The cyclic sieving phenomenon - an introduction Bruce Sagan - - PowerPoint PPT Presentation

The cyclic sieving phenomenon - an introduction

Bruce Sagan Department of Mathematics Michigan State University East Lansing, MI 48824-1027 sagan@math.msu.edu www.math.msu.edu/˜sagan July 5, 2011

Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof

Outline

Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof

Suppose S is a set and let G be a finite cyclic group acting on S.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t}

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity. Finally, suppose f(q) is a polynomial in q with coefficients in Z.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity. Finally, suppose f(q) is a polynomial in q with coefficients in Z.

Definition (Reiner-Stanton-White, 2004)

The triple (S, G, f(q)) exhibits the cyclic sieving phenomenon (c.s.p.) if, for all g ∈ G, we have #Sg = f(ω). where ω is chosen so that o(ω) = o(g).

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity. Finally, suppose f(q) is a polynomial in q with coefficients in Z.

Definition (Reiner-Stanton-White, 2004)

The triple (S, G, f(q)) exhibits the cyclic sieving phenomenon (c.s.p.) if, for all g ∈ G, we have #Sg = f(ω). where ω is chosen so that o(ω) = o(g).

• Notes. 1. At first blush, this is a surprising equation.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity. Finally, suppose f(q) is a polynomial in q with coefficients in Z.

Definition (Reiner-Stanton-White, 2004)

The triple (S, G, f(q)) exhibits the cyclic sieving phenomenon (c.s.p.) if, for all g ∈ G, we have #Sg = f(ω). where ω is chosen so that o(ω) = o(g).

• Notes. 1. At first blush, this is a surprising equation.
• 2. The case #G = 2 was first studied by Stembridge.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity. Finally, suppose f(q) is a polynomial in q with coefficients in Z.

Definition (Reiner-Stanton-White, 2004)

The triple (S, G, f(q)) exhibits the cyclic sieving phenomenon (c.s.p.) if, for all g ∈ G, we have #Sg = f(ω). where ω is chosen so that o(ω) = o(g).

• Notes. 1. At first blush, this is a surprising equation.
• 2. The case #G = 2 was first studied by Stembridge.
• 3. Dozens of c.s.p.’s have been found.

Suppose S is a set and let G be a finite cyclic group acting on

• S. If g ∈ G, we let

Sg = {t ∈ S : gt = t} and o(g) = order of g in G. Note: if d = o(g) then d | #G, the cardinality of G. We also let ω = ωd = primitive dth root of unity. Finally, suppose f(q) is a polynomial in q with coefficients in Z.

Definition (Reiner-Stanton-White, 2004)

The triple (S, G, f(q)) exhibits the cyclic sieving phenomenon (c.s.p.) if, for all g ∈ G, we have #Sg = f(ω). where ω is chosen so that o(ω) = o(g).

• Notes. 1. At first blush, this is a surprising equation.
• 2. The case #G = 2 was first studied by Stembridge.
• 3. Dozens of c.s.p.’s have been found.
• 4. Three proof techniques: evaluation, representation theory,

and combinatorics.

Running example. Let [n] = {1, 2, . . . , n}

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.
• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Let Cn = (1, 2, . . . , n).

• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Let Cn = (1, 2, . . . , n). Now g ∈ Cn acts on T = {t1, . . . , tk} by gT = {g(t1), . . . , g(tk)}.

• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Let Cn = (1, 2, . . . , n). Now g ∈ Cn acts on T = {t1, . . . , tk} by gT = {g(t1), . . . , g(tk)}.

• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Also C4 = (1, 2, 3, 4) = {e, (1, 2, 3, 4), (1, 3)(2, 4), (1, 4, 3, 2)}.

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Let Cn = (1, 2, . . . , n). Now g ∈ Cn acts on T = {t1, . . . , tk} by gT = {g(t1), . . . , g(tk)}.

• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Also C4 = (1, 2, 3, 4) = {e, (1, 2, 3, 4), (1, 3)(2, 4), (1, 4, 3, 2)}. For g = (1, 3)(2, 4)

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Let Cn = (1, 2, . . . , n). Now g ∈ Cn acts on T = {t1, . . . , tk} by gT = {g(t1), . . . , g(tk)}.

• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Also C4 = (1, 2, 3, 4) = {e, (1, 2, 3, 4), (1, 3)(2, 4), (1, 4, 3, 2)}. For g = (1, 3)(2, 4) we have (1, 3)(2, 4)12 = 34,

Running example. Let [n] = {1, 2, . . . , n} and S = [n] k

• = {T ⊆ [n] : #T = k}.

Let Cn = (1, 2, . . . , n). Now g ∈ Cn acts on T = {t1, . . . , tk} by gT = {g(t1), . . . , g(tk)}.

• Ex. Suppose n = 4 and k = 2. We have

S = [4] 2

• = {12, 13, 14, 23, 24, 34}.

Also C4 = (1, 2, 3, 4) = {e, (1, 2, 3, 4), (1, 3)(2, 4), (1, 4, 3, 2)}. For g = (1, 3)(2, 4) we have (1, 3)(2, 4)12 = 34, (1, 3)(2, 4)13 = 13, (1, 3)(2, 4)14 = 23, (1, 3)(2, 4)23 = 14, (1, 3)(2, 4)24 = 24, (1, 3)(2, 4)34 = 12.

return 1 return 2

Let [n]q = 1 + q + q2 + · · · + qn−1

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q.

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

• Ex. Consider n = 4, k = 2. So

4 2

• q

= [4]q! [2]q![2]q! = 1 + q + 2q2 + q3 + q4.

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

Theorem (Reiner-Stanton-White)

The c.s.p. is exhibited by [n] k

• , Cn,

n k

• q
• .
• Ex. Consider n = 4, k = 2. So

4 2

• q

= [4]q! [2]q![2]q! = 1 + q + 2q2 + q3 + q4.

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

Theorem (Reiner-Stanton-White)

The c.s.p. is exhibited by [n] k

• , Cn,

n k

• q
• .
• Ex. Consider n = 4, k = 2. So

4 2

• q

= [4]q! [2]q![2]q! = 1 + q + 2q2 + q3 + q4. For g = (1, 3)(2, 4) we have o(g) = 2 and ω = −1

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

Theorem (Reiner-Stanton-White)

The c.s.p. is exhibited by [n] k

• , Cn,

n k

• q
• .
• Ex. Consider n = 4, k = 2. So

4 2

• q

= [4]q! [2]q![2]q! = 1 + q + 2q2 + q3 + q4. For g = (1, 3)(2, 4) we have o(g) = 2 and ω = −1 so 4 2

• −1

= 1 − 1 + 2 − 1 + 1

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

Theorem (Reiner-Stanton-White)

The c.s.p. is exhibited by [n] k

• , Cn,

n k

• q
• .
• Ex. Consider n = 4, k = 2. So

4 2

• q

= [4]q! [2]q![2]q! = 1 + q + 2q2 + q3 + q4. For g = (1, 3)(2, 4) we have o(g) = 2 and ω = −1 so 4 2

• −1

= 1 − 1 + 2 − 1 + 1 = 2

Let [n]q = 1 + q + q2 + · · · + qn−1 and [n]q! = [1]q[2]q · · · [n]q. Define the Gaussian polynomials or q-binomial coeffiecients by n k

• q

= [n]q! [k]q![n − k]q!.

Theorem (Reiner-Stanton-White)

The c.s.p. is exhibited by [n] k

• , Cn,

n k

• q
• .
• Ex. Consider n = 4, k = 2. So

4 2

• q

= [4]q! [2]q![2]q! = 1 + q + 2q2 + q3 + q4. For g = (1, 3)(2, 4) we have o(g) = 2 and ω = −1 so 4 2

• −1

= 1 − 1 + 2 − 1 + 1 = 2 = #S(1,3)(2,4).

Outline

Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof

Lemma

If m ≡ n (mod d) and ω = ωd,

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d),

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else.

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d.

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1)

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]ω.

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]ω.

If n ≡ 0 (mod d) then [n]ω = 0

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]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 = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]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 = ℓd and m = kd,

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]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 = ℓd and m = kd, so [m]q [n]q = (1 + q + · · · + qd−1)(1 + qd + q2d + · · · + q(k−1)d) (1 + q + · · · + qd−1)(1 + qd + q2d + · · · + q(ℓ−1)d) .

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]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 = ℓd and m = kd, so [m]q [n]q = (1 + q + · · · + qd−1)(1 + qd + q2d + · · · + q(k−1)d) (1 + q + · · · + qd−1)(1 + qd + q2d + · · · + q(ℓ−1)d) . Cancelling and plugging in ω = ωd gives lim

q→ω

[m]q [n]q = k ℓ

Lemma

If m ≡ n (mod d) and ω = ωd, then lim

q→ω

[m]q [n]q =   

m n

if n ≡ 0 (mod d), 1 else. Proof Let m, n have remainder r modulo d. So for ω = ωd: [m]ω =

• (1 + ω + · · · + ωd−1) + · · · + (1 + ω + · · · + ωr−1) = [n]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 = ℓd and m = kd, so [m]q [n]q = (1 + q + · · · + qd−1)(1 + qd + q2d + · · · + q(k−1)d) (1 + q + · · · + qd−1)(1 + qd + q2d + · · · + q(ℓ−1)d) . Cancelling and plugging in ω = ωd gives lim

q→ω

[m]q [n]q = k ℓ = m n .

Corollary

If ω = ωd and d|n

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Lemma

Let g ∈ Sn (symmetric group) have disjoint cycle decomposition g = g1 · · · gk.

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Lemma

Let g ∈ Sn (symmetric group) have disjoint cycle decomposition g = g1 · · · gk. Let T ⊆ [n].

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Lemma

Let g ∈ Sn (symmetric group) have disjoint cycle decomposition g = g1 · · · gk. Let T ⊆ [n]. Then gT = T ⇐ ⇒ T = gi1 ∪ · · · ∪ gim for some i1, . . . , im.

return 1 return 2

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Lemma

Let g ∈ Sn (symmetric group) have disjoint cycle decomposition g = g1 · · · gk. Let T ⊆ [n]. Then gT = T ⇐ ⇒ T = gi1 ∪ · · · ∪ gim for some i1, . . . , im.

return 1 return 2

• Ex. If g = (1, 3, 4)(2, 6)(5) then the T ∈

[6]

3

• with gT = T are

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Lemma

Let g ∈ Sn (symmetric group) have disjoint cycle decomposition g = g1 · · · gk. Let T ⊆ [n]. Then gT = T ⇐ ⇒ T = gi1 ∪ · · · ∪ gim for some i1, . . . , im.

return 1 return 2

• Ex. If g = (1, 3, 4)(2, 6)(5) then the T ∈

[6]

3

• with gT = T are

T = {1, 3, 4}

Corollary

If ω = ωd and d|n then, n k

• ω

=      n/d k/d

• if d|k,

else.

Lemma

Let g ∈ Sn (symmetric group) have disjoint cycle decomposition g = g1 · · · gk. Let T ⊆ [n]. Then gT = T ⇐ ⇒ T = gi1 ∪ · · · ∪ gim for some i1, . . . , im.

return 1 return 2

• Ex. If g = (1, 3, 4)(2, 6)(5) then the T ∈

[6]

3

• with gT = T are

T = {1, 3, 4} and T = {2, 5, 6}.

Proposition

If S = [n]

k

• and g ∈ Cn has o(g) = d

Proposition

If S = [n]

k

• and g ∈ Cn has o(g) = d then

#Sg =      n/d k/d

• if d|k,

Proposition

If S = [n]

k

• and g ∈ Cn has o(g) = d then

#Sg =      n/d k/d

• if d|k,

else.

Proposition

If S = [n]

k

• and g ∈ Cn has o(g) = d then

#Sg =      n/d k/d

• if d|k,

else. Proof If g ∈ Cn and o(g) = d then g = g1 · · · gn/d where #g1 = . . . = #gn/d = d.

Proposition

If S = [n]

k

• and g ∈ Cn has o(g) = d then

#Sg =      n/d k/d

• if d|k,

else. Proof If g ∈ Cn and o(g) = d then g = g1 · · · gn/d where #g1 = . . . = #gn/d = d. So, by the second lemma , T ∈ [n]

k

• satisfies gT = T iff T is a union of k/d of the n/d

cycles gi.

Outline

Definitions and an example Proof by evaluation Proof by representation theory A combinatorial proof

If G acts on S = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ C for all i}.

If G acts on S = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ C for all i}. Element g ∈ G corresponds to an invertible linear map [g].

If G acts on S = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ 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 = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ 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 = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ 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 = S3 acts on S = {1, 2, 3} and so on

CS = {c11 + c22 + c33 : c1, c2, c3 ∈ C}.

If G acts on S = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ 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 = S3 acts on S = {1, 2, 3} and so on

CS = {c11 + c22 + c33 : c1, c2, c3 ∈ 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 = {s1, . . . , sk} then G also acts on the vector space V = CS = {c1s1 + · · · + cksk : ci ∈ 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 = S3 acts on S = {1, 2, 3} and so on

CS = {c11 + c22 + c33 : c1, c2, c3 ∈ 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 [(1, 2)(3)]S =   1 1 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 CS 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 CS is χ(g) = tr[g]S = #Sg. (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 CS is χ(g) = tr[g]S = #Sg. (1) If G is cyclic, then there will be a basis B for CS such that every g ∈ G satisfies [g]B = diag(

m0

1, . . . , 1,

m1

• ω, . . . , ω,

m2

• ω2, . . . , ω2, . . .)

where ω = ωo(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 CS is χ(g) = tr[g]S = #Sg. (1) If G is cyclic, then there will be a basis B for CS such that every g ∈ G satisfies [g]B = diag(

m0

1, . . . , 1,

m1

• ω, . . . , ω,

m2

• ω2, . . . , ω2, . . .)

where ω = ωo(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 CS is χ(g) = tr[g]S = #Sg. (1) If G is cyclic, then there will be a basis B for CS such that every g ∈ G satisfies [g]B = diag(

m0

1, . . . , 1,

m1

• ω, . . . , ω,

m2

• ω2, . . . , ω2, . . .)

where ω = ωo(g). Thus χ(g) = tr[g]B =

• i≥0

miωi

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 CS is χ(g) = tr[g]S = #Sg. (1) If G is cyclic, then there will be a basis B for CS such that every g ∈ G satisfies [g]B = diag(

m0

1, . . . , 1,

m1

• ω, . . . , ω,

m2

• ω2, . . . , ω2, . . .)

where ω = ωo(g). Thus χ(g) = tr[g]B =

• i≥0

miωi = f(ω). (2) where f(q) =

i≥0 miqi.

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 CS is χ(g) = tr[g]S = #Sg. (1) If G is cyclic, then there will be a basis B for CS such that every g ∈ G satisfies [g]B = diag(

m0

1, . . . , 1,

m1

• ω, . . . , ω,

m2

• ω2, . . . , ω2, . . .)

where ω = ωo(g). Thus χ(g) = tr[g]B =

• i≥0

miωi = f(ω). (2) where f(q) =

i≥0 miqi. Now (1) and (2) imply f(ω) = #Sg so

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 CS is χ(g) = tr[g]S = #Sg. (1) If G is cyclic, then there will be a basis B for CS such that every g ∈ G satisfies [g]B = diag(

m0

1, . . . , 1,

m1

• ω, . . . , ω,

m2

• ω2, . . . , ω2, . . .)

where ω = ωo(g). Thus χ(g) = tr[g]B =

• i≥0

miωi = f(ω). (2) where f(q) =

i≥0 miqi. Now (1) and (2) imply f(ω) = #Sg so

we have the c.s.p. To get the S = [n]

k

• example, one uses the

kth 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) =

• T∈S

wt T. (3)

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) =

• T∈S

wt T. (3) 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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|ω = 1. (II) For i > #Sg we have #Bi > 1 and wt Bi|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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|ω = 1. (II) For i > #Sg we have #Bi > 1 and wt Bi|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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|ω = 1. (II) For i > #Sg we have #Bi > 1 and wt Bi|ω = 0. We then have the c.s.p. since for each g ∈ G f(ω) =

• T∈S

wt T|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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|ω = 1. (II) For i > #Sg we have #Bi > 1 and wt Bi|ω = 0. We then have the c.s.p. since for each g ∈ G f(ω) =

• T∈S

wt T|ω =

• i

wt Bi|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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|ω = 1. (II) For i > #Sg we have #Bi > 1 and wt Bi|ω = 0. We then have the c.s.p. since for each g ∈ G f(ω) =

• T∈S

wt T|ω =

• i

wt Bi|ω =

#Sg

• 1 + · · · + 1 +0+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) =

• T∈S

wt T. (3) If B ⊆ S we let wt B =

T∈B wt T. For each g ∈ G we then find

a partition of S π = πg = {B1, B2, . . .} satisfying, the following two criteria where ω = ωo(g): (I) For 1 ≤ i ≤ #Sg we have #Bi = 1 and wt Bi|ω = 1. (II) For i > #Sg we have #Bi > 1 and wt Bi|ω = 0. We then have the c.s.p. since for each g ∈ G f(ω) =

• T∈S

wt T|ω =

• i

wt Bi|ω =

#Sg

• 1 + · · · + 1 +0+0+· · · = #Sg.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

.

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

.

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

So T : 12 13 14 23 24 34,

• T wt T

= q0 + q1 + q2 + q2 + q3 + q4

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

.

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

So T : 12 13 14 23 24 34,

• T wt T

= q0 + q1 + q2 + q2 + q3 + q4 = 4 2

• q

.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

. Suppose g ∈ Cn with o(g) = d,

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

So T : 12 13 14 23 24 34,

• T wt T

= q0 + q1 + q2 + q2 + q3 + q4 = 4 2

• q

.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

. Suppose g ∈ Cn with o(g) = d, so g = g1 . . . gn/d where #g1 = . . . = #gn/d = d.

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

So T : 12 13 14 23 24 34,

• T wt T

= q0 + q1 + q2 + q2 + q3 + q4 = 4 2

• q

.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

. Suppose g ∈ Cn with o(g) = d, so g = g1 . . . gn/d where #g1 = . . . = #gn/d = d. Suppose h ∈ Sn satisfies h = h1 . . . hn/d where #h1 = . . . = #hn/d = d.

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

So T : 12 13 14 23 24 34,

• T wt T

= q0 + q1 + q2 + q2 + q3 + q4 = 4 2

• q

.

Theorem (Reiner, Stanton, White)

The c.s.p. is exhibited by the triple [n] k

• , Cn,

n k

• q
• .

Proof (Roichman & S) For T ∈ [n]

k

• let wt T = q
• t∈T t−(k+1

2 ).

∴

• T

wt T = n k

• q

. Suppose g ∈ Cn with o(g) = d, so g = g1 . . . gn/d where #g1 = . . . = #gn/d = d. Suppose h ∈ Sn satisfies h = h1 . . . hn/d where #h1 = . . . = #hn/d = d. Then, by the second lemma , # [n]

k

g = # [n]

k

h.

• Ex. If n = 4 and k = 2 then wt{t1, t2} = qt1+t2−3.

So T : 12 13 14 23 24 34,

• T wt T

= q0 + q1 + q2 + q2 + q3 + q4 = 4 2

• q

.

Let h = (1, 2, . . . , d)(d + 1, d + 2, . . . , 2d) · · ·

Let h = (1, 2, . . . , d)(d + 1, d + 2, . . . , 2d) · · · = h1 · · · hn/d.

Let h = (1, 2, . . . , d)(d + 1, d + 2, . . . , 2d) · · · = h1 · · · hn/d. Ex: n = 4, k = 2, g = (1, 3)(2, 4).

Let h = (1, 2, . . . , d)(d + 1, d + 2, . . . , 2d) · · · = h1 · · · hn/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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 0 < ℓ < d. ∴ wt B = wt T + wt hiT + · · · + wt hd−1

i

T 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 0 < ℓ < d. ∴ wt B = wt T + wt hiT + · · · + wt hd−1

i

T ∴ wt B|ω = ωj + 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 0 < ℓ < d. ∴ wt B = wt T + wt hiT + · · · + wt hd−1

i

T ∴ wt B|ω = ωj + ωj+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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 0 < ℓ < d. ∴ wt B = wt T + wt hiT + · · · + wt hd−1

i

T ∴ wt B|ω = ωj + ωj+ℓ + · · · + ωj+(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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 0 < ℓ < d. ∴ wt B = wt T + wt hiT + · · · + wt hd−1

i

T ∴ wt B|ω = ωj + ωj+ℓ + · · · + ωj+(d−1)ℓ = ωj 1 − ωdℓ 1 − ωℓ 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, . . . , 2d) · · · = h1 · · · hn/d. For any T ∈ [n]

k

• define the block B of π containing T:

(a) If hT = T then B = {T}. (b) If hT = T, then find the smallest index i such that 0 < #(T ∩ hi) < d and let B = {T, hiT, h2

i T, . . . , hd−1 i

T}. Proof of (II) If ω = ωd, wt T = qj, ℓ = #(T ∩hi) so 0 < ℓ < d. ∴ wt B = wt T + wt hiT + · · · + wt hd−1

i

T ∴ wt B|ω = ωj + ωj+ℓ + · · · + ωj+(d−1)ℓ = ωj 1 − ωdℓ 1 − ωℓ = 0 since ωd = 1 and ωℓ = 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.

THANKS FOR LISTENING!