compact topological minors in graphs

Compact TopologicalMinors in Graphs

Tao Jiang

DEPARTMENT OF MATHEMATICSMIAMI UNIVERSITY, OXFORD, OH 45056

E-mail: [email protected]

Received May 29, 2009; Revised May 6, 2010

Published online 2 September 2010 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jgt.20522

Abstract: Let � be a real number such that 0<�<12 and t a positive

integer. Let n be a sufficiently large positive integer as a function of t and �.We show that every n-vertex graph with at least n1+� edges contains asubdivision of Kt in which each edge of Kt is subdivided less than 10/�times. This refines the main result in [A. Kostochka and Pyber, Combina-torica 8 (1988), 83–86] and resolves an open question raised there. We alsopose some questions. � 2010 Wiley Periodicals, Inc. J Graph Theory 67: 139–152, 2011

Keywords: topological clique; Turan number; subdivision; minor

1. INTRODUCTION

We consider only simple graphs in this article unless otherwise specified. Our workis motivated by the Turan problem and the study of graph minors. In extremalgraph theory we generally study how edge density affects the existence of certainsubstructures. The classic Turan number ex(n,H) of a family of graphs H is definedas the maximum number of edges in an n-vertex graph not containing any member

Contract grant sponsor: National Security Agency; Contract grant number:H98230-07-1-027.Journal of Graph Theory� 2010 Wiley Periodicals, Inc.

139

140 JOURNAL OF GRAPH THEORY

of H as a subgraph. When H consists of a single graph H, we simply write ex(n,H)for ex(n,H). Turan determined the exact value of ex(n,Kr). The celebrated Erdos–Stone–Simonovits Theorem determined ex(n,H) asymptotically for all non-bipartiteH, showing that ex(n,H)∼ (1−(1 / (p−1)))

(n2

), where p=�(H) is the chromatic

number of p.Subdividing an edge uv in a graph, as usual, means replacing uv by a path uwv

through a new vertex w. A graph G is a subdivision of another graph H if it can beobtained from H via edge subdivisions. For instance, a cycle of any length may beviewed as a subdivision of a C3. If a graph G contains a subgraph that is a subdivisionof H, we also say that G contains H as a topological minor. The name topologicalminor derives from another notion called minor. A graph G contains another graph Has a minor if G has a subgraph L that can be turned into H by contracting its edgesand deleting redundant multiple edges and loops.

Mader was among the first to explore Turan-type questions for minors and subdi-visions. We have seen from the Erdos–Stone–Simonovits Theorem that to force anygiven non-bipartite H to appear as a subgraph of an n-vertex graph G, we need �(n2)edges. Indeed, Kn/2,n/2 has n2 /4 edges and avoids any non-bipartite H as a subgraph.Contrasting this phenomenon for subgraphs, Mader showed that it only takes O(n)edges in an n-vertex host graph to force a subdivision of H, as implied by the followingtheorem of his.

Theorem 1.1 (Mader [5]). Given a positive integer t, there exists a positive constantct depending only on t such that every n-vertex graph G with at least ctn edges containsa subdivision of Kt.

Bollobas and Thomason [3] and independently Komlos and Szemeredi [12] laterdetermined the growth rate of the optimal constant ct, showing that Theorem 1.1 holdsfor some ct ≤ct2, where c is an absolute constant.

Turan’s theorem and Mader’s theorem indicate that there is a big gap betweenrequired edge-densities for forcing a graph H directly as a subgraph versus as a topo-logical minor (subdivision), with the former much higher. What seems to make it easierto force a subdivision of H is that we allow as many subdivisions on an edge of Has needed and this gives us a lot of leeway. This is reminiscent of forcing cycles: it ismuch easier to force a cycle of length at least a given length k than to force a cycle oflength at most k; for the former O(n) edges suffice while for the latter �(n1+1/(k+1))edges are required (see [2] Chapter 3).

This brings up a natural question: what if we require that each edge in the originalH is only allowed to be subdivided some bounded number p times? What edge densityis needed in a host graph G to force such a more restricted/compact subdivisionof H?

Definition 1.2. Let p be a positive integer. A subdivision L of a graph H is called ap-subdivision of H if in forming L each edge of H is subdivided at most p−1 times.

Note that a 1-subdivision of H is just H itself. We pose the following generalquestion.

Question 1.3. Fix a positive integer p and a graph H, how many edges does ann-vertex graph G need to have to force a p-subdivision of H?

Journal of Graph Theory DOI 10.1002/jgt

COMPACT TOPOLOGICAL MINORS IN GRAPHS 141

We may rephrase Question 1 in terms of Turan numbers as follows. Given a positiveinteger p and a graph H, let H(≤p) denote the family of p-subdivisions of H (i.e. graphsobtainable from H by subdividing each edge of H at most p−1 times).

Problem 1.4. Given positive integers p, n and a graph H, determine ex(n,H(≤p)).

In this article, we focus on the case where H =Kt, the complete graph, as it providesnatural bounds for the general H. Another motivation for our work in this article comesfrom the following work of Kostochka and Pyber [14]. It is well known that anyn-vertex graph with more than 3n−6 edges contains a non-planar subgraph. However,if we want to force a non-planar subgraph of a bounded (constant) order then it is easyto see that O(n) edges are not sufficient. Erdos [5] asked the following: Is it true thatevery n-vertex graph with at least n1+� edges contains a non-planar subgraph of orderat most c(�), where c(�) is a constant depending only on �? Kostochka and Pyber [14]answered the question in the affirmative, proving the following more general result(with the t=5 case answering Erdos’s question).

Theorem 1.5 (Kostochka and Pyber [14]). Let � be a positive real such that 0<�<1.Let n, t be positive integers. Every n-vertex graph with at least 4t2n1+� edges containsa subdivision of Kt on at most 7t2 ln t /� vertices.

It is well known that for each 0<�<1 there are n-vertex graphs with �(n1+�) edgesand girth at least 1 /� (see [2] Chapter 3). In such a graph, any subdivision of Kt, wheret≥3, must contain at least �(t2 /�) vertices. Kostochka and Pyber [14] asked whethertheir O(t2 ln t /�) bound can be improved to the optimal O(t2 /�). We answered thisquestion in the affirmative in our main theorem below, which also makes a nontrivialprogress toward Problem 1.4. For �≥ 1

2 , Proposition 2.3 in the next section alreadyshows every n-vertex graph with �(n3/2) edges contains a 2-subdivision of Kt (whichis best possible). We thus restrict our attention to �< 1

2 .

Theorem 1.6. Let t be a positive integer and 0<�< 12 be a real. There exists a function

n0(t,�) such that for all integers n≥n0(t,�), if G be an n-vertex graph with at leastn1+� edges, then G contains a �10/��-subdivision of Kt.

Remark 1.7. Due to the girth result mentioned above, the �10/�� bound cannot beimproved to be smaller than � 1

3 (�1/��−1)� (see Proposition 1.9 below).

Theorem 1.6 and Proposition 2.3 yield

Corollary 1.8. Let p≥2, t≥3 be fixed positive integers. We have ex(n,K(≤p)t )=

O(n1+(10/p)).

As mentioned before, there are n-vertex graph with �(n1+1/g) edges and girth atleast g. On the other hand, a p-subdivision of Kt with t≥3 contains a cycle of length atmost 3p. So, there are n-vertex graphs with �(n1+(1/(3p+1))) edges and no p-subdivisionsof Kt. Hence, we have

Proposition 1.9. Let p≥2, t≥3 be fixed positive integers. As a function of n, we haveex(n,K(≤p)

t )=�(n1+(1/(3p+1))).



In this article, we prove Theorem 1.6. We will incorporate ideas in [10] and [14] andsome fresh new ideas. The rest of the article is organized as follows. In Section 2, weprove some preliminary lemmas. In Section 3, the most technical section, we solve themain case. Then in Section 4, we put all the pieces together. We conclude the articlewith some remarks.

2. PRELIMINARIES

The study of 2-subdivisions is already implicit in the work of Alon et al. [1] and inFuredi [11]. In particular, Alon et al. proved the following more general result.

Proposition 2.1 (Alon et al. [1]). Let H be a bipartite graph with maximum degreer on one side. Then exists a constant cH>0, depending on H, such that ex(n,H)≤cHn2−(1/r).

From the proof of Theorem 2.2 in [1], one can easily check that cH ≤n(H) whenr=2. Thus, we have

Corollary 2.2. Let H be a bipartite graph with h vertices such that vertices in onepartite set all have degree at most 2, then ex(n,H)≤ (h /2)n3/2.

Let K(2)t denote the graph obtained from Kt by subdividing each edge exactly once.

Observe that K(2)t is a bipartite graph on t+( t

2

)<t2 vertices in which the vertices used to

subdivide the edges form one part X and the branching vertices form the other part Y .Further, each vertex in X has degree 2. By Corollary 2.2, we have

Proposition 2.3. For each positive integer t, we have ex(n,K(≤2)t )≤ex(n,K(2)

t )≤(t2 /2) ·n3/2.

Proposition 2.3 gives the correct order of magnitude for ex(n,K(≤2)t ) and ex(n,K(2)

t )because of the well-known fact that there are n-vertex graphs with �(n3/2) edges andno C3 or C4. By Proposition 2.3, for �≥ 1

2 every n-vertex graph with at least (t2 /2)n1+�

edges contains a 2-subdivision of Kt. Thus, we henceforth restrict our attention to �< 12 .

The next two technical lemmas will be used in Section 3.

Lemma 2.4. Let a,m,q be positive integers. Let A1, . . . ,Am be a collection of sets ofsize a. Suppose each element of A=⋃

i Ai lies in at most q different Ai’s. Then for eachi∈ [m], there exists Bi ⊆Ai of size �a /q� such that B1, . . . ,Bm are pairwise disjoint.

Proof. Let p=�a /q�. Create a bipartite graph H with a bipartition (X,A) where|X|=mp as follows. Label the vertices of X by x1

1, . . . ,xp1,x1

2, . . . ,xp2, . . . ,x1

m, . . . ,xpm. For

each i∈ [m] and y∈A, if y∈Ai then we add edges between y and x1i , . . . ,xp

i . By ourconstruction, each vertex in X has degree a. Also, since each y∈A lies in at mostq different Ai’s, each vertex in A has degree at most pq≤a in H. By Hall’s Theorem,H has a matching M saturating all of X. For each i∈ [m], the elements of A that x1

i , . . . ,xpi

are matched to by M are elements of Ai. Hence, we obtain disjoint B1, . . . ,Bm each ofsize p=�a /q�, with Bi ⊆Ai for each i∈ [m]. �



The depth of a vertex x in a rooted tree T is the distance between the root and x.The depth of the tree is the maximum depth of a vertex in T . If T ′ is a tree rooted atu in which for every pair of leaves x and x′ the unique x,x′-path goes through u, thenwe call T ′ a spider with center u. The paths from u to the leaves are called the legs ofthe spider.

Lemma 2.5. Let p,m be positive integers. Let T be a tree rooted at u of depth m. LetS be a set of pm vertices in V(T)−u. Then there exists a subset S′ ⊆S of size p anda spider T ′ in T whose set of leaves is S′. Furthermore, each leg of T ′ has length atmost m.

Proof. We use induction on the depth m. The claim holds trivially when m=1.For the induction step, assume that m>1 and that the claim holds for trees of depthat most m−1. For each child x of u, let Tx denote the subtree of T under x. If forat least p different children x of u, V(Tx)∩S �=∅, then let S′ be a set of p verticesof S, each from a different Tx. The union of the paths from u to these verticesforms a spider with center u whose set of leaves is S′. Each leg clearly has length atmost m.

Next, we may assume that |V(Tx)∩S| �=∅ for fewer than p children x of u. Then forsome child z of u, |V(Tz)∩S−z|≥|S| / (p−1)−1≥pm−1. Since Tz is a rooted tree ofdepth m−1. We can apply induction hypothesis to find the desired S′ and T ′. �

3. FORCING COMPACT SUBDIVISIONS IN GRAPHS WITH NO

DENSE SUBGRAPHS OF SMALL RADIUS

To prove Theorem 1.6, we first prove in this section that if a graph is reasonably denseitself but contains no large dense subgraph of small radius, then we can find a compactsubdivision of Kt. The following notion will be used frequently in our proofs.

Definition 3.1. Let c,� be positive reals where 0<�<1. A graph H is called (c,�)-dense if e(H)≥c[n(H)]1+�.

In our proofs, we will often use the well-known fact that a graph G satisfyinge(G)≥d ·n(G) contains an induced subgraph of minimum degree at least d. The nexttechnical theorem provides the most crucial ingredient of our proof of Theorem 1.6.To facilitate our proof, we need more definitions. Let p,q be two positive integers. Twovertices u,v in a graph are (p,q)-linked if there exist p internally disjoint u,v-paths oflength at most q. We will also say that u is (p,q)-linked to v and vice versa. Giventwo disjoint subsets A,B⊆V(G), we use e(A,B) to denote the number of edges in Gwith one end in A and the other end in B.

Theorem 3.2. Let c,� be positive reals where 0<�< 12 and c≥ 1

5 . Let D be a positivereal depending on �. Let p be a positive integer. Let l(�)= 1/4��+ 1.8 /��+ 1/��+1.There exists a function n1(�,p) such that for all integers n≥n1(�,p) if G is an n-vertexgraph satisfying that (1) �(G)≥cn�, (2) �(G)≤Dn�, and (3) G contains no (c /27,�)-dense subgraph that has radius at most l(�) and order at least n0.9�, then G containsat least n1.51 pairs of vertices u,v that are (p, l(�))-linked.



Proof. First we note that the lack of large dense subgraph of small radius forcesG to have good expansion properties. Let u be any vertex in G. For each i, let Ui denotethe set of vertices at distance at most i from u. In particular, U0 ={u}.Claim 1. We have |U1|≥cn�. For each 1≤ i≤ l(�)−1, we have |Ui+1|≥10n�/(1+�) ·|Ui|1/(1+�).

Proof of Claim 1. Since �(G)≥cn�, we have |U1|≥cn�. For i≥1, there are at least12 |Ui|·cn� edges in G incident to Ui, all of which lie in G[Ui+1]. Since G[Ui+1] hasradius at most i+1≤ l(�) and order at least |U1|≥cn� ≥n0.9�, by our assumption, it isnot (c /27,�)-dense. So

1

2|Ui|·cn� ≤e(G[Ui+1])≤ c

27 |Ui+1|1+�.

Solving the inequality for |Ui+1| and using 26/(1+�) ≥10, we get the claim. �

Define a sequence b1, . . . ,bl(�) by letting b1 =� and for each 2≤ i≤ l(�), bi =� / (1+�)+(1 / (1+�))bi−1 =bi−1 +(� / (1+�))(1−bi−1). We have |U1|≥cn� ≥ 1

5 nb1 . For i≥2, byClaim 1, one can show that |Ui|≥2nbi . Solving the recurrence for bi, we get

bi =1− 1−b1

(1+�)i−1=1− 1−�

(1+�)i−1,

for 1≤ i≤ l(�). Let m=min{i : |Ui|≥2n0.2}. If �≥0.21, then clearly m=1 for largeenough n. Suppose 0<�<0.21. Then it is easy to check that 1+�>e0.9�. We have

bi =1− 1−�2

(1+�)i≥1− 1

(1+�)i≥1− 1

e0.9�i .

When i= 1/4��, we have bi ≥1−(1 /e0.9�/4�)≥0.2. Hence, m≤ 1/4��. Note that thisalso holds when �≥0.21 as m=1 in this case.

Let Au =Um −Um−1. So, Au is the set of vertices at distance exactly m from u.If m=1, then clearly |Au|=|Um|−1≥n0.2. Suppose m≥2. By Claim 1, |Um|≥10n�/(1+�)|Um−1|1/(1+�) ≥10|Um−1|�/(1+�) ·|Um−1|1/(1+�) ≥10|Um−1|. Hence |Au|≥9

10 |Um|≥n0.2. Let Tu be a tree rooted at u of depth m that results from performing thebreath-first search from u until all vertices at distance at most m from u are included.In particular, Au is a set of leaves of Tu.

The plan for the rest of the proof is to show that vertices in Tu collectively are (p, l(�))-linked to many vertices in G. Toward that goal, we will make use of good expansionproperties of G. First, we need the following claim, which generalizes Claim 1 and hasa similar proof.

Claim 2. Let H be a subgraph of G of radius at most l(�)−1, where n(H)≤2na anda<1. Let W be a set of neighbors of V(H) in G outside V(H) such that e(W,V(H))≥(c /16)na+�. Then |W|≥2na+(�/(1+�))(1−a).

Proof of Claim 2. Let F denote the subgraph of G induced by V(H)∪W. Since Hhas radius at most l(�)−1, F has radius at most l(�). Also, since n(H)≤2na and V(H)



sends at least (c /16)na+� edges to W, we have n(F)≥|W|≥ (c /32)n� ≥n0.9�. By ourassumption about G, F is not (c /27,�)-dense. Suppose n(F)=nb. We have

c

16na+� ≤e(F)≤ c

27 (nb)1+� (1)

From this, we get

nb ≥81/(1+�)n(a+�)/(1+�) ≥4na+ �1+� (1−a).

Hence, |W|=n(F)−n(H)≥4na+ �1+� (1−a)−2na ≥2na+ �

1+� (1−a). �

We iteratively define a sequence of disjoint sets L1,L2, . . . ,Ll(�)−m+1 starting withL1 =Au. During our procedure, we will maintain the following conditions:

(1) Each Li is called a level and will be designated as strong or weak. Set L1 willbe strong.

(2) Each Li is partitioned into some di subsets Lji called sectors of equal size. If Li

is a strong level then each sector consists only of a single vertex and di =|Li|≥|Li−1|≥di−1. If Li is a weak level, then 1

8 di−1 ≤di ≤di−1.(3) Each vertex in a strong level Li, i>1, has neighbors in at least r=pm sectors of

Li−1.(4) If Li is a weak level, then there exists an injection f from the collection of sectors

of Li into the collection of sectors of Li−1 such that if Lji is a sector of Li then

each vertex in Lji has at least one neighbor in f (Lj

i). We call f (Lji) the parent

sector of Lji in Li−1. Furthermore, |Lj

i|≥2|f (Lji)|.

(5) For each i, suppose |Li|=nai and di =n�i . Then |Li+1|≥2|Li|·n(�/(1+�))(1−ai).Thus, ai+1 ≥ai +(� / (1+�))(1−ai). Further, if Li+1 is a weak level, then |Li+1|≥|Li|·n(�/(1+�))(1−ai+0.99�i).

To start, let L1 =Au. Designate L1 as a strong level and designate each vertexin L1 as a sector on its own. Fix 2≤ i≤ l(�)−m, and suppose for all j≤ i we havedefined Lj that satisfy (1)–(5). Let Ui−1 =V(Tu)∪L1 ∪·· ·∪Li−1. By (5), for each j≤ i,|Lj|≥2|Lj−1|·n(�/(1+�))(1−aj−1) ≥2|Li−1|. Also, |L1|=|Au|≥9|V(T)−Au|≥2|V(T)−Au|.So, |Ui−1|≤|Li−1|(1+ 1

2 + 122 +·· ·)≤2|Li−1|≤|Li|. Let Hi denote the subgraph of G

induced by Ui−1 ∪Li. Then n(Hi)≤|Ui−1|+|Li|≤2|Li|, and from (3) and (4) it is easyto see that Hi has radius at most i+m−1≤ l(�)−1. Also, n(Hi)≥|V(Tu)|≥cn� ≥n0.9�.

Let d denote the minimum degree of G. We have d≥cn�. We say a vertex x in Liis bad if at least d /2 of its neighbors lie inside Hi. Otherwise we say it is good. Notethat a good vertex has at least d /2 neighbors outside Hi. Let Bi denote the set of badvertices in Li and Ci the set of good vertices in Li. If |Bi|> 1

4 |Li|, then |Bi|≥ 18 n(Hi) and

e(Hi)≥ 12 |Bi|·d /2≥ 1

16 n(Hi) ·(c /2)n� ≥ (c /32)[n(Hi)]1+�, contradicting G containing no(c /27,�)-subgraph with radius at most l(�) and order at least n0.9�. Hence |Bi|≤ 1

4 |Li|and |Ci|≥ 3

4 |Li|.Let W denote the set of vertices outside Hi that have neighbors in Li. We now

analyze the vertices in Li ∪W and the edges between Li and W. We say a vertex y inW is heavy if it has neighbors in at least r=pm different sectors Lj

i of Li. Otherwisewe say y is light. Let W+ denote the set of heavy vertices in W and W− the set oflight vertices in W.



Recall that each vertex x in Ci sends at least d /2 edges to W. If at least d /4 of theseedges go to W+ we say that x is heavy-leaning. Otherwise at least d /4 of these edgesgo to W− and we say that x is light-leaning. Let C+

i denote the set of heavy-leaningvertices in Ci and C−

i the set of light-leaning vertices in Ci. We consider two cases.Recall that |Li|=nai .

Case 1. |C+i |≥|Ci| /2.

In this case, we have |C+i |≥ 3

8 |Li|. We have e(V(Hi),W+)≥e(C+i ,W+)≥ 3

8 |Li|·d /4≥(3c /32)nai+�. Since n(Hi)≤2|Li|=2nai and Hi has radius at most l(�)−1, by Claim 2,|W+|≥2nai+(�/(1+�))(1−ai) =2|Li|·n(�/(1+�))(1−ai). Let Li+1 =W+ and designate Li+1 asa strong level. Let each vertex be a sector by itself. It is straightforward to see thatLi+1 satisfies (1)–(5).

Case 2. |C−i |≥|Ci| /2.

This case is more involved. Our plan is to define Li+1 inside W−. Roughlyspeaking, sectors of Li+1 will be defined using neighborhoods of certain sectorsof Li in W−. We will use Lemma 2.4 to handle the overlaps between theseneighborhoods.

First, we decide which sectors of Li to use. We know |C−i |≥|Ci| /2≥ 3

8 |Li|. Let

J ={j : |Lji ∩C−

i |≥|Lji| /4}. Since all the di sectors Li

j’s of Li have the same size and∑

j |Lji ∩C−

i |≥ 38 |Li|, |J|≥di /8. We will use the neighborhoods of Lj

i in W− for thosej∈J.

For each j∈J, let Wji =N(Lj

i)∩W−. Suppose |Lji|=nai,j . First we argue that there is

a subgraph Tji of order at most 2nai,j and radius at most l(�)−1 containing Lj

i. If Li is a

strong level then this is trivial, since Lji consists of a single vertex and we can let Tj

i be

that vertex. If Li is a weak level, then by (4), each vertex in Lji has some neighbor in its

parent sector f (Lji) in Li−1. Furthermore, |f (Lj

i)|≤ 12 |Lj

i|. We can continue backtracking,moving up the levels, until we hit a strong level (where a sector is just a single vertex).This gives us a subgraph Tj

i of order at most |Lji|(1+ 1

2 + 122 +·· ·)≤2nai,j that has radius

at most i≤ l(�)−1. Now, since j∈J, we have e(Lji,W

ji )=e(Lj

i,W−)≥|Lj

i ∩C−i |·(d /4)≥

14 |Lj

i|·(d /4)≥ (c /16)nai,j+�. By Claim 2, |Wji |≥2nai,j+(�/(1+�))(1−ai,j). Let Yj

i be a subset

of Wji of size 2nai,j+(�/(1+�))(1−ai,j). Since |Lj

i|=2nai,j is the same for all j, |Yji | is the

same for all j∈J.Note that

⋃i∈J Yj

i ⊆W− and by the definition of W− each element of W− lies in at

most r different Yji ’s. By Lemma 2.4, for each j∈J, there exists a subset Aj

i ⊆Yji of size

|Yji | /r such that the Aj

i’s so obtained are pairwise disjoint. We let Li+1 =⋃j∈J Aj

i. Let

di+1 =|J|. The Aji’s for j∈J form a partition of Li+1 into di+1 many subsets of equal

size. We define them to be the sectors of Li+1 and designate Li+1 as a weak level. Itremains to verify that (2), (4), and (5) hold for Li+1. We have di+1 =|J|≥ 1

8 di. So (2)

holds. For (4), for each j∈J, we let f (Aji)=Lj

i. Finally, we rename the Aji’s, j∈J as

L1i+1, . . . ,Ldi+1

i+1 .

Recall that |Li|=nai and di =n�i . By (2), di ≥d1 /8i−1 ≥n0.199, for large n. So, �i ≥0.199. For each j∈J, |Li

j|=|Li| /di =nai−�i . Thus, ai,j =ai −�i. Since |J|≥di /8, and

for each j∈J, |Aij|=2nai,j+(�/(1+�))(1−ai,j) /r=2|Li

j|·n(�/(1+�))(1−ai,j) /r, we have |Li+1|≥di /



8r ·2|Lij|n(�/(1+�))(1−ai,j) = (1 /4r)|Li|· n(�/(1+�))(1−ai +�i) ≥|Li|·n(�/(1+�))(1−ai +0.99�i), for

large n. Thus, (5) holds for Li+1.We have now constructed the sets L1,L2, . . . ,Ll(�)−m+1 that satisfy conditions (1)–(5).

Note that by (5), |Li| is non-decreasing in i.

Claim 3. Each vertex x in a strong level Li is joined to Au by at least r paths oflength i−1 every two of which share only x in common.

Proof of Claim 3. By (3), x has neighbors in at least r sectors of Li−1. Lety1

i−1, . . . ,yri−1 be neighbors of x in Li−1, all from different sectors. If Li−1 is a weak

level, then by condition (4), the r sectors involving y1i−1, . . . ,yr

i−1 all have different

parent sectors in Li−2. So, for each j∈ [r], we can find a neighbor yji−2 of yj

i−1 in Li−2

so that y1i−2, . . . ,yr

i−2 are from different sectors of Li−2. If Li−1 is a strong level instead

and i−1>1, then by (3) each yji−1, j∈ [r], has neighbors in at least r different sectors of

Li−2. In this case, it is easy to see that we can still find y1i−2, . . . ,yr

i−2, all from different

sectors of Li−2, such that yji−2 is a neighbor of yj

i−1 for each j∈ [r]. We can continuethis process and build r paths of length i−1 from x back to Au =L1 so that every twoof them share only x in common. �

Claim 4. Each vertex x in a strong level Ls is (p,m+s−1)-linked to some vertex win Tu −Au.

Proof of Claim 4. Recall that Tu is a tree rooted at u of depth m≤ 1/4�� withAu =L1 being its set of leaves. By Claim 3, x sends r internally disjoint paths P1, . . . ,Prof length s−1 to r different vertices y1, . . . ,yr ∈Au, respectively.

Let S={y1, . . . ,yr}. Since r=pm and Tu has depth m, by Lemma 2.5, there exists asubset S′ ⊆S of size p and a spider T ′ whose set of leaves is S′. Furthermore, the lengthof each leg of T ′ is at most m. Without loss of generality suppose S′ ={y1, . . . ,yp} andsuppose w is the center of T ′. For each i∈ [p], let Qi denote the unique w,yi-path. Now,P1 ∪Q1, . . . ,Pp ∪Qp is a collection of p internally disjoint x,w-paths of length at mostm+s−1. Hence x is (p,m+s−1)-linked to w. �

Claim 5. For some s≤ l(�)−m+1, Ls is a strong level with |Ls|≥n0.81.

Proof of Claim 5. Recall that for each i, |Li|=nai . We first find an i for whichai ≥0.81. By Condition (5), ai+1 ≥ai +(� / (1+�))(1−ai). Solving the recurrence, we getai ≥1−((1−a1) / (1+�)i−1). Also, a1 ≥0.2. So ai ≥1−(0.8 / (1+�)i−1). For ai ≥0.81, itsuffices that 1−(0.8 / (1+�)i−1)≥0.81, or equivalently i≥ (ln(0.8 /0.19) / ln(1+�))+1.Since 0<�<0.5, ln(1+�)≥0.8�. So, it suffices if i≥ (ln(0.8 /0.19) /0.8�)+1, which holdsif i≥ (1.8 /�)+1. Let t=min{i :ai ≥0.81}. Then t≤ 1.8 /��+1≤ l(�)−m− 1/��.

If Ls is a strong level for some s∈{t+1, . . . , t+ 1/��+1}, then s≤ l(�)−m+1 and|Ls|≥|Lt|≥n0.81 and we are done. So, suppose for each i∈{t+1, . . . , t+ 1/��+1}, Li isa weak level. By (5), |Li| / |Li−1|≥n(�/(1+�))(1−ai+0.99�i)>n(�/(1+�))(0.19), for large n (recallthat �i ≥0.199). But now, this yields |Lt+ 1/��+1|>|Lt|·n(�/(1+�))(0.19)·((1/�)+1)>n0.81 ·n0.19 =n, a contradiction. �

Now, fix an s that satisfies Claim 5. By Claim 4, each vertex in Ls is (p,m+s−1)-linked to some vertex w in Tu−Au. Since m+s−1≤l(�) and |Ls|≥n0.81, this gives us at



least n0.81 pairs (w,x) that are (p, l(�))-linked, where w∈Tu−Au and x /∈Tu−Au. Sincevertices in Tu−Au are at distance at most �1/4�� from u and �(G)≤Dn�, each w liesin at most 1+Dn�+(Dn�)2+·· ·+(Dn�)�1/4��≤n0.26 different Tu−Au, for large n. Sinceeither one of the two endpoints of a (p, l(�))-linked pair can play the role of w, thenumber of (p, l(�))-linked pairs in G is at least n ·n0.81 /2n0.26≥n1.51, for large n. �

Note that to use Theorem 3.2, we need the graph G to be “almost regular”. For thispurpose, we recall the following old lemma of Erdos and Simonovits [9].

Lemma 3.3 (Erdos and Simonovits [9]). Let � be a real satisfying 0<�<1. Let n bea positive integer that is sufficiently large as a function of �. Let G be an n-vertexgraph with e(G)≥n1+�. Then G contains a subgraph G′ on m≥n�((1−�)/(1+�)) verticessuch that e(G′)≥ 2

5 m1+� and �(G′) /�(G′)≤c�, where c� =10 ·21/�2 +1.

Theorem 3.4. Let t be a positive integer and � a fixed real, where 0<�< 12 . Let l(�)=

1/4��+ 1.8 /��+ 1/��+1. There exists a function n2(�, t) such that for all integersn≥n2(�, t) if G is a (1,�)-dense graph on n vertices that contains no ( 1

5·27 ,�)-dense

subgraph that has radius at most l(�) and order at least n�3/5 then G contains a2l(�)-subdivision of Kt.

Proof. By Lemma 3.3, G contains a subgraph G′ on m≥n�((1−�)/(1+�)) ≥n�2/3

vertices such that e(G′)≥ 25 m1+� and �(G′) /�(G′)≤c�, where c� =10 ·21/�2 +1.

Suppose e(G′)= 25 m1+�′ , where �′ ≥�. Then G′ has average degree 4

5 m�′ and so

�(G′)≤ (4c� /5)m�′ . Iteratively remove vertices of degree less than 15 m�′ from G′

until we get stuck. Call the remaining subgraph G′′. Then �(G′′)≥ 15 m�′ and �(G′′)≤

(4c� /5)m�′ . Since fewer than 15 m1+�′ edges are removed from G′ to form G′′, we have

e(G′′)≥ 15 m1+�′ . So, n(G′′)≥2e(G′′) /�(G′′)≥ m

2c�≥n�2/4, for large n. Let N =n(G′′).

Then �(G′′)≤ (4c� /5)m�′ ≤ (4c� /5)(2c�)�′ ·N�′ =D ·N�′ , where D= (4c� /5)(2c�)�

′. Also,

�(G′′)≥ 15 m�′ ≥cN�′ , where c= 1

5 . So G′′ is (c,�′)-dense and thus (c,�)-dense.Suppose G′′ contains a ( 1

5·27 ,�)-dense subgraph F that has radius at most l(�) and

order at least N0.9�. Then since n(F)≥N0.9� ≥n�3/5, this contradicts our assumptionabout G. Hence G′′ has no ( c

27 ,�)-dense subgraph that has radius at most l(�) and order

at least N0.9�. Let p= t2 ·l(�). Since N ≥n�2/4, for large enough n we have N ≥n1(�,p).By Theorem 3.2, G′′ contains at least N1.51 many (p, l(�))-linked pairs.

Define a new graph H with V(H)=V(G′′) such that xy∈E(H) if and only if x and y are(p, l(�))-linked in G′′. Then, e(H)≥N1.51 ≥ (t2 /2)N3/2, for large N. By Proposition 2.3,H contains a 2-subdivision F of Kt. We now describe how to obtain in G′′ an l(�)-subdivision of F, which yields the desired 2l(�)-subdivision of Kt. Note that F hast+( t

2

)<t2 vertices and 2

( t2

)<t2 edges. By our definition of H and F, for every edge

xy∈E(F), vertices x and y are (p, l(�))-linked in G′′. We can build an l(�)-subdivision ofF in G′′ by replacing each xy∈E(F) with a x,y-path of length at most l(�) in G′′ whileensuring that these paths are pairwise vertex disjoint outside V(F). This is possiblesince when any pair (x,y) is processed fewer than t2 ·l(�)=p vertices have been usedto embed earlier pairs and since x,y are (p, l(�))-linked we can choose a x,y-path oflength at most l(�) that is internally disjoint from all these vertices. �



4. PROOF OF THEOREM 1.6

In this section, we prove Theorem 1.6. Let us first recall some facts from [14]. For avertex u in a graph G, and a positive integer i, let Du

i (G)={x∈V(G) :distG(u,x)= i}.In other words, Du

i (G) is the set of vertices at distance i from u in G. When the hostgraph G is clear, we write Du

i for Dui (G). The following lemma was essentially proved

in [14]. The statement we give here is more general, but the proof is essentially thesame.

Lemma 4.1 (Kostochka and Pyber [14]). Let G be a connected graph with averagedegree M and u be a vertex in G. Then there exists an i such that G[Du

i ∪Dui+1] has

average degree at least M /2.

Proof. For each i, let di =|Dui ∪Du

i+1|. Note that∑

i di ≤2n(G). Suppose no i exists

that satisfies the conclusion of the lemma. Then e(G[Dui ∪Du

i+1])< 12 (M /2)di =Mdi /4

for all i and hence

e(G)≤∑

ie(G[Du

i ∪Dui+1])<

M

4

∑

idi ≤ M ·n(G)

2=e(G),

a contradiction. �

The following simple observation somewhat surprisingly played a key role in [14].Observation 4.2. Let H be a bipartite graph with a bipartition (X,Y). Suppose there

is a path P of 2m vertices in H, where the vertices on P are a1,b1,a2,b2, . . . ,am,bm inorder. Then either all the ai’s or all the bi’s are contained in X.

Recall that a graph with average degree at least 2m contains a path on 2m vertices [7].

Proof of Theorem 1.6. Recall that 0<�< 12 . Let l(�)= 1/4��+ 1.8 /��+ 1/��+1

as in Theorems 3.2 and 3.4. By considering �<0.25 or �≥0.25, one can check thatl(�)≤4.8 /�<5/�. It is well known that G contains a spanning bipartite subgraph G′ withe(G′)≥ 1

2 e(G)≥ 12 n1+�. Let G0 =G′ and n0 =n(G0). We iteratively define a sequence of

subgraphs of G′ as follows. Since G′ is bipartite, all these graphs will be bipartite. Forconvenience, let �= 1

5·28 . If such a subgraph exists, let G1 be a (2�,�)-dense subgraph

of G0 that has radius at most 5 /� and order n1 ≥n�3/60 . By our assumption, G1 has

average degree at least 4�n�1. Let u1 be a vertex in the center of G1. By Lemma 4.1,

for some two consecutive distance classes X1,Y1 from u1 in G1 (where X1 denotesthe class closer to u1) the subgraph H1 induced by them has average degree at least2�n�

1. We have n(H1)≥2�n�1 and e(H1)≥�[n(H1)]1+�. So H1 is (�,�)-dense. Since G1

is bipartite, X1 and Y1 are independent sets in G′. So (X1,Y1) is a bipartition of H1.In general, suppose we have defined G1,H1,G2,H2, . . . ,Gi,Hi. If such a subgraph

exists, let Gi+1 be a (2�i+1,�)-dense subgraph of Hi that has radius at most 5 /� andorder ni+1 ≥ [n(Hi)]�

3/6. Let ui+1 be a vertex in the center of Gi+1. By Lemma 4.1, thesubgraph Hi+1 of Gi+1 induced by some two consecutive distance classes Xi+1,Yi+1from ui (with Xi+1 being the class closer to ui+1) has average degree at least 2�i+1n�

i+1.

Thus, n(Hi+1)≥2�i+1n�i+1 and Hi+1 is (�i+1,�)-dense. Furthermore, Hi+1 is bipartite

with (Xi+1,Yi+1) being a bipartition.



Let s= t(t−1) if we can carry on this process for t(t−1) steps without getting stuck.Otherwise, let s be the largest index i<t(t−1) such that Gi,Hi are defined but nosubgraph Gi+1 of Hi exists that fits the description. We consider two cases.

Case 1. s= t(t−1).Since Hs is (�s,�)-dense, Hs has average degree has at least 2�s[n(Hs)]�. By our

definitions of the Gi’s and Hi’s, one can see that for fixed �, t, n(Hs)→∞ as n→∞.When n is sufficiently large we can ensure 2�s[n(Hs)]� ≥2t. So, Hs contains a pathP on 2t vertices. Since H1 ⊇H2 ⊇·· ·⊇Hs, P lies in all of H1, . . . ,Hs.

Let a1,b1, . . . ,at,bt be the vertices on P in order. For each i∈ [s], since (Xi,Yi) is abipartition of Hi, by Observation 4, we have either a1, . . . ,at ∈Xi or b1, . . . ,bt ∈Xi. LetI denote the set of indices i for which the former occurs and I′ the set of indices forwhich the latter occurs. Without loss of generality, we may assume that |I|≥s /2=( t

2

).

For each i∈ I, we can connect any two ax,ay ∈{a1, . . . ,at} by a path Qx,y of length atmost 2rad(Gi)≤2(5 /�) such that V(Qx,y)∩V(Gi+1)={ax,ay}. Using Gi, i∈ I, we canfind such Qx,y for all x,y∈{1, . . . , t},x �=y. These paths have no common internal verticesand their union is a subdivision of Kt in which each edge of Kt is replaced by a pathof length at most 2(5 /�)=10/�.

Case 2. s<t(t−1).By our assumption, Hs is (�s,�)-dense but contains no (2�s+1,�)-dense subgraph

that has radius at most 5 /� and order at least [n(Hs)]�3/6, since otherwise Gs+1 would

have been defined. For convenience, let m=n(Hs) and choose �>0 such that �sm1+� =m1+�−� =m1+�′ , where �′ =�−�. Note that �→0 and �′ →� as n→∞. By our assump-tion, Hs is (1,�′)-dense. Suppose Hs contains a (2�,�′)-dense subgraph F that has orderat least m(�′)3/5 and radius at most l(�′). For large enough n, we have n(F)≥m(�′)3/5>m�3/6

and l(�′)≤4.8 /�′ ≤5/�. Also, e(F)≥2�[n(F)]1+�′ =2�[n(F)]1+�−� ≥2�[n(F)]1+� /m� =2�s+1[n(F)]1+�. Thus F is (2�s+1,�)-dense and has radius at most 5 /� and order atleast m�3/6. This contradicts our assumption about Hs. So no such F exists. Notethat 2�= 1

5·27 . Thus, Hs is an m-vertex (1,�′)-dense graph that has no ( 15·27 ,�′)-dense

subgraph with radius at most l(�′) and order at least m(�′)3/5. Note that m→∞ asn→∞. By Theorem 3.4, Hs contains 2l(�′)-subdivision of Kt. For large n, we havel(�′)≤5/� and thus Hs contains a 10/�-subdivision of Kt. �

5. CONCLUDING REMARKS

Given a graph H and a positive integer p, define H(p) to be the graph obtained fromH by subdividing each edge of H exactly p−1 times. The function ex(n,H(p)) might beeven harder to determine than ex(n,H(≤p)). One earlier result of this type was obtainedby Faudree and Simonovits [10], described as below. Let Cm,k be the graph obtainedfrom joining two vertices x and y by m internally disjoint paths of length k. Faudreeand Simonovits [10] showed for fixed m and k that ex(n,Cm,k)=O(n1+1/k), extendingthe classic result of Bondy and Simonovits [4] that ex(n,C2k)=O(n1+1/k). We mayview Cm,k as obtained from the graph consisting of two vertices joined by m multipleedges by subdividing each of those edges k−1 times.

In general, there are very few results on graphs whose Turan number is O(n1+�)for � close to 0. For a graph H with v vertices and e edges, Erdos and Renyi [8]



showed that ex(n,H)>cHn2−(v/e), where cH is a constant depending on H. Hence, ifex(n,H)=O(n1+�), where � is close to 0, then H must have average degree close to 2.Since removing degree 1 vertices does not affect the Turan number substantially, onemay restrict one’s attention to graphs H with minimum degree at least 2. For such H tohave average degree close to 2, most of its vertices should have degree 2 and so wemay view H as a subdivision of another graph. This observation somewhat indicatesthat studying p-subdivisions is a meaningful step toward the study of graphs whoseTuran number is O(n1+�) for small �.

ACKNOWLEDGMENTS

The author thanks Sasha Kostochka for suggesting Lemma 2.4 in place of an old lemmaand the referees for helpful comments.

Note added in proof: The corresponding problem concerning the function ex(n,H(p))will be studied in detail in a forthcoming paper by the author and Robert Seiver.

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