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 |: XT = −p4 + 3p1p3 − 3p2

1p2 + p4 1

Stanley asks (1995): Does XT determine T?

Why is the Tree Conjecture so hard?

For a general tree, there are a lot of coefficients: pn p1pn−1 p2pn−2 p3pn−3 · · · p2

1pn−2

p1p2pn−3 p1p3pn−4 · · · p2

2pn−4

p2p3pn−5 · · · p2

3pn−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 pkpn−k terms. A slightly more general family of trees (so-called 2-spiders) can be distinguished by looking only at terms of the form pkpn−k, p1pkpn−k−1, and p4

1pn−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 XT, 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 XG :=

• F⊂E

 (−1)|F|

• K∈K(F)

p|V (K)|

 .

This is the subsets-of-edges formula. The important part: pn represents the power sum symmetric function pn :=

• k∈N

xn

k .

Reorganization

Since XG sits inside the ring ΛZ of symmetric functions, we can “reorganize” XG by changing basis. For example, we could use the elementary symmetric functions, which relates to some other open questions about XG: e3 := x1x2x3 + x1x2x4 + · · · + x1x3x4 + x1x3x5 + · · · + x2x3x4 + · · · 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: S4 := The chromatic symmetric function of a star is easy to write: sn := XSn =

n

• k=0

(−1)kpn−k

1

pk+1

• n

k

• .

The set {sn}∞

n=0 forms an algebraic basis—the star basis—for ΛZ.

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

• n the right, it either vanishes or becomes a leaf edge!

So, if we carry out unweighted deletion-contraction on each vertex

• f 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 s0) while applying rule (1) always introduces a factor of −s0. Thus no cancellation will occur when we apply UWDC to expand XT in the star basis.

An easy computation

= (0) − (1) + (2) Let T := Apply each rule to each non-leaf edge: XT = − + = s2

2 − s0s4 + s5.

In the p-basis (power sum symmetric function basis), XT = −p6 + 4p1p5 + p2

3 − 6p2 1p4 − 4p1p2p3 + 10p3 1p3 − 5p4 1p2 + p6 1.

A less easy computation

One can write an analogue to the subsets-of-edges formula for XT 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 XT will have 35 = 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)

= s2

6

Turn rule (2)s into rule (1)s (first separately on each component): 1022 → −s0s5 · s6 2012 → s6 · −s0s5 2011 → s6 · s2

0s4

A less easy computation, continued

= (0) − (1) + (2) Turn rule (2)s into rule (1)s (first separately on each component): 1022 → −s0s5 · s6 2012 → s6 · −s0s5 2011 → s6 · s2

0s4

Turning (2)s into (1)s on both components gives us a product s6,1s6,2 = (s6 − s0s5)(s6 − s0s5 + s2

0s4),

where sm,n :=

n

• k=0
• n

k

• (−s0)ksm−k.

A less easy computation, continued

= (0) − (1) + (2) That is, the term s6,1s6,2 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 sm,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, XT = s13,5 + s1,0s11,3 + s2,0s10,3 + s4,1s8,2 + s6,1s6,2 + · · ·

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 XT.

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).