Expanding permutation statistics as sums of permutation patterns

Transcription

1 Expanding permutation statistics as sums of permutation patterns Petter Branden Stockholm University Anders Claesson Reykjavik University 1 / 23

2 Permutations/patterns as functions Think of 2 S as a function : S! N that counts occurrences of Example 1 = j j 21 = inv 2 / 23

3 Statistics as linear combinations of patterns Any function stat : S! C may be represented uniquely as a (typically innite) sum stat = X 2S () where f()g 2S C Example lmax = X ( 1) jj 1 = S (jj)=1 3 / 23

4 The incidence algebra I(P ) Let Q be a locally nite poset. The incidence algebra, I(Q), is the C-algebra of all functions Q Q! C with multiplication X (F G)(x; z) = xyz F (x; y)g(y; z) and identity (x; y) = ( 1 if x = y; 0 if x 6= y 4 / 23

5 The incidence algebra I(S) Dene in S if () > 0 Dene P 2 I(S) by P (; ) = () : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : / 23

6 The incidence algebra I(S) Dene in S if () > 0 Dene P 2 I(S) by P (; ) = () P is invertible because P (; ) = 1. Therefore, for any stat : S! C, there are unique scalars f()g 2S C such that stat = X 2S (): (1) Indeed, I(S) acts on the right of C S by (f F )() = X f()f (; ): Thus (1) i stat = P i = stat P 1. 5 / 23

7 lmax = Why? 6 / 23

8 lmax = Why? des; maj; exc; x; : : : How do we expand them? 6 / 23

9 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

10 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

11 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

12 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

13 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

14 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

15 Mesh patterns A mesh pattern is a pair p = (; R) with 2 S k and R [0; k] [0; k] Example p = 3241; f(0; 2); (1; 3); (1; 4); (4; 2); (4; 3)g = 7 / 23

16 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R 8 / 23

17 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

18 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

19 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

20 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

21 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

22 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

23 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example With p = (1; f(0; 1)g) = p( ) = p 0 we have 1 C A = 3 8 / 23

24 What is an occurrence of a mesh pattern? p = (; R) : S! N p() is the number of \classical" occurrences of in such that no elements of are in the shaded regions dened by R Example (Chayne Homsberger) # consecutive adjacent entries in =! + () 8 / 23

25 p = (; R) 2 S k [0; k] 2 classic: R = ; barred: R = f(i 1; (i) 1)g = segment: R = [1; k 1] 2 ddes = dashed/vincular: R = [ vertical strips 23-1 = bivincular: R = [ vertical and horizontal strips 9 / 23

26 The reciprocity theorem Theorem (Reciprocity) For any mesh pattern p = (; R) we have p = X 2S ( 1) jj jj p? () Denition (Dual pattern) For p = (; R) let p? = (; R c ), where R c = [0; jj] 2 n R:!? = 10 / 23

27 The reciprocity theorem Theorem (Reciprocity) For any mesh pattern p = (; R) we have p = X 2S ( 1) jj jj p? () Denition (Dual pattern) For p = (; R) let p? = (; R c ), where R c = [0; jj] 2 n R:!? = 10 / 23

28 We can now explain why lmax = We have lmax = Thus and? =. lmax = X 2S () where () = ( 1) jj 1 () = ( ( 1) jj 1 if (jj) = 1; 0 otherwise 11 / 23

29 Also, des = We have des = Thus and? =. des = X 2S () where () = ( 1) jj () = ( ( 1) jj if (1) > (jj); 0 otherwise 12 / 23

30 Babson and Steingrmsson classied Mahonian statistics using patterns. For instance maj = (21) + (1-32) + (2-31) + (3-21) = Thus we may write maj = P 2S () where ( 1) j j ( ) = This last expression simplies to () = where n = jj 8 >< >: 1 if = 21 ( 1) n if (2) < (n) < (1) ( 1) n+1 if (1) < (n) < (2) 0 otherwise 13 / 23

31 If a 1 : : : a n 2 S n, then a i is called a strong xed point if j < i =) a j < a i j > i =) a j > a i. and 14 / 23

32 If a 1 : : : a n 2 S n, then a i is called a strong xed point if Let j < i =) a j < a i j > i =) a j > a i. and sx() = # strong xed points of ssx() = # strong xed points of r (skew sx) 14 / 23

33 If a 1 : : : a n 2 S n, then a i is called a strong xed point if Let j < i =) a j < a i j > i =) a j > a i. and sx() = # strong xed points of ssx() = # strong xed points of r (skew sx) Theorem sx = X ( 1) jj 1 ssx() Proof. sx = and ssx = 14 / 23

34 x Q 2 (; x) Q 1 (; x) Q 3 (; x) Q 4 (; x) j Q 1 (; x) = f (i) : i > j; (i) > x g Q 2 (; x) = f (i) : i < j; (i) > x g Q 3 (; x) = f (i) : i < j; (i) < x g Q 4 (; x) = f (i) : i > j; (i) < x g The point x in is a xed point if jq 2 (; x)j = jq 4 (; x)j an excedance if jq 4 (; x)j > jq 2 (; x)j 15 / 23

35 A corollary to the reciprocity theorem Recall that P (; ) = (). Theorem (Inverse) The inverse of P in I(S) is given by P 1 (; ) = ( 1) j j jj P (; ) 16 / 23

36 A corollary to the reciprocity theorem Recall that P (; ) = (). Theorem (Inverse) The inverse of P in I(S) is given by P 1 (; ) = ( 1) j j jj P (; ) Proof. For 2 S k, let p = (; [0; k] [0; k]). Then p? = (; ;) and Thus p() = X 2S( 1) jj jj p? ()() (reciprocity) (; ) = X ( 1) jj jj P (; )P (; ) 16 / 23

37 By the inverse theorem (but not trivially): x = X exc = X ( 1) jj 1 ( 1) jj 2 X x2ssf() X x2ssf()!1 jj 1 A x 1!1 jj 2 A x 2 where SSF() is the set of skew strong xed points in 17 / 23

38 Alternating permutations and the Euler numbers A permutation 2 S n is said to be alternating if (1) > (2) < (3) > (4) < Alternating permutations are those that avoid ; and In 1879, Andre showed that the number of alternating permutations in S n is the Euler number E n given by X n0 E n x n =n! = sec x + tan x: 18 / 23

39 Simsun permutations A permutation 2 S n is simsun if for all i 2 [1; n], after removing the i largest letters of, the remaining word has no double descents. A permutation is simsun if and only if it avoids the pattern simsun = The number of simsun permutations in S n is E n / 23

40 Andre permutations Andre permutations of various kinds were introduced by Foata and Schutzenberger and further studied by Foata and Strehl. If 2 S n and x = (i) 2 [1; n] let (x); (x) [1; n] be dened as follows. Let (0) = (n + 1) = 1. (x) = f(k) : j 0 < k < ig where j 0 = maxfj : j < i and (j) < (i)g, and (x) = f(k) : i < k < j 1 g where j 1 = minfj : i < j and (j) < (i)g. Then is an Andre permutation of the rst kind if max (x) max (x) for all x 2 [1; n], where max ; = 1. In particular, has no double descents and (n) = n. 20 / 23

41 Andre permutations Andre permutations of various kinds were introduced by Foata and Schutzenberger and further studied by Foata and Strehl. If 2 S n and x = (i) 2 [1; n] let (x); (x) [1; n] be dened as follows. Let (0) = (n + 1) = 1. (x) = f(k) : j 0 < k < ig where j 0 = maxfj : j < i and (j) < (i)g, and (x) = f(k) : i < k < j 1 g where j 1 = minfj : i < j and (j) < (i)g. Then is an Andre permutation of the rst kind if max (x) max (x) for all x 2 [1; n], where max ; = 1. In particular, has no double descents and (n) = n. 20 / 23

42 Andre permutations Fact: There are exactly E n Andre permutations in S n. Theorem Let 2 S n. Then is an Andre permutation of the rst kind if and only if it avoids and = andre Corollary S n (andre) = E n+1 21 / 23

43 A Wilf-equivalence Corollary The two patterns simsun = and andre = are Wilf-equivalent 22 / 23

44 A Wilf-equivalence Corollary The two patterns simsun = and andre = are Wilf-equivalent Moreover, these are the only essentially dierent patterns in this Wilf-class 22 / 23

45 We have barely scratched the surface. For n = 3 we have # patterns = # 2 element sets of patterns = # 3 element sets of patterns = # 4 element sets of patterns = p = (; R) 2 S k [0; k] 2 Restrict R? R as a relation: reexive, symmetric, transitive,... R as a digraph: acyclic, rooted tree, tournament, / 23