A Chromatic Symmetric Function Conjecture

Transcription

1 A Chromatic Symmetric Function Conjecture p. A Chromatic Symmetric Function Conjecture Richard P. Stanley M.I.T.

2 A Chromatic Symmetric Function Conjecture p. Basic notation G: simple graph with d vertices V : vertex set of G E: edge set of G Coloring of G: any κ: V P = {1, 2,... } Proper coloring: uv E κ(u) κ(v)

3 A Chromatic Symmetric Function Conjecture p. The chromatic symmetric function X G = X G (x 1, x 2,... ) = x κ, proper κ: V P the chromatic symmetric function of G, where x κ = v V x κ(v) = x #κ 1 (1) 1 x #κ 1 (2) 2.

4 A Chromatic Symmetric Function Conjecture p. The chromatic symmetric function X G = X G (x 1, x 2,... ) = x κ, proper κ: V P the chromatic symmetric function of G, where x κ = v V x κ(v) = x #κ 1 (1) 1 x #κ 1 (2) 2. X G (1 n ) := X G (1, 1,..., 1) = χ }{{} G (n), n 1 s the chromatic polynomial of G.

5 A Chromatic Symmetric Function Conjecture p. Example of a monomial x κ = x 2 1 x 2 x 2 3 x 5

6 A Chromatic Symmetric Function Conjecture p. Simple examples X point = x 1 + x 2 + x 3 + = e 1. More generally, let e k = 1 i 1 < <i k x i1 x ik, the kth elementary symmetric function. Then X Kn = n! e n X G+H = X G X H.

7 A Chromatic Symmetric Function Conjecture p. Acyclic orientations Acyclic orientation: an orientation o of the edges of G that contains no directed cycle.

8 A Chromatic Symmetric Function Conjecture p. Acyclic orientations Acyclic orientation: an orientation o of the edges of G that contains no directed cycle. Theorem (RS, 1973). Let a(g) denote the number of acyclic orientations of G. Then a(g) = ( 1) d χ G ( 1). Easy to prove by induction, by deletioncontraction, bijectively, geometrically, etc.

9 A Chromatic Symmetric Function Conjecture p. Fund. thm. of symmetric functions Write λ d if λ is a partition of d, i.e., λ = (λ 1, λ 2,... ) where λ 1 λ 2 0, λi = d. Let e λ = e λ1 e λ2. Fundamental theorem of symmetric functions. Every symmetric function can be uniquely written as a polynomial in the e i s, or equivalently as a linear combination of e λ s.

10 A Chromatic Symmetric Function Conjecture p. A refinement of a(g) Note that if λ d, then e λ (1 n ) = ( n ) λ i, so e λ (1 n ) n= 1 = ( ) 1 = ( 1) d. λ i Hence if X G = λ d c λe λ, then a(g) = λ d c λ.

11 A Chromatic Symmetric Function Conjecture p. Sinks Sink of an acylic orientation (or digraph): vertex for which no edges point out (including an isolated vertex). a k (G): number of acyclic orientations of G with k sinks l(λ): length (number of parts) of λ

12 The sink theorem Theorem. Let X G = λ d c λe λ. Then c λ = a k (G). λ d l(λ)=k

13 The sink theorem Theorem. Let X G = λ d c λe λ. Then c λ = a k (G). λ d l(λ)=k Proof based on quasisymmetric functions.

14 The sink theorem Theorem. Let X G = λ d c λe λ. Then c λ = a k (G). λ d l(λ)=k Proof based on quasisymmetric functions. Open: Is there a simpler proof?

15 The claw Example. Let G be the claw K 13. Then X G = 4e 4 + 5e 31 2e 22 + e 211. Thus a 1 (G) = 1, a 2 (G) = 5 2 = 3, a 3 (G) = 1, a(g) = 5.

16 The claw Example. Let G be the claw K 13. Then X G = 4e 4 + 5e 31 2e 22 + e 211. Thus a 1 (G) = 1, a 2 (G) = 5 2 = 3, a 3 (G) = 1, a(g) = 5. When is X G e-positive (i.e., each c λ 0)?

17 3 + 1 Let P be a finite poset. Let denote the disjoint union of a 3-element chain and 1-element chain:

18 (3 + 1)-free posets P is (3+1)-free if it contains no induced (3+1) free not

19 The main conjecture inc(p): incomparability graph of P (vertices are elements of P ; uv is an edge if neither u v nor v u)

20 The main conjecture inc(p): incomparability graph of P (vertices are elements of P ; uv is an edge if neither u v nor v u) Conjecture. If P is (3 + 1)-free, then X inc(p) is e-positive.

21 Two comments Suggests that for incomparability graphs of (3 + 1)-free posets, c λ counts acyclic orientations of G with l(λ) sinks and some further property depending on λ. Open: What is this property?

22 Two comments Suggests that for incomparability graphs of (3 + 1)-free posets, c λ counts acyclic orientations of G with l(λ) sinks and some further property depending on λ. Open: What is this property? True if P is 3 free, i.e., X G is e-positive if G is the complement of a bipartite graph. More generally, X G is e-positive if G is the complement of a triangle-free (or K 3 free) graph.

23 A simple special case Fix k 2. Define P d = i 1,...,i d x i1 x id, where i 1,..., i d ranges over all sequences of d positive integers such that any k consecutive terms are distinct.

24 A simple special case Fix k 2. Define P d = i 1,...,i d x i1 x id, where i 1,..., i d ranges over all sequences of d positive integers such that any k consecutive terms are distinct. Conjecture. P d is e-positive.

25 The case k = 2 P d = x i1 x id, i 1,...,i d where i j 1, i j i j+1. Theorem (Carlitz). Pd t d = i 0 e it i 1 i 1 (i 1)e it i.

26 The case k = 2 P d = x i1 x id, i 1,...,i d where i j 1, i j i j+1. Theorem (Carlitz). Pd t d = i 0 e it i 1 i 1 (i 1)e it i. Corollary. P d is e-positive for k = 2.

27 The case k = 3 Ben Joseph (2001) probably had a complicated Inclusion-Exclusion proof.

28 The case k = 3 Ben Joseph (2001) probably had a complicated Inclusion-Exclusion proof. Pd t d = numerator 1 (2e 3 t 3 + 6e 4 t e 5 t 5 + (64e 6 + 6e 51 e 33 )t 6 + ).

29 Schur functions Schur functions {s λ } forms a linear basis for symmetric functions. e λ is s-positive. (Gasharov) X G is s-positive if G is the incomparability graph of a (3 + 1)-free poset. Conjecture (Gasharov). If G is claw-free, then X G is s-positive. (Need not be e-positive).

30 A Chromatic Symmetric Function Conjecture p. 2 A final word When G is a unit interval graph (special case of incomparability graphs of (3 + 1)-free posets), then Haiman found a close connection with Verma modules and Kazhdan-Lustzig polynomials.

31 A Chromatic Symmetric Function Conjecture p. 2