Europ. J. Combinatorics (1996) 17 , 727 – 730
An Erdo
Us – Ko – Rado Theorem for Direct Products
P. FRANKL
Let ni , ki be positive integers, i 5 1 , . . . , d , satisfying ni > 2ki . Let X 1 , . . . , Xd be pairwise
disjoint sets with uXi u 5 ni . Let * be the family of those (k1 1 ? ? ? 1 kd )-element sets which have
exactly ki elements in Xi , i 5 1 , . . . , d . It is shown that if ^ ’ * is an intersecting family then
u^ u / u* u < maxi ki / ni , and this is best possible. The proof is algebraic, although in the d 5 2 case
a combinatorial argument is presented as well.
÷ 1995 Academic Press Limited
1. INTRODUCTION
Let X be an n -element set, and for k , n let ( Xk ) denote the family of all k -subsets of
X. A family ^ ’ ( Xk ) is called t -intersecting if for all F , F 9 P ^ one has uF > F 9u > t.
Also, intersecting stands for 1-intersecting.
One of the oldest and most important results in extremal set theory is the following.
ERDOU S – KO – RADO THEOREM (cf. [2, 3, 8]). Suppose that n > (k 2 t 1 1)(t 1 1) and let
^ ’ ( Xk ) be a t -intersecting family. Then
u^ u <
Snk 22 ttD
(1)
holds.
Over the years, many extensions and sharpenings of this result have proved: see [1]
for survey. The following problem arose in connection with a recent result of Sali [7].
Suppose that n 5 n 1 1 ? ? ? 1 nd , k 5 k 1 1 ? ? ? 1 kd , and X 5 X 1 < ? ? ? < Xd , with
uXi u 5 ni . Define
H SXk D: F > X 5 k for i 5 1, . . . , dJ.
*5 FP
u
iu
i
What is the maximal size of an intersecting subfamily of * ?
For an arbitrary element x P X the family *x 5 hH P * : x P H j is obviously
intersecting. Moreover, if x P Xi then u*x u / u* u 5 ki / ni .
THEOREM 1.
Suppose that ^ ’ * is intersecting and kd / nd < ? ? ? < kd / nd < 1 / 2 . Then
u^ u / u* u < k 1 / n 1 holds .
The proof of this result, which is presented in the next section, is based on the
eigenvalue argument of Lovasz [5]. In Section 3, a combinatorial argument is provided
for the case d 5 2. An application of this result is given in [4]. It is based on the cyclic
permutation method of Katona [6].
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÷ 1996 Academic Press Limited
728
P . Frankl
2. PROOF
OF
THEOREM 1
Let Mi denote the symmetric 0 – 1 matrix the rows and columns of which are indexed
by the ki -subsets of Xi , and the entry being 1 iff the corresponding two sets are disjoint.
It is well-known that the eigenvalues of Mi are
(21)j
Sn k2 2k j2 jD
i
i
i
with corresponding multiplicities
Snj D 2 Sj 2n jD,
i
i
j 5 0 , . . . , ki .
Let M 5 M1 % ? ? ? % Md be the tensor product of the matrices Mi . Then if a row
(column) of M is indexed by (F1 , . . . , Fd )((G1 , . . . , Gd )) , respectively, then the
corresponding entry is 1 or 0 according to whether or not the intersection of
F1 < ? ? ? < Fd and G1 < ? ? ? < Gd is empty. That is, M has its rows and columns indexed
by the members of * , and the corresponding entry is 1 or 0 according to whether or
not the intersection of the sets is empty. Since the eigenvalues of M are the products of
those of Mi , the largest eigenvalue of M is
l5
Sn 2k k D 3 ? ? ? 3 Sn k2 k D,
i
1
d
i
d
d
it corresponds to the all-one vector. The smallest eigenvalue is
m 52
Sn k2 k2 12 1DSn k2 k D 3 ? ? ? 3 Sn k2 k D.
1
1
2
1
2
d
d
2
d
Let I (J ) denote the identity (all-one) matrices of order u* u , respectively. Then
N 5 M 2 mI 2
l2m
J
u* u
is positive semidefinite.
Let ^ ’ * be an intersecting family and let y 5 y (^ ) be its characteristic vector: it is
a 0 – 1 vector of length u* u, with entries indexed by the members of * , in the same
order as the rows and columns of M.
Since N is positive semidefinite, we have the following inequality:
0 < y Ny t 5 y My t 2 my Iy t 2
Consequently,
u^ u / u* u < 2m / (l 2 m ) 5
l 2m
l2m 2
y Jy t 5 2m u^ u 2
u^ u .
u* u
u* u
(2)
Sn k2 k2 12 1DYSSn k2 k D 1 Sn k2 k2 12 1DD 5 k /n .
1
1
1
1
1
1
1
1
1
1
1
Using the proof of (1) given by Wilson [8], one obtains the following, more general,
result in the same way.
THEOREM 2. Let ni , ki and ti be positiy e integers satisfying ni > (ki 2 ti 1 1)(ti 1 1) ,
i 5 1 , . . . , d. Suppose that ^ ’ * has the property that , for ey ery F , G P ^ , there exists
An Erdo
U s – Ko – Rado theorem for direct products
729
1 < i < d such that uF > G > Xi u > ti holds. Then
Skn 22 tt DYSkn D,
u^ u / u* u < max
i
i
i
i
i
i
i
and this is best possible.
3. THE COMBINATORIAL PROOF
OF
THEOREM 1
IN THE
CASE d 5 2
Let x1 , . . . , xm be a cyclic permutation, 1 , r , m an integer. Consider a collection 5
of cyclic intervals of length r . Let 59 be the collection of those intervals of length r 2 1
which are contained in some member of 5. It is easy to see that u59u > minhm , u5u 1 1j
holds. Applying this argument repeatedly implies that, for every 1 < i , r , the number
of cyclic intervals of length i which are contained in some member of 5 is at least
minhm , u5u 1 r 2 i j. This simple result can be called the Circular Kruskal – Katona
Theorem.
Recall that two families of sets are called cross-intersecting if each member of one
intersects each member of the other. We need the following.
PROPOSITION. Suppose that # and $ are a cross -intersecting families of cyclic intery als
of respectiy e lengths c and d , c 1 d < m. Then the following hold :
(a ) u# u 1 u$ u < m
(b ) u# u 1 u$ u < c 1 d if both # and $ are non -empty.
PROOF. Set r 5 m 2 c and let 5 consist of the complements of the members of # .
Let & consist of those cyclic intervals of length d which are contained in some member
of 5. By the cross-intersecting property & > $ 5 [ , and by the Circular Kruskal –
Katona Theorem,
u$ u < m 2 minhm , u# u 1 m 2 c 2 d j .
Using u$ u > 1 ,
u# u 1 u$ u < c 1 d
follows.
Now (a) follows from (b) unless # and $ is empty, in which case it is trivial.
Next we turn to the proof of Theorem 1, d 5 2 .
Let x1 , . . . , xni be a cyclic ordering of the elements of Xi , i 5 1 , 2. Let !i be the
collection of the ni cyclic intervals of length ki . Define ! 5 hA1 < A2: Ai P !i j.
LEMMA. i ! > ^ u < k 1 n 2 .
PROOF. Define @ 5 hB P ! 1: 'A2 P ! 2 , (A1 < A2) P ^ j. We distinguish two cases,
as follows.
(i) u@u < 2ki . Choose some family @9, such that @ ’ @9 ’ ! 1 , u@9u 5 2k 1 . Let @9 consist
of the sets B1 , B2 , . . . , B2k1 in this order. Let bi denote the number of those A P ! 2 for
which (bi < A) P ^ holds. Since Bi > Bi1k1 5 [ , bi 1 bi 1k < n 2 follows for 1 < i < k1
from the proposition. Consequently,
u! > ^ u 5 b 1 1 b 2 1 ? ? ? 1 b 2k1 < k 1 n 2
holds.
(ii) u@u . 2k 1 . For A P ! 1 , let $ (A) be the collection of those D P ! 2 for which
730
P . Frankl
(A < D ) P ^ holds. Note that, if A > A9 5 [ , then $ (A) and $ (A9) are crossintersecting. Let d1 , d2 , . . . , dn1 be the numbers u$ (A)u , A P ! 1 , in decreasing order.
CLAIM. dj 1 dn1112j < 2k 2 , 1 < j < n 1 .
To prove the claim it is sufficient to consider the case 2j < n 1 . First observe that
every cyclic interval of length ki intersects exactly 2ki 2 1 cyclic intervals of length ki .
Thus d1 > 2k 2 would imply d2k2 5 0 , contrary to our assumptions. Consequently,
d1 , 2k 2. Thus the claim is true if dn1112j 5 0 . Therefore we suppose that dn1112j > 1.
Let Ai be the cyclic interval corresponding to the number di . In view of the
proposition hA1 , . . . , Aj j and hA1 , . . . , An1112jj cannot be cross-intersecting. That is,
there exist 1 < u < j and 1 < y < n1 1 1 2 j such that Au > Ay 5 [—a contradiction. By
the proposition, we have dj 1 dn1112j < du 1 dy < 2k 2, as desired.
Summing up the inequality in the claim for 1 < j < n 1 gives 2 u! > ^ u 5 2(d 1 1 ? ? ? 1
dn1) < 2k2 n 1 < 2k 1 n2, concluding the proof of the lemma.
The lemma says that for each pair of cyclic permutations out of the possible n1 n 2
sets, at most k1 n 2 are in ^ : that is, a proportion of at most k 1 / n 1. Therefore, by
averaging (see, e.g., [6]), u^ u / u* u < k 1 / n 1 follows.
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Us – Ko – Rado Theorem is true for n 5 ckt , Coll. Soc . Math. J. Bolyai , 18 (1978),
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Us – Ko – Rado Theorem, J. Combin. Theory , Ser. B , 13 (1972),
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Us – Ko – Rado Theorem, Combinatorica , 4 (1984), 247 – 257.
Receiy ed 7 June 1990 and accepted 23 June 1995
P. FRANKL
Uniy ersite´ Paris 6 ,
Combinatoire ,
4 place Jussieu , 75005 Paris , France