chromatic symmetric functions on graphs and polytopes
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Chromatic symmetric functions on graphs and polytopes 30th FPSAC, Hanover NH, Darmouth College Ra ul Penagui ao University of Zurich July 16th, 2018 Ra ul Penagui ao (University of Zurich) Kernel problems July 16th, 2018 1 / 23


  1. Chromatic symmetric functions on graphs and polytopes 30th FPSAC, Hanover NH, Darmouth College Ra´ ul Penagui˜ ao University of Zurich July 16th, 2018 Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 1 / 23

  2. Introduction CF on graphs The chromatic symmetric function on graphs A colouring on a graph G is a map f : V ( G ) → N . It is proper if f ( v 1 ) � = f ( v 2 ) when { v 1 , v 2 } ∈ E ( G ) . Figure: Example of a proper colouring f of a graph � x f ( v ) . We have x f = x 2 1 x 2 Set x f = 2 x 4 in the figure. v Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 2 / 23

  3. Introduction CF on graphs The chromatic symmetric function on graphs � The chromatic symmetric function (CSF) of G is Ψ G ( G ) = x f . f proper This is a Hopf algebra morphism between G = span { all graphs } and Sym . Example: Figure: The line graph P 2 and the path P 3 Their CSF are     � � � x 2  . Ψ G ( P 2 ) = 2 x i x j , Ψ G ( P 3 ) = 6 x i x j x k  + i x j  1 ≤ i<j 1 ≤ i<j<k i � = j Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 3 / 23

  4. Introduction CF on graphs Tree conjecture on graphs Evaluating x 1 = · · · = x t = 1 and x i = 0 for i > t we obtain the chromatic polynomial χ G ( t ) . With the CSF , we can compute the number of connected components , compute the degree sequence for trees, etc... , but Figure: Non-isomorphic graphs with the same CSF 1 Conjecture (Tree conjecture - Stanley and Stembridge) Any two non-isomorphic trees T 1 , T 2 have distinct CSF . Think about the chromatic polynomial 1 Rose Orelanna and Scott Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 4 / 23

  5. Introduction CF on graphs CF on graphs - The kernel problem Question (The kernel problem on graphs) Compute generators of ker Ψ G . I.e. describe all linear relations of the form � a i Ψ G ( G i ) = 0 . i Theorem (RP-2017) The space ker Ψ G is spanned by the modular relations and isomorphism relations. Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 5 / 23

  6. Introduction CF on graphs Outline Introduction 1 CF on graphs Kernel problem on graphs 2 CF on polytopes 3 Generalised permutahedra Kernel problem on nestohedra Tree conjecture 4 Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 6 / 23

  7. Kernel problem on graphs Graphs terminology The edge deletion of a graph: H \ { e } . The edge addition of a graph: G + { e } . Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 7 / 23

  8. Kernel problem on graphs Modular relations � Ψ G ( G ) = x f . f proper on G Proposition (Modular relations - Guay-Paquet, Orellana, Scott, 2013) Let G be a graph that contains an edge e 3 and does not contain e 1 , e 2 such that the edges { e 1 , e 2 , e 3 } form a triangle. Then, Ψ G ( G ) − Ψ G ( G + { e 1 } ) − Ψ G ( G + { e 2 } ) + Ψ G ( G + { e 1 , e 2 } ) = 0 . Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 8 / 23

  9. Kernel problem on graphs The kernel problem For G 1 , G 2 isomorphic graphs, we have G 1 − G 2 ∈ ker Ψ G . These are called isomorphism relation . Theorem (RP-2017) The kernel of Ψ G is generated by modular relations and isomorphism relations. Let M = � modular relations , isomorphism relations � . Goal: ker Ψ G = M . Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 9 / 23

  10. Kernel problem on graphs Idea of proof - Rewriting graph combinations Condition to be a modular relation: e 3 ∈ G ⇒ G − ( G + { e 1 } ) − ( G + { e 2 } ) + ( G + { e 1 , e 2 } ) ∈ M . � Take z = G i a i in the kernel of Ψ G . i Goal: by working on ker Ψ G / M , show that z ∈ M . Some of the G i can be rewritten as graphs with more edges (through modular relation). We call them extendible . The non-extendible graphs { H 1 , H 2 , · · · } are not a lot, and { Ψ G ( H 1 ) , Ψ G ( H 2 ) , · · · } is linearly independent. Linear algebra ‘magic’ ⇒ a theorem is born. Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 10 / 23

  11. Kernel problem on graphs Idea of proof - Rewriting graph combinations e 3 ∈ G ⇒ G − ( G + { e 1 } ) − ( G + { e 2 } ) + ( G + { e 1 , e 2 } ) ∈ M . Proposition (Non-extendible graphs) A graph is non-extendible if and only if any connected component of G c , the complement graph of G , is a complete graph. Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 11 / 23

  12. Kernel problem on graphs Idea of proof - Linear algebra magic So, always working on ker Ψ G / M , we car rewrite: � K c z = λ a λ ∈ ker Ψ G , λ ∈P n Apply Ψ G to get � Ψ G ( K c 0 = λ ) a λ ⇒ a λ = 0 . λ ∈P n Possible to show: the set { Ψ G ( K c λ ) } λ ∈P n is linearly independent. So z = 0 , as desired. Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 12 / 23

  13. CF on polytopes Generalised permutahedra Polytopes Fix a dimension n . A polytope is a bounded set of the form q = { x ∈ R n | Ax ≤ b } . Given a colouring f : [ n ] → N of the coordinates , the face q f is n � q f = arg min x i f ( i ) . x ∈ q i =1 Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 13 / 23

  14. CF on polytopes Generalised permutahedra Polytopes: Examples Simplexes and its dilations: Consider J ⊆ [ n ] non empty. λ s J = conv { λe i | i ∈ J } . Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 14 / 23

  15. CF on polytopes Generalised permutahedra The permutahedron and its generalisations The n order permutahedron: per = conv { ( σ (1) , . . . , σ ( n )) | σ ∈ S n } . Is ( n − 1) -dimensional. Figure: The 4 -permutahedron 2 2https://en.wikipedia.org/wiki/Permutohedron Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 15 / 23

  16. CF on polytopes Generalised permutahedra The permutahedron and its generalisations A generalised permutahedron is a polytope q of the form      M �  M �  − M  , q = a J s J b J s J J � = ∅ J � = ∅ A nestohedron is only the positive part: M � q = a J s J . J � = ∅ Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 16 / 23

  17. CF on polytopes Generalised permutahedra Chromatic function and zonotopes We define the chromatic quasisymmetric function (CF) as � Ψ GP ( q ) = x f . q f =pt Given a graph G , its zonotope is defined as M � Z ( G ) = s e . e ∈ E ( G ) These are all Hopf algebra morphisms from the Hopf algebra GP = span { generalised permutahedra in R n , n ≥ 0 } . Also, Ψ G = Ψ GP ◦ Z . Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 17 / 23

  18. CF on polytopes Generalised permutahedra Some relations in nestohedra Proposition (Modular relations on nestohedra) Consider a nestohedron q , { B j | j ∈ T } a family of subsets on { 1 , · · · n } and { a j | j ∈ T } some positive scalars. Suppose “some magic”   M � T ⊆ J ( − 1) # T Ψ GP �  = 0 .  q + M a j s B j happens. Then, j ∈ T Additionally, there are also the so called simple relations - describe that we only care about which coefficients are positive, not how big they are. Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 18 / 23

  19. CF on polytopes Generalised permutahedra Some relations on nestohedra - Example An example of a modular relation: Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 19 / 23

  20. CF on polytopes Kernel problem on nestohedra K c π parallel and conclusion of proof Theorem (RP 2017) The modular relations, the isomorphism relations and the simple relations span the kernel of the restriction of Ψ GP to the nestohedra. Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 20 / 23

  21. Tree conjecture Tree conjecture on graphs The following: � � q # monochromatic edges in f of colour i χ ′ ( G ) = x f i f i is a graph invariant, where the sum runs over all colourings. If we consider the projection of this invariant modulo the relations q i ( q i − 1) 2 = 0 , then the modular relations are in ker χ ′ . We obtain ker Ψ G = ker χ ′ . Conjecture (Tree conjecture - χ ′ formulation) Any two non-isomorphic trees T 1 , T 2 have distinct χ ′ . Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 21 / 23

  22. Conclusion Further questions From nestohedra to generalised permutahedra? The image of the CF on graphs Ψ G is spanned by { Ψ G ( K c λ ) } λ , which forms a basis of im Ψ G . Combinatorial meaning of the coefficients? Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 22 / 23

  23. Conclusion Thank you Ra´ ul Penagui˜ ao (University of Zurich) Kernel problems July 16th, 2018 23 / 23

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