Combinatorial and colorful proofs of cyclic sieving phenomena Bruce - PowerPoint PPT Presentation

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T .

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) :

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1.

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1. (II) For i > # S g we have # B i > 1 and wt B i | ω = 0.

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1. (II) For i > # S g we have # B i > 1 and wt B i | ω = 0. We then have the c.s.p. since for each g ∈ C f ( ω )

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1. (II) For i > # S g we have # B i > 1 and wt B i | ω = 0. We then have the c.s.p. since for each g ∈ C � f ( ω ) = wt T | ω T ∈ S

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1. (II) For i > # S g we have # B i > 1 and wt B i | ω = 0. We then have the c.s.p. since for each g ∈ C � � f ( ω ) = wt T | ω = wt B i | ω T ∈ S i

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1. (II) For i > # S g we have # B i > 1 and wt B i | ω = 0. We then have the c.s.p. since for each g ∈ C # S g � � � �� � f ( ω ) = wt T | ω = wt B i | ω = 1 + · · · + 1 + 0 + 0 + · · · T ∈ S i

Most proofs of the c.s.p. involve either explicitly evaluating polynomials at roots of unity or representation theory. We have given the first purely combinatorial proof. To combinatorially prove ( S , C , f ( q )) exhibits the c.s.p., first find a weight function wt : S → R [ q ] such that � f ( q ) = wt T . (1) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ C we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) : (I) For 1 ≤ i ≤ # S g we have # B i = 1 and wt B i | ω = 1. (II) For i > # S g we have # B i > 1 and wt B i | ω = 0. We then have the c.s.p. since for each g ∈ C # S g � � � �� � 1 + · · · + 1 + 0 + 0 + · · · = # S g . f ( ω ) = wt T | ω = wt B i | ω = T ∈ S i

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k )

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 .

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � 4 � � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T = . 2 q

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Suppose g ∈ C n with o ( g ) = d , say g = ( 1 , . . . , n ) n / d Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � 4 � � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T = . 2 q

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Suppose g ∈ C n with o ( g ) = d , say g = ( 1 , . . . , n ) n / d so g = ( 1 , 1 + n / d , 1 + 2 n / d , . . . )( 2 , 2 + n / d , 2 + 2 n / d , . . . ) · · · Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � 4 � � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T = . 2 q

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Suppose g ∈ C n with o ( g ) = d , say g = ( 1 , . . . , n ) n / d so g = ( 1 , 1 + n / d , 1 + 2 n / d , . . . )( 2 , 2 + n / d , 2 + 2 n / d , . . . ) · · · Let g i = ( i , i + n / d , i + 2 n / d , . . . ) for 1 ≤ i ≤ n / d . Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � 4 � � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T = . 2 q

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Suppose g ∈ C n with o ( g ) = d , say g = ( 1 , . . . , n ) n / d so g = ( 1 , 1 + n / d , 1 + 2 n / d , . . . )( 2 , 2 + n / d , 2 + 2 n / d , . . . ) · · · So T ∈ S g Let g i = ( i , i + n / d , i + 2 n / d , . . . ) for 1 ≤ i ≤ n / d . iff T can be written as T = g i 1 ⊎ g i 2 ⊎ · · · Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � 4 � � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T = . 2 q

Theorem (Reiner, Stanton, White) � n � � [ n ] � � � The c.s.p. is exhibited by the triple , C n , . k k q � [ n ] � t ∈ T t − ( k + 1 2 ) . P Combinatorial Proof. For T ∈ let wt T = q k � n � � wt T = . ∴ k q T ∈ ( [ n ] k ) Suppose g ∈ C n with o ( g ) = d , say g = ( 1 , . . . , n ) n / d so g = ( 1 , 1 + n / d , 1 + 2 n / d , . . . )( 2 , 2 + n / d , 2 + 2 n / d , . . . ) · · · So T ∈ S g Let g i = ( i , i + n / d , i + 2 n / d , . . . ) for 1 ≤ i ≤ n / d . iff T can be written as T = g i 1 ⊎ g i 2 ⊎ · · · Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So T : 12 13 14 23 24 34 , � 4 � � = q 0 + q 1 + q 2 + q 2 + q 3 + q 4 T wt T = . 2 q If g = ( 1 , 3 )( 2 , 4 ) then S g = { 13 , 24 } .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # k k Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. k Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 }

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 ,

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . ∴ wt B = wt T + wt h i T + · · · + wt h d − 1 T i Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . ∴ wt B = wt T + wt h i T + · · · + wt h d − 1 T i ∴ wt B | ω = ω j + Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . ∴ wt B = wt T + wt h i T + · · · + wt h d − 1 T i ∴ wt B | ω = ω j + ω j + ℓ + Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . ∴ wt B = wt T + wt h i T + · · · + wt h d − 1 T i ∴ wt B | ω = ω j + ω j + ℓ + · · · + ω j +( d − 1 ) ℓ Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . ∴ wt B = wt T + wt h i T + · · · + wt h d − 1 T i ∴ wt B | ω = ω j + ω j + ℓ + · · · + ω j +( d − 1 ) ℓ = ω j 1 − ω d ℓ 1 − ω ℓ Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · · and h i = ( id + 1 , id + 2 , . . . , ( i + 1 ) d ) for 0 ≤ i < n / d . Since g and � g = # � h . � [ n ] � [ n ] h have the same cycle type, # For any k k � [ n ] � T ∈ define the block B of π containing T by as follows. If k hT = T then B = { T } . If hT � = T , then find the smallest index i such that 0 < #( T ∩ h i ) < d and let B = { T , h i T , h 2 i T , . . . , h d − 1 T } . i Proof of (II): If ω = ω d , wt T = q j , ℓ = | T ∩ h i | then 0 < ℓ < d . ∴ wt B = wt T + wt h i T + · · · + wt h d − 1 T i ∴ wt B | ω = ω j + ω j + ℓ + · · · + ω j +( d − 1 ) ℓ = ω j 1 − ω d ℓ 1 − ω ℓ = 0 since ω d = 1 and ω ℓ � = 1. Ex: n = 4, k = 2, g = ( 1 , 3 )( 2 , 4 ) . So h = ( 1 , 2 )( 3 , 4 ) , and π : { 12 } , { 34 } , { 13 , ( 1 , 2 ) 13 } = { 13 , 23 } , { 14 , ( 1 , 2 ) 14 } = { 14 , 24 } . wt { 12 }| − 1 = ( − 1 ) 0 = 1 , wt { 13 , 23 }| − 1 = ( − 1 ) 1 + ( − 1 ) 2 = 0 , wt { 34 }| − 1 = ( − 1 ) 4 = 1 , wt { 14 , 24 }| − 1 = ( − 1 ) 2 + ( − 1 ) 3 = 0 .

Outline Definitions and an example A combinatorial proof A colorful proof Future work

A triangulation , T , is a subdivision of a regular polygon P into triangles using noncrossing diagonals.

1 1 1 1 ❜ q q q q ✂ ❇ ❇ ✂ ❇ ✂✂ ❇ ❜ ✂ ❇ ✂ ❇ ❇ ❜ ❇ ✂ ❜ ❇ ✂ ❇ ❇ ❇ ✂ 2 q 2 2 q 2 q q ❅ � ❅ � ❅ ❇ � ❅ ❇ ✂ � ❅ ❇ � ❅ ❇ � ✂ q q 1 1 Figure: Two triangulations: proper (left) and improper (right) A triangulation , T , is a subdivision of a regular polygon P into triangles using noncrossing diagonals.

1 1 1 1 ❜ q q q q ✂ ❇ ❇ ✂ ❇ ✂✂ ❇ ❜ ✂ ❇ ✂ ❇ ❇ ❜ ❇ ✂ ❜ ❇ ✂ ❇ ❇ ❇ ✂ 2 q 2 2 q 2 q q ❅ � ❅ � ❅ ❇ � ❅ ❇ ✂ � ❅ ❇ � ❅ ❇ � ✂ q q 1 1 Figure: Two triangulations: proper (left) and improper (right) A triangulation , T , is a subdivision of a regular polygon P into triangles using noncrossing diagonals. Let T n be the set of all triangulations of an n -gon.

1 1 1 1 ❜ q q q q ✂ ❇ ❇ ✂ ❇ ✂✂ ❇ ❜ ✂ ❇ ✂ ❇ ❇ ❜ ❇ ✂ ❜ ❇ ✂ ❇ ❇ ❇ ✂ 2 q 2 2 q 2 q q ❅ � ❅ � ❅ ❇ � ❅ ❇ ✂ � ❅ ❇ � ❅ ❇ � ✂ q q 1 1 Figure: Two triangulations: proper (left) and improper (right) A triangulation , T , is a subdivision of a regular polygon P into triangles using noncrossing diagonals. Let T n be the set of all triangulations of an n -gon. Then � 2 n � 1 # T n + 2 = . n + 1 n

1 1 1 1 ❜ q q q q ✂ ❇ ❇ ✂ ❇ ✂✂ ❇ ❜ ✂ ❇ ✂ ❇ ❇ ❜ ❇ ✂ ❜ ❇ ✂ ❇ ❇ ❇ ✂ 2 q 2 2 q 2 q q ❅ � ❅ � ❅ ❇ � ❅ ❇ ✂ � ❅ ❇ � ❅ ❇ � ✂ q q 1 1 Figure: Two triangulations: proper (left) and improper (right) A triangulation , T , is a subdivision of a regular polygon P into triangles using noncrossing diagonals. Let T n be the set of all triangulations of an n -gon. Then � 2 n � 1 # T n + 2 = . n + 1 n Let C n be the group of rotations of a regular n -gon.

1 1 1 1 ❜ q q q q ✂ ❇ ❇ ✂ ❇ ✂✂ ❇ ❜ ✂ ❇ ✂ ❇ ❇ ❜ ❇ ✂ ❜ ❇ ✂ ❇ ❇ ❇ ✂ 2 q 2 2 q 2 q q ❅ � ❅ � ❅ ❇ � ❅ ❇ ✂ � ❅ ❇ � ❅ ❇ � ✂ q q 1 1 Figure: Two triangulations: proper (left) and improper (right) A triangulation , T , is a subdivision of a regular polygon P into triangles using noncrossing diagonals. Let T n be the set of all triangulations of an n -gon. Then � 2 n � 1 # T n + 2 = . n + 1 n Let C n be the group of rotations of a regular n -gon. Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by the triple � 2 n � � � 1 T n + 2 , C n + 2 , . [ n + 1 ] q n q

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . .

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . . Call a triangulation proper if it contains no monochromatic triangle.

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . . Call a triangulation proper if it contains no monochromatic triangle. Let P n be the set of proper triangulations of a regular n -gon.

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . . Call a triangulation proper if it contains no monochromatic triangle. Let P n be the set of proper triangulations of a regular n -gon. Theorem (S) We have  � 3 m � 2 m  if n = 2 m,   2 m + 1 m   # P n + 2 = � 3 m + 1 �  2 m + 1   if n = 2 m + 1 .   2 m + 2 m

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . . Call a triangulation proper if it contains no monochromatic triangle. Let P n be the set of proper triangulations of a regular n -gon. Theorem (S) We have  � 3 m � 2 m  if n = 2 m,   2 m + 1 m   # P n + 2 = � 3 m + 1 �  2 m + 1   if n = 2 m + 1 .   2 m + 2 m Note that for n odd, rotation does not preserve properness.

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . . Call a triangulation proper if it contains no monochromatic triangle. Let P n be the set of proper triangulations of a regular n -gon. Theorem (S) We have  � 3 m � 2 m  if n = 2 m,   2 m + 1 m   # P n + 2 = � 3 m + 1 �  2 m + 1   if n = 2 m + 1 .   2 m + 2 m Note that for n odd, rotation does not preserve properness. If n = 2 m then let � + 2 ⌈ m / 2 ⌉− 1 � − [ 2 ] ⌈ m / 2 ⌉− 1 � 3 m ( 1 + q 2 ) [ 2 ] m − 1 � q q p n ( q ) = . m [ 2 m + 1 ] q q

Label (color) the vertices of P cyclically 1 , 2 , 1 , 2 , . . . Call a triangulation proper if it contains no monochromatic triangle. Let P n be the set of proper triangulations of a regular n -gon. Theorem (S) We have  � 3 m � 2 m  if n = 2 m,   2 m + 1 m   # P n + 2 = � 3 m + 1 �  2 m + 1   if n = 2 m + 1 .   2 m + 2 m Note that for n odd, rotation does not preserve properness. If n = 2 m then let � + 2 ⌈ m / 2 ⌉− 1 � − [ 2 ] ⌈ m / 2 ⌉− 1 � 3 m ( 1 + q 2 ) [ 2 ] m − 1 � q q p n ( q ) = . m [ 2 m + 1 ] q q Theorem (Roichman-S) If n = 2 m then ( P n + 2 , C n + 2 , p n ( q )) exhibits the c.s.p.

Outline Definitions and an example A combinatorial proof A colorful proof Future work

I. Is there a combinatorial proof of the Reiner-Stanton-White theorem about (uncolored) triangulations?

I. Is there a combinatorial proof of the Reiner-Stanton-White theorem about (uncolored) triangulations? The first difficulty is to find a weight function wt : T n → R [ q ] such that (a) we have � 2 n � 1 � wt T = , [ n + 1 ] q n q T ∈T n + 2

I. Is there a combinatorial proof of the Reiner-Stanton-White theorem about (uncolored) triangulations? The first difficulty is to find a weight function wt : T n → R [ q ] such that (a) we have � 2 n � 1 � wt T = , [ n + 1 ] q n q T ∈T n + 2 (b) and wt T is well behaved with respect to rotation.

I. Is there a combinatorial proof of the Reiner-Stanton-White theorem about (uncolored) triangulations? The first difficulty is to find a weight function wt : T n → R [ q ] such that (a) we have � 2 n � 1 � wt T = , [ n + 1 ] q n q T ∈T n + 2 (b) and wt T is well behaved with respect to rotation. Note that there are various other families of combinatorial objects (Dyck paths, 2-rowed standard Young tableaux) with a weighting giving the q -Catalan numbers.

I. Is there a combinatorial proof of the Reiner-Stanton-White theorem about (uncolored) triangulations? The first difficulty is to find a weight function wt : T n → R [ q ] such that (a) we have � 2 n � 1 � wt T = , [ n + 1 ] q n q T ∈T n + 2 (b) and wt T is well behaved with respect to rotation. Note that there are various other families of combinatorial objects (Dyck paths, 2-rowed standard Young tableaux) with a weighting giving the q -Catalan numbers. The hope is that one of these can be reformulated in terms of triangulations in a way that (b) above will be satisfied.

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals.

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals. So if k = n − 3 then we have a triangulation.

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals. So if k = n − 3 then we have a triangulation. We have � n + k �� n − 3 � 1 # D n , k = . n + k k + 1 k

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals. So if k = n − 3 then we have a triangulation. We have � n + k �� n − 3 � 1 # D n , k = . n + k k + 1 k There is an action of C n on dissections just as on triangulations.

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals. So if k = n − 3 then we have a triangulation. We have � n + k �� n − 3 � 1 # D n , k = . n + k k + 1 k There is an action of C n on dissections just as on triangulations. Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by the triple � n + k � n − 3 � � � � 1 D n , k , C n , . k + 1 k [ n + k ] q q q

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals. So if k = n − 3 then we have a triangulation. We have � n + k �� n − 3 � 1 # D n , k = . n + k k + 1 k There is an action of C n on dissections just as on triangulations. Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by the triple � n + k � n − 3 � � � � 1 D n , k , C n , . k + 1 k [ n + k ] q q q Burstein-Roichman-S are investigating proper dissections (no monochromatic sub-polygon) even for q = 1.

II. Let D n , k be the set of all dissections of a regular n -gon using k noncrossing diagonals. So if k = n − 3 then we have a triangulation. We have � n + k �� n − 3 � 1 # D n , k = . n + k k + 1 k There is an action of C n on dissections just as on triangulations. Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by the triple � n + k � n − 3 � � � � 1 D n , k , C n , . k + 1 k [ n + k ] q q q Burstein-Roichman-S are investigating proper dissections (no monochromatic sub-polygon) even for q = 1. So far we have proved a formula for triangulations with a different coloring scheme which involves a new basis for the algebra of symmetric functions.