Saturated and weakly saturated hypergraphs


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1 Satuated and weakly satuated hypegaphs Algebaic Methods in Combinatoics, Lectues 67 Satuated hypegaphs Recall the following Definition. A family A P([n]) is said to be an antichain if we neve have A B fo any two distinct A, B A. The wellknown LYM inequality states that if A P([n]) is an antichain, then n A [n] (i) ( n. i) i=0 Lubell poved this by obseving that the lefthand side is a pobability: it is simply the pobability that a maximal chain, chosen unifomly at andom, intesects A. ( The LYM inequality implies that an antichain in P([n]) has size at most n n/2 ), the size of the middle laye in P([n]). (This can also be poved by patitioning P([n]) into ( n n/2 ) disjoint chains.) Bollobás Inequality is a useful extension of the LYM inequality. Theoem (Bollobás Inequality). Let A, A 2,..., A m, B, B 2,..., B m [n] such that A i B i = i [m], but Then A i B j i j. m i= ( Ai + B i A i ). Remak. Note that the above inequality does not involve n, the size of the goundset. Remak. Taking B i = A c i fo each i, we ecove the LYM inequality. Lubell genealized his method to pove Bollobás Inequality, by showing that the lefthand side is the pobability that in a pemutation (odeing) of {, 2,..., n} chosen unifomly at andom, all of A i comes befoe all of B i, fo some i. Recall that a maximal chain is a chain of size n + in P([n]), i.e. a family of the fom {, {x }, {x, x 2 }, {x, x 2, x 3 },..., [n]}.
2 Definition. We say that a pemutation σ S n is compatible with a pai of sets (A i, B i ) if all the elements of A i come befoe all the elements of B i in the odeing σ(), σ(2),..., σ(n). Poof of Bollobás Inequality. Obseve that if A, A 2,..., A m, B, B 2,..., B m [n] satisfy the hypotheses of Bollobás Inequality, then a pemutation σ S n cannot be compatible with two distinct pais (A i, B i ) and (A j, B j ). (Indeed, suppose the pemutation σ S n wee compatible with both (A i, B i ) and (A j, B j ). Wite σ as an odeing, σ(), σ(2),..., σ(n). Without loss of geneality, we may assume that max{k : σ(k) A i } < max{k : σ(k) A j }. Since all the elements of B j come afte all the elements of A j in σ, all the elements of B j come afte all the elements of A i in σ, so A i B j =, a contadiction.) Moeove, the pobability that a unifom andom pemutation σ is compatible with a fixed pai (A i, B i ) is pecisely ( Ai + B i A i since any A i subset of A i B i is equally likely to fom the fist A i elements of A i B i in σ. Hence, the pobability that a unifom andom pemutation in S n is compatible with one of {(A i, B i ) : i [m]} is pecisely m i= ), ( Ai + B i A i this is at most, poving Bollobás Inequality. Bollobás Inequality has seveal applications in extemal combinatoics. It can be used, fo example, to detemine the minimum numbe of edges of a tsatuated gaph on n vetices. Recall the following Definition. An unifom hypegaph (o gaph) is a pai (V, E), whee V is a set, and E V () is a collection of element subsets of V. We call V the vetexset and E the edgeset. We denote by K () t ); the complete unifom hypegaph on t vetices. Definition. An gaph is said to be tsatuated if it is maximal K () t fee. In othe wods, it contains no K () t, but the addition of any set poduces a K () t. Fo example, a (2) gaph is tsatuated if it is maximal K t fee. What is the minimum possible numbe of edges in a tsatuated gaph on n vetices? When t = 3, we ae looking at maximal tianglefee gaphs. Any complete bipatite gaph is maximal tianglefee; in paticula, the sta on [n] with edgeset {{, i} : 2 i n} is a maximal tianglefee gaph with n edges. It is easy to see that any maximal tianglefee gaph on n vetices has at least n edges. In geneal, we have the following Theoem 2 (Bollobás Satuated Hypegaph Theoem). Let G = (V, E) be a tsatuated gaph on n vetices. Then ( ) ( ) n n t + E. 2
3 Remak. This is best possible: the gaph [n] () \ Y (), whee Y = n t +, is tsatuated and has size ( ) ( n n t+ ). Poof. Suppose E [n] () is a tsatuated gaph on [n]. Let A = {A,..., A m } = [n] () \E be the set of nonedges. Fo each nonedge A i A, let C i be the vetexset of a copy of K () t fomed by the addition of A i to E. Then we have A i C i fo each i [m], but A j C i fo all j i. Let B i = Ci c, so that A i B i = fo each i [m], but A j B i fo all j i. By Bollobás Inequality, we have A ( +n t ) = ( Ai + B i A i ), and theefoe so poving the theoem. E ( ) n t + A, ( ) n ( n t + ), 2 Weakly satuated hypegaphs Definition. We say that an gaph G = (V, E) is weakly tsatuated if thee is an odeing of V () \ E (the nonedges), say A, A 2,..., A m, such that if we add the nonedges to E one by one, in this ode, at each step we poduce at least one moe copy of K () t. Fo example, a (2)gaph is weakly 3satuated if the nonedges can be added, one by one, in such a way that each addition ceates at least one new tiangle. It is easy to see that a weakly 3satuated gaph on n vetices has at least n edges. Moeove, an (n )edge gaph is weakly 3satuated if and only if it is a tee. In geneal, evey tsatuated gaph is weakly tsatuated: we can add the nonedges in any ode. In paticula, the gaph [n] () \ Y (), whee Y = n t +, is a weakly tsatuated gaph on n vetices, with ( ) ( n n t+ ) edges. It tuns out that again, this is the minimum possible numbe of edges: Theoem 3. Let G = (V, E) be a tsatuated gaph on n vetices. Then ( ) ( ) n n t + E. To pove this, let C j be the vetexset of a K () t fomed when A j is added. Then we have A j C j fo each j, but A k C j fo all k > j. Let B j = Cj c, so that A j B j = j [m], A k B j k > j. Bollobás Inequality does not always hold unde this weake condition (execise!), but since all the A j s have size, and all the B j s have size n t, it suffices theefoe to pove the following 3
4 Theoem 4 (Lovász / Fankl / Kalai / Alon). Let A, A 2,..., A m [n] () and B, B 2,..., B m [n] (s) such that A j B j = fo all j [m], and A k B j fo all k > j. Then ( ) + s m. We will pove this in two ways. The fist uses only elementay linea algeba, but with some ingenuity. The second uses the machiney of the exteio algeba, but once this has been absobed, the poof almost wites itself. Poof : Ou aim is to exhibit a linealy independent set of m objects in a vecto space of dimension ( ) +s. To each set Bj, we will associate a function f j : R + R, and to each set A j, we will associate a point u j R + ; the functions f j and the points u j will satisfy the stictly uppetiangula condition we had befoe, f j (u j ) 0 j [m]; f j (u k ) = 0 k > j. It will follow immediately that the f j s ae linealy independent as ealvalued functions. We will then show that they live in a vectospace of dimension ( ) +s, completing the poof. The constuction of ou functions and ou points is slightly moe involved than befoe. Take a set Y of n points in geneal position in R +, meaning that evey ( + )subset of Y is linealy independent. (We can do this, fo example, by taking Y to consist of n distinct points on the moment cuve {(, a, a 2,..., a ) : a R} in R +. The fact that the Vandemonde deteminant is nonzeo implies that these points ae in geneal position.) Identify the gound set [n] with the points of Y ; fom now on, we think of the A i s and the B i s as subsets of Y R +. Since any subset of Y is linealy independent, fo each k [m], we have dim(span(a k )) =, and theefoe dim(span(a k ) ) =. Choose any nonzeo vecto u k Span(A k ). Since the points of Y ae in geneal position, fo any y Y, we have y A k y Span(A k ) y, u k = 0. Hence, A k B j if and only if thee exists some b B j with b, u k = 0. In view of this, fo each j [m], define f j : R + R by f j (z,..., z + ) = b B j b, z. Note that fo any z R +, we have f j (z) = 0 if and only if thee exists some b B j such that b, z = 0. Hence, f j (u k ) = 0 if and only if thee exists some b B j with b, u k = 0, i.e. if and only if A k B j. Theefoe, we have f j (u j ) 0 j [m]; f j (u k ) = 0 k > j, as desied. This implies that the f j s ae linealy independent. Note that the f j s can be consideed as (multivaiate) polynomials in R[X,..., X + ], of total degee exactly s. The vecto space of these polynomials is spanned by 4
5 the set of monomials in X,..., X + of total degee exactly s. The numbe of these is exactly ( ) + s ; they ae in  coespondence with the set of sequences of s dots and bas, whee the numbe of dots between the ith and the (i + )th bas coesponds to the powe of X i. Hence, m ( ) +s, as equied. We now sketch the idea behind the second poof. It tuns out that we can constuct an algeba 2 E ove R, equipped with a poduct, such that we can associate a vecto v A E with any set A [n] (), and a vecto v B E with any sset B [n] (s), whee v A v B = 0 A B. This immediately implies that the v Aj s ae linealy independent in E. Indeed, suppose m c j v Aj = 0 j= fo some c,..., c m R; we claim that all the c j s ae zeo. Suppose not, and let l be minimal such that c l 0. Take the poduct of the lefthand side with v Bl : m c j (v Aj v Bl ) = 0. j= Since A k B l fo all k > l, we have v Ak v Bl = 0 fo all k > l, so this educes to c l (v Al v Bl ) = 0. Since A l B l =, we have v Al v Bl 0, so c l = 0, a contadiction. This poves the claim. Finally, {v A : A [n] () } will lie in a vecto subspace of E with dimension ( ) +s. This will immediately imply Lovasz Theoem. Now fo the constuction of ou algeba. It will be the exteio algeba of the eal vecto space R +s. We may as well define the exteio algeba of an abitay vecto space. Let V be a vecto space ove a field F, and let T (V ) denote the tenso algeba of V, T (V ) := F V (V V ) (V V V )... = k 0 V k, equipped with the multiplication. Let I denote the 2sided ideal of T (V ) geneated by the set {v v : v V } V V. The exteio algeba Λ(V ) of V is defined to be the quotient of the algeba T (V ) by I: Λ(V ) := T (V )/I. 2 Recall that an algeba E ove a field F is an Fvecto space equipped with a multiplication opeation,, which is distibutive ove addition ((x + y) z = x z + y z, z (x + y) = z x + z y fo any x, y, z E), and satisifes λ(x y) = (λx) y = x (λy) fo any x, y E and λ F. Ou algebas will all be associative: x (y z) = (x y) z fo any x, y, z E. 5
6 Since I V k is a subspace of V k, we may define Λ k (V ) = V k /(I V k ). Obseve that Λ(V ) decomposes as a diect sum of these subspaces: Λ(V ) = k 0 Λ k (V ). The subspace Λ (k) (V ) is called the kth exteio powe of V ; we have Λ 0 (V ) = F and Λ (V ) = V. The multiplication on Λ(V ) (inheited fom T (V )) is witten : (v + I) (w + I) := v w + I. It is called the wedgepoduct, and tuns Λ(V ) into an algeba ove F; in paticula, is multilinea. It is associative, since is. By constuction, x x = 0 fo all x Λ(V ). This implies that is anticommutative, meaning that y x = x y fo all x, y Λ(V ). It follows that is altenating, meaning that fo any x,..., x k Λ(V ), and any pemutation σ S k, we have x σ() x σ(2)... x σ(k) = sign(σ)(x x 2... x k ). Moeove, it follows that x x 2... x k = 0 wheneve the sequence x, x 2,..., x k contains a epetition. This is the fist popety of Λ(V ) that we need. We want to deal with the linea dependence of sequences (whee we allow epetitions), athe than of sets, so it will be convenient to use the following (slightly fussy) definition Definition. Let V be a vecto space ove a field F. We say that a sequence (x,..., x k ) in V is linealy dependent if it contains a epetition, o is linealy dependent as a set in othe wods, if thee exist scalas λ, λ 2,..., λ k F, not all zeo, such that k λ i x i = 0. i= Othewise, it is said to be linealy independent. It follows fom the multilineaity and anticommutativity of that x x 2... x k = 0 wheneve the sequence (x,..., x k ) is linealy dependent. Cucially, it tuns out that this is an if and only if : Fact. Fo any x,..., x k V, we have x x 2... x k = 0 if and only if the sequence (x,..., x k ) is linealy dependent. 6
7 Remak. Thus, the wedgepoduct of vectos in F N can be consideed as a vectovalued vesion of the deteminant function. Indeed, if x,..., x N F N, then x x 2... x N = det(x x 2... x N )(e e 2... e N ), whee {e, e 2,..., e N } denotes the standad basis of unit vectos in F N, and (x x 2... x N ) denotes the N N matix with jth column x j. Fact follows fom Lemma 5. Let U be an Ndimensional dimensional vectospace ove F, and let B = {e,..., e N } be a basis of U. Then fo each k {0,,..., N}, the set is a basis fo Λ k (U). {e i e i2... e ik : i < i 2 <... < i k N} This is intuitively plausible; we give a poof in the appendix. In paticula, we have ( ) N dim(λ k (U)) =. k This is the second (and final) popety we need; ou v A s will lie in the th exteio powe Λ (R +s ). Remak. Since Λ k (U) = 0 fo all k > N, we have and theefoe the set Λ(U) = N Λ k (U), k=0 {e i e i2... e ik : 0 k N, i < i 2 <... < i k N} is a basis fo Λ(U), which has dimension 2 N. Hence, the natual basis of Λ k (F N ) is in  coespondence with [N] (k), and the natual basis of Λ(F N ) is in  coespondence with P([N]). We ae now eady to give the second poof of Theoem 4. Ou algeba E will be the exteio algeba of the eal vecto space R +s, Λ(R +s ). Second poof of Theoem 4. Take a set of n points Y = {y,..., y n } R +s, in geneal position. Fo each subset S = {i < i 2 <... < i k } [n], define v S = y i y i2... y ik Λ(R +s ). Then fo each A [n] (), v A lies in Λ (R +s ), which has dimension ( ) +s, and moeove, v A v B = 0 if and only if the sequence A B (A concatenated with B) is linealy dependent. Since the points of Y ae in geneal position in R +s, this holds if and only if the sequence A B contains a epetition, i.e. if and only if A B. By the above discussion, Theoem 4 follows immediately. 7
8 3 Appendix Poof of Lemma 5: We aleady know the if pat of Fact. It follows that if U is a finitedimensional vecto space ove F, and B = {e,..., e N } is a basis of U, then fo each k {0,,..., N}, the set {e i e i2... e ik : i < i 2 <... < i k N} spans Λ k (U). It tuns out that dim(λ k (U)) = ( ) N k, so the above set is a basis fo Λ k (U). To pove igouously that dim(λ k (U)) = ( ) N k, we ague as follows. Let W denote the Fvecto space spanned by all finite fomal Flinea combinations of sequences (e i, e i2,..., e ik ), whee i < i 2 <... < i k N. Define a map by wheneve b i = b j fo some i j, and f : B k W f(b,..., b k ) = 0 f(e iσ(),..., e iσ(k) ) = sign(σ)(e i,..., e ik ) wheneve i < i 2 <... < i k N and σ S k. Extend this to a multilinea map f : U k W ; then f is altenating, meaning that f (u σ(),..., u σ(k) ) = sign(σ)f (u,..., u k ) u i U, σ S k. Let f : U k W denote the induced linea map on U k. It is easy to see that I U k ke( f), and theefoe Hence, dim(i U k ) dim(ke( f)) = dim(u k ) dim(im( f)) = N k dim(w ) ( ) N = N k. k dim(λ k (U)) = dim(u k /(I U k )) = N k dim(i U k ) Theefoe, dim(λ k (U)) = ( N k ), as equied. ( ) N. k 8
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