Classification and construction of calibrated representations

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 3 March 2013

Classification and construction of calibrated representations

The following theorem classifies and constructs all irreducible calibrated representations of the affine Hecke algebra Hn. It shows that the theory of standard Young tableaux plays an intrinsic role in the combinatorics of the representations of the affine Hecke algebra. The construction given in Theorem 4.1 is a direct generalization of A. Young’s classical “seminormal construction” of the irreducible representations of the symmetric group [You1931,You1934]. Young’s construction was generalized to Iwahori-Hecke algebras of type A by Hoefsmit [Hoe1974] and Wenzl [Wen1988] independently, to Iwahori-Hecke algebras of types B and D by Hoefsmit [Hoe1974] and to cyclotomic Hecke algebras by Ariki and Koike [AKo1994]. It can be shown that all of these previous generalizations are special cases of the construction for affine Hecke algebras given here. Recently, this construction has been generalized even further [Ram1998], to affine Hecke algebras of arbitrary Lie type. Some parts of Theorem 4.1 are due, originally, to I. Cherednik, and are stated in [Che1987, §3].

Garsia and Wachs [GWa1989] showed that the theory of standard Young tableaux and Young’s constructions play an important role in the combinatorics of the skew representations of the symmetric group. At that time it was not known that these representations are actually irreducible as representations of the affine Hecke algebra!!

Theorem 4.1. Let (c,λ/μ) be a placed skew shape with n boxes. Define an action of Hn on the vector space

H(c,λ/μ)= -span{vLLis a standard tableau of shapeλ/μ}

by the formulas

xivL = q2c(L(i)) vL, TivL = (Ti)LL vL+ (q-1+(Ti)LL) vsiL,

where siL is the same as L except that the entries i and i+1 are interchanged,

(Ti)LL= q-q-1 1- q 2 ( c(L(i))- c(L(i+1)) ) ,vsiL=0 ifsiL is not a standard tableau,

and L(i) denotes the box of L containing the entry i.

  1. H(c,λ/μ) is a calibrated irreducible Hn-module.
  2. The modules H(c,λ/μ) are non-isomorphic.
  3. Every irreducible calibrated Hn-module is isomorphic to H(c,λ/μ) for some placed skew shape (c,λ/μ).

Step 1. The given formulas for the action of H(c,λ/μ) define an Hn-module.

Proof.

If L is a standard tableau then the entries i and i+1 cannot appear in the same diagonal in L. Thus, for all standard tableaux L, c(L(i))c(L(i+1)) and for this reason the constant (Ti)LL is always well defined.

Let L be a standard tableau of shape λ/μ. Then (Ti)LL+ (Ti)siL,siL =q-q-1 and so

Ti2vL = ( (Ti)LL2+ ( q-1+ (Ti)LL ) ( q-1+ (Ti)siL,siL ) ) vL + ( q-1+ (Ti)LL ) ( (Ti)LL+ (Ti)siL,siL ) vsiL = (Ti)LL ( (Ti)LL+ (Ti)siL,siL ) vL+q-1 ( q-1+ (Ti)LL+ (Ti)siL,siL ) vL + ( q-1+ (Ti)LL ) (q-q-1) vsiL = (Ti)LL (q-q-1)vL+ ( q-1+ (Ti)LL ) (q-q-1) vsiL+q-1 (q-1+q-q-1) vL = ( (q-q-1) Ti+1 ) vL.

The calculations to check the identities (1.1a), (1.1f) and (1.5) are routine. Checking the identity TiTi+1Ti= Ti+1TiTi+1 is more involved. One can proceed as follows. According to the formulas for the action, the operators Ti and Ti+1 preserve the subspace S spanned by the vectors vQ indexed by the standard tableaux Q in the set { L,siL,si+1L, sisi+1L, si+1siL,si si+1siL } . Depending on the relative positions of the boxes containing i,i+1,i+2 in L, this space is either 1, 2, 3 or 6 dimensional. Representative cases are when these boxes are positioned in the following ways.

Case (1) Case (2) Case (3) Case (4)

In Case (1) the space S is one dimensional and spanned by the vector vQ corresponding to the standard tableau

a b c

where a=i, b=i+1, and c=i+2. The action of Ti and Ti+1 on S is given by the matrices

ϕS(Ti)= (q),and ϕS(Ti+1) =(q),

respectively. In case (2) the space S is two dimensional and spanned by the vectors vQ corresponding to the standard tableaux

a b c a c b

where a=i, b=i+1, and c=i+2. The action of Ti and Ti+1 on S is given by the matrices

ϕS(Ti)= ( q0 0-q-1 ) and ϕS(Ti+1)= ( q-q-1 1-q4 q-q-5 1-q-4 q-q3 1-q4 q-q-1 1-q-4 ) ,

In case (3) the space S is three dimensional and spanned by the vectors vQ corresponding to the standard tableaux

a b c a c b b c a

where a=i, b=i+1, and c=i+2. The action of Ti and Ti+1 on S is given by the matrices

ϕS(Ti) = ( q00 0 q-q-1 1-q2(c1-c3) q-q2(c3-c1)-1 1-q2(c3-c1) 0 q-q2(c1-c3)-1 1-q2(c1-c3) q-q-1 1-q2(c3-c1) ) and ϕS(Ti+1) = ( q-q-1 1-q2(c2-c3) q-q2(c3-c2)-1 1-q2(c3-c2) 0 q-q2(c2-c3)-1 1-q2(c2-c3) q-q-1 1-q2(c3-c2) 0 00q )

respectively, where c1=c(L(i)), c2=c(L(i+1)) and c3=c(L(i+2)). In case (4) the space S is six dimensional and spanned by the vectors vQ corresponding to the standard tableaux

a b c b a c a c b b c a c a b c b a

where a=i, b=i+1, and c=i+2. The action of Ti and Ti+1 on S is given by the matrices

ϕS(Ti) = ( q-q-1 1-q2d12 q-q2d21-1 1-q2d21 0000 q-q2d12-1 1-q2d12 q-q-1 1-q2d21 0000 00 q-q-1 1-q2d13 q-q2d31-1 1-q2d31 00 00 q-q2d13-1 1-q2d13 q-q-1 1-q2d31 00 0000 q-q-1 1-q2d23 q-q2d32-1 1-q2d32 0000 q-q2d23-1 1-q2d23 q-q-1 1-q2d32 )

and

ϕS(Ti+1) = ( q-q-1 1-q2d23 0 q-q2d32-1 1-q2d32 000 0 q-q-1 1-q2d13 00 q-q2d31-1 1-q2d31 0 q-q2d23-1 1-q2d23 0 q-q-1 1-q2d32 0 0 0 0 0 0 q-q-1 1-q2d12 0 q-q2d21-1 1-q2d21 0 q-q2d13-1 1-q2d13 0 0 q-q-1 1-q2d31 0 0 0 0 q-q2d12-1 1-q2d12 0 q-q-1 1-q2d21 )

where dk=c (L(i+k-1))-c (L(i+l-1)). In each case we compute directly the products ϕS(Ti) ϕS(Ti+1) ϕS(Ti) and ϕS(Ti+1) ϕS(Ti) ϕS(Ti+1) and verify that they are equal. (This proof of the braid relation is, in all essential aspects, the same as that used by Hoefsmit [Hoe1974], Wenzl [Wen1988] and Ariki and Koike [AKo1994]. For a more elegant but less straightforward proof of this relation see the proof of Theorem 3.1 in [Ram1998].)

Step 2. The module H(c,λ/μ) is irreducible.

Proof.

Let L be a standard tableaux of shape λ/μ and define

πL= i=1n PL xi-q2c(P(i)) q2c(L(i))- q2c(P(i)) ,

where the second product is over all standard tableaux P of shape λ/μ which are not equal to L. Then πL is an element of Hn such that

πLvQ= δLQvL,

for all standard tableaux Q of shape λ/μ. This follows from the formula for the action of xi on H(c,λ/μ) and the fact that the sequence ( q2c(L(1)) ,, q2c(L(n)) ) completely determines the standard tableau L (Lemma 2.2).

Let N be a nonzero submodule of H(c,λ/μ) and let v=QaQvQ be a nonzero element of N. Let L be a standard tableau such that the coefficient vL is nonzero. Then πLv=aLvL and so vLN.

By Proposition 2.1 we may identify the set λ/μ with an interval in Sn (under Bruhat order). Under this identification the minimal element is the column reading tableau C and there is a chain C<si1C<<sipsi1C=L such that all elements of the chain are standard tableaux of shape λ/μ. Then, by the definition of the τi-operators,

τi1 τipvL= κvC,

for some constant κ*. It follows that vCN.

Let Q be an arbitrary standard tableau of shape λ/μ. Again, there is a chain C<sj1C<<sjpsj1C=Q of standard tableaux in λ/μ and we have

τjp τj1vC= κvQ,

for some κ*. Thus vQN. It follows that N=H(c,λ/μ). Thus H(c,λ/μ) is irreducible.

Step 3. The modules H(c,λ/μ) are nonisomorphic.

Proof.

Each of the modules H(c,λ/μ) has a unique basis (up to multiplication of each basis vector by a constant) of simultaneous eigenvectors for the xi. Each basis vector is determined by its weight, the sequence of eigenvalues (t1,,tn) given by

xivt=tivt, for1in.

By the definition of the action of the xi, a weight of H(c,λ/μ) is equal to ( q2c(L(1)) ,, q2c(L(n)) ) for some standard tableau L. By Lemma 2.2, both the standard tableau L and the placed skew shape (c,λ/μ) are determined uniquely by this weight. Thus no two of the modules H(c,λ/μ) can be isomorphic.

Step 4. If t=(t1,,tn) is the weight of a calibrated Hn-module M then t= ( q2c(L(1)) ,, q2c(L(n)) ) for some standard tableau L of placed skew shape.

Proof.

Let mt be a weight vector in M of weight t=(t1,,tn), i.e.

ximt=ti mt,for all1 in.

We want L such that ( q2c(L(1)) ,, q2c(L(n)) ) =(t1,,tn). We shall show that if ti=tj for i<j then there exist k and such that i<k<<j, tk=q±2ti and t=q2ti. This will show that if there are two adjacent boxes of L in the same diagonal then these boxes must be contained in a complete 2×2 block, i.e. if there is a configuration in L of the form

i j thenLmust contain i k j or i k j .

This is sufficient to guarantee that L is of skew shape.

Let j>i be such that tj=ti and j-i is minimal. The argument is by induction on the value of j-i.

Case 1: j-i=1. Then mt and Timt are linearly independent. If they were not we would have Timt=amt which would give

tiamt = xiTimt = ( Tixi+1- (q-q-1) xi+1 ) mt = ( ati+1- (q-q-1) ti+1 ) mt = (a-(q-q-1)) timt.

Since q-q-10, this equation implies that ti=0 which is a contradiction. Now the relations (1.1d) and (1.4) show that

xiTimt = ti ( Timt- (q-q-1) mt ) , xi+1Timt = ti+1 ( Timt- (q-q-1) mt ) , xjTimt = tjTimt, for allji, i+1.

It follows that Timt is an element of Mtgen but not an element of Mt. This is a contradiction to the fact that M is calibrated. So this is not possible, i.e. ti+1 cannot equal ti.

Case 2: j-i=2. Since titi+1 and mt is a weight vector, the vector

msit= ( Ti- (q-q-1)ti+1 ti+1-ti ) mt

is a weight vector of weight t=sit (see (3.7a)). Then ti=ti+1 and so, by Case 1, msit=0. This implies that

Timt= (q-q-1)ti+1 ti+1-ti mt.

By equation (1.3), all eigenvalues of Ti are either q or -q-1. Thus Timt=±q±1mt and so ti=q±2ti+1. A similar argument shows that

msi+1t= ( Ti+1- (q-q-1)ti+2 ti+2-ti+1 ) mt

must be 0 and thus that

Ti+1mt= (q-q-1)ti+2 ti+2-ti+1 mt= (q-q-1)ti ti-ti+1 mt=q1mt.

From Timt=±q±1mt and Ti+1mt=q1mt we get

±q±1mt = TiTi+1Timt = Ti+1TiTi+1 mt = q1mt.

This is impossible since q is not a root of unity. So this case is not possible, i.e. ti+2 cannot equal ti.

Induction step. Assume that i and j are such that ti=tj and the value j-i is minimal such that this is true.

If tj-1q±2tj then the vector

msjt= ( Tj- (q-q-1)tj tj-1-tj ) mt

is a weight vector of weight t=sjt and by (3.7b) this vector is nonzero. Since ti=ti=tj=tj-1 we can apply the induction hypothesis to conclude that there are k and with i<k<<j-1 such that tk=q±2ti and t=q2ti. This implies that tk=q±2ti and t=q2ti.

Similarly, if tiq±2ti+1 then the vector

msit= ( Ti= (q-q-1)ti+1 ti+1-ti ) mt

is a weight vector of weight t=sit and by (3.7b) this vector is nonzero. Since ti+1=ti=tj=tj we can apply the induction hypothesis to conclude that there are k and with i+1<k<<j such that tk=q±2tj and t=q2tj. This implies that tk=q±2ti and t=q2ti.

If we are not in either of the previous cases then ti+1=q2ti or ti+1=q-2ti and tj-1=q2tj or tj-1=q-2tj. We cannot have ti+1=tj-1 since the i and j are such that j-i is minimal such that ti=tj. Thus q±2ti+1= q2tj-1= ti.

Step 5. Suppose that M is an irreducible calibrated Hn-module and that mt is a weight vector in M with weight t=(t1,,tn) such that ti=q±2ti+1. Then τimt=0.

Proof.

Assume that msit=τimt0. Then, by the second identity in Proposition 3.2 (b), τimsit=0. Since M is irreducible there must be some sequence of τ-operators such that

τi1 τip msit= κmt,

with κ*. Assume that τi1τip is a minimal length sequence such that this is true. We have si1sipsit=t.

Assume that si1sipsi1. Then there must be 1i<jn such that ti=tj (because some nontrivial permutation of the ti fixes t). Since si1sipsi is of minimal length such that it fixes t it must be a transposition (i,j) for some i<j such that ti=tj. Furthermore there does not exist i<k<j such that ti=tk. The element si1sipsi switches the ti and the tj in t. In the process of doing this switch by a sequence of simple transpositions there must be some point where ti and tj are adjacent and thus there must be some such that si ( si+1 sipsit ) =si+1 sipsit. Then

mt= τi+1 τiτi mt

is a nonzero weight vector in M of weight t=si+1sipsit. Since sit=t it follows that ti=ti+1. Since M is calibrated this is a contradiction to (Case 1 of) step 4.

So si1sipsi=1. Let k be minimal such that si1sik is not reduced. Assume p>1. Then we can use the braid relations (the third and fourth lines of Proposition 3.2 (b)) to write

κmt=τj1 τjk-2 τik τik τik+1 τipτimt,

for some j1,,jk-2. Then, by the second line of Proposition 3.2 (b),

κmt=τj τjk-2 τik+1 τipτimt,

for some κ*. This is a contradiction to the minimality of the length of the sequence τi1τip.

So p=1, ip=i and τiτimt=κmt. This is a contradiction since the second identity in Proposition 3.2 (b) and the assumption that ti=q±2ti+1 imply that τiτi=0. So τimt=0.

Step 6. An irreducible calibrated Hn-module M is isomorphic to H(c,λ/μ) for some placed skew shape (c,λ/μ).

Proof.

Let mt be a nonzero weight vector in M. Since M is calibrated step 4 implies that there is a placed skew shape (c,λ/μ) and a standard tableau L of shape λ/μ such that t= ( q2c(L(1)) ,, q2c(L(n)) ) , Let us write mL in place of mt. Let C be the column reading tableau of shape λ/μ. It follows from Proposition 2.1 that there is a chain C,si1C,,sipsi1C=L of standard tableaux of shape λ/μ. By (3.7b), all of the τij in this sequence are bijections and so

mC=τi1 τipmL

is a nonzero weight vector in M. Similarly, if Q is any other standard tableau of shape λ/μ then there is a chain C,sj1C,,sjpsj1C=Q and so

mQ=τjp τj1mC

is a nonzero weight vector in M. Finally, by step 5, τimQ=0 if siQ is not standard (since q2c(Q(i))= q±2 q2c(Q(i+1)) ) and so the span of the vectors {mQQa standard tableau of shapeλ/μ} is a submodule of M. Since M is irreducible this must be all of M and the map

M H(c,λ/μ) τvmC τvvC

is an isomorphism of Hn-modules.

This completes the proof of Theorem 4.1.

Notes and References

This is an excerpt of the preprint entitled Skew shape representations are irreducible authored by Arun Ram in 1998.

Research supported in part by National Science Foundation grant DMS-9622985, and a Postdoctoral Fellowship at Mathematical Sciences Research Institute.

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