On diamond-free subposets of the Boolean lattice

Transcription

1 On diamond-free subposets of the Boolean lattice Lucas Kramer, Ryan R Martin, Michael Young 2 Department of Mathematics, Iowa State Uniersity, Ames, Iowa 500 Abstract The Boolean lattice of dimension two, also known as the diamond, consists of four distinct elements with the following property: A B, C D A diamondfree family in the n-dimensional Boolean lattice is a subposet such that no four elements form a diamond Note that elements B and C may or may not be related There is a diamond-free family in the n-dimensional Boolean lattice of size 2 o n n/2 In this paper, we proe that any diamond-free family in the n- dimensional Boolean lattice has size at most o n n/2 Furthermore, we show that the so-called Lubell function of a diamond-free family in the n- dimensional Boolean lattice which contains the empty set is at most o, which is asympotically best possible AMS 200 Subject Classification: Primary 06A07; Secondary 05D05, 05C35 Keywords: algebras forbidden subposets, extremal set theory, diamond-free, flag Introduction and Background In 928, Sperner [] proed that the largest family of subsets of [n] def {, 2,, n} with the property that no one of these sets contains another has size n n/2 Sperner s Theorem can be deried from the YBLM or LYM inequality which states that if A is an antichain in B n, then n A A A See, for example, De Bonis and Katona [4] The n-dimensional Boolean lattice, B n, denotes the partially ordered set or poset 2 [n],, where, for eery finite set S, 2 S denotes the set of subsets of S The poset B n is also called Q n in other literature, such as [] addresses: ljkramer@iastateedu Lucas Kramer, rymartin@iastateedu Ryan R Martin, myoung@iastateedu Michael Young Corresponding author This author s research partially supported by NSF grant DMS This author s research partially supported by a fellowship funded by NSF grant DMS Preprint submitted to Elseier October 26, 202

2 We say that poset P P, is a weak subposet of poset P P, if there exists an injectie function f : P P that preseres the partial ordering That is, if u in P then fu f in P See Stanley [2] Consider a family F of members of 2 [n] Then F can be iewed as a subposet of B n ia the inherited relation If F contains no P as a weak subposet we say F is P-free To correspond to Turán notation, we denote Lan, P to be the size of the largest family F of elements of B n that is P-free A layer of B n is a family of sets of the same cardinality, denoted [n] k for some integer k A chain in a poset is a set of elements, all of which are pairwise related A chain with k elements is said to hae length k and is denoted P k So, Sperner s theorem says that a subposet in B n with no P 2 has size at most n n/2 Erdős [6] generalized this result to show that, for any k 2, a Pk -free family in B n has size at most k n n/2 Furthermore, the size is exactly the sum of the k largest layers For many other subposets, P, the size of the largest P-free family in the n-dimensional Boolean lattice is not known, een asymptotically The simplest example, the diamond poset, consisting of only four elements, still has a wide gap between the largest known families and a proen upper bound In this paper, we proe that eery diamond-free family in the Boolean lattice has size at most o n n/2 The lower bound of 2 o n n/2 is found by simply taking the layers [n] n/2 and [n] n/2 + If n is a fixed integer, let Bn, k denote the collection of subsets of the k middle sizes, that is, the collection of sets of size n k + /2,, n + k /2 or n k + /2,, n + k /2 depending on the parity of n If n is a fixed integer and n, k denotes the sum of the k largest binomial coefficients of the form n l, then if Pk denotes a chain with k elements, Erdős result is as follows: Theorem Erdős [6] For n k, Lan, P k n, k Moreoer, the P k -free families of maximum size in B n are gien by Bn, k Asymptotically, Theorem gies that for any fixed k, Lan, P k k o n n/2 There are a number of results on the problem of finding Lan, P for seeral other posets P In particular, there are results regarding asymptotic bounds on Lan, P/ n n/2 A number of these results are catalogued in the work of Griggs, Li and Lu [7] We will highlight a few here: Let V r be the r-fork poset defined to be the poset that has elements A < B, B 2,, B r for r 2 Katona and Tarján [9] began studying these for r 2 and obtained bounds on Lan, V 2 and Katona and DeBonis [4] expanded these bounds for general V r for r 2 They showed that + r n + Ω n 2 Lan, V r + 2 r n + O n 2 n n/2 From this we can see that Lan, V r n n/2 2

3 There hae been a number of results that establish the asymptotic alue of Lan, P/ n n/2 for certain posets P See, for instance, [3, 8, 5, 4] In each case, the asymptotic alue of Lan, P/ n n/2 is determined and was found to be an integer As a result, Griggs and Lu [8] stated the following conjecture: Conjecture 2 For eery finite poset P, the limit πp def lim exists and is an integer n Lan, P/ n n/2 This conjecture was erified by Griggs and Lu [8] for tree posets of height 2 as well as for the crown poset O 2k, k 2, when k is een and k 4 The crown is the poset of height 2 that is a cycle of length 2k in the Hasse diagram Howeer, when k is odd, Griggs and Lu were only able to show that O 2k / n n/2 is at most o Griggs, Li and Lu [7] relate that Mike Saks and Peter Winkler obsered that known alues of πp are all equal to ep The expression ep denotes the maximum m such that, for all n, the union of the m middle layers in the n-dimensional Boolean lattice does not contain P as a poset 3 Bukh [3] erified both Conjecture 2 as well as the Saks-Winkler obseration that πt et for any tree poset T meaning that the Hasse diagram of T is a tree Griggs, Li and Lu [7] further erified the stronger conjecture in the case where P is a k-diamond poset for infinitely many alues of k The k-diamond poset consists of k + 2 distinct sets with the property A B,, B k C Howeer, a alue of k for which they do not determine the π alue exactly is the case of k 2, which we just call the diamond This poset can also be iewed as the 2-dimensional Boolean lattice, B 2 Hence, the diamond is a poset with four distinct sets A, B, C, D such that A B, C D Griggs, Li and Lu were, howeer, able to proe that, if the limit exists, then πb In Theorem 3, we proe that, if the limit exists, then πb In other words, Theorem 3 Let F be a diamond-free poset in the n-dimensional boolean lattice, B n Then, n F o n/2 The proof of one of our lemmas Lemma 8, stated below was found using Razboro s flag algebra method [0] After finding an upper bound that can be expressed as a graph inariant, we use the method to show that this inariant is bounded asymptotically by 225 Knowing flag algebras is important to understanding the origins and motiations behind the proof, but it is not necessary for the reader to understand flag algebras in order to understand our proof In fact, we do not use flag algebra terminology, per se 3 The notation ep has been used in seeral places in extremal poset theory It is not to be confused with the notation for the number of edges in graph G, denoted eg, which we will use later in the paper 3

4 2 Proof of Theorem 3 The proof of Theorem 3 follows from three lemmas: Lemma 5 proen in Section 3, Lemma 7 proen in Section 4 and Lemma 8 proen in Section 5 We first use a ariant of the YBLM argument that proes Sperner s theorem So, we define the Lubell function [7] of a family in a Boolean lattice: Definition 4 If F is a family of sets in the n-dimensional Boolean lattice, the Lubell function of that family is defined to be Λn, F def n F F F Let Λ * n, P be the maximum of Λn, F oer all families F that are both P-free and contain the empty set Furthermore, set Λ * P def lim sup {Λ * n, P} n The Lubell function of a family F gies the aerage number of times a random full chain will meet F This will yield an upper bound for the size of a family Our main result is that the Lubell function of a B 2 -free poset which contains the empty set is at most o, which immediately implies Theorem 3 by way of Lemma 5 Griggs, Li and Lu [7] address Lemma 5 Here we proe it in detail for completeness Lemma 5 Let B 2 denote the diamond If F is B 2 -free in B n then F Λ * B 2 + o n n/2 That is, Lan, B2 Λ * B 2 + o n n/2 The expression Λ * B 2 proides us with an upper bound on the alue of Lan, B 2 but calculating it in general can be difficult We next introduce a method to compare the alue of Λ * n, B 2 to a graph inariant that we can calculate more directly First we introduce some definitions Definition 6 For a graph G, let α i α i G denote the number of three-ertex subgraphs that induce exactly i edges for i 0,, 2, 3 and let β j β j G denote the number of four-ertex subgraphs that induce exactly j edges for j 0,, 6 If X, Y is an ordered bipartition of V G, then let ex denote the number of nonedges in the subgraph induced by X and ey denote the number of nonedges in the subgraph induced by Y Recall that for eery nonnegatie integer k and real number n, n k nn n k + for k and n 0 Now we introduce our primary method for bounding the Λ * n, F function which will, in turn, bound the size of our extremal function for the size of the family of sets F Using the Pochhammer notation, we hae the following bound: Lemma 7 For eery B 2 -free family F in B n with F, there exist the following: a graph G V, E on n ertices and 4

5 a set W {w +,, w n } such that, for each w W, X w, Y w is an ordered bipartition of V ; for which where, with the notation as aboe, fn, G, W 2α G 2α 2 G n 3 Λn, F 2 + fn, G, W, + 6β 0G + n 4 w W [ Xw Y w + 4eY ] w 2eX w n 3 Een though the function fn, G, W is somewhat complicated, Lemma 8 gies that f is bounded asymptotically by /4, regardless of the graph G or the set W Lemma 8 For any integer n, graph G V, E on n ertices and a set W, of n bipartitions of V G, fn, G, W 4 + O n Theorem 3 follows because Lemma 7 and then Lemma 8 gie that the Lubell function of any B 2 -free family in B n that contains the emptyset is at most o Lemma 5 then gies that F o n n/2 We note that Griggs-Li-Lu [7] gie two constructions of diamond-free families that both contain the empty set and hae Lubell function alues of 2 + Thus it is not possible to obtain a result better than F nn n o n n/2 by using Lemma 5 3 Proof of Lemma 5 We first draw upon some conclusions from a paper by Axenoich, Manske and the second author [] By Lemma in [] we know that k n/2 n 2/3 n k 2 Ωn/3 n n/2 We may assume, therefore that all the elements of F are close to the middle layer that is, they are subsets of [n] with size between n/2 n 2/3 and n/2 + n 2/3 This is because the total number of the remaining elements is o n n/2 For an eent A, let A be the indicator ariable of that eent Let C be the set of full chains in B n For a diamond-free family F, let F min denote the minimal elements of F For any F F min, let C F denote the full chains that contain F and let C 0 denote the full chains that contain no member of F min 5

6 Our goal is to bound F by using the triial inequality: F n n n/2 F F F By counting the number of times a full chain contains an element F in F we see that n F C F n! F F F F C C F C + F C n! n! C C 0 F F F F min C C F F F Since F does not hae a 4-chain this would imply a copy of B 2, it can be partitioned into at most three antichains: the minimal elements, the maximal elements that are not also minimal and whateer remains Hence if C C 0, it is the case that F F F C 2 because no chain in C 0 can contain more than two members of F So we hae the following: F F n 2 F n! C 0 + n! F F min C C F F F F C One can partition full chains contained in C F into collections such that one collection contains all extensions of one chain from F to [n] The size of each of these collections is F! So we focus our attention on the interal [F, [n]] def {S : F S [n]} The full chains in this interal are denoted C F Therefore, C C F F F F C F! C C F F F F C and so, F F F n 2 F n! C 0 + n! 2 n! C n! C 0 + n F! F! n F! F F min F F min F F min n F n F! C C F F F n F Λ * n F, B 2, F C C C F F F F C because F may be regarded as the empty set in the interal [F, [n]] Next we make the obseration that F min is an antichain and, as such, obeys 6

7 YBLM Set M n def max k { Λ * n k, B 2 : k n/2 < n 2/3} F F n 2 F n! C n! C 0 + n F M n F F min C 0 M n n! It is obious that Λ * k, B 2 is at least 2 for k 2 and so n M n max {Λ * n k, B 2 : k n/2 < n 2/3} F k F F Λ * B 2 + o By returning to, we see that F Λ * B 2 + o n n/2 This concludes the proof of Lemma 5 4 Proof of Lemma 7 Let us define a partition of the all the chains in B n as follows For a family of sets F B n, denote Ψ i Ψ i F to be the set of full chains that contain exactly i members of F for i 0,,, n + Note that Ψ 0 0 since F and Ψ i 0 for i 4 since F has no 4-chain Hence i {, 2, 3} Since eery chain can only hit, 2 or 3 elements in a B 2 -free poset F we hae that n! Ψ 3 + Ψ 2 + Ψ Then for each i {, 2, 3} the total number of times all the chains in Ψ i contains an element of F is i Ψ i Recall that Λn, F is the aerage number of times a chain contains an element of F, and hence we hae that Λn, F 3 Ψ3 +2 Ψ2 + Ψ n! Therefore Λn, F 2 + Ψ3 Ψ n! Let W {w [n] : {w} F}; ie, the set of singletons in F Let V [n] W and without loss of generality let W {w +, w +2,, w n }, where V Define the graph G V, EG, where EG {{, } :, V and {, } F} For eery w W, there is an ordered partition of V called X w, Y w where X w {x V : {x, w} F} and Y w V X w Obsere that if w, w are distinct members of W, then {w, w } / F, else we would hae a B 2 with, {w}, {w }, {w, w } We proceed by placing bounds on both Ψ and Ψ 3 First, we find a lower bound on Ψ We hae the following 4 members of Ψ : 7

8 a Let z, z 2, z V be a set of 3 ertices that induces exactly two edges with z z 2 being the nonedge in G Then Ψ includes chains of the form, {z }, {z, z 2 }, {z, z 2, z },, [n] and chains of the form, {z 2 }, {z 2, z }, {z 2, z, z },, [n] There are 2α 2 n 3! such chains b Let w W, y Y w and z V {y} such that yz is an edge in G Then Ψ includes chains of the form, {y}, {y, w}, {y, w, z },, [n] There are w W y Y w degyn 3! such chains, where degy is the degree of y in the graph induced by V c Let w W, x X w and z V {x} such that xz is a nonedge in G Then Ψ includes chains of the form, {x}, {x, z }, {x, z, w},, [n] There are w W x X w degxn 3! such chains, where degx is the degree of x in the complement of the graph induced by V ie, the nondegree of x d Let w W, w W {w} and y Y w Then Ψ includes chains of the form, {y}, {y, w}, {y, w, w },, [n] There are w W Y w W n 3! such chains It is clear that there is no chain counted twice among the chains a, b, c and d Therefore, Ψ n 3! 2α 2 + w W y Y w degy + w W x X w degx + Y w W w W Let us simplify y Y w degy + x X w degx Fix w W Then each edge in Y w gets counted twice Each non-edge in Y w does not get counted Each edge in X w does not get counted The non-edges in X w each get counted twice Each edge from X w to Y w gets counted once since there is a y Y w incident to the edge Each non-edge from Y w to X w gets counted once since there is a x X w incident to each non-edge Then we hae y Y w degy + x X w degx X w Y w + 2eY w + 2eX w Hence, Ψ n 3! 2α 2 + [ X w Y w + 2eY w + 2eX w + W Y w ] 2 w W Now we turn our attention to Ψ 3 It is important to distinguish two types of members of F: We will count separately the types of chains that contain a singleton that is, chains of the form, {w},, [n] for some w W and those that do not contain a singleton Let T be the minimal nonempty elements of F and U be F { } T We then hae the following 4 possibilities for members of Ψ 3 i Let w W and x X w Then Ψ 3 contains chains of the form, {w}, {x, w},, [n] We hae a total of w W X w n 3! of this type of chain 8

9 ii Let w W, U U and U 3 such that w U Then Ψ 3 contains chains of the form, {w},, U,, [n] There are at most Yw 2 Y w 2eY w n 3! of these types of chains This w W bound is proen in Claim 0 iii Let, 2, V be a set of 3 ertices that induces exactly one edge with 2 being the edge in G Then Ψ 3 contains chains of the form, { }, {, 2 }, {, 2, },, [n] and chains of the form, { 2 }, { 2, }, { 2,, },, [n] The number of chains of this type is 2α n 3! i Let U V be a member of U with U 4 where U does not contain an edge Then Ψ 3 contains chains of the form,, U,, [n] Note that only one additional member of F can be in the interal [, U], hence at most a / U fraction of these chains contain three members 6 of F There are at most n 3 β 0n 3! such chains This bound will be proen in Claim Claim 9 The chains found in i, ii, iii, and i is an exhaustie list of the chains in Ψ 3 Proof of Claim 9 Suppose not Let C be a chain containing three elements of F Eery chain contains and in order to hae 3 elements of F in a full chain, that full chain contains an element T T Suppose T {w} for w W All such chains are counted in i and ii Suppose T 2 Then there must be a U U aboe T in C If T 2 then T induces an edge in the graph G and we hae a chain of type iii If T > 2 then the chain gets counted as a chain of type i Therefore, we hae counted all possible chains that contain 3 elements of F Claim 0 For a fixed w W the number of chains of type ii is bounded aboe by Y w 2 Y w 2eY w n 3! Hence, the total number of such chains is at most w W Yw 2 Y w 2eY w n 3! Proof of Claim 0 For fixed w W, we can bound the number of chains of type ii as follows: U!n U! U U:U w, U 3 n 3! U U:U w, U 3 n 3! Y w Y w Yw Y w U U n 3 U 3 Yw Y w 2 U 3 U n 3 U 3 U U:U w, U 3 9

10 Since Y w n, we hae Yw 2 U 3 n 3 U 3 Furthermore, consider subsets of Y w Note that U {w} Y w, else we hae a chain of length 4 in F The family {U {w} : U U, U w, U 3} E G Yw forms an antichain in B Yw First, no edge y y 2 can be a subset of one of those sets U {w}, otherwise {w}, {y, y 2 } U Second, any two sets U U 2 in this family would form a chain of length 4 in F, namely, {w} U U 2 So, the YBLM inequality gies ey w Y w 2 + U U:U w, U 3 U U:U w, U 3 Yw U Thus, U! U!n U! Yw n 3! Y w Y w U U U:U w, U 3 Yw n 3! Y w Y w ey w 2 n 3! Y w 2 Y w 2eY w This concludes the proof of Claim 0 Claim The number of chains of type i is at most 6 n 3 β 0n 3! Proof of Claim If we fix some T T with T 3, then we will count these chains by replacing the members of U that are supersets of T T with supersets of T that coer T A set U coers set T if there is a u U such that T U {u} This approach is also discussed in [] Let UT denote the members of U that are supersets of T and U 0 T denote the family of all sets U such that U coers T but there is no other member of T that U coers Note that members of U 0 T need not be members of U Note further that, by definition, each U 0 T is an antichain If we consider the family U T T U 0T, then { } T U has no B 2 This is because, first, no U U can be a superset of distinct T, T 2 T by definition Second, if U, U 2 U such that U U 2, then there are distinct T, T 2 such that T U and T 2 U 2, but then U 2 T, a contradiction to the definition of U 0 T 2 Furthermore, any chain that contains T and some U UT has the property that there is some U 0 U 0 T such that this chain also contains U 0 If not, then the chain contains some X that coers T for which T X U The only reason that X U 0 T is that there is another T T such that X T Thus, T, T U, a B 2 0

11 So, we can bound the number of chains of type i as follows: U U,U V, U 4 n 3! U!n U! U U U,U V, U 4 4 n 3! n 3 U U U,U V, U 4 U! n 3 U 3 4 U 4 U U n 4 U 4 We hae that n, hence 4 U 4 n 4 U 4 Furthermore, / U /4 for all U in the sum Therefore, U U,U V, U 4 U U!n U! n 3! 4 n 3 U U,U V, U 4 4 U The members of {U U : U V, U 4} together with quadruples of ertices that induce at least one edge form an antichain Hence, the YBLM inequality gies that 4 β0 4 + U Hence, U U,U V, U 4 U U!n U! n 3! 4 n 3 This concludes the proof of Claim 4 n 3! 4n 3 β n 3! n 3 β 0 U U,U V, U 4 4 U Returning to the proof of Lemma 7, we can combine the bounds on Ψ and

12 Ψ 3 : Ψ 3 Ψ n 3! 2α + 6 n 3 β 0 + [ X w n 2 + Y w Y w 2eY w ] w W 2α 2 w W 2α α n 3 β 0 + w W [ X w Y w + 2eY w + 2eX w + W Y w ] [ X w n 2 + Y w 2 Y w W X w Y w 4eY w 2eX w ] 2α α n 3 β 0 + w W [ X w n 2 + Y w Y w W X w Y w 2 Y w 2 + 4eY w 2eX w ] We can collect terms and obsere that Y w Y w W X w 2 Y w + 2 Y w n 2 The expression then greatly simplifies to gie us the following: Ψ 3 Ψ n 3! 2α α n 3 β 0 + w W [ X w Y w n 2 + 4eY w 2eX w ] If we diide by n 3 we get the bound on 3 Ψ 3 +2 Ψ 2 + Ψ 2+ Ψ 3 Ψ, as needed This concludes the proof of Lemma 7 5 Proof of Lemma 8 The first thing we do is compute fn, G, W by a summation on the set of induced subgraphs on 4 ertices In order to do so, howeer, we must eliminate the cases of when 3 If 3, then α, X w 3 and ey w 3 So triially, fn, G, W 2 n 3 + n n 3 which is at most /4 if n Now we assume that 4 Let H 4 V 4, the set of all 4-tuples of ertices For ease of notation, if H H 4, we will use H to mean both that set of 4 ertices and the subgraph induced by them, 2

13 We obsere that H H 4 α i H 3α i G for i 0,, 2, 3 and for any set S V, we hae H H 4 es H 2 es and H H 4 S H 3 S So, we can rewrite fn, G, W as follows: fn, G, W 2α 2α 2 + 6β [ 0 Xw Y w + 4eY ] w 2eX w n 3 n 3 4 H H 4 + n 4 w W [ 3 αh α 2 H 3n w W + 4 4n 4 β 0 H X w H Y w H eY w H 2eX w H 2n 3 6 ] We will see in Claim 3 that in most cases of H H 4, the largest alue of the summand occurs when X w H H So, we will rearrange the terms: fn, G, W [ 3 αh α 2 H + 4 n β 0 H + 3n n 4 H H 4 where + n 2 eh 6n 3 + w W ɛn, w, G, H ], 3 ɛn, w, G, H 2 Y w H + 2 2eH + 4eY w H 2eX w H 4 2n 3 6 At this point, we need notation preiously used in [] to describe the eleen distinct nonisomorphic graphs on exactly 4 ertices as follows Notation 2 For i 0,, 5, 6, the 4-ertex graph with exactly i edges is denoted H i The graph with exactly two edges that are incident is H and the graph with exactly two edges that are nonincident is H Their complements are H Q and H, respectiely The graph inducing a star with three edges the claw is H, the graph inducing a triangle is H and the path with three edges is H We use to denote that two graphs are isomorphic Claim 3 bounds the ɛ term in 3 Note that it is 0 for all but three of the eleen 4-ertex graphs 3

14 Claim 3 For any integer n, graph G V, E on < n ertices, ordered bipartition X w, Y w of V and H H 4, 3 max{0, 2 3 }, if H H 0 ; 5 2n ɛn, w, G, H 2 max{0, 4 5 }, if H H ; 5 3 max{0, 9 0 }, if H H ; 0, otherwise Proof of Claim 3 We note that we hae defined ɛn, w, G, H to be zero if Y w H For each case of Y w H, we choose the set of ertices that makes the expression 4eY w H 2eX w H as large as possible We look at each indiidual possibility for Y w H First suppose eh 3 In this case, ey w H is at most eh if Y w H {3, 4} In all such cases, ɛn, w, G, H 0, see Table A for the table of terms in the case eh 3 We can now consider the remaining four possibilities for H The case H is displayed in Table A2, in which it is shown that ɛn, w, G, H 0 The case H is displayed in Table A3, in which it is shown that ɛn, w, G, H 0 unless Y w H 3 and is large enough The case H is displayed in Table A4 in which ɛn, w, G, H is either at most 0 or at most the case where Y w H 4 The case H 0 is displayed by Table A5 in which ɛn, w, G, H 0 is either at most 0 or at most the case where Y w H 4 This concludes the proof of Claim 3 Note that for fixed n, G and W, we hae expressed fn, G, W as H 4 dh for some density function d of H in Equation 3 Moreoer, we can use the bounds for the ɛ terms in Claim 3 in order to get upper bounds for dh, call them d H As such, we hae Table A6, which details the alues of d H for the eleen different nonisomorphic graphs on 4 ertices Case 4 2n /3 In this case, the expressions inoling the max function are zero These simplified alues of d H are displayed in Table A7 It is triial to see that, if 2n /3 and H {H, H, H, H Q, H 5, H 6 } that d H n For the rest of the graphs H, the expressions in Table A7 4

15 can be simplified as follows: d H 0 n d H n + 2 4n n 7 4n 2 + 2n n 3 6 5n 2 d H n + 2 2n 3 9 8n 2 d H n d H n + 2 3n 3 3 2n n 3 3 2n 2 With this simplification, it is easy to see that d H < n if 2n /3 and H {H, H, H, H } As to H 0, the expression 2 +4n 7 4n 2 + 2n+6 increases in for all n 4 Hence, its maximum alue for 2n /3 is at most 8n 2 0n + 06/9, which is negatie for all n 4 Therefore, if 4 2n /3 and n 4, then fn, G, W max {d n H} H H 4 n 2 4 Case 2 2n/3 and 4 We will upper bound fn, G, W by adding a nonnegatie term to 4 d H We use N to denote the neighborhood of ertex, N[] N {} to denote the closed neighborhood of ertex For any ertex-subset S, we denote S V S Now we can bound fn, G, W, where the coefficient γ 0 will be chosen later: fn, G, W d H 4 H H 4 d H 4 H H 4 + γ 4 + γ 4 z,z 2:z z 2 EG z,z 2:z z 2 EG Nz N[z 2 ] N[z ] Nz Nz Nz 2 N[z ] N[z 2 ] 2 5 So, the expression 4 is a sum oer ordered pairs that form nonedges in G and so z z 2 in this summation The expression 5 is a sum oer ordered pairs that form edges in G The reader familiar with Razboro s flag algebra method [0] will see it at work here There are two types the edges and nonedges The flags are subgraphs induced by three ertices For each type, the 5

16 positie semidefinite matrix is a 4 4 matrix that is a multiple of a matrix with two diagonal entries equal to, two off-diagonal entries equal to and the remaining twele entries being equal to 0 For more on flag algebras, we refer the reader to Baber and Talbot [2] who gie a nice discussion of an application to hypergraphs The key obseration regarding expressions 4 and 5 is that these terms can be rewritten so that they count an expression we call γch for each subgraph H on 4 ertices, as well as some terms corresponding to subgraphs on 3 ertices For example, the expression 4 assigns 6γ for each subgraph isomorphic to H, because it assigns a alue of 2γ for each ordered pair of distinct nonadjacent ertices, of which there are eight The expression 5 assigns +8γ for each subgraph isomorphic to H, because it assigns a alue of +2γ for each ordered pair of distinct adjacent ertices, of which there are four This gies a net alue of 8γ We compute all alues of γch in this way and display them in Table A8 This gies the expression: fn, G, W d H + γ ch + γ 4 H H 4 4 H H 4 46α G + 6α 3 G 6 d H + γch + 24γ H H 4 We choose the alue of γ so as to ensure that d H 6 +γch 6 n +24γ is exactly /4 Hence, γ 96 n 24 Since 2n/3, this choice of γ is nonnegatie as long as n 7 It remains to show that d H + γch /4 for the remaining choices of H In Table A9, we list all of the expressions for d H + γch and simplify it to obtain Table A0, which gies d H + γ ch /4 for each of the eleen graphs H on 4 ertices We now only need to see that these expressions are at most 0 First, we see in Table A0 that d H 0 + γch 0 /4 0 As to the rest, the easiest approach is to express the functions in Table A0 in terms of x /n, ignoring the lower-order terms and obtaining a function g H x Since the numerators are polynomials in and n and denominators are polynomials in n, then g H x 0 implies that d H 0 + γch 0 /4 O/n The functions g H x are gien in Table A along with their maximum alue oer the range 2/3 x They are strictly negatie for all H {H, H } and at most zero for those This concludes the proof of Lemma 8 6 Concluding remarks Griggs, Li and Lu conjectured that the Lubell function for a diamond-free family in B n is at most 2 + For n large enough, we hae shown this in the case where < 2n 3 n 2 4 6

17 In the case of 2n 3, for all H {H, H }, the fact that g H x is always strictly negatie Table A gies that d H + γch /4 0 for n large enough Furthermore, it can be shown that d H + γch /4 0 and d H + γch /4 0 with equality if and only if n Coupling this with the fact that 24γ n 3, the O/n term in the statement of the lemma can be seen to be as small as /4n 2 for n large enough 96 n 24 n Howeer, is either 4n or 4n, depending on the parity of n We beliee that the bound by Griggs, Li and Lu is correct That is, if F is a diamond free family, then Λn, F 2 + nn n Acknowledgements The authors would like to thank Peter Keeash and John Talbot for suggesting the method of flag algebras References [] Maria Axenoich, Jacob Manske, and Ryan R Martin Q 2 -free families of the boolean lattice Order, 29:77 9, 202 [2] Rahil Baber and John Talbot Hypergraphs do jump Combin Probab Comput, 202:6 7, 20 [3] Boris Bukh Set families with a forbidden subposet Electron J Combin, 6:Research Paper 42,, 2009 [4] Annalisa De Bonis and Gyula O H Katona Largest families without an r-fork Order, 243:8 9, 2007 [5] Annalisa De Bonis, Gyula O H Katona, and Konrad J Swanepoel Largest family without A B C D J Combin Theory Ser A, 2:33 336, 2005 [6] Paul Erdős On a lemma of Littlewood and Offord Bull Amer Math Soc, 5: , 945 [7] Jerrold R Griggs, Wei-Tian Li, and Linyuan Lu Diamond-free families J Combin Th Ser A, 92:30 322, 202 [8] Jerrold R Griggs and Linyuan Lu On families of subsets with a forbidden subposet Combin Probab Comput, 85:73 748, 2009 [9] Gyula O H Katona and Tamás Gy Tarján Extremal problems with excluded subgraphs in the n-cube In Graph theory Lagów, 98, olume 08 of Lecture Notes in Math, pages Springer, Berlin, 983 7

18 [0] Alexander A Razboro Flag algebras J Symbolic Logic, 724: , 2007 [] Emanuel Sperner Ein Satz über Untermengen einer endlichen Menge Math Z, 27: , 928 [2] Richard P Stanley Enumeratie combinatorics Vol, olume 49 of Cambridge Studies in Adanced Mathematics Cambridge Uniersity Press, Cambridge, 997 With a foreword by Gian-Carlo Rota, Corrected reprint of the 986 original [3] Hai Tran Thanh An extremal problem with excluded subposet in the Boolean lattice Order, 5:5 57, 998 Appendix A Tables Appendix A Tables of alues to proe Claim 3 Y w H maximum alue of ɛn, w, G, H if eh eH 2n eh eH+4 2n eH+4eH 2n eh+2 6 eh eH+4eH 2n eh 2 0 < 0 0 < 0 Table A: Maximum alue of ɛn, w, G, H if eh 3 Y w H maximum alue of ɛn, w, G, H n n n n < 0 < 0 < 0 0 Table A2: Maximum alue of ɛn, w, G, H 8

19 Y w H maximum alue of ɛn, w, G, H n n n n < 0 0 Table A3: Maximum alue of ɛn, w, G, H Y w H maximum alue of ɛn, w, G, H n n n n Table A4: Maximum alue of ɛn, w, G, H Y w H maximum alue of ɛn, w, G, H n n n n Table A5: Maximum alue of ɛn, w, G, H 0 9

20 Appendix A2 Tables for the proof of Lemma 8 H d H H n 4 H 3 6 n 3 H 3 2 n 3 H 3 3 n 3 H n 3 H H 3 4 n 3 H 3 n 3 H Q 3 4 n 3 H n 3 H 6 + n + n + n + n + n n + n + n + n + n n n 3 5 6n 3 2 3n 3 2 3n 3 2n 3 2n 3 2n 3 3n 3 3n 3 6n 3 + 3n + 5n 2 + 5n 3 { max { max max 0, 2 3 0, 4 5 } } { 0, 9 0 } Table A6: The alues of d H for the eleen distinct nonisomorphic graphs on 4 ertices 20

21 H H 0 H H H H H H H H Q H 5 H 6 d H n n 4 n 3 n n 3 5 6n 3 n n 3 2 3n 3 n n 3 2 3n 3 n n 3 2n 3 n 2n 3 n n 3 2n 3 n 3 n 3 3n 3 n 3 4 n 3 3n 3 n 3 2 n 3 6n 3 n Table A7: The alues of d H in the case of 4 2n /3 H γch H H 0 + 4γ 4γ H 4γ + 0 4γ H 6γ + 8γ 8γ H H 4γ + 0 4γ H 2γ + 2γ 0 H H Q 0 + 4γ 4γ H γ 4γ H γ 24γ Table A8: The alues of γch for the eleen distinct nonisomorphic graphs on 4 ertices The first term is the contribution from the expression 4, the second term from the expression in 5 Their sum is the last term 2

22 H d H + γch H n 4 H 3 6 n 3 H 3 2 n 3 H 3 3 n 3 H n 3 H H 3 4 n 3 H 3 n 3 H Q 3 4 n 3 H n 3 H 6 + n + n + n + n + n n + n + n + n + n n 3 5 6n 3 2 3n 3 2 3n 3 2n 3 2n 3 2n 3 3n 3 3n 3 + 3n + 5n 2 + 5n 3 { max { max max 0, 2 3 0, 4 5 } } { 0, 9 0 } n 3 +4 n +24 Table A9: The alues of d H + γch in the case of 2n/3 96 n n n n n n n 24 H d H + γch /4 H H H H n 4 3 n 3 3 n 3 3 n 3 3 n 3 + n n 3 + n 6n 3 max{5n 5 5, 0 7n + 4} + n 6n 3 max{5n 4 6, 6 4n + 2} + n 4 6 3n 3 + n 3 2n 3 H H n 7n 3 3 n 3 H H 4 3 H Q H H 6 0 n 3 3 n 3 3 n 3 6n 3 + n n 3n 5 3n 3 + n 7 2 6n 3 + n 5n 9 6n 3 Table A0: Simplified alues of d H + γch /4 in the case of 2n/3 22

23 H g H x max oer 2 3 x H H H H H H H H Q H x3 6 + xx 6 max{5 5x, 0x 7} x x3 2 + xx 6 max{5 4x, 6x 4} 7 x x3 3 + xx4 2x 3 x 4 3x3 4 + xx2 x 2 35 x xx7 3x 6 23 x x3 4 + xx2 x 2 x 4 x3 + xx3 x 3 2 x x3 4 + xx7 2x 6 0 x x3 2 + xx5 x 6 27 x 2 3 Table A: The g H x functions that determine the asymptotic alue of d H+γcH /4 23