The full spectrum of random walks on complete finite dary trees

In the present paper, we determine the full spectrum of the simple random walk on finite, complete d-ary trees. We also find an eigenbasis for the transition matrix. As an application, we apply our results to get a lower bound for the interchange process on complete, finite d-ary trees, which we conjecture to be sharp.


Introduction
Finding the spectrum of a transition matrix is a very popular subject in graph theory and Markov chain theory. There are only a few techniques known to describe the exact spectrum of a Markov chain, and they usually work under very specific conditions, such as when the Markov chain is a random walk on a finite group, generated only by a conjugacy class [12]. Most well-known examples where a transition matrix has been diagonalized usually rely on combination of advanced representation theory, Fourier analysis, and combinatorial arguments [3,5,6,7,13,15,17,22]. But even in most of these cases, there is no description of what an eigenbasis of the transition matrix would look like, which in general is needed as well in order to understand the transition matrix.
In this work, we present the full spectrum of the simple random walk on complete, finite d-ary trees and a corresponding eigenbasis, and we use this information to produce a lower bound for the interchange process on the trees, which we conjecture is sharp.
Consider the complete, d-ary tree T h of height h, which has n := 1 + d + · · · + d h = d h+1 −1 d−1 vertices. We study the simple random walk on T h , whose transition matrix is denoted by Q h , according to which when we are at the root we stay fixed with probability 1/(d + 1), x + 1 xd , is a solution of one of the following h + 1 equations: d h+1 x 2h+2 = 1 (1.2) and d k+2 x 2k+4 − d k+2 x 2k+3 + dx − 1 = 0, for 0 ≤ k ≤ h − 1. Both x and 1 xd correspond to the same λ. The correspondence between x and λ is 2-to-1. , an eigenvector f λ with respect to λ is given by the formula , an eigenvector f v,j,j+1 with respect to λ is given by the formula Spectrum of random walks (c) The collection of these eigenvectors together with the all-1 vector form an eigenbasis of Q h .
In Lemma 2.1 and Figure 1, we describe and illustrate the eigenvectors in more detail.
The idea behind the proof is to consider appropriate projections of the random walk.
For example, let X t be the state of the random walk at time t and let Y t be the distance of X t from the root. Then Y t is a Markov chain on [0, h], whose eigenvalues are also eigenvalues of Q h . Also, the eigenvectors of Y t lift to give the eigenvectors presented in (1.4). This computation is not going to give us the full spectrum, however. For example, in the case of the binary tree, another type of projection to consider is as follows. We consider the process W t , which is equal to −Y t if X t ∈ T ρ 1 and equal to Y t otherwise. The second largest eigenvalue can be derived by this new process, while the eigenvectors are of the form presented in (1.5). The reason why this is the right process to study is hidden in the mixing time of the random walk on T h . A coupling argument roughly says that we have to wait until X t reaches the root ρ. The first time that X t hits ρ is captured by W t , since W t is a Markov chain on [−h, h], where the bias is towards the ends and away from zero. The projections that we consider form birth and death processes, whose spectral and mixing properties have been studied by Chen and Saloff-Coste [9,10,11] and Ding, Lubetzky, and Peres [14]. To capture the entire spectrum, our method is to find in each eigenspace a well-structured eigenvector, which occurs by considering an appropriate projection, which gives rise to a birth and death process. Our analysis has immediate applications to card shuffling, namely the interchange process on T h , and to the exclusion process. We enumerate the nodes in V (T h ) and we assign cards to the nodes. At time zero, card i is assigned to node i. The interchange process on T h chooses an edge uniformly at random and then flips a fair coin. If heads, interchange the cards on the ends of e; if tails, stay fixed. A configuration of the deck corresponds to an element of the symmetric group, which we denote by S n where we recall that n is the number of vertices of T h . Let g ∈ S n . Let P be the transition matrix of the interchange process on the complete, finite d-ary tree T h and let P t id (g) be the probability that we are at g after t steps, given that we start at the identity. We define the total variation distance between P t id and the uniform measure U to be A celebrated result concerning the interchange process was the proof of Aldous conjecture [8, Theorem 1.1], which states that the spectral gap of P is the same as the spectral gap of the Markov chain that the ace of spades performs. Adjusting our computations, we now get the following result.

Theorem 1.2.
Consider the interchange process on the complete d-ary tree of depth h.
We have that (a) the spectral gap of the transition matrix is where ε is any positive constant. Here, the implicit constants do not depend on d, h, n.
The order for the mixing time of the interchange process on T h has been known to be of the form C(d)n 2 log n (see Corollary 8 of [2] and Theorem 1.4a of [18]). However, no explicit bounds on constant C(d) have been stated. We conjecture that the lower bound in part (b) of Theorem 1.2 is sharp and that the interchange process on T h exhibits cutoff at 1 d−1 n 2 log n. This would imply that this process mixes much faster than the interchange process on the path, another card shuffle that uses n − 1 transpositions, which Lacoin [19] recently proved exhibits cutoff at 1 2π 2 n 3 log n.
We can get lower bounds for the mixing time of another well studied process, the exclusion process on the complete d-ary tree. This is a famous interacting particle system process, according to which at time zero, k ≤ n/2 nodes of the tree are occupied by indistinguishable particles. At time t, we pick an edge uniformly at random and we flip the two ends. Similar computations to the ones of the proof of Theorem 1.2 give that if where ε > 0 is a constant. Combining Oliveira's result [21] with Theorem 1.2 (b), we get that the order of the mixing time of the exclusion process on the complete d-ary tree is n 2 log k.
As potential open questions, we suggest trying to find the spectrum or just the exact value of the spectral gap for the simple random on finite Galton-Watson trees or for the frog model as presented in [16].

The spectrum of Q h
This section is devoted to the proof of Theorem 1.1, which presents the full spectrum of Q h and the corresponding eigenvectors.
Let λ be an eigenvalue of Q h and let E(λ) be the corresponding eigenspace. We first show that there exists an eigenvector in E(λ) that has the form described in Theorem Lemma 2.1. The eigenspace E(λ) contains an eigenvector f that has one of the following forms.
In this case we call f completely symmetric for T h .
We call such f pseudo anti-symmetric.
The following illustrations explain what the described eigenvectors look like for binary trees.
Proof. Assume that E(λ) does not contain a completely symmetric eigenvector. Let f be a nonzero element of E(λ). Since f is not completely symmetric, there exist vertices of the same level at which f takes different values. Let v be a vertex with the largest (v) such that there are at least two of its children, say the i-th and j-th children, at which f has different values. For example, if there are two leaves u and w at which f (u) = f (w) that have the same parent v then we simply take v to be v . By Thus, f takes the same value at all grandchildren of u.
Repeating this argument shows that f | T u is completely symmetric. Consider the vector g obtained from f by switching its values on T v i and T v j . More , and g = f elsewhere. By the symmetry of the tree and the matrix Q h , g also belongs to E(λ). So is f − g, which we denote by h. Observe that h is an eigenvector that is 0 everywhere except on Moreover, h is completely symmetric when restricted to T v i and T v j because both f and g are, as seen above. Thus, h ∈ E(λ) and is pseudo anti-symmetric.

Completely symmetric eigenvectors
In this section, we describe completely symmetric eigenvectors. We show that the completely symmetric eigenvectors of Q h are given by the formula (1.4) and correspond to λ and x satisfying (1.1) and (1.2) as in Theorem 1.1. Since a completely symmetric eigenvector of Q h has the same value at every node of the same level (see Figure 2), we can project it onto the path [0, h] and obtain an eigenvector of the corresponding random walk on the path. Proof. Each symmetric eigenvector of Q h corresponds one-to-one to an eigenvector of the following projection onto the path [0, h] with transition matrix R h : Since R h is a reversible transition matrix with stationary distribution . Therefore, A is diagonalizable and so is R h . In other words, R h has h + 1 independent real eigenvectors. This implies that Q h has h + 1 linearly independent completely symmetric eigenvectors.
is a non-real solution of the equation This equation has exactly 2h such solutions. If x is a solution, so is 1 xd . There is a 2-to-1 correspondence between x and λ. An eigenvector y = (y 0 , y 1 , . . . , y h ) of R h with respect to λ is given by The vector f : T h → R that takes value y i at all nodes of depth i is an eigenvector of Q h with respect to λ.
Proof. Let λ be an eigenvalue of R h and y = (y 0 , y 1 , . . . , y h ) be an eigenvector corresponding to λ. We have Since y is not the zero vector, the above equations imply that y 0 = 0. Without loss of generality, we assume y 0 = 1. Let x 1 , x 2 be the solutions to the characteristic equation of (R2): Indeed, assuming that (2.3) holds for y 0 , . . . , y i for some 1 ≤ i ≤ h − 1 then by (2.1), Thus, by (R2), and so 3) also holds for y i+1 and hence, for all y 0 , . . . , y h .
Similarly, by (R3), we get Thus, as they are both equal to y h .
This equation has 2h non-real solutions and 2 real solutions ± 1 √ d . For each non-real solution x 1 , observe that x 2 := 1 dx1 is also a non-real solution. Note that x 1 = x 2 and by setting λ and y as in (2.2) and (2.3) with α 1 and α 2 as in (2.6), one can check that y is indeed an eigenvector corresponding to λ. Thus, these 2h non-real solutions correspond to exactly h eigenvalues λ = 1 of R n . Since R n has exactly h + 1 eigenvalues, these are all.

Pseudo anti-symmetric eigenvectors
In this section, we describe the pseudo anti-symmetric eigenvectors. We show that the pseudo anti-symmetric eigenvectors of Q h are given by the formula (1.5) and correspond to λ and x satisfying (1.1) and (1.3) as in Theorem 1.1.
Consider a pseudo anti-symmetric eigenvector f with node v and indices i, j as described in Lemma 2.1 (see Figure 1) Figure 1 and Figure 3 below, let y = (y 0 , y 1 , . . . , y k ) where y 0 is the value of f at the i-th child of v, which is denoted by u, y 1 is the value of f at the children of u and so on. With these notations, we also write f as f y,v,i,j . Observe that y is an eigenvector of the following matrix S k : Conversely, for any eigenvector y of S k , for any node v at depth h − k − 1 and for any choice of i, j ∈ [1, d] with i = j, we can lift it to a pseudo anti-symmetric eigenvector f y,v,i,j . Proof. Since S k differs from R k only at the (0, 0) entry, it also satisfies the equation π(x)S k (x, y) = π(y)S k (y, x) where π = [1, d, d 2 , . . . , d k ]. Thus, like R k , the matrix DS k D −1 is symmetric where D is the diagonal matrix with D(x, x) = π(x) 1/2 . By symmetry, DS k D −1 has k + 1 eigenvalues and so does S k .
For each eigenvector y of S k , we create d − 1 linearly independent vectors f y,v,i,i+1 for 1 ≤ i ≤ d − 1. It is clear that any f y,v,i,j can be written as a linear combination of these vectors. This completes the proof.
We now describe the eigenvectors of S k . Lemma 2.5. Each of the k + 1 eigenvalues λ of S k is of the form This equation has 2k + 2 solutions that differ from 1 √ d . If x is a solution, so is 1 dx . There is a 2-to-1 correspondence between x and λ. An eigenvector y = (y 0 , y 1 , . . . , y k ) of S k with respect to λ is given by Proof. Let λ be an eigenvalue of S k and y = (y 0 , y 1 , . . . , y k ) be an eigenvector corresponding to λ. We have (S1) d d+1 y 1 = λy 0 , d+1 y k−1 + d d+1 y k = λy k . As before, we let x 1 , x 2 be the solutions to the equation By exactly the same argument as in the proof of Lemma 2.3, we derive by setting y 0 = 1 Note that, x = ± 1 √ d are solutions of (2.7). The remaining 2k + 2 solutions split into pairs (x, 1 dx ) of distinct components. For each of these pairs, let x 1 := x and x 2 := 1 dx . We have x 1 = x 2 and by setting λ and y as in (2.2) and (2.3) with α 1 = dx 2 dx 2 −1 and α 2 = 1 one can check that y is indeed an eigenvector corresponding to λ. Thus, these 2k + 2 solutions correspond to exactly k + 1 eigenvalues λ of S k . Since S k has exactly k + 1 eigenvalues, these are all of them.

Proof of Theorem 1.1
The following lemma shows that we can retrieve all eigenvectors of Q h from completely symmetric and pseudo anti-symmetric eigenvectors. Let A S k be the eigenbasis of S k as described in Lemma 2.5 and B be a collection of h + 1 independent completely symmetric eigenvectors of Q h as in Lemma 2.3. Let That gives the second part. Finally, the third part follows from Lemma 2.6.
Before proving Lemma 2.6, we make the following simple observation. For a rootedtree T that is not necessarily regular, recall that a vector f : T → R is said to be completely symmetric if f (u) = f (v) for all pairs of vertices u, v at the same level. A vector f is said to be energy-preserving if for all levels l, We now prove that the vectors considered are linearly independent. Assume that there exist coefficients c y,v,i and c g such that where the first sum runs over . We need to show that c y,v,i and c g are all 0.
Since pseudo anti-symmetric vectors are energy-preserving on T h , the sum is both completely symmetric and energy-preserving. Thus, by Observation 2.7, c y,v,i f y,v,i,i+1 = g∈B c g g = 0. (2.8) By the independence of vectors in B, we conclude that c g = 0 for all g ∈ B.
Spectrum of random walks To complete the proof of the lemma, we now prove by induction on the vertices of v ∈ V (T h−1 ) and i ∈ [d − 1] that c y,v,i = 0 for all y ∈ A S h− (v)−1 . For this induction, we use the natural ordering of pairs (v, i) as follows.
By the independence of vectors in A S h−1 , we conclude that c y,ρ,1 = 0 for all y ∈ A S h−1 , establishing the base case.
For the induction step, assume that for some (v, i), it is proven that c y,w,k = 0 for all (w, k) < (v, i) and y ∈ A S h− (w)−1 . We now show that c y,v,i = 0 for all y ∈ A S h− (v)−1 . By this assumption, the left-most side in (2.8) reduces to c y,u,j f y,u,j,j+1 = 0 (2.9) where the sum runs over all (u, j) ≥ (v, i). Our argument now is similar to the base case.
Similarly to the base case, when restricting on the subtree T v i , F v,i is both completely symmetric and energy-preserving on T v j . By Observation 2.7, F v,i = 0 on T v j . This leads By the independence of vectors in A S h− (v)−1 , we conclude that c y,v,i = 0 for all y ∈ A S h− (v)−1 , establishing the induction step and thus finishing the proof.

Proof of Theorem 1.2
In this section, we prove Theorem 1.2, which gives the spectral gap for the interchange process on T h and a lower bound for its mixing time.

Proof of Theorem 1.2(a)
Consider the interchange process on T h . Let Q h be the transition matrix of the ace of spades. In other words, Q h is the transition matrix of any fixed card on the tree. By Aldous' spectral gap conjecture proved in [8,Theorem 1.1], the spectral gap of the interchange process on the complete d-ary tree of depth h is the same as the spectral gap of Q h . We note that And therefore, the spectral gap of Q h is d+1 2(n−1) times the spectral gap of Q h .
Thus Theorem 1.2 (a) is deduced from the following.
To prove Lemma 3. (3.5) Furthermore, for k = h, the eigenvalue that corresponds to this x satisfies Assuming Lemma 3.2, we conclude that for sufficiently large h, the largest x that satisfies one of the equations (3.4) for some k in [1, h] Since the right-hand side of (1.1) is increasing in x for x ≥ 1 √ d , the second largest eigenvalue λ 2 of Q h corresponds to such x and so it satisfies (3.6), proving (3.2). By the first part of Lemma 3.2, there are no eigenvalues of Q h whose absolute value is larger than λ 2 . This proves Lemma 3.1.
Thus, for k = h, These bounds together with the equation n = d h+1 −1

Proof of Theorem 1.2 (b)
For the proof of the lower bound, we use Wilson's lemma. Lemma 3.3 (Lemma 5, [25]). Let ε and R be positive numbers and 0 < γ < 2 − √ 2. Let F : X → R be a function on the state space X of a Markov chain (C t ) such that .
Then the total variation distance from stationarity at time t is at least 1 − ε.
Proof of Theorem 1.2 (b). Let 0 < x < 1 be a solution of (1.3) (for k = h) satisfying (3.5) and λ be the eigenvalue of Q h corresponding to x. In particular, Let f : T h → R be an eigenvector of Q h corresponding to λ. As in the proof of Lemma 2.5, we can choose f as follows. (3.8) We now consider the interchange process on the d-ary tree T h . Fix an arbitrary enumeration of the vertices of T h by 1, 2, . . . , n. Let σ ∈ S n . We define F (σ) = n v=1 f (v)f (σ(v)). Then, we have that By direct computation, we obtain e E (f (σ t+1 (v))|σ t , e) = n − d 2 − 3 2 + λ(d + 1) 2 f (σ t (v)).

Spectrum of random walks
So, we have For n sufficiently big, we have that where the first inequality arises from keeping only the leaves of T ρ 1 ∪ T ρ 2 and the last inequality follows from Lemma 3.2 (b).
Finally, we consider what happens if we change a configuration σ t by transposing an edge e, which connects two vertices u and v. We have that For all vertices w, we have by definition of f , |f (w)| ≤ d and by Lemma 3.2 (b), assuming wlog that (u) = (v) + 1 =: l + 1, The definition of f given by (3.8) gives that the left-hand side is 0 if neither u nor v belongs to T ρ 1 ∪ T ρ 2 . And so, . Using Wilson's lemma, we have that if t ≤ t 0 := log(F (id)) + 1 2 log((1 − λ )ε/(4R)) − log(λ ) then the total variation distance is at least 1 − ε. By Lemma 3.1, which gives We get t 0 = log n 2 + log ε 2 + O(log d) This completes the proof.