On Chromatic Polynomial and Ordinomial

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1 On Chromatic Polynomial and Ordinomial Amir Barghi Department of Mathematics and Statistics Rochester Institute of Technology Rochester, NY Master s Thesis under Supervision of Dr Hossein Shahmohamad 1

2 1 Introduction 2 1 Introduction Suppose Γ is a graph with V Γ = n For λ a positive integer, let [λ] = {1, 2,, λ} be a set of λ distinct colors A λ-coloring of Γ is a mapping f from V Γ to [λ] Whenever for every two adjacent vertices u and v, fu fv, we will call f a proper coloring of Γ; otherwise, improper When a proper λ-coloring exists, we call Γ a λ-colorable graph The chromatic number of Γ, denoted by χγ, is defined as the smallest λ such that Γ is λ-colorable, and if that is the case, we call Γ a λ-chromatic graph As we are only interested in proper colorings of graphs using [λ] as our color set, we will drop the term proper and the prefix λ from proper λ-coloring throughout this thesis, unless stated otherwise Colorings f and g are considered distinct, if there exists v V Γ, such that fv gv The chromatic function of a graph, CΓ; λ, is the number of distinct colorings of Γ Theorem 11: CΓ; λ is a degree n monic polynomial of λ Proof: Let r be a positive integer and m r Γ denote the number of distinct r-color-partitions; an r-color-partition of V Γ is a partition of vertices into r nonempty subsets, known as color-classes, such that no two vertices in a subset are adjacent Clearly, for r greater than n, m r Γ = 0 For r less than or equal to n, we can color each r-color-partition in λ r = λλ 1λ 2 λ r + 1 ways; Hence, as there are m r Γ such partitions, we have CΓ; λ = n m r Γ λ r 11 r=1 It is clear that for every r, λ r is a polynomial of λ which implies CΓ; λ is also a polynomial of λ Furthermore, as there is one color-partition of V Γ into n color-classes, the coefficient m n Γ of λ n a polynomial of degree n that has the highest degree among λ r is equal to 1 This proves that CΓ; λ is monic with degree n From now on, we will refer to CΓ; λ as the chromatic polynomial of graph Γ Furthermore, whenever Γ contains a loop, no color-partition is possible and we have m r Γ = 0, for 1 r n; Hence, by using the explicit expression 11, known as the factorial form of CΓ; λ, CΓ; λ = 0 In addition to that, as multiple edges have no effect on color-partitions, we will assume that graphs under consideration have no multiple edges Due to these observations, graphs of interest are loop-less graphs with no multiple edges Conventionally, we will assume that CΓ; 0 = 0 which implies λ is a factor of chromatic polynomial of any graph Moreover, Γ is λ-colorable if and only if CΓ; λ > 0 From this remark, it is obvious that χγ = min {λ N CΓ; λ > 0} and the four-color theorem can be rephrased as follows:

3 1 Introduction 3 If Γ is a planar graph, then CΓ; 4 > 0 One important remark which we have to make here is: by N we mean the set of all positive integers, while by N 0 the set of all natural numbers is meant Clearly, N = N 0 {0} We will find chromatic polynomials of some familiar graphs and introduce methods that are useful in finding them Our next proposition deals with chromatic polynomial of complete graphs: Proposition 12: If K n denotes the complete graphs of order n, CK n ; λ = λ n Proof: Because every two vertices in K n are adjacent, there is no color-partition of this graph into r color-classes, for 1 r < n, by using the Pigeon-hole Principle; In other words, m r K n = 0 On the other hand, we can uniquely color-partition K n into n color-classes, each containing only one vertex, so m n K n = 1; Therefore, CK n ; λ = n r=1 m rk n λ r = λ n Let Γ be a graph having two components, Γ 1 and Γ 2 As V Γ 1 and V Γ 2 are disconnected, we can color them independently; Hence, Similarly, we have CΓ; λ = CΓ 1 ; λ CΓ 2 ; λ Theorem 13: Let Γ be a graph with k components, denoted by Γ 1, Γ 2,, Γ k Then, CΓ; λ = k CΓ i ; λ This implies the following two corollaries: Corollary 14: If N n denotes the null graph of order n, then CN n ; λ = λ n Corollary 15: Let Γ be a graph with c components Then λ c is factor of CΓ; λ Now, we will state a very important theorem which enables us to compute the chromatic polynomial of a graph by computing chromatic polynomial of smaller and simpler graphs In order to do so, we need to introduce some notation: suppose e is an edge in Γ By deleting e from Γ, we will obtain a graph, denoted by Γ e, which has the same vertices as Γ but edge-set EΓ {e} By contracting e in Γ, we mean the graph which results from Γ e by identifying the two vertices which e had them as its ends in Γ; Let Γ e denote this graph Using the above notation, we have the so-called deletion-contraction theorem:

4 1 Introduction 4 e Figure 11: Γ, Γ e, Γ e Theorem 16: For every edge e in any graph Γ, CΓ; λ = CΓ e ; λ CΓ e ; λ 12 Proof: Let u and v be the two ends of e in Γ Let C 1 and C 2, respectively, be the set of colorings of Γ e in which u and v are colored differently and have the same colors Clearly, there is a 1-1 correspondence from C 1 to colorings of Γ; Also, from C 2 to colorings of Γ e As we have CΓ e ; λ = C 1 + C 2 = CΓ; λ + CΓ e ; λ, thus the result Corollary 17: Let Γ be a graph and u and v two non-adjacent vertices in Γ If Γ + uv and Γ uv denote graphs obtained by, respectively, linking and contracting u and v in Γ, then CΓ; λ = CΓ + uv; λ + CΓ uv ; λ In using the deletion-contraction theorem, whenever a multiple edge occurs, as we previously made this remark on multiple edges, we will delete all edges but one in a multiple edge As applications of the deletion-contraction method, the following two propositions are quite classic: Proposition 18: For every tree T of order n, CT ; λ = λλ 1 n 1 Proof: Proof by induction on n: When n = 2, it is clear that CT ; λ = CK 2 ; λ = λ 2 = λλ 1 Now, assume for every tree of order n 1 it is true, and let T be a tree of order n Let v be a leaf in T and e the edge incident to v It is obvious that T e has two components: the isolated vertex v and a tree of degree n 1 By the induction hypothesis and Theorem 13, CT e ; λ = λ λλ 1 n 2 = λ 2 λ 1 n 2 On the other hand, T e is also a tree of order n 1 and CT e ; λ = λλ 1 n 2 Now, by using the deletion-contraction method, we have CT ; λ = CT e ; λ CT e ; λ = λ 2 λ 1 n 2 λλ 1 n 2 = λλ 1 n 1 By using Theorem 13 and Proposition 18, we have the following:

5 1 Introduction 5 Corollary 19: If F is a forest of order n with c components, then CF ; λ = λ c λ 1 n c Proposition 110: If C n denotes the cycle of order n, then CC n ; λ = λ 1 n + 1 n λ 1 Proof: Proof by induction on n: When n = 3, it is clear that CC 3 ; λ = CK 3 ; λ = λλ 1λ 2 = λ 1 3 λ 1 By assuming that this is true for C n 1, we will show it is also true for C n Let e be an edge in C n By deleting and contracting e, we will have P n 1, path of length n 1, and C n 1, respectively By the induction hypothesis, the deletion-contraction method, and the fact that every path is a tree, we have CC n ; λ = CP n 1 ; λ CC n 1 ; λ = λλ 1 n 1 [λ 1 n n 1 λ 1] = λλ 1 n + 1 n λ 1 We will introduce another useful method, but before that, some notation is required Let Γ 1 and Γ 2 be two graphs and r be a natural number less than or equal to minimum of ωγ 1 and ωγ 2 note that ωγ denotes the clique number of a graph Γ It follow that both these graphs have a copy of K r as a subgraph and the graph Γ obtained from Γ 1 and Γ 2 by identifying these two copies of K r is called a K r -gluing of Γ 1 and Γ 2 Clearly, when r is equal to zero, the gluing is just the disjoint union of the two graphs By using the aforementioned notation, we have the following theorem: Theorem 111: If Γ is a K r -gluing of the graphs Γ 1 and Γ 2, then CΓ; λ = CΓ 1; λ CΓ 2 ; λ 13 CK r ; λ Proof: When r is equal to zero, the validity of this formula has been verified in Theorem 13, for k = 2 Now, assume that r is a positive integer The number of distinct ways to color Γ with λ colors is equal to CΓ 1 ; λ times CΓ 2 ; λ/ck r ; λ, the number of ways we can extend distinct λ-colorings of K r to distinct λ-colorings of Γ 2 This proves the result Corollary 112: Let Γ be graph containing K r, for r a positive integer For graph Γ obtained from Γ by adding a new vertex v and connecting v to vertices of K r, we have CΓ; λ = CΓ ; λλ r Let Γ be a graph and for k 4, a copy of C k in Γ, say Γ, is called a pure cycle when there are no edges in Γ linking two non-consecutive vertices in Γ If such an edge exists, it is called a chord in Γ A graph Γ in which every cycle has a chord is called a chordal graph see Figure 12 and it can be shown that any chordal graph of order n is constructed

6 1 Introduction 6 recursively from N 1 by applying repeatedly the rule we have used in Corollary 112, n 1 times This implies that if Γ is a chordal graphs and for 0 i k = χγ 1, r i N such that k i=0 r i = n, then CΓ; λ = λ r 0 λ 1 r 1 λ 2 r2 λ k r k There is a well-known example that the converse of the above remark is not true see Figure 13 Figure 12: An Example of a Chordal Graph Figure 13: Non-chordal Graph with Chromatic Polynomial λλ 1λ 2λ 3 3 λ 4 For q N, we will define the class of q-trees recursively as follows: i K q is a q-tree; ii A q-tree of order n + 1 is obtained from Γ, a q-tree of order n, provided that n q, by adding a new vertex and linking it to a K q in Γ Clearly, as one may see how a q-tree of order n is constructed, every q-tree is a chordal graph and we have the following proposition on their chromatic polynomial using induction: If Γ is a q-tree of order n n q, then CΓ; λ = λλ 1 λ q + 1λ q n q

7 1 Introduction 7 One interesting example that we will return to later in Section 5 is L n = K 2 P n, for n N see Figure 14 We know from Proposition 110 that CC 4 ; λ = λ λ 1 = λλ 1λ 2 3λ + 3 On the other hand, it is easy to verify that L 1 is the cycle C 4 In general, as L n is an edge-gluing of L n 1 and C 4, chromatic polynomial of L n is equal to λλ 1λ 2 3λ + 3 n, using induction and Theorem 111 Figure 14: L n Another interesting example is the graph n = K 3 P n, for n N see Figure 15 First, we will find the chromatic polynomial of 1 see Figure 16 and then as n is a K 3 -gluing of n 1 and 1, we will use 13 to write a recursive formula for C n ; λ By using Theorem 111 a few times, C 1 ; λ = Figure 15: n CK2 ; λ CK 3 ; λ CK 1 ; λ 2 CK 3; λ CK 1 ; λ + 2 CK 3; λ CK 2 ; λ 1 CK 3 ; λ = λ 3 5λ 2 + 9λ 5 CK 3 ; λ = λλ 1λ 2λ 3 5λ 2 + 9λ 5 On the other had, for n 2, C n ; λ = C 1; λ C n 1 ; λ CK 3 ; λ = λ 3 5λ 2 + 9λ 5 C n 1 ; λ = λ 3 5λ 2 + 9λ 5 n 1 C 1 ; λ = λλ 1λ 2λ 3 5λ 2 + 9λ 5 n Hence, for n N, we have C n ; λ = λλ 1λ 2λ 3 5λ 2 + 9λ 5 n

8 1 Introduction 8 Deletion Contraction = = Figure 16 Theorem 113: Let Γ be a connected graph,and u and v two non-adjacent distinct vertices of Γ Let Γ be the graph obtained from Γ by adding a uv-path of length k Then, CΓ; λ = CC k+1; λ CΓ ; λ λλ k CΓ uv; λ Proof: From Corollary 17, we have CΓ; λ = CΓ + uv; λ + CΓ uv ; λ Clearly, Γ + uv is an edge-gluing of Γ + uv and C k+1 On the other hand, Γ uv is a vertex-gluing of Γ uv and C k As so, by using Theorem 111, we have the following: CΓ; λ = CC k+1; λ CΓ + uv; λ λλ 1 + CC k; λ CΓ uv ; λ λ = CC k+1 ; λ [ CΓ ; λ CΓ uv ; λ ] + CC k; λ CΓ uv ; λ = λλ 1 λ CC k+1 ; λ CΓ ; λ + CΓ CCk ; λ λλ 1 uv; λ CC k+1; λ = λ λλ 1 CC k+1 ; λ CΓ ; λ λλ 1 +

9 1 Introduction 9 λ 1 CΓ k + 1 k λ 1 uv; λ λ 1k k+1 λ 1 = λ λλ 1 CC k+1 ; λ CΓ ; λ + 1 k CΓ λλ 1 uv; λ For r, s, t N, a theta graph θr, s, t is constructed from N 2, by connecting the two vertices by three disjoint paths of length r, s, and t see Figure 17 By applying Theorem 113, we have Cθr, s, t; λ = CC r+1; λ CC s+t ; λ λλ 1 r s Figure 17: θr, s, t t + 1 r CC s; λ CC t ; λ λ More generally, for k 3 and s 1, s 2,, s k N, a theta graph θs 1, s 2,, s k is similarly constructed from N 2, but this time, by using k number of disjoint paths of length s 1, s 2,, s k By using induction and Theorem 113, we can easily prove the following: Cθs 1, s 2,, s k ; λ = k CC s i +1; λ λ k 1 λ 1 k 1 + k CC s i ; λ λ k 1 Let Γ 1 and Γ 2 be two disjoint graphs Suppose V Γ i = n i, for i = 1, 2 By the join graph of these two graphs, denoted by Γ 1 + Γ 2, we mean the graph obtained by connecting every vertex in one graph to all vertices in the other Now, assume that CΓ 1 ; λ and CΓ 2 ; λ are expressed in factorial form, n 1 r=1 m rγ 1 λ r and n 2 s=1 m sγ 2 λ s, respectively The umbral product of these two forms, denoted by CΓ 1 ; λ CΓ 2 ; λ, is also a factorial form, obtained by applying the standard polynomial product, as if treating the factorials λ r as powers λ r ; more precisely, λ r λ s = λ r+s Theorem 114: Let Γ 1 and Γ 2 be graphs with chromatic polynomials expressed in factorial form The chromatic polynomial of their join, Γ 1 + Γ 2, is the umbral product of their chromatic polynomials CΓ 1 + Γ 2 ; λ = CΓ 1 ; λ CΓ 2 ; λ

10 1 Introduction 10 Proof: To prove this, first we will prove that m t Γ 1 t n 1 + n 2, the color-partitions of Γ, is equal to r+s=t m rγ 1 m s Γ 2, for 1 r n 1 and 1 s n 2 A color-class of Γ is either a color-class of Γ 1 or a color-class of Γ 2, since every vertex in Γ 1 is adjacent to every vertex in Γ 2, and vice versa Due to this fact, as stated before, the number of colorpartitions of Γ into t color-classes is equal to the sum of color-partitions of Γ 1 and Γ 2 into r and s color-classes, respectively, provided that t = r + s Now, we can write n 1 n 2 CΓ 1 ; λ CΓ 2 ; λ = m r Γ 1 λ r m s Γ 2 λ s = n 1 n 2 r=1 s=1 r=1 m r Γ 1 m s Γ 2 λ r λ s = n 1 +n 2 t=1 n 1 n 2 r=1 s=1 s=1 m t Γ λ t = CΓ 1 + Γ 2 ; λ m r Γ 1 m s Γ 2 λ r+s = Theorem 115: Let Γ be a graph For N 1 + Γ and N 2 + Γ which are called the cone and suspension of Γ, respectively, we have CN 1 + Γ; λ = λ CΓ; λ 1 and CN 2 + Γ; λ = λ CΓ; λ 1 + λλ 1 CΓ; λ 2 Proof: For the former, we can write, λ 1 CN 1 + Γ; λ = CN 1 ; λ CΓ; λ = n m r Γ λ r = r=1 n m r Γ λ r+1 = r=1 n r=1 m r Γ λ 1 λ r = n m r Γ λ λ 1 r = r=1 n λ m r Γλ 1 r = λ CΓ; λ 1 r=1 By using the identity λ 2 = λ 1 + λ 2, the proof of the latter is as follows: n r=1 CN 2 + Γ; λ = CN 2 ; λ CΓ; λ = λ 1 + λ 2 m r Γ λ 1 λ r + n m r Γ λ r = r=1 n r=1 m r Γ λ 2 λ r =

11 1 Introduction 11 n n m r Γ λ r+1 + m r Γ λ r+2 = r=1 n m r Γ λ λ 1 r + r=1 r=1 n m r Γ λλ 1 λ 2 r = r=1 n n λ m r Γ λ 1 r + λλ 1 m r Γλ 2 r = r=1 r=1 λ CΓ; λ 1 + λλ 1 CΓ; λ 2 In the same fashion, we can prove that for Γ a graph and k N, CN k + Γ; λ = k λ i CΓ; λ i As an application of Theorem 115, we will find C n K k i + N 2 ; λ: 1 λ n λ n 1 1 λ n 1 n C K ki + N 2 ; λ = n n λ C K ki ; λ 1 + λλ 1 C K ki ; λ 2 = λ n n n CK ki ; λ 1 + λλ 1 λ n λ 1 ki + λλ 1 λ 1 ki + λλ 1 ki + 1 λ n 1 n λ ki +1 i+1 n CK ki ; λ 2 = n λ 2 ki = 1 λ n λ 1 n λ n 1 λ 1 n 1 1 λ n 1 λ 1 n 1 + n 1 λ n 1 λ 1 n 1 1 n λ 1 n 1 λ λ n 1 λ 1 n 1 ki +1 + n λ 2 ki = λλ 1λ 2 ki = n λ ki +2 n λ ki +2 = =

12 1 Introduction 12 1 n λ 1 n 1 λ λ n 1 λ 1 n 1 ki +1 + n λ k i +1 λ 1 n 1 + λ n 1 λ 1 n 1 n λ k i 1λ ki +1 = n λ k i 1 Figure 17: W 8, Π 9, B 9 By wheel pyramid and double pyramid of order n, respectively, we mean the cone of C n 1 and suspension of C n 2, denoted by W n and Π n By using Theorem 115 and Proposition 110, it is clear that CW n ; λ = λλ 2 n n 1 λλ 2, CΠ n ; λ = λλ 1λ 3 n 2 + λλ 2 n n λλ 2 3λ + 1 The bi-wheel of order n, denoted by B n, is the graph K 2 + C n 2 and by using Theorem 16, we have CB n ; λ = CΠ n ; λ CW n 1 ; λ = λλ 1[λ 3 n n 2 λ 3] Proposition 116: Let n, k, m 1, m 2,, m k N with m 1 + m m k = n 1 By a broken wheel of W n, denoted by W n m 1, m 2,, m k, we mean the graph obtained from W n by deleting n 1 k spokes and so there are k spokes left such that the number of rim edges between existing successive spokes are m 1, m 2,, m k By letting Q r λ being equal to 1 λλ 1 CC r+2; λ = 1 λ [λ 1r r ], we have CW n m 1, m 2,, m k ; λ = λ k Q mi λ + 1 n 1 λλ 2 14

13 1 Introduction 13 Proof: We will prove this proposition using induction on k For k = 1, we have m 1 = n 1 and W n n 1 is a vertex-gluing of C n 1 and K 2 Hence, by Theorem 111, CW n n 1; λ is equal to 1 λ CC n 1; λ CK 2 ; λ It is an easy task to verify that CW n n 1; λ is equal to λ Q n n 1 λλ 2 It can be verified that when the wheel is not broken, namely when for every 1 i k, m i = 1, 14 is valid If that is the case, k = n 1, W n m 1, m 2,, m k is basically W n, and CW n m 1, m 2,, m k ; λ is equal to λλ 2 n n 1 λλ 2 On the other hand, λ k Q m i λ + 1 n 1 λλ 2 = λ n 1 Q 1λ + 1 n 1 λλ 2 By substituting Q 1 λ = λ 2 and simplifying, we will have λλ 2 n n 1 λλ 2, as expected Now, assume that for k, the proposition is true and we will prove it is also true for k + 1 Let W n m 1, m 2,, m k+1 be a a broken wheel with m 1 + m m k+1 = n 1 Without lacking any generality, we can assume that m k+1 is greater than 1, unless for every 1 i k + 1, m i = 1 which we talked about in the previous paragraph Now, let u and v be the non-adjacent vertices at the consecutive existing spokes that are connected by P, this path of length m k+1 It is clear that W n m 1, m 2,, m k+1 uv = W n mk m 1, m 2,, m k and our induction hypothesis implies that 14 is valid for this graph On the other hand, if W denotes the graph from which W n m 1, m 2,, m k+1 can be constructed by connecting u and v using P, then we have the following, using Theorem 113: CC mk+1 +1; λ CW ; λ λλ 1 CW n m 1, m 2,, m k+1 ; λ = + 1 m k+1 CW n m 1, m 2,, m k+1 uv ; λ It is not not hard to see that CW ; λ = [ k CC m i +2; λ ]/[λλ 1] k 1 This fact enables us to write CW n m 1, m 2,, m k+1 ; λ = CC mk+1 +1; λ k CC m i +2; λ + 1 m k+1 λ λ k λ 1 k CC mk+1 +1; λ k CC mk+1 +1; λ CC mi +2; λ λλ 1 k Q mi + 1 m k+1 λ + 1 m k+1 λ k Q mi λ + 1 n mk+1 1 λλ 2 = k Q mi λ + 1 n 1 λλ 2 = k Q mi λ + 1 n 1 λλ 2 = k CC mk+1 +1; λ + 1 m k+1 λ Q mi λ + 1 n 1 λλ 2 =

14 1 Introduction 14 λ Q mk+1 k Q mi λ + 1 n 1 λλ 2 = k+1 λ Q mi λ + 1 n 1 λλ 2 We will finish this section by indicating this remark that, although in this section we assumed λ to be an positive integer representing the number of colors available to properly color the vertices of a graph Γ with, but from now on, we will assume λ to be any complex number, and as a result, chromatic polynomial of any graph will be an element of C[λ]

15 2 Ordinal Numbers and Their Arithmetic 15 2 Ordinal Numbers and Their Arithmetic In order to introduce the algebraic machinery which will be used in the fourth section, we first need to give a brief review of ordinal numbers We will start with some definitions which are required to define ordinal numbers We will assume that X, < denotes an ordered set X with order < an order is a relation on X that is irreflexive, antisymmetric and transitive An ordered set has a total order when the condition of trichotomy is satisfies: for all x, y X, one and only one of x < y, x = y, and y < x holds An element x X is called a least element when for all y X, x y When an element x X exists such that for all y X, y x implies y = x, we call x a minimal element Regarding least and minimal elements we have the following theorem: Theorem 21: Let X, < be an ordered set i If a least element exists in X, then it is the unique minimal element ii If X is totally ordered, then any minimal element of X is a least element We call a set X, < with a total order a well-ordered set when every non-empty subset of X has a least element with respect to < Let X, < be a totally ordered set and an element of X, say y, is given A section X y is defined as the set of all element in X less than y with respect to <: X y = {x X x < y} An ordinal is a well-ordered set X, < such that for all y X, X y = y The following is a very important theorem, proving that if ordinals exist, they are unique up to isomorphism - in set-theoretic terminology, an isomorphism between two ordered sets is a bijection which preserves the order both ways Theorem 22: Any well-ordered set is isomorphic to a unique ordinal Now, we will show that ordinals exist The starting ordinals are defined as follows which are identified with natural numbers: 0 = 1 = { } = {0} 2 = {, { }} = {0, 1} 3 = {, { }, {, { }}} = {0, 1, 2} n + 1 = {0, 1, 2,, n} To expand our list of ordinal numbers we need the following theorem:

16 2 Ordinal Numbers and Their Arithmetic 16 Theorem 23: i If α is an ordinal number, so is α {α} ii The union of a set of ordinals is an ordinal Based on this theorem we define three type of ordinals The first type consists only of zero The second type are called successor ordinals which are constructed by using i in the above theorem If α is an ordinal, then the successor of α, denoted by sα, is the smallest ordinal bigger than α So, natural number n + 1 is the successor of n and the smallest ordinal bigger than n Finally, the third type of ordinals are those constructed by using ii A non-zero ordinal which is the union of all its predecessors is called a limit ordinal If γ is a limit ordinal, then γ = α<γ α The smallest ordinal number after the natural numbers is the union of all natural numbers denoted by ω and the next ordinal after ω is ω {ω} = sω With regard to this classification, we have the following theorem: Theorem 24: Any non-zero ordinal is either a successor ordinal or a limit ordinal A worth-mentioning remark that might clarify confusions regarding ordinals is that for ordinals α and β, α < β, α β, and α β are equivalent In addition to that, for any ordinals α and β exactly one of α < β, α = β, and β < α holds In order to make expanding our list of ordinals much easier, we will define addition, multiplication, and exponentiation for ordinal numbers Let α and β be ordinal numbers Addition: α + 0 = α, α + sβ = sα + β, If γ is a limit ordinal, α + γ = β<γ α + β Multiplication: α 0 = 0, α sβ = α β + α, If γ is a limit ordinal, α γ = β<γ α β Exponentiation: α 0 = 1, α sβ = α β α, If γ is a limit ordinal, α γ = β<γ αβ Using transfinite induction which we will not discuss in this thesis, the first three properties in the following theorem can be proved The fourth and fifth properties are proved by using apropos set-theoretic isomorphisms

17 2 Ordinal Numbers and Their Arithmetic 17 Theorem 25: For ordinals α, β, and γ, α + β + γ = α + β + γ; α + β γ = α γ + β γ; α β+γ = α β α γ ; If γ + α = γ + β, then α = β; If γ α = γ β and γ 0, then α = β It is note-worthy that addition and multiplication of ordinals are not commutative operations For example, it is not difficult to verify that 1 + ω = ω ω + 1 and 2 ω = ω ω 2 = ω + ω In addition to that, left cancelation laws are not necessarily valid for addition and multiplication As an example, from 1 + ω = 2 + ω = ω and 1 ω = 2 ω = ω, we can not cancel ω from the left and conclude that 1 = 2, which clearly is false Now, with quite an ease, we have 0, 1, 2, 3,, n, ω, ω + 1, ω + 2, ω + 3,, ω + n, ω + ω = ω 2, ω 2 + 1, ω 2 + 2, ω 2 + 3,, ω 2 + n, ω 2 + ω = ω + ω + ω = ω 3 ω n ω 2 = ω ω = n N ω n, ω2 + 1, ω 2 + 2,, ω 2 + n, ω 2 + ω = n N ω2 + n, ω 2 + ω + 1, ω 2 + ω + 2,, ω 2 + ω + n, ω 2 + ω 2 = ω 2 + ω + ω ω n ω ω = n N ωn, ω ω + 1, ω ω + 2,, ω ω + n, ω ω + ω = n N ωω + n

18 2 Ordinal Numbers and Their Arithmetic 18 ω ω + ω 2 ω ω + ω n ω ω + ω 2, ω ω + ω 2 + 1, ω ω + ω 2 + 2,, ω ω + ω 2 + n, ω ω + ω 2 + ω ω ω + ω ω = ω ω 2 ω ω n ω ω+1 = ω ω ω = n N ωω n ω ω ω = ω ω2 ω ωn ω ωω = n N ωωn We can extend this list even further by defining towers of ω as follows Note: γ is a limit ordinal: ω 0 = ω; ω sα = ω ωα ; ω γ = α<γ ω α

19 2 Ordinal Numbers and Their Arithmetic 19 As one may see, the infinite tower ω ω = ω ωω, known also as ɛ 0, is the smallest solution to the equation ω ɛ = ɛ Other solutions of the aforementioned equation can be found by using the limit operation and the so-called epsilon function which is not in the scope of this thesis Moreover, the unimaginably monstrous ordinal ω ωω is a solution to the equation ω ɛ = ɛ The following theorem and lemma can be proved by applying transfinite induction: Theorem 26: i Any infinite ordinal can be written in the form γ + n, where γ is a limit ordinal and n a natural number ii Any limit ordinal can be written in the form ω α for some ordinal α Corollary 27: Any ordinal number can be written in the form ω α + n, where α is an ordinal and n a natural number Lemma 28: For any ordinal α > 0, 1 + ω α = ω α Theorem 29: For every natural number k and every infinite ordinal γ, we have k + γ = γ Proof: Using the above result, there exist α > 0 and n N 0 such that γ = ω α + n Our proof is by induction on k When k = 0, the result is obvious Now, assume that the result is true for k, we will show that it is also true for k + 1: k γ = k ω α + n = k ω α + n = k ω α + n = k + ω α + n = ω α + n = γ More generally, by using Corollary 27, we have the following theorem: Theorem 210: For every ordinal numbers α and β, if β α, then there exist ordinal number γ such that β + γ = α We will finish this section by stating an interesting theorem which is an assertion to avoid a variant of Russell s Paradox, known as Burali-Forti paradox: Theorem 211: The ordinal numbers do not form a set but a class, denoted by O

20 3 The Ring of Ordinomials 20 3 The Ring of Ordinomials In this section, we will step-by-step develop an algebraic structure which we will utilize in section 5 in order to define partial and universal chromatic ordinomials of infinite graphs Please be informed that by O we mean O {0} First, we will define a free multiplication on the set B = {λ ζ α ζ C, α O } {1} of formal objects by putting these building blocks next to each other, satisfying the following axioms: A 1 x i B x 1 x 2 = x 2 x 1 ; A 2 x i B x 1 x 2 x 3 = x 1 x 2 x 3 ; A 3 x B x 1 = x ; A 4 ζ C α i O α 1 α 2 λ ζ α 1 λ ζ α 2 = λ ζ α 2 λ ζ α 1 = λ ζ α 1+α 2 To clarify how A 4 works, we will have a few examples: i For k, l N such that k l, λ ζ k λ ζ l = λ ζ l λ ζ k = λ ζ k+l ; ii For k N, λ ζ ω λ ζ k = λ ζ k λ ζ ω = λ ζ k+ω = λ ζ ω ; iii λ ζ ω λ ζ ω 2 = λ ζ ω 2 λ ζ ω = λ ζ ω+ω 2 = λ ζ ω 3 ; iv λ ζ ω λ ζ ω2 = λ ζ ω2 λ ζ ω = λ ζ ω+ω2 = λ ζ ω2 ; v By Theorem 29, for k N and γ an infinite ordinal, λ ζ γ = λ ζ k+γ = λ ζ k λ ζ γ The set B is totally ordered in the sense that, for all ζ and α, 1 < λ ζ α, and if α 1 < α 2, then λ ζ α 1 < λ ζ 2 α 2 In addition to that, if ζ 1 < ζ 2, then for every α 1, α 2 O, λ ζ 1 α 1 < λ ζ 2 α 2 ; we say ζ 1 = a + i b < ζ 2 = c + i d, when either a < c or when a = c, then b < d Our first theorem in this section is Theorem 31: If B[λ] = { n x i n N, x i B}, then B[λ] is a free commutative monoid Proof: Let n and m be arbitrary positive integer and for 1 i n and 1 j m, x i, y j B[λ] We have n m n+m x i y j = z k, j=1 in which, for 1 k n, z k = x k and for n + 1 k n + m, z k = y k n It is clear from axioms A 1 -A 3 that B[λ] is associative and commutative and has 1 as its identity k=1

21 3 The Ring of Ordinomials 21 An element x of B[λ] is called a reduced element, when x is either 1 or n λ ζ i α i, for distinct ζ i s By reduced form of y B[λ], we mean a reduced element x, such that y is equal to, by applying axioms A 1, A 3, and A 4, finitely many times Consequently, reduced elements n λ ζ i α i and m j=1 λ ξ i β j are equal, when n = m and there exist σ S n such that ζ i = ξ σi and α i = β σi, for i {1, 2,, n} By convention, we will assume that 1 is not equal to any other reduced element Theorem 32: Every element of B[λ] has a reduced form which is unique Proof: As we made a clear remark in previous paragraph, uniqueness follows from existence Let x be an element of B[λ] equal to n x i, with x i B Our proof is based on induction on n When n = 1, either x = 1 or x = λ ζ α, which in either case by definition x is a reduced element Now, assume that every element k y i has a reduced form We will show that, every element k+1 x i also has a reduced form For 1 i k + 1, we can assume that x i = λ ζ i α i, because if there exist i {1, 2,, k + 1} such that x i = 1, without loss of generality assume x k+1 = 1, then x = k+1 x i = k x i which by induction hypothesis has a reduced form Thus, two cases may occur: i For i {1, 2,, k + 1}, ζ i s are distinct complex numbers which by definition x is a reduced element ii For i and j in {1, 2,, k + 1}, ζ i = ζ j By using A 1 and A 4, finitely many times, we will decrease the number of x i s occurring in x and then by induction hypothesis x has a reduced form From now on, we will assume every x B[λ] is a reduced element For every x B[λ], 1 x and if x = n λ ζ i α i, then for β O and ζ C, λ ζ β x, when there exist i {1, 2,, n}, such that ζ = ζ i and β α i Let y = m j=1 y j, then y x, if m n and y j x, for every j By convention, if x 1, then x = 1 If x, y B[λ], then we will define the greatest common divisor of x and y, denoted by x, y, as the element z such that z x and z y, and if w x and w y, then w z Assume that for y = m j=1 λ ξ j β j and x = n λ ζ i α i, y x So, by definition m n and λ ξ j β j n λ ζ i α i, for every j Without loss of generality, assume that ξ j = ζ j and as a result, β j α j Then, by Theorem 210, there exist γ j such that β j + γ j = α j By using these fact, we have x = y m λ ζ i γ i n i=m+1 λ ζ i α i So, when y x, there exist z such that x = yz Our next goal is to define a degree function, denoted by degx, for every element x of B[λ] Conventionally, deg1 = 0 For x = n λ ζ i α i, let β 1, β 2,, β n be the rearrangement of α 1, α 2,, α n, such that β 1 β 2 β n Then, degx is equal to the sum β 1 + β 2 + β β n We will call x purely infinite, when β 1 is an infinite ordinal, which clearly

22 3 The Ring of Ordinomials 22 implies degx is infinite As deg is now a well-defined mapping from B[λ] to O, it is clear that for x, y B[λ], if degx degy, then x y Now assume, for x B[λ], degx is infinite It is easy to see that x can be uniquely decomposed into the form yz such that y is purely infinite, degz N 0, and y, z = 1 In the case that x is purely infinite, it is clear that the decomposition would be x 1 Now it is time to equip B[λ] with a complex scalar product, defined as τ x = τ, x C B[λ] and having the following properties: P 1 τ C x B[λ] degx N 0 τ x = τx C[λ] ; Particularly, τ 1 = τ and to be more specific, 0 = 0 1 and 1 = 1 1 P 2 x B[λ] 0 x = 0 1 x = x ; P 3 τ C x B[λ] τ x = 0 τ = 0 ; P 4 τ i C x i B[λ] τ 1 x 1 = τ 2 x 2 τ 1 = τ 2 x 1 = x 2 [ C = C {0} ] We should assert that, from P 1 and P 3, it is clear that 1 = = 0 We will define a free addition and a multiplication on elements of C B[λ], having the following properties in order to introduce the desired free ring: Addition: P + 1 τ i C x i B[λ] τ 1 x 1 + τ 2 x 2 = τ 2 x 2 + τ 1 x 1 ; P 2 + τ i C x i B[λ] τ 1 x 1 + τ 2 x 2 + τ 3 x 3 = τ 1 x 1 + τ 2 x 2 + τ 3 x 3 ; P 3 + τ i C x B[λ] τ 1 x + τ 2 x = τ 1 + τ 2 x ; P 4 + τ C x B[λ] τ x + 0 = τ x ; P 5 + τ C x B[λ] τ x + τ x = 0 Multiplication: P 1 τ i C x i B[λ] τ 1 x 1 τ 2 x 2 = τ 1 τ 2 x 1 x 2 ;

23 3 The Ring of Ordinomials 23 P 2 τ i C x i B[λ] τ 1 x 1 τ 2 x 2 + τ 3 x 3 = τ 1 x 1 τ 2 x 2 + τ 1 x 1 τ 3 x 3, τ 1 x 1 + τ 2 x 2 τ 3 x 3 = τ 1 x 1 τ 3 x 3 + τ 2 x 2 τ 3 x 3 ; P 3 τ C x B[λ] τ x 1 = 1 τ x = τ x ; P 4 τ i C x i B[λ] τ 1 x 1 τ 2 x 2 τ 3 x 3 = τ 1 x 1 τ 2 x 2 τ 3 x 3 ; P 5 τ i C x i B[λ] τ 1 x 1 τ 2 x 2 = τ 2 x 2 τ 1 x 1 It is important to remark that one can deduce P + 5 from P + 1, and P 4 and P 5 from P 1 and properties of B[λ] and C It is easy to check that for every τ C and x B[λ], 0 τ x = 0 The free ring we are looking for is defined as { n O [λ] = } τ i x i n N, τi C, x i B[λ] Theorem 33: O[λ] is a free commutative ring having a multiplicative identity Proof: Let n τ i x i and m j=1 φ j y j be elements of O [λ] We will first prove that both addition is a closed operation We have n m τ i x i + φ j y j = n+m k=1 ψ k z j, in which ψ k = τ k and z k = x k, for 1 k n, and ψ k = φ k n and z k = y k n, for n + 1 k n + m Being an associative and commutative operation and having 0 as the additive identity element are inherited from P 1 +, P 2 +, and P 4 + Finally, for every element n τ i x i, there exist an additive inverse, namely n τ i x i Now we will show that multiplication is a closed operation: n m τ i x i φ j y j = j=1 nm k=1 ψ k z k, such that, for 1 l n and 1 k nm, when l 1m+1 k lm, then ψ k = τ l φ k l 1m and z k = x l y k l 1m

24 3 The Ring of Ordinomials 24 Being a distributive operation with respect to addition and having associative and commutative properties are inherited from properties of P 2, P 4, and P 5 Finally, 1 = 1 1 is the identity of multiplication From now on, an element of O [λ] will be called an ordinomial and the ring itself will be the ring of ordinomials as it is a free ring constructed in a similar fashion as the familiar polynomial rings were and it also utilizes ordinal arithmetic Theorem 34: C[λ] is a subring of O [λ] Proof: The fundamental theorem of algebra states that every monic polynomial of degree n with complex coefficients has n complex roots not necessarily all distinct So, we can write every element pλ of C[λ] in the form τλ ξ 1 λ ξ 2 λ ξ n with ξ i, τ C After rewriting it in the form τλ ζ 1 n 1 λ ζ 2 n2 λ ζ m n m m, n 1, n 2,, n m N and ζ i s all distinct complex numbers, it is clear from P 1 that pλ O [λ], and thus C[λ] O [λ] Now, define g : C[λ] O [λ] that maps every element pλ = τ n λ n + + τ 2 λ 2 + τ 1 λ + τ 0 n N, τ i C and τ n 0 of C[λ] to τ n λ n + + τ 2 λ 2 + τ 1 λ + τ 0 1 of O [λ] It is easy to check that g is a well-defined mapping and for every pλ and qλ, gpλ + qλ = gpλ + gqλ and gpλqλ = gpλ gqλ Let s assume that for x = n τ i x i, F x {1, 2,, n} is the set of all i s such that degx i is finite If F x is non-empty, then for i F x, τ i x i is in C[λ] and so is i F x τ i x i Without loss of generality, assume that, for 1 k n, F x = { k, k + 1, n} As i F x τ i x i is in C[λ], three cases may occur: it is equal to zero, a non-zero complex number, or a polynomial If it is equal to zero, then we will omit it from x and rewrite x as k 1 τ i x i Clearly, when k = 1, x = 0 If it is a non-zero complex number, say τ, we will rewrite x as k 1 τ i x i + τ 1 Obviously, when k = 1, x = τ 1 = τ Finally, if it is a polynomial τλ ζ 1 n 1 λ ζ 2 n2 λ ζ m n m in which m, n 1, n 2,, n m N and ζ i s are distinct complex numbers, then clearly y = λ ζ 1 n 1 λ ζ 2 n2 λ ζ m nm is a reduced element of B[λ] and we can rewrite x as n τ x i = k 1 τ x i + τ y Surely, when k = 1, x = τ y With regard to the above observation, we call an ordinomial x polynomially reduced, when the set F x defined above is either empty or a singleton From definition of 0 = 0 1, F 0 = {1} By polynomially reduced form of y, we mean a polynomially reduced ordinomial x which y is equal to using the above process The above lines proves the following theorem: Theorem 35: Every element of O [λ] is equal to a polynomially reduced ordinomial

25 3 The Ring of Ordinomials 25 Now, let x = n τ i x i be a polynomially reduced ordinomial such that either F x is empty, or if not, n > 1 So, either degx i is infinite for all i or for all except one index which we assume it is equal to n If the former happens, we know that every x i can be decomposed into y i z i such that degy i is infinite, degz i N 0, and y i, z i = 1 If the latter happens, we will do the same except for x n If all y i s are distinct, we will call x a reduced ordinomial Now, assume they are not all distinct We will partition the set of indices into S 1, S 2,, S m while S i contains all the indices i such that y i s are equal; in latter case, we will assume that S m = {n} Then, we have x = i S 1 τ i x i + i S 2 τ i x i + + i S m τ i x i We assume for 1 j m, y j = y i such that i S i ; in latter case, we will assume that y m = 1 So, we can rewrite x as y 1 i S 1 τ i z i + y 2 i S 2 τ i z i + + y m i S m τ i z i For 1 j m, i S j τ i z i is a member of C[λ]: if it is zero, then y j i S j τ i z i = 0 ; If it is a non-zero complex φ j, then y j i S j τ i z i = y j φ j = φ j y j ; Finally, it is equal to a polynomial φ j z j, then y j i S j τ i z i = y j φ j z j = φ j y jz j = φ j y jz j in which z j is an element of B[λ] such that y jz j is the reduced form of y jz j and y j, z j = 1 By rewriting x as described, x will either be equal to zero or a reduced ordinomial When n = 1 and F x is not empty, then x is a polynomially reduced element of C[λ] which would be a reduced ordinomial by definition; Hence, 0 is a reduced ordinomial By the reduced form of an ordinomial y, we mean a reduced ordinomial x such that y is equal to x by going through the processes described above If y becomes zero, we will call y a degenerate ordinomial; otherwise, it is called a non-degenerate ordinomial The above lines proves the following theorem: Theorem 36: Every element of O [λ] is equal to a reduced ordinomial Let x = n τ i x i be a non-degenerate ordinomial such that F x is empty and for every i, x i is a purely infinite element of B[λ]; then, we will call x a purely infinite ordinomial When, only F x is empty, x is just called an infinite ordinomial If n τ i x i is a non-degenerate ordinomial, then so is n τ i x i Moreover, It is clear that when one is purely infinite or just infinite, so would be the other Theorem 37: Every element of O [λ] is either equal to 0 or a unique non-degenerate ordinomial up to permutation of indices Proof: From the previous theorem, we know that every ordinomial is equal to a reduced element and from definition, its is either degenerate or not So, we only have to prove that if non-degenerate ordinomials x = n τ i x i and y = m j=1 φ j y j are equal, then n = m and there exist σ S n such that τ i = φ σi and x i = y σi for all i Not only these two ordinomials are reduced, but also they are polynomially reduced Three cases may occur: 1 F x and F y are both empty which is the case when they are both infinite

26 3 The Ring of Ordinomials 26 ordinomials; 2 F x and F y are non-empty sets and we assume F x = {n} and F y = {m}; 3 Just one of F x or F y is empty We start with the first case: Let n τ i x i and m j=1 φ j y j be infinite non-degenerate ordinomials such that x i = x ix i and y j = y jy j are the unique decomposition of x i and y j ; clearly, x i, x i = 1 and y j, y j = 1 Now assume that x i s and y j s are all distinct elements of B[λ] It is not hard to see that the sum n τ i x i + m j=1 φ j y j = n τ i x ix i + m j=1 φ j y jy j is also an infinite non-degenerate ordinomial Now, assume that infinite non-degenerate ordinomials n τ i x i and m j=1 φ j y j are equal to each other Let x i, x i, y j and y j be defined as was defined in the previous paragraph We can write n τ i x i + m j=1 φ j y j = n τ i x ix i + m j=1 φ j y jy j = 0, and clearly there exist i 1 and j 1, such that x i 1 = y j 1 Because if x i s and y j s were all distinct, then one side of the equation was degenerated, while the other side is non-degenerate which clearly is not possible As x i s and y j s are all distinct, x i 1 is only equal to y j 1, and vise versa Without loss of generality, assume i 1 = n and j 1 = m and z = x n = y m So, n 1 τ i x ix i + m 1 j=1 φ j y jy j + z τ n x n + φ m y n = 0 If τ n x n + φ m y m is not equal to zero, because z is distinct from the rest of x i s and y j s, the sum cannot be degenerated to 0 So, τ n x n = φ m y m and as a result τ n = φ m and x n = y m; Hence, x n = y m n 1 τ i x ix i + m 1 j=1 φ j y jy j = 0 By repeating the argument given above we can prove that n = m and there exist σ S n such that y i = x σi and φ i = τ σi Now, assume for x = n τ i x i and m j=1 φ j y j, we have x = y, F x = {n}, and F y = {m} So, we can write n 1 τ i x i + m 1 j=1 φ j y j + τ n x n + φ m y m = 0 Assume, τ n x n and φ m y m are not equal Then, on one side of the above equation we have a nondegenerate ordinomial, while the other side is zero which is impossible It follows that τ n x n = φ m y m and as a result τ n = φ m and x n = y m So, we can rewrite the above equation as n 1 τ i x i + m 1 j=1 φ j y j = 0 If n and m are both greater than 1, then we can argue as we did for the first case; If one was 1 while the other one was greater than 1, we would have a non-degenerate ordinomial equal to 0 which surely is not possible; Finally, if they are both equal to one we are not left with anything to argue about At the end, for non-degenerate ordinomials x and y, assume that x = y and F x = {n} while F y is empty We will show this case is not possible We have n 1 τ i x i + m 1 j=1 φ j y j + τ n x n = 0 We know τ n 0, but then we have a non-degenerate ordinomial equal to 0 which is not possible From now on, by an ordinomial we mean either 0 or a non-degenerate ordinomial The degree of an ordinomial x = n τ i x i, denoted by degx, is defined as the largest ordinal occurring in the set {degx i i = 1, 2,, n} and as 0 = 0 1, have deg0 = 0 Theorem 37 asserts that deg is a well-defined mapping from O [λ] to O So, for x and y ordinomials,

27 3 The Ring of Ordinomials 27 if degx degy, then x y An ordinomial x is called a monic ordinomial, when for all i, τ i = 1 and degx > 0 The subset {1 x x B[λ]} of O [λ] denoted by B[λ] and armed with, is a monoid isomorphic to B[λ] From P 3 it is clear that if 1 x and 1 y are elements of B[λ], then 1 x 1 y = 1 xy is not equal to zero It is clear from definition that every element of B[λ] is a monic ordinomial On the other hand, if M[λ] is the set all monic polynomials in C[λ], we know that with polynomial multiplication it is a submonoid of C[λ] homomorphic to B[λ] This observation will be used in section 5 when we talk about chromatic ordinomials In general, as one may see in the following example O [λ] is not an integral domain: λ ζ ω3 λ ζ ω λ ζ ω2 = λ ζ ω+1 λ ζ ω3 + 1 λ ζ ω2 λ ζ ω3 = λ ζ ω+1+ω3 + 1 λ ζ ω2 +ω 3 = λ ζ ω3 + 1 λ ζ ω3 = 0 For x O [λ], 1 x, and if x = n τ i x i, then y B[λ] divides x, if y x i, for every i By convention, for every x B[λ], x 0 Assume that for a non-degenerate ordinomial x and y B[λ], y x = n τ x i By definition for every i, y x, and as a result there exist z i such that x = yz i Therefore, x = y n τ z i Finally, if F x is not empty and by assumption equal to {n}, then by definition y x n and as x n is a polynomial, degy N 0 We will now develop a limit operation which would be the foundation of how chromatic ordinomials will be defined for infinite graphs in section 5 Let f : N N f : ω ω be an order preserving mapping, meaning that if k 1 < k 2, then fk 1 fk 2 We will define lim k ω λ ζ fk as follows: One may clearly see that lim λ k ω ζfk = limλ ζ fk = λ ζ fk k<ω k<ω i lim k ω λ ζ fk = λ ζ k 0, when there exist M N such that for k > M and k 0 N, fk = k 0 ; ii lim k ω λ ζ fk = λ ζ ω, otherwise This limit operation has the properties lim τ k ω n λ ζ i f ik = τ n lim λ ζ i f ik, k ω

28 3 The Ring of Ordinomials 28 for distinct complex numbers ζ i and order preserving mappings f i : N N, and lim k ω n m i τ i λ ζ ij fijk = j=1 n m lim τ j i λ ζ ij fijk k ω j=1, such that for every i, ζ ij are distinct complex numbers and f ij : N N are order preserving mappings But, if fk can be decomposed into the form f 1 k + f 2 k in which f 1, f 2 : N N are two other order preserving mappings, then following is not necessarily valid lim λ k ω ζfk = lim λ k ω ζf 1k lim λ n ω ζf 2k For example, we know lim k ω λ ζ 2n = λ ζ ω On the other hand, we can decompose 2n into the form n + n So, we have lim k ω λ ζ n lim n ω λ ζ n = λ ζ ω λ ζ ω = λ ζ ω 2 which is not equal to λ ζ ω More generally, we can define a limiting operation for all limit ordinals Let γ be a limit ordinal, β an ordinal such that γ β and Φ : γ β an order preserving map, meaning that for any ordinals α 1, α 2 < γ, if α 1 < α 2, then Φα 1 Φα 2 < β Our generalized limit operation will be defined as follows: lim λ α γ ζφα = limλ ζ Φα = λ ζ Φα α<γ α<γ The generalized limit has the properties lim τ α γ n λ ζ i Φ iα = τ n lim λ ζ i Φ iα, α γ for distinct complex numbers ζ i and order preserving mappings Φ i : γ β, and lim α γ n m i τ i λ ζ ij Φ ijα = j=1 n lim τ i α γ m i j=1 λ ζ ij Φ ijα such that for i, ζ ij are distinct complex numbers, and Φ ij : γ β order preserving mappings,

29 3 The Ring of Ordinomials 29 But again, if the order preserving mapping Φ : γ β can be decomposed into the form Φ 1 + Φ 2 in which Φ 1, Φ 2 are two other order preserving mappings from limit ordinal γ to ordinal β such that γ β, then following is not necessarily valid lim λ α γ ζφα = lim λ α γ ζφ 1α We would finish this section with the following question: lim λ α γ ζφ 2α Question: What would happen, if we replace A 4 with the axiom below? A 4 ζ C α i O α 2 α 1 λ ζ α 1 λ ζ α 2 = λ ζ α 2 λ ζ α 1 = λ ζ α 1+α 2

30 4 Chromatic Polynomial of Some Chains of Graphs 30 4 Chromatic Polynomial of Some Chains of Graphs In the first section we saw the chromatic polynomial of some families of graphs such as null graphs, trees, cycles, complete graphs, chordal graphs, q-trees, n-ladders, n graphs, theta graphs, wheels, double-pyramids, bi-wheels, and broken wheels In this section we will first define what a sequence of graphs is and afterwards try to find chromatic polynomial of some interesting sequences of graphs, either in closed or recursive form By a sequence of graphs, we mean the family {Γ n } n N in which Γ n is a finite graph We call a sequence, a chain of graphs provided that Γ 1 Γ 2 Γ n and for all n N, V Γ n = fn in which fn is a strictly increasing function from N to N When for all n N, Γ n is connected, the sequence is called a sequence of connected graphs On the other hand, if there exists M N such that for all n N, Γ n < M, then we call {Γ n } n N a finite-degree sequence of graphs It is obvious from our definition that {N n } n N is a finite-degree chain of disconnected graphs while {K n } n N is a non-finite-degree chain of connected graphs Finally, a chain {Γ n } n N is called non-oscillating provided that for all n N and for all v V Γ n /V Γ n 1, for every m > n, d Γm v = d Γn+1 v A chain is called chromatically conformal, when for m N, there exist distinct ζ 1, ζ 2,, ζ m in C such that for all n N, CΓ n ; λ = λ ζ 1 f 1n λ ζ 2 f 2n λ ζ m f mn, in which for 1 i m, f i n : N N is order preserving and fn = f 1 n + f 2 n + + f m n Furthermore, for r N a chain is called chromatically recursive of degree r, when for all n N, CΓ n+r ; λ = p 1 λ CΓ n+r 1 ; λ + p 2 λ CΓ n+r 2 ; λ + + p r λ CΓ n ; λ, in which for 1 i r, p i n is an element of C[λ] which are not dependent on n, and r the smallest natural number such a recursion holds Moreover, we will assume that p 1 λ + p 2 λ + + p r λ is not just a complex number As one may see, {L n } n N is a finite-degree chain of connected graphs which is chromatically conformal and chromatically recursive of degree 1: CL n ; λ = λλ 1λ 3 + i 3 2 n λ 3 i 3 n, 2 CL n+1 ; λ = λ 2 3λ + 3CL n ; λ The first example we would discuss in this section is defined as follows: Assume Ψ 1 is the graph in Figure 41a, and Ψ n is recursively built from Ψ n 1 by identifying the vertices a n 1, b n 1, c n 1, d n 1 in Ψ n 1 with vertices a n 1, b n 1, c n 1, d n 1 of Ψ n, respectively One may also see in Figure 41b how Ψ n looks like We know that CΨ 1 ; λ = CC 8; λ CC 4 ; λ 2 λ 2 λ 1 2 = λλ 1λ 2 3λ+3 2 λ 6 7λ 5 +21λ 4 35λ 3 +35λ 2 21λ+7

31 4 Chromatic Polynomial of Some Chains of Graphs 31 a 1 a n-1 a n ~ a n-1 a ~ n-1 b 1 b n-1 b n ~ b n-1 ~ b n-1 c 1 c n-1 c n c~ n-1 ~ c n-1 d 1 d n-1 d n ~ ~ d d n-1 n-1 a Ψ 1, Ψ n, Ψ n, Ψ n b Ψ n Figure 41 In order to find the chromatic polynomial of Ψ n for n 2, we need to find the chromatic polynomial of Ψ n and Ψ n which are built from Ψ n 1 by identifying vertices a n 1, b n 1, c n 1, d n 1 with vertices ã n 1, b n 1, c n 1, d n 1 of Ψ n and Ψ n see Figure 41a, respectively In Figure 42a and 42b, we have drawn Ψ n and Ψ n, for n 2 Ψ 1 and Ψ 1 can be found in Figure 42c For n 2, we have the following relations between the chromatic polynomial of Ψ n and Ψ n CΨ n; λ = CΨ n 1; λ CΩ 3 ; λ λλ 1 CΨ n; λ = CΨ n 1; λ CΩ 5 ; λ λλ 1 + CΨ n 1; λcω 4 ; λ, λ + CΨ n 1; λcω 6 ; λ, λ in which Ω i, for 1 i 6, are graphs in Figure 43a and 43b

32 4 Chromatic Polynomial of Some Chains of Graphs 32 a Ψ n b Ψ n c Ψ 1, Ψ 1 Figure 42 The recursive relations above can be written in matrix form [ ] [ CΨ CΩ3 ;λ CΩ 4 ;λ n ; λ CΨ λλ 1 λ = n; λ CΩ 5 ;λ λλ 1 CΩ 6 ;λ λ ] [ CΨ n 1; λ CΨ n 1; λ ],

33 4 Chromatic Polynomial of Some Chains of Graphs 33 and as a result, [ CΨ n ; λ CΨ n; λ ] = [ CΩ3 ;λ λλ 1 CΩ 5 ;λ λλ 1 CΩ 4 ;λ λ CΩ 6 ;λ λ ] n 1 [ CΨ 1; λ CΨ 1; λ ] On the other hand, it easily can be seen that for n 2, CΨ n ; λ = CΨ n 1; λcω 1 ; λ λλ 1 + CΨ n 1; λcω 2 ; λ, λ which can be rewritten as CΨ n ; λ = [ CΩ1 ;λ λλ 1 CΩ 2 ;λ λ ] [ CΨ n 1; λ CΨ n 1; λ ] These observations enable us to write the chromatic polynomial of Ψ n in the matrix from [ ] [ ] CΩ 3 ;λ CΩ 4 ;λ n 2 [ ] CΨ n ; λ = CΩ1 ;λ CΩ 2 ;λ λλ 1 λ CΨ 1; λ λλ 1 λ CΨ 45 1; λ CΩ 5 ;λ λλ 1 CΩ 6 ;λ λ In order to compute CΩ i ; λ, for 1 i 6, we will apply Theorem 113 So, we have CΩ 1 ; λ = CC 6; λ CΩ 7 ; λ λλ 1 CΩ 2 ; λ = CC 6; λ CΩ 8 ; λ λλ 1 CΩ 3 ; λ = CC 4; λ CΩ 7 ; λ λλ 1 CΩ 4 ; λ = CC 4; λ CΩ 8 ; λ λλ 1 CΩ 5 ; λ = CC 3; λ CΩ 7 ; λ λλ 1 CΩ 6 ; λ = CC 3; λ CΩ 8 ; λ λλ CΩ 9 ; λ, CΩ 10 ; λ, CΩ 9 ; λ, CΩ 10 ; λ, CΩ 9 ; λ, CΩ 10 ; λ, in which Ω 9 and Ω 10 are graphs in Figure 43c To compute CΩ 9 ; λ and CΩ 10 ; λ one may notice that CΩ 9 ; λ = CΩ 8 ; λ CΩ 10 ; λ and CΩ 10 ; λ = CΩ 11 ; λ+cω 12 ; λ See Figure 44

34 4 Chromatic Polynomial of Some Chains of Graphs 34 a Ω 1, Ω 3, Ω 5, Ω 7 b Ω 2, Ω 4, Ω 6, Ω 8 By using Theorem 111, we have c Ω 9, Ω 10, Ω 11, Ω 12 Figure 43 CΩ 11 ; λ = CK 3; λ 2 CK 4 ; λ λ 2 λ 1 2 = λλ 1λ 2 3 λ 3, CΩ 12 ; λ = CK 2; λ 2 CK 3 ; λ λ 2 = λλ 1 3 λ 2,

35 4 Chromatic Polynomial of Some Chains of Graphs 35 CΩ 7 ; λ = CC 4; λ 3 λ 2 λ 1 2 = λλ 1λ2 3λ + 3 3, CΩ 8 ; λ = CC 4; λ 2 CC 3 ; λ λ 2 λ 1 2 = λλ 1λ 2λ 2 3λ By substituting these polynomials back in the expressions CΩ 9 ; λ = CΩ 8 ; λ CΩ 10 ; λ and CΩ 10 ; λ = CΩ 11 ; λ + CΩ 12 ; λ we had above, we will have CΩ 10 ; λ = λλ 1λ 2λ 3 6λ λ 11, CΩ 9 ; λ = λλ 1λ 2 3 λ 2 3λ + 5 Finally, in order to be able to compute CΨ n ; λ, we need to plug the following polynomials into 45: CΩ 6 ; λ λ = λ 1λ 2λ 5 8λ λ 3 54λ λ 29, CΩ 5 ; λ λλ 1 = λ 2λ6 9λ λ 4 88λ λ 2 133λ + 47, CΩ 4 ; λ = λ 1λ 2 2 λ 5 7λ λ 3 38λ λ 19, λ CΩ 3 ; λ λλ 1 = λ8 12λ λ 6 217λ λ 4 683λ λ 2 408λ + 121, CΩ 2 ; λ λ = λ 1λ 2 2 λ 7 9λ λ 5 89λ λ 3 134λ λ 28, CΩ 1 ; λ λλ 1 = λ10 14λ λ 8 361λ λ λ λ λ λ 2 759λ + 175, CΨ 1 = CC 6; λ CC 4 ; λ 2 λ 2 λ 1 2 = λλ 1λ 2 3λ λ 4 5λ λ 2 10λ + 5, CΨ 1; λ = CC 5; λ CC 4 ; λ 2 λ 2 λ 1 2 = λλ 1λ 2λ 2 3λ λ 2 3λ + 2 Now, we would look at the chromatic polynomial of the chain Θ n = C 4 P n see Figure 44a By applying Theorem 16 on Θ n more precisely, deleting and contracting the edges connecting Θ n 1 to last copy of C 4 one can write chromatic polynomial of this graph in