Saturated and weakly saturated hypergraphs
Algebraic Methods in Combinatorics, Lectures 6-7
1
Saturated hypergraphs
Recall the following
Definition. A family A ⊂ P([n]) is said to be an antichain if we never have
A⊂B for any two distinctA, B∈ A.
The well-known LYM inequality states that ifA ⊂ P([n]) is an antichain,
then n X i=0 |A ∩[n](i)| ¡n i ¢ ≤1.
Lubell proved this by observing that the left-hand side is a probability: it is simply the probability that a maximal chain,1 chosen uniformly at random,
intersectsA.
The LYM inequality implies that an antichain in P([n]) has size at most
¡ n
bn/2c
¢
, the size of the ‘middle layer’ in P([n]). (This can also be proved by partitioningP([n]) into¡bn/n2c¢disjoint chains.)
Bollob´as’ Inequality is a useful extension of the LYM inequality.
Theorem 1 (Bollob´as’ Inequality). Let A1, A2, . . . , Am, B1, B2, . . . , Bm ⊂[n] such that Ai∩Bi=∅ ∀i∈[m], but Ai∩Bj6=∅ ∀i6=j. Then m X i=1 1 ¡|A i|+|Bi| |Ai| ¢≤1.
Remark. Note that the above inequality does not involve n, the size of the ground-set.
Remark. TakingBi=Aci for each i, we recover the LYM inequality.
Lubell generalized his method to prove Bollob´as’ Inequality, by showing that the left-hand side is the probability that in a permutation (ordering) of {1,2, . . . , n} chosen uniformly at random, all ofAi comes before all of Bi, for some i.
1Recall that amaximal chainis a chain of sizen+ 1 inP([n]), i.e. a family of the form
Definition. We say that a permutation σ ∈ Sn is compatible with a pair of sets (Ai, Bi)if all the elements of Ai come before all the elements of Bi in the orderingσ(1), σ(2), . . . , σ(n).
Proof of Bollob´as’ Inequality. Observe that ifA1, A2, . . . , Am, B1, B2, . . . , Bm⊂ [n] satisfy the hypotheses of Bollob´as’ Inequality, then a permutation σ ∈Sn cannot be compatible with two distinct pairs (Ai, Bi) and (Aj, Bj). (Indeed, suppose the permutation σ ∈ Sn were compatible with both (Ai, Bi) and (Aj, Bj). Write σ as an ordering, σ(1), σ(2), . . . , σ(n). Without loss of gen-erality, we may assume that max{k: σ(k)∈Ai}<max{k: σ(k)∈Aj}. Since all the elements of Bj come after all the elements of Aj in σ, all the elements ofBj come after all the elements ofAi in σ, soAi∩Bj =∅, a contradiction.)
Moreover, the probability that a uniform random permutationσis compat-ible with a fixed pair (Ai, Bi) is precisely
1
¡|A
i|+|Bi|
|Ai|
¢,
since any|Ai|-subset ofAitBi is equally likely to form the first|Ai| elements ofAitBi inσ. Hence, the probability that a uniform random permutation in
Sn is compatible with one of{(Ai, Bi) : i∈[m]}is precisely m X i=1 1 ¡|A i|+|Bi| |Ai| ¢; this is at most 1, proving Bollob´as’ Inequality.
Bollob´as’ Inequality has several applications in extremal combinatorics. It can be used, for example, to determine the minimum number of edges of a
t-saturatedr-graph onnvertices. Recall the following
Definition. An r-uniform hypergraph (or r-graph) is a pair (V, E), where V
is a set, andE⊂V(r) is a collection of r-element subsets of V. We callV the
‘vertex-set’ and E the ‘edge-set’.
We denote byKt(r)the completer-uniform hypergraph ont vertices.
Definition. An r-graph is said to be t-saturatedif it is maximalKt(r)-free. In other words, it contains noKt(r), but the addition of anyr-set produces aKt(r). For example, a (2-) graph is t-saturated if it is maximal Kt-free. What is the minimum possible number of edges in a t-saturated graph on n vertices? When t = 3, we are looking at maximal triangle-free graphs. Any complete bipartite graph is maximal triangle-free; in particular, thestaron [n] with edge-set{{1, i}: 2≤i≤n}is a maximal triangle-free graph withn−1 edges. It is easy to see that any maximal triangle-free graph onnvertices has at leastn−1 edges. In general, we have the following
Theorem 2 (Bollob´as’ Saturated Hypergraph Theorem). LetG= (V, E) be a
t-saturatedr-graph onn vertices. Then |E| ≥ µ n r ¶ − µ n−t+r r ¶ .
Remark. This is best possible: ther-graph[n](r)\Y(r), where|Y|=n−t+r,
ist-saturated and has size¡nr¢−¡n−rt+r¢.
Proof. SupposeE⊂[n](r)is at-saturatedr-graph on [n]. LetA={A
1, . . . , Am}= [n](r)\Ebe the set of non-edges. For each non-edgeA
i∈ A, letCibe the vertex-set of a copy ofKt(r)formed by the addition ofAitoE. Then we haveAi⊂Ci for each i∈[m], but Aj *Ci for all j 6=i. Let Bi =Cic, so thatAi∩Bi =∅ for eachi∈[m], butAj∩Bi6=∅for allj6=i. By Bollob´as’ Inequality, we have
|A| ¡r+n−t r ¢ = ¡|A 1 i|+|Bi| |Ai| ¢ ≤1, and therefore |A| ≤ µ n−t+r r ¶ , so |E| ≥ µ n r ¶ − µ n−t+r r ¶ ,
proving the theorem.
2
Weakly saturated hypergraphs
Definition. We say that an r-graphG= (V, E)is weaklyt-saturatedif there is an ordering of V(r)\E (the non-edges), sayA
1, A2, . . . , Am, such that if we
add the non-edges to E one by one, in this order, at each step we produce at least one more copy of Kt(r).
For example, a (2-)graph is weakly 3-saturated if the non-edges can be added, one by one, in such a way that each addition creates at least one new triangle. It is easy to see that a weakly 3-saturated graph onnvertices has at leastn−1 edges. Moreover, an (n−1)-edge graph is weakly 3-saturated if and only if it is a tree.
In general, everyt-saturatedr-graph is weaklyt-saturated: we can add the non-edges in any order. In particular, the r-graph [n](r)\Y(r), where |Y| =
n−t+r, is a weakly t-saturated r-graph on n vertices, with ¡n r
¢
−¡n−t+r r
¢
edges. It turns out that again, this is the minimum possible number of edges:
Theorem 3. Let G= (V, E) be at-saturated r-graph onnvertices. Then |E| ≥ µ n r ¶ − µ n−t+r r ¶ .
To prove this, letCj be the vertex-set of aKt(r)formed whenAj is added. Then we haveAj ⊂Cj for eachj, butAk*Cj for allk > j. LetBj =Cjc, so that
Aj∩Bj =∅ ∀j∈[m], Ak∩Bj6=∅ ∀k > j.
Bollob´as’ Inequality does not always hold under this weaker condition (exer-cise!), but since all the Aj’s have size r, and all the Bj’s have size n−t, it suffices therefore to prove the following
Theorem 4(Lov´asz / Frankl / Kalai / Alon). LetA1, A2, . . . , Am⊂[n](r)and
B1, B2, . . . , Bm⊂[n](s) such thatAj∩Bj =∅for allj ∈[m], andAk∩Bj 6=∅ for allk > j. Then
m≤ µ r+s r ¶ .
We will prove this in two ways. The first uses only elementary linear algebra, but with some ingenuity. The second uses the machinery of theexterior algebra, but once this has been absorbed, the proof almost ‘writes itself’.
Proof 1: Our aim is to exhibit a linearly independent set of m objects in a vector space of dimension ¡r+rs¢. To each setBj, we will associate a function
fj : Rr+1 → R, and to each set Aj, we will associate a point uj ∈ Rr+1; the functions fj and the points uj will satisfy the ‘strictly upper-triangular’ condition we had before,
fj(uj)6= 0∀j∈[m]; fj(uk) = 0∀k > j.
It will follow immediately that the fj’s are linearly independent as real-valued functions. We will then show that they live in a vector-space of dimension¡r+rs¢, completing the proof.
The construction of our functions and our points is slightly more involved than before. Take a set Y of n points in general position in Rr+1, meaning
that every (r+ 1)-subset of Y is linearly independent. (We can do this, for example, by taking Y to consist of n distinct points on the moment curve {(1, a, a2, . . . , ar) : a∈R} in Rr+1. The fact that the Vandermonde
determi-nant is non-zero implies that these points are in general position.)
Identify the ground set [n] with the points ofY; from now on, we think of theAi’s and theBi’s as subsets ofY ⊂Rr+1.
Since anyr-subset of Y is linearly independent, for each k ∈[m], we have dim(Span(Ak)) =r, and therefore dim(Span(Ak)⊥) = 1. Choose any non-zero vector uk ∈Span(Ak)⊥. Since the points of Y are in general position, for any
y∈Y, we have
y∈Ak ⇔ y∈Span(Ak) ⇔ hy, uki= 0.
Hence,Ak∩Bj6=∅if and only if there exists someb∈Bj withhb, uki= 0. In view of this, for eachj∈[m], definefj :Rr+1→Rby
fj(z1, . . . , zr+1) =
Y
b∈Bj
hb, zi.
Note that for anyz ∈ Rr+1, we have f
j(z) = 0 if and only if there exists some b∈Bj such that hb, zi= 0. Hence,fj(uk) = 0 if and only if there exists someb∈Bj withhb, uki= 0, i.e. if and only ifAk∩Bj 6=∅. Therefore, we have
fj(uj)6= 0∀j∈[m]; fj(uk) = 0∀k > j, as desired. This implies that thefj’s are linearly independent.
Note that thefj’s can be considered as (multivariate) polynomials inR[X1, . . . , Xr+1],
the set of monomials in X1, . . . , Xr+1 of total degree exactlys. The number of these is exactly µ r+s r ¶ ;
they are in 1-1 correspondence with the set of sequences ofs‘dots’ andr‘bars’, where the number of dots between the ith and the (i+ 1)th bars corresponds to the power ofXi. Hence, m≤
¡r+s
r
¢
, as required.
We now sketch the idea behind the second proof. It turns out that we can construct an algebra2 E over R, equipped with a product∧, such that we can
associate a vectorvA ∈E with anyr-set A∈[n](r), and a vector vB ∈E with anys-set B∈[n](s), where
vA∧vB = 0 ⇔ A∩B6=∅.
This immediately implies that thevAj’s are linearly independent inE. Indeed,
suppose
m
X
j=1
cjvAj = 0
for somec1, . . . , cm ∈R; we claim that all the cj’s are zero. Suppose not, and letl be minimal such thatcl6= 0. Take the product of the left-hand side with
vBl:
m
X
j=1
cj(vAj∧vBl) = 0.
Since Ak ∩Bl 6=∅ for all k > l, we have vAk∧vBl = 0 for all k > l, so this
reduces to
cl(vAl∧vBl) = 0.
SinceAl∩Bl=∅, we havevAl∧vBl6= 0, socl= 0, a contradiction. This proves
the claim. Finally,
{vA: A∈[n](r)} will lie in a vector subspace ofE with dimension¡r+s
r
¢
. This will immediately imply Lovasz’ Theorem.
Now for the construction of our algebra. It will be the exterior algebra of the real vector spaceRr+s.
We may as well define the exterior algebra of an arbitrary vector space. Let
V be a vector space over a fieldF, and letT(V) denote thetensor algebraofV,
T(V) :=F⊕V ⊕(V ⊗V)⊕(V ⊗V ⊗V)⊕. . .=M k≥0
V⊗k,
equipped with the multiplication ⊗. Let I denote the 2-sided ideal of T(V) generated by the set{v⊗v: v∈V} ⊂V ⊗V. The exterior algebraΛ(V) ofV
is defined to be the quotient of the algebraT(V) byI: Λ(V) :=T(V)/I.
2Recall that analgebraEover a fieldFis anF-vector space equipped with a multiplication
operation,·, which is distributive over addition ((x+y)·z=x·z+y·z, z·(x+y) =z·x+z·y
for anyx, y, z∈E), and satisifesλ(x·y) = (λx)·y=x·(λy) for anyx, y∈Eandλ∈F. Our algebras will all beassociative:x·(y·z) = (x·y)·zfor anyx, y, z∈E.
Since I∩V⊗k is a subspace of V⊗k, we may define Λk(V) =V⊗k/(I∩V⊗k). Observe that Λ(V) decomposes as a direct sum of these subspaces:
Λ(V) =M k≥0
Λk(V).
The subspace Λ(k)(V) is called thekth exterior powerofV; we have Λ0(V) =F
and Λ1(V) =V.
The multiplication on Λ(V) (inherited fromT(V)) is written∧: (v+I)∧(w+I) :=v⊗w+I.
It is called thewedge-product, and turns Λ(V) into an algebra overF; in partic-ular,∧is multilinear. It is associative, since⊗is.
By construction, x∧x = 0 for all x∈ Λ(V). This implies that ∧ is anti-commutative, meaning that y∧x=−x∧y for allx, y ∈Λ(V). It follows that ∧isalternating, meaning that for anyx1, . . . , xk ∈Λ(V), and any permutation
σ∈Sk, we have
xσ(1)∧xσ(2)∧. . .∧xσ(k)= sign(σ)(x1∧x2∧. . .∧xk). Moreover, it follows that
x1∧x2∧. . .∧xk= 0
whenever the sequence x1, x2, . . . , xk contains a repetition. This is the first property of Λ(V) that we need.
We want to deal with the linear dependence of sequences (where we allow repetitions), rather than of sets, so it will be convenient to use the following (slightly fussy) definition
Definition. Let V be a vector space over a field F. We say that a sequence (x1, . . . , xk) inV islinearly dependent if it contains a repetition, or is linearly dependent as a set — in other words, if there exist scalars λ1, λ2, . . . , λk ∈F, not all zero, such that
k
X
i=1
λixi= 0. Otherwise, it is said to belinearly independent.
It follows from the multilinearity and anticommutativity of∧that
x1∧x2∧. . .∧xk= 0
whenever the sequence (x1, . . . , xk) is linearly dependent. Crucially, it turns out that this is an ‘if and only if’:
Fact 1. For anyx1, . . . , xk ∈V, we have
x1∧x2∧. . .∧xk= 0
Remark. Thus, the wedge-product of vectors in FN can be considered as a ‘vector-valued version’ of the determinant function. Indeed, ifx1, . . . , xN ∈FN, then
x1∧x2∧. . .∧xN = det(x1|x2|. . .|xN)(e1∧e2∧. . .∧eN),
where {e1, e2, . . . , eN} denotes the standard basis of unit vectors in FN, and (x1|x2|. . .|xN)denotes the N×N matrix withjth column xj.
Fact 1 follows from
Lemma 5. Let U be an N-dimensional dimensional vector-space over F, and letB={e1, . . . , eN} be a basis ofU. Then for eachk∈ {0,1, . . . , N}, the set
{ei1∧ei2∧. . .∧eik : 1≤i1< i2< . . . < ik≤N}
is a basis for Λk(U).
This is intuitively plausible; we give a proof in the appendix. In particular, we have dim(Λk(U)) = µ N k ¶ .
This is the second (and final) property we need; our vA’s will lie in the rth exterior power Λr(Rr+s).
Remark. Since Λk(U) = 0 for allk > N, we have Λ(U) =
N
M
k=0
Λk(U), and therefore the set
{ei1∧ei2∧. . .∧eik: 0≤k≤N, 1≤i1< i2< . . . < ik≤N}
is a basis forΛ(U), which has dimension2N. Hence, the natural basis ofΛk(FN) is in 1-1 correspondence with [N](k), and the natural basis of Λ(FN)is in 1-1 correspondence withP([N]).
We are now ready to give the second proof of Theorem 4. Our algebra E
will be the exterior algebra of the real vector spaceRr+s, Λ(Rr+s).
Second proof of Theorem 4. Take a set ofnpointsY ={y1, . . . , yn} ⊂Rr+s, in general position. For each subsetS={i1< i2< . . . < ik} ⊂[n], define
vS =yi1∧yi2∧. . .∧yik ∈Λ(R
r+s). Then for each A∈[n](r),v
A lies in Λr(Rr+s), which has dimension
¡r+s
r
¢
, and moreover, vA∧vB = 0 if and only if the sequenceA B (A concatenated with
B) is linearly dependent. Since the points ofY are in general position inRr+s, this holds if and only if the sequenceA Bcontains a repetition, i.e. if and only ifA∩B6=∅. By the above discussion, Theorem 4 follows immediately.
3
Appendix
Proof of Lemma 5: We already know the ‘if’ part of Fact 1. It follows that ifU
is a finite-dimensional vector space over F, and B ={e1, . . . , eN} is a basis of
U, then for eachk∈ {0,1, . . . , N}, the set
{ei1∧ei2∧. . .∧eik : 1≤i1< i2< . . . < ik≤N}
spans Λk(U). It turns out that dim(Λk(U)) =¡N k
¢
, so the above set is a basis for Λk(U). To prove rigourously that dim(Λk(U)) =¡N
k
¢
, we argue as follows. Let
W denote theF-vector space spanned by all finite formalF-linear combinations of sequences (ei1, ei2, . . . , eik), where 1≤i1< i2< . . . < ik ≤N. Define a map
f :Bk →W
by
f(b1, . . . , bk) = 0 wheneverbi=bj for somei6=j, and
f(eiσ(1), . . . , eiσ(k)) = sign(σ)(ei1, . . . , eik)
whenever 1≤i1< i2< . . . < ik ≤N andσ∈Sk. Extend this to a multilinear map
f0:Uk →W; thenf0 is alternating, meaning that
f0(uσ(1), . . . , uσ(k)) = sign(σ)f0(u1, . . . , uk) ∀ui∈U, σ∈Sk. Let
˜
f :U⊗k→W
denote the induced linear map onU⊗k. It is easy to see thatI∩U⊗k⊂ker( ˜f), and therefore dim(I∩U⊗k) ≤ dim(ker( ˜f)) = dim(U⊗k)−dim(Im( ˜f)) = Nk−dim(W) = Nk− µ N k ¶ . Hence,
dim(Λk(U)) = dim(U⊗k/(I∩U⊗k)) =Nk−dim(I∩U⊗k)≥
µ N k ¶ . Therefore, dim(Λk(U)) =¡N k ¢ , as required.
