On distinguishing special trees by their chromatic symmetric functions

# On distinguishing special trees by their chromatic symmetric functions

Melanie Gerling
July 13, 2019
###### Abstract

In 1995, Stanley introduced the well-known chromatic symmetric function of a graph . It is a sum of monomial symmetric functions such that for each vertex coloring of there is exactly one of these summands. The question, whether distinguishes nonisomorphic trees with the same number of vertices, is still open in general. For special trees it has already been shown. In 2008, Martin, Morin and Wagner proved it for spiders and some caterpillars. We decompose a tree by separating the leafs and their neighbors and do the same to the remaining forest until there remains a forest with vertices of degree not greater than . For nonisomorphic trees and with the same number of vertices and special properties concerning their number of leafs and leaf neighbors for each subgraph in their leaf decomposition we prove that the chromatic symmetric function distinguishes the graphs. Our idea is to find independent partitions of and with respectively a block of maximal cardinality for as well as for . These cardinalities are different for graphs and with special properties of their leaf decompositions. Additionally, we give explicit formulas for the cardinality of such a maximal block of an independent partition of star connections and spiders.

## 1 Introduction

Let be a set and let denote the power set of . For a natural number let . Because we will exclude certain types of graphs, a simplified definition of a graph suffices. We define a graph as a tuple of two sets and with . The members of are called vertices and the members of edges of . This definition excludes the occurrence of loops, i. e. members of , and multiple edges, which are irrelevant for our considerations. Two vertices are called adjacent to each other, if . For a vertex the number of all vertices that are adjacent to in is called the degree of in and denoted by .
A graph is called a subgraph of , if and . We also write . A subgraph is an induced subgraph of , if . For the graph is the subgraph induced by the set and for we set .
Let be a finite graph and . A function with a set of colors is called a vertex coloring. It is called a proper vertex coloring, if for all edges the relation holds. The number of all proper vertex colorings for a graph in at most colors was introduced as a polynomial by  and was called the chromatic polynomial by . Now, we turn to a similar function.

###### Definition 1

(, p. 168)
Let be (commuting) indeterminates and . Then the symmetric function generalization of the chromatic polynomial of is defined as

 XG(x1,x2,…)=∑κxκ(v1)xκ(v2)…xκ(vn)

where is a proper vertex coloring.

This function has the following connection with the chromatic polynomial.

###### Remark 2

(, p. 169)
For we have .

For the next Remark we need some definitions.

###### Definition 3

Let be a graph. We call a partition of a partition of . denotes the set of all partitions of . Each member of a partition is called a block.

We will only be interested in partitions which are defined as follows.

###### Definition 4

Non adjacent vertices and edges which don’t have any vertex in common are called independent of each other. A set or is said to be independent if its members are pairwise independent of each other.
An independent partition of denotes, as defined in , section 3, a partition of , such that each block is independent. We denote the set of all independent partitions by .

The next obersvation will be very useful to answer the question for special types of trees, whether two not isomorphic trees with the same number of vertices can be distinguished by their symmetric function generalization of the chromatic polynomial or not.

###### Remark 5

We can identify a vertex coloring of a graph with an independent partition of , such that the vertices of each block are colored in the same color and vertices from different blocks are colored in different colors. As Definition 1 shows, each summand contains a product , if there are vertices with , which means that they are in the same block. So, shows the cardinality of that block.

For special types of non-isomorphic trees with the same number of vertices it will be easy to show, that the cardinalities of their blocks with maximal cardinality are different from each other. So the chromatic symmetric functions of these trees differ of course, too.

We call a graph with vertex set and edge set with and pairwise distinct a path of length from to .
A non-empty graph is called connected, if for all there is a path from to , i. e. a path with and .
A graph without any cycles is called a forest, a connected forest is called a tree and a vertex of degree one is called a leaf.

###### Definition 6

Let be a tree, the vertices its leaves and all vertices that are adjacent to a leaf. Of course, and is a forest. For we do the same and continue in this manner. We get a subgraph relation , such that is a forest with vertex degrees . For each there are leaves and all possible vertices , that are adjacent to a leaf. If in the forest all vertices have degree , we have with . But if there are vertices of degree with , we set and . Note, that is always even. Then holds. For a tree we call such a relation a leaf decomposition and denote it by .

###### Remark 7

For every tree there is exactly one , , such that is a leaf decomposition of as in Definition 6.
Each vertice is of the form or with , and . So, is either a leaf or a neighbor of a leaf in the associated graph .

Note, that two non-isomorphic trees are not determined by their tree decomposition as the following example shows.

###### Example 8

In the following picture there are two non-isomorphic trees and with the same leaf decomposition. For we define the leaf decomposition and for the leaf decomposition . We have , and .

## 2 The number of leaves and further properties

###### Theorem 9

Let . For a tree with there is a maximum with , , such that for all leaves we have .

Proof:
For all leaves there is a vertex with , i. e. of course . From all such pairs we may choose at most one vertex for our block because of its independence and because of the maximality of we have to chose at least one. We choose always the leaf because the number of neighbors of leafs is equal to or smaller then the number of leafs. So we have for all leaves .
q.e.d.

The number of leaves can be an important hint to decide, whether two trees can be distinguished by their symmetric function generalization of the chromatic polynomial as the next theorem shows.

###### Theorem 10

Let and be trees with . The number of leaves of for is denoted by and . Let be the number of all vertices with for a leaf and let . Let the set of vertices which contribute to and let be a path. Furthermore, we assume one of the following cases:

1. ,

2. ,

3. and ,

4. and for all with and
.

Then .

Proof:
Because of Theorem 9 we compare independent partitions of and , such that there is respectively one block with maximum and for all leaves . We set

 Mi = max{|Bi||Bi∈σi∧σi∈ΠI(Ti)}.

For each case we will show . Note, that we can choose from the vertices exactly further vertices for the block because we did not choose any neighbor of a leaf and because of the path structure of .

1. Case is even. Because of and we have

 M1=b1+ρ12 > b2+ρ22=M2.
2. Case is odd. Like in the first case we get

 M1=b1+ρ1+12 > b2+ρ2+12=M2.
1. Case and are even. Because of and it leads to

 M1=b1+ρ12 > b2+ρ22=M2.
2. Case and are odd. Like above the following holds:

 M1=b1+ρ1+12 > b2+ρ2+12=M2.
3. Case is odd and is even. Similar as above we get

 M1=b1+ρ1+12 > b2+ρ22=M2.
4. Case is even and is odd. Like above and because of it leads to

 M1=b1+ρ12 > b2+ρ2+12=M2.
1. Case and are even. Hence,

 b1−b2 > ⌈ρ2−ρ12⌉ M1=b1+ρ12 > b2+ρ22=M2.
2. Case and are odd. Similary we get

 b1−b2 > ⌈ρ2−ρ12⌉ M1=b1+ρ1+12 > b2+ρ2+12=M2.
3. Case is odd and is even. Like above we have

 b1−b2 > ρ2−ρ12−12 M1=b1+ρ1+12 > b2+ρ22=M2.
4. Case is even and is odd. Thus,

 b1−b2 >⌈ρ2−ρ12⌉ b1−b2 > ρ2−ρ12+12 M1=b1+ρ12 > b2+ρ2+12=M2.
1. First we show that holds. As assumed, we have

 ρ2−ρ1 ≥ kk−2 ⇔(k−2)(ρ2−ρ1)−k ≥ 0 ⇔(2−k)ρ2+(k−2)ρ1−k ≥ 0 ⇔2ρ2−2ρ1 ≥ kρ2−kρ1+k ⇔ρ2−ρ1k ≥ ρ2−ρ12+12.
1. Case and are even. Then we get

 b1−b2 ≥ ⌈ρ2−ρ1k⌉≥⌈ρ2−ρ12⌉ M1=b1+ρ12 ≥ b2+ρ22=M2.
2. Case and are odd. It leads to

 b1−b2 ≥ ⌈ρ2−ρ1k⌉≥⌈ρ2−ρ12⌉ b1−b2 ≥ ρ2+12−ρ1+12 M1=b1+ρ1+12 ≥ b2+ρ2+12=M2.
3. Case is odd and is even. Similary as above we have

 b1−b2 ≥ ⌈ρ2−ρ1k⌉≥⌈ρ2−ρ12⌉ b1−b2 ≥ ρ2−ρ12−12 M1=b1+ρ1+12 ≥ b2+ρ22=M2.
4. Case is even and is odd. Hence,

 b1−b2 ≥ ⌈ρ2−ρ1k⌉≥⌈ρ2−ρ12⌉ b1−b2 ≥ ρ2−ρ12+12 M1=b1+ρ12 ≥ b2+ρ2+12=M2.

q.e.d.

We show an example for Theorem 10. It contains two graphs which are special cases of the graphs claimed in Theorem 10. This special graph type was already introduced by . They already proved that so-called caterpillars can be distinguished by their chromatic symmetric functions.

###### Example 11

Let . We consider the following two graphs. On the left side there is the graph and on the right side we see the graph . We have .

For the graph we have , and , while we have , and for the graph . Because of and Theorem 10 leads to as we can see easily.

Now, we will pay attention to a special situation which is interesting for our further considerations.

###### Situation 12

Let and be trees with and the leaf decompositions and . We set . Let , and for there exist respectively leafs and neighbor vertices like in Definition 6. Analogously, let with and for like in Definition 6.
We set and , where and or and become zero for from a certain or respectively on.

We turn to some special cases. The next theorem shows the first one.

###### Theorem 13

We regard Situation 12 for trees and and assume as well as for all . We except the case that there is equality everywhere. Then .

Proof:
By Theorem 9, for there is a partition with and maximal, such that for all leaves we have . Let as in the proof of Theorem 10. Then we have , for all . Because we have at least one strict inequality the relation holds because of and so .
q.e.d.

###### Example 14

Let . We consider the following two graphs. On the left side there is a tree and on the right side we see another tree . We have .

We get the following relations:

 b11=5 > b21=4 η11=2 < η21=4 b12=1 > b22=0 η12=0 = η22=0,

such that Theorem 13 holds. Indeed,

 |M1|=5+1=6 > |M2|=4+0=4.

In the following Theorem we introduce another case.

###### Theorem 15

We consider again Situation 12. Let , and for all and for all . Furthermore,

 is∑α=i1(b2α−b1α) < ∑j∈{1,…,r}∖{i1,…,is}(b1j−b2j) (1)

holds. Then we have .

Proof:
As in the proof of Theorem 13, we look at for with and maximal, such that for all leaves we have . Again, , for all . Because of inequality 1 we have and hence .
q.e.d.

## 3 Maximal blocks of independent partitions of star connections

In this section we will be concerned with maximal blocks of independent partitions of a special type of graph. It is constructed from stars, which are renownedly defined as follows.

###### Definition 16

A star for a natural number is a graph with vertex set and edge set .
The vertex is called the center of the star.

Now, we introduce two operations. The first one is a graph operation and the second one a set operation.
Let and be two graphs. Then we can create a new graph . Note, that the sets and (and thus of course possibly and ) are not necessarily disjoint.
An intersection of sets, such that the are pairwise distinct from each other, will be denoted by . For a special type of graph we give the following Definition.

###### Definition 17

Let be stars with vertices for with and . We define a tree with the following properties: Let be the set of all centers of stars in and

 {v1,…,vt}= ⎧⎨⎩v∈V∖C | ∃ Sni1,…,Snis with i1,…,is∈{1,…,r}, {v}=∙⋂l∈{i1,…,is}Snl⎫⎬⎭.

Then we assume for all pairwise distinct and that no center belongs to more than one star as a center or another vertex. Hence, for each we have and for with we have . Then we call the graph a star connection and denote it by .

Before we continue with the symmetric function generalization of the chromatic polynomial, we observe the following property of a star connection.

###### Theorem 18

Like in the proof of Theorem 10, we define

 M = max{|Bj||Bj∈σj∧σj∈ΠI(T)}.

In the situation of Definition 17 we have

 M = r∑k=1(nk−1)−t∑i=1(degT(vi)−1).

Proof:
We will prove the Theorem by complete induction.
For the graph is a star connection with exactly one common vertex , that is not a center of a star. Thus, . The greatest set of pairwise independent vertices is of course the set of all vertices without the two centers. Additionally, we may count only one time. Hence, for we have

 M = n1+n2−3 = n1+n2−2−(2−1) = 2∑k=1(nk−1)−1∑i=1(degT(vi)−1).

For the induction step we assume that the claim is true for an . Let . We want to connect with a star to a new star connection . Then there is a leaf which is also a leaf of stars in with and . We consider the maximal set of pairwise independent vertices in the graph . Because is not a center we have . By induction, for we get

 MT = r∑k=1(nk−1)−t∑i=1(degT(vi)−1).

All the other leaves of are not leaves of other stars. Hence, there are new vertices in that are contained in the maximal set of pairwise independent vertices in . We have .
Case 1: is identified in with a leaf of that is not a connection vertex in . We set . Then for all while and , such that

 MH = MT+(nr+1−2) = r∑k=1(nk−1)−t∑i=1(degT(vi)−1)+(nr+1−2) = r+1∑k=1(nk−1)−t∑i=1(degT(vi)−1)−1 = r+1∑k=1(nk−1)−t∑i=1(degH(vi)−1)−(degH(vt+1)−1) = r+1∑k=1(nk−1)−t+1∑i=1(degH(vi)−1).

Case 2: is identified in with a vertex that is not a center and not a leaf of . Then that vertex is already a connection vertex in , say . As in case 1, for all and , such that

 MH = MT+(nr+1−2) = r∑k=1(nk−1)−t∑i=1(degT(vi)−1)+(nr+1−2) = r+1∑k=1(nk−1)−t∑i=1(degT(vi)−1)−1 = r+1∑k=1(nk−1)−t∑i=1(degH(vi)−1).

q.e.d.

Now we will give an example for Theorem 18.

###### Example 19

We look at the graph with vertices, the number of connection vertices and the number of centers which is depicted below.

In this graph we have , and for the connection vertices we can see that . This leads to

 M = r∑k=1(nk−1)−t∑i=1(degT(vi)−1) = (3+4+2+3)−(1+1+1) = 9.

Now we can easily compare the symmetric function generalizations of the chromatic polynomial of two non-isomorphic star connections with the same number of vertices and get the following result.

###### Theorem 20

Let and two star connections with and . Then .

To prove Theorem 20 we need a Lemma.

###### Lemma 21

In the situation of Definition 17 the two following relations hold.

• .

• .

Proof:

• Because of we have just to sum up all for all and to substract all multiple countings of vertices that connect different stars.

• We show .
For each vertex with