Let [ n ] q = 1 + q + q 2 + · · · + q n − 1 and [ n ] q ! = [ 1 ] q [ 2 ] q · · · [ n ] q . Define the Gaussian polynomials or q-binomial coeffiecients by � n � [ n ] q ! = [ k ] q ![ n − k ] q ! . k q Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by � n � � [ n ] � � � , C n , . k k q Ex. Consider n = 4, k = 2. So � 4 � [ 4 ] q ! [ 2 ] q ![ 2 ] q ! = 1 + q + 2 q 2 + q 3 + q 4 . = 2 q For g = ( 1 , 3 )( 2 , 4 ) we have o ( g ) = 2 and ω = − 1 so � 4 � = 1 − 1 + 2 − 1 + 1 2 − 1
Let [ n ] q = 1 + q + q 2 + · · · + q n − 1 and [ n ] q ! = [ 1 ] q [ 2 ] q · · · [ n ] q . Define the Gaussian polynomials or q-binomial coeffiecients by � n � [ n ] q ! = [ k ] q ![ n − k ] q ! . k q Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by � n � � [ n ] � � � , C n , . k k q Ex. Consider n = 4, k = 2. So � 4 � [ 4 ] q ! [ 2 ] q ![ 2 ] q ! = 1 + q + 2 q 2 + q 3 + q 4 . = 2 q For g = ( 1 , 3 )( 2 , 4 ) we have o ( g ) = 2 and ω = − 1 so � 4 � = 1 − 1 + 2 − 1 + 1 = 2 2 − 1
Let [ n ] q = 1 + q + q 2 + · · · + q n − 1 and [ n ] q ! = [ 1 ] q [ 2 ] q · · · [ n ] q . Define the Gaussian polynomials or q-binomial coeffiecients by � n � [ n ] q ! = [ k ] q ![ n − k ] q ! . k q Theorem (Reiner-Stanton-White) The c.s.p. is exhibited by � n � � [ n ] � � � , C n , . k k q Ex. Consider n = 4, k = 2. So � 4 � [ 4 ] q ! [ 2 ] q ![ 2 ] q ! = 1 + q + 2 q 2 + q 3 + q 4 . = 2 q For g = ( 1 , 3 )( 2 , 4 ) we have o ( g ) = 2 and ω = − 1 so � 4 � = 1 − 1 + 2 − 1 + 1 = 2 = # S ( 1 , 3 )( 2 , 4 ) . 2 − 1
Outline Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof
Lemma If m ≡ n ( mod d ) and ω = ω d ,
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else.
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d .
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) [ m ] ω =
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω =
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω = If n �≡ 0 ( mod d ) then [ n ] ω � = 0
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω = If n �≡ 0 ( mod d ) then [ n ] ω � = 0 and [ m ] ω / [ n ] ω = 1, proving the ”else” case.
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω = If n �≡ 0 ( mod d ) then [ n ] ω � = 0 and [ m ] ω / [ n ] ω = 1, proving the ”else” case. If n ≡ 0 ( mod d ) then n = ℓ d and m = kd ,
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω = If n �≡ 0 ( mod d ) then [ n ] ω � = 0 and [ m ] ω / [ n ] ω = 1, proving the ”else” case. If n ≡ 0 ( mod d ) then n = ℓ d and m = kd , so = ( 1 + q + · · · + q d − 1 )( 1 + q d + q 2 d + · · · + q ( k − 1 ) d ) [ m ] q ( 1 + q + · · · + q d − 1 )( 1 + q d + q 2 d + · · · + q ( ℓ − 1 ) d ) . [ n ] q
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω = If n �≡ 0 ( mod d ) then [ n ] ω � = 0 and [ m ] ω / [ n ] ω = 1, proving the ”else” case. If n ≡ 0 ( mod d ) then n = ℓ d and m = kd , so = ( 1 + q + · · · + q d − 1 )( 1 + q d + q 2 d + · · · + q ( k − 1 ) d ) [ m ] q ( 1 + q + · · · + q d − 1 )( 1 + q d + q 2 d + · · · + q ( ℓ − 1 ) d ) . [ n ] q Cancelling and plugging in ω = ω d gives [ m ] q = k lim [ n ] q ℓ q → ω
Lemma If m ≡ n ( mod d ) and ω = ω d , then m if n ≡ 0 ( mod d ) , [ m ] q n lim = [ n ] q q → ω 1 else. Proof Let m , n have remainder r modulo d . So for ω = ω d : 0 � �� � ( 1 + ω + · · · + ω d − 1 ) + · · · + ( 1 + ω + · · · + ω r − 1 ) = [ n ] ω . [ m ] ω = If n �≡ 0 ( mod d ) then [ n ] ω � = 0 and [ m ] ω / [ n ] ω = 1, proving the ”else” case. If n ≡ 0 ( mod d ) then n = ℓ d and m = kd , so = ( 1 + q + · · · + q d − 1 )( 1 + q d + q 2 d + · · · + q ( k − 1 ) d ) [ m ] q ( 1 + q + · · · + q d − 1 )( 1 + q d + q 2 d + · · · + q ( ℓ − 1 ) d ) . [ n ] q Cancelling and plugging in ω = ω d gives [ m ] q = k ℓ = m lim n . [ n ] q q → ω
Corollary If ω = ω d and d | n
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else.
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else. Lemma Let g ∈ S n (symmetric group) have disjoint cycle decomposition g = g 1 · · · g k .
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else. Lemma Let g ∈ S n (symmetric group) have disjoint cycle decomposition g = g 1 · · · g k . Let T ⊆ [ n ] .
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else. Lemma Let g ∈ S n (symmetric group) have disjoint cycle decomposition g = g 1 · · · g k . Let T ⊆ [ n ] . Then gT = T ⇐ ⇒ T = g i 1 ∪ · · · ∪ g i m for some i 1 , . . . , i m . return 1 return 2
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else. Lemma Let g ∈ S n (symmetric group) have disjoint cycle decomposition g = g 1 · · · g k . Let T ⊆ [ n ] . Then gT = T ⇐ ⇒ T = g i 1 ∪ · · · ∪ g i m for some i 1 , . . . , i m . return 1 return 2 � [ 6 ] � Ex. If g = ( 1 , 3 , 4 )( 2 , 6 )( 5 ) then the T ∈ with gT = T are 3
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else. Lemma Let g ∈ S n (symmetric group) have disjoint cycle decomposition g = g 1 · · · g k . Let T ⊆ [ n ] . Then gT = T ⇐ ⇒ T = g i 1 ∪ · · · ∪ g i m for some i 1 , . . . , i m . return 1 return 2 � [ 6 ] � Ex. If g = ( 1 , 3 , 4 )( 2 , 6 )( 5 ) then the T ∈ with gT = T are 3 T = { 1 , 3 , 4 }
Corollary If ω = ω d and d | n then, � n / d � � n if d | k, � k / d = k ω 0 else. Lemma Let g ∈ S n (symmetric group) have disjoint cycle decomposition g = g 1 · · · g k . Let T ⊆ [ n ] . Then gT = T ⇐ ⇒ T = g i 1 ∪ · · · ∪ g i m for some i 1 , . . . , i m . return 1 return 2 � [ 6 ] � Ex. If g = ( 1 , 3 , 4 )( 2 , 6 )( 5 ) then the T ∈ with gT = T are 3 T = { 1 , 3 , 4 } T = { 2 , 5 , 6 } . and
Proposition � [ n ] � If S = and g ∈ C n has o ( g ) = d k
Proposition � [ n ] � If S = and g ∈ C n has o ( g ) = d then k � n / d � if d | k, # S g = k / d
Proposition � [ n ] � If S = and g ∈ C n has o ( g ) = d then k � n / d � if d | k, # S g = k / d 0 else.
Proposition � [ n ] � If S = and g ∈ C n has o ( g ) = d then k � n / d � if d | k, # S g = k / d 0 else. Proof If g ∈ C n and o ( g ) = d then g = g 1 · · · g n / d where # g 1 = . . . = # g n / d = d .
Proposition � [ n ] � If S = and g ∈ C n has o ( g ) = d then k � n / d � if d | k, # S g = k / d 0 else. Proof If g ∈ C n and o ( g ) = d then g = g 1 · · · g n / d where # g 1 = . . . = # g n / d = d . So, by the second lemma , � [ n ] � T ∈ satisfies gT = T iff T is a union of k / d of the n / d k cycles g i .
Outline Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } .
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } . Element g ∈ G corresponds to an invertible linear map [ g ] .
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } . Element g ∈ G corresponds to an invertible linear map [ g ] . If B is a basis for V then let [ g ] B be the matrix of [ g ] in B .
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } . Element g ∈ G corresponds to an invertible linear map [ g ] . If B is a basis for V then let [ g ] B be the matrix of [ g ] in B . In particular, [ g ] S is the permutation matrix for g acting on S .
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } . Element g ∈ G corresponds to an invertible linear map [ g ] . If B is a basis for V then let [ g ] B be the matrix of [ g ] in B . In particular, [ g ] S is the permutation matrix for g acting on S . Example. G = S 3 acts on S = { 1 , 2 , 3 } and so on C S = { c 1 1 + c 2 2 + c 3 3 : c 1 , c 2 , c 3 ∈ C } .
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } . Element g ∈ G corresponds to an invertible linear map [ g ] . If B is a basis for V then let [ g ] B be the matrix of [ g ] in B . In particular, [ g ] S is the permutation matrix for g acting on S . Example. G = S 3 acts on S = { 1 , 2 , 3 } and so on C S = { c 1 1 + c 2 2 + c 3 3 : c 1 , c 2 , c 3 ∈ C } . For g = ( 1 , 2 )( 3 ) and basis S : ( 1 , 2 )( 3 ) 1 = 2 , ( 1 , 2 )( 3 ) 2 = 1 , ( 1 , 2 )( 3 ) 3 = 3 .
If G acts on S = { s 1 , . . . , s k } then G also acts on the vector space V = C S = { c 1 s 1 + · · · + c k s k : c i ∈ C for all i } . Element g ∈ G corresponds to an invertible linear map [ g ] . If B is a basis for V then let [ g ] B be the matrix of [ g ] in B . In particular, [ g ] S is the permutation matrix for g acting on S . Example. G = S 3 acts on S = { 1 , 2 , 3 } and so on C S = { c 1 1 + c 2 2 + c 3 3 : c 1 , c 2 , c 3 ∈ C } . For g = ( 1 , 2 )( 3 ) and basis S : ( 1 , 2 )( 3 ) 1 = 2 , ( 1 , 2 )( 3 ) 2 = 1 , ( 1 , 2 )( 3 ) 3 = 3 . And so 0 1 0 . [( 1 , 2 )( 3 )] S = 1 0 0 0 0 1
A G-module is any C -vector space V where G acts by invertible linear transformations.
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] .
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis.
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1)
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1) If G is cyclic, then there will be a basis B for C S such that every g ∈ G satisfies m 2 m 0 m 1 � �� � � �� � � �� � ω 2 , . . . , ω 2 , . . . ) [ g ] B = diag ( 1 , . . . , 1 , ω, . . . , ω, where ω = ω o ( g ) .
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1) If G is cyclic, then there will be a basis B for C S such that every g ∈ G satisfies m 2 m 0 m 1 � �� � � �� � � �� � ω 2 , . . . , ω 2 , . . . ) [ g ] B = diag ( 1 , . . . , 1 , ω, . . . , ω, where ω = ω o ( g ) . Thus χ ( g ) = tr [ g ] B
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1) If G is cyclic, then there will be a basis B for C S such that every g ∈ G satisfies m 2 m 0 m 1 � �� � � �� � � �� � ω 2 , . . . , ω 2 , . . . ) [ g ] B = diag ( 1 , . . . , 1 , ω, . . . , ω, where ω = ω o ( g ) . Thus � m i ω i χ ( g ) = tr [ g ] B = i ≥ 0
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1) If G is cyclic, then there will be a basis B for C S such that every g ∈ G satisfies m 2 m 0 m 1 � �� � � �� � � �� � ω 2 , . . . , ω 2 , . . . ) [ g ] B = diag ( 1 , . . . , 1 , ω, . . . , ω, where ω = ω o ( g ) . Thus � m i ω i = f ( ω ) . χ ( g ) = tr [ g ] B = (2) i ≥ 0 where f ( q ) = � i ≥ 0 m i q i .
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1) If G is cyclic, then there will be a basis B for C S such that every g ∈ G satisfies m 2 m 0 m 1 � �� � � �� � � �� � ω 2 , . . . , ω 2 , . . . ) [ g ] B = diag ( 1 , . . . , 1 , ω, . . . , ω, where ω = ω o ( g ) . Thus � m i ω i = f ( ω ) . χ ( g ) = tr [ g ] B = (2) i ≥ 0 i ≥ 0 m i q i . Now (1) and (2) imply f ( ω ) = # S g so where f ( q ) = � we have the c.s.p.
A G-module is any C -vector space V where G acts by invertible linear transformations. The character of G on V is the function χ : G → C given by χ ( g ) = tr [ g ] . Note that χ is well defined in that the trace is independent of the basis. For a group G acting on S , the character on C S is χ ( g ) = tr [ g ] S = # S g . (1) If G is cyclic, then there will be a basis B for C S such that every g ∈ G satisfies m 2 m 0 m 1 � �� � � �� � � �� � ω 2 , . . . , ω 2 , . . . ) [ g ] B = diag ( 1 , . . . , 1 , ω, . . . , ω, where ω = ω o ( g ) . Thus � m i ω i = f ( ω ) . χ ( g ) = tr [ g ] B = (2) i ≥ 0 i ≥ 0 m i q i . Now (1) and (2) imply f ( ω ) = # S g so where f ( q ) = � � [ n ] � we have the c.s.p. To get the S = example, one uses the k k th exterior power of a vector space V with dim V = n .
Outline Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T .
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G we then find a partition of S π = π g = { B 1 , B 2 , . . . } satisfying, the following two criteria where ω = ω o ( g ) :
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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.
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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.
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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 ∈ G f ( ω )
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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 ∈ G � f ( ω ) = wt T | ω T ∈ S
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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 ∈ G � � f ( ω ) = wt T | ω = wt B i | ω T ∈ S i
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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 ∈ G # S g � � � �� � f ( ω ) = wt T | ω = wt B i | ω = 1 + · · · + 1 + 0 + 0 + · · · T ∈ S i
To combinatorially prove ( S , G , f ( q )) exhibits the c.s.p., first find a weight function wt : S → Z [ q ] such that � f ( q ) = wt T . (3) T ∈ S If B ⊆ S we let wt B = � T ∈ B wt T . For each g ∈ G 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 ∈ G # 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 ) . � Proof (Roichman & S) 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So : 12 34 , T 13 14 23 24 � = 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So : 12 34 , T 13 14 23 24 � 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T Suppose g ∈ C n with o ( g ) = d , Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So : 12 34 , T 13 14 23 24 � 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T Suppose g ∈ C n with o ( g ) = d , so g = g 1 . . . g n / d where # g 1 = . . . = # g n / d = d . Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So : 12 34 , T 13 14 23 24 � 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T Suppose g ∈ C n with o ( g ) = d , so g = g 1 . . . g n / d where # g 1 = . . . = # g n / d = d . Suppose h ∈ S n satisfies h = h 1 . . . h n / d where # h 1 = . . . = # h n / d = d . Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So : 12 34 , T 13 14 23 24 � 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 ) . � Proof (Roichman & S) For T ∈ let wt T = q k � n � � wt T = . ∴ k q T Suppose g ∈ C n with o ( g ) = d , so g = g 1 . . . g n / d where # g 1 = . . . = # g n / d = d . Suppose h ∈ S n satisfies h = h 1 . . . h n / d where # h 1 = . . . = # h n / d = d . � g = # � h . � [ n ] � [ n ] Then, by the second lemma , # k k Ex. If n = 4 and k = 2 then wt { t 1 , t 2 } = q t 1 + t 2 − 3 . So : 12 34 , T 13 14 23 24 � 4 � � = q 0 q 1 q 2 q 2 q 3 q 4 T wt T + + + + + = . 2 q
Let h = ( 1 , 2 , . . . , d )( d + 1 , d + 2 , . . . , 2 d ) · · ·
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