4 The Sperner property.

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1 4 The Sperer property. I this sectio we cosider a surprisig applicatio of certai adjacecy matrices to some problems i extremal set theory. A importat role will also be played by fiite groups. I geeral, extremal set theory is cocered with fidig (or estimatig) the most or least umber of sets satisfyig give settheoretic or combiatorial coditios. For example, a typical easy problem i extremal set theory is the followig: What is the most umber of subsets of a -elemet set with the property that ay two of them itersect? (Ca you solve this problem?) The problems to be cosidered here are most coveietly formulated i terms of partially ordered sets, or posets for short. Thus we begi with discussig some basic otios cocerig posets. 4.1 Defiitio. A poset (short for partially ordered set) P is a fiite set, also deoted P, together with a biary relatio deoted satisfyig the followig axioms: (P1) (reflexivity) x x for all x P (P2) (atisymmetry) If x y ad y x, the x = y. (P3) (trasitivity) If x y ad y z, the x z. Oe easy way to obtai a poset is the followig. Let P be ay collectio of sets. If x, y P, the defie x y i P if x y as sets. It is easy to see that this defiitio of makes P ito a poset. If P cosists of all subsets of a -elemet set S, the P is called a (fiite) boolea algebra of rak ad is deoted by B S. If S = {1, 2,..., }, the we deote B S simply by B. Boolea algebras will play a importat role throughout this sectio. There is a simple way to represet small posets pictorially. The Hasse diagram of a poset P is a plaar drawig, with elemets of P draw as dots. If x < y i P (i.e., x y ad x = y), the y is draw above x (i.e., with a larger vertical coordiate). A edge is draw betwee x ad y if y covers x, i.e., x < y ad o elemet z is i betwee, i.e., o z satisfies x < z < y. By the trasitivity property (P3), all the relatios of a fiite 17

2 poset are determied by the cover relatios, so the Hasse diagram determies P. (This is ot true for ifiite posets; for istace, the real umbers R with their usual order is a poset with o cover relatios.) The Hasse diagram of the boolea algebra B 3 looks like Ø We say that two posets P ad Q are isomorphic if there is a bijectio (oe-to-oe ad oto fuctio) ϕ : P Q such that x y i P if ad oly if ϕ(x) ϕ(y) i Q. Thus oe ca thik that two posets are isomorphic if they differ oly i the ames of their elemets. This is exactly aalogous to the otio of isomorphism of groups, rigs, etc. It is a istructive exercise to draw Hasse diagrams of the oe poset of order (umber of elemets) oe (up to isomorphism), the two posets of order two, the five posets of order three, ad the sixtee posets of order four. More ambitious readers ca try the 63 posets of order five, the 318 of order six, the 2045 of order seve, the of order eight, the of order ie, the of order te, the of order eleve, the of order twelve, the of order thirtee, ad the of order fourtee. Beyod this the umber is ot curretly kow. A chai C i a poset is a totally ordered subset of P, i.e., if x, y C the either x y or y x i P. A fiite chai is said to have legth if it has + 1 elemets. Such a chai thus has the form x 0 < x 1 < < x. We say that a fiite poset is graded of rak if every maximal chai has legth. (A chai is maximal if it s cotaied i o larger chai.) For istace, the boolea algebra B is graded of rak [why?]. A chai y 0 < y 1 < < y j is said to be saturated if each y i+1 covers y i. Such a chai eed ot be maximal sice there ca be elemets of P smaller tha y 0 or greater tha y j. If P is graded of rak ad x P, the we say that x has rak j, deoted α(x) = j, if some (or equivaletly, every) saturated chai of P with top elemet x has legth j. Thus [why?] if we let P j = {x P : α(x) = j}, the P is a 18

3 disjoit uio P = P 0 P 1 P, ad every maximal chai of P has the form x 0 < x 1 < < x where α(x j ) = j. We write p j = P j, the umber of elemets of P of rak j. For example, if P = B the α(x) = x (the cardiality of x as a set) ad ( ) p j = #{x {1, 2,, } : x = j} =. j (Note that we use both S ad #x for the cardiality of the fiite set S.) We say that a graded poset P of rak (always assumed to be fiite) is rak-symmetric if p i = p i for 0 i, ad rak-uimodal if p 0 p 1 p j p j+1 p j+2 p for some 0 j. If P is both rak-symmetric ad rak-uimodal, the we clearly have p 0 p 1 p m p m+1 p, if = 2m p 0 p 1 pm = p m+1 p m+2 p, if = 2m + 1. We also say that the sequece p 0, p 1,..., p itself or the polyomial F (q) = p0 + p 1 q + + p q is symmetric or uimodal, as the case may be. For istace, B is rak-symmetric ad rak-uimodal, ( ) ( ) ( ) sice it is well-kow (ad easy to prove) that the sequece,,..., (the th row of Pas 0 1 cal s triagle) is symmetric ad uimodal. Thus the polyomial (1 + q) is symmetric ad uimodal. A few more defiitios, ad the fially some results! A atichai i a poset P is a subset A of P for which o two elemets are comparable, i.e., we ca ever have x, y A ad x < y. For istace, i a graded poset P the levels P j are atichais [why?]. We will be cocered with the problem of fidig the largest atichai i a poset. Cosider for istace the boolea algebra B. The problem of fidig the largest atichai i B is clearly equivalet to the followig problem i extremal set theory: Fid the largest collectio of subsets of a -elemet set such that o elemet of the collectio cotais aother. A good guess would be to take all the subsets of cardiality ( ) /2 (where x deotes the greatest iteger x), givig a total of sets i all. But how ca we actually prove there is o larger /2 collectio? Such a proof was first give by Emmauel Sperer i 1927 ad is kow as Sperer s theorem. We will give two proofs of Sperer s theorem 19

4 i this sectio; oe proof uses liear algebra ad will be applied to certai other situatios, while the other proof is a elegat combiatorial argumet due to David Lubell i 1966, which we preset for its cultural value. Our extesio of Sperer s theorem to certai other situatios will ivolve the followig crucial defiitio. 4.2 Defiitio. Let P be a graded poset of rak. We say that P has the Sperer property or is a Sperer poset if max{ A : A is a atichai of P } = max{ P i : 0 i }. I other words, o atichai is larger tha the largest level P i. Thus Sperer s theorem is equivalet to sayig that B has the Sperer property. Note that if P has the Sperer property there may still be atichais of maximum cardiality other tha the biggest P i ; there just ca t be ay bigger atichais. 4.3 Example. A simple example of a graded poset that fails to satisfy the Sperer property is the followig: We ow will discuss a simple combiatorial coditio which guaratees that certai graded posets P are Sperer. We defie a order-matchig from Pi to P i+1 to be a oe-to-oe fuctio µ : P i P i+1 satisfyig x < µ(x) for all x P i. Clearly if such a order-matchig exists the p i p i+1 (sice µ is oe-to-oe). Easy examples show that the coverse is false, i.e., if p i p i+1 the there eed ot exist a order-matchig from P i to P i+1. We similarly defie a order-matchig from P i to P i 1 to be a oe-to-oe fuctio µ : P i P i 1 satisfyig µ(x) < x for all x P i. 4.4 Propositio. Let P be a graded poset of rak. Suppose there exists a iteger 0 j ad order-matchigs P 0 P 1 P2 Pj P j+1 Pj+2 P. (17) The P is rak-uimodal ad Sperer. 20

5 Proof. Sice order-matchigs are oe-to-oe it is clear that Hece P is rak-uimodal. p 0 p 1 p j p j+1 p j+2 p. Defie a graph G as follows. The vertices of G are the elemets of P. Two vertices x, y are coected by a edge if oe of the order-matchigs µ i the statemet of the propositio satisfies µ(x) = y. (Thus G is a subgraph of the Hasse diagram of P.) Drawig a picture will covice you that G cosists of a disjoit uio of paths, icludig sigle-vertex paths ot ivolved i ay of the order-matchigs. The vertices of each of these paths form a chai i P. Thus we have partitioed the elemets of P ito disjoit chais. Sice P is rak-uimodal with biggest level P j, all of these chais must pass through Pj [why?]. Thus the umber of chais is exactly p j. Ay atichai A ca itersect each of these chais at most oce, so the cardiality A of A caot exceed the umber of chais, i.e., A p j. Hece by defiitio P is Sperer. It is ow fially time to brig some liear algebra ito the picture. For ay (fiite) set S, we let RS deote the real vector space cosistig of all formal liear combiatios (with real coefficiets) of elemets of S. Thus S is a basis for RS, ad i fact we could have simply defied RS to be the real vector space with basis S. The ext lemma relates the combiatorics we have just discussed to liear algebra ad will allow us to prove that certai posets are Sperer by the use of liear algebra (combied with some fiite group theory). 4.5 Lemma. Suppose there exists a liear trasformatio U : RP i RPi+1 (U stads for up ) satisfyig: U is oe-to-oe. For all x P i, U(x) is a liear combiatio of elemets y P i+1 satisfyig x < y. (We the call U a order-raisig operator.) The there exists a order-matchig µ : P i P i+1. 21

6 Similarly, suppose there exists a liear trasformatio U : RP i RP i+1 satisfyig: U is oto. U is a order-raisig operator. The there exists a order-matchig µ : P i+1 P i. Proof. Suppose U : RP i RP i+1 is a oe-to-oe order-raisig operator. Let [U] deote the matrix of U with respect to the bases P i of RP i ad P i+1 of RP i+1. Thus the colums of [U] are idexed by the elemets x 1,..., x pi of Pi (i some order) ad the rows by the elemets y 1,..., y pi+1 of P i+1. Sice U is oe-to-oe, the rak of [U] is equal to p i (the umber of colums). Sice the colum rak of a matrix equals its row rak, [U] must have p i liearly idepedet rows. Say we have labelled the elemets of P i+1 so that the first pi rows of [U] are liearly idepedet. Let A = (a ij ) be the p i p i matrix whose rows are the first p i rows of [U]. (Thus A is a square submatrix of [U].) Sice the rows of A are liearly idepedet, we have det(a) = ±a π(1),1 a π(pi ),p i = 0, where the sum is over all permutatios λ of 1,..., p i. Thus some term ±a π(1),1 aπ(p i ),p i of the above sum i ozero. Sice U is order-raisig, this meas that [why?] x k < y π(k) for 1 k p i. Hece the map µ : P i P i+1 defied by µ(x k ) = y π(k) is a order-matchig, as desired. The case whe U is oto rather tha oe-to-oe is proved by a completely aalogous argumet. We ow wat to apply Propositio 4.4 ad Lemma 4.5 to the boolea algebra B. For each 0 i <, we eed to defie a liear trasformatio Ui : R(B ) i R(B ) i+1, ad the prove it has the desired properties. We simply defie U i to be the simplest possible order-raisig operator, amely, 22

7 for x (B ) i, let U i (x) = y. (18) y (B) i+1 y>x Note that sice (B ) i is a basis for R(B ) i, equatio (18) does ideed defie a uique liear trasformatio U i : R(B ) i R(B ) i+1. By defiitio U i is order-raisig; we wat to show that U i is oe-to-oe for i < /2 ad oto for i /2. There are several ways to show this usig oly elemetary liear algebra; we will give what is perhaps the simplest proof, though it is quite tricky. The idea is to itroduce dual operators D i : R(B ) i (B ) i 1 to the U i s (D stads for dow ), defied by D i (y) = x, (19) x (B) i 1 x<y for all y (B ) i. Let [U i ] deote the matrix of U i with respect to the bases (B ) i ad (B ) i+1, ad similarly let [D i ] deote the matrix of D i with respect to the bases (B ) i ad (B ) i 1. A key observatio which we will use later is that [D i+1 ] = [U i ] t, (20) i.e., the matrix [D i+1 ] is the traspose of the matrix [U i ] [why?]. Now let Ii : R(B ) i R(B ) i deote the idetity trasformatio o R(B ) i, i.e., I i (u) = u for all u R(B ) i. The ext lemma states (i liear algebraic terms) the fudametal combiatorial property of B which we eed. For this lemma set U = 0 ad D 0 = 0 (the 0 liear trasformatio betwee the appropriate vector spaces). 4.6 Lemma. Let 0 i. The D i+1 U i U i 1 D i = ( 2i)I i. (21) (Liear trasformatios are multiplied right-to-left, so AB(u) = A(B(u)).) Proof. Let x (B ) i. We eed to show that if we apply the left-had side of (21) to x, the we obtai ( 2i)x. We have D i+1 U i (x) = D i+1 y 23 y =i+1 x y

8 = z. y =i+1 z =i x y z y If x, z (B ) i satisfy x z < i 1, the there is o y (B ) i+1 such that x y ad z y. Hece the coefficiet of z i D i+1 U i (x) whe it is expaded i terms of the basis (B ) i is 0. If x z = i 1, the there is oe such y, amely, y = x z. Fially if x = z the y ca be ay elemet of (B ) i+1 cotaiig x, ad there are i such y i all. It follows that D i+1 U i (x) = ( i)x + z. (22) z =i x z =i 1 By exactly aalogous reasoig (which the reader should check), we have for x (B ) i that U i 1 D i (x) = ix + z. (23) z =i x z =i 1 Subtractig (23) from (22) yields (D i+1 U i U i 1 D i )(x) = ( 2i)x, as desired. 4.7 Theorem. The operator U i defied above is oe-to-oe if i < /2 ad is oto if i /2. Proof. Recall that [D i ] = [U i 1 ] t. From liear algebra we kow that a (rectagular) matrix times its traspose is positive semidefiite (or just semidefiite for short) ad hece has oegative (real) eigevalues. By Lemma 4.6 we have D i+1 U i = U i 1 D i + ( 2i)I i. Thus the eigevalues of D i+1 U i are obtaied from the eigevalues of U i 1 D i by addig 2i. Sice we are assumig that 2i > 0, it follows that the eigevalues of D i+1 U i are strictly positive. Hece D i+1 U i is ivertible (sice it has o 0 eigevalues). But this implies that U i is oe-to-oe [why?], as desired. The case i /2 is doe by a dual argumet (or i fact ca be deduced directly from the i < /2 case by usig the fact that the poset B is selfdual, though we will ot go ito this). Namely, from the fact that U i D i+1 = D i+2 U i+1 + (2i + 2 )I i+1 24

9 we get that U i D i+1 is ivertible, so ow U i is oto, completig the proof. Combiig Propositio 4.4, Lemma 4.5, ad Theorem 4.7, we obtai the famous theorem of Sperer. 4.8 Corollary. The boolea algebra B has the Sperer property. It is atural to ask whether there is a less idirect proof of Corollary 4.8. I fact, several ice proofs are kow; we give oe due to David Lubell, metioed before Defiitio 4.2. Lubell s proof of Sperer s theorem. First we cout the total umber of maximal chais Ø = x 0 < x 1 < < x = {1,..., } i B. There are choices for x 1, the 1 choices for x 2, etc., so there are! maximal chais i all. Next we cout the umber of maximal chais x 0 < x 1 < < x i = x < < x which cotai a give elemet x of rak i. There are i choices for x 1, the i 1 choices for x 2, up to oe choice for x i. Similarly there are i choices for x i+1, the 2 choices for x i+2, etc., up to oe choice for x. Hece the umber of maximal chais cotaiig x is i!( i)!. Now let A be a atichai. If x A, the let C x be the set of maximal chais of B which cotai x. Sice A is a atichai, the sets C x, x A are pairwise disjoit. Hece C x = C x = (α(x))!( α(x))! Sice the total umber of maximal chais i the C x s caot exceed the total umber! of maximal chais i B, we have (α(x))!( α(x))!! Divide both sides by! to obtai 1 ( ) 1. λ(x) 25

10 ( ) Sice is maximized whe i = /2, we have i 1 1 ( ) ( ), /2 for all x A (or all x B ). Thus or equivaletly, ( ) A. /2 ( ) Sice is the size of the largest level of B, it follows that B is Sperer. /2 λ(x) 1 ( ) 1, /2 I view of the above elegat proof of Lubell, the reader may be woderig what was the poit of givig a rather complicated ad idirect proof usig liear algebra. Admittedly, if all we could obtai from the liear algebra machiery we have developed was just aother proof of Sperer s theorem, the it would have bee hardly worth the effort. But i the ext sectio we will show how Theorem 4.7, whe combied with a little fiite group theory, ca be used to obtai may iterestig combiatorial results for which simple, direct proofs are ot kow. 26