the heat kernel of homesick random walks on k … · 2016-06-15 · the heat kernel of homesick...

THE HEAT KERNEL OF HOMESICK RANDOM WALKS ON k-REGULARTREES

ERIC P. LOWNES1, STRATOS PRASSIDIS1,2, AND STEFAN P. SABO1

Abstract. Certain types of homesick random walks were introduced by Lyons in [5] to estimate

the growth of groups. For such a random walk on a k-regular tree, we compute its Laplacian and

its heat kernel. Our methods are based on the use of combinatorial coverings, as introduced by

Chung and Yau in [4], to compute the spectrum of the Laplacian.

Contents

1. Introduction 12. Preliminaries 23. Finding the Heat Kernel of the k-Regular Tree 53.1. Calculating ht,1(a,b) 63.2. Calculating ht,2(a,b) 9References 25

1. Introduction

Gromov asked to what extent the spectrum of a Cayley graph of a group determines the geometryof the group at infinity. For example, amenable groups are completely characterized by the propertythat the spectral radius of the random walk its Cayley graph is maximal. Along these lines, Lyonsin [5] introduced a homesick random walk on Cayley graphs of groups and used its recurrenceproperties to estimate the growth of the underlying group. The goal of this project is to explainLyons’ result from the point of view of the spectral properties of the Markov operator introducedby Lyons.

In this paper, we consider the simplest case of the k-regular tree, the Cayley graph of the freegroup. The main purpose of the paper is the calculation of the heat kernel corresponding to thecombinatorial Laplacian for Lyons’ homesick random walk.

Date: July-August 2007.1 Partially Supported by an N.S.F. R.E.U. Grant.2 Partially Supported by a Canisius College Summer Research Grant.

1

Combinatorial Laplacians for directed graphs were introduced in [3], generalizing the definitionsin [4]. Also in [4], the authors used combinatorial covers to compute the spectrum of the Lapla-cian and the heat kernel associated with it. We generalize their arguments using the version ofcombinatorial covers for weighted directed graphs given in [1]. More precisely, we start with thehomesick random walk on the k-regular tree, considering it as a directed weighted graph. Thenwe consider combinatorial balls centered at the root of the tree. That allows us to reduce theproblem to finite graphs. For the final computation we will take the limit as the radius of the ballgoes to infinity. As in [1] and [4], we realize the finite ball as a combinatorial cover of a weightedray. Then the spectrum of the Laplacian can be computed, partially, from the spectrum of theLaplacian of the ray. The calculations of this spectrum follow along the lines of [4]. Contrary to theundirected case, the calculations here become more complicated because there are eigenvalues ofthe ball that do not come from the ray. That phenomenon appeared in [1]. For these eigenvalues,we proceed by induction, moving inside the branches of the ball. For each branch we proceed bydirectly calculating the spectrum and the eigenvectors.

Regular trees are universal covers of regular graphs, in particular Cayley graphs. The authorsintend to continue their line of research by reproving and strengthening Lyons’ results using thespectral calculations developed in this paper.

Finally, the first and the third author would like to extend their thanks to Canisius College forhosting the R.E.U. Program “Geometry and Physics on Graphs” during the summer of 2007, whenthis paper was completed. All three authors would like to thank Terry Bisson for discussions duringthe preparation of the project.

2. Preliminaries

In this section, we define some terminology that will be used throughout the paper. Let G be asimple undirected graph and let z ∈ V (G) be a fixed vertex called the root. Define the neighborhoodof z as:

N(z) = u ∈ V (G)|u ∼ z

Also define the combinatorial ball of radius r centered at z by:

Br(z) = u ∈ V (G)|d(u, z) ≤ r

where d(u, z) denotes the combinatorial distance. The combinatorial sphere of radius r centered atz is given by:

Sr(z) = u ∈ V (G)|d(u, z) = r

By definition, S0(z) = z.2

The following weighting function was defined by Lyons in [5].

w(u, v) :=

µ

du + (µ− 1)d−u, v ∈ S|u|−1(z)

1du + (µ− 1)d−u

, otherwise

where d−u = |N(u) ∩ S|u|−1(z)| and µ is the homesick parameter.

Remark 2.1.

(1) When µ = 1, we have a simple random walk. For the purposes of this paper, we areconcerned with the case when µ > 1.

(2) Notice that w(u, v) ≥ 0, for all u, v ∈ V. Also, u ∼ v ⇐⇒ w(u, v) > 0.(3) The degree of a vertex du on a weighted directed graph is defined as

du =∑v∈V

w(u, v)

(4) The transition probability matrix P of a graph is defined as:

P (u, v) =w(u, v)du

.

(5) The Frobenius–Perron Theorem implies that there is a unique positive eigenvector φ of Pwith eigenvalue 1 and length 1. In other words, φP = φ, φ(v) > 0 and φ has length 1. Thevector φ is called the Frobenius–Perron vector of P .

The following theorem is the main result in [5]. For Cayley graphs of groups, it connects thehomesick parameter with the growth of the group.

Theorem 2.2. If µ < gr(G), then RWµ is transient. If µ > gr(G), then RWµ is positive recurrent.

In this theorem, gr(G) refers to the “growth” of G and is defined as:

gr(G) = limn→∞

|Sn|1/n.

On a random walk, “transient” means that at some point you will never return to the origin of thewalk. “Positive recurrent” means that you will always eventually return to the origin of the walk.

In a directed graph G, the combinatorial Laplacian L is defined as ([3], [1]):

L = I − Φ12PΦ−

12 + Φ−

12P ∗Φ

12

2.

Here Φ is the diagonal matrix with Φ(v, v) = φ(v), where φ is the Frobenius–Perron vector of P .Also, P ∗ denotes the conjugate transpose of P . For our purposes, by the spectrum of the graph wewill mean the spectrum of its combinatorial Laplacian. For g an eigenfunction of the Laplacian L,its associated harmonic eigenfuction f is defined as: f = g · Φ−

12 .

3

The Heat Equation on a directed graph is the differential equation:∂ht∂t

= −Lht, with h0 = I and t ≥ 0.

The solution of the heat equation is given by

ht = e−tL =∞∑r=0

(−1)rtrLr

r!= I − tL +

t2

2L2 − . . .

The classical theory ([2]) implies that for any two vertices u and v,

ht(u, v) =∑i

e−λit · φi(u) · φi(v)

where the φi’s are orthonormal eigenfunctions of the Laplacian L. Since L is symmetric, ht issymmetric, ht(u, v) = ht(v, u).

The following definition was given in [1] generalizing the definion in [4].

Definition 2.3. Let G and G be two weighted directed graphs. We say G is a combinatorialcovering of G (or G is covered by G) if there is a mapping π : V (G) → V (G) satisfying thefollowing two properties:

(i) There is an n ∈ R, called the index of π, such that for u, v ∈ V (G), we have∑x∈π−1(u)

∑y∈π−1(v)

w(x, y) = nw(u, v), and∑

x∈π−1(u)

∑y∈π−1(v)

w(y, x) = nw(v, u).

(ii) For x, y ∈ V (G) with π(x) = π(y) and v ∈ V (G), we have∑z∈π−1(v)

w(z, x) =∑

z′∈π−1(v)

w(z′, y), and∑

z∈π−1(v)

w(x, z) =∑

z′∈π−1(v)

w(y, z′).

The connections between the eigenvalues in a combinatorial covering were proved in [1], gener-alizing the results in [4]:

Lemma 2.4. If G is a covering of G, then an eigenvalue of G is an eigenvalue of G.

Lemma 2.5. Suppose G is a covering of G with respect to the mapping π of index n. If a harmoniceigenfunction f of G, associated with an eigenvalue λ, has a nontrivial image in G, then λ is alsoan eigenvalue for G.

Let Tk denote the k-regular tree. We fix a point z in Tk as its root. Denote the Ω-ball of thek-regular tree by T (Ω)

k . We restrict the homewsick random walk on T (Ω)k but we normalize it at the

ends:

wT

(Ω)k

(x, y) :=

1k , x = z, y ∈ S1(z)

1k + µ− 1

, x ∈ Sm(z) y ∈ Sm+1(z)

1, x ∈ SΩ(z) y ∈ SΩ−1(z)µ

k + µ− 1, x ∈ Sm(z) y ∈ Sm−1(z)

4

where 1 ≤ m < Ω.Let P+ denote the graph such that V (P+) = N ∪ 0 and i ∼ j ⇐⇒ |i − j| = 1. There is a

natural map

π : V (Tk)→ V (P+), π(x) = d(z, x).

The map π preserves adjacencies. We define weights P+ by summing the weights of their preimages:

wP+(i, j) =∑

x∈π−1(i)

y∈π−1(j)

w(x, y).

Let P (Ω)+ be the subgraph with V (P (Ω)

+ ) = 0, 1, 2, . . . ,Ω. We consider the induced weights onP

(Ω)+ normalized so that they define a random walk. More presicely:

wP

(Ω)+

(u, v) :=

1, u = 0, v = 1

k(k − 1)m

k + µ− 1, u = m, v = m+ 1

k(k − 1)Ω−1, u = Ω, v = Ω− 1

µk(k − 1)m−1

k + µ− 1, u = m, v = m− 1

Proposition 2.6. The map π induces a combinatorial cover of P (Ω)+ by T (Ω)

k .

Proof. Condition (i) is clearly satisfied by this construction, where n = 1. To verify condition(ii), first notice that, for x ∈ Sm(z), y ∈ Sn(z), π(x) = π(y) iff m = n, because only verticesfrom the same l-sphere of the tree are projected onto the same vertex on the ray. The desiredsummations in Part (ii) will hold true since the weights between vertices are constant betweenadjacent l-spheres.

Therefore, the k-regular tree cropped at a given length Ω is a combinatorial cover of the weightedray of the same length, and so harmonic eigenfunctions can be lifted from the ray to the tree forthe purpose of computing the heat kernel of the k-regular tree, as will be done in the followingsection.

3. Finding the Heat Kernel of the k-Regular Tree

Each of the eigenvectors of the Laplacian L(Ω) has a component associated with each vertex ofthe tree. Let L2(T (Ω)

k ) be the L2-space generated by the vertices of T (Ω)k . The set of the eigenvalues

of L(Ω) is split into parts. Let E1 be the set of eigenvalues that admit an eigenfunction that itsz-component is not zero and E2 the eigenvalues whose eigenfunctions have z-component 0. Then,

L2(T (Ω)k ) = V1 ⊕ V2, Vi =

⊕λ∈Ei

Vλ, i = 1, 2.

5

From Lemma 2.5, the eigenfunctions in V1 are those which can be “lifted” from P(Ω)+ since the

harmonic eigenfunctions will have a nontrivial image in the ray. The heat kernel can be split intotwo summands:

ht(a, b) = ht,1(a, b) + ht,2(a, b)

where ht,1(a, b) = ht(a, b)|V1 and ht,2(a, b) = ht(a, b)|V2 .

3.1. Calculating ht,1(a,b). Let σ(Ω)λn

be an eigenvector of P (Ω)+ associated with the eigenvalue λn.

These eigenfunctions and eigenvalues were determined in [1].

λ0 = 0

λn = 1−2√µ · (k − 1)µ+ k − 1

cosπn

Ω, for n = 1, . . . ,Ω− 1

λΩ = 2

Eigenvectors associated with λ0 = 0 are given by:

σ(Ω)0 (0) = 1

σ(Ω)0 (p) =

√(µ+ k − 1) · (k − 1)p−1

µp, for 1 ≤ p ≤ Ω− 1

σ(Ω)0 (Ω) =

√(k − 1)Ω−1

µΩ−1

Eigenvectors associated with λn, where 1 ≤ n ≤ Ω− 1 are given by:

σ(Ω)λn

(0) =

√µ+ k − 1k − 1

sinπn

Ω

σ(Ω)λn

(p) = sinπn(p+ 1)

Ω− µ

k − 1sin

πn(p− 1)Ω

, for 1 ≤ p ≤ Ω− 1

σ(Ω)λn

(Ω) = −√µ · (µ+ k − 1)

k − 1sin

πn

Ω

Eigenvectors associated with λΩ = 2 are given by:

σ(Ω)2 (0) = 1

σ(Ω)2 (p) = (−1)p

√(µ+ k − 1) · (k − 1)p−1

µp, for 1 ≤ p ≤ Ω− 1

σ(Ω)2 (Ω) = (−1)Ω

√(k − 1)Ω−1

µΩ−1

6

Denote a basis of eigenvectors of V1 as ψ(Ω)λn

. These must be normalized before calculating theheat kernel. Therefore,

ht,1(a, b) =Ω∑n=0

ψ(Ω)λn

(a) · ψ(Ω)λn

(b)∥∥∥ψ(Ω)λn

∥∥∥2 · e−λnt

For vertices a and b on the tree, let la = d(z, a) and lb = d(z, b). Let fR be a harmonic eigenfunctionfor the ray. Since the tree covers the weighted ray, we can “lift” the harmonic eigenfunctions fromthe ray to the tree such that fT (a) = fR(la). Let φT be the Frobenius–Perron vector of the tree,and let φR be the Frobenius-Perron vector of the ray. Let ΦT be the diagonal matrix such thatΦT (a, a) = φT (a) and let ΦR be the diagonal matrix such that ΦR(la, la) = φR(la). From thedefinition of harmonic eigenfunctions, we have

fT (a) = ψ(Ω)λn

(a) · Φ−12

T (a, a) and fR(la) = σ(Ω)λn

(la) · Φ− 1

2R (la, la).

This imlpies:

ψ(Ω)λn

(a) = σ(Ω)λn

(la) · Φ− 1

2R (la, la) · Φ

12T (a, a) = σ

(Ω)λn

(la) · φ− 1

2R (la) · φ

12T (a).

Then, from [1] we have

φR(la) =

1, la = 0ρla−1(1 + ρ), 1 ≤ la ≤ Ω− 1ρΩ−1, la = Ω

and φT (a) =

1, la = 0(1 + ρ)k · µla−1

, 1 ≤ la ≤ Ω− 1

1k · µΩ−1

, la = Ω

where ρ =k − 1µ

. Using the functions defined above, we show that:

ψ(Ω)λn

(a) =

σ

(Ω)λn

(la), la = 0

σ(Ω)λn

(la)√k · (k − 1)la−1

, 1 ≤ la ≤ Ω

There are k · (k − 1)la−1 vertices a distance la from the root of the tree, and all the componentsof the eigenvectors are the same along each level ([1]). We use this fact to calculate the magnitude:∥∥∥ψ(Ω)

λn

∥∥∥2=µ+ k − 1k − 1

sin2 πn

Ω+

Ω−1∑p=1

(sin

πn(p+ 1)Ω

− µ

k − 1sin

πn(p− 1)Ω

)2+

µ(µ+ k − 1)(k − 1)2

sin2 πn

Ω

where 1 ≤ n ≤ Ω− 1. Likewise,∥∥∥ψ(Ω)λ0

∥∥∥2=∥∥∥ψ(Ω)

λΩ

∥∥∥2= 1 +

Ω−1∑p=1

(µ+ k − 1) · (k − 1)p−1

µp

+(k − 1)Ω−1

µΩ−1

7

The first sum is calculated in [4]:∥∥∥ψ(Ω)λn

∥∥∥2=

Ω(µ+ k − 1)2

2(k − 1)2

[1− 4µ(k − 1)

(µ+ k − 1)2cos2 nπ

Ω

]This is all we need to determine ht,1(a, b):

ht,1(z, z) =1∥∥∥ψ(Ω)λ0

∥∥∥2 +e−2t∥∥∥ψ(Ω)λΩ

∥∥∥2 + 2(k − 1) ·Ω−1∑n=1

e−t[1− 2

√µ·(k−1)

µ+k−1cos πn

Ω] sin2 nπ

Ω

Ω(µ+ k − 1)[1− 4µ(k−1)(µ+k−1)2 cos2 nπ

Ω ]

From [1], the random walk is positive recurrent exactly when ρ < 1. This occurs when µ > k − 1.In this case,

limΩ→∞

∥∥∥ψ(Ω)λ0

∥∥∥2= lim

Ω→∞

∥∥∥ψ(Ω)λΩ

∥∥∥2=

2µµ− k + 1

If µ ≤ k − 1, the limit diverges. So, in the limiting case as Ω→∞:

ht,1(z, z) =(µ− k + 1)(1 + e−2t)

2µ+

2(k − 1)π

∫ π

0

e−t[1− 2

√µ·(k−1)

µ+k−1cosx] sin2 x

(µ+ k − 1)[1− 4µ(k−1)(µ+k−1)2 cos2 x]

dx

(If µ ≤ k − 1, then the first term is not present.) Similarly, when la > 0 and µ > k − 1:

ht,1(z, a) =(µ− k + 1)

√µ+k−1µla

[1 + (−1)la · e−2t

]2µ√k

+2(k − 1)2

π√k · (k − 1)la

∫ π

0

e−t[1− 2

√µ·(k−1)

µ+k−1cosx]

(√µ+ k − 1 sinx

)(sin [(la + 1)x]− µ

k − 1sin [(la − 1)x]

)(µ+ k − 1)2

[1− 4µ(k−1)

(µ+k−1)2 cos2 x] dx

(Again, the first term is not present when µ ≤ k − 1.)Finally, when la, lb > 0 and µ > k − 1:

ht,1(a, b) =(µ− k + 1)(µ+ k − 1)(

√µ)−(la+lb)

[1 + (−1)(la+lb) · e−2t

]2µk

+∫ π

0

e−t[1− 2

√µ·(k−1)

µ+k−1cosx]

(sin [(la + 1)x]− µ

k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)

(µ+k−1)2 cos2 x]dx

(Once again, the the first term is not present when µ ≤ k − 1.)It is important to note that ht(z, z) = ht,1(z, z) and ht(z, a) = ht,1(z, a) so for these special cases

we have determined the heat kernel entirely.8

3.2. Calculating ht,2(a,b). To find the eigenvectors of the Laplacian L(Ω) of T (Ω)k , it will be

necessary to define certain subsets of V (T (Ω)k ), which we will call branches. First we provide a

visually intuitive definition of a branch. A branch with root v is the set of all vertices including vwhich can be reached from v in a walk on the tree without moving backwards toward the root ofthe tree.

Definition 3.1. Let v be a vertex of T (Ω)k such that the distance d(z, v) = y, where z is the root

of the tree. Then the branch of length Ω− y with root v, denoted B(Ω−y)(v), is defined as:

B(Ω−y)(v) = w ∈ T (Ω)k | d(0, w) ≥ y and v ∈ P(0,w),

where P(0,w) is the set of vertices on the unique path from 0 to w.

Let v(1)1 , v

(1)2 , · · · , v(1)

k be the vertices of distance 1 from the root of the tree. It was shown in [1],that for the set of eigenfunctions which vanish at the root, the system of equations correspondingto φ · L(Ω) = λφ separates into k identical systems corresponding to the branches of length Ω− 1.Therefore, to find the eigenvectors in V2 we need to find the eigenvectors of the submatrix of theLaplacian corresponding to a branch of length Ω− 1. The connection between the solutions on thebranches B(Ω−1)(v(1)

1 ), B(Ω−1)(v(1)2 ), · · · , B(Ω−1)(v(1)

k ) is given by the following relationship

φ(v(1)1 ) + φ(v(1)

2 ) + · · ·+ φ(v(1)k ) = 0

Let L2(B(Ω−y)) be the L2-space generated by the vertices of a branch of length Ω−y. As before,L2(B(Ω−y)) = U1⊕U2, where U1 is generated by eigenfunctions for the submatrix of the LaplacianL(Ω) that are not zero at the root of the branch, and U2 is generated by eigenfunctions which vanishat the root. Let ΛΩ−y be the submatrix of L(Ω) corresponding to a branch of length Ω− y. From[1] we have:

ΛΩ−y(a, b) :=

1, a = b

β, a ∼ b, la, lb < Ω− yγ, a ∼ b, la = Ω− y or lb = Ω− y0, otherwise

where

β = −√µ

k + µ− 1and γ = − 1√

k + µ− 1

The eigenfunctions in U1 have a nice property which makes the calculations easier; the componentsof the eigenfunctions corresponding to vertices the same distance from the root will be the same.For each eigenvector ψ, this condition greatly simplifies the system of ((k− 1)Ω−y+1− 1) · (k− 2)−1

equations, corresponding to ψ · ΛΩ−y = λψ, into a system of Ω − y + 1 equations. Then we only9

need to find the eigenvalues and eigenvectors of the (Ω− y + 1)×(Ω− y + 1) matrix:

1 β 0 · · · 0 0(k − 1)β 1 β · · · 0 0

0 (k − 1)β 1. . . 0 0

......

. . . . . ....

...0 0 0 · · · 1 γ

0 0 0 · · · (k − 1)γ 1

In the limiting case as Ω→∞, the eigenvalues and eigenvectors will approach those of the matrix:

1 β 0 · · · 0 0(k − 1)β 1 β · · · 0 0

0 (k − 1)β 1. . . 0 0

......

. . . . . ....

...0 0 0 · · · 1 β

0 0 0 · · · (k − 1)β 1

For reasons that will become apparent later, we will only need the eigenvalues and eigenvectors ofthis matrix. Thus we need to calculate the determinant of the following matrix:

1− λ β 0 · · · 0 0(k − 1)β 1− λ β · · · 0 0

0 (k − 1)β 1− λ . . . 0 0...

.... . . . . .

......

0 0 0 · · · 1− λ β

0 0 0 · · · (k − 1)β 1− λ

Divide the matrix by β

√k − 1 and let x =

1− λ2β√k − 1

. Then our matrix becomes:

An =

2x1√k − 1

0 · · · 0 0√k − 1 2x

1√k − 1

· · · 0 0

0√k − 1 2x

. . . 0 0...

.... . . . . .

......

0 0 0 · · · 2x1√k − 1

0 0 0 · · ·√k − 1 2x

10

Set gn(x) = det(An). This satisfies the recursive relation:

gn(x) = 2x · gn−1(x)− gn−2(x)

The last recurrence relations are exactly the ones that define the Chebyshev polynomials of secondkind. Moreover,

g1(x) = 2x, g2(x) = 4x2 − 1

which are the base cases for the Chebyshev polynomials of second kind. The roots of gn are givenby:

xj = cosjπ

n+ 1, where 1 ≤ j ≤ n and n = Ω− y + 1.

This gives us the eigenvalues

λj = 1− 2β√k − 1 cos

jπ

n+ 1, 1 ≤ j ≤ n and n = Ω− y + 1.

Denote the components of the eigenvectors of this matrix by ϑ(Ω−y)λj

(m), where m is the distanceof a vertex from the root of the branch. Then we get the following system of equations:

(1− λj)ϑ(Ω−y)λj

(0) = (k − 1)βϑ(Ω−y)λj

(1)


(1) = βϑ(Ω−y)λj

(0) + (k − 1)βϑ(Ω−y)λj

(2)


(2) = βϑ(Ω−y)λj

(1) + (k − 1)βϑ(Ω−y)λj

(3)· · ·(1− λj)ϑ(Ω−y)

λj(Ω− y − 1) = βϑ

(Ω−y)λj

(Ω− y − 2) + (k − 1)βϑ(Ω−y)λj

(Ω− y)


(Ω− y) = βϑ(Ω−y)λj

(Ω− y − 1)

Next, we substitute our values for λj into the system, and it becomes:

2√k − 1 cos jπ

n+1ϑ(Ω−y)λj

(0) = (k − 1)ϑ(Ω−y)λj

(1)

2√k − 1 cos jπ

n+1ϑ(Ω−y)λj

(1) = ϑ(Ω−y)λj

(0) + (k − 1)ϑ(Ω−y)λj

(2)

2√k − 1 cos jπ

n+1ϑ(Ω−y)λj

(2) = ϑ(Ω−y)λj

(1) + (k − 1)ϑ(Ω−y)λj

(3)· · ·

2√k − 1cos

jπ

n+ 1ϑ

(Ω−y)λj

(Ω− y − 1) = ϑ(Ω−y)λj

(Ω− y − 2) + (k − 1)ϑ(Ω−y)λj

(Ω− y)

2√k − 1cos

jπ

n+ 1ϑ

(Ω−y)λj

(Ω− y) = ϑ(Ω−y)λj

(Ω− y − 1)11

Now, let ϑ(Ω−y)λj

(0) = 1, for convenience. Then the system simplifies to:

ϑ(Ω−y)λj

(0) = 1

ϑ(Ω−y)λj

(1) =2√k − 1

cosjπ

n+ 1

ϑ(Ω−y)λj

(m) =2√k − 1

cosjπ

n+ 1ϑ

(Ω−y)λj

(m− 1)− 1k − 1

ϑ(Ω−y)λj

(m− 2), for 2 ≤ m ≤ Ω− y − 1

ϑ(Ω−y)λj

(Ω− y) =1

2√k − 1 cos jπ

n+1

ϑ(Ω−y)λj

(Ω− y − 1)

Then, for 2 ≤ m ≤ Ω − y − 1, we have a linear recurrence relation for the eigenvectors with thecharacteristic polynomial:

x2 − 2√k − 1

cosjπ

n+ 1x+

1k − 1

with roots2√k−1

cos jπn+1 ±

√4

k−1cos2 jπn+1 −

4k−1

2=

cos jπn+1 ± ı sin

jπn+1√

k − 1So, for 2 ≤ m ≤ Ω− y − 1,

ϑ(Ω−y)λj

(m) = c1 ·

(cos jπ

n+1 + ı sin jπn+1√

k − 1

)m+ c2 ·

(cos jπ

n+1 − ı sinjπn+1√

k − 1

)mwhere c1 and c2 are constants. Next, we find the first few values of the recursion by hand and usethese to find the constants:

ϑ(Ω−y)λj

(2) =4

k − 1cos2 jπ

n+ 1− 1k − 1

, ϑ(Ω−y)λj

(3) =8 cos3 jπ

n+1 − 4 cos jπn+1

(√k − 1)3

Thus

c1 =cos jπ

n+1 + ı sin jπn+1

2ı sin jπn+1

, c2 =− cos jπ

n+1 + ı sin jπn+1

2ı sin jπn+1

Substituting the constants back into the expression, we have:

ϑ(Ω−y)λj

(m) =

(cos jπ

n+1 + ı sin jπn+1

2ı sin jπn+1

)·

(cos jπ

n+1 + ı sin jπn+1√

k − 1

)m+

(− cos jπ

n+1 + ı sin jπn+1

2ı sin jπn+1

)·

(cos jπ

n+1 − ı sinjπn+1√

k − 1

)mWith a little manipulation, the expression simplifies to:

ϑ(Ω−y)λj

(m) =sin (m+1)jπ

n+1

(√k − 1)m sin jπ

n+1

, 2 ≤ m ≤ Ω− y − 1.

12

Summarizing,

ϑ(Ω−y)λj

(0) = 1

ϑ(Ω−y)λj

(1) =2√k − 1

cosjπ

Ω− y + 2

ϑ(Ω−y)λj

(m) =sin (m+1)jπ

Ω−y+2

(√k − 1)m sin jπ

Ω−y+2

, for 2 ≤ m ≤ Ω− y − 1

ϑ(Ω−y)λj

(Ω− y) =sin (Ω−y)jπ

Ω−y+2

(√k − 1)Ω−y sin 2jπ

Ω−y+2

There will be Ω− y + 1 eigenvectors in U1, denoted by ψ(Ω−y)λj

.

Remark 3.2. Be sure to distinguish between eigenvectors of V1, denoted ψ(Ω)λj

, and eigenvectors of

U1, denoted ψ(Ω−y)λj

. We have chosen to use the “ψ” notation for both cases since neither vanishat their respective roots. Likewise, we have consistently used “φ” to denote eigenvectors which dovanish.

For a vertex a on the tree which is a distance la from the root of the tree, it will be a distancela − y from the root of the branch. In the limiting case,

limΩ→∞

ψ(Ω−y)λj

(a) = limΩ→∞

ϑ(Ω−y)λj

(la − y)

This can be more easily realized by remembering the matrix limit shown earlier.This implies that:

limΩ→∞

Ω−y+1∑j=1

ψ(Ω−y)λj

(a) · ψ(Ω−y)λj

(b)∥∥∥ψ(Ω−y)λj

∥∥∥2 · e−λjt = limΩ→∞

Ω−y+1∑j=1

ϑ(Ω−y)λj

(la − y) · ϑ(Ω−y)λj

(lb − y)∥∥∥Θ(Ω−y)λj

∥∥∥2 · e−λjt

where ∥∥∥Θ(Ω−y)λj

∥∥∥2=

Ω−y∑m=0

(k − 1)m[ϑ

(Ω−y)λj

(m)]2

This comes from the fact that the number of vertices a distance m from the root of the branch is(k − 1)m. Thus∥∥∥Θ(Ω−y)

λj

∥∥∥2= 1 + 4 cos2 jπ

Ω− y + 2+

Ω−y−1∑m=2

sin2 (m+1)jπΩ−y+2

sin2 jπΩ−y+2

+sin2 (Ω−y)jπ

Ω−y+2

sin2 2jπΩ−y+2

This expression can be simplified to:∥∥∥Θ(Ω−y)λj

∥∥∥2= 1 + 4 cos2 jπ

Ω− y + 2+

Ω− y − 22 sin2 jπ

Ω−y+2

−sin (Ω−y−2)jπ

Ω−y+2 cos (Ω−y+3)jπΩ−y+2

2 sin3 jπΩ−y+2

+sin2 (Ω−y)jπ

Ω−y+2

sin2 2jπΩ−y+2

13

Let us now introduce a new function η(Ω−y) for later convenience:

η(Ω−y)(a, b) =Ω−y+1∑j=1

ψ(Ω−y)λj



∥∥∥2 · e−λjt

Now let:

η∞(la − y, lb − y) = limΩ→∞

η(Ω−y)(a, b) = limΩ→∞

Ω−y+1∑j=1

ϑ(Ω−y)λj

(la − y) · ϑ(Ω−y)λj

(lb − y)∥∥∥Θ(Ω−y)λj

∥∥∥2 · e−λjt

We calculate the following limits to determine the function η∞. We will show that ht,2(a, b) can beexpressed as a finite sum of these functions:

η∞(0, 0) =2π

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin2 x dx

η∞(0, 1) =4

π√k − 1

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin2 x cosx dx

η∞(0,m) =2

π(√k − 1)m

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin [(m+ 1)x] sinx dx, where m > 1

η∞(1, 1) =8

π(k − 1)

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin2 x cos2 x dx

η∞(1,m) =4

π(√k − 1)m+1

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sinx sin [(m+ 1)x] cosx dx, where m > 1

η∞(m,n) =2

π(√k − 1)m+n

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin [(m+ 1)x] sin [(n+ 1)x] dx, where m,n > 1

For the set of eigenvectors in U2, the ones which vanish at the root of the branch, the equations splitinto k− 1 identical systems corresponding to the subbranches of length Ω− y− 1. The connectionbetween the solutions on the branches is given by

φ(u(1)1 ) + φ(u(1)

2 ) + · · ·+ φ(u(1)k−1) = 0

where u(1)1 , u

(1)2 , · · · , u(1)

k−1 are the vertices of distance 1 from the root of the branch. Now, let usintroduce a notation for eigenvectors of the branch. An eigenvector φΩ−y of the branch of length

14

Ω− y can be written as:

φΩ−y(u(0))φΩ−y|

BΩ−y−1(u(1)1 )

φΩ−y|BΩ−y−1(u

(1)2 )

φΩ−y|BΩ−y−1(u

(1)3 )

...φΩ−y|

BΩ−y−1(u(1)k−1)

where φΩ−y(u(0)) is the component of the eigenvector corresponding to the root of the branch andφΩ−y|

BΩ−y−1(u(1)i )

is the vector corresponding to the branch BΩ−y−1(u(1)i ).

Suppose that the eigenvectors of the branch of length Ω − y − 1 are known. Then there are((k − 1)Ω−y − 1) · (k − 2)−1 linearly independent eigenvectors in total for the branch of this lengthin U2. There are Ω − y eigenvectors of the branch of length Ω − y − 1 which are nonzero at theroot, denoted ψ

(Ω−y−1)λj

. For each of them, k − 2 orthogonal eigenvectors of the branch of lengthΩ− y can be constructed as:

0ψ

(Ω−y−1)λj

−ψ(Ω−y−1)λj~0~0...~0

,

0ψ

(Ω−y−1)λj

ψ(Ω−y−1)λj

−2ψ(Ω−y−1)λj~0...~0

,

0ψ

(Ω−y−1)λj

ψ(Ω−y−1)λj

ψ(Ω−y−1)λj

−3ψ(Ω−y−1)λj...~0

, · · · · · · ,

0ψ

(Ω−y−1)λj

ψ(Ω−y−1)λj

ψ(Ω−y−1)λj

ψ(Ω−y−1)λj

...−(k − 2)ψ(Ω−y−1)

λj

For instance, in the first eigenvector depicted above, 0 is the value at the root, ψ(Ω−y−1)

λjis

the eigenvector with components corresponding to the branch B(Ω−y−1)(u(1)1 ), −ψ(Ω−y−1)

λjis the

eigenvector with components corresponding to the branch B(Ω−y−1)(u(1)2 ), and ~0 is the vector

with components corresonding to branches B(Ω−y−1)(u(1)3 ), · · · , B(Ω−y−1)(u(1)

k−1) (~0 is the vector of((k − 1)Ω−y − 1) · (k − 2)−1 zeros). These constructed eigenvectors will satisfy the condition:

φ(u(1)1 ) + φ(u(1)

2 ) + · · ·+ φ(u(1)k−1) = 0

There are ((k−1)Ω−y−1) · (k−2)−1−Ω+y eigenvectors of the branch of length Ω−y−1 whichvanish at the root. Denote them by φ(Ω−y−1)

λj ,m, where 1 ≤ m ≤ ((k− 1)Ω−y − 1) · (k− 2)−1 −Ω + y.

Since the component at the root is zero, k−1 orthogonal eigenvectors of the branch of length Ω−y15

can be constructed as:

0φ

(Ω−y−1)λj ,m

~0~0~0...~0

,

0~0

φ(Ω−y−1)λj ,m

~0~0...~0

,

0~0~0


~0...~0

, · · · · · · ,

0~0~0~0~0...


The constructed eigenvectors will satisfy the desired condition:

φ(u(1)1 ) + φ(u(1)

2 ) + · · ·+ φ(u(1)k−1) = 0

It is easy to check that the eigenvectors constructed this way are orthogonal. All of these eigenvec-tors will be in U2, corresponding to the branch of length Ω − y. So, we have found (k − 2) · (Ω −y) + (k − 1) · [((k − 1)Ω−y − 1) · (k − 2)−1 − Ω + y] eigenvectors in U2, considering the ones whichare nonzero at the root and the ones that vanish. Now, if we count the Ω − y + 1 eigenvectors inU1, we have ((k − 1)Ω−y+1 − 1) · (k − 2)−1 orthogonal eigenvectors in total as desired. Therefore,we have established a method for finding all the linearly independent eigenvectors of a branch oflength Ω− y, given the eigenvectors of a branch of length Ω− y − 1.

We assume we have determined the eigenvectors of a branch of length Ω − 1. Then for each ofthe eigenvectors which do not vanish at the root, denoted ψ

(Ω−1)λj

, we can find the following k − 1orthogonal eigenvectors for the tree:

0ψ

(Ω−1)λj

−ψ(Ω−1)λj~0~0...~0

,

0ψ

(Ω−1)λj

ψ(Ω−1)λj

−2ψ(Ω−1)λj~0...~0

,

0ψ

(Ω−1)λj

ψ(Ω−1)λj

ψ(Ω−1)λj

−3ψ(Ω−1)λj...~0

, · · · · · · ,

0ψ

(Ω−1)λj

ψ(Ω−1)λj

ψ(Ω−1)λj

ψ(Ω−1)λj

...−(k − 1)ψ(Ω−1)

λj

These constructed eigenvectors satisfy the condition:

φ(v(1)1 ) + φ(v(1)

2 ) + · · ·+ φ(v(1)k ) = 0

16

For each of the eigenvectors of the branch of length Ω − 1 which vanish at the root, denotedφ

(Ω−1)λj ,m

, we can find the following k orthogonal eigenvectors for the tree:

0φ

(Ω−1)λj ,m

~0~0~0...~0

,

0~0

φ(Ω−1)λj ,m

~0~0...~0

,

0~0~0

φ(Ω−1)λj ,m

~0...~0

, · · · · · · ,

0~0~0~0~0...

φ(Ω−1)λj ,m

Again, it is easy to check these constructions provide all the linearly independent eigenvectors ofV2.

Now we will arrange all this information to determine ht,2(a, b). Let:

V2 = U(1)1 ⊕ U (1)

2

where U (1)1 is the set of eigenvectors which have at least one nonzero component corresponding to

vertices a distance 1 from the root of the tree, and U(1)2 is the set of eigenvectors such that the

components corresponding to vertices of distance 1 from the root vanish. Let U (1)2 = U

(2)1 ⊕ U (2)

2

where U (2)1 is the set of eigenvectors which have at least one nonzero component at distance 2 from

the root of the tree, and U (2)2 is the set of eigenvectors such that the components corresponding to

vertices of distance 2 from the root vanish. Continuing this process yields the following recursiverelationship:

U(y)2 = U

(y+1)1 ⊕ U (y+1)

2

Therefore, ht,2(a, b) will split into two summands, denoted ht,2(a, b)|U

(1)1

and ht,2(a, b)|U

(1)2

, where

ht,2(a, b)|U

(1)i

is the restriction of ht,2(a, b) to U(1)i , i = 1, 2. We can continue splitting the heat

kernel into two summands:

ht,2(a, b)|U

(y)2

= ht,2(a, b)|U

(y+1)1

+ ht,2(a, b)|U

(y+1)2

Now suppose that for a and b on the tree, that the smallest branch which includes both a and bis of length Ω−D. If a and b are on different branches of length Ω− 1, let D = 0.

Proposition 3.3. For a and b on the tree such that the smallest branch which includes both a andb is of length Ω−D, ht,2(a, b)|

U(D+1)2

= 0.

Proof. First consider the case where D > 0 (a similar argument applies when D = 0). Considerthe components of the eigenvectors in U (D+1)

2 corresponding to a branch of length Ω−D. If thesecomponents are not all zero, they must form eigenvectors for the submatrix of the Laplacian L(Ω)

corresponding to a branch of length Ω − D. For eigenvectors in U(D+1)2 , denoted φ

U(D+1)2

, the17

components corresponding to vertices a distance D+ 1 or less from the root of the tree are 0. Theroot of the branch will be a distance D from the root of the tree, so the eigenvectors for the branchwill have the property that the components at the root of the branch and at distance 1 from theroot of the branch will be 0. We have already determined that the eigenvectors of the submatrixwith this property take the form:

0φ

(Ω−D−1)λj ,m

~0~0~0...~0

,

0~0

φ(Ω−D−1)λj ,m

~0~0...~0

,

0~0~0


~0...~0

, · · · · · · ,

0~0~0~0~0...


where each φ(Ω−D−1)

λj ,mdenotes an eigenvector for the branch of length Ω−D − 1 which vanishes at

the root. For all of these eigenvectors in U(D+1)2 , ~0 is the vector with components on all but one

of the subbranches of length Ω−D − 1. Now, a and b cannot be on the same subbranch of lengthΩ−D − 1, as we assumed that the smallest branch which contains them both is of length Ω−D.From this we can see that φ

U(D+1)2

(a) = 0 or φU

(D+1)2

(b) = 0. Therefore φU

(D+1)2

(a) · φU

(D+1)2

(b) = 0

and so ht,2(a, b)|U

(D+1)2

= 0.

So, we have the following:

Corollary 3.4. With the notation as in Proposition 3.3,

ht,2(a, b) = ht,2(a, b)|U

(1)1

+ ht,2(a, b)|U

(2)1

+ · · ·+ ht,2(a, b)|U

(D+1)1

To calculate ht,2(a, b), it will be necessary to consider two cases: when a and b are on differentbranches of length Ω− 1, and when they are on the same branch of length Ω− 1.

Proposition 3.5. If a and b are on different branches of length Ω− 1, then

ht,2(a, b) = −1kη∞(la − 1, lb − 1)

Proof. Suppose a and b are on different branches of length Ω − 1, then ht,2(a, b) = ht,2(a, b)|U

(1)1

.

For eigenvectors in U (1)1 , denoted ψ, we know that at least one of the components of the eigenvector

associated with vertices a distance 1 from the root of the tree is nonzero. We know the eigenvectorsof the submatrix of the Laplacian corresponding to a branch of length Ω− 1 which don’t vanish at

18

the root are ψ(Ω−1)λj

. Therefore, the eigenvectors in U(1)1 take the form:

0ψ

(Ω−1)λj

−ψ(Ω−1)λj~0~0...~0

,

0ψ

(Ω−1)λj

ψ(Ω−1)λj

−2ψ(Ω−1)λj~0...~0

,

0ψ

(Ω−1)λj

ψ(Ω−1)λj

ψ(Ω−1)λj

−3ψ(Ω−1)λj...~0

, · · · · · · ,

0ψ

(Ω−1)λj

ψ(Ω−1)λj

ψ(Ω−1)λj

ψ(Ω−1)λj

...−(k − 1)ψ(Ω−1)

λj

Now suppose a ∈ B(Ω−1)(v(1)

m ) and b ∈ B(Ω−1)(v(1)n ) where 1 ≤ m < n ≤ k, then for each ψ

(Ω−1)λj

,

there exist k − 1 eigenvectors ψi in U(1)1 such that:

ψi(a) · ψi(b) = 0, where 1 ≤ i < n− 1ψi(a) · ψi(b) = −(n− 1)ψ(Ω−1)

λj(a) · ψ(Ω−1)

λj(b), where i = n− 1

ψi(a) · ψi(b) = ψ(Ω−1)λj

(a) · ψ(Ω−1)λj

(b), where n− 1 < i ≤ k − 1

and∥∥∥ψi∥∥∥2

= [i2 + i] ·∥∥∥ψ(Ω−1)

λj

∥∥∥2. Therefore:

ht,2(a, b)|U

(1)1

=

[−(n− 1)

(n− 1)2 + (n− 1)+k−1∑i=n

1i2 + i

]·

Ω∑j=1

ψ(Ω−1)λj

(a) · ψ(Ω−1)λj


∥∥∥2 · e−λjt

so

ht,2(a, b)|U

(1)1

=

[−(n− 1)

(n− 1)2 + (n− 1)+k−1∑i=n

1i2 + i

]η(Ω−1)(a, b) = −1

kη(Ω−1)(a, b).

Taking the limit as Ω→∞, we arrive at:

ht,2(a, b) = −1kη∞(la − 1, lb − 1)

Case 2: a and b on the same branch of length Ω− 1.Recall the relationship:

ht,2(a, b) = ht,2(a, b)|U

(1)1

+ ht,2(a, b)|U

(2)1

+ · · ·+ ht,2(a, b)|U

(D+1)1

We will divide the next calculation into three steps. First we calculate ht,2(a, b)|U

(1)1

. Then we

calculate ht,2(a, b)|U

(y)1

, where 2 ≤ y ≤ D. Finally we calculate ht,2(a, b)|U

(D+1)1

, which depends onwhether la = D or lb = D, or if la, lb 6= D.

19

Step 1: Calculate ht,2(a,b)|U

(1)1

.

We know the eigenvectors in U(1)1 take the form shown above in Case 1.

If a, b ∈ B(Ω−1)(v(1)m ) where 1 ≤ m ≤ k, then for each ψ

(Ω−1)λj

, there are k − 1 eigenvectors ψi in

U(1)1 where 1 ≤ i ≤ k − 1 such that:

ψi(a) · ψi(b) = 0, where 1 ≤ i < m− 1

ψi(a) · ψi(b) = (m− 1)2ψ(Ω−1)λj

(a) · ψ(Ω−1)λj

(b), where i = m− 1

ψi(a) · ψi(b) = ψ(Ω−1)λj

(a) · ψ(Ω−1)λj

(b), where m− 1 < i ≤ k − 1

and ∥∥∥ψi∥∥∥2= [i2 + i] ·

∥∥∥ψ(Ω−1)λj

∥∥∥2

Therefore, if m = 1:

ht,2(a, b)|U

(1)1

=[ k−1∑i=1

1i2 + i

]·

Ω∑j=1

ψ(Ω−1)λj

(a) · ψ(Ω−1)λj


∥∥∥2 · e−λjt =k − 1k

η(Ω−1)(a, b)

and if m > 1:

ht,2(a, b)|U

(1)1

=[ (m− 1)2

(m− 1)2 + (m− 1)+k−1∑i=m

1i2 + i

]·

Ω∑j=1

ψ(Ω−1)λj

(a) · ψ(Ω−1)λj


∥∥∥2 · e−λjt =k − 1k

η(Ω−1)(a, b)


ht,2(a, b)|U

(1)1

=k − 1k

η∞(la − 1, lb − 1)


(y)1

, where 2 ≤ y ≤ D.

For each ψ(Ω−y)λj

, there are eigenvectors in U(y)1 that take the form:

0φ(Ω−1)

~0~0...~0

,

0~0

φ(Ω−1)

~0...~0

,

0~0~0

φ(Ω−1)

...~0

, · · · · · · ,

0~0~0~0...

φ(Ω−1)

20

where

φ(Ω−1) =

0φ(Ω−2)

~0~0...~0

,

0~0

φ(Ω−2)

~0...~0

,

0~0~0

φ(Ω−2)

...~0

, · · · · · · ,

0~0~0~0...

φ(Ω−2)

...

......

......

φ(Ω−y+2) =

0φ(Ω−y+1)

~0~0...~0

,

0~0

φ(Ω−y+1)

~0...~0

,

0~0~0

φ(Ω−y+1)

...~0

, · · · · · · ,

0~0~0~0...

φ(Ω−y+1)

φ(Ω−y+1) =

0ψ

(Ω−y)λj

−ψ(Ω−y)λj~0~0...~0

,

0ψ

(Ω−y)λj

ψ(Ω−y)λj

−2ψ(Ω−y)λj~0...~0

,

0ψ

(Ω−y)λj

ψ(Ω−y)λj

ψ(Ω−y)λj

−3ψ(Ω−y)λj...~0

, · · · · · · ,

0ψ

(Ω−y)λj

ψ(Ω−y)λj

ψ(Ω−y)λj

ψ(Ω−y)λj

...−(k − 2)ψ(Ω−y)

λj

By our assumption, a and b are on the same subbranch of length Ω − y. Consider the subbranchof length Ω − y + 1 which contains both a and b. From the construction, it is clear that for eachψ

(Ω−y)λj

, there are k − 2 eigenvectors in U(y)1 such that the vector of components corresponding to

this subbranch is not ~0. By a similar calculation to that of Step 1, we can show that:

ht,2(a, b)|U

(y)1

=[ (m− 1)2

(m− 1)2 + (m− 1)+k−2∑i=m

1i2 + i

]·

Ω∑j=1

ψ(Ω−y)λj



∥∥∥2 · e−λjt =k − 2k − 1

η(Ω−y)(a, b)


ht,2(a, b)|U

(y)1

=k − 2k − 1

η∞(la − y, lb − y)


(D+1)1

.21

First suppose la = D or lb = D. In U(D+1)1 , the components of the eigenvectors corresponding

to vertices a distance D from the origin are 0. Therefore:

ψU

(D+1)1

(a) = 0 or ψU

(D+1)1

(b) = 0

soht,2(a, b)|

U(D+1)1

= 0

Now suppose la, lb 6= D. The eigenvectors in U(D+1)1 take the form shown above in Step 2, for

y = D + 1. Using the assumption that a and b are on different branches of length Ω−D − 1, anda calculation similar to the one in Case 1, we determine that:

ht,2(a, b)|U

(D+1)1

=[ −(n− 1)

(n− 1)2 + (n− 1)+k−2∑i=n

1i2 + i

]·

Ω∑j=1

ψ(Ω−D−1)λj

(a) · ψ(Ω−D−1)λj

(b)∥∥∥ψ(Ω−D−1)λj

∥∥∥2 · e−λjt

so

ht,2(a, b)|U

(D+1)1

=[ −(n− 1)

(n− 1)2 + (n− 1)+k−2∑i=n

1i2 + i

]η(Ω−D−1)(a, b) = − 1

k − 1η(Ω−D−1)(a, b)


ht,2(a, b)|U

(D+1)1

= − 1k − 1

η∞(la −D − 1, lb −D − 1)

Now we have completely determined ht,2(a, b):

If a, b are on different branches of length Ω− 1:

limΩ→∞

ht,2(a, b) = −1kη∞(la − 1, lb − 1)

If a, b are on the same branch of length Ω− 1, and la, lb 6= D:

limΩ→∞

ht,2(a, b) =k − 1k

η∞(la − 1, lb − 1)− 1k − 1

η∞(la −D − 1, lb −D − 1)+D∑y=1

k − 2k − 1

η∞(la − y, lb − y)

If a, b are on the same branch of length Ω− 1, and la = D or lb = D:

limΩ→∞

ht,2(a, b) =k − 1k

η∞(la − 1, lb − 1) +D∑y=1

k − 2k − 1

η∞(la − y, lb − y)

22

Theorem 3.6. Let la = d(0, a) and lb = d(0, b), where 0 denotes the root of the tree. Let S be theset of branch roots for branches which contain both a and b. Let D = maxd(0, s)|s ∈ S. D = 0if S is the empty set. Then the heat kernel of the k-regular tree is given by:

I: If µ > k − 1 :i:

ht(0, 0) =(µ− k + 1)(1 + e−2t)

2µ+

2(k − 1)π

∫ π

0

e−t[1− 2

√µ·(k−1)


(µ+ k − 1)[1− 4µ(k−1)(µ+k−1)2 cos2 x]

dx

ii:

ht(0, a) =(µ− k + 1)

√µ+ k − 1

µla

[1 + (−1)la · e−2t

]2µ√k

+2(k − 1)2

π√k · (k − 1)la

∫ π

0

e−t[1− 2

√µ·(k−1)

µ+k−1cosx](√

µ+ k − 1 sinx)(

sin [(la + 1)x]− µ

k − 1sin [(la − 1)x]

)(µ+ k − 1)2[1− 4µ(k−1)


iii: if D = 0

ht(a, b) = −1kη∞(la − 1, lb − 1) +

(µ− k + 1)(µ+ k − 1)(√µ)−(la+lb)

[1 + (−1)(la+lb) · e−2t

]2µk

+∫ π

0

e−t[1− 2

√µ·(k−1)

µ+k−1cosx]( sin [(la + 1)x]− µ

k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)


iv: if D > 0 and la, lb 6= D

ht(a, b) =k − 1k

η∞(la − 1, lb − 1)− 1k − 1

η∞(la −D − 1, lb −D − 1) +D∑y=1

k − 2k − 1

η∞(la − y, lb − y)

+(µ− k + 1)(µ+ k − 1)(

√µ)−(la+lb)

[1 + (−1)(la+lb) · e−2t

]2µk

+∫ π

0

e−t[1− 2

√µ·(k−1)


k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)


23

v: if D > 0 and la = D or lb = D

ht(a, b) =k − 1k

η∞(la − 1, lb − 1) +D∑y=1

k − 2k − 1

η∞(la − y, lb − y)

+(µ− k + 1)(µ+ k − 1)(

√µ)−(la+lb)

[1 + (−1)(la+lb) · e−2t

]2µk

+∫ π

0

e−t[1− 2

√µ·(k−1)


k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)


II: If µ ≤ k − 1:i:

ht(0, 0) =2(k − 1)

π

∫ π

0

e−t[1− 2

√µ·(k−1)


(µ+ k − 1)[1− 4µ(k−1)(µ+k−1)2 cos2 x]

dx

ii: ht(0, a) =

2(k − 1)2

π√k · (k − 1)la

∫ π

0

e−t[1− 2

√µ·(k−1)

µ+k−1cosx](√

µ+ k − 1 sinx)(

sin [(la + 1)x]− µ

k − 1sin [(la − 1)x]

)(µ+ k − 1)2[1− 4µ(k−1)


iii: if D = 0

ht(a, b) = −1kη∞(la − 1, lb − 1)

+∫ π

0

e−t[1− 2

√µ·(k−1)


k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)


iv: if D > 0 and la, lb 6= D

ht(a, b) =k − 1k

η∞(la − 1, lb − 1)− 1k − 1

η∞(la −D − 1, lb −D − 1) +D∑y=1

k − 2k − 1

η∞(la − y, lb − y)

24

+∫ π

0

e−t[1− 2

√µ·(k−1)


k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)


v: if D > 0 and la = D or lb = D

ht(a, b) =k − 1k

η∞(la − 1, lb − 1) +D∑y=1

k − 2k − 1

η∞(la − y, lb − y)

+∫ π

0

e−t[1− 2

√µ·(k−1)


k−1 sin [(la − 1)x])(

sin [(lb + 1)x]− µk−1 sin [(lb − 1)x]

)kπ

2(k−1)2

√(k − 1)la+lb−2(µ+ k − 1)2[1− 4µ(k−1)


where

η∞(0, 0) =2π

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin2 x dx

η∞(0, 1) =4

π√k − 1

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin2 x cosx dx

η∞(0,m) =2

π(√k − 1)m

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin [(m+ 1)x] sinx dx, where m > 1

η∞(1, 1) =8

π(k − 1)

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin2 x cos2 x dx

η∞(1,m) =4

π(√k − 1)m+1

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sinx sin [(m+ 1)x] cosx dx, where m > 1

η∞(m,n) =2

π(√k − 1)m+n

∫ π

0e−t[1+

2√µ·(k−1)

µ+k−1cosx] sin [(m+ 1)x] sin [(n+ 1)x] dx, where m,n > 1

References

[1] C. E. Brasseur, R. E. Grady, and S. Prassidis, Coverings, Laplacians, and Heat Kernels of Directed Graphs,

(2006).

[2] F. R. K. Chung, Spectral Graph Theory, CBMS Regional Conference Series in Mathematics, 92. Published for

the Conference Board of the Mathematical Sciences, Washington, D.C.; by the American Mathematical Society,

Providence, RI, 1997.

[3] F. R. K. Chung, Laplacians and the Cheeger inequality for directed graphs, Ann. of Comb., 9 (2005), 1–19.

[4] F. R. K. Chung and S.- T. Yau, Coverings, heat kernels, and spanning trees, Electron. J. Combin., 6 (1999),

Research Paper 12.

25

[5] R. Lyons, Random walks and the growth of groups, C. R. Acad. Sci. Paris Ser. I Math., 320 (1995), 1361–1366.

E.P.L.: North Carolina State University, Raleigh, NC 27695

E-mail address: [email protected]

S.P.: Department of Mathematics, Canisius College, Buffalo, NY 14208


S.P.S.: University of Pennsylvania, Philadelphia, PA 19104


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