GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC

NUMBER

CHEUK TO TSUI

Abstract. This paper investigates graphs that have large girth and largechromatic number. We first give a construction of a family of graphs thatdo not contain cycles of length 3, but have unbounded chromatic number.Next, we prove the existence of graphs that have large girth and large chro-matic number. With existence proven, we give an explicit construction of suchgraphs.

Contents

1. Introduction 12. Triangle-Free Graphs with Large Chromatic Number 23. Existence of Graphs with Large Girth and Large Chromatic Number 34. Construction of Graphs with Large Girth and Large Chromatic Number 54.1. Lower Bound on Girth of X

p

84.2. Lower Bound on Chromatic Number of Xp,q 11Acknowledgments 13References 14

1. Introduction

Finding a lower bound for the chromatic number of a given graph is, in general,di�cult to do. There are few techniques that show that a certain number of colorsare not enough to color a graph. Because of this, it is interesting to consider howwe may construct graphs with large chromatic number. One way to do so is toinclude many cycles; since every even cycle requires at least 2 colors and everyodd cycle requires at least 3, it is easy to construct graphs with large chromaticnumber as long as cycles are permitted (take the complete graphs, for instance).But the matter becomes more complicated when we do not permit cycles below agiven length, and want to construct graphs with large girth and large chromaticnumber. In the remainder of Section 1, we define the terminology that allows us tobegin addressing this problem. In section 2, we address the simplified problem ofshowing that there are triangle-free graphs with arbitrarily large chromatic number.In section 3, we show more generally that there exist graphs that have girth � land chromatic number � k for any k and l. In section 4, we explicitly constructgraphs with large girth and chromatic number, show the basic method of finding alower bound for their girth, and find a lower bound for their chromatic number.

1

2 CHEUK TO TSUI

Definition 1.1. A graph G = (V,E) consists of a vertex set V and an edge setE, where elements of the edge set E are unordered pairs {v

i

, vj

}, with vi

6= vj

andvi

, vj

2 V .

Definition 1.2. Two vertices vi

and vj

are adjacent if there is an edge betweenthem, that is, {v

i

, vj

} 2 E. We can also write vi

⇠ vj

.

Definition 1.3. A cycle is a sequence of vertices that begins and ends at the samevertex, such that successive vertices are adjacent and no vertex except the firstrepeats. The number of distinct vertices in the sequence is called the length of thecycle. A cycle of length k is denoted C

k

.

Definition 1.4. The girth of a graph is the length of the shortest cycle containedin it. If a graph contains no cycles, its girth is defined to be 1.

Definition 1.5. A graph G is triangle-free if it does not contain a cycle of length3.

Definition 1.6. A set of vertices S is independent if no two vertices in S areadjacent.

Definition 1.7. The independence number ↵(G) of a graph G is the maximumsize of an independent vertex set.

Definition 1.8. A (proper) k-coloring of a graph is a function f : V ! {1, 2, ..., k}such that if v

i

and vj

are adjacent, then f(vi

) 6= f(vj

).

Definition 1.9. The chromatic number �(G) of a graph G is the minimum valueof k such that a k-coloring exists.

2. Triangle-Free Graphs with Large Chromatic Number

Before addressing the general question of constructing graphs with large girthand chromatic number, we first consider the simpler problem of constructing triangle-free graphs with large chromatic number.

Theorem 2.1. For any positive integer k, there exists a triangle-free graph withchromatic number k.

Proof. For k = 1, a single point will work. For k = 2, two adjacent vertices willwork, and for k = 3, C5 will work. For chromatic numbers larger than 3, wegive a construction that, when applied to a triangle-free graph, produces a newtriangle-free graph that increases the chromatic number by 1.

Construction 2.2 (Mycielski. See [1]). Given a graph G, label its vertices vi

, i 2{1, 2, ..., n}. Create n new vertices u

i

and an additional vertex w, and draw an edgebetween each u

i

and w. Then, if {vi

, vj

} is an edge, draw the two edges {ui

, vj

}and {v

i

, uj

}. Call this new graph G0.

First, we show that if G is triangle-free, then G0 is triangle-free. Suppose thereexists a C3 in G0. Vertex w cannot be a vertex of a C3, because it is adjacent onlyto all the u

i

, which are not pairwise adjacent. There are also no C3’s containing twoof the u

i

, since no two ui

are adjacent. Therefore, the C3 must contain two adjacentvi

and one of the ui

. Let the vertices be vj

, vk

and ul

. They must be pairwiseadjacent, meaning that {v

j

, ul

} and {vk

, ul

} are edges of G0. By the construction,this would be the case only if {v

j

, vl

} and {vk

, vl

} were edges of G. But this would

GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 3

mean that vj

, vk

, vl

, vj

would be a C3 in G, which contradicts the fact that G istriangle-free. Thus there exists no C3 in G0.

Now, we show that �(G0) = �(G) + 1. Let �(G) = k. This means that the vi

must be colored by at least k colors. We claim that in any coloring of G0, the ui

must also be colored by at least k colors. Suppose that the ui

can be colored byat most k � 1 colors. so at least one of the v

i

are colored by color k, while noneof the u

i

are. Observe that for all i, vi

and ui

are adjacent to exactly the samevj

’s in G. Take the vi

that are colored by color k, and change them to the colorof the corresponding u

i

. This new coloring of the vi

would still be valid, becauseeach v

i

has the same neighbours as the corresponding ui

, meaning that none ofthe v

j

’s adjacent to each vi

have the same color as the corresponding ui

. But thisnew coloring (of the v

i

) uses at most k � 1 colors, contradicting the fact that Gmust be colored by at least k colors. Thus, the u

i

must be colored by at least kcolors as well. Since w is adjacent to all the u

i

, it must be colored by a (k + 1)stcolor. Therefore, �(G0) � �(G) + 1. Furthermore, G0 can be colored by �(G) + 1colors: First color the v

i

with k colors, then color each ui

the same color as thecorresponding v

i

. Then color w with a (k + 1)st color. So �(G0) = �(G) + 1.Therefore, to obtain a triangle-free graph with chromatic number k > 3, we

iteratively apply the construction to C5 k � 3 times.⇤

3. Existence of Graphs with Large Girth and Large Chromatic

Number

Theorem 2.1 shows that the chromatic number of a graph can be made arbitrarilylarge, even if it does not contain cycles of length 3. Now we consider the moregeneral question of whether it is possible to avoid cycles below a given length, yetstill have an arbitrarily large chromatic number. As preparation, we present oneprobabilistic result and define some more notation.

Lemma 3.1 (Markov’s Inequality). If X is a non-negative discrete random variableand a > 0, then

P [X � a] E[X]

a.

Proof. We have

E[X] =X

x�0

xP (x)

=X

x�a

xP (x) +X

x<a

xP (x)

�X

x�a

xP (x)

� aX

x�a

P (x)

= aP [X � a].

⇤

4 CHEUK TO TSUI

Definition 3.2 ([2, p. 38]). We write G ⇠ G(n, p) if G is a random graph on n ver-tices chosen by picking each pair of vertices as an edge randomly and independentlywith probability p.

Definition 3.3 (little-oh notation). We write an

= o(bn

) if

limn!1

an

bn

= 0.

Remark 3.4. an

= o(1) means that limn!1 a

n

= 0.

Theorem 3.5 (Erdos 1959. See [2, pp. 38-39]). For all k, l there exists a graph Gwith girth(G) > l and �(G) > k.

Proof. Let ✓ < 1/l, and let G ⇠ G(n, p) with p = n✓�1. Let X be the number ofcycles of length at most l in G, and let X

i

be the number of cycles of size exactlyi. By linearity of expectation, we have

E[X] =lX

i=3

E[Xi

].

To determine E[X], we count the possible number of cycles of length i. There are�n

i

�ways to choose the vertices of the cycle, and (i�1)!

2 ways to order them (we canstart with any vertex). We must divide by 2, because the reverse of any orderinggives the same cycle. So the total number of possible cycles is

✓n

i

◆(i� 1)!

2=

(n)i

2i,

where (n)i

= (n)(n� 1)...(n� i+ 1) (falling factorial).The probability of each of these cycles occuring is pi, so by linearity of expecta-

tion, E[Xi

] = (n)i

2i pi. This, along with the bound (n)i

ni, gives us

E[X] =lX

i=3

(n)i

2ipi

lX

i=3

ni

2ini(✓�1) =

lX

i=3

n✓i

2i n✓l

lX

i=3

1

2i.

Now, since ✓l < 1 by assumption, (n✓l

Pl

i=312i )/n = n✓l�1

Pl

i=312i goes to 0 as n

goes to 1. So E[X] = o(n). Using this fact and Markov’s Inequality, we get

P [X � n/2] E[X]/(n/2) = o(1).

We now use the independence number ↵(G) to estimate the chromatic number�(G). Let us compute P [↵(G) � x]. Any set of x vertices are independent if all

�x

2

�

edges between them are not drawn, which occurs with probability (1�p)(x

2). Thus,the probability P [↵(G) � x] that at least one set of x vertices is independent is at

most�n

x

�(1�p)(

x

2) by the union bound. By the bounds�n

x

� nx and (1�p)x e�px,

we obtain

P [↵(G) � x] ✓n

x

◆(1� p)(

x

2) [ne�p(x�1)/2]x

.

GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 5

We want to set x such that P [↵(G) � x] goes to 0 as n gets large. To do this, wecan require that ne�p(x�1)/2 < 1p

n

. This is equivalent to

n32 < ep(x�1)/2

lnn <p(x� 1)

33

plnn < x� 1

1 +3

plnn < x.

As n gets large, the constant 1 becomes negligible, so we can simply set x = d 3p

lnneto ensure P [↵(G) � x] = o(1). We can verify this: We have, using d 3

p

lnne <3p

lnn+ 1,

P [↵(G) � x] [ne�p(x�1)/2]x

< [ne�p( 3p

lnn)/2]3p

lnn+1

= [n� 12 ]

3n

✓�1 lnn+1

= n� 32n

1�✓ lnn+ 12

= o(1).

Now let n be so large that P [X � n/2] < 0.5 and P [↵(G) � x] < 0.5, which ispossible since both are o(1). Then there exists a graph G with less than n/2 cyclesof length at most l and with ↵(G) < 3n1�✓ lnn. Removing a vertex from each cycleof length at most l leaves a graph G⇤ with at least n/2 vertices and girth greaterthan l. We also have ↵(G⇤) ↵(G), since every independent set of G⇤ is also oneof G. This allows us to give a lower bound for the chromatic number, using thefollowing lemma.

Lemma 3.6. For any graph G, �(G)↵(G) � n.

Proof. Let �(G) = k and ↵(G) = a. Since all the vertices colored by the same colorform an independent set, there can be at most a vertices of each color. G can bek-colored, and k colors can color at most ka vertices, so ka � n. ⇤

Denote the number of vertices of G⇤ by |G⇤|. We have, by the lemma,

�(G⇤) � |G⇤|↵(G⇤)

� n/2

3n1�✓ ln(n)=

n✓

6 lnn.

Since n

✓

6 lnn

is unbounded, we can make �(G⇤) larger than k by choosing a largeenough n. Thus G⇤ has the desired properties.

⇤

4. Construction of Graphs with Large Girth and Large Chromatic

Number

The probabilistic argument given in the proof of Theorem 3.5 proves the existenceof graphs with large girth and large chromatic number. The next question to askis how such graphs can be explicitly constructed. We address this question in thissection by constructing graphs called Ramanujan graphs. We then give the basicidea behind obtaining a lower bound for the girth of these graphs, and obtain a

6 CHEUK TO TSUI

lower bound for the chromatic number. First, we state some definitions from grouptheory.

Definition 4.1. A group (G, ·) consists of a set of elements G and an operation ·with the following properties:

Closure: For all a, b in G, ab is in G.Associativity: For all a, b, c in G, a(bc) = (ab)c.Identity: There exists an element 1

G

in G such that for a in G, 1G

a = a1G

= a.Inverse: For all a in G, there exists an element a�1 in G such that

a · a�1 = a�1 · a = 1G

Definition 4.2. A subset H of G is a subgroup of G, denoted H G, if H formsa group under the same operation.

Definition 4.3. A subgroup N of G is a normal subgroup, denoted N C G, if forall a in G,

a�1Na = N.

Definition 4.4. The general linear group of degree n GL(n,F) is the group of n⇥nnonsingular matrices over a field F, with the matrix multiplication operation.

Definition 4.5. The special linear group of degree n SL(n,F) is the group of n⇥nmatrices with determinant 1 over a field F, with the matrix multiplication operation.

Definition 4.6. Given a subgroup H G, and x 2 G, definexH = {xh : h 2 H} and Hx = {hx : h 2 H}. A subset of G that is in the form xHfor some x 2 G is a left coset of H. A subset of G in the form Hx for some x 2 Gis a right coset of H.

Definition 4.7. Given a group G and N C G, we denote by G/N the set of all(right) cosets of N . This set is called the quotient set. Define an operation ⇤ by

Na ⇤Nb = Nab.

Then (G/N, ⇤) is called the quotient group of N in G.

Example 4.8 (Modulo d residue classes). Examples of quotient groups are themodulo d residue classes, denoted Z/dZ, where dZ = {dk : k 2 Z} is a subgroup ofZ.

Definition 4.9. The center of a group G, denoted Z(G), is defined as Z(G) ={a 2 G : 8b 2 G, ab = ba}.

The following result will be used in the construction.

Proposition 4.10 (Center of the general linear group). The center of GL(n,F) isthe set of scalar matrices except for the zero matrix, i.e. {�I : � 2 F\{0}}.

The next result is necessary for the following two definitions.

Proposition 4.11. The center of a group is a normal subgroup.

Definition 4.12. The projective general linear group of degree n is defined asPGL(n,F) = GL(n,F)/Z(GL(n,F)).

Definition 4.13. The projective special linear group of degree n is defined asPSL(n,F) = SL(n,F)/Z(SL(n,F)).

GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 7

The following definition provides a way to generate a graph from a group and asubset.

Definition 4.14. The Cayley graph of a group G with respect to a subset S,denoted �(G,S), is defined as follows:

V (�) = GFor all g 2 G, g ⇠ gh if and only if h 2 S.

We state the following theorem without proof.

Theorem 4.15 (Jacobi’s Four Squares Theorem. See [4, p. 262]). The number ofways to express a positive integer n as a sum of 4 squares is

r4(n) = 8X

d|n4-d

d.

Let X = Xn,k

be a k-regular graph on n vertices. Let AX

be its adjacencymatrix with eigenvalues �1 � �2 � . . . � �

n

. We know that |�i

| k for 1 i n.Let �(X) denote the absolute value of the largest eigenvalue (in absolute value) ofA

X

that is not equal to ± k.

Definition 4.16. Xn,k

is called a Ramanujan graph if

�(X) 2pk � 1.

We now construct Ramanujan graphs that will be shown to have large girth andchromatic number.

Construction 4.17 ([4]). Choose two prime numbers p, q that are congruent to1 mod 4, and choose an integer i such that i2 ⌘ �1 (mod q). By Jacobi’s FourSquares Theorem, the number of solutions ↵ = (a0, a1, a2, a3) to the equation

a02 + a1

2 + a22 + a3

2 = p

is 8(p+1). It can be proven that p+1 of the solutions satisfy the conditions that a0is positive and odd, and a1, a2, a3 are even. For each of these solutions ↵, associatethe matrix

↵ =

a0 + ia1 a2 + ia3�a2 + ia3 a0 � ia1

�.

We can then form the Cayley graph of PGL(2,Z/qZ) with respect to the set ofthese p + 1 matrices. The next two lemmas give the number of vertices that thisCayley graph has.

Lemma 4.18. |GL(2,Z/qZ)| = (q2 � q)(q2 � 1)

Proof. The size of GL(2,Z/qZ) is the number of 2 by 2 nonsingular matrices witheach entry an integer between 0 to q� 1 inclusive. There are q2 � 1 ways to choosethe first column, since a column with all 0s cannot occur in a nonsingular matrix.The second column can be anything except for the multiples of the first column, ofwhich there are q, so the second column can be chosen in q2 � q ways. Thus, thereare a total of (q2 � q)(q2 � 1) matrices in GL(2,Z/qZ). ⇤

Lemma 4.19. |PGL(2,Z/qZ)| = q(q2 � 1)

8 CHEUK TO TSUI

Proof. The elements of PGL(2,Z/qZ) are the cosets of Z(G) in GL(2,Z/qZ), whichare the subsets in the form {�A : � 2 {1, 2, ..., q � 1}}, where A 2 GL(2,Z/qZ).Thus, each coset contains q � 1 elements of GL(2,Z/qZ), and there are

|GL(2,Z/qZ)|/(q � 1) = q(q2 � 1)

cosets in PGL(2,Z/qZ) by the previous lemma. ⇤

It can be similarly shown that |SL(2,Z/qZ)| = q(q2 � 1) and |PSL(2,Z/qZ)| =q(q2 � 1)/2.

From the two above lemmas, we can see that the Cayley graph has n = q(q2�1)vertices. It can also be shown that the graph is (p+1)-regular. If p is not a perfectsquare modulo q, then this graph can be shown to be connected. However, considerthe case when p is a (non-zero) perfect square modulo q. The determinants of thegenerators ↵ are

(a0 + ia1)(a0 � ia1)� (�a2 + ia3)(a2 + ia3) = a02 + a1

2 + a22 + a3

2 = p,

which is a perfect square by assumption. This means that each ↵ is a scalarmultiple of a matrix in PSL(2,Z/qZ), i.e. a matrix with determinant 1. So allthe generators lie in the subgroup PSL(2,Z/qZ). But then the component of theCayley graph of PGL(2,Z/qZ) with respect to the set of generators that containsthe identity matrix will only include vertices corresponding to matrices that arein PSL(2,Z/qZ). Thus, if p is a perfect square mod q, the Cayley graph is notconnected.

Because of this, if p is not a perfect square mod q, we defineXp,q to be the Cayleygraph of PGL(2,Z/qZ) with respect to the p + 1 generators. If p is a (non-zero)perfect square mod q, we define Xp,q to be the Cayley graph of PSL(2,Z/qZ) withrespect to the generators, which can be shown to be connected, and in addition,non-bipartite. It can be shown (see [4]) that the Xp,q are Ramanujan graphs (i.e.they satisfy Definition 4.16).

The next step is to find a lower bound on the girth of the Ramanujan graphsXp,q. The result is given below.

Theorem 4.20 ([4]). If p is not a perfect square mod q, then girth(Xp,q) � 4logp

q�log

p

4. If p is a non-zero perfect square mod q, then girth(Xp,q) � 2logp

q.

Since logp

q is can be made arbitrarily large, the girth of Xp,q can be made aslarge as desired.

The proof of Theorem 4.20 is omitted, but the basic idea behind obtaining alower bound is as follows: A cycle is formed by two di↵erent walks from one vertexto another. We can represent a walk as the multiplication of a sequence of matrices.Using the matrix norm, we can find how long two walks must at least be for themto start and end at the same vertex (thus forming a cycle). This will then give usa bound on the girth. We illustrate this idea by estimating the girth of a simplerCayley graph X

p

that is defined below.

4.1. Lower Bound on Girth of Xp

. [5]If H is a subset of a group G, define a word W over H to be a finite sequence

f1, f2, ..., fn such that for all 1 i n, either fi

or fi

�1 is in H. Define a word Wto be reduced if f

i+1 6= fi

�1 for all 1 i n� 1.

GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 9

Definition 4.21. A map ' : G ! L for groups G and L, is called a homomorphismif for all x, y 2 G, we have

'(xy) = '(x)'(y).

Consider the matrices A =

1 20 1

�and B =

1 02 1

�. It is known that there is

no nontrivial multiplicative relation between A and B in SL(2,Z), which meansthat any two reduced words over the set {A,B} will define di↵erent matrices inSL(2,Z), where the matrix defined by a word is just f1f2 · · · fn.

Given a prime p, denote the group SL(2,Z/pZ) by Gp

. Define the homomor-phism '

p

from SL(2,Z) to Gp

which maps each matrix X 2 SL(2,Z) to the matrix'p

(X), obtained by taking each entry of X mod p. Then, define Ap

= 'p

(A), Bp

='p

(B), Ap

�1 = 'p

(A�1), and Bp

�1 = 'p

(B�1). Note that this notation is justified,

since we have 'p

(A)'p

(A�1) = 'p

(AA�1) = I, and so 'p

(A)�1 = 'p

(A�1). Nowdefine the set U

p

= {Ap

, Bp

, Ap

�1, Bp

�1} and denote the Cayley graph �(Gp

,Up

)by X

p

.The goal is to find a lower bound for girth(X

p

). To do so, we will find a lowerbound for d(X

p

), which we define as the smallest integer such that there exist two

walks in Xp

of lengths d(Xp

) starting at I =

1 00 1

�that end at the same vertex.

In this context, a walk means a sequence of adjacent vertices x0, x1, ..., xk

such thatxi�1 6= x

i+1. We have the following relationship between girth(Xp

) and d(Xp

):

Lemma 4.22. girth(Xp

) � 2d(Xp

)� 1.

Proof. Let D(Xp

) be the largest integer such that any two walks in Xp

of lengths D(X

p

) starting at I end at di↵erent vertices. By the definitions of d(Xp

) andD(X

p

), d(Xp

) = D(Xp

) + 1. Let gI

be the length of the shortest cycle in Xp

thatcontains I. We show that either g

I

= 2D(Xp

) + 1 or gI

= 2D(Xp

) + 2.Suppose that g

I

2D(Xp

). Then there exists a cycle x1 = I, x2, ..., xk

, x1, wherek 2D(X

p

). But the walks x1, x2, ..., xdk/2e and x1, xk

, xk�1, ..., xdk/2e both have

length D(Xp

), start at I and have a common endpoint, which contradicts thedefinition of D(X

p

).Now suppose that g

I

� 2D(Xp

)+3. Then any two walks of length D(Xp

)+1beginning at I must end at di↵erent vertices, otherwise they will form a cycle oflength at most 2D(X

p

)+2. But D(Xp

) is the largest integer with this property, sothis is a contradiction.

Therefore, gI

= 2D(Xp

) + 1 or gI

= 2D(Xp

) + 2. Equivalently, gI

= 2d(Xp

)� 1or g

I

= 2d(Xp

). Now, Xp

is vertex-transitive, so this result applies to all verticesof X

p

, that is, the length of the shortest cycle that contains any vertex is either2d(X

p

)� 1 or 2d(Xp

). So girth(Xp

) � 2d(Xp

)� 1 as desired. ⇤

Suppose that we have two walks P = (p0, p1, ..., pr) and S = (s0, s1, ..., st) thatboth start at I = p0 = s0 and end at the same vertex p

r

= st

. Since Xp

=�(G

p

,Up

), we have pi

= pi�1vi and s

j

= sj�1wj

, 1 i r, 1 j t, wherevi

, wj

2 Up

. From this, we obtain pi

= v1 · ... · vi and sj

= w1 · ... · wj

. We havepr

= st

, so

(4.23) v1 · ... · vr = w1 · ... · wt

.

10 CHEUK TO TSUI

Now define the word V = (v1, ..., vr) by

vi

=

8>>>><

>>>>:

A if vi

= Ap

B if vi

= Bp

A�1 if vi

= Ap

�1

B�1 if vi

= Bp

�1

and W = (w1, ..., wr

) similarly. Since P and S are walks, we have pi+1 6= p

i�1, butpi+1 = p

i�1vivi+1, implying that vi+1 6= v

i

�1. So V = (v1, ..., vr), and hence V , isa reduced word. The same is true for W = (w1, ..., wt

) and hence W . But P andS are di↵erent walks and both begin at I, so V and W are di↵erent, and so V andW are di↵erent. Since V and W are di↵erent reduced words over {A,B}, we have

(4.24) v1 · ... · vr 6= w1 · ... · wt

because there is no nontrivial multiplicative relation between A and B. But by(4.23), we have

'p

(v1 · ... · vr) = 'p

(w1 · ... · wt

).

This means that the entries of the matrix v1 · ... · vr are equal mod p to the entriesof the matrix w1 · ... · wt

. It follows that all entries of v1 · ... · vr � w1 · ... · wt

aredivisible by p. In addition, it follows from (4.24) that at least one of the entriesmust be nonzero. Next, we have a result regarding the norm of a matrix, wherethe norm of a square matrix M is defined by

kMk = supx 6=0

kMxkkxk ,

and the norm of the column vector x = (x1, x2, ..., xn

)T ispx1

2 + x22 + ...+ x

n

2.

Lemma 4.25. If M = (mij

) is a n⇥n square matrix and mij

= k for some k 2 R,then kMk � |k|.

Proof. It su�ces to find a unit vector x in Rn with kMxk � |k|. Let

x = (0, 0, · · · , 1, 0, · · · , 0)T ,

where the 1 is in the jth column. Then Mx is just the jth column of M , which hasnorm

kMxk =qm1j

2 + ...+mij

2 + ...+mnj

2 �qm

ij

2 =��m

ij

�� = |k| .

⇤

It follows from the lemma that

kv1 · ... · vr � w1 · ... · wt

k � p.

Using the fact that kMk+kLk �kM + Lk for all n⇥ n matrices M and L, we get

kv1 · ... · vrk+kw1 · ... · wt

k �kv1 · ... · vr � w1 · ... · wt

k � p.

This implies

(4.26) max {kv1 · ... · vrk ,kw1 · ... · wt

k} � p/2

GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 11

The norms of the matrices A, B, A�1 and B�1 are ↵ = 1 +p2 (calculation

omitted). The matrix norm is submultiplicative, meaning kMkkLk �kMLk for alln⇥ n matrices M and L. Thus by (4.26), we have ↵max{r,t} � p/2, or

(4.27) max{r, t} � log↵

(p/2).

What (4.27) shows is that given two walks that start at I and have a commonendpoint, one of them must have length at least log

↵

(p/2). Recall that d(Xp

) isdefined as the smallest integer such that there exist two walks with length d(X

p

)that start at I and have a common endpoint. Thus we must have

d(Xp

) � log↵

(p/2).

By Lemma 4.22, we have

girth(Xp

) � 2 log↵

(p/2)� 1.

Thus we have obtained a lower bound for the girth of Xp

. This is the basic ideabehind finding a lower bound for the girth of the constructed Ramanujan graphsXp,q, which is given in Theorem 4.20.

4.2. Lower Bound on Chromatic Number of Xp,q. We now find a lower boundon the chromatic number of Xp,q by using a relationship between the chromaticnumber of a graph and the eigenvalues of its adjacency matrix.

Theorem 4.28 ([6]). Let �1 �2 ... �n

be the eigenvalues of a graph G, andlet k = �(G). Then

�1 + ...+ �k�1 ��

n

.

Proof. Since k = �(G), we can color G with colors {1, 2, ..., k}. Let mi

be thenumber of points with color i. By labelling the m1 vertices with color 1 1, 2, ...,m1,the m2 vertices with color 2 m1 + 1,m1 + 2, ...,m1 +m2 and so on, the adjacencymatrix A of G can be written in the form2

66664

0 A12 · · · A1k

A21 0 · · · A2k...

.... . .

...A

k1 Ak2 · · · 0

3

77775,

where Aij

has dimension mi

⇥mj

and Aji

= Aij

T .Let v be an eigenvector corresponding to �

n

. We can break v into pieces v1, v2,. . . , v

k

in Rm1 , Rm2 , . . . , Rm

k respectively. Let us define

wi

=

2

66664

|vi

|0...0

3

77775

9>>>>=

>>>>;

mi

entries, w =

2

66664

w1

w2...

wk

3

77775.

Let Bi

be any orthogonal matrix such that

Bi

wi

= vi

for i = 1, ..., k. Note that such an orthogonal matrix always exists, since we canconstruct one by setting the first column of the matrix to be v

i

/kvi

k, extendingvi

/kvi

k to an orthonormal basis of Rm

i using the Gram-Schmidt process, and

12 CHEUK TO TSUI

setting the other columns to be the other vectors in the orthonormal basis. Nowlet

B =

2

66664

B1 0B2

. . .

0 Bk

3

77775.

Then

Bw =

2

66664

B1 0B2

. . .

0 Bk

3

77775

2

66664

w1

w2...

wk

3

77775=

2

66664

B1w1

B2w2...

Bk

wk

3

77775=

2

66664

v1

v2...vk

3

77775= v,

and

B�1ABw = B�1Av = �n

B�1v = �n

w,

which means that w is an eigenvector of B�1AB. B�1AB is in the form

2

66664

0 B1�1A12B2 · · · B1

�1A1kBk

B2�1A21B1 0 · · · B2

�1A2kBk

......

. . ....

Bk

�1Ak1B1 B

k

�1Ak2B2 · · · 0

3

77775.

Since the Bi

are orthogonal, we have Bi

�1Aij

Bj

T

= Bj

TAji

Bi

�1T = Bj

�1Aji

Bi

,so B�1AB is symmetric.

Form a k⇥k symmetric submatrix, call it D, by choosing the entry in the upperleft corner of each of the k2 submatrices B

i

�1Aij

Bj

(Aii

= 0). Consider the vector

u =

2

66664

|v1||v2|...

|vk

|

3

77775.

We claim that u is an eigenvector of D. We can obtain D from B�1AB by deletingthe rows and columns that do not contain the entries in the upper left corner of eachsubmatrix B

i

�1Aij

Bj

. Let c1, c2, ..., ck be the columns of B�1AB that contain the”upper left corner entries”, and let r1, r2, ..., rk be the rows. The non-zero entries|v

i

| of w are in the rows ri

. Thus, the vector B�1ABw = �n

w is just

(?) |v1| c1 +|v2| c2 + ...+|vk

| ck

.

Let c01, c02, ..., c

0k

be the columns c1, c2, ..., ck with all the rows not r1, r2, ..., rkdeleted. We write D =

⇥c01 c02 · · · c0

k

⇤, and Du is equal to

|v1| c01 +|v2| c02 + ...+|vk

| c0k

.

GRAPHS WITH LARGE GIRTH AND LARGE CHROMATIC NUMBER 13

This expression is equal to the vector obtained by taking the rows ri

of (?), whichis 2

66664

�n

|v1|�n

|v2|...

�n

|vk

|

3

77775= �

n

u.

So Du = �n

u, and �n

is an eigenvalue of D.Let µ1 µ2 ... µ

k

be the eigenvalues of D. D has all 0’s in its maindiagonal, so we have tr(D) = 0, which implies µ1 +µ2 + ...+µ

k

= 0, using the factthat the trace of a matrix is equal to the sum of its eigenvalues. This is equivalentto µ1 + ... + µ

k�1 = �µk

. Now �n

is an eigenvalue of D, so �n

µk

. By theInterlacing Theorem, we have �

i

µi

for i = 1, ..., k � 1. Therefore,

�1 + ...+ �k�1 µ1 + ...+ µ

k�1 = �µk

��n

.

⇤

Corollary 4.29 (Ho↵man. See [4, p. 276]). If the eigenvalues of the adjacencymatrix of a graph G are �1 . . . �

n

, then

�(G) � 1� �n

�1.

Proof. We have

(�(G)� 1)�1 �1 + ...+ ��(G)�1 ��

n

by Theorem 4.28. Since �n

is greater than or equal to the average degree of G,which is positive (except when G is an empty graph), and �1 + . . .+ �

n

= 0 by thetrace, we must have �1 < 0. Dividing by �1 gives the result. ⇤

We now use Corollary 4.29 to obtain a lower bound on the chromatic numberof constructed Ramanujan graphs. Let X

n,k

be a non-bipartite Ramanujan graphwith eigenvalues �1 � . . . � �

n

(note the change in order from the statement ofthe corollary). Since X

n,k

is regular, we know that �1 = k. Since �n

= �k if andonly if X

n,k

is bipartite, we have �n

6= �k. By Definition 4.16, |�n

| 2pk � 1, so

��n

2pk � 1 and 1

��

n

� 12pk�1

. By Corollary 4.29, we have

(4.30) �(Xn,k

) � 1� �1

�n

� k

2pk � 1

.

If we let p be a perfect square mod q, then Xp,q is Ramanujan and non-bipartite,and is (p + 1)-regular on q(q2 � 1)/2 vertices. So Xp,q can be written, in thenotation of Definition 4.16, as X

q(q2�1)/2,p+1. Therefore, by (4.30), its chromaticnumber can be made arbitrarily large. Along with Theorem 4.20, this shows thatthe Ramanujan graphs constructed in Construction 4.17 have large girth and largechromatic number.

Acknowledgments. I thank Prof. Babai for suggesting the topic for this paper,providing and discussing the material which the paper is based on, and giving helpwhenever needed. I also thank Yi Guo for her suggestions on the writing andformatting of the paper and help on the mathematical content. Finally, I thankProf. May for organizing the REU and providing helpful comments for revision.

14 CHEUK TO TSUI

References

[1] Gabriel Lugo. Mycielski’s Theorem. http://www.math.uri.edu/ eaton/mycielski.pdf[2] Noga Alon and Joel H. Spencer. The Probabilistic Method. John Wiley & Sons, Inc. 2000.[3] Jirı Matousek and Jan Vondrak. The Probabilistic Method Lecture Notes. Department of Ap-

plied Mathematics, Charles University. 2008.[4] A. Lubotzky, R. Phillips and P. Sarnak, Ramanujan Graphs, Combinatorica 8, 1988, 261-277.[5] G.A. Margulis. Graphs without short cycles, Combinatorica 2, 1982, 71-78.[6] Laszlo Lovasz. Combinatorial Problems and Exercises. Akademiai Kiado. 1979.