The core tree

Let T be a tree on n vertices, and let ∆ > 2 be fixed. Then we say that a vertex x of T

is ∆-core if every edge e incident to x has we(x) 6 (1 − 1/∆)n. We call the subgraph of

T induced by ∆-core vertices of T the core tree of T with parameter ∆, and denote it by T∆. With this definition, for any tree T , the core tree T∆ is the same as the ∆-heart of T considered by H¨aggkvist and Thomason in [14]. The following proposition states the most important properties of the core tree. These properties are also noted in Section 3 of [14], but we include the proof for completeness.

Proposition 2.19 LetT be a tree onnvertices and let∆ > 2. Then:

(i) T∆is a tree containing at least one vertex.

(ii) we(x) > n/∆ife = xy is an edge ofT∆.

(iii) ∆(T∆) 6 ∆.

Proof. For (i), note that since ∆ > 2, for any edge e = uv of T at most one of we(u) >

(1_{− 1/∆)n and w}e(v) > (1− 1/∆)n holds. Since T has more vertices than edges, there must therefore be some vertexv _{∈ T such that w}e(v) 6 (1− 1/∆)n for every edge e incident to v, and so v _{∈ T}∆. It remains to show that T∆ is connected. Observe that if u, v, w are distinct vertices of T such that there is an edge between u and v and an edge between v and w, then wuv(u) > wvw(v). Now, suppose x, y ∈ T∆, and letx = v1, v2, . . . , vr = y be the vertices of the path fromx to y in T (in order). Suppose for a contradiction that some vi is not inT∆. Then for some neighbourz of vi,wviz(vi) > (1− 1/∆)n. If z 6= vi+1, then for eachi 6 j 6 r− 1 we

havewvjvj+1(vj+1) > (1− 1/∆)n, and so y /∈ T∆, giving a contradiction. On the other hand, if

z = vi+1, then for each2 6 j 6 i, wv_j−1vj(vj−1) > (1− 1/∆)n, and so x /∈ T∆, again giving a

contradiction.

Now, (ii) is immediate from the fact that if e = xy is an edge of T then we(x) + we(y) = n. Then (iii) follows directly from (ii), as the sum ofwe(v) over all edges incident to v is n− 1.

Finally, for (iv), observe that for any suchT0 _{there isu∈ T}0_{,v ∈ T}

∆withe = uv an edge of T . Suppose that_|T0_{| > n/∆. Then w}e(v) > |T0| > n/∆, and so we(u) 6 (1− 1/∆)n. But since

we0(u) < w_e(v) 6 (1− 1/∆)n for every other edge e0 incident tou, this means that u ∈ T_∆,

giving a contradiction. 

Note thatT∆is an undirected tree obtained from an undirected treeT . However we often refer to the core tree of a directed treeT ; this means the directed tree formed by taking the core tree T∆ of the underlying graphTunder (an undirected tree) and directing each edge of T∆ as it is directed inT .

The idea behind this definition is that the core tree is a bounded degree tree. The general technique we use to work with a treeT of unbounded maximum degree (in both this and later

T − T∆, making use of the fact that each such component is small.

The following proposition is needed in the proof of Lemma 2.21, which we shall use in the next chapter. Essentially the latter states that if trees T1 _andT2 _{almost partition a treeT , then the} core treeT∆is not much larger thanT∆1 ∪ T∆2.

Proposition 2.20 Let T be a tree on n vertices, let x be a leaf of T, and let ∆ > 2. Then

|(T − x)∆| > |T∆| − 1.

Proof. Let y be a vertex of T∆ − (T − x)∆, and let z be an arbitrary vertex of (T − x)∆. Then for some edge e incident to y we have we(y) > (1 − 1/∆)(n − 1) in T − x. Since by Proposition 2.19(iv) the component of(T _{− x) − (T − x)}∆ containingy contains at most

(n_{− 1)/∆ vertices, this edge must in fact be the first edge of the path in T from y to z. If e is}

also the first edge of the path inT from y to x then we have we(y) > (1− 1/∆)(n − 1) + 1 >

(1− 1/∆)n in T , and so y /∈ T∆, giving a contradiction. Soy must lie on the path in T from x toz. Since y ∈ T∆we must havewe(y) 6 (1− 1/∆)n in T , and so in T we have

(1₋ 1 ∆)n− 1 6 (1 − 1 ∆)(n− 1) < we(y) 6 (1− 1 ∆)n.

Clearly this can hold for at most one vertexy on the path from x to z. So|T∆− (T − x)∆| 6 1,

as desired. 

Lemma 2.21 Let T be a tree onn vertices, let ∆ > 2 and let γ, α > 0. Also letT1 _andT2

be subtrees ofT such that _|T1 _{∪ T}2_{| > (1 − γ)n}. Suppose also that _|T∆1_{|, |T∆}2_{| 6 αn}. Then

|T∆| 6 γn + 2αn + 2n/∆.

Proof. Arbitrarily choose verticesx1 ∈ T∆1 andx2 ∈ T∆2, and letP be the path from x1 tox2 (soP is also a subtree of T ). Then let T∗ := T1_{∪ P ∪ T}2_{, so|T}∗_{| > (1 − γ)n. Furthermore,}

|T | − |T∗_{| > |T}

∆| − |T∆∗|, and so

|T∆| 6 |T | − |T∗| + |T∆∗| 6 γn − |P − (T1∪ T2)| + |T∆∗|. (2.22)

LetT∗

c := T∆1∪ P ∪ T∆2. We claim thatT∆∗ ⊆ Tc∗. Indeed, suppose for a contradiction that there exists a vertex y _{∈ T}∗

∆− Tc∗. Since Tc∗ is a subtree of T , every vertex of Tc∗ lies in the same componentC of T∗_{− y. Note that T}∗_{− C is a tree. Now, T}1

∆andT∆2 are subtrees ofC, so by Proposition 2.19(iv)T∗_{−C contains at most |T}1_{|/∆ vertices of T}1_{and at most|T}2_{|/∆ vertices} ofT2_{. Lete be the edge of T}∗_{betweeny and C. Then since y∈ T}∗

∆,we(y) 6 (1− 1/∆)|T∗| in

T∗. So at least_|T∗_{|/∆ vertices of T}∗lie in components ofT∗− y other than C. As every vertex

ofP lies in C, either at least|T1_{|/∆ vertices of T}1_{lie in components ofT}∗_{− y other than C, or} at least_|T2_{|/∆ vertices of T}2 lie in components ofT∗− y other than C. In the former case this

implies thatT∗_{− C contains more that |T}1_{|/∆ vertices of T}1_{, and in the latter case this implies} thatT∗_{− C contains more that |T}2_{|/∆ vertices of T}2_{. In either case this yields a contradiction.}

Now,_|Tc∗_{| 6 2αn+|P −(T∆}1_∪T∆2)|. Since (P ∩T1_)−T1

∆is contained within a single component ofT1_{− T}1

∆,|(P ∩ T1)− T∆1| 6 |T1|/∆, by Proposition 2.19(iv). Similarly |(P ∩ T2)− T∆2| 6

|T2_{|/∆. So} |T∆∗| 6 |Tc∗| 6 2αn + |T 1_{| + |T}2_| ∆ +|P − (T 1_{∪ T}2_)|. So by (2.22) |T∆| 6 γn + |T 1_{| + |T}2_| ∆ + 2αn 6 γn + 2n ∆ + 2αn.