The Erdős-Moser Conjecture

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1 The Erdős-Moser Conjecture An Application of Linear Algebra to Combinatorics Richard P. Stanley M.I.T.

2 The function f(s, α) Let S R, #S <, α R. f(s, α) = #{T S : i T i = α} NOTE. i i = 0

3 The function f(s, α) Let S R, #S <, α R. f(s, α) = #{T S : i T i = α} NOTE. i i = 0 Example. f({1, 2, 4, 5, 7, 10}, 7) = 3: 7 = =

4 The conjecture for S R Example. α f({1, 2,π, 10, 100},α) = 0, 1

5 The conjecture for S R Example. α f({1, 2,π, 10, 100},α) = 0, 1 Erdős-Moser Conjecture. #S = 2n + 1 f(s,α) f({ n, n + 1,...,n}, 0) #S = 2n f(s,α) f({ n + 1, n + 2,...,n}, 0)

6 The conjecture for S R + Let R + = {i R : i > 0}.

7 The conjecture for S R + Let R + = {i R : i > 0}. Weak Erdős-Moser Conjecture. S R +, #S = n ( f(s,α) f {1, 2,...,n}, 1 2 ( ) ) n + 1 2

8 The conjecture for S R + Let R + = {i R : i > 0}. Weak Erdős-Moser Conjecture. S R +, #S = n ( f(s,α) f {1, 2,...,n}, NOTE ( n+1 ) 2 = 1 2 ( n) ( ) ) n + 1 2

9 Posets A poset (partially ordered set) is a set P with a binary relation satisfying: Reflexivity: t t Antisymmetry: s t, t s s = t Transitivity: s t, t u s u

10 Graded posets chain: u 1 < u 2 < < u k

11 Graded posets chain: u 1 < u 2 < < u k Assume P is finite. P is graded of rank n if P = P 0 P 1 P n, such that every maximal chain has the form t 0 < t 1 < < t n, t i P i.

12 Diagram of a graded poset... P n P n 1 P 2 P 1 P 0

13 Rank-symmetry and unimodality Let p i = #P i. Rank-generating function: F P (q) = n p i q i i=0

14 Rank-symmetry and unimodality Let p i = #P i. Rank-generating function: F P (q) = Rank-symmetric: p i = p n i i n i=0 p i q i

15 Rank-symmetry and unimodality Let p i = #P i. Rank-generating function: F P (q) = Rank-symmetric: p i = p n i i n i=0 p i q i Rank-unimodal: p 0 p 1 p j p j+1 p n for some j

16 Rank-symmetry and unimodality Let p i = #P i. Rank-generating function: F P (q) = Rank-symmetric: p i = p n i i n i=0 p i q i Rank-unimodal: p 0 p 1 p j p j+1 p n for some j rank-unimodal and rank-symmetric j = n/2

17 The Sperner property antichain A P : s,t A, s t s = t

18 The Sperner property antichain A P : s,t A, s t s = t NOTE. P i is an antichain

19 The Sperner property antichain A P : NOTE. P i is an antichain s,t A, s t s = t P is Sperner (or has the Sperner property) if max A #A = max i p i

20 An example rank-symmetric, rank-unimodal, F P (q) = 3 + 3q

21 An example rank-symmetric, rank-unimodal, F P (q) = 3 + 3q not Sperner

22 The boolean algebra B n : subsets of {1, 2,...,n}, ordered by inclusion

23 The boolean algebra B n : subsets of {1, 2,...,n}, ordered by inclusion p i = ( ) n i, FBn (q) = (1 + q) n rank-symmetric, rank-unimodal

24 Diagram of B φ

25 Sperner s theorem, 1927 Theorem. B n is Sperner.

26 Sperner s theorem, 1927 Theorem. B n is Sperner. Emanuel Sperner 9 December January 1980

27 The poset M(n) M(n): subsets of {1, 2,...,n}, ordered as follows:

28 The poset M(n) M(n): subsets of {1, 2,...,n}, ordered as follows: Let n a 1 > a 2 > > a k 1 n b 1 > b 2 > > b j 1 S = {a 1,...,a k }, T = {b 1,...,b j }

29 The poset M(n) M(n): subsets of {1, 2,...,n}, ordered as follows: Let n a 1 > a 2 > > a k 1 n b 1 > b 2 > > b j 1 S = {a 1,...,a k }, T = {b 1,...,b j } Define S T if k j and a 1 b 1, a 2 b 2,..., a j b j.

30 M(1), M(2), M(3) φ 1 φ 2 1 φ

31 M(4) φ

32 Rank function of M(n) Easy: M(n) k #M(n) k = {S {1,...,n} : i S = f({1,...,n},k) i = k} F P (q) = (1 + q)(1 + q 2 ) (1 + q n )

33 Rank function of M(n) Easy: M(n) k #M(n) k = {S {1,...,n} : i S = f({1,...,n},k) i = k} F P (q) = (1 + q)(1 + q 2 ) (1 + q n ) Rank-symmetry clear: p i = #M(n) i = p ( n+1 2 ) i

34 Rank function of M(n) Easy: M(n) k #M(n) k = {S {1,...,n} : i S = f({1,...,n},k) i = k} F P (q) = (1 + q)(1 + q 2 ) (1 + q n ) Rank-symmetry clear: p i = #M(n) i = p ( n+1 2 ) i Rank-unimodality is unclear (no combinatorial proof known).

35 Lindström s observation Theorem. If M(n) is Sperner, then the weak Erdős-Moser conjecture holds for #S = n.

36 Lindström s observation Theorem. If M(n) is Sperner, then the weak Erdős-Moser conjecture holds for #S = n. Weak Erdős-Moser Conjecture. S R +, #S = n ( f(s,α) f ({1, 2,...,n}, 1 2 ( ) ) n + 1 2

37 Lindström s observation Theorem. If M(n) is Sperner, then the weak Erdős-Moser conjecture holds for #S = n. Proof. Suppose S = {a 1,...,a k }, a 1 > > a k. Let a i1 + + a ir = a j1 + + a js, where i 1 > > i r, j 1 > > j s.

38 Conclusion of proof Now {i 1,...,i r } {j 1,...,j s } in M(n) r s, i 1 j 1,...,i s j s a i1 b j1,...,a is b js r = s, a ik = b ik k.

39 Conclusion of proof Now {i 1,...,i r } {j 1,...,j s } in M(n) r s, i 1 j 1,...,i s j s a i1 b j1,...,a is b js r = s, a ik = b ik k. Thus a i1 + + a ir = b j1 + + b js {i 1,...,i r } and {j 1,...,j s } are incomparable or equal in M(n) ( ( ) ) 1 n + 1 #S max #A = f {1,...,n}, A 2 2

40 Linear algebra to the rescue! To prove: M(n) is Sperner.

41 Linear algebra to the rescue! To prove: M(n) is Sperner. P = P 0 P m : graded poset QP i : vector space with basis Q U : QP i QP i+1 is order-raising if U(s) span Q {t P i+1 : s < t}

42 Order-matchings Order matching: µ: P i P i+1 : injective and µ(t) < t

43 Order-matchings Order matching: µ: P i P i+1 : injective and µ(t) < t P i +1 P i µ

44 Order-raising and order-matchings Key Lemma. If U : QP i QP i+1 is injective and order-raising, then there exists an order-matching µ: P i P i+1.

45 Order-raising and order-matchings Key Lemma. If U : QP i QP i+1 is injective and order-raising, then there exists an order-matching µ: P i P i+1. Proof. Consider the matrix of U with respect to the bases P i and P i+1.

46 Key lemma proof P i s 1. s m P i+1 {}}{ t 1 t m t n det 0

47 Key lemma proof P i s 1. s m P i+1 {}}{ t 1 t m t n det 0 s 1 < t 1,...,s m < t m

48 Minor variant Similarly if there exists surjective order-raising U : QP i QP i+1, then there exists an order-matching µ: P i+1 P i.

49 A criterion for Spernicity P = P 0 P n : finite graded poset Proposition. If for some j there exist order-raising operators QP 0 inj. QP 1 inj. inj. QP j surj. QP j+1 surj. surj. QP n, then P is rank-unimodal and Sperner.

50 A criterion for Spernicity P = P 0 P n : finite graded poset Proposition. If for some j there exist order-raising operators QP 0 inj. QP 1 inj. inj. QP j surj. QP j+1 surj. surj. QP n, then P is rank-unimodal and Sperner. Proof. Glue together the order-matchings.

51 Gluing example j

52 Gluing example j

53 Gluing example j

54 Gluing example j

55 Gluing example j

56 Gluing example j

57 A chain decomposition P = C 1 C pj (chains) A = antichain,c = chain #(A C) 1 #A p j.

58 Back to M(n) Since M(n) has rank ( ) n+1 2 and is self-dual, suffices to find injective order-raising operators ( ) n + 1 U : QM(n) i QM(n) i+1, i < 2

59 How to find U?

60 How to find U?

61 Definition of U For s M(n) i define U(s) = t M(n) i+1 s<t t

62 Definition of U For s M(n) i define U(s) = t M(n) i+1 s<t t We don t know how to choose µ(s), so we make all possible choices at once (a quantum matching).

63 How to prove injectivity?

64 How to prove injectivity?

65 A lowering operator Define D: QP i QP i 1 by D(t) = c(s, t)s, s M(n) i 1 s<t where c(s,t) is given as follows. s = {a 1 > > a j } {1,...,n} t = {b 1 > > b k } {1,...,n} There is a unique r for which a r = b r 1 (and a m = b m for all other m). In the case b r = 1 we set a r = 0.

66 Two examples Example. s = {8, 7, 4, 2}, t = {8, 7, 5, 2} r = 3

67 Two examples Example. s = {8, 7, 4, 2}, t = {8, 7, 5, 2} r = 3 Example. s = {5, 4}, t = {5, 4, 1} r = 3

68 Definition of c(s, t) c(s,t) = D(t) = { ( n+1 ) 2, if ar = 0 (n a r )(n + a r + 1), if a r > 0. s M(n) i 1 s<t c(s,t)s, t M(n) i

69 Why this choice of U and D? Lemma. D i+1 U i U i 1 D i = (( ) n ) 2i I i. (Subscripts denote level at which operator acts.)

70 Why this choice of U and D? Lemma. D i+1 U i U i 1 D i = (( ) n ) 2i I i. (Subscripts denote level at which operator acts.) Proof. Straightforward computation.

71 Injectivity Claim: D i+1 U i : QM(n) i QM(n) i has positive eigenvalues for i < ( 1 n+1 ) 2 2.

72 Injectivity Claim: D i+1 U i : QM(n) i QM(n) i has positive eigenvalues for i < ( 1 n+1 ) 2 2. Proof: Induction on i. Trivial to check i = 0: the matrix is [1].

73 Injectivity Claim: D i+1 U i : QM(n) i QM(n) i has positive eigenvalues for i < ( 1 n+1 ) 2 2. Proof: Induction on i. Trivial to check i = 0: the matrix is [1]. Recall from linear algebra: V, W : finite-dimensional vector spaces A: V W, B: W V : linear transformations x dimw det(i xba) = x dimv det(i xab)

74 Eigenvalues of D i+1 I i Thus AB and BA have same nonzero eigenvalues. Recall: D i+1 U i U i 1 D i = (( ) ) n+1 2 2i Ii.

75 Eigenvalues of D i+1 I i Thus AB and BA have same nonzero eigenvalues. Recall: D i+1 U i U i 1 D i = (( ) ) n+1 2 2i Ii. Assume induction hypothesis for i 1, i.e., D i U i 1 has positive eigenvalues.

76 Eigenvalues of D i+1 I i Thus AB and BA have same nonzero eigenvalues. Recall: D i+1 U i U i 1 D i = (( ) ) n+1 2 2i Ii. Assume induction hypothesis for i 1, i.e., D i U i 1 has positive eigenvalues. Thus U i 1 D i has nonnegative eigenvalues.

77 Eigenvalues of D i+1 I i Thus AB and BA have same nonzero eigenvalues. Recall: D i+1 U i U i 1 D i = (( ) ) n+1 2 2i Ii. Assume induction hypothesis for i 1, i.e., D i U i 1 has positive eigenvalues. Thus U i 1 D i has nonnegative eigenvalues. Eigenvalues of D i+1 U i exceed those of U i 1 D i by ( ) n+1 2 2i > 0.

78 Completion of proof We showed: D i+1 U i has positive eigenvalues, i < ( 1 n+1 ) 2 2. Thus: D i+1 U i is invertible. If v ker(u i ) then v ker(d i+1 U i ), so U i is injective.

79 Completion of proof We showed: D i+1 U i has positive eigenvalues, i < ( 1 n+1 ) 2 2. Thus: D i+1 U i is invertible. If v ker(u i ) then v ker(d i+1 U i ), so U i is injective. weak Erdős-Moser conjecture is true!

80 Completion of proof We showed: D i+1 U i has positive eigenvalues, i < ( 1 n+1 ) 2 2. Thus: D i+1 U i is invertible. If v ker(u i ) then v ker(d i+1 U i ), so U i is injective. weak Erdős-Moser conjecture is true!

81 The original conjecture Recall weak conjecture was for S R +, original conjecture for S R. Original conjecture can be proved in several ways:

82 The original conjecture Recall weak conjecture was for S R +, original conjecture for S R. Original conjecture can be proved in several ways: Combinatorial argument using the weak conjecture (Kleitman) Spernicity of M(n) M(n) using same methods, where M(n) is the dual of M(n). Note: M(n) = M(n) Use results on preservation of Spernicity under product.

83 How were U and D found? QM(n) = H (SO(2n + 1, C)/P; Q)

84 How were U and D found? QM(n) = H (SO(2n + 1, C)/P; Q) = V spin, the weight space of the spin representation of SO(2n + 1, C).

85 How were U and D found? QM(n) = H (SO(2n + 1, C)/P; Q) = V spin, the weight space of the spin representation of SO(2n + 1, C). Reduced to linear algebra by Proctor.

86 What next?

87 What next?

arxiv: v1 [math.co] 16 Feb 2018

arxiv: v1 [math.co] 16 Feb 2018 CHAIN POSETS arxiv:1802.05813v1 [math.co] 16 Feb 2018 IAN T. JOHNSON Abstract. A chain poset, by definition, consists of chains of ordered elements in a poset. We study the chain posets associated to two