Generators of Hecke Ring for Gelfand Pair (S(2n),...
Transcript of Generators of Hecke Ring for Gelfand Pair (S(2n),...
Why o why? Introduction Overview Proof of the generation theorem
Generators of Hecke Ring for Gelfand Pair(S(2n), B(n))
Kürsat Aker, Feza Gürsey Institute - M. B. Can, Tulane Uni.
ODTÜ - Bilkent Algebraic Geometry Seminar2010-12-17 Friday
Why o why? Introduction Overview Proof of the generation theorem
REPRESENTATIONS: WHAT IS THE PURPOSE OF LIFE?
Representation Theory
I is used in the form Fourier series and Fourier Transform insignal processing,
I helps us multiply matrices faster,I let us analyze random walks on groups, e.g. shuffling cards
(the symmetric group on 52 letters), encoding (to producerandom orthogonal matrices),
I makes it possible to evaluate integrals on unitary,orthogonal, symplectic groups.
Why o why? Introduction Overview Proof of the generation theorem
REPRESENTATIONS: FUN FACT
I The character table of the Monster group (the final piece ofthe puzzle in the classification of finite simple groups) wasconstructed before the group itself!
I Number of elements of the Monster is ∼ 8 · 1053,I Its character table is 194-by-194 ( ∼ 4 · 104 ).
Why o why? Introduction Overview Proof of the generation theorem
REPRESENTATIONS: AN EXAMPLE
Definition (Perfect Matching)
A (perfect) matching of the set {1, 2, . . . , 2n} is a partition of theset {1, 2, . . . , 2n} into two-element subsets.
Example
I {{1, 2}, {3, 4}, . . . , {2n− 1, 2n}},I {{1, 3}, {2, 7}, {4, 8}, {5, 6}}.
Set of Matchings
LetM be the set of all matching of the set {1, 2, . . . , 2n}.The symmetric group S(2n) acts onM.
Why o why? Introduction Overview Proof of the generation theorem
REPRESENTATIONS: ANOTHER EXAMPLE
I Let C[M] the complex vector space with basisM.I The symmetric group S(2n) acts on C[M] as well. Denote
this action by X.I LetHn be the set of C-linear operators on C[M] which
commute with the action of the symmetric group S(2n):
T ∈ Hn ⇐⇒ T · X(w) = X(w) · T
for all w ∈ S(2n).
Why o why? Introduction Overview Proof of the generation theorem
AIM OF THIS TALK
Let B(n) be the stabilizer of the matching
{{1, 2}, {3, 4}, . . . , {2n− 1, 2n}}.
Then,Hn = EndS(2n)1S(2n)B(n) .
Construct a set of generators ofHn.
Why o why? Introduction Overview Proof of the generation theorem
JUMP TO THE VERY END. . .Can you find a “nice” set of ring generators for the Hecke ringHn := Hecke(S(2n),B(n)) ?
I S(2n), symmetric group on 2n letters,I B(n), centralizer of the element
t := (1, 2)(3, 4) · · · (2n− 1, 2n),I eB(n) := 1
|B(n)|∑
w∈B(n) w,
I Hn := Hecke(S(2n),B(n)) := eB(n)CS(2n)eB(n).
Theorem (A. - Can, 2010)
The ringHn is generated by the elements
Hi :=∑
w has exactly i circuits
w for i = 1, . . . ,n.
Why o why? Introduction Overview Proof of the generation theorem
GRAPHS ATTACHED TO ELEMENTS w ∈ S(2n)
w = (1 2 3 4 5 6 7 8 9 10 11 121 7 2 8 6 4 10 3 9 5 12 11
)
8
5
9
12
10
8
6
4
3
2
111
7
1
7
2
610
3
9
5
12
4
11
Figure: Coset Type (3, 2, 1) with 3 circuits
Why o why? Introduction Overview Proof of the generation theorem
THIS IS A DIRECT ANALOGUE OF. . .
Theorem (Farahat-Higman, 1959)
Let Zn be the the center of integral symmetric group ring, ZS(n).Then, the ring Zn is generated by the elements
Zi :=∑
w has exactly i cycles
w for i = 1, . . . ,n.
I (1, 3)(5, 2, 6, 4) ∈ S(6) is of cycle type (4, 2) and has 2 cycles,I (1, 3)(5, 2, 6, 4) ∈ S(7) is of cycle type (4, 2, 1) and has 3
cycles, since (1,3)(5,2,6,4)=(1,3)(5,2,6,4)(7).
Why o why? Introduction Overview Proof of the generation theorem
WHY THE CENTER?
I Complex Representations of a finite group GI Complex Characters of GI Center of the group ring CG := {
∑g agg}
Why o why? Introduction Overview Proof of the generation theorem
TWO INCARNATIONS
(C(G) := {f : G→ C}, ?) (CG, ·)Cl(G) = Class/Central Functions on G Z(CG)Character χ
∑g χ(g)g
1C conjugacy class sum of CCl(G) = ⊕χ∈Irr(G)Cχ Z(ZG) = ⊕CZC
Character Table of G = Change of Basis Matrix from IrreducibleCharacter Basis to Class Sum Basis
Why o why? Introduction Overview Proof of the generation theorem
MULTIPLICATION IN THE CENTER OF ZG
Given two conjugacy classes/class sums Cλ,Cµ in Z(ZG), theproduct CλCµ has the following expansion:
CλCµ =∑
ν
aνλµCν ,
where, Cν runs over all possible conjugacy classes.Fix any element z ∈ Cν , then
aνλµ := {(x, y)|xy = z, x ∈ Cλ, y ∈ Cµ}.
Why o why? Introduction Overview Proof of the generation theorem
WHAT ABOUT THE CENTER OF ZG AS A RING?
Theorem (Farahat-Higman, 1959)
Let Zn be the the center of integral symmetric group ring, ZS(n).Then, the ring Zn is generated by the elements
Zi :=∑
w has exactly i cycles
w for i = 1, . . . ,n.
RemarkDimension/Rank of Zn as a Z-module is the number ofpartitions of integer n. Number of ring generators is just n.
Idea of the proof: Let n→∞ and get rid of accidentalrelations.
Why o why? Introduction Overview Proof of the generation theorem
FARAHAT-HIGMAN AND LATER DEVELOPMENTS
I Farahat-HigmanA new proof of Nakayama’s conjecture for modularrepresentations of S(n)
I Jucys 1974: Zi = en−i(J1, J2, . . . , Jn) for n = 1, 2, . . . ,n.
Jk :=∑
transpositions in S(k)−∑
transpositions in S(k− 1).
I Murphy 1984:Reworked Young’s Orthonormal and SeminormalConstruction of irreducible representations of S(n) usingYJM elements
Why o why? Introduction Overview Proof of the generation theorem
YJM ELEMENTS
I Vershik-Okounkov 1995:Gelfand-Zetlin algebra GZ(n) := 〈J1, . . . , Jn〉 ⊂ CS(n) playsthe role of Cartan subalgebra in a semisimple complex Liealgebra g.
I Post 1995:I Asymptotic Representation Theory and Free ProbabilityI Weingarten formulas for Unitary and Orthogonal Groups
Why o why? Introduction Overview Proof of the generation theorem
CENTER→ HECKE: GELFAND PAIRS
Definition (Gelfand Pairs)
A finite group G with a subgroup K is called a Gelfand pair ifany irreducible representation of G appears at most once in theinduced representation 1G
K(= C[G/K]). Equivalently, EndG(1GK)
is commutative. We denote this algebra by Hecke(G,K).
Example (Gelfand Pairs)
I (G× G,∆G): In this case, Hecke(G× G,∆G) ∼= Z(ZG).I (S(2n),B(n)):
1S(2n)B(n)
∼= ⊕λ`nS2λ
Why o why? Introduction Overview Proof of the generation theorem
SETUP
I G(n) = S(2n), or S(n)× S(n),I K(n) = B(n), or ∆S(n),I Hn, the corresponding Hecke ring, (the centers),I G(∞) = S(2∞), or S(∞)× S(∞),I K(∞) = B(∞), or ∆S(∞).
Why o why? Introduction Overview Proof of the generation theorem
MULTIPLICATION IN A HECKE RING
Given two double coset sums Kλ,Kµ in a Hecke ring, theproduct KλKµ has the following expansion:
KλKµ =∑
ν
bνλµKν ,
where, Kν runs over all possible double cosets.Fix any element z ∈ Kν , then
bνλµ := {(x, y)|xy = z, x ∈ Kλ, y ∈ Kµ}.
Why o why? Introduction Overview Proof of the generation theorem
IDEA OF THE PROOF (FARAHAT-HIGMAN):
I Construct a universal ring U to dominate allHn,I Show that U is a free polynomial ring in some elements Ui
for i ∈ N.I Then, the images of Ui will generate each Hecke ringHn.
Why o why? Introduction Overview Proof of the generation theorem
CONTROL THE DOUBLE COSETS!
I Lemmax and y are in the same K(∞)-double coset if and only ifthey have the same reduced type, λ.
I DefinitionFor any partition λ, denote by Kλ the correspondingK(∞)-double coset in G(∞).Set Kλ(n) := Kλ ∩ G(n).
I LemmaI Kλ(n) is non-empty if and only if |λ|+ `(λ) ≤ n.I In that case, Kλ(n) is a G(n)-double coset.
Why o why? Introduction Overview Proof of the generation theorem
PRODUCTS Kλ(n)Kµ(n)
Denote the structure coefficients of the algebraHn with respectto the basis Kµ(n) by bν
λµ(n):
Kλ(n)Kµ(n) =∑
ν
bνλµ(n)Kν(n).
Let B be the subring of Q[t] so that f ∈ B iff f (n) ∈ Z for n ∈ Z.The structure constants bν
λµ(n) obey the tricohotomy:
bνλµ(n) =
0 , if |ν| > |λ|+ |µ|,positive integer indep. of n , if |ν| = |λ|+ |µ|,f νλµ(n) , if |ν| ≤ |λ|+ |µ|,
where f νλµ ∈ B.
Why o why? Introduction Overview Proof of the generation theorem
AN IMPORTANT CALCULATION
TheoremAssume |ν| = |µ|+ r for partition ν. Then,
Kλ(n)K(r)(n) =∑
ρ
b(r+|ρ|)∪λ−ρλ(r) K(r+|ρ|)∪λ−ρ(n) + lower order terms,
where the sum runs over those partitions ρ ⊆ λ so that `(ρ) ≤ r + 1.In this case, the coefficient bν
λ(r) is given by
bνλ(r) =
(mr+|ρ|(λ) + 1)(r + |ρ|+ 1)r!∏i≥0 mi(ρ)!
, (1)
where m0(ρ) := r + 1− `(ρ).
Why o why? Introduction Overview Proof of the generation theorem
THE UNIVERSAL RING U
I As a B-module: U := ⊕λBKλ.I Multiplication:
KλKµ :=∑
|ν|≤|λ|+|µ|
f νλµKν .
I Surjection onto allHn for all n:
∑fµKµ 7→
∑fµ(n)Kµ(n).
I Setting the degree of the generator Kλ as |λ|, the universalring U is filtered.
Why o why? Introduction Overview Proof of the generation theorem
UNIVERSAL RING U IS A FREE POLYNOMIAL RING!
Theorem (Farahat-Higman)
The universal ring U is isomorphic to the polynomial ringB[U1,U2, . . .], where for i = 1, 2, . . .,
Ui :=∑|λ|=i
Kµ.
Why o why? Introduction Overview Proof of the generation theorem
GENERATION THEOREMS
Theorem (Farahat-Higman, 1959)
Let Zn be the the center of integral symmetric group ring, ZS(n).Then, the ring Zn is generated by the elements
Zi :=∑
w has exactly i cycles
w for i = 1, . . . ,n.
I Proof:The universal ring U surjects onto Zn for all n.Recall that ∑
fµKµ 7→∑
fµ(n)Kµ(n).
If |λ| ≥ n, then Kλ 7→ 0. So Un,Un+1, . . . 7→ 0.But, Ui 7→ Zn−i for i = 0, . . . ,n− 1.
Why o why? Introduction Overview Proof of the generation theorem
GENERATION THEOREMS, CONTINUED. . .
Theorem (A. - Can, 2010)
The ringHn is generated by the elements
Hi :=∑
w has exactly i circuits
w for i = 1, . . . ,n.
I Proof:Similarly, Un,Un+1, . . . 7→ 0.But, Ui 7→ Hn−i for i = 0, . . . ,n− 1.
Why o why? Introduction Overview Proof of the generation theorem
IN TERMS OF YJM ELEMENTS
Recall that Jucys proved that
Zi = en−i(J1, J2, . . . , Jn)
for n = 1, 2, . . . ,n.Recently, Zinn-Justin and Matsumoto proved that
Hi = en−i(J1, J3, . . . , J2n−1)eB(n)
for n = 1, 2, . . . ,n.