The chromatic symmetric function of almost complete graphs

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1 The chromatic symmetric function of almost complete graphs Gus Wiseman Department of Mathematics, University of California, One Shields Ave., Davis, CA Faculty Sponsor: Anne Schilling Department of Mathematics, University of California, One Shields Ave., Davis, CA Abstract The chromatic symmetric function of a graph is a symmetric function that generalizes the chromatic polynomial. Its investigation has largely been motivated by the existence of an open problem, the poset-chain conjecture, which is equivalent to the assertion that for certain graphs, the coefficients in the expansion of the chromatic symmetric function in terms of elementary symmetric functions, known as the e-coefficients of the graph, are nonnegative. In this paper we study how the e-coefficients of a graph G are effected by taking the disjoint union of G and a complete graph, plus all edges between the vertices of G and the vertices of the complete graph. We call such a graph an almost complete graph. Our main result is a description of the chromatic symmetric function of almost complete graphs that can be used to study their e-coefficients. 1 Introduction The chromatic symmetric function X G (x 1, x 2,...) = X G of a graph G was first studied by Stanley [6]. However, the open problem that has been the motivation for much subsequent work (including the present work) was proposed earlier by Stanley and Stembridge [8]. This problem is the so-called posetchain conjecture, which is equivalent to the assertion that the chromatic symmetric function of certain graphs (incomparability graphs of free posets) is e-positive (i.e., the coefficients in its expansion in terms of elementary symmetric functions are nonnegative). That this conjecture remains unresolved illustrates the fact that the coefficients in the expansion of the chromatic symmetric function of a graph in terms of elementary symmetric functions (henceforth we will call these the e-coefficients of the graph) are poorly understood. In fact, there are some very simple sets of graphs (such as strips of triangles) for which it is not known whether the e-coefficients are nonnegative. The e-coefficients of the complete graph K n are trivial: we have X Kn = n!e n. However, it is not clear how the e-coefficients are affected by removing a small number of edges from a complete graph. More formally, given a graph G, we are interested in the effect of taking the disjoint union of G and a complete graph, plus all edges between the vertices of G and the vertices of the complete graph. Intuitively, we think of G as being small and the complete graph as being large, and we call such a graph an almost complete graph. One would suspect that the resulting chromatic symmetric function would in some sense be close to the chromatic symmetric function of a complete graph, and (due to the conjugate triangularity of the transition matrix between monomial and elementary symmetric functions) this is in fact the case. In particular, the number of nonzero e-coefficients of an almost complete graph is no more than the number of partitions of the number of vertices of G. These coefficients are the subject of the present work. Actually, we will not work with the e-coefficients directly. Instead, we will work with the p-coefficients (the coefficients in the expression of the chromatic symmetric function in terms of power sum symmetric 1

2 functions). Our main result (Theorem 6) expresses these coefficients in a way which, in light of Theorem 1, can be used to calculate the e-coefficients. Though the chromatic polynomial is our only example, we prefer to work with a more general object that we call a polynomial set map of binomial type (PSMBT). The definition of a PSMBT is a simple generalization of the definition of a polynomial sequence of binomial type (PSBT); instead of being indexed by positive integers (as in a sequence), the elements of a PSMBT are indexed by finite subsets of some fixed set. There is a natural way to associate to any PSMBT a symmetric function (set map) that we call the symmetric function counterpart of the PSMBT. The symmetric function counterpart of the chromatic polynomial (set map) of a graph is just the chromatic symmetric function (set map) of the graph. In the special case of PSBT s, the symmetric function counterpart has been studied by Loeb [3]. It should be noted that the definition of a PSMBT does not actually suggest any connection to graph theory. It is indeed a surprising fact that the important properties of the chromatic polynomial still hold for this much more general object. While, as noted by Stanley ([6], p. 187), there does not appear to be much hope for finding a nice property of graphs which is equivalent to e-positivity of the chromatic symmetric function, this may not be true for PSMBT s. Hence the study of PSMBT s may be relevant to the poset-chain conjecture. Let us conclude this introduction with a summary of what follows. In Section 2 we establish the notation used throughout the rest of the paper. In Section 3 we prove a theorem in symmetric function theory giving a new relationship between elementary symmetric functions and power sum symmetric functions. In Section 4 we give the definition of a PSMBT, the principal object of our investigation, and study its basic properties. We also introduce the symmetric function counterpart of a PSMBT. In Section 5 we prove our main results (Theorems 4 and 6). Finally, in Section 6 we show how our results are related to the chromatic symmetric function of almost complete graphs. 2 Notation For a positive integer n define [n] = {1, 2,..., n}. We use the notation (x) n for the falling factorial x(x 1)(x 2) (x n + 1). We will work extensively with finite sequences. The elements of a finite sequence α are called parts, and the i th part is denoted α i. The length l(α) of a finite sequence α is the number of parts. The weight α is the sum of the parts. We will denote the empty sequence by. A composition is a finite sequence of positive integers. If α is a composition and α = n, then we say that α is a composition of n. Define ε α = ( 1) α l(α). If α and β are compositions, we define α + β to be their concatenation. A partition is a composition that is weakly decreasing. If λ is a partition and λ = n > 0, we write λ n. If α is a composition, we denote by t(α) the partition obtained by ordering the parts of α so that they are weakly decreasing. A set partition is a set of disjoint nonempty sets. The elements of a set partition are called blocks. The length l(σ) of a set partition σ is the number of blocks. If the union of the blocks is S, then we say that σ is a set partition of S and write σ = S. We denote by t(σ) the partition whose parts are the sizes of the blocks of σ. The relation π σ means every block of σ is contained in some block of π. Definition 1. Let λ and ϕ be compositions. Define (λ) ϕ to be the number of s i : [l(ϕ)] [l(λ)] such that λ ik = ϕ k (where i k = i(k)) for all k [l(ϕ)]. In particular, for a positive integer n, (λ) n is the number of times n appears in λ. Also define (λ) = 1. Define the notation ϕ λ to mean (λ) ϕ > 0. Lemma 1. Let λ and ϕ be compositions such that ϕ λ. Then (λ) ϕ = j 1 (λ) j! ((λ) j (ϕ) j )!. Proof. Let k 1 < k 2 <... < k (ϕ)j be the positions such that ϕ km = j. There are choose i k1, i k2,..., i k(ϕ)j such that λ ikm = j. (λ) j! ((λ) j (ϕ) j)! ways to 2

3 One important consequence of this lemma is that (λ) λ = i 1(λ) i!. 3 Symmetric Functions For background on the theory of symmetric functions, see [4] or [7]. In addition to monomial symmetric functions m λ and elementary symmetric functions e λ, we will also use their augmented counterparts, m λ and ẽ λ, defined by m λ = (λ) λ m λ and l(λ) ẽ λ = λ i! e λ. The purpose of the augmented symmetric functions is to ensure that certain coefficients are integral. Definition 2. Let η µ λ be the coefficient of p µ in ẽ λ. That is, ẽ λ = µ n η µ λ p µ, where n = λ. Also define η µ = η µ µ, where µ denotes the partition whose only part is µ. Lemma 2. We have l(λ) η λ = ε λ λ! λ 1 i (λ) 1 λ. Proof. See [7] section 7.7. From the definition of η λ we know that η λ p λ = n!e n. This next theorem generalizes this. Theorem 1. Let n m be positive integers, and let ϕ be a composition of m. Then l(ϕ) (λ) ϕ η λ n! p λ = ε ϕ ϕ 1 i p ϕ ẽ n m. (n m)! Proof. We have λ n λ n (λ) ϕ η λ p λ = λ n λ n m (λ + ϕ) ϕ η t(λ+ϕ) p t(λ+ϕ). (1) For compositions λ and µ we have (λ + µ) j = (λ) j + (µ) j. Hence from Lemma 1 we obtain (λ + ϕ) ϕ = ((λ) j + (ϕ) j )! ( ) (λ)j + (ϕ) j = (ϕ) ϕ. (λ) j! (λ) j j 1 From Lemma 2 we have l(λ) η t(λ+ϕ) = ε λ ε ϕ n! l(ϕ) λ 1 i ϕ 1 i j 1 ( ) n ( ) 1 (λ + ϕ) i! 1 = η λ η ϕ (λ)i + (ϕ) i. m (λ) i i 1 i 1 3

4 Substituting into (1), we obtain (λ) ϕ η λ p λ = λ n Now expand η ϕ using Lemma 2. ( ) n m (ϕ) ϕ η ϕ p ϕ λ n m η λ p λ = ( ) n (ϕ) ϕ η ϕ p ϕ ẽ n m. m Corollary 1. Let µ be a partition of n whose largest part is n m, and let µ be µ with its largest part removed. Then (n m)! ẽ µ = η λ p λ (λ) ϕ η ϕ µε l(ϕ) ϕ ϕ i. n! Proof. We have ẽ µ = ẽ µ ẽ n m = ϕ m Now interchange the order of summation. λ n η ϕ µp ϕ ẽ n m = ϕ m ϕ m η ϕ µε ϕ (n m)! n! l(ϕ) 4 Polynomial Set Maps of Binomial Type ϕ i λ n(λ) ϕ η λ p λ. Definition 3. Let M be a countable set, and let f be a map whose domain is the set of finite subsets of M. Then we call f a set map on M, or simply a set map, and we use the notation f(s) = f S. If the range of f is the polynomial ring Q[x], then we write f S = f S (x) and we call f a polynomial set map. If f also satisfies f S (x + y) = f A (x)f B (y) (2) A B=S for every finite subset S M, where the sum is over all ordered pairs A, B of disjoint sets whose union is S, then we say that f is a polynomial set map of binomial type (PSMBT). If f is a PSMBT such that f S (n) = f T (n) whenever S = T, we define f k (n) = f S (n), where S is any subset of M such that S = k. We have f k (x + y) = k i=0 ( ) k f i (x)f k i (y), i so in this case a PSMBT is a polynomial sequence of binomial type. Lemma 3. Let f be a PSMBT. Then either f is trivial (i.e., f S (x) = 0 for all finite S M) or 1. For every finite non-empty subset S M, f S (0) = f (x) = 1. Proof. We have f (0) = f (0 + 0) = f (0) 2 (because is disjoint from itself) so f (0) = 0 or 1. Now, if S is a set with one element, then f S (0) = f S (0 + 0) = f S (0)f (0) + f (0)f S (0) = 2f S (0)f (0) so we must have f S (0) = 0. Now suppose this is true for 0 < S < k. Then if S = k we have f S (0) = f S (0+0) = A B=S f A(0)f B (0) = f S (0)f (0)+0+f (0)f S (0) = 2f S (0)f (0) so f S (0) = 0. Furthermore, f S (x) = f S (x + 0) = A B=S f A(x)f B (0) = f S (x)f (0), so if f (0) = 0 then f is trivial. Now assume f is nontrivial so that f (0) = 1. Then for any positive integer t we have f (x) = f ( x t )t (for example, f (x) = f ( x 3 + x 3 + x 3 ) = f ( x 3 )f ( x 3 + x 3 ) = f ( x 3 )f ( x 3 )f ( x 3 )). Hence if the degree of f (x) is greater than 0, then f (x) cannot be a polynomial. Since f (0) = 1, we must have f (x) = 1. 4

5 Lemma 4. Let f be a nontrivial polynomial set map on M. Then f is a PSMBT if and only if f (x) = 1 and for every nonempty subset S M and finite sequence of rational numbers λ, f S ( λ ) = f T (λ i(t) ). (3) σ =S Proof. If f satisfies (3), then we can take λ = (x, y) to obtain (2). Hence f is a PSMBT. For the converse, suppose f is a PSMBT. Suppose also that the Lemma is true for l(λ) < k (it is certainly true for l(λ) = 0). Then for l(λ) = k we have f S ( λ ) = f S (λ k + λ λ k ) = f A (λ k )f B ( λ λ k ), which by the inductive hypothesis is equal to f S (λ k ) + f A (λ k ) π =B But this is the same as (3). A B=S B A B=S i:π [k 1] f T (λ i(t) ). Theorem 2. Let F be the set of all polynomial set maps on M, and let H be the set of all set maps on M with range Q. The maps θ, φ : H F defined by (θ(h)) S (x) = (x) l(σ) h T, (4) σ =S (φ(h)) S (x) = σ =S T π x l(σ) h T, (5) and (θ(h)) (x) = (φ(h)) (x) = 1 are bijections between H and the set of all nontrivial PSMBT s on M, and (θ 1 (f)) S = f S (1), (φ 1 (f)) S = f S(0). Proof. Let h H. Let a n (x) be any polynomial sequence of binomial type, and define a polynomial set map f by f S (x) = a l(σ) (x) h T. σ =S Then f S (x + y) = a l(σ) (x + y) h T = a l(a) (x)a l(b) (y) h T h T. σ =S σ =S A B=σ T A T B Interchanging the order of summation, we see that this becomes a l(σ) (x) h T a l(σ) (y) h T = A B=S σ =A σ =B A B=S f A (x)f B (y). Therefore, f is a PSMBT. Since a n (x) = x n and a n (x) = (x) n are polynomial sequences of binomial type, it follows that θ(h) and φ(h) are PSMBT s. Now let f be a PSMBT. We will show that f = θ(θ 1 (f)) = φ(φ 1 (f)). For any positive integer t we have f S (x) = f S (t x t ) = f S( x t + x t + + x t ) = (t) l(σ) f T ( x t ) (6) σ =S 5

6 by Lemma 4. If x is a positive integer, we can let t = x so that f S (x) = (x) l(σ) f T (1) = (θ(θ 1 (f))) S (x). σ =S Since this is true when x is a positive integer, it is true in general. If we now let t in (6), we have (t) l(σ) t l(σ) so f S (x) = lim t σ =S tf T ( x t ) = lim tf T( x t t ) = x l(σ) f T(0), σ =S σ =S which is (φ(φ 1 (f))) S (x). Finally, let h H. We have (1) k = 0 for each integer k > 1, and (1) 1 = 1. Hence (θ(h)) S (1) = h S, which implies that h = θ 1 (θ(h)). Also, the coefficient of x in (φ(h)) S (x) is h S, so (φ(h)) S (0) = h S. Hence h = φ 1 (φ(h)). Definition 4. Let f be a PSMBT. For each finite subset S M define (f) S = (θ 1 (f)) S = f S (1). That is, (f) S is the sum of the coefficients in f S (x). Also define f S = (φ 1 (f)) S = f S (0). That is, fs is the coefficient of x in f S (x). We will make frequent use of the equations f S (x) = σ =S (x) l(σ) (f) T (7) and f S (x) = σ =S x l(σ) f T. (8) If f S (x) depends only on k = S, then letting f(x, t) = k=0 1 k! f k(x)t k and g(t) = k=0 1 k! fk t k = d dx f(x, t) x=0 (where f k = f k (0)), we have f(x, t) = e xg(t) by applying the compositional formula (see [7], Chapter 5) to (8). In the terminology of umbral calculus [5], this means that f k (x) is the associated sequence of the compositional inverse of g(t). Theorem 3. Let S be the set of set maps on M with range Λ Q, the algebra of symmetric functions with coefficients in Q. Define maps Θ, Φ : H S by (Θ(h)) S = m t(σ) h T (9) σ =S and (Φ(h)) S = σ =S p t(σ) h T. (10) Then Θ θ 1 = Φ φ 1. Proof. We have because (Θ(θ 1 (f))) S = m t(σ) T = σ =S (f) m t(σ) σ =S π σ T π (f) T = f T (1) = τ =T R τ f R f T 6

7 by (8), and f R = f T. R τ π σ T π τ =T Now interchange the order of summation to obtain (Θ(θ 1 (f))) S = π =S σ π m t(σ) f T = T π π =S p t(π) f T = (Φ(φ 1 (f))) S T π because (see [2] for a proof) p t(π) = σ π m t(σ). Definition 5. Let f be a PSMBT. We call Θ(θ 1 (f)) = Φ(φ 1 (f)) the symmetric function counterpart of f, and we denote it by f. Also define f S λ to be the coefficient of p λ in f S. The symmetric function counterpart of a polynomial sequence of binomial type has been studied by Loeb [3]. 5 An Extension Definition 6. Let M and Q be countable disjoint sets. Let f be a PSMBT on M, and let g be a PSMBT on Q. Define the join f + g of f and g to be the PSMBT on M Q determined by 1 if S = and K = (f) ((f + g)) S K = S if S and K = (g) K if S = and K 0 otherwise for finite subsets S M and K Q. In this definition we have defined the PSMBT f + g by specifying θ 1 (f + g). That f + g is well defined follows from Theorem 2. Definition 7. Let f be a PSMBT. Define For the rest of this section, we will assume that ˆf S (x) = ( 1) S x x S f S( x + S ). 1. M and Q are disjoint sets, M is finite, Q is countably infinite, and m = M. 2. S M, K Q, s = S and k = K. 3. f is a PSMBT on M. 4. c is the PSMBT on Q defined by c K (x) = (x) K. 5. F = f + c. Lemma 5. We have F S K = ( 1) k 1 (k 1)!f S ( k) = η s+k k! (s + k)! ˆf S (s + k), where the s + k in η s+k denotes the one-part partition (s + k). 7

8 Proof. By definition F S K is the coefficient of x in F S K (x). From (7) we have F S K (x) = (x) l(π) (F) T. π =S K But for all K Q with more than one element, the definition of c implies that (c) K = c K (1) = (1) K = 0 (and if K has one element then (c) K = 1). Hence The coefficient of x is or But Therefore, F S K (x) = σ =S T π (x) l(σ)+k (f) T. σ =S( 1) l(σ)+k 1 (l(σ) + k 1)! ( 1) k 1 (k 1)! σ =S l(σ) (l(σ) + k 1)! ( 1) (k 1)! (f) T, (f) T. l(σ) (l(σ) + k 1)! ( 1) = ( k)( k 1) ( k l(σ) + 1) = ( k) l(σ). (k 1)! F S K = ( 1) k 1 (k 1)! σ =S ( k) l(σ) (f) T = ( 1) k 1 (k 1)!f S ( k), which proves the first equality. The second equality follows from the fact that which follows from Lemma 2. η s+k = ( 1) s+k 1 (s + k 1)!, Theorem 4. Let N = S K and n = N = s + k, and let λ be a partition of n. Then Fλ S K = η λ k! (s + k)! σ =S λ i(t) T where F is the symmetric function counterpart of F = f + c. Proof. We have F S K λ = π =S K t(π)=λ ˆf T (λ i(t) ), F T. (11) T π Now, to obtain a set partition π of S K such that t(π) = λ, we can: 1. Choose a set partition σ of S. 2. Choose an i : σ [l(λ)] such that λ i(t) T. 3. Enlarge the blocks of the set partition σ to obtain a set partition τ of a subset of S K containing S so that τ T = λ i(t), where τ T is the block of τ that enlarges T σ. 4. Choose a set partition ρ of the remainder of K such that t(τ ρ) = λ, and set π = τ ρ. 8

9 Let r : σ N be defined by r(t) = λ i(t), and let ϕ be the partition whose parts are λ i(t) for T σ. Steps 3 and 4 depend only on r (rather than i), so for a given partition λ there are (λ) ϕ ways to obtain the same set partition π. Now, if T σ we use Lemma 5 to write and also if U ρ, Using the fact that F τt = η τt ( τ T T )! τ T! η λ i(t) ˆf T ( τ T ) = η λ i(t) (λ i(t) T )! λ i(t)! F U = η U. l(λ) η U = η λj = 1 l(λ) n! ηλ (λ) λ λ j!, U ρ we can write F T = F τt F U = 1 l(λ) n! ηλ (λ) λ λ j! T π j=1 U ρ j=1 Since this depends only on λ, σ, and i, let us define z σ λ,i = T π F T, j=1 ˆf T (λ i(t) ), (λ i(t) T )! λ i(t)! ˆf T (λ i(t) ). (12) so that F S K λ = σ =S λ i(t) T 1 (λ) ϕ z σ λ,i Aσ λ,i, (13) where A σ λ,i is the number of ways to perform steps 3 and 4 above. Let t = ϕ, and let γ = λ ϕ (i.e., γ is the unique partition satisfying t(γ +ϕ) = λ). An enlargement of σ to a set partition τ as in step 3 corresponds to a map b : σ 2 K (where 2 K denotes the set of subsets of K) such that the sets b(t) are disjoint (they map be empty) and such that b(t) = λ i(t) T (set τ T = T b(t)). To construct such a map we first choose a t s element subset V of K. Then we divide the elements of V into blocks (possibly empty) whose sizes are λ i(t) T for T σ. The number of ways to do this is ( ) k (t s)! t s 1 (λ i(t) T )! = k! (s + k t)! 1 (λ i(t) T )!. (14) is The number of ways to choose a set partition ρ of the s+k t remaining elements such that t(ρ) = γ ( ) l(γ) l(λ) 1 1 (s + k t)! (γ) γ γ j! = (s + k t)! 1 1 λ i(t)!. (15) (γ) γ λ j! j=1 Combining (12), (13), (14), and (15) and performing cancellations, we obtain F N λ = k! ηλ (s + k)! σ =S λ i(t) T j=1 (λ) λ (λ) ϕ (γ) γ ˆf T (λ i(t) ). 9

10 Now, so (γ) γ = (γ) j! = j (ϕ) j )!, j 1 j 1((λ) (λ) λ = (λ) j! (γ) γ ((λ) j (ϕ) j )! = (λ) ϕ. j 1 Theorem 5. There are unique functions α S ϕ : Z Q, defined for all partitions ϕ (including ϕ = ) such that for all partitions λ with λ s, F S K λ η λ = (λ) ϕ α S ϕ( λ ), ϕ ϕ <s where K Q is any set such that K = λ s. Proof. Uniqueness follows from the independence of {(λ) ϕ : ϕ is a partition} and ϕ in the space of functions from the set of all partitions (of integers greater than or equal to S) to Q. To show existence we use Theorem 3 to write F S K = m t(π) m t(σ)+1 λ s (f) T. π =S K T π(f) T = σ =S Now, it is well known (see for example [7] section 7.5) that m λ = µ λ δ µ λ e µ, where λ denotes the conjugate of λ, denotes the dominance order on partitions, and δ µ λ Q. But if µ (t(σ) + 1 λ s ), then the largest part of µ is greater than λ s. If we apply this in combination with Corollary 1, we obtain an expansion of the desired form. Corollary 2. Let f be a PSMBT. Then ˆf is also a PSMBT. Proof. It follows from Theorem 5 that if each part of λ is greater than or equal to s, we have FS K λ η λ (because ϕ λ and ϕ < s implies ϕ = ). Hence from Theorem 4, ˆf T (λ i(t) ) ˆf S ( λ ) = σ =S = FS K λ η λ as long as each part of λ is greater than or equal to s. Since ˆf S (x) is a polynomial, this must be true in general. The result now follows from Lemma 4. Definition 8. Define N S λ where K Q is any set such that K = λ s. = (s + k)! k! F S K λ η λ, From Theorem 4 we have N S λ = σ =S λ i(t) T ˆf T (λ i(t) ). (16) 10

11 Theorem 6. We have N S λ = σ =A S ( 1) l(σ) λ i(t) < T ˆf S A ( λ t) ˆf T (λ i(t) ), where t = λ i(t). Proof. We apply the principle = < to each of the l(σ) appearances of beneath the second sum in (16) to obtain Nλ S = ( ) ( ) ( 1) l(a) ˆf T (λ ia(t)) ˆf T (λ ib(t)). σ =S A B=σ T A T B i A:A [l(λ)] λ i(t) < T Interchanging the order of summation yields Nλ S = ( ) ( 1) l(σa) ˆfT (λ ia(t)) A A B=S σ A =A i A:σ A [l(λ)] λ i(t) < T Since ˆf is a PSMBT, we can use Lemma 4 to obtain N S λ = A B=S σ A =A ( 1) l(σa) i A:σ A [l(λ)] λ i(t) < T i B:B [l(λ)] i A(A) i B(B)= σ B =B i B:σ B [l(λ)] i A(A) i B(B)= ( ( ) ˆfT (λ ia(t)) ˆf B ( λ t). A B ˆfT (λ ib(t)) ). Note that we can also write Theorem 6 as Nλ S = ( 1) l(σ) σ =A S ϕ:σ N ϕ(t)< T (λ) ϕ ˆfS A (n ϕ ) ˆf T (ϕ(t)), which allows us to calculate the α s of Theorem 5. 6 Almost Complete Graphs Our primary example of a PSMBT is the chromatic polynomial (see [1] for background). Traditionally, the chromatic polynomial is regarded as a function of a graph and a number (of colors). Here we will instead assume that G is fixed and define the chromatic PSMBT as a polynomial defined for each subset of the vertices of G. Definition 9. Let G be a graph with vertex set V. If S V, define χ S (x) = χ G S (x), where χ G S (x) is the traditional chromatic polynomial of G restricted to the vertex set S. We call χ the chromatic PSMBT of G. Theorem 7. The polynomial set map χ is of binomial type. Proof. If we have a coloring of G with x + y colors, then we can divide V into two disjoint subsets depending on whether a vertex is colored with one of the first x colors or with the remaining y colors. Furthermore, if we restrict the coloring to one of the subsets, it will still be proper. Conversely, if we choose disjoint subsets and proper colorings of each, the first using only the first x colors and the second using only the remaining y colors, the induced coloring of G will be proper. 11

12 In particular, (χ) S = χ S (1) is the number of proper colorings of G S with just one color. Therefore, (χ) S = 1 if G S has no edges and (χ) S = 0 otherwise. For a combinatorial interpretation of χ S, see [9]. Theorem 8. Let G and H be graphs, and let χ and ψ be their respective PSMBT s. Then the PSMBT of the join G + H of G and H (i.e., the graph consisting of the disjoint union of G and H plus all edges between the vertices of G and the vertices of H) is χ + ψ. Proof. Let f be the PSMBT of G+H. Then (f) S = f S (1) = 0 if S contains vertices from both G and H because G + H contains edges between these vertices. Therefore, if S and K are subsets of the vertices of G and H respectively, 1 if S = and K = (χ) (f) S K = S if S and K =, (ψ) K if S = and K 0 otherwise so f = χ + ψ. For background on the chromatic symmetric function X G, see [6]. Theorem 9. Let χ be the chromatic PSMBT of G, and let X be its symmetric function counterpart. Then X S = X G S, where X G S is the chromatic symmetric function of G restricted to the vertex set S. Proof. Write X S = σ =S m t(σ) (χ) T. Now, (χ) T = 0 unless the restriction of G to T contains no edges for all T σ, and (χ) T = 1 otherwise. Hence this is precisely the expansion of X G S in terms of stable partitions [6]. Definition 10. Let G be a graph. Define G to be the join of the graph G and a complete graph with countably infinitely many vertices. That is, G is the graph G with countably infinitely many additional vertices and all edges connecting the new vertices to the vertices of G. We call the restriction of G to a finite vertex set (usually containing all the vertices of G) an almost complete graph. Theorem 10. Let G be a graph with vertex set V, and let Q be a countably infinite set disjoint from V. Let χ be the chromatic PSMBT of G, and let c be the PSMBT on Q defined by c K (x) = (x) K. Then the chromatic PSMBT of G (where the new vertices are the elements of Q) is χ + c. Proof. This follows from the fact that c is the chromatic PSMBT of a complete graph with vertex set Q. References [1] Norman Biggs. Algebraic graph theory. Cambridge Mathematical Library. Cambridge University Press, Cambridge, second edition, [2] Peter Doubilet. On the foundations of combinatorial theory. VII. Symmetric functions through the theory of distribution and occupancy. Studies in Appl. Math., 51: , [3] Daniel E. Loeb. Sequences of symmetric functions of binomial type. Stud. Appl. Math., 83(1):1 30, [4] I. G. Macdonald. Symmetric functions and Hall polynomials. Oxford Mathematical Monographs. The Clarendon Press Oxford University Press, New York, second edition, With contributions by A. Zelevinsky, Oxford Science Publications. [5] Steven Roman. The umbral calculus, volume 111 of Pure and Applied Mathematics. Academic Press Inc. [Harcourt Brace Jovanovich Publishers], New York,

13 [6] Richard P. Stanley. A symmetric function generalization of the chromatic polynomial of a graph. Adv. Math., 111(1): , [7] Richard P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, With a foreword by Gian-Carlo Rota and appendix 1 by Sergey Fomin. [8] Richard P. Stanley and John R. Stembridge. On immanants of Jacobi-Trudi matrices and permutations with restricted position. J. Combin. Theory Ser. A, 62(2): , [9] Gus Wiseman. A partition of connected graphs. Electron. J. Combin., 12:N1, 8 pp.,