Non-existence of bipartite graphs of diameter at least 4 and defect 2

Non-existence of bipartite graphs of diameter at least 4 and defect 2

Guillermo Pineda-Villavicencio

Received: 3 January 2010 / Accepted: 8 November 2010 / Published online: 20 November 2010

© Springer Science+Business Media, LLC 2010

Abstract The Moore bipartite bound represents an upper bound on the order of a bipartite graph of maximum degreeΔand diameterD. Bipartite graphs of maximum degreeΔ, diameterDand order equal to the Moore bipartite bound are called Moore bipartite graphs. Such bipartite graphs exist only ifD=2,3,4 and 6, and forD= 3,4,6, they have been constructed only for those values ofΔsuch thatΔ−1 is a prime power.

The scarcity of Moore bipartite graphs, together with the applications of such large topologies in the design of interconnection networks, prompted us to investigate what happens when the order of bipartite graphs misses the Moore bipartite bound by a small number of vertices. In this direction the first class of graphs to be studied is naturally the class of bipartite graphs of maximum degreeΔ, diameterD, and two vertices less than the Moore bipartite bound (defect 2), that is, bipartite(Δ, D,−2)- graphs.

ForΔ≥3 bipartite(Δ,2,−2)-graphs are the complete bipartite graphs with par- tite sets of orders Δ andΔ−2. In this paper we consider bipartite (Δ, D,−2)- graphs forΔ≥3 andD≥3. Some necessary conditions for the existence of bipartite (Δ,3,−2)-graphs forΔ≥3 are already known, as well as the non-existence of bi- partite(Δ, D,−2)-graphs withΔ≥3 andD=4,5,6,8. Furthermore, it had been conjectured that bipartite(Δ, D,−2)-graphs forΔ≥3 andD≥4 do not exist. Here, using graph spectra techniques, we completely settle this conjecture by proving the non-existence of bipartite(Δ, D,−2)-graphs for allΔ≥3 and allD≥6.

Keywords Degree/diameter problem·Moore bipartite bound·Moore bipartite graphs, defect·Dickson polynomials of the second kind

G. Pineda-Villavicencio (

Centre for Informatics and Applied Optimization, University of Ballarat, Mount Helen, Victoria 3353, Australia

e-mail:work@guillermo.com.au

1 Introduction

The ever increasing need for the design of interconnection networks has motivated the study of large graphs of given maximum degree and diameter. While the modelling of a network by such graphs does not take into account implementation factors of the network, it does provide an effective means of abstraction to study many relevant network properties [6,9,12,19,20].

Since the features of an interconnection network depend considerably on the par- ticular application, it is then understandable that many interpretations about the “op- timality” of a network coexist. One possible interpretation is presented as follows;

see [9, p. 18], [12, p. 168], and [19, p. 91].

An optimal network contains the maximum possible number of nodes, given a limit on the number of connections attached to a node and a limit on the diameter of the network.

This interpretation is encapsulated by the so-called degree/diameter problem [15], which can be stated as follows.

Degree/diameter problem: Given natural numbersΔ≥2 andD≥2, find the largest possible numberN_Δ,Dof vertices in a graph of maximum degreeΔand diameterD.

It is known that the Moore boundM_Δ,D, defined below, provides an upper bound forNΔ,D. Regular graphs of degreeΔ, diameterDand orderMΔ,Dare called Moore graphs. Non-trivial Moore graphs (that is, those withΔ≥3 andD≥2) exist only for diameter 2, in which case their degree is 2, 3, 7 or possibly 57; see [15].

M_Δ,D=

1+Δ^(Δ−_Δ¹⁾₋^D₂⁻¹ ifΔ >2

2D+1 ifΔ=2

In the design of interconnection networks with bidirectional communication chan- nels, networks subject to further topological restrictions have been also widely con- sidered, for instance, planar networks and bipartite networks [6,19]. Planar graphs are universally used as topologies in the design of printed circuits, such as VLSI cir- cuits [19,20], while bipartite graphs model several interconnection networks, such as the mesh and the hypercube [9,19]. In this paper, henceforth we consider bipartite networks.

It is not difficult to see that the degree/diameter problem can be reformulated to consider only graphs of a given class. For instance, the degree/diameter problem for bipartite graphs [15] can be stated as follows.

Degree/diameter problem for bipartite graphs: Given natural numbersΔ≥2 andD≥2, find the largest possible number N_Δ,D^b of vertices in a bipartite graph of maximum degreeΔand diameterD.

It is straightforward to verify thatN_Δ,D^b is well defined forΔ≥2 andD≥2.

Similar to the general case of the Moore bound, an upper bound forN_Δ,D^b is given by the Moore bipartite bound, which is denoted byM_Δ,D^b and is defined below.

M_Δ,D^b =1+Δ+Δ(Δ−1)+ · · · +Δ(Δ−1)^D−2+(Δ−1)^D−1

=2(1+(Δ−1)+ · · · +(Δ−1)^D−1)

=

2^(Δ−_Δ¹⁾₋^D₂⁻¹ ifΔ >2

2D ifΔ=2

A bipartite graph of maximum degreeΔ, diameterDand orderM_Δ,D^b is called a Moore bipartite graph. Moore bipartite graphs are necessarily regular of degreeΔ, and have turned out to be very rare. They exist only when the diameter is 2, 3, 4 or 6 [15,17], and in the cases ofD=3,4 and 6, they have been produced only for those values ofΔsuch thatΔ−1 is a prime power [3,15].

With the exception ofN_3,5^b =M_3,5^b −6, settled in [13], the other known values of N_Δ,D^b are those for which there is a Moore bipartite graph.

In this context it is natural to investigate what happens when the order of bipartite graphs misses the Moore bipartite bound by a small number of vertices. So we are prompted to consider bipartite graphs of maximum degreeΔ, diameterDand order M_Δ,D^b −, that is, bipartite(Δ, D,−)-graphs, where the parameteris called the defect.

Conditions for under which a bipartite (Δ, D,−)-graph must be regular of degreeΔwere established in [8], and one of them is stated below.

Proposition 1.1 [8] For <1+(Δ−1)+(Δ−1)²+ · · · +(Δ−1)^D−2,Δ≥3 and D≥3, a bipartite(Δ, D,−)-graph is regular.

By Proposition1.1, a bipartite (Δ, D,−1)-graph with Δ≥3 and D≥3 must be regular, implying its two partite sets have the same number of vertices, which is impossible becauseM_Δ,D^b −1 is odd. Thus, there is no bipartite(Δ, D,−1)-graph withΔ≥3 andD≥3.

In this paper we analyze the case of defect 2. Since bipartite(2, D,−2)-graphs must be paths of length 2D−3, with the exception ofD=3 they clearly do not exist forD≥2. In the case ofD=3 the path of length 3 is the only such graph.

WhenΔ≥3, bipartite(Δ,2,−2)-graphs need not be regular; they are the com- plete bipartite graphs with partite sets of ordersΔandΔ−2. In the following, assume Δ≥3 andD≥3.

Concerning bipartite (Δ, D,−2)-graphs, the paper [8] considered bipartite (Δ,3,−2)-graphs, deriving some necessary conditions for their existence, and prov- ing the uniqueness of the two known bipartite(Δ,3,−2)-graphs (both graphs are depicted in Fig.1). Results about bipartite(Δ, D,−2)-graphs forΔ≥3 andD≥4 were first obtained in [7], where it was proved that the eigenvalues other than±Δ of such graphs are the roots of the polynomialsH_D−1(x)±1, withH_D−1(x)being the Dickson polynomial of the second kind with parameterΔ−1 and degreeD−1 [14]. Moreover, [7] provided some necessary conditions for the existence of bipar- tite(Δ, D,−2)-graphs, ruled out the existence of bipartite(Δ, D,−2)-graphs for all

Fig. 1 Two known bipartite (Δ, D,−2)-graphs for Δ ≥ 3 and D ≥3, the unique bipartite (3,3,−2)-graph (a) and the unique bipartite(4,3,−2)-graph (b)

Δ≥3 andD=4,5,6 and 8, and conjectured that bipartite(Δ, D,−2)-graphs for allΔ≥3 and allD≥4 do not exist.

This paper is a follow-up of [7]. Here we prove the aforementioned conjecture by settling the non-existence of bipartite(Δ, D,−2)-graphs for all Δ≥3 and all D≥6. In our proof we are influenced by the reasoning used in the proofs of the non-existence of Moore graphs for Δ≥3 and D≥3 [1], the non-existence of regular graphs of degree Δ≥3, even girth g≥8 and order M_Δ,g/2^b +2 [5], and the non-existence of regular graphs of degree Δ≥3, odd girth g≥5 and order M_Δ,(g−1)/2+1 [2]. We first prove that the multiplicities of the eigenvalues of a hypo- thetical graph satisfy certain inequalities, and based on these inequalities, we derive that certain sums of two eigenvalues must be integer. But, from another point of view, we can prove that those sums must be in the open interval(0,1), and thus arriving at a contradiction.

As a consequence, forΔ≥3 andD≥4 whenever there exists no Moore bipartite graph, it follows thatN_Δ,D^b ≤M_Δ,D^b −4.

It is worth acknowledging that some of the computations in the paper were per- formed with the help of the software Wolfram Mathematica^®[18].

2 Preliminaries

The terminology used in this paper is standard and consistent with that employed in [4].

The set of edges in a graphΓ joining a vertexx inX⊆V (Γ )to a vertexy in Y⊆V (Γ )is denoted byE(X, Y ).

LetΓ be a bipartite graph of diameterD, anduvan edge ofΓ. Also, for 0≤i≤ D−1 define the setsU_i andV_ias follows:

U_i=

z∈V (Γ )|d(u, z)=i, d(v, z)=i+1

V_i=

z∈V (Γ )|d(v, z)=i, d(u, z)=i+1

The decomposition ofΓ into the setsU_i andV_i is called the standard decomposition for a graph of even girth with respect to the edgeuv[5].

SinceΓ is bipartite, the setsU_i andV_i are disjoint for 0≤i≤D−1.

From now on,Γ denotes a bipartite(Δ, D,−2)-graph forΔ≥3 andD≥3,n= M_Δ,D^b −2 denotes its order, andAdenotes its adjacency matrix. By Proposition1.1, Γ is regular of degreeΔ. Considering the girth ofΓ, denoted g(Γ ), it is known that a graph of degreeΔand girth 2Dhas order at leastM_Δ,D^b [4, Proposition 23.1(2)], so g(Γ )≤2D−2. But, if g(Γ )≤2D−4 then the order ofΓ would be at most M_Δ,D^b −2(Δ−1)−2, a contradiction. Therefore, g(Γ )=2D−2.

Proposition 2.1 Every vertexuofΓ is contained in exactly one cycleC_uof length 2D−2.

Proof Consider the standard decomposition for Γ with respect to the edge uv.

Suppose that there are at least two edges joining vertices at U_D−2 to vertices at VD−2, that is,|E(UD−2, VD−2)| ≥2. In such a case,|UD−1| ≤(Δ−1)^D−1−2 and

|V_D−1| ≤(Δ−1)^D−1−2. Therefore,

|V (Γ )| =

D−1 i=0

|U_i| +

D−1 i=0

|V_i|

≤2

1+(Δ−1)+(Δ−1)²+ · · · +(Δ−1)^D−2 +2

(Δ−1)^D−1−2

=2

1+(Δ−1)+(Δ−1)²+ · · · +(Δ−1)^D−1

−4

=M_Δ,D^b −4

which is a contradiction. Consequently, 0≤ |E(U_D−2, V_D−2)| ≤1.

Suppose |E(UD−2, VD−2)| =1. Then, since |UD−1| =(Δ−1)^D−1 −1 and

|V_D−1| =(Δ−1)^D−1−1, we obtain our unique cycleC_u.

Suppose |E(XD−2, YD−2)| =0. Then, since |UD−1| =(Δ−1)^D−1 −1 and

|V_D−1| =(Δ−1)^D−1−1, there must exist exactly one vertexx∈U_D−1such that

|E({x}, UD−2)| =2, and exactly one vertexy∈VD−1such that|E({y}, VD−2)| =2.

This argument shows the existence of the unique cycleC_u, which containsuandx.

We call the unique vertex onC_uat distanceD−1 fromuthe repeat ofu, and we denote it by rep(u). From now on setr=^{g(Γ )}₂ =D−1.

Therefore, we have the following lemma.

Lemma 2.1 [7] If a bipartite (Δ, D,−2)-graph exists, then (2D−2)divides its ordern.

The fact that rep is an automorphism ofΓ was proved in [8]. The permutation matrix associated withrepis called the defect matrix ofΓ and plays an important role in the study of the structure ofΓ (see [7]).

In [7] it was proved that the eigenvalues ofΓ, other than±Δ, satisfy (1).

Theorem 2.1 [7] Ifθ (= ±Δ)is an eigenvalue of the adjacency matrixAofΓ, then

H_r(θ )−ε=0 (1)

whereε= ±1.

The polynomials Hr(x) satisfy Recurrence Equation (2) [7, 17], and as noted in [7], they are the Dickson polynomials of the second kind with parameterΔ−1 and degreer[14].

⎧⎪

⎨

⎪⎩

H₀(x)=1 H1(x)=x

H_i+1(x)=xH_i(x)−(Δ−1)Hi−1(x) fori≥1

(2)

The roots ofH_r(x), obtained in [17], are 2√

Δ−1 cos_r^iπ₊₁ fori=1, . . . , r. This result suggests settingx= −2√

Δ−1 cosϕ, 0< ϕ < π, inH_r(x), from which we obtain

H_r(x)=(−s)^rsin(r+1)ϕ

sinϕ , withs=√

Δ−1 (3)

For the rest of the paper lets=√ Δ−1.

Now we make the change of variableϕ=^iπ_r+⁻₁^α, as suggested in [2,5]. Then, by using (3), equation (1) transforms into the following function inα.

sinα−η_is^−rsin

iπ−α r+1

=0, whereη_i=ε(−1)^r+i+1 (4) We note that the polynomialH_r(x)equals the polynomialE_r+1(x)(E0(x)=0, E1(x)=1 andEi+1(x)=xEi(x)−(Δ−1)Ei−1(x)fori≥1) from [5, p. 4]. There- fore, by substitutingrforr+1 in Lemma 3.3 from [5], we obtain the following result (the bounds forαare derived from the proof of Lemma 3.3).

Lemma 2.2 (Modification of Lemma 3.3 from [5]) For either value ofε, (1) hasr distinct rootsθ₁< θ₂<· · ·< θ_r, withθ_i = −2scosϕ_i (0< ϕ_i< π ). Furthermore, if we setϕ_i=^iπ_r+⁻₁^αi then

0< α_i<min

s^−rϕ_i, s^−r(π−ϕ_i)

ifη_i=1 max

−s^−rϕ_i,−s^−r(π−ϕ_i)

< α_i<0 ifη_i= −1 and consequently

iπ

r+1+s^−r < ϕ_i< iπ

r+1 ifη_i=1 iπ

r+1< ϕ_i< iπ

r+1−s^−r ifη_i= −1

By Theorem2.1and Lemma2.2, it follows that the polynomial ψ (x)=

x²−Δ²

Hr(x)−1

Hr(x)+1 has only simple roots and is a multiple of the minimal polynomial ofΓ.

From (2) and (3) we obtain the following assertion.

Proposition 2.2 The roots of the polynomialψ (x)=(x²−Δ²)(H_r(x)−1)(Hr(x)+ 1)are symmetric with respect to 0, that is,θis a root ofψ (x)if, and only if,−θis a root ofψ (x).

Proof Supposer is even. Then, from (2) and (3) it follows that Hr(−x)=Hr(x), that is,θ is a root ofH_r(x)−εif, and only if, −θ is a root ofH_r(x)−ε, where ε= ±1.

Supposeris odd. Then, from (2) and (3) it follows thatHr(−x)= −Hr(x), that is,θis a root ofH_r(x)−εif, and only if,−θis a root ofH_r(x)+ε, whereε= ±1.

3 Multiplicities of eigenvalues

In this section we compute the multiplicities of the eigenvalues different from±Δ ofΓ. First some lemmas and some definitions are needed.

Lemma 3.1 (Lemma 3.4 from [10]) Letθ be a simple root of the polynomialf (x), and putf_θ(x)= ^{f (x)}_x−θ. If M is a matrix satisfying f (M)=0 then ^tr(f_f^θ(M))

θ(θ ) is the multiplicity ofθ as an eigenvalue ofM, and so is rational, where tr(M)stands for the trace ofM.

LetGbe a Moore bipartite graph of degreeΔand diameterD(and of girth 2D), and letBDbe the(D+1)×(D+1)intersection matrix ofG. Then, the matrixBD

is defined as follows; see [4, p. 182].

BD=

⎛

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎜⎜

⎝

0 1

Δ 0 1 0

Δ−1 0 1

Δ−1 0 1 . .. . .. . .. 1

0 Δ−1 0 Δ

Δ−1 0

⎞

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎟⎟

⎠

LetTΔ denote the infiniteΔ-regular tree, and for a graphGand a vertex uof G, denote byN_G^q(u) the number of closed walks of lengthq starting at u. Also, let (B)_i,j denote the entry(i, j )in the matrixB.

Lemma 3.2 (Corollary of Proposition 21.2 from [4]) LetGbe a Moore bipartite graph of degreeΔand diameter D(and of girth 2D). Then,GhasD+1 distinct eigenvalues, which are the eigenvalues of the matrixBDofG.

Lemma 3.3 [11] In a Δ-regular graph G the number N_G^q(u) of closed walks of lengthq <g(G)starting (and ending) at any vertexuofGequals the numberN_T^q

Δ(u) of such walks starting (and ending) at any vertex uof the infinite Δ-regular tree.

Furthermore, the entry(B^qg(G) 2

)0,0gives this number.

Recall that the number of closed walks of lengthqinΓ is given by tr(A^q).

Observation 3.1 Letube a vertex ofΓ. Since g(Γ )=2r, inΓ the numberN_Γ^q(u) of closed walks of lengthq(q=1, . . . ,2r−1) starting atuis the same as the number N_T^q

Δ(u)of closed walks of lengthq (q=1, . . . ,2r−1) starting atuin the infinite Δ-regular treeT_Δrooted atu. By Lemma3.3,N_T^q

Δ(u)=(B^q_r+1)0,0.

Furthermore, asΓ is bipartite,Γ contains no closed walk of odd length, assertion that is also true inT_Δ.

Thus, it follows that tr(A^q)=n(B^q_r+1)0,0forq=1, . . . ,2r−1,2r+1.

Theorem 3.1 The multiplicity m(θ )ofθ,θ= ±Δ, as an eigenvalue ofΓ is m(θ )=nΔ(Δ−1)Hr−1(θ )

2H_r(θ )(Δ²−θ²) − nθ

εH_r(θ )(Δ²−θ²) (5) whereH_r(x)is the derivative ofH_r(x),ε= ±1 andH_r(θ )−ε=0.

Proof To compute the multiplicity of an eigenvalueθ ofΓ, we follow the method suggested in [2]. Considerψ (x)=(x²−Δ²)(Hr(x)−1)(Hr(x)+1)and setψθ(x)=

ψ (x)

x−θ. Then, asψ (A)=0, by Lemma3.1m(θ )=^tr(ψ_ψ^θ(A))

θ(θ ) .

As deg(Hr(x))=r, deg(ψθ(x))=2r+1, where deg(p(x))stands for the degree of the polynomialp(x).

Let us assume thatψ_θ(x)=x^2r+1+a_2rx^2r+ · · · +a₁x+a₀. Then, by virtue of Proposition2.2and Viète’s formulae, we obtain thata_2r =θ. Furthermore,

tr ψ_θ(A)

=tr A^2r+1

+a_2rtr A^2r

+ · · · +a₁tr(A)+a₀tr(In)

Let nowBi+1, fori≥0, be the intersection matrix representing a Moore bipartite graph of girth 2i+2 and degreeΔ(in particular,B1=_0Δ

Δ0

).

By Observation3.1, we have tr(A^q)=n(B_r^q₊₁)0,0forq=1, . . . ,2r−1,2r+1.

Furthermore, since every vertexuofΓ is contained in exactly one cycle of length 2r,N_Γ^2r(u)=N_T^2r

Δ(u)+2. Thus tr

ψθ(A)

=n

ψθ(Br+1)

0,0+2a2r

=n

ψ_θ(Br+1)

0,0+2θ

As(x²−Δ²)H_r(x)is the minimal polynomial ofBr+1(see [17]), we have ψ_θ(Br+1)= −B²_r+1−Δ²In

Br+1−θIn

SettingL_i+1(x)=^x2_x⁻₋^Δ_θ²(H_i(x)−H_i(θ ))fori=0, . . . , r, we get L_r+1(Br+1)= −H_r(θ )B_r²₊₁−Δ²In

Br+1−θIn

= −εB²_r+1−Δ²In

Br+1−θIn

Therefore,ψ_θ(Br+1)=εL_r+1(Br+1).

By using((x−θ )ψ_θ(x))=((x²−Δ²)(H_r²(x)−1)), where the primeindicates the derivative of the corresponding function, we haveψ_θ(θ )=2εH_r(θ )(θ²−Δ²).

Thus

m(θ )=n(L_r+1(Br+1))_0,0

2H_r(θ )(θ²−Δ²)+ nθ εH_r(θ )(θ²−Δ²)

We are now interested in finding a recurrence relation for the expression (L_i+1(Bi+1))_0,0. In fact

Li+1(x)=x²−Δ² x−θ

Hi(x)−Hi(θ )

=x²−Δ² x−θ

xHi−1(x)−(Δ−1)Hi−2(x)

−

θ Hi−1(θ )−(Δ−1)Hi−2(θ )

=x²−Δ² x−θ

xH_i−1(x)−θ H_i−1(θ )

−(Δ−1)L_i−1(x)

=x²−Δ² x−θ

xHi−1(x)−θ Hi−1(x)

+θ Li(x)−(Δ−1)Li−1(x)

=

x²−Δ²

H_i−1(x)+θ L_i(x)−(Δ−1)Li−1(x) Settingx=Bi, we haveL_i+1(Bi)=θ L_i(Bi)−(Δ−1)Li−1(Bi).

As the Moore bipartite graphs represented byBi+1,Bi,Bi−1have girths 2i+2, 2i and 2i−2, respectively,(B_i+^q ₁)_0,0=(B_i^q)_0,0=(B_i−^q ₁)_0,0forq=0, . . . ,2i−3 and i≥2 (this can be deduced by reasoning as in Observation3.1).

Since deg(Li+1(x))=i+1,(L_i−1(Bi))_0,0=(Li−1(Bi−1))_0,0and(L_i+1(Bi+1))_0,0

=(L_i+1(Bi))_0,0.

Thus, (L_i+1(Bi+1)_0,0 =θ (L_i(Bi))_0,0 −(Δ −1)(Li−1(Bi−1))_0,0, and setting l_i+1=(L_i+1(Bi+1))0,0, we have the desired recurrence relation

l₀=l₁=0 l₂=Δ−Δ²

l_i+1=θ l_i−(Δ−1)li−1 fori≥2

Hence, we obtain thatl_i+1=(Δ−Δ²)H_i−1(θ )fori≥1, and the theorem follows, that is,

m(θ )=nΔ(Δ−1)Hr−1(θ )

2H_r(θ )(Δ²−θ²) − nθ

εH_r(θ )(Δ²−θ²) 3.1 Multiplicities as functions of cosϕ

Next we express m(θ ) withθ= −2scosϕ as a function of cosϕ. But before, we define the following functionsf (z)andg(z).

f (z)=4s²(1−z²) Δ²−4s²z² g(z)=Δ(Δ−1)(

1−s^−2r(1−z²)+s^−rz)−4s^2−rz (r+1)

1−s^−2r(1−z²)+s^−rz These functions will receive some attention from now on.

Lemma 3.4 For either value ofε, if we setθi= −2scosϕi fori=1, . . . , r then m(θi)= n

4s²f (cosϕi)g(ηicosϕi) wheref andgare defined as above.

Proof By (3), we have dH_r(θ )

dθ dθ dϕ = d

dϕ

(−s)^rsin(r+1)ϕ sinϕ

=(−s)^r sinϕ

(r+1)cos(r+1)ϕ−cotϕsin(r+1)ϕ

We now evaluate ^dH_dθ^{r(θ )d}_dϕ^θ atϕ_i=^iπ_r+⁻₁^αi fori=1, . . . , r. dHr(θ )

dθ dθ dϕ

ϕ_i=iπ−αi

r+1

=(−s)^r sinϕ_i

(r+1)cos(iπ−α_i)−cotϕ_isin(iπ−α_i)

=(−s)^r(−1)ⁱ sinϕ_i

(r+1)cosα_i+cotϕ_isinα_i

Therefore

H_r(θ_i)=(−s)^r(−1)ⁱ 2ssin²ϕ_i

(r+1)cosα_i+cotϕ_isinα_i

and by (4), we have

H_r(θi)=(−s)^r(−1)ⁱ 2ssin²ϕi

(r+1)cosαi+ηis^−rcosϕi

(6)

SubstitutingH_r−1(θ_i)=(−s)^r−1(−1)^{i+1 sin(ϕ}_sinⁱ_ϕ⁺_i^αi) andH_r(θ_i)in (5), we obtain m(θi)= nsinϕ_i

(Δ²−θ_i²)

Δ(Δ−1)sin(ϕ_i+α_i)+2θiη_is^1−rsinϕ_i (r+1)cosα_i+η_is^−rcosϕ_i Since sin(ϕi+α_i)=sinϕ_i(cosα_i+η_is^−rcosϕ_i)(by (4)), we have

m(θi)= nsin²ϕ_i (Δ²−θ_i²)

Δ(Δ−1)(cosα_i+η_is^−rcosϕ_i)+2θiη_is^1−r (r+1)cosαi+ηis^−rcosϕi

By (4) and Lemma2.2, as Δ≥3 and r≥2, it follows that ifη_i =1 then 0<

αi <^π₂ and that ifηi= −1 then−^π₂ < αi<0. Therefore, cosαi >0, and cosαi= 1−s^−2r(1−cos²ϕ_i)by (4).

Consequently m(θi)= n

4s²

4s²(1−cos²ϕ_i) (Δ²−4s²cos²ϕ_i)

×Δ(Δ−1)(

1−s^−2r(1−cos²ϕi)+ηis^−rcosϕi)−4ηis^2−rcosϕi

(r+1)

1−s^−2r(1−cos²ϕ_i)+η_is^−rcosϕ_i and takingf (z)=^4s_Δ2²−⁽¹4s^−2z2z²⁾ (as suggested in [2,5]) and

g(z)=Δ(Δ−1)(

1−s^−2r(1−z²)+s^−rz)−4s^2−rz (r+1)

1−s^−2r(1−z²)+s^−rz ,

we obtain the desired formula m(θi)= n

4s²f (cosϕ_i)g(η_icosϕ_i) Corollary 3.1 The polynomialψ (x)=(x²−Δ²)(H_r(x)−1)(Hr(x)+1)is the min- imal polynomial ofΓ.

Proof It is not difficult to see that for|z|<1, we havef (z) >0. To see thatg(z) >0 for|z|<1, we multiply both the numerator and the denominator ofg(z)bys^r. Then,

g(z)=Δ(Δ−1)(s^r

1−s^−2r(1−z²)+z)−4s²z s^r(r+1)

1−s^−2r(1−z²)+z

=Δ(Δ−1)√

s^2r−1+z²+(Δ(Δ−1)−4s²)z (r+1)√

s^2r−1+z²+z

In the last expression ofg(z)we can readily see that, for|z|<1,g(z) >0.

Therefore, settingθ_i= −2scosϕ_iwithi=1, . . . , rfor either value ofε, it follows that m(θi) >0. Thus,θ_i is an eigenvalue ofΓ.

4 Properties of the functionsf andg

In the previous section we expressed the multiplicity of an eigenvalue θ of Γ by means of certain functionsf andg. In this section we obtain several properties and relationships between these functions.

Lemma 4.1 (See Lemma 3.5 from [5]) ForΔ≥3 and|z|<1 the functionf (z)is even and concave down.

Lemma 4.2 ForΔ≥3,r≥5 and|z|<1, the monotonicity ofg(z)behaves as fol- lows.

(i) ForΔ=3,4 the functiong(z)is monotonic decreasing; and (ii) forΔ≥5g(z)is monotonic increasing

Proof To prove thatg(z)is monotonic increasing (decreasing) for|z|<1, it suffices to show thatg(z)is positive (negative) along the interval.

g(z)= − s^−r(−1+s^2r)(Δ−1)(4(1+r)−rΔ) 1+s^−2r(−1+z²)(z+(1+r)s^r

1+s^−2r(−1+z²))² From the expression ofg(z), we can verify thatg(z)is negative forΔ=3,4 and

r≥5, while it is positive forΔ≥5 andr≥5.

Lemma 4.3 If cosϕ2<−cosϕ_r andr≥5, the following relationships between the functionsf andghold.

(i) IfΔ=3,4 then

f (cosϕ₂)

f (cosϕ_r)>g(cosπ ) g(cos 0)

with the exception of the pairs(Δ, r)=(3,5),(3,6),(3,7), and(3,8); and (ii) ifΔ≥5 then

f (cosϕ2)

f (cosϕ_r)> g(cos 0) g(cosπ )

Proof Sinceϕ₂∈(0,^π₂)andϕ_r∈(^π₂, π )by Lemma2.2,τ =π−ϕ_r ∈(0,^π₂)and τ < ϕ2(since cosϕ2<−cosϕr=cosτ).

From the expression off (cosϕ), we have f (cosϕ₂)

f (cosϕr)=f (cosϕ₂)

f (cosτ ) =1+cos 2τ−cos 2ϕ2

2 sin²τ

Δ²−4s² Δ²−4s²cos²ϕ2

By the mean value theorem, we have^{cos 2τ}_2τ⁻₋^{cos 2ϕ}_2ϕ ²

2 = −sin 2γfor someγ∈(τ, ϕ₂).

Consequently

cos 2τ−cos 2ϕ2

2 sin²τ =(ϕ2−τ )sin 2γ

sin²τ =2(ϕ2−τ )sinγcosγ sin²τ

Also, since 0< τ < γ < ϕ₂<^π₂, we have sinγ >sinτ and cosγ >cosϕ₂, and thus cos 2τ−cos 2ϕ2

2 sin²τ >2(ϕ2−τ )cosϕ₂ sinτ

By Lemma2.2, fori=1, . . . , r ifηi =1 then _r+1^iπ_+s−r < ϕi <_r^iπ₊₁, and _r^iπ₊₁<

ϕ_i< ^iπ

r+1−s^−r otherwise. Therefore, in any case iπ

r+1+s^−r < ϕ_i< iπ r+1−s^−r and consequently

ϕ2−τ=ϕ2+ϕ_r−π > 2π

r+1+s^−r + rπ

r+1+s^−r −π= π(−1+s^r) 1+(1+r)s^r Furthermore, cosϕ2>cos_r+1^2π_−s−r >cos^2π₅ = ^√5₄⁻¹ >¹₄ (sincer≥5), and since

|sinx| ≤ |x|for allx∈R, we have

sinτ ≤τ =π−ϕ_r< π− rπ

r+1+s^−r = π(1+s^r) 1+(1+r)s^r Therefore

f (cosϕ2)

f (cosτ ) >1+ 2π(−1+s^r) (1+(1+r)s^r)

1+(1+r)s^r 4π(1+s^r)

4(Δ²−4s²) 4Δ²−s²

=1+2(−1+s^r) 1+s^r

Δ²−4s²

4Δ²−s² (7)

Proof of Claim (i) From the expression ofg(z), we obtain g(−1)

g(1) =1+ 2s^r(4(r+1)−rΔ)

(−1+s^r(r+1))(Δ(1+s^r)−4) (8) Therefore, considering (7) and (8), it suffices to show that, forΔ=3,4 andr≥5, with the exceptions of the pairs(Δ, r)=(3,5),(3,6),(3,7), and(3,8), the following inequality holds:

−1+s^r 1+s^r

Δ²−4s²

4Δ²−s²> s^r(4(r+1)−rΔ) (−1+s^r(r+1))(Δ(1+s^r)−4) or equivalently that

h(r)=−1+s^r 1+s^r

Δ²−4s²

4Δ²−s²− s^r(4(r+1)−rΔ)

(−1+s^r(r+1))(Δ(1+s^r)−4)>0 Indeed, usings=√

Δ−1, forΔ=3 andr≥9 we have

h(r)=−1−(131+33r)2^r2 −143·2^r−19r2^1+r+(3+3r)2^3r2 34(1+2^r/2)(−1+3·2^r2)(−1+(1+r)2^r2) >0

(As the numerator ofh(r) is clearly monotonic increasing, with value−248321+ 238912√

2>0 atr=9.) While forΔ=4 andr≥4 we have h(r)=−57−4r3^r2 −23·3^1+r2 +(4+4r)3^r

61(1+3^r/2)(−1+(1+r)3^r2) >0

(As the numerator ofh(r)is clearly monotonic increasing, with value 798 atr=4.)

Thus we obtain the claim.

Proof of Claim (ii) It is proved analogously to Claim (i). A complete proof of this

claim can be found in [16, p. 112].

This completes the proof of the lemma.

5 Main result

In this section we prove the main result of this paper (Theorem5.1).

Theorem 5.1 Bipartite(Δ, D,−2)-graphs forΔ≥3 andD≥6 do not exist.

To prove the theorem we prepare two more lemmas, and some definitions.

Henceforth, letλ1< λ2<· · ·< λ_r be the roots ofH_r(x)−1=0, and letρ1<

ρ2<· · ·< ρ_r be the roots ofH_r(x)+1=0.

Lemma 5.1 Letλ1, . . . , λ_r andρ1, . . . , ρ_r be defined as above. Then the following assertions hold forΔ≥3 andr≥5.

(i) Ifr is even then m(λi)=m(λ1+r−i)and m(ρi)=m(ρ1+r−i), whereas if r is odd then m(λi)=m(ρ1+r−i), for 1≤i≤r; and

(ii) Ifr is even then m(ρ1) <m(ρi), whereas ifr is odd then m(λ1) <m(λi), for i=2, . . . , r−1, and any pair(Δ, r)=(3,5), (3,6), (3,7), (3,8); and

(iii) m(λr) <m(λi)fori=2, . . . , r−1 Proof Next we prove each claim in order.

Proof of Claim (i) Ifr is even, it follows thatHr(−x)=Hr(x)(see (3)) and that λ_i+λ_1+r−i=ρ_i+ρ_1+r−i=0, and thus, by checking (5), m(λi)=m(λ1+r−i)and m(ρi)=m(ρ1+r−i)for 1≤i≤r. If insteadris odd,H_r(−x)= −H_r(x); therefore, λ_i+ρ1+r−i=0 and m(λi)=m(ρ1+r−i)for 1≤i≤r.

Proof of Claim (ii) In the proof of this claim we consider the trigonometric form of the multiplicity of an eigenvalue ofΓ (Lemma3.4).

Assumeris even,ε= −1, andρ_i= −2scosϕ_ifori=1, . . . , r. Thenη_i=(−1)ⁱ. By (i),−cosϕ1=cosϕr. Then, since −cosϕ1=cosϕr <cosϕi <cosϕ1 (for i∈ {2, . . . , r−1}), by Lemma4.1we have

f (cosϕ₁) < f (cosϕ_i) fori=2, . . . , r−1

First supposeΔ≥5 andr≥5. Since cosϕ_i<|cosϕ₁|, we obtain thatg(−cosϕ₁)

< g(±cosϕ_i)fori=2, . . . , r−1 (by Lemma4.2(ii)), and thus, m(ρ1) <m(ρi).

Suppose Δ=3,4 and r ≥ 5. In this case g is monotonic decreasing (by Lemma4.2(i)), so we cannot use the same argument as before. Here we also want to prove

f (cosϕ1)g(−cosϕ1) < f (cosϕi)g(ηicosϕi) fori=2, . . . , r−1 Since cos 0>cosϕ_i>cosπ, we haveg(cos 0) < g(cosϕ_i) < g(cosπ ). Then

f (cosϕ1)g(−cosϕ1) < f (cosϕ1)g(cosπ )andf (cosϕ_i)g(cos 0)

< f (cosϕi)g(ηicosϕi)

Consequently, it suffices to prove that, for any pair(Δ, r) other than(3,5), (3,6), (3,7), (3,8),

f (cosϕi)

f (cosϕ1)>g(cosπ ) g(cos 0)

Besides, sincef (cosϕ2)≤f (cosϕi)fori=2, . . . , r−1, we can equivalently prove that

f (cosϕ₂)

f (cosϕ_r)>g(cosπ ) g(cos 0)

and as cosϕ2<cosϕ1= −cosϕ_r, such an inequality follows from Lemma4.3(i).

Therefore, m(ρ1) <m(ρi).

Now assume thatr is odd,ε=1, λ_i= −2scosϕ_i, and ρ_i = −2scosσ_i for i= 1, . . . , r. Thenηi=(−1)ⁱ.

Since λ1< λi, ηicosϕi >−cosϕ1 for ηi = −1. Moreover, since r is odd, we obtain thatρi+λ1+r−i=0 by virtue of (i). We next prove thatηicosϕi>−cosϕ1

forηi=1. Since cosϕ1>0 by Lemma2.2, we only consider the case of cosϕi <0.

By Lemma2.2 iπ

r+1+s^−r < ϕ_i< iπ

r+1 and iπ

r+1 < σ_i< iπ r+1−s^−r

As a consequence,ϕi< σi, and sinceϕi, σi ∈(0, π ), it follows that cosϕi>cosσi. Sinceρr=2scosϕ1≥ρi= −2scosσi>−2scosϕi, cosϕi>−cosϕ1. That is

−cosϕ₁<cosϕ_i<cosϕ₁ fori=2, . . . , r Then, by Lemma4.1,f (cosϕ₁) < f (cosϕ_i)fori=2, . . . , r.

First supposeΔ≥5 andr≥5. By Lemma4.2(ii),g(−cosϕ1) < g(±cosϕ_i)for i=2, . . . , r−1, and thus, m(λ1) <m(λi).

SupposeΔ=3,4 andr≥5. Since cos 0>cosϕi >cosπ, by Lemma4.2(i) we haveg(cos 0) < g(cosϕi) < g(cosπ ). Therefore, as above, we only need to show that

f (cosϕi)

f (cosϕ1)>g(cosπ ) g(cos 0)