Chromatic Bases and Stanleys Tree Conjecture Jake Huryn joint work - PowerPoint PPT Presentation

Chromatic Bases and Stanley’s Tree Conjecture Jake Huryn joint work with Noah Donald, Eric Fawcette, Rushil Raghavan, and Ishaan Shah under Sergei Chmutov The Ohio State University October 19, 2019

What is the Tree Conjecture? Take a tree T = ( V , E ): T : = Consider all the subgraphs induced by subsets of E : Forget the “internal structure” of each subgraph, treating each one as a partition of | V | : X T = − p 4 + 3 p 1 p 3 − 3 p 2 1 p 2 + p 4 1 Stanley asks (1995): Does X T determine T ?

Why is the Tree Conjecture so hard? For a general tree, there are a lot of coefficients: p n · · · p 1 p n − 1 p 2 p n − 2 p 3 p n − 3 p 2 · · · 1 p n − 2 p 1 p 2 p n − 3 p 1 p 3 p n − 4 p 2 2 p n − 4 p 2 p 3 p n − 5 · · · p 2 3 p n − 6 · · · . . . Specifically, the coefficients correspond to partitions of | V | .

Why is the Tree Conjecture so hard? The simpler partitions are easier to work with. For example, the family of spiders can be distinguished by looking only at p k p n − k terms. A slightly more general family of trees (so-called 2-spiders ) can be distinguished by looking only at terms of the form p 4 p k p n − k , p 1 p k p n − k − 1 , and 1 p n − 4 . Unfortunately, this approach is (without significant rethinking) doomed to failure, since it ignores the complicated “inner” coefficients. The dream: A way of reorganizing the information in X T , to get at the inner information indirectly.

The Chromatic Symmetric Function More formally: Let G : = ( V , E ) be a graph, and let K ( F ) denote the set of connected components of the graph ( V , F ) for any F ⊂ E . Then the chromatic symmetric function of G can be expressed as    ( − 1) | F | � � X G : = p | V ( K ) |  . F ⊂ E K ∈ K ( F ) This is the subsets-of-edges formula. The important part: p n represents the power sum symmetric function � x n p n : = k . k ∈ N

Reorganization Since X G sits inside the ring Λ Z of symmetric functions, we can “reorganize” X G by changing basis . For example, we could use the elementary symmetric functions , which relates to some other open questions about X G : e 3 : = x 1 x 2 x 3 + x 1 x 2 x 4 + · · · + x 1 x 3 x 4 + x 1 x 3 x 5 + · · · + x 2 x 3 x 4 + · · · The fundamental idea: Use a basis constructed from chromatic symmetric functions of trees !

Star Graphs A leaf of a graph is a vertex of degree one. A star is a tree with exactly one non-leaf vertex: S 4 : = The chromatic symmetric function of a star is easy to write: n � � n � ( − 1) k p n − k s n : = X S n = p k + 1 1 k . k = 0 The set { s n } ∞ n = 0 forms an algebraic basis— the star basis —for Λ Z .

Unweighted Deletion-Contraction The chromatic symmetric function satisfies the following unweighted deletion-contraction relation: = − + Note that the edge on the left is (possibly) a non-leaf edge—but on the right, it either vanishes or becomes a leaf edge! So, if we carry out unweighted deletion-contraction on each vertex of a graph, we get a sum of (chromatic symmetric functions) of stars.

Lush Trees = − + (0) (1) (2) For the remainder of the talk we suppose our trees are lush : every non-leaf vertex has a leaf attached. If T is lush, applying rules (0) or (2) does not introduce isolated vertices (i.e., factors of s 0 ) while applying rule (1) always introduces a factor of − s 0 . Thus no cancellation will occur when we apply UWDC to expand X T in the star basis.

An easy computation = − + (0) (1) (2) Let T : = Apply each rule to each non-leaf edge: = s 2 X T = 2 − s 0 s 4 + s 5 . − + In the p -basis (power sum symmetric function basis), X T = − p 6 + 4 p 1 p 5 + p 2 3 − 6 p 2 1 p 4 − 4 p 1 p 2 p 3 + 10 p 3 1 p 3 − 5 p 4 1 p 2 + p 6 1 .

A less easy computation One can write an analogue to the subsets-of-edges formula for X T in the star basis as a sum over tripartitions of the edge set, but this is messy and unenlightening. So, let’s find a nicer way to compute the star basis expansion. Suppose T : = T has 5 non-leaf edges, so the star basis expansion for X T will have 3 5 = 243 terms. Let’s not do that.

A new approach = − + (0) (1) (2) Here’s the idea: we first look at all ways of only doing rules (0) and (2). Then, looking at a particular way of doing only rules (0) and (2), we look at all possible ways of turning rule (2) into rule (1) .

A less easy computation, continued = − + (0) (1) (2) Apply only rule (0) and rule (2) to T (in the pattern 2022): rule (0) rule (2) = s 2 6 Turn rule (2)s into rule (1)s (first separately on each component): 1022 → − s 0 s 5 · s 6 2012 → s 6 · − s 0 s 5 2011 → s 6 · s 2 0 s 4

A less easy computation, continued = − + (0) (1) (2) Turn rule (2)s into rule (1)s (first separately on each component): 1022 → − s 0 s 5 · s 6 2012 → s 6 · − s 0 s 5 2011 → s 6 · s 2 0 s 4 Turning (2)s into (1)s on both components gives us a product s 6 , 1 s 6 , 2 = ( s 6 − s 0 s 5 )( s 6 − s 0 s 5 + s 2 0 s 4 ) , where n � � n � ( − s 0 ) k s m − k . s m , n : = k k = 0

A less easy computation, continued = − + (0) (1) (2) That is, the term s 6 , 1 s 6 , 2 corresponds to the following subset of edges (taking the subset to be the edges to which we apply rule (1)): Our new invariant, in terms of s m , n , is a sum over subsets of non-leaf edges. The first subscript number of edges in a given component and the second gives the number of non-leaf edges in this component. So, X T = s 13 , 5 + s 1 , 0 s 11 , 3 + s 2 , 0 s 10 , 3 + s 4 , 1 s 8 , 2 + s 6 , 1 s 6 , 2 + · · ·

Some nice things The condensed star-basis expression can be thought of as an amalgamation of the p - and star-basis expressions. It is strictly better than both in the number of terms needed. Ignoring the first subscripts of each term gives the p -basis expansion of the leaf-supressed graph. However, the question of whether the condensed expression can, a priori, be derived from the star basis is open.

Other bases Recall the unweighted deletion-contraction rule: = − + (0) (1) (2) Solving for term (2), we get a new rule that turns leaf-edges into non-leaf edges. Applying this new rule to each non-leaf edge gives the path basis expansion. Other bases that have been considered include those constructed from modified stars, hypergraphs, and more.

Other bases One of the main problems behind working with other bases is coming up with “combinatorial interpretations” analogous to the subsets-of-edges formula. This can be done much more generally than by using unweighted deletion-contraction. In particular, for certain bases, the combinatorial interpretation depends on an ordering of the vertices . In general, the idea is to consider a large family of bases and collect the information that each basis tells about X T .

References Richard P. Stanley. “A Symmetric Function Generalization of the Chromatic Polynomial of a Graph”. In: Advances in Mathematics 111.1 (1995), pp. 166–194. J. L. Martin, M. Morin, and J. D. Wagner. “On distinguishing trees by their chromatic symmetric functions”. In: Journal of Combinatorial Theory, Series A 115.2 (2008), pp. 237–253. S. Noble and D. Welsh. “A weighted graph polynomial from chromatic invariants of knots”. In: Annales de l’Institut Fourier 49.3 (1999), pp. 1057–1087. S. Cho and S. Willigenburg. “Chromatic bases for symmetric functions”. In: The electronic journal of combinatorics 23 (2015).