Descent sets of cyclic permutations Sergi Elizalde Dartmouth - PowerPoint PPT Presentation

Definitions Main result Non-bijective proof Final remarks Descent sets of cyclic permutations Sergi Elizalde Dartmouth College AMS Fall Eastern Section Meeting Special Session on Algebraic Combinatorics Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Permutations [ n ] = { 1 , 2 , . . . , n } , π ∈ S n Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Permutations [ n ] = { 1 , 2 , . . . , n } , π ∈ S n π = 2517364 = (1 , 2 , 5 , 3)(4 , 7)(6) = (5 , 3 , 1 , 2)(6)(7 , 4) � �� � � �� � � �� � one line notation cycle notation cycle notation Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Permutations [ n ] = { 1 , 2 , . . . , n } , π ∈ S n π = 2517364 = (1 , 2 , 5 , 3)(4 , 7)(6) = (5 , 3 , 1 , 2)(6)(7 , 4) � �� � � �� � � �� � one line notation cycle notation cycle notation C n ⊂ S n cyclic permutations |C n | = ( n − 1)! C 3 = { (1 , 2 , 3) , (1 , 3 , 2) } = { 231 , 312 } Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Permutations [ n ] = { 1 , 2 , . . . , n } , π ∈ S n π = 2517364 = (1 , 2 , 5 , 3)(4 , 7)(6) = (5 , 3 , 1 , 2)(6)(7 , 4) � �� � � �� � � �� � one line notation cycle notation cycle notation C n ⊂ S n cyclic permutations |C n | = ( n − 1)! C 3 = { (1 , 2 , 3) , (1 , 3 , 2) } = { 231 , 312 } The descent set of π ∈ S n is D ( π ) = { i : 1 ≤ i ≤ n − 1 , π ( i ) > π ( i + 1) } . D (25 · 17 · 36 · 4) = { 2 , 4 , 6 } Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Origin Descents of cyclic permutations come up when determining the smallest number of symbols needed to realize a permutation by shifts. Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Origin Descents of cyclic permutations come up when determining the smallest number of symbols needed to realize a permutation by shifts. E.g., the permutation 4217536 can be realized using three symbols:  2102212210 . . . 4    102212210 . . . 2     02212210 . . . 1   lexicographic order 2212210 . . . 7 of the shifted sequences  212210 . . . 5     12210 . . . 3     2210 . . . 6 Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Origin Descents of cyclic permutations come up when determining the smallest number of symbols needed to realize a permutation by shifts. E.g., the permutation 4217536 can be realized using three symbols:  2102212210 . . . 4    102212210 . . . 2     02212210 . . . 1   lexicographic order 2212210 . . . 7 of the shifted sequences  212210 . . . 5     12210 . . . 3     2210 . . . 6 The number of symbols needed is related to the descents of the cycle (4 , 2 , 1 , 7 , 5 , 3 , 6). Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Descent sets of 5-cycles C 5 C 5 (1 , 2 , 3 , 4 , 5) = 2345 · 1 (2 , 3 , 1 , 4 , 5) = 4 · 3 · 15 · 2 (2 , 4 , 3 , 1 , 5) = 5 · 4 · 13 · 2 (2 , 1 , 3 , 4 , 5) = 3 · 145 · 2 (3 , 2 , 1 , 4 , 5) = 4 · 125 · 3 (4 , 2 , 3 , 1 , 5) = 5 · 3 · 124 (4 , 3 , 2 , 1 , 5) = 5 · 1234 (1 , 4 , 2 , 3 , 5) = 4 · 35 · 2 · 1 (2 , 1 , 4 , 3 , 5) = 4 · 15 · 3 · 2 (1 , 3 , 2 , 4 , 5) = 34 · 25 · 1 (1 , 4 , 3 , 2 , 5) = 45 · 23 · 1 (2 , 3 , 4 , 1 , 5) = 5 · 34 · 12 (3 , 4 , 2 , 1 , 5) = 5 · 14 · 23 (3 , 1 , 2 , 4 , 5) = 24 · 15 · 3 (3 , 1 , 4 , 2 , 5) = 45 · 123 (4 , 2 , 1 , 3 , 5) = 3 · 15 · 24 (4 , 3 , 1 , 2 , 5) = 25 · 134 (1 , 3 , 4 , 2 , 5) = 35 · 4 · 2 · 1 (3 , 4 , 1 , 2 , 5) = 25 · 4 · 13 (1 , 2 , 4 , 3 , 5) = 245 · 3 · 1 (2 , 4 , 1 , 3 , 5) = 345 · 12 (4 , 1 , 3 , 2 , 5) = 35 · 2 · 14 (4 , 1 , 2 , 3 , 5) = 235 · 14 (3 , 2 , 4 , 1 , 5) = 5 · 4 · 2 · 13 Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result Example Non-bijective proof Final remarks Descent sets of 5-cycles C 5 S 4 C 5 S 4 (1 , 2 , 3 , 4 , 5) = 2345 · 1 1234 (2 , 3 , 1 , 4 , 5) = 4 · 3 · 15 · 2 3 · 2 · 14 (2 , 4 , 3 , 1 , 5) = 5 · 4 · 13 · 2 4 · 2 · 13 (2 , 1 , 3 , 4 , 5) = 3 · 145 · 2 2 · 134 (3 , 2 , 1 , 4 , 5) = 4 · 125 · 3 3 · 124 (4 , 2 , 3 , 1 , 5) = 5 · 3 · 124 4 · 3 · 12 (4 , 3 , 2 , 1 , 5) = 5 · 1234 4 · 123 (1 , 4 , 2 , 3 , 5) = 4 · 35 · 2 · 1 3 · 24 · 1 (2 , 1 , 4 , 3 , 5) = 4 · 15 · 3 · 2 2 · 14 · 3 (1 , 3 , 2 , 4 , 5) = 34 · 25 · 1 13 · 24 (1 , 4 , 3 , 2 , 5) = 45 · 23 · 1 14 · 23 (2 , 3 , 4 , 1 , 5) = 5 · 34 · 12 4 · 23 · 1 (3 , 4 , 2 , 1 , 5) = 5 · 14 · 23 4 · 13 · 2 (3 , 1 , 2 , 4 , 5) = 24 · 15 · 3 23 · 14 (3 , 1 , 4 , 2 , 5) = 45 · 123 34 · 12 (4 , 2 , 1 , 3 , 5) = 3 · 15 · 24 3 · 14 · 2 (4 , 3 , 1 , 2 , 5) = 25 · 134 24 · 13 (1 , 3 , 4 , 2 , 5) = 35 · 4 · 2 · 1 14 · 3 · 2 (3 , 4 , 1 , 2 , 5) = 25 · 4 · 13 24 · 3 · 1 (1 , 2 , 4 , 3 , 5) = 245 · 3 · 1 124 · 3 (2 , 4 , 1 , 3 , 5) = 345 · 12 134 · 2 (4 , 1 , 3 , 2 , 5) = 35 · 2 · 14 34 · 2 · 1 (4 , 1 , 2 , 3 , 5) = 235 · 14 234 · 1 (3 , 2 , 4 , 1 , 5) = 5 · 4 · 2 · 13 4 · 3 · 2 · 1 Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks Main theorem Theorem For every n there is a bijection ϕ : C n +1 → S n such that if π ∈ C n +1 and σ = ϕ ( π ) , then D ( π ) ∩ [ n − 1] = D ( σ ) . Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks The bijection ϕ : C n +1 → S n . First step Given π ∈ C n +1 , write it in cycle form with n + 1 at the end: π = (11 , 4 , 10 , 1 , 7 , 16 , 9 , 3 , 5 , 12 , 20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17 , 21) ∈ C 21 Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks The bijection ϕ : C n +1 → S n . First step Given π ∈ C n +1 , write it in cycle form with n + 1 at the end: π = (11 , 4 , 10 , 1 , 7 , 16 , 9 , 3 , 5 , 12 , 20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17 , 21) ∈ C 21 Delete n + 1 and split at the “left-to-right maxima”: σ = (11 , 4 , 10 , 1 , 7)(16 , 9 , 3 , 5 , 12)(20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17) ∈ S 20 . Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks The bijection ϕ : C n +1 → S n . First step Given π ∈ C n +1 , write it in cycle form with n + 1 at the end: π = (11 , 4 , 10 , 1 , 7 , 16 , 9 , 3 , 5 , 12 , 20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17 , 21) ∈ C 21 Delete n + 1 and split at the “left-to-right maxima”: σ = (11 , 4 , 10 , 1 , 7)(16 , 9 , 3 , 5 , 12)(20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17) ∈ S 20 . This map π �→ σ is a bijection Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks The bijection ϕ : C n +1 → S n . First step Given π ∈ C n +1 , write it in cycle form with n + 1 at the end: π = (11 , 4 , 10 , 1 , 7 , 16 , 9 , 3 , 5 , 12 , 20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17 , 21) ∈ C 21 Delete n + 1 and split at the “left-to-right maxima”: σ = (11 , 4 , 10 , 1 , 7)(16 , 9 , 3 , 5 , 12)(20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17) ∈ S 20 . This map π �→ σ is a bijection, but unfortunately it does not always preserve the descent set: π (7) = 16 > π (8) = 13 but σ (7) = 11 < σ (8) = 13 . Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks The bijection ϕ : C n +1 → S n . First step Given π ∈ C n +1 , write it in cycle form with n + 1 at the end: π = (11 , 4 , 10 , 1 , 7 , 16 , 9 , 3 , 5 , 12 , 20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17 , 21) ∈ C 21 Delete n + 1 and split at the “left-to-right maxima”: σ = (11 , 4 , 10 , 1 , 7)(16 , 9 , 3 , 5 , 12)(20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17) ∈ S 20 . This map π �→ σ is a bijection, but unfortunately it does not always preserve the descent set: π (7) = 16 > π (8) = 13 but σ (7) = 11 < σ (8) = 13 . We say that the pair { 7 , 8 } is bad . We will fix the bad pairs. Sergi Elizalde Descent sets of cyclic permutations

Definitions Main result The bijection Non-bijective proof The inverse Final remarks The bijection ϕ : C n +1 → S n . Fixing bad pairs = (11 , 4 , 10 , 1 , 7 , 16 , 9 , 3 , 5 , 12 , 20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17 , 21) π σ = (11 , 4 , 10 , 1 , 7)(16 , 9 , 3 , 5 , 12)(20 , 2 , 6 , 14 , 18 , 8 , 13 , 19 , 15 , 17) Sergi Elizalde Descent sets of cyclic permutations