Combinatorics with quantum matrices. St´ ephane Launois

Reims, 27 November 2008

The algebra of 2 × 2 quantum matrices.

"

Oq (M2(C)) := C

a b c d

#

with

ba = q −1ab

dc = q −1cd

ca = q −1ac

db = q −1bd

cb = bc da = ad − (q − q −1)bc Oq (M2(C)) is an iterated Ore extension that we can write Oq (M2(C)) = C[a][b; σb][c; σc][d; σd, δd]. Hence Oq (M2(C)) is a noetherian domain.

The algebra of m × p quantum matrices. 

Y1,1 . . .  R = Oq (Mm,p(C)) := C  ... ··· Ym,1 . . .



Y1,p  ... , Ym,p

where each 2×2 sub-matrix is a copy of Oq (M2(C)).

Oq (Mm,p(C)) is an iterated Ore extension with the indeterminates Yi,α adjoined in the lexicographic order

The prime and primitive spectra of Oq (Mm,p(C)).

We now assume that q is a non-zero complex number which is not a root of unity, and we set R = Oq (Mm,p(C)).

Theorem 1 (Goodearl-Letzter) Prime ideals of R are completely prime.

The torus H := (C∗)m+p acts by automorphisms on R via : (a1, . . . , am, b1, . . . , bp).Yi,α = aibαYi,α. This action of H on R induces an action of H on Spec(R). We denote by H-Spec(R) the set of those prime ideals in R which are H-invariant.

Theorem 2 (Goodearl-Letzter) R has at most 2mp H-primes.

Stratification Theorem (Goodearl-Letzter) :

If J ∈ H-Spec(R), then we set SpecJ (R) := {P ∈ Spec(R) |

\

h.P = J}.

h∈H

1. Spec(R) =

G

SpecJ (R)

J∈H-Spec(R)

2. For all J ∈ H-Spec(R), SpecJ (R) is homeomorphic to the prime spectrum of a (commutative) Laurent polynomial ring in n(J) indeterminates over C. 3. The primitive ideals of R are precisely the primes maximal in their H-strata.

Parametrisation of H-Spec(R). We denote by W the set of Cauchon diagrams with m rows and p columns.

∈W

∈ /W

Theorem 3 (Cauchon) There exists an (explicit) bijection from W onto H-Spec(R). If w is a Cauchon diagram, then we denote by Jw the unique H-prime associated to w.

Theorem 4 ht(Jw ) is equal to the number of black boxes in w.

2 × 2 Cauchon diagrams :

Direct graph associated to a diagram Joint work with Jason Bell.

To a Cauchon diagram C, one can associate a direct graph G(C) and and skew-symmetric matrix AC as follows. The vertices of G(C) are the white boxes of C labeled 1 to N . We draw an arrow between two vertices in the same column (going from North to South) or on the same row (going from West to East). AC is the N × N skew-symmetric matrix whose coefficient aij (i < j) is the number of arrows going from the vertex labeled i to the vertex labeled j.

An example

C=

G(C) : •1



•4 /

/

       AC =       

•3

/4

•6

/

•8



•5

•7



/*

•2





0 1 1 1 0 0 0 −1 0 1 0 1 0 0 −1 −1 0 0 0 1 0 −1 0 0 0 1 1 0 0 −1 0 −1 0 1 0 0 0 −1 −1 −1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 −1 −1

0 0 1 0 0 1 1 0

              

Dimension of the strata 1 Theorem 5 The H-stratum of Oq (Mm,p(C)) associated to C is a Laurent polynomial ring over C in dim ker AC indeterminates. Corollary 1 The dimension of each H-stratum is less than or equal to min(m, p). Objective : Compute Q(q) :=

min(m,p) X

bk q k ∈ N[q],

k=0

where bk :=| {C | dim ker AC = k} | .

Dimension of the strata 2 Our starting motivation was to understand the Hprimes that are primitive. It follows from previous results that a Cauchon diagram corresponds to a primitive H-prime iff det(AC ) 6= 0. As AC is skewsymmetric, this implies that a Cauchon diagram corresponds to a primitive H-prime iff Pfaffian(AC) 6= 0. Now one can use combinatorial tools to compute the Pfaffian, in particular perfect matchings. Definition 1 Given a labeled Cauchon diagram C, we say that π = {{i1, j1}, . . . , {im, jm}} is a perfect matching of C if : 1. i1, j1, . . . , im, jm are distinct ; 2. {i1, . . . , im, j1, . . . , jm} is precisely the set of labels which appear in C ; 3. ik < jk for 1 ≤ k ≤ m ; 4. for each k the white in the same row or white square labeled

square labeled ik is either the same column as the jk .

An example

C=

G(C) : •1



•4 /

/

/*

•3

/4

•6

/

•8

•2



•5

•7

C has 4 perfect matchings : {{1, 2}, {3, 6}, {4, 5}, {7, 8}}, {{1, 3}, {2, 5}, {4, 6}, {7, 8}}, {{1, 4}, {2, 3}, {5, 6}, {7, 8}}, {{1, 4}, {2, 5}, {3, 6}, {7, 8}}.





Pfaffian and perfect matchings We let P M C denote the collection of perfect matchings of C. Given a perfect matching π of C, we define sgn(π) := sgn

1 2 3 4 · · · 2m − 1 2m i1 j1 i2 j2 · · · im jm

!

.

Then Pfaffian(AC ) =

X

sgn(π).

π∈P M C

In particular, if C has no perfect matchings, then Pfaffian(C) = 0.

Observe that a Cauchon diagram with an odd number of white boxes has no perfect matchings and so cannot be primitive. Hence a primitive H-prime corresponds to a Cauchon diagram with an even number of white boxes, so that the corresponding quotient algebra has even Gelfand-Kirillov dimension.

Dimension of the strata 3

Theorem 6 Let n be a positive integer. Then the number of primitive H-prime ideals in the ring Oq (M2,n) is 3n+1 − 2n+1 + (−1)n+1 + 2 . 4 Corollary 2 3n+1 − 2n+1 + (−1)n+1 + 2 Q(q) = 4 +(3n − 2n−1)q 3n + (−1)n − 2 2 + q 4 n = 2 : P (q) = 5 + 7q + 2q 2.

Dimension of the strata 4

Conjecture 1 The number of primitive H-primes in the ring Oq (M3,n) is given by 1 ·(15 · 4n − 18 · 3n + 13 · 2n − 6 · (−1)n + 3 · (−2)n) . 8 More generally we believe the following holds. Conjecture 2 Let m ≥ 1 be a positive integer and let P (m, n) denote the number of primitive H-prime ideals in m × p quantum matrices. Then there exist rational constants cm+1, cm, . . . , c1−m such that P (m, n) =

m+1 X

cj j n

j=1−m

for all positive integers n. Moreover, cm+1 = 1 · 3 · 5 · · · (2m − 1)/2m.

Decomposition of a diagram

Let C be an m × n labeled diagram with labels {1, 2, . . . , k}. We say that (V, H) is a decomposition of C if : 1. V ∪ H = {1, 2, . . . , k} and V ∩ H = ∅ ; 2. For 1 ≤ j ≤ n, # {i ∈ V | square i is in column j} is even. We write (V, H) ` C when (V, H) is a decomposition of C.

A labeled diagram with decomposition V = {1, 5, 6, 8, 9, 10, 13, 15} , H = {2, 3, 4, 7, 11, 12, 14} .

1 5 12

2 6 9

3

4 7 8 10 11 13 14 15

Excess Let C be an m × n labeled diagram and let S be a subset of the labels. We define the excess of S to be the vector excess(S) := (s1 + 2Z, s2 + 2Z, . . . , sm + 2Z) ∈ Zm 2, where si = # {k ∈ S | square k is in the ith row of C}.

Example : A labeled diagram with decomposition V = {1, 5, 6, 8, 9, 10, 13, 15} , H = {2, 3, 4, 7, 11, 12, 14} .

1 5 12

2 6 9

3

4 7 8 10 11 13 14 15

In this example, we have : excess(V ) = (1, 1, 0, 0) and excess(H) = (1, 1, 1, 0).

Excess and Pfaffian We define the inversions of S to be inv(S) = X

δ(i < j and square i is southwest of square j).

i,j∈S

Proposition 1 Pfaffian(C) =

X (V,H)`C excess(H)=(0,0,...,0)

(−1)inv(V )+inv(V |H)

Primitivity

Proposition 2 Let C be an m × n diagram. Then Pfaffian(C) is either 0 or ±2k for some nonnegative integer k. Sketch of proof : We use previous proposition to show that Pfaffian(C) = 2−l

X

(−1)P (x1,...,xN )

(x1 ,...,xN )∈{0,1}N

where P is a quadratic form over Z2.

Corollary 3 Pfaffian(C) = 0 if and only if Pfaffian(C) ≡ 0 (mod 3)

Semigroup structure on Dm

Let C1 and C2 be respectively m × n1 and m × n2 diagrams. We define C1 ?C2 to be the m× (n1 + n2) diagram obtained by placing C2 immediately after C1 .

If C1 =

and C2 =

,

then C1 ? C2 =

The empty diagram is the identity and it is clear that ? is associative. We let Dm denote the semigroup consisting of diagrams with exactly m rows. We let Cm denote the subset of Dm consisting of Cauchon diagrams with exactly m rows.

Excess group

Our goal is to determine which diagrams in Dm are primitive. To do this we introduce a related object, which we call the Excess group. Let m be a natural number. We define the mth excess group Exm = {(v, h, ε) | v, h ∈ (Z/2Z)m, ε ∈ {±1} , v · 1 = 0} . with multiplication given by (v, h, ε) · (v0, h0, ε0) = (v + v0, h + h0, ε00) 0 0 1≤i 1, then the commutator subgroup Ex0m of Exm is {(0, 0, ±1)} ; 3. Exm has 22m−1 + 2 conjugacy classes ; 4. if K is an algebraically closed field whose characteristic is not equal to 2, then ∼ K 22m−1 × M m−1 (K) × M m−1 (K). K[Exm] = 2 2 As a corollary, we obtain an (explicit) isomorphism Φ : Z3[Ex3]/J3 :→ M4(Z3) × M4(Z3).

Image of D3 under [ · ] Let G3 ⊆ Z3 [Ex3] /J3 be the image of D3 under [ · ]. Proposition 5 G3 is a group of order 384. Sketch of Proof. For a subset S ⊆ {1, 2, 3}, we take C(S) to be the 3 × 1 column which has white squares precisely in the rows indexed by S ; these 8 elements are the eight columns C(∅) =

C({1}) =

C({1, 2}) =

, C({2}) =

, C({3}) =

, C({1, 3}) =

and C({1, 2, 3}) =

,

,

, C({2, 3}) =

.

Since D3 is a semigroup and [ · ] : D3 → Z3 [Ex3] /J3 is a semigroup homomorphism, G3 is generated by the 8 columns {[C(S)] | S ⊆ {1, 2, 3}}, it is sufficient to check that each of these 8 elements has an inverse. We check that in M4(Z3) × M4(Z3).

Further simplification Let N := {[D] | D ∈ D3 and [D] = ±(v, 0, ε) with

v

o 3 ∈ Z2, ε ∈ {±1} .

N is a normal subgroup of G3 and |N | = 8.. Proposition 6 If D1, D2 ∈ D3 and [D1] N = [D2] N then either D1, D2 are both primitive or both fail to be primitive. Thus we can look at the image of diagrams in G3/N instead of in G3 to determine whether they are primitive or not. We find that N has order 8 and so G3/N is a group of order 48.

The group G3/N Proposition 7 We have an isomorphism ψ : G3/N → S4 × {±1} given by [C({1, 2, 3})] 7→ ((124), −1), [C({1, 2})] 7→ ((1243), 1), [C({1, 3})] 7→ ((1324), 1), [C({2, 3})] 7→ ((1234), 1), [C({1})] 7→ ((13)(24), −1), [C({2})] 7→ ((12)(34), −1), [C({3})] 7→ ((14)(23), −1), [C(∅)] 7→ (id, 1). Moreover, the elements of G3/N in the preimage ψ −1((S4 × {1}) are precisely the images of the diagrams C ∈ D3 with an even number of white squares. Proposition 8 The primitive diagrams in D3 are the preimage under H of the three conjugacy classes with representatives (id, 1), ((1, 2, 3), 1), and ((1, 2, 3, 4), 1).

3 × n primitive Cauchon diagrams We now explain how we will use these techniques to enumerate the primitive 3 × n Cauchon diagrams. Note that the semigroup morphism H : c : C [D ] [[t]] → D3 → S4 × {±1} extends to a map H 3 C [S4 × {±1}] [[t]]. One can explicitely compute an expression for the generating series of the Cauchon diagrams in C [D3] [[t]] as a rational function. Let

C(t) =

X

C · t# columns of C ∈ C [D3] [[t]]

C∈C3

be the generating function for Cauchon diagrams.

Generating series of Cauchon diagrams Proposition 9 Then

C(t) 



= F1(t) ? 1 + (C({2, 3}) + C({1, 2, 3}))t ? F4(t) 



+F2(t)? 1+(C({3})+C({2, 3})+C({1, 2, 3}))t?F4(t) 



+F3(t) ? 1 + (C({2, 3}) + C({1, 2, 3}))t ? F4(t) where F1(t) =

1 , 1 − (C(∅) + C({1}))t

and the other Fi(t) are too long for my slides ! c C(t)). Using this result, we can easily compute H( We are now ready to enumerate the 3 × n primitive Cauchon diagrams.

3 × n primitive Cauchon diagrams Let P : S4 × {±1} → C be the class function that sends the conjugacy classes containing (id, +1), ((123), +1), and ((1234), +1) to 1 and all other conjugacy classes to 0. Then P is a linear combination of characters. Using the orthogonality relations, we find 15 P = (χ1 ⊗ ∆0 + χ1 ⊗ ∆1) 48 3 + (χ2 ⊗ ∆0 + χ2 ⊗ ∆1) 48 6 − (χ3 ⊗ ∆0 + χ3 ⊗ ∆1) 48 3 − (χ4 ⊗ ∆0 + χ4 ⊗ ∆1) 48 9 + (χ5 ⊗ ∆0 + χ5 ⊗ ∆1), 48 where the characters χ1, χ2, χ3, χ4, χ5 are the characters of S4 and ∆0, ∆1 are the characters of {±1} ; and χi ⊗ ∆j ((σ, ε)) = χ(σ)∆(ε).

Character tables of S4 and {±1}

χ1 χ2 χ3 χ4 χ5

∆0 ∆1

{id}

{(12)}

{(12)(34)}

{(123)}

{(1234)}

1 1 2 3 3

1 −1 0 1 −1

1 1 2 −1 −1

1 1 −1 0 0

1 −1 0 −1 1

{+1}

{−1}

1 1

1 −1

Conclusion Then the class functions P, χ1 ⊗ ∆0, . . . , χ5 ⊗ ∆0, χ1 ⊗ ∆1, . . . , χ5 ⊗ ∆1 can each be extended to functions on C [S4 × {±1}] [[t]]. Using the actual matrix representations corresponding to these ten irreducible characters and then computing the traces, we find c C(t))) = χ1 ⊗ ∆0(H(

6 6 1 − + , 1 − 4t 1 − 3t 1 − 2t

c C(t))) = 1, χ1 ⊗ ∆1(H( c C(t))) = χ2 ⊗ ∆0(H(

c C(t))) = χ2 ⊗ ∆1(H(

1 , 1 − 2t

6 6 +1− , 1 + 2t 1+t

c C(t))) = χ3 ⊗ ∆0(H(

3 1 − , 1 − 3t 1 − 2t

c C(t))) = χ3 ⊗ ∆1(H(

3 − 1, 1+t

−3 c χ4 ⊗ ∆0(H(C(t))) = + 6, 1−t

c C(t))) = χ4 ⊗ ∆1(H(

3 , 1−t

c C(t))) = χ5 ⊗ ∆0(H(

3 , 1−t

−3 c + χ5 ⊗ ∆1(H(C(t))) = 1−t

6 . 1 − 2t

Using the expression for P as a linear combination of the irreducible characters, we find c C(t))) = P (H(

X

t#columns of C

C∈C3 C primitive

1 = · 15/(1 − 4t) − 18/(1 − 3t) + 13/(1 − 2t) + 1 8 



−6/(1 + t) + 3/(1 + 2t) . c C(t))) is the number The coefficient of tn of P (H( of 3 × n Cauchon diagrams that are primitive by construction. On the other hand, the coefficient of tn for n ≥ 1 of the previous rational function is given by

1 · 15 · 4n − 18 · 3n + 13 · 2n − 6 · (−1)n + 3 · (−2)n . 8 



Bon Anniversaire Jacques !