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Two Enumerative Tidbits p. Two Enumerative Tidbits Richard P. Stanley M.I.T.

Two Enumerative Tidbits p. The first tidbit The Smith normal form of some matrices connected with Young diagrams

Two Enumerative Tidbits p. Partitions and Young diagrams λ is a partition of n: λ = (λ 1,λ 2,...), λ 1 λ 2 0, λ i = n

Two Enumerative Tidbits p. Partitions and Young diagrams λ is a partition of n: λ = (λ 1,λ 2,...), λ 1 λ 2 0, λ i = n Example. λ = (5, 3, 3, 1) = (5, 3, 3, 1, 0, 0,... ). Young diagram:

Two Enumerative Tidbits p. Extended Young diagrams λ: a partition (λ 1,λ 2,...), identified with its Young diagram (3,1)

Two Enumerative Tidbits p. Extended Young diagrams λ: a partition (λ 1,λ 2,...), identified with its Young diagram (3,1) λ : λ extended by a border strip along its entire boundary

Two Enumerative Tidbits p. Extended Young diagrams λ: a partition (λ 1,λ 2,...), identified with its Young diagram (3,1) λ : λ extended by a border strip along its entire boundary (3,1)* = (4,4,2)

Two Enumerative Tidbits p. Initialization Insert 1 into each square of λ /λ. 1 1 1 1 1 1 (3,1)* = (4,4,2)

Two Enumerative Tidbits p. M t Let t λ. Let M t be the largest square of λ with t as the upper left-hand corner.

Two Enumerative Tidbits p. M t Let t λ. Let M t be the largest square of λ with t as the upper left-hand corner. t

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1.

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1. 1 1 1 1 1 1

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1. 2 1 1 1 1 1 1

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1. 2 1 2 1 1 1 1 1

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1. 2 1 3 2 1 1 1 1 1

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1. 5 2 1 3 2 1 1 1 1 1

Two Enumerative Tidbits p. Determinantal algorithm Suppose all squares to the southeast of t have been filled. Insert into t the number n t so that detm t = 1. 9 5 2 1 3 2 1 1 1 1 1

Two Enumerative Tidbits p. Uniqueness Easy to see: the numbers n t are well-defined and unique.

Two Enumerative Tidbits p. Uniqueness Easy to see: the numbers n t are well-defined and unique. Why? Expand det M t by the first row. The coefficient of n t is 1 by induction.

Two Enumerative Tidbits p. λ(t) If t λ, let λ(t) consist of all squares of λ to the southeast of t.

Two Enumerative Tidbits p. λ(t) If t λ, let λ(t) consist of all squares of λ to the southeast of t. t λ = (4,4,3)

Two Enumerative Tidbits p. λ(t) If t λ, let λ(t) consist of all squares of λ to the southeast of t. t λ = (4,4,3) λ( t ) = (3,2)

Two Enumerative Tidbits p. 1 u λ u λ = #{µ : µ λ}

Two Enumerative Tidbits p. 1 u λ Example. u (2,1) = 5: u λ = #{µ : µ λ} φ

Two Enumerative Tidbits p. 1 u λ Example. u (2,1) = 5: u λ = #{µ : µ λ} φ There is a determinantal formula for u λ, due essentially to MacMahon and later Kreweras (not needed here).

Two Enumerative Tidbits p. 1 Carlitz-Scoville-Roselle theorem Berlekamp (1963) first asked for n t (mod 2) in connection with a coding theory problem. Carlitz-Roselle-Scoville (1971): combinatorial interpretation of n t (over Z). Theorem. n t = u λ(t).

Two Enumerative Tidbits p. 1 An example 7 3 2 1 2 1 1 1 1 1

Two Enumerative Tidbits p. 1 An example 7 3 2 1 2 1 1 1 1 1 φ

Two Enumerative Tidbits p. 1 Smith normal form A: n n matrix over commutative ring R (with 1) Suppose there exist P,Q GL(n,R) such that PAQ = B = diag(d 1 d 2 d n,d 1 d 2 d n 1,...,d 1 ), where d i R. We then call B a Smith normal form (SNF) of A.

Two Enumerative Tidbits p. 1 Existence of SNF If R is a PID, such as Z or K[x] (K = field), then A has a unique SNF up to units.

Two Enumerative Tidbits p. 1 Existence of SNF If R is a PID, such as Z or K[x] (K = field), then A has a unique SNF up to units. Otherwise A typically does not have a SNF but may have one in special cases.

Two Enumerative Tidbits p. 1 Algebraic interpretation of SNF R: a PID A: an n n matrix over R with det(a) 0 and rows v 1,...,v n R n diag(e 1,e 2,...,e n ): SNF of A

Two Enumerative Tidbits p. 1 Algebraic interpretation of SNF R: a PID A: an n n matrix over R with det(a) 0 and rows v 1,...,v n R n diag(e 1,e 2,...,e n ): SNF of A Theorem. R n /(v 1,...,v n ) = (R/e 1 R) (R/e n R).

Two Enumerative Tidbits p. 1 An explicit formula for SNF R: a PID A: an n n matrix over R with det(a) 0 diag(e 1,e 2,...,e n ): SNF of A

Two Enumerative Tidbits p. 1 An explicit formula for SNF R: a PID A: an n n matrix over R with det(a) 0 diag(e 1,e 2,...,e n ): SNF of A Theorem. e n i+1 e n i+2 e n is the gcd of all i i minors of A. minor: determinant of a square submatrix. Special case: e n is the gcd of all entries of A.

Two Enumerative Tidbits p. 1 Many indeterminates For each square (i,j) λ, associate an indeterminate x ij (matrix coordinates).

Two Enumerative Tidbits p. 1 Many indeterminates For each square (i,j) λ, associate an indeterminate x ij (matrix coordinates). x x x x 11 12 13 x 21 22

Two Enumerative Tidbits p. 1 A refinement of u λ u λ (x) = µ λ (i,j) λ/µ x ij

Two Enumerative Tidbits p. 1 A refinement of u λ u λ (x) = µ λ (i,j) λ/µ x ij a b c c d e d e λ µ λ/µ (i,j) λ/µ x ij = cde

Two Enumerative Tidbits p. 1 An example a d b e c abcde+bcde+bce+cde +ce+de+c+e+1 bce+ce+c +e+1 c+1 1 de+e+1 e+1 1 1 1 1 1

Two Enumerative Tidbits p. 2 A t A t = (i,j) λ(t) x ij

Two Enumerative Tidbits p. 2 A t A t = t (i,j) λ(t) x ij a b c d e f g h i j k l m n o

Two Enumerative Tidbits p. 2 A t A t = t (i,j) λ(t) x ij a b c d e f g h i j k l m n o A t = bcdeghiklmo

Two Enumerative Tidbits p. 2 The main theorem Theorem. Let t = (i,j). Then M t has SNF diag(a ij,a i 1,j 1,...,1).

Two Enumerative Tidbits p. 2 The main theorem Theorem. Let t = (i,j). Then M t has SNF diag(a ij,a i 1,j 1,...,1). Proof. 1. Explicit row and column operations putting M t into SNF. 2. (C. Bessenrodt) Induction.

Two Enumerative Tidbits p. 2 An example a d b e c abcde+bcde+bce+cde +ce+de+c+e+1 bce+ce+c +e+1 c+1 1 de+e+1 e+1 1 1 1 1 1

Two Enumerative Tidbits p. 2 An example a d b e c abcde+bcde+bce+cde +ce+de+c+e+1 bce+ce+c +e+1 c+1 1 de+e+1 e+1 1 1 1 1 1 SNF = diag(abcde,e, 1)

Two Enumerative Tidbits p. 2 A special case Let λ be the staircase δ n = (n 1,n 2,...,1). Set each x ij = q.

Two Enumerative Tidbits p. 2 A special case Let λ be the staircase δ n = (n 1,n 2,...,1). Set each x ij = q. u δn 1 (x) xij counts Dyck paths of length 2n by =q (scaled) area, and is thus the well-known q-analogue C n (q) of the Catalan number C n.

Two Enumerative Tidbits p. 2 A q-catalan example C 3 (q) = q 3 + q 2 + 2q + 1

Two Enumerative Tidbits p. 2 A q-catalan example C 3 (q) = q 3 + q 2 + 2q + 1 C 4 (q) C 3 (q) 1 + q C 3 (q) 1 + q 1 1 + q 1 1 SNF diag(q 6,q, 1)

Two Enumerative Tidbits p. 2 A q-catalan example C 3 (q) = q 3 + q 2 + 2q + 1 C 4 (q) C 3 (q) 1 + q C 3 (q) 1 + q 1 1 + q 1 1 SNF diag(q 6,q, 1) x x x

Two Enumerative Tidbits p. 2 q-catalan determinant previously known SNF is new

Two Enumerative Tidbits p. 2 q-catalan determinant previously known SNF is new END OF FIRST TIDBIT

Two Enumerative Tidbits p. 2 The second tidbit A distributive lattice associated with three-term arithmetic progressions

Two Enumerative Tidbits p. 2 Numberplay blog problem New York Times Numberplay blog (March 25, 2013): Let S Z, #S = 8. Can you two-color S such that there is no monochromatic three-term arithmetic progression? bad: 1,2,3,4,5,6,7,8 1, 4, 7 is a monochromatic 3-term progression

Two Enumerative Tidbits p. 2 Compatible pairs Elkies proof is related to the following question: Let 1 i < j < k n and 1 a < b < c n. {i,j,k} and {a,b,c} are compatible if there exist integers x 1 < x 2 < < x n such that x i,x j,x k is an arithmetic progression and x a,x b,x c is an arithmetic progression.

Two Enumerative Tidbits p. 2 An example Example. {1, 2, 3} and {1, 2, 4} are not compatible. Similarly 124 and 134 are not compatible.

Two Enumerative Tidbits p. 2 An example Example. {1, 2, 3} and {1, 2, 4} are not compatible. Similarly 124 and 134 are not compatible. 123 and 134 are compatible, e.g., (x 1,x 2,x 3,x 4 ) = (1, 2, 3, 5).

Two Enumerative Tidbits p. 3 Elkies question What subsets S ( [n] 3 ) have the property that any two elements of S are compatible?

Two Enumerative Tidbits p. 3 Elkies question What subsets S ( [n] 3 ) have the property that any two elements of S are compatible? Example. When n = 4 there are eight such subsets S:, {123}, {124}, {134}, {234}, {123, 134}, {123, 234}, {124, 234}. Not {123, 124}, for instance.

Two Enumerative Tidbits p. 3 Another example Example. For n = 5 one example is S = {123, 234, 345, 135} M 5, achieved by 1 < 2 < 3 < 4 < 5.

Two Enumerative Tidbits p. 3 Conjecture of Elkies Conjecture. #M n = 2 (n 1 2 ).

Two Enumerative Tidbits p. 3 Conjecture of Elkies Conjecture. #M n = 2 (n 1 2 ). Proof (with Fu Liu).

Two Enumerative Tidbits p. 3 Conjecture of Elkies Conjecture. #M n = 2 (n 1 2 ). Proof (with Fu Liu ).

Two Enumerative Tidbits p. 3 A poset on M n Jim Propp: Let Q n be the subposet of [n] [n] [n] (ordered componentwise) defined by Q n = {(i,j,k) : i + j < n + 1 < j + k}. antichain: a subset A of a poset such that if x,y A and x y, then x = y There is a simple bijection from the antichains of Q n to M n induced by (i,j,k) (i,n + 1 j,k).

Two Enumerative Tidbits p. 3 The case n = 4 134 224 124 234 133 124 123 134 ( i, j, k ) ( i, 5 j, k) antichains:, {123}, {124}, {134}, {234}, {123, 134}, {123, 234}, {124, 234}.

Two Enumerative Tidbits p. 3 Order ideals order ideal: a subset I of a poset such that if y I and x y, then x I There is a bijection between antichains A of a poset P and order ideals I of P, namely, A is the set of maximal elements of I.

Two Enumerative Tidbits p. 3 Join-irreducibles join-irreducible of a finite lattice L: an element y such that exactly one element x is maximal with respect to x < y (i.e., y covers x) Theorem (FTFDL). If L is a finite distributive lattice with the subposet P of join-irreducibles, then L = J(P).

Two Enumerative Tidbits p. 3 Why distributive lattices? Two distributive lattices L and L are isomorphic if and only if their posets P and P of join-irreducibles are isomorphic. L and L may be large and complicated, but P and P will be much smaller and (hopefully) more tractable.

Two Enumerative Tidbits p. 3 The case n = 4 124 234 P = Q 4 123 134 124 J(P) = M 4 234 123 134

Two Enumerative Tidbits p. 3 A partial order on M n Recall: there is a simple bijection from the antichains of Q n to M n induced by (i,j,k) (i,n + 1 j,k). Also a simple bijection from antichains of a finite poset to order ideals.

Two Enumerative Tidbits p. 4 Semistandard tableaux T : semistandard Young tableau of shape of shape δ n 1 = (n 2,n 3,...,1), maximum part n 1 1 1 2 3 3 4 4 5 2 5

Two Enumerative Tidbits p. 4 Semistandard tableaux T : semistandard Young tableau of shape of shape δ n 1 = (n 2,n 3,...,1), maximum part n 1 1 1 2 3 3 4 4 5 2 5 L n : poset of all such T, ordered componentwise (a distributive lattice)

Two Enumerative Tidbits p. 4 L 4 and M 4 compared 23 3 124 22 3 13 3 234 12 3 13 2 123 134 11 3 12 2 M 4 L 4 11 2

Two Enumerative Tidbits p. 4 L n = M n Theorem. L n = M n ( = J(Q n )).

Two Enumerative Tidbits p. 4 L n = M n Theorem. L n = M n ( = J(Q n )). Proof. Show that the poset of join-irreducibles of L n is isomorphic to Q n.

Two Enumerative Tidbits p. 4 #L n Theorem. #L n = 2 (n 1 2 ) (proving the conjecture of Elkies).

Two Enumerative Tidbits p. 4 #L n Theorem. #L n = 2 (n 1 2 ) (proving the conjecture of Elkies). Proof. #L n = s δn 2 (1, 1,..., 1). Now use }{{} n 1 hook-content formula.

Two Enumerative Tidbits p. 4 #L n Theorem. #L n = 2 (n 1 2 ) (proving the conjecture of Elkies). Proof. #L n = s δn 2 (1, 1,..., 1). Now use }{{} n 1 hook-content formula. In fact, s δn 2 (x 1,...,x n 1 ) = 1 i<j n 1 (x i + x j ).

Two Enumerative Tidbits p. 4 Maximum size elements of M n f(n): size of largest element S of M n.

Two Enumerative Tidbits p. 4 Maximum size elements of M n f(n): size of largest element S of M n. Example. Recall M 4 = {, {123}, {124}, {134}, {234}, {123, 134}, {123, 234}, {124, 234}}. Thus f(4) = 2.

Two Enumerative Tidbits p. 4 Maximum size elements of M n f(n): size of largest element S of M n. Example. Recall M 4 = {, {123}, {124}, {134}, {234}, Thus f(4) = 2. {123, 134}, {123, 234}, {124, 234}}. Since elements of M n are the antichains of Q n, f(n) is also the maximum size of an antichain of Q n.

Two Enumerative Tidbits p. 4 Evaluation of f(n) Easy result (Elkies): { m 2, n = 2m + 1 f(n) = m(m 1), n = 2m.

Two Enumerative Tidbits p. 4 Evaluation of f(n) Easy result (Elkies): { m 2, n = 2m + 1 f(n) = m(m 1), n = 2m. Conjecture #2 (Elkies). Let g(n) be the number of antichains of Q n of size f(n). (E.g., g(4) = 3.) Then g(n) = { 2 m(m 1), n = 2m + 1 2 (m 1)(m 2) (2 m 1), n = 2m.

Two Enumerative Tidbits p. 4 Maximum size antichains P : finite poset with largest antichain of size m J(P): lattice of order ideals of P D(P) := {x J(P) : x covers m elements} (in bijection with m-element antichains of P ) Easy theorem (Dilworth, 1960). D(P) is a sublattice of J(P) (and hence is a distributive lattice)

Two Enumerative Tidbits p. 4 Example: M 4 124 124 234 234 123 134 123 134 Q 4 M = J(Q ) 4 4 D(Q ) = J(R ) 4 4 R 4

Two Enumerative Tidbits p. 4 Application to Conjecture 2 Recall: g(n) is the number of antichains of Q n of maximum size f(n). Hence g(n) = #D(Q n ). The lattice D(Q n ) is difficult to work with directly, but since it is distributive it is determined by its join-irreducibles R n.

Two Enumerative Tidbits p. 4 Examples of R n R 6 R 7 = ~ Q + Q 4 4

Two Enumerative Tidbits p. 5 Structure of R n n = 2m + 1: R n = Qm+1 + Q m+1. Hence ( g(n) = #J(R n ) = 2 2) ) 2 (m = 2 m(m 1), proving the Conjecture 2 of Elkies for n odd.

Two Enumerative Tidbits p. 5 Structure of R n n = 2m + 1: R n = Qm+1 + Q m+1. Hence ( g(n) = #J(R n ) = 2 2) ) 2 (m = 2 m(m 1), proving the Conjecture 2 of Elkies for n odd. n = 2m: more complicated. R n consists of two copies of Q m+1 with an additional cover relation, but can still be analyzed.

The last slide Two Enumerative Tidbits p. 5