walk, where we move in each time step to a neighbor chosen uniformly at random.The claim follows from the well-known fact-see equation ( 4) in [10]-that this discrete-time simple random walk satisfies and we can conclude with a time-change argument.Since for any tree G, we have this yields and the second inequality in (28) follows.
Note that the bound in Proposition 3.1 does not use the assumption that the root is a δ-center of mass.In fact, we see that (28) holds for an arbitrarily chosen root.Next, we state a lower bound on the mixing time, following the ideas of Lemma 5.4 in [7], which does indeed require that the root is a δ-center of mass.
Proposition 3.2 Assume that the root o is a δ-center of mass for some δ > 0. Let be the maximum degree in G. Then for all ε ≤ δ, Proof Let v ∈ V with v = o be fixed and recall that π denotes the uniform distribution on V (G).Moreover, recall the trees T and T in (5) from the definition of the δ-center of mass and assume that v ∈ V (T ).We now claim that The first inequality in (34) follows since η t mix (ε/2) ∈ V ( T ) implies that the root was visited before t mix (ε/2), the second inequality uses the definition of the mixing time and the third inequality follows since π(V ( T )) ≥ δ ≥ ε.In words, (34) says that with probability at least ε/2, the random walk hits the root by time t mix (ε/2).Using the Markov property of (η t ) t≥0 , since (34) holds for any v ∈ V (T ), we can iterate (34) to get that the probability of hitting the root for the first time by time k • t mix (ε/2) is at least the probability that a Geometric-(ε/2)-random variable takes a value ≤ k.This yields Since v ∈ T was arbitrary, we obtain the first inequality in (33).For the second inequality, recall (31) and (32), and use a comparison with a discrete-time simple random walk, where we choose in each step one neighbor uniformly at random, and speed the walk up by a factor of .
So far, we gathered techniques in order to give upper and lower bounds on the mixing and relaxation time.The next seminal result by Basu et al. gives a characterization of cutoff for the simple random walks on trees; see [7].

Proof of the Main Criteria for Cutoff
We now use the preliminary bounds from the previous two sections in order to establish our main criteria on cutoff for the simple random walk on trees, i.e., we prove Theorems 1.3, 1.4 and 1.6.

A General Lower Bound on the Relaxation Time
We start with a lower bound on the relaxation time.Recall that for a family of graphs (G n ), the root is chosen to be a δ-center of mass for some fixed δ > 0.

Proposition 4.1
The constant A from Lemma 2.2 satisfies In particular, Lemma 2.2 implies that Proof We show the lower bound on A by considering an explicit function g e * in the definition of A in ( 14) and (15).Fix some e * ∈ E(G) which maximizes the right-hand side of (36

Cutoff for v-Retractions of Trees
In this section, we prove Theorem 1.4.We use Proposition 2.3 with a specific choice of edge weights (a e ) improving the choice of weights leading to Corollary 2.6.
Lemma 4.2 Suppose that assumptions (7) and (8) of Theorem 1.4 hold and recall that ( Gn ) n≥1 denotes the sequence of v * n -retractions of (G n ) n≥1 .Then, we have where tn rel is the relaxation time corresponding to Gn .
Proof Let k := |v * n | and let ( Ti ) i∈{0,...,k} denote the trees in Gn attached along the segment (v * n ), ordered according to their distances from the root, which are used in the construction of the v * n -retraction.We consider the following choice of the weights (a e ) e∈E(G n ) .For e ∈ (v * n ), we let a e := |e| −1/2 .For e ∈ E( Ti ), we set In order to apply Proposition 2.3, we will now give an upper bound for for any possible choice of the edge e ∈ E( Gn ).We claim that without loss of generality, it suffices to consider e ∈ (v * n ).To see this, consider e ∈ Ti for i ≥ 1 and let e i be the corresponding edge in (v * n ) with |e i | = i.For every v ∈ Ti , we have Together with the fact that Ti has a binary structure, we get that Similarly, for e ∈ T0 , we see that A e ≤ 5| T0 | ≤ 5|V (G n )|, which is of smaller order than the right-hand side in (38) by assumption (7).Hence, it suffices to bound A e for all edges e within (v * n ) to conclude.From (41), we see that the right-hand side of (40) is bounded from above by It remains to show that We see that for all e ∈ (v * n ), using the Cauchy-Schwarz inequality for the first step, as well as the fact that by assumptions ( 8) and ( 7) for the second step.Taking square roots of both sides of (43) finishes the proof.
Proof of Theorem 1. 4 For the lower bound on the mixing time from Proposition 3.2, note that the leading order does not decrease when we replace (G n ) by ( Gn ) for v * n from (7), and that the root is still a δ-center of mass for some δ > 0. Theorem 1.4 follows together with the product criterion from Theorem 3. holds for some constant α.The above choice of binary trees satisfies (44) with α = 2. Depending on (8), we may also allow α to go to infinity with n sufficiently slow.

A Sufficient Condition for Cutoff on Trees
In this section, we give an upper bound on the relaxation time, which will allow us to give a sufficient condition for cutoff on trees.For a finite rooted tree G = (V , E, o), we let S := {B ⊆ V : B = ∅, o / ∈ B}, and define for all B ∈ S where the infimum is taken over functions f : E → R. For the following bound on the relaxation time, we use the ideas of Evans et al. for proving a Hardy inequality on continuous weighted trees [17].A corresponding Hardy inequality for discrete weighted trees was obtained by Miclo; see Proposition 16 in [27].We will now provide a similar result on the relaxation time of the simple random walk.
Proposition 4.4 Recall (9).For any finite tree G, we have that Proof Recall the trees T , T for G from (5).For the first inequality in (46), it suffices by Lemma 2.2 to show that for every function g : E(T ) → R with g ≡ 0, we have and a similar statement with respect to T .We will now show (47) only for T .Fix a nonnegative function g (otherwise consider |g|), and define for all i ∈ Z the set g(e) < 2 i+1 .
We let I := {i ∈ Z : B i = ∅} and note that B i ∈ S when i ∈ I. Observe that holds.Hence, it suffices now to prove that for all i ∈ I is satisfied in order to conclude (47).To see that (48) indeed holds, consider for all e ∈ E(T ).Using the definition of the sets (B i ) i∈Z and the fact that g is nonnegative, we have for all v ∈ B i e∈ (v) Plugging g i into the definition (45) of ν B i yields (48), and hence the first inequality in (46).For the second inequality in (46), observe that for all B ∈ S f (e) 2 : holds, where the infimum is attained when f (e) = |v| −1 1 {e∈ (v)} .We then optimize in (49) over the choice of B ∈ S to conclude.
Proof of Theorem 1.6 Similar to the proof of Theorem 1.3, recall that by Theorem 3.3, we have cutoff if and only if the relaxation time is of strictly smaller order than the mixing time.Note that by Proposition 3.2 we obtain a lower bound on the mixing time, while Proposition 4.4 gives us an upper bound on the relaxation time.We conclude taking into account assumption (10).
As a direct consequence of Theorem 1.6, we see that the simple random walk on the following family of trees constructed by Peres and Sousi [31] exhibits cutoff: Fix k and consider the tree G N for N = N (k) which is constructed as follows: Set n i = 2 2 i and start with a segment of length n k , rooted at one of the endpoints.At the root, we attach a binary tree of size N := n 3 k .Then at distance n i from the root, attach a binary tree of size N /n i for all i ∈ {k/2, . . ., k}.We reindex the sequence of trees (G N ) to obtain a family of trees (G n ) n≥1 for all n ≥ 1.A visualization of these trees can be found in [31].Note that in this construction, the root is always a δ-center of mass for some δ > 0 which does not depend on n.

Corollary 4.5 The family of trees described by Peres and Sousi has a mixing time of order N k and a relaxation time of order N , and hence exhibits cutoff provided that k → ∞ with N → ∞.
Proof We first consider the ε-mixing time for some ε > 0. In Proposition 3.2, we choose v to be a leaf in the tree attached at distance n k from the root and get for some constant c = c(ε) > 0, when summing only over the sites in the attached binary trees for (33).Since N /n i n i for all i ≤ k, a corresponding upper bound follows from Proposition 3.1.While a lower bound on t rel of order N is immediate from Proposition 4.1, we obtain a corresponding upper bound of the same order by Proposition 4.4, by a straight-forward count of the sites of distance at least m from the root.We conclude that cutoff occurs by Theorem 3.3.

Cutoff for SRW on Trees is Atypical
We now use the previous estimates and ask about cutoff for the simple random walk on the following three families of trees from Theorem 1.1: infinite spherically symmetric trees truncated at level n, supercritical Galton-Watson trees conditioned on survival and truncated at height n, and families of combinatorial trees converging to the Brownian continuum random tree (CRT).The latter includes critical Galton-Watson trees conditioned to have exactly n sites.In all three cases, we will in the following verify that cutoff does not occur.

Spherically Symmetric Trees
Let G = (V , E, o) be a rooted tree.We say that G is spherically symmetric if deg(v) = deg(v ) holds for all v, v ∈ V with |v| = |v |.Examples for such trees are regular trees.We write deg k for the degree of the vertices at generation k provided that V k = ∅ holds (recall (9)).

Proposition 5.1 Consider an infinite spherically symmetric tree G of maximum degree
. For every n ∈ N, let G n the tree induced by G by restricted to the sites V \ V n .Then, the simple random walk on (G n ) n≥1 does not exhibit cutoff.
When G = N the claim is well known; see [22].Otherwise, for all n ∈ N large enough, we choose the root of G n to be the first branching point in G, i.e., the vertex closest to the root with degree at least 3.In particular, note that the root of G n will be a 1  4 -center of mass for all n sufficiently large.By Propositions 3.1 and 3.2, we see that the 1  8 -mixing time of G n satisfies We will now show a lower bound on the relaxation time of the same order by using a comparison to birth-and-death chains.Our strategy will be similar to Evita and Oanh [29], where Nestoridi and Nguyen study eigenvalues and eigenfunctions for the d-regular tree.More precisely, we will exploit that certain eigenfunctions of birthand-death chains can be converted to eigenfunctions of the simple random walk on spherically symmetric trees.Without loss of generality, we will assume that deg(o) > 1 holds since removing the sites before the first branching point in G changes the relaxation time by at most a constant factor.This can be seen using the characterization of the spectral gap in Lemma 2.1.Let (X t ) t≥0 be the continuous-time birth-and-death chain on the segment {1, . . ., 2n − 1} with nearest neighbor transition rates (r (x, y)) given by for all {x, y} ∈ E(G n ).We make the following observation on the spectral gap of (X t ) t≥0 .
Proof Note that there exists an eigenfunction g corresponding to the spectral gap of an irreducible birth-and-death chain which is monotone and non-constant; see Lemma 22.17 in [22].By the symmetry of the transition rates, we see that the function h given by h(x) = g(2n − x) for all x ∈ {1, . . ., 2n − 1} is also an eigenfunction for (X t ) t≥0 with respect to λ.Since g(1) = g(2n − 1), we see that the function f := g − h is an eigenfunction corresponding to λ which has the desired properties.
With these observations, we have all tools to give the proof of Proposition 5.1.
Proof of Proposition 5.1 Note that we can extend the eigenfunction f of (X t ) t≥0 from Lemma 5.2 belonging to the spectral gap to an eigenfunction F with eigenvalue λ of the random walk on G n .To see this, let x 1 , x 2 be two sites adjacent to o, and consider the function F : V → R given by The fact that F is an eigenfunction of the simple random walk on G n follows by a direct verification using the generator.Hence, in order to give a lower bound on the relaxation time of the same order as in (50), it suffices to bound the spectral gap λ of (X t ) t≥0 .Note that the stationary distribution π of (X t ) t≥0 satisfies for all x ∈ {1, . . ., 2n − 1}, where Z is a normalization constant.Recalling that G has maximum degree and using Theorem 4.2 in [12], we see that This gives the desired lower bound on the relaxation time of the tree G n which is of the same order as the upper bound on the mixing time in (50).Hence, using Theorem 3.3, we see that no cutoff occurs.Remark 5.3 Using Proposition 5.1, we can see that taking v * n -retractions in Theorem 1.4 is necessary for cutoff.More precisely, consider the spherically symmetric tree G with for all i ∈ N. The corresponding trees (G n ) n≥0 truncated at level n satisfy (8), but due to Proposition 5.1, the simple random walk on (G n ) does not exhibit cutoff.

Galton-Watson Trees Conditioned on Survival
In this section, we consider a family of random trees (G n ) n≥1 which we obtain by truncating a supercritical Galton-Watson tree conditioned on survival.More precisely, let μ be an offspring distribution and assume that , i.e., we have a supercritical Galton-Watson process whose offspring distribution has finite variance.In the following, we take the genealogical tree G of a realization of the Galton-Watson process conditioned on survival.We then obtain the trees (G n ) n≥1 by restricting G onto the sites V \ V n (recalling ( 9)).We keep the sequence of trees fixed and perform simple random walks on (G n ) n≥1 .We denote the law of G by P (P depends on μ, but we will not write it).We now have the following result on cutoff for the simple random walk on (G n ) n≥1 .
Proposition 5.4 For P-almost all trees G, the family of simple random walks on (G n ) n≥1 does not exhibit cutoff.
Proof From the proof of Theorem 1.4(b) in [20], we know that the relaxation time is P-almost surely of order m n for all n sufficiently large.Hence, by Theorem 3.3, it suffices to give an upper bound on the mixing time of order m n for the random walk on G n in order to exclude cutoff.Providing an upper bound on the mixing time of order m n answers a question of Jonasson [20].Recall that for any supercritical Galton-Watson tree G = (V , E) with offspring distribution μ satisfying (56), is a martingale which converges P-almost surely and in L 2 , due to the Kesten-Stigum theorem.For all v ∈ V and N ≥ 0 let k(v, N ) be the number of sites in the tree T v with distance at most |v| + N from the root.From (57), we see that for all v ∈ V sup P-almost surely for some random variable Y v .It is easy to show, applying Doob's inequality to the martingale in (57), that Y v has a finite second moment.
Recall that Proposition 3.1 does not require the root to be a δ-center of mass, and hence holds P-almost surely for all T > 0 and some constant C > 0. It remains to give a bound on the random variables Y v .Note that the random variables Y v do not depend on the number of sites in the generation |v| in the Galton-Watson tree.Hence, writing E for the expectation corresponding to P, we see that is satisfied for all s ∈ N and some constant c > 0. Together with (59) and the first Borel-Cantelli lemma, we see that P-almost surely holds for some c > 0 depending on G and all n sufficiently large.Note that a bound of the same form follows when conditioning the underlying Galton-Watson tree on survival, using its representation as a multi-type Galton-Watson process with one child in every generation having a size-biased offspring distribution; see Chapter 12 of Lyons and Peres [25].Together with the corresponding bound on the relaxation time for the random walk on (G n ) of order m n from Johan [20], this finishes the proof.
Remark 5. 5 For critical or subcritical Galton-Watson trees, i.e., when m ≤ 1 holds in (56), one can consider the family of random walks on (G n ) which we obtain from the associated Kesten tree truncated at generation n; see [21] for a formal definition of the tree and [19] for a more comprehensive discussion.Note that the resulting tree consists of a segment of size n with finite trees attached.Using Theorem 1.3, one can show that there is P-almost surely no cutoff.

Combinatorial Trees Converging to the CRT
In the previous two examples, we considered an infinite tree, which we truncated at generation n in order to obtain the family of trees (G n ) n≥1 .We now take a different perspective and study the simple random walk on a sequence of random trees, where we assume that G n has exactly n sites.
For each tree (G n ), we assign a labeling to the n sites chosen uniformly at random, and declare the vertex with label 1 to be the root of (G n ).Let s : {1, . . ., 2n −1} → V n be the contour function of G n , which is given as the walk on V n when performing depth-first search on G n , giving priority to sites with a smaller label; see also Sect.2.6 in [5].Intuitively, we obtain s by embedding G n into the plane such that the shortest path distance is preserved and sites with a common ancestor are ordered increasingly according to their labels; see Fig. 2. We will write |s(•)| for the distance of s(•) to the root, and call |s| : {1, . . ., 2n − 1} → N again the contour function, with a slight abuse of notation.For c > 0, consider the normalized contour function s : [0, 1] → R given by for all i ∈ {1, . . ., 2n − 1}, sn (0) = sn (1) = 0 and linear interpolation between these values.We define that a family of random trees (G n ) n≥1 converges to the CRT for c if (s n ) n≥1 converges in distribution (in the space C([0, 1])) to the Brownian excursion (e(t)) t∈ [0,1] .Note that the terminology CRT refers to the Brownian continuum random tree, which can be seen as the limit object of (G n ) n≥1 ; see [3][4][5] for an introduction and equivalent definitions, and [18,26] for examples of trees converging to the CRT.Arguably the most prominent example of such a sequence of trees are independently chosen critical Galton-Watson trees (G n ) n≥1 conditioned on having exactly n sites, where we assume that holds, i.e., the number of offspring has mean 1, finite variance and is not supported on a sublattice of the integers.Note that the assumption of having m = 1 is not a restriction as we can transform any given offspring distribution with positive mean into a critical offspring distribution without changing the law of the resulting graphs when conditioning on a given number of sites; see Sect.2.1 in [4].The following lemma is a classical result due to Aldous.
Note that for critical Galton-Watson trees, mixing and relaxation times were studied by Jonasson in [20], relying on estimates of the effective resistance between the root and the leaves.We now present an alternative proof of his results, which extends to more general families of combinatorial trees.In the following, we first choose the trees (G n ) n≥1 , keep them fixed and then perform simple random walks on (G n ) n≥1 .We denote the law of (G n ) n≥1 by P. Proposition 5.7 Let (G n ) n≥1 be a family of random trees converging to the Brownian CRT for some c.Then, we have that P-almost surely, the family of simple random walks on (G n ) n≥1 does not exhibit cutoff.
Intuitively, the absence of cutoff is explained by the fact that the law of the random walk on (G n ) n≥1 converges to the law of a Brownian motion on the CRT, indicating a smooth decay of the total-variation distance for the random walk at a time of order N 3/2 .In order to prove Proposition 5.7, we will show that P-almost surely, (35) is not satisfied.First, we will prove that the mixing time and the relaxation time are, for all n sufficiently large, with positive probability both of order n 3/2 .We start with the following upper bound on the mixing time.We now use Lemma 5.9 to show the following bound on the relaxation time.
Lemma 5.10 For all ε > 0, there exists some constant c ε > 0 such that for n sufficiently large, Proof Recall the contour function s and |s|.Fix some ε > 0 and let θ, δ > 0 be the constants from Lemma 5.9.We start the proof by first choosing a new root o * of G n as follows: Recall the constant c > 0 from the convergence of (G n ) to the CRT.We set We claim that whenever the event B n 1 given by occurs then o * is a δ-center of mass.To see this, observe that the subtree T = T o * , i.e., the largest subtree containing the new root o * and only sites of distance at least |o * | from the old root o, must have at least n/2 sites.This is due to the fact that the contour in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material.If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

A Proof of the Existence of a Center of Mass
Proof of Proposition 1.2 Note that each tree must have a vertex separator x ∈ V (G), that means that when x is removed from the tree, all of the remaining connected components contain at most |V (G)|/2 many sites.This is a well-known fact from graph theory.We will now show that each vertex separator of a tree must also be a (v) denote the set of edges in the shortest path of a vertex v ∈ V from the root o and write |v| := | (v)|.For every edge e ∈ E, we let e − , e + ∈ V with |e − | < |e + | denote the endpoints of the edge and set |e| := |e + |.Further, let T v be the largest subtree of G rooted at v which consists only of sites with distance at least |v| from o, and we use the conventions that T e := T e + and |T e | := |V (T e + )| for all e ∈ E. We say that a vertex x is a δ-center of mass if there are two trees T and T

Fig. 2
Fig. 2 Visualization of the linear interpolation of the contour function |s|: {1, . . ., 27} → N for the graph G given on the left-hand side with n = 14 sites

Fig. 3 2 has probability at least 1 − ε/ 4 .
Fig. 3 Visualization of the events B 1 and B 2 for the Brownian excursion

1 3 -
center of mass in the above sense.For a given vertex separator x, let S 1 , S 2 , . . ., S d for some d ≥ 2 denote the connected components of the tree when removing x.Note that when d = 2, we take S 1 ∪ {x} and S 2 ∪ {x} as vertex sets for the trees T and T .When d = 3, associate T with the largest component S i , which must contain at least (|V (G)| − 1)/3, but at most |V (G)|/2 many sites.Combine the remaining two components for T .For d ≥ 4, we describe now a procedure to reduce the problem of finding the trees T and T for a vertex separator x with d components (S i ), to a problem of finding the trees T and T for d − 1 components ( Si ) which have all size at most |V (G)|/2.Hence, solving the problem of finding the trees T and T for x recursively will allow us to conclude the proof.Let S and S be the smallest and second smallest component in (S i ), respectively.By the pigeonhole principle, we have that |S | + |S | ≤ cases.If the left-hand side in (71) is at least (|V (G)| − 1)/3, then we associate T with S ∪ S and T with its complement, and we are done.If the left-hand side in (71) is at most (|V (G)| − 1)/3, then remove S and S from (S i ) and replace it by S ∪ S to obtain ( Si ).

)
Next, we have the following bound by choosing the weights a e = |T e | for all e ∈ E(T ) proportional to size of the tree T e attached tree to e.
a family of trees with ε-mixing times (t n mix ) and relaxation times (t n rel ) for the simple random walk on (G n ), respectively.The simple random walk on (G n ) exhibits cutoff if and only if In other words, we see cutoff whenever then product criterion (4) holds.In view of Corollary 2.6, Proposition 3.2 and Theorem 3.3, we aim at verifying cutoff by giving weights (a e ) in Proposition 2.3, which lead to a strictly improved bound over the weights a e = |T e |.
(6)ng the definition of A in Lemma 2.2, we see that (36) holds.Recall that by Theorem 3.3, there is no cutoff when the mixing time and the relaxation time are of the same order.Note that Proposition 3.1 yields an upper bound on the mixing time, while Proposition 4.1 establishes a lower bound on the relaxation time.Both bounds are of the same order, due to assumption(6), which finishes the proof.
* .We define g e * : E(G) → R to be g e * (e) := * (e) 2 ≥ v∈T e * e∈ (v) g e * (e) 2 = |T e * | Note that instead of binary trees, we may consider other trees Ti rooted at v i and attached to the segment at distance i from õ for which 3, and Lemma 4.2.Remark 4.3