2.4 Embedding trees of unbounded maximum degree in a robust outexpander
2.4.6 Proof of Lemma 2.18
To prove Lemma 2.18, we first prove a stronger result for directed trees T whose core tree T∆ is large. This is the next lemma, which is also used in Chapter 3.
Lemma 2.32 Suppose that1/n 1/∆ µ ν η γ β 1. LetT be a directed
tree onnvertices such that|T∆| > βn, and letGbe a robust(µ, ν)-outexpander tournament on
at least(2− γ)nvertices, withδ0(G) > η|G|. ThenGcontains a copy ofT.
To prove Lemma 2.32, we apply Lemma 2.25 to find a subtree Text of T and a subset H ⊆
V (Text). Then we find a cluster cycle C in G such that |C| is slightly larger than |Text|. We then embedTextintoC using Lemma 2.27, restricting H to a set U of vertices of C which have many inneighbours and outneighbours outsideC. Finally we use this property of U to embed the vertices ofT − TextinV (G)\ V (C) and thereby complete the embedding.
Proof. We begin by introducing new constants∆∗, M, M0, ε, d and α which satisfy
1 n 1 ∆∗ 1 M 1 M0, 1 ∆ ε d µ ν η γ α β 1.
If|G| > 3n, then G contains a copy of T by Theorem 1.2. So we may assume that |G| < 3n. Now, since G is a robust (µ, ν)-outexpander with δ0(G) > η|G|, Lemma 2.10 implies that
G contains an ε-regular d-dense cycle of cluster tournaments on clusters V1, . . . , Vk each of equal size between (1− ε)|G|/k > (1 − ε)(2 − γ)m > 2(1 − γ)m and |G|/k 6 3m, where m := n/k and M0 6 k 6 M. So we may remove vertices from each Vi to obtain a2ε-regular
(d/2)-dense cycle of cluster tournaments G0 on clustersV10, . . . , Vk0 each of size2(1− γ)m. So
|G0| = 2(1 − γ)n. Let
δ := dαβ 160k.
Choose any vertext1 ∈ T∆as the root ofT . Then by Lemma 2.25, we may choose a subtree
TextofT and a subset H ⊆ V (Text) satisfying the following properties.
(i) T∆⊆ Text.
(ii) ∆(Text) 6 ∆∗.
(iii) For any edgee between T − TextandText, the endvertex ofe in Textlies inH.
(iv) The number of verticesv ∈ Textwhich satisfy1 6 d(v,Pk(H)) 6 k3 is at mostδβn.
(v) |H| 6 n/∆k1/δβ 6δβn/7k.
LetT1+, . . . , T+
r be the outcomponents of Text in T and let T1−, . . . , Ts− be the incomponents of Text in T . Then for each i let v+i be the vertex ofTi+ with an inneighbour in Text and let
vi− be the vertex ofTi−with an outneighbour in Text. By (i) and Proposition 2.19(iv) each Ti+ and eachTi− contains at mostn/∆ vertices. Let x = |Text|, let y = |T1+∪ · · · ∪ Tr+| and let
z =|T1−∪ · · · ∪ T−
s |, so x + y + z = n.
SinceT∆ ⊆ Text, we havex > βn. Also, all but at most 2y + x− αn/2 vertices of G have at leasty + x/2− αn/4 outneighbours, and all but at most 2z + x − αn/2 vertices of G have at
leastz + x/2−αn/4 inneighbours. So at least (2−γ)n−2y −2z −2x+αn > αn/2 vertices of G satisfy both of these conditions. Let U0 be the set of these vertices, so|U0| > αn/2, and each
v ∈ U0has at leasty + x/2− αn/4 outneighbours and at least z + x/2 − αn/4 inneighbours.
From each clusterV0
i ofG0 choose a setXiof(1 + α)x/k vertices uniformly at random, and let
X := X1∪ · · · ∪ Xk. Then|X| = (1 + α)x. For any single vertex u ∈ G0, the probability thatu is included inX is (1 + α)x/|G0| > x/2n, so by Proposition 2.6, with probability at least 2/3 the setU := X∩ U0 satisfies|U| > αx/5 > αβn/5. Also, for any vertex v ∈ U, the expected
number of outneighbours ofv outside X is at least y +x 2 − αn 4 1− (1 + α)x |G0| >y− αn 4 + x 2 − (1 + α)xy 2(1− γ)n − (1 + α)x2 4(1− γ)n >y− αn 4 + 2xn− 2xy − x2− 2γxn − 2αxy − αx2 4(1− γ)n >y + x 2 4n − 2αn > y + β2n 4 − 2αn > y + 2αn,
where in the first inequality of the third line we used the fact that2n− 2y − x > x. A similar
calculation shows that for eachv ∈ U, the expected number of inneighbours of v outside X is
at leastz + 2αn. So by Proposition 2.6 we find that with probability at least 2/3, every vertex v ∈ U has at least y + αn outneighbours outside X and at least z + αn inneighbours outside X. Fix a choice of X such that both these events of probability at least 2/3 occur.
Since every vertex ofU has either at least (|G| − |X|)/2 > y + z + αn inneighbours outside X or at least y + z + αn outneighbours outside X, we may choose a set U0 ⊆ U of size
|U0| > |U|/2 > αβn/10 such that either
(α1) every v ∈ U0 has at least y + αn outneighbours outside X and at least y + z + αn inneighbours outsideX, or
(α2) every v ∈ U0 has at least y + z + αn outneighbours outside X and at least z + αn inneighbours outsideX.
SoG0[X] is a (3ε/β)-regular (d/2)-dense cycle of cluster tournaments on clusters X
1, . . . , Xk of size(1 + α)x/k, and U0 ⊆ X
1 ∪ · · · ∪ Xk has size|U0| > αβx/10. Also Text is a directed tree onx vertices rooted at t1 and with∆(Text) 6 ∆∗, andH ⊆ V (Text) has|H| 6 δβn/7k 6
δx/7k and|{t ∈ Text: 1 6 d(t,Pk(H)) 6 k3}| 6 δβn 6 δx. So by Lemma 2.27 (with αβ/10,
∆∗ andd/2 in place of λ, ∆ and d respectively), G0[X] contains a copy of Textin which every vertex ofH is embedded to a vertex of U0.
In either case, let Vext be the set of vertices of G to which Text is embedded. We may now complete the embedding ofT in G. If U0satisfies (a), then we first proceed through the treesT+ i
in turn. For eachTi+, letu+i be the inneighbour of vi+ inText (so u+i ∈ H by (iii)). Then u+i has been embedded to some vertex v ∈ U0. This v ∈ U0 has at least y + αn outneighbours outsideVext, of which at mosty have been used for embedding the trees Tj+ forj < i. So there are at leastαn outneighbours of v outside Vext available to embed Ti+, and so since |Ti+| 6
n/∆ 6 αn/3, by Theorem 1.2 we can embed Ti+ among these vertices. In this way we may embed each of the T+
i . We then proceed through the Ti− similarly. For each Ti− let u−i be the inneighbour of v−i in Text (so u−i ∈ H by (iii)). Then u−i has been embedded to some vertex v ∈ U0. Thisv ∈ U0 has at least y + z + αn inneighbours outside Vext, of which at mosty + z have been used for embedding the trees T1+, . . . , T+
r and the trees Tj− for j < i. So there are at least αn inneighbours of v outside Vext available to embed Ti−, and so since
|T−
i | 6 n/∆ 6 αn/3, again by Theorem 1.2 we can embed Ti− among these vertices. IfU0 satisfies (b) we can embedT similarly, first embedding the Ti−, and then theTi+. Either way we
have completed the embedding ofT in G.
We can now give the proof of Lemma 2.18, which is restated below.
Lemma 2.18 Suppose that1/n µ ν η α, that G is a tournament on 2(1 + α)n
vertices which is a robust(µ, ν)-outexpander with δ0(G) > ηn and that T is a directed tree
onn vertices. Then G contains a copy of T .
Proof. Introduce new constants∆, γ and β with
1
n
1
If|T∆| > βn, then G contains a copy of T by Lemma 2.32. So we may assume that |T∆| < βn. As in the proof of the previous lemma, letT1+, . . . , T+
r be the outcomponents ofT∆ inT , and for eachi let vi+be the vertex of Ti+with an inneighbour inT∆. Similarly, letT1−, . . . , Ts− be the incomponents ofT∆inT , and for each i let vi−be the vertex ofTi−with an outneighbour in
T∆. By (i) and Proposition 2.19(iv) eachTi+and each Ti− contains at mostn/∆ vertices. Let
x =|T∆|, let y = |T1+∪ · · · ∪ Tr+| and let z = |T1−∪ · · · ∪ Ts−|, so x + y + z = n.
Now, all but at most2y + x + αn/2 vertices of G have at least y + x/2 + αn/4 outneighbours in G, and all but at most 2z + x + αn/2 vertices of G have at least z + x/2 + αn/4 inneighbours
in G. So at least 2(1 + α)n− 2y − 2z − 2x − αn = αn vertices of G satisfy both of these
conditions. Choose anyαn/8 of these vertices to form U0. Then|U0| = αn/8, and each v ∈ U0 has at leasty + x/2 + αn/8 outneighbours outside U0 and at leastz + x/2 + αn/8 inneighbours outside U0. Since every vertex v of G has either d+(v) > (1 + α)n− 1 > y + z + αn or
d−(v) > (1 + α)n− 1 > y + z + αn, we can choose a set U0 ⊆ U
0of size|U0| > αn/16 which satisfies either
(a) everyv ∈ U0 has at leasty + αn/8 outneighbours outside U0 and at leasty + z + αn/8 inneighbours outsideU0, or
(b) everyv ∈ U0 has at leasty + z + αn/8 outneighbours outside U0 and at leastz + αn/8 inneighbours outsideU0.
Since|T∆| < βn 6 |U0|/3, and G[U0] is a tournament, by Theorem 1.2 we may embed T∆ in
G[U0]. We may then proceed to embed all the T+
i and Tj− exactly as in the previous lemma,