“Garnir relations” and an analogue of Young’s natural basis

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

Last update: 3 March 2013

“Garnir relations” and an analogue of Young’s natural basis

Each of the modules H(c,λ/μ) constructed in Theorem 4.1 has two natural bases:

  1. The “seminormal basis” {vLLλ/μ} which, up to multiplication of each basis vector by a constant, is given by vL=τw vC, (5.1) where C is the column reading tableau of shape λ/μ, w is the permutation such that L=wC and τw is as in (3.3).
  2. The “natural basis” {nLLλ/μ} given by nL=TwvC, (5.2) where C is the column reading tableau of shape λ/μ, w is the permutation such that L=wC and Tw is as defined in (1.6).

Proposition 5.3. Let (c,λ/μ) be a placed skew shape with n boxes and let H(c,λ/μ) be the Hn-module defined in Theorem 4.1. Let C be the column reading tableau of shape λ/μ and let wC denote the tableau C with the entries permuted according to the permutation w. For each standard tableau L let nL be defined by formula (5.2). Then {nLLλ/μ} is a basis of H(c,λ/μ).

Proof.

If L is a standard tableau of shape λ/μ let vL be as given in (5.1). It follows from the formulas defining the module H(c,λ/μ) that the basis {vLLλ/μ} is simply a renormalized version of the basis {vLLλ/μ}, i.e there are constants κL* such that vL=κLvL.

Let si1sip=w be a reduced word for w. Then, with notations as in Theorem 4.1,

vL= τi1τip vC= ( Ti1- (Ti1)L2L2 ) ( Ti2- (Ti1)L3L3 ) ( Tip- (Tip)LpLp ) nC,

where Lj=sij+1sipC. Expanding this expression yields

vL= ( Tw+u<w buTu ) nC=nL+ u<wbu nuC,

for some constants bu. The second equality is a consequence of the fact that, by Proposition 2.1, the tableaux uC,u<w are always standard. Since {vLLλ/μ} is a basis it follows from the triangular relation above that {nLLλ/μ} is also a basis of H(c,λ/μ).

The construction of H(c,λ/μ) in Theorem 4.1 makes the notational assumption that vL=0, whenever L is not a standard tableau. Formula (5.1) can be used as a definition of vL even in the case when L is not standard and, from the definition of the action in Theorem 4.1,

vL=0,if Lis not standard. (5.4)

Theorem 5.5 proves that these relations, when expanded in terms of the basis {nLLλ/μ} are exactly the classical Garnir relations!!

Let λ/μ be a skew shape. A pair of adjacent boxes in the same row of λ/μ determines a snake in λ/μ consisting of the boxes in the pair, all the boxes above the righthand box of this pair, and all the boxes below the lefthand box of the pair. See the picture in Theorem 5.5 (b).

Theorem 5.5. ("Garnir relations") Let (c,λ/μ) be a placed skew shape and let H(c,λ/μ) be the Hn-module defined in Theorem 4.1. Let {nLLλ/μ} be the basis of H(c,λ/μ) defined by Proposition 5.3 and let C be the column reading tableau of shape λ/μ.

  1. If i and i+1 are entries in the same column of C then TinC=-q-1 nC.
  2. Fix a snake in λ/μ. Let P be the standard tableau which has all entries the same as C except that the entries in the snake are entered in row reading order instead of in column reading order. C= j+1 j+2 -1 i i+1 j-1 j P= i i+1 k-1 k+1 k k+2 -1 Let SA,SB and SAB be the subgroups of Sn consisting of the permutations of A={i,i+1,,j}, B= {j+1,,-1,} and AB, respectively, and let SAB/(SA×SB) be the set of minimal length coset representatives of cosets of SA×SB in SAB. The elements of SAB/(SA×SB) are sometimes called the “shuffles” of A and B. Then 0 = ( uSAB/(SA×SB) (-q) (wAB)- (wAwB)- (u) Tu ) nC = TknP+ uP (-q) (wAB)- (wAwB)- (u) nuC, where (wAB)= (-i+12,) (wAwB)= (j-i+12) (-j2) and the last sum is over all standard tableaux uC which are obtained from C by permuting entries which are in the snake.

Proof.

Part (a) follows immediately from the definition of H(c,λ/μ) in Theorem 4.1.

(b) The subgroups SA,SB and SAB have longest elements

wA= ( ii+1j-1j jj-1i+1i ) ,wB= ( j+1j+2-1 -1j+2j+1 ) ,

wAB= ( ii+1-1 -1i+1i ) ,

with lengths (wA)= (j-i+12), (wB)= (-j2), and (wAB)= (-i+12), respectively, and wAwB is the longest element of SA×SB SAB. Let t=(t1,,tn)= ( a,aq2,aq4,, aq2(n-1) ) where a= q 2 ( c(C(j)) -(i-1) ) *. The positions of C(i),,C() in λ/μ are such that

( q2c(C(i)) ,, q2c(C()) ) = wAwB ( q2c(C(j)), q2c(C(j-1)), , q2c(C(i)), q2c(C()), q2c(C(-1)), , q2c(C(j+1)), ) = wAwB ( q2c(C(j)), q2c(C(j)) q2,, q2c(C(j)) q2(-i+1) ) = wAwB (ti,,t).

By Proposition 3.6 and the fact that TsnC=-q-1nC for all is,sj,

τwAwB tnC= ( wSA×SB (-q) (wAwB) -(w) Tw ) nC= [j-i+1]! [-j]!nC,

where [r]!=[r] [r-1][2] [1] and [p]= (qp-q-p)/ (q-q-1).

Let π be the permutation such that P=πC and let vP=τπvC. Then siπwAwB=wAB and, by (3.5),

τivP = τiτπvC= τiτπ wAwBt nC = 1 [j-i+1]! [-j]! τiτπ wAwBt τwAwB tnC = 1 [j-i+1]! [-j]! τiτπ τwAwB tnC = 1 [j-i+1]! [-j]! τwAB tnC = 1 [j-i+1]! [-j]! ( wSAB (-q) (wAB) -(w) Tw ) nC.

Each element wSAB has a unique expression w=uv such that vSA×SB and (w)=(u)+(v). The left factor u is the minimal length representative of the coset w(SA×SB) in SAB. Then

wSAB (-q) (wAB) -(w) Tw = ( uSAB/(SA×SB) (-q) (wAB)- (wAwB)-(u) Tu ) × ( vSA×SB (-q) (wAwB)- (v) Tv ) .

It follows from the formulas for the action on H(c,λ/μ) that the element vP=τπvC is a nonzero multiple of the basis element vP. Furthermore, by (5.4), τkvP=0. So

0 = τivP = 1 [j-i+1]! [-j]! ( uSAB/(SA×SB) (-q) (wAB)- (wAwB)-(u) Tu ) × ( vSA×SB (-q) (wAwB)- (v) Tv ) nC = ( uSAB/(SA×SB) (-q) (wAB)- (wAwB)-(u) Tu ) nC

For each uSAB/(SA×SB) except u=wABwAwB, the tableau uC is standard. In fact these are exactly the standard tableaux Q which are obtained by permuting entries of C which are in the snake. The tableau wABwAwB C=skP. Note that (wABwAwB) =(wAB) -(wAwB). Thus we have

0=TknP+ uP (-q) (wAB)- (wAwB)- (u) nuC,

where the sum over all standard tableaux uC which are obtained from C by permuting entries which are in the snake.

Proposition 5.6. Let (c,λ/μ) be a placed skew shape and let {nLLis a standard tableau of shapeλ/μ} be the basis of the Hn-module H(c,λ/μ) which is defined by Proposition 5.3.

  1. If wSn and L is a standard tableau of shape λ/μ then TwnL=Q bQnQ, with coefficientsbQ [q,q-1].
  2. Assume that the content function c takes values in . If 1in and L is a standard tableau of shape λ/μ then xinL=Q bQnQ, with coefficientsbQ [q,q-1].

Proof.

(a) It is sufficient to show that for all 1dn and all standard tableaux L of shape λ/μ we have

TdnL= QbQnQ,

with coefficients bQ[q,q-1]. For notational convenience let us identify each standard tableau L of shape λ/μ with the permutation wSn such that wC=L, where C is the column reading tableau of shape λ/μ.

From the definitions

TdnL = nsdL, if(sdL)= (L)+1and sdLis standard, TdnL = (q-q-1)nL +nsdL, if(sdL)= (L)-1.

Let L be a standard tableau such that (sdL)=(L)+1 and sdL is not standard. The pair of boxes where the nonstandardness in sdL occurs in either a row or a column:

sdL= a ^ b ^ ^ c ^ d > d+1 ^ e ^ ^ f sdL= d+1 d Case (1) Case (2)

Case (1): The two boxes in sdL where the nonstandardness occurs determine a snake in the shape λ/μ consisting of all the boxes above the entry d and all the boxes below the entry d+1. Use the same notation as in Theorem 5.5 so that P is the standard tableau which is the same as the column reading tableau C except that the entries in the snake are in row reading order. Then skP is nonstandard and there exists a (unique) vSn such that sdL=vskP and (sdL)=(v)+(skP). The permutation v is given by

v= ( ii+1k-1 kk+1k+2 abcd d+1ef )

and v(x)=L(x) for all x{i,i+1,,}. Thus

TdnL=TdTL nC=TvTkTP nC=TvTknP.

Let (wAB)=(-i+12). By Theorem 5.5

TknP=-uP (-q) (wAB)- (wAwB)- (u) TunC,

and so

TdnL=-Tv ( uP (-q) (wAB)- (wAwB)- (u) TunC ) .

The permutations v and uP are such that (vu)=(v)+(u) and so TvTu=Tvu for all uP. Furthermore, the tableaux vuP, for uP, are exactly the standard tableaux which are obtained by permutations of the entries in L which are in the snake. Thus

TdnL=-u P (-q) (wAB)- (wAwB)- (u) nvuC.

Case (2): Suppose that the boxes containing d and d+1 in L are the boxes containing k and k+1 in the column reading tableau C.

C= k k+1

Then Lsk=sdL and

TdnL=TdTL nC=TLTknC =-q-1TLnC =-q-1nL,

by Theorem 5.5 (a).

(b) Let L be a standard tableau of shape λ/μ and let w be the permutation such that L=wC, where C is the column reading tableau of shape λ/μ. By repeatedly using the relations (1.4) we obtain

xinL=xiTw nC=Twxw-1(i) nC+v<wTv pv(x1,,xn) nC,

where the pv(x1,,xn) are polynomials in x1,,xn with coefficients in [q,q-1]. If xi acts on vC by integral powers of q then we have

xinL=Tw q2c(C(w-1(i))) nC+v<wTv pv ( q2c(C(1)) ,, q2c(C(n)) ) nC=vw bvnvC,

with coefficients bv in [q,q-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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