Degenerate Tur\'an problems for hereditary properties

0.1 MB·

by  Vladimir Nikiforov

Degenerate Turán problems for hereditary properties
Vladimir Nikiforov Michael Tait† Craig Timmons‡
∗
7 January 27, 2017
1
0
2

n
a Abstract
J
Let H be a graph and t s 2 be integers. We prove that if G is an n-vertex graph
6 ≥ ≥
2 with no copy of H and no induced copy of K , then λ(G) = O n1 1/s where λ(G) is
s,t −

the spectral radius of the adjacency matrix of G. Our results are(cid:0)motivat(cid:1)ed by results of
]
O Babai, Guiduli, andNikiforovboundingthemaximumspectralradiusofagraphwithno
C copy (not necessarily induced) of K .
s,t
.
h
t
a
1 Introduction
m
[

Many questions in extremal graph theory start from the classical Turán-type question:
1
v given a forbidden subgraph H, what is the maximum number of edges in an n-vertex
3 graphthatdoesnotcontain H asasubgraph? Thismaximumisdenotedbyex(n,H). The
9
Erdo˝s-Stone Theorem gives an asymptotic formula for ex(n,H) when χ(H) 3 which
6
≥
7 is quadratic in n. On the other hand, the Ko˝vari-Sós-Turán Theorem [7] implies that for
0 anybipartite graph H,thereisapositiveconstantδ suchthatex(n,H) = O n2 δ . Turán
. −
1
problems forbidding bipartite graphs are often called degenerate and have(cid:0)been(cid:1)studied
0
7 extensively [5].
1
In many cases, theorems in classical extremal graph theory may be strengthened via
:
v spectralgraphtheory. Foran n-vertexgraph G,letλ λ λ betheeigenvalues
1 2 n
i ≥ ≥ ··· ≥
X of its adjacency matrix. Write λ(G) = λ for the spectral radius of G. Since the spectral
1
r radius satisfies the inequality 2e(G)/n λ(G), any upper bound on λ(G) implies an
a ≤
upper bound on e(G). For example, in [10], Nikiforov improved a result of Babai and
Guiduli [1] as follows:
Theorem (Nikiforov). Let s t 2, and let G be a K -free graph of order n. If t = 2, then
s,t
≥ ≥
λ(G) 1/2+ (s 1)(n 1)+1/4.
≤ q − −
DepartmentofMathematicalSciences,UniversityofMemphis,[email protected]
∗
†Department of Mathematical Sciences, Carnegie Mellon University, [email protected]. Research supported
byNationalScienceFoundation grantDMS-1606350.
‡Department of Mathematics and Statistics, California State University Sacramento,
[email protected]. This work was supported by a grant from the Simons Foundation (#35419,
CraigTimmons)
1
If t 3, then
≥
λ(G) (s t+1)1/tn1−1/t+(t 1)n1−2/t+t 2.
≤ − − −
Inview of2e(G)/n λ(G),theabove theoremimplies alsoFüredi’simprovementof
≤
the Ko˝vari-Sós-Turán Theorem[3].
In this paper, we consider a modification of the Turán-type question, where one for-
bidsinducedcopies ofasubgraph F. Withoutadditionalrestrictions,thisproblemistrivial
if F is not complete, because K has (n) edges and no induced F, whereas the problem is
n 2
thoroughlyinvestigatedif F isacompletegraph. Moreprecisely,westudythemaximum
spectral radius that a graph may have if it has no induced K and no copy (not neces-
s,t
sarily induced) of a fixed forbidden subgraph H. We note that if χ(H) 3, then one
≥
may forbid H and have quadratically many edges, or one may forbid an induced copy
of K and have quadratically many edges. One of the main theorems in [8] shows that
s,t
when one forbids both H and K -induced at the same time, then a graph may not have
s,t
quadratically many edges.
Theorem (Loh, Tait, Timmons [8]). Let s and t be integers and H be a graph. Then there is a
constant C depending on s, t, and H such that if G is a graph on n vertices which has no copy of
H as a subgraph and nocopy of K as an induced subgraph, then
s,t
e(G) < Cn2 1/s.
−
Our main theorems are spectral strengthenings of the above theorem via the same
inequality 2e(G)/n λ(G).
≤
Finally, we note that this problem could be discussed in a more general context. A
hereditary graph property is a family of graphs which is closed under isomorphisms and
taking induced subgraphs. Given a hereditary property , let denote the set of n-
n
P P
vertex graphs in . One may ask to find ex(n, ) := max e(G) and λ(n, ) :=
P P G∈Pn P
max λ(G). Nikiforov[11] found the asymptotics of theseparameterssimilarly to the
G∈Pn
Erdo˝s-Simonovits theorem for monotone graph properties. Let us note that both Erdos-
Simonovits’s and Nikiforov’s theorems are informative only for problems with dense
extremal graphs. Not surprisingly, extremal problems leading to sparse extremal graphs
are harder and need special methods. As before, we call such problems degenerate.
In this note we focus on a degenerate extremal problem that we feel is quintessential
for the area; our main result reads as follows.
Theorem 1. Let t s 3 be integers, H be a graph, and K (R(H,K ))2/sR(H,K ). If G is
t s
≥ ≥ ≥
an H-free graph oforder n and
λ(G) Kn1 1/s, (1)
−
≥
then G contains an induced copy of K .
s,t
Here,andthroughouttherestofthepaper, R(H,G)istheRamseynumberof H vs. G.
One can question why in the premises of Theorem 1 the parameter s is at least 3, while
2 seems a more natural value. The reason is that for s = 2 we can prove a somewhat
strongerestimate as stated in the theorembelow.
2
Theorem 2. Let r 2, t 2 be integers, and K R(K ,K ). If G is a K -free graph of order
r t r+1
≥ ≥ ≥
n and
λ(G) Kn1/2,
≥
then G contains an induced copy of K .
2,t
We note that Theorem 2 may be made more general by forbidding an arbitrary sub-
graph H instead of K . For ease of exposition, we will prove Theorem 2 only when
r+1
H = K . It is clear from the proof of Theorem 1 how to generalize the result. The
r+1
proofs of the two theorems differ at several points, so we shall keep them separate; they
are proved in Section 2.
Finally, in Section 3 we will consider the specific case when H = C and when an
5
induced copy of K is forbidden. This will serve as an example of how, when more in-
2,t
formationabout H is known,onecan obtain close totightestimatesonthemultiplicative
constant. We will prove the following theorem.
Theorem 3. Let t 2 be an integer. If G is a C -free graph with no inducedcopy of K , then
5 2,t
≥
λ(G) 2t+0.3751/2n1/2+O(n3/8).
≤
p
Foreachintegertandprimepowerqforwhich t 1dividesq2 1,thereisabipartite
− −
q2 1
K2,t-free graph that is q-regular and has t−1 vertices in each part (see Füredi[4]). Such a
−
graph will have spectral radius 1(t 1)n+1 where n is the number of vertices and so
q2 −
Theorem3 is best possible up to a multiplicative factor of at most 2.
Before concluding our introduction, we mention the following corollary to Theorems
1 and 2 which implies one of the main resultsof [8] mentioned above.
Corollary4. Let H beagraphandt s 2beintegers. If G isagraphoforder nthatis H-free
≥ ≥
andhas nocopy of K as an induced subgraph, then
s,t
1
e(G) (R(H,K ))2/sR(H,K )n2 1/s.
t s −
≤ 2
1.1 Some notation
If G and H are graphs, we write H G to indicate that H is an induced subgraph of G.
≺
Given a graph G, we write:
- V(G) for the vertex set of G and v(G) for V(G) ;
| |
- E(G) for the edge set of G and e(G) for E(G) ;
| |
- Γ(X) for the set of vertices joined to all vertices of a set X V(G) and d(X) for
⊂
Γ(X) ;
| |
- G[X] for the subgraph of G induced by a set X V(G);
⊂
- I (M) for the set of independentr-sets of G and i (G) for I (G) ;
r r r
| |
- λ(G) for the largest eigenvalue of the adjacency matrix of G;
- C (G) for the number of 4-cycles of G;
4
We also write:
- (V) for the set of r-sets of a set V;
r
- R(H,G) for the Ramsey number of H vs. G and R for the Ramsey number of K
p,q p
vs. K .
q
3
2 Proofs of the main results
The following technical statement will be used in the proofs of both Theorems1 and 2.
Proposition 5. If G is agraph oforder n,then
1
∑ d2(X) λ4(G) nλ2(G) . (2)
≥ 2 −
(cid:16) (cid:17)
X (V)
∈ 2
Proof. Set for short V := V(G), and observe that
d(X)
∑ = 2C (G).
4
(cid:18) 2 (cid:19)
X (V)
∈ 2
Hence,
d(v)
∑ d2(X) = 4C + ∑ d(X) = 4C + ∑
4 4
(cid:18) 2 (cid:19)
X (V) X (V) v V
∈ 2 ∈ 2 ∈
1
= 4C + ∑ d2(v) e(G). (3)
4
2 −
v V
∈
On the other hand, writing CW (G) for the number of closed walks of length 4, it is
4
known that
CW (G) = 8C (G)+2 ∑ d2(i) 2e(G).
4 4
−
i V
∈
Hence inequality (3) implies that
2 ∑ d2(X) = 8C + ∑ d2(i) 2e(G) = CW (G) ∑ d2(v). (4)
4 4
− −
X (V) i V v V
∈ 2 ∈ ∈
Finally, in view of the identity
CW (G) = λ4(G)+ +λ4 (G)
4 1 ··· n
and Hofmeister’s bound
1
λ2(G) ∑ d2(v),
≥ n
v V
∈
inequality (2) follows immediately from (4).
Proof of Theorem 2. Supposethat t,r,K, and G satisfythepremisesofthetheorem. Note
that for any pair X (V), the graph G[Γ(X)] is K -free; hence, Turán’s theorem implies
∈ 2 r
that
(r 2)
k (G[Γ(X)]) − d2(X),
2
≤ 2(r 1)
−
and therefore,
d(X) (r 2) 1 1
i (G[Γ(X)]) − d2(X) = d2(X) d(X).
2
≥ (cid:18) 2 (cid:19)− 2(r 1) 2(r 1) − 2
− −
4
Summing this inequality over all pairs X (V) and applying Proposition 5, we obtain
∈ 2
d(I) 1 1
∑ ∑ d2(X) ∑ d(X)
(cid:18) 2 (cid:19) ≥ 2(r 1) − 2
I∈I2(G) − X∈(V2) X∈(V2)
1 1 d(v)
λ4(G) nλ2(G) ∑ .
≥ 4(r 1) − − 2 (cid:18) 2 (cid:19)
− (cid:16) (cid:17) v V
∈
Using Hofmeister’s bound and some algebra, we find that
d(I) 1 1
∑ λ4(G) nλ2(G) nλ2(G)
(cid:18) 2 (cid:19) ≥ 4(r 1) − − 4
(cid:16) (cid:17)
I∈I2(G) −
1
= λ2(G) λ2(G) rn
4(r 1) −
− (cid:0) (cid:1)
K2 K3 n n
K2 r n2 > K .
≥ 4(r 1) − ≥ 2 (cid:18)2(cid:19) (cid:18)2(cid:19)
− (cid:0) (cid:1)
That is to say, there is an I I (G), such that d(I) > K R . Since G[Γ(I)] is K -free,
2 r,t r
∈ ≥
it follows that K G[Γ(I)], and so K G, completing the proof.
t 2,t
≺ ≺
TheproofofTheorem1 issimilar totheproofofTheorem2, butneedsamoretechni-
cal approach; in particular, Turán’s theorem does not apply as above. The focal point of
the proof is the fact that if
d(I) R(H,K ) n
∑ t
, (5)
(cid:18) 2 (cid:19) ≥ (cid:18) 2 (cid:19)(cid:18)s(cid:19)
I∈Is(G)
then G has an independent s-set I such that d(I) R(H,K ); since G[Γ(I)] is H-free, it
t
≥
follows that K G[Γ(I)], and hence K G.
t s,t
≺ ≺
In turn, we deduce (5) along the following lines: we show that the premises of the
theoremimply that G contains many copies of K , albeit not necessarily induced. How-
2,s
ever, the fact that G is H-free implies that for a positive proportion of thesesubgraphs of
G their part of size s is an independent set. The requirement K (R(H,K ))2/sR(H,K )
t s
≥
is sufficient to obtain (5) eventually.
To make this argument precise, we need three additional technical statements.
Proposition 6. If k 2 and G isa graph of order n with λ(G) √n, then
≥ ≥
∑ dk(X) 1 λk(G) λ2(G) n k/2.
≥ 2nk 2 −
X (V) − (cid:0) (cid:1)
∈ 2
Proof. The Power Mean inequality and inequality (2) imply that
1/k 1/2
1 1
 n − ∑ dk(X)  n − ∑ d2(X)
(cid:18)2(cid:19) ≥ (cid:18)2(cid:19)
X (V) X (V)
 ∈ 2   ∈ 2 
1/2 1/2
n − 1 λ4(G) nλ2(G) .
≥ (cid:18)2(cid:19) (cid:18)2 (cid:16) − (cid:17)(cid:19)
5
Hence, after simple algebra, we get
1 k/2 k/2
∑ d2(X) n − 1 λ4(G) nλ2(G)
≥ (cid:18)2(cid:19) (cid:18)2 − (cid:19)
(cid:16) (cid:17)
X (V)
∈ 2
k/2
n2 k2k/2 12 k/2 λ4(G) nλ2(G)
− − −
≥ −
(cid:16) (cid:17)
= 1 λk(G) λ2(G) n k/2.
2nk 2 −
− (cid:0) (cid:1)
Proposition 7. Let K 2, s 3, and n s 1. If G is a graph of order n with λ(G)
≥ ≥ ≥ − ≥
Kn1 1/s, then G contains at least
−
n
Ks
(cid:18)s(cid:19)
copies of K .
2,s
Proof. Proposition 6 implies that
∑ ds(X) 1 λs(G) λ2(G) n s/2 > Ksns−1 K2n2 2/s n s/2
−
X (V) ≥ 2ns−2 (cid:0) − (cid:1) 2ns−2 (cid:16) − (cid:17)
∈ 2
Ks s/2
n 3n2−2/s > 2Ksns.
≥ 2
(cid:16) (cid:17)
Next, we find that
d(X) 1 (s 1)ns 1 1
∑ ∑ ds(X) (s 1)ds−1(X) = − − + ∑ ds(X)
(cid:18) s (cid:19) ≥ s! − − − s! s!
(cid:16) (cid:17)
X (V) X (V) X (V)
∈ 2 ∈ 2 ∈ 2
ns (s 1)ns 1 ns n
2Ks − − Ks > Ks .
≥ s! − s! ≥ s! (cid:18)s(cid:19)
>
To complete the proof, it is enough to note that if s 2, the sum
d(X)
∑
(cid:18) s (cid:19)
X (V)
∈ 2
is precisely the number of K copies in G.
2,s
Proposition 8. Let s 2. If G is a H-free graph of order n, then
≥
R(H,Ks) −1 n
i (G) 1. (6)
s
≥ (cid:18) s (cid:19) (cid:18)s(cid:19)−
Proof. If n < R(H,K ), then the inequality is trivial, so assume n R(H,K ). Since G
s s
≥
is H-free, any set of R(H,K ) vertices must contain an independent set of size s. Each
s
independentset of size s may be contained in at most
n s
−
(cid:18)R(H,K ) s(cid:19)
s
−
6
independentsets of size R(H,K ). Therefore,
s
n n s −1 n R(H,Ks) −1
i (G) − = .
s
≥ (cid:18)R(H,K )(cid:19)(cid:18)R(H,K ) s(cid:19) (cid:18)s(cid:19)(cid:18) s (cid:19)
s s
−
Armed with the above propositions, we encounter no difficulty in proving Theorem
1.
Proof of Theorem 1. Suppose that s,t,H,K, and G satisfy the premises of the theorem.
Note that n > s 1, because n > λ(G) Kn1 1/s and therefore n > Ks > 2s > s 1.
−
− ≥ −
Further, for any X (V), the graph G[Γ(X)] is H-free; hence, Proposition 8 implies that
∈ 2
i (G[Γ(X)]) R(H,Ks) −1 d(X) 1.
s
≥ (cid:18) s (cid:19) (cid:18) s (cid:19)−
Summing this inequality over all pairs X (V) and double counting, we obtain
∈ 2
∑ d(I) n + R(H,Ks) −1 ∑ d(X) . (7)
(cid:18) 2 (cid:19) ≥ −(cid:18)2(cid:19) (cid:18) s (cid:19) (cid:18) s (cid:19)
I∈Is(G) X∈(V2)
On the other hand, Proposition 7 implies that G contains at least
n
R(H,K )2R(H,K )s
t s
(cid:18)s(cid:19)
copies of K , that is to say,
2,s
d(X) n
∑ R(H,K )2R(H,K )s .
t s
(cid:18) s (cid:19) ≥ (cid:18)s(cid:19)
X (V)
∈ 2
Combining this inequality with (7), we find that
∑ d(I) n + R(H,Ks) −1R(H,K )2R(H,K )s n > R(H,Kt) n .
t s
(cid:18) 2 (cid:19) ≥ −(cid:18)2(cid:19) (cid:18) s (cid:19) (cid:18)s(cid:19) (cid:18) 2 (cid:19)(cid:18)s(cid:19)
I∈Is(G)
Therefore,inequality (5) holds; as shown above, it implies Theorem1.
3 Forbidding C and induced K
5 2,t
Webeginthissectionwithagenerallemmathatgivesanupperboundonλ(G)thatholds
whenever G is H-free and has no induced K . Because we will be working with eigen-
2,t
vectors,itwill beconvenienttoassumethroughoutthissectionthatV(G) = 1,2,...,n .
{ }
Furthermore, given a pair of vertices i,j , we will write d(i,j) rather than d( i,j ) and
{ } { }
we do the same for Γ( i,j ).
{ }
7
Lemma9. Let t 2beaninteger and H beagraph with h 2vertices. If G isan H-free graph
≥ ≥
of order n with noinduced copy of K , then for any vertex x V(H),
2,t
∈
1/2
ω(G) 1 1/2
λ(G)2 (R(H x,K )+1)n+ ∑ d(i,j)2 − .
t
≤ − (cid:18) 2ω(G) (cid:19)
i,j E(G)
{ }∈ 
Additionally, if x and y is a pair of nonadjacent vertices in H, then R(H x,K ) can be replaced
t
−
with R(H x y,K ) in the estimate above.
t
− −
Proof. Let x = (x ,...,x ) be a non-negative eigenvector for the eigenvalue λ := λ(G)
1 n
scaled to have 2-norm equal to 1. We have
2
n n n n
λ2 = λ2∑x2 = ∑(λx )2 = ∑ ∑ x  = ∑d(i)x2+2 ∑ d(i,j)x x .
i i j i i j
i=1 i=1 i=1 j Γ(i) i=1 1 i<j n
∈  ≤ ≤
If 1 i < j n and i,j / E(G), then for any vertex x V(H), the common neighbor-
≤ ≤ { } ∈ ∈
hood Γ(i,j) cannot contain a copy of H x or an independentset of size t, otherwise we
−
find a copy of H or an induced copy of K . Therefore,
2,t
d(i,j) < R(H x,K ). (8)
t
−
Using this inequality, we have for any vertex x V(H),
∈
n
λ2 = ∑d(i)x2+2 ∑ d(i,j)x x
i i j
i=1 1 i<j n
≤ ≤
n
< n∑x2+2 ∑ d(i,j)x x +2 ∑ d(i,j)x x
i i j i j
i=1 i,j /E(G) i,j E(G)
{ }∈ { }∈
< n+2R(H x,K ) ∑ x x +2 ∑ d(i,j)x x
t i j i j
−
i,j /E(G) i,j E(G)
{ }∈ { }∈
n n
n+R(H x,K )∑ ∑ x x +2 ∑ d(i,j)x x .
t i j i j
≤ −
i=1j=1 i,j E(G)
{ }∈
The double sum
n n
∑ ∑
x x
i j
i=1j=1
is at most n. This follows from two applications of the Cauchy-Schwarz inequality and
the fact that x = 1. Therefore,
k k
λ2 (1+R(H x,K ))n+2 ∑ d(i,j)x x .
t i j
≤ −
i,j E(G)
{ }∈
By Cauchy-Schwarz,
1/2 1/2
∑ d(i,j)x x  ∑ d(i,j)2  ∑ x2x2 . (9)
i j ≤ i j
i,j E(G) i,j E(G) i,j E(G)
{ }∈ { }∈  { }∈ 
8
Since G is H-free, G does not contain a complete graph on h := V(H) vertices. As
| |
∑n x2 = 1, we can apply the Motzkin-Straus inequality [9] to get
i=1 i
ω(G) 1
∑ x2x2 − . (10)
i j ≤ 2ω(G)
i,j E(G)
{ }∈
Combining (9) and (10), we have
1/2
ω(G) 1 1/2
∑ d(i,j)x x  ∑ d(i,j)2 − .
i j ≤ (cid:18) 2ω(G) (cid:19)
i,j E(G) i,j E(G)
{ }∈ { }∈ 
We conclude that for any vertex x V(H),
∈
1/2
ω(G) 1 1/2
λ2 (R(H x,K )+1)n+ ∑ d(i,j)2 − .
t
≤ − (cid:18) 2ω(G) (cid:19)
i,j E(G)
{ }∈ 
If H contains a pair of nonadjacent vertices x and y, then (8) can be replaced with
d(i,j) < R(H x y,K )
t
− −
and the restof the proof is the same.
We may use Lemma 9 to be more precise than Theorem 2 in the case that we can say
something about the number of triangles in the graph.
Proof of Theorem 3. Lett 2beanintegerand G beaC -freegraphwithnoinducedcopy
5
≥
of K . We must show that
2,t
λ(G) 2t+0.3751/2n1/2+O(n3/8).
≤
p
Using the fact that ω(G) 4, we have by Lemma 9,
≤
1/2
1/2
3
λ(G)2 (R(P ,K )+1)n+ ∑ d(i,j)2 .
3 t
≤ (cid:18)8(cid:19)
i,j E(G)
{ }∈ 
By [12], R(P ,K ) = 2(t 1)+1 = 2t 1. Now
3 t
− −
d(i,j)
∑ d(i,j)2 = 2 ∑ + ∑ d(i,j)
(cid:18) 2 (cid:19)
i,j E(G) i,j E(G) i,j E(G)
{ }∈ { }∈ { }∈
d(i,j)
= 2 ∑ +3k (G)
3
(cid:18) 2 (cid:19)
i,j E(G)
{ }∈
d(i,j)
2 ∑ +cn3/2
≤ (cid:18) 2 (cid:19)
i,j E(G)
{ }∈
for some constant c. In the last line we have used a result of Bollobas and Györi [2]
d(i,j)
∑
which bounds the number of triangles in a C -free graph. The sum
5
(cid:18) 2 (cid:19)
i,j E(G)
{ }∈
9
counts pairs of vertices, say z ,z V(G), such that there is an edge i,j E(G) for
1 2
{ } ⊂ { } ∈
which z ,z Γ(i,j). Supposethat this sum counts the same pair more than once. Let
1 2
{ } ⊂
z ,z Γ(i,j) Γ(x,y). Without loss of generality, we may assume that i, j, and x are
1 2
{ } ⊂ ∩
all distinct vertices. In this case, ijz xz i is a cycle of length 5 which is a contradiction.
1 2
Thus,
d(i,j) n
2 ∑ 2 n2.
(cid:18) 2 (cid:19) ≤ (cid:18)2(cid:19) ≤
i,j E(G)
{ }∈
We conclude that
1/2
λ(G)2 2tn+ (n2+cn3/2) (3/8)1/2 (2t+0.3751/2)n+O(n3/4).
≤ ≤
(cid:16) (cid:17)
Taking square roots completes the proof of Theorem3.
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m n
10