Skew shapes, ribbons, conjugation, etc. in type A

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

Last update: 13 November 2012

Skew shapes, ribbons, conjugation, etc. in type A

In this section we shall explain how the definitions in Section 5.1 correspond to classical notions in Young tableaux theory. As in the previous section let R be the root system of Type An-1 as given in Section 5.2. For clarity, we shall state all of the results in this section for placed shapes (γ,J) such that γ is dominant and integral, i.e., γ=(γ1,,γn) with γ1γn and γi. This assumption is purely for notational clarity.

6.1. Axial distance

Let (γ,J) be a local region such that γ is dominant and integral. Let wp(γ,J) and let p be the corresponding standard tableau as defined by the map in Theorem 5.9. Then it follows from the definitions of γ and wp in (5.6) and (5.4) that

wγ,εi= γ,wp-1 ei =c(boxwp-1(i)) =c(p(i)), (6.2)

where p(i) is the box of p containing the entry i.

In classical standard tableau theory the axial distance between two boxes in a standard tableau is defined as follows. Let λ be a partition and let p be a standard tableau of shape λ. Let 1i,jn and let p(i) and p(j) be the boxes which are filled with i and j, respectively. Let (ri,ci) and (rj,cj) be the positions of these boxes, where the rows and columns of λ are numbered in the same way as for matrices. Then the axial distance from j to i in p is

dji(p)= cj-ci+ri -rj,

(see [Wen1988]). Rewriting this in terms of the local region (γ,J) determined by (5.7),

dji(p)=c (p(j))-c (p(i))= wpγ,εj- εi =dεj-εi (w),

where wp(γ,J) is the permutation corresponding to the standard tableau p and dα(wp)is the α-axial distance defined in (f) of Section 5.1. This shows that the axial distance defined in (f) of Section 5.1 is a generalization of the classical notion of axial distance. These numbers are crucial to the classical construction of the seminormal representations of the symmetric group given by Young (see Remark (3) of Section 5.1).

6.3. Skew shapes

The following proposition shows that, in the case of a root system of type A, the definition of skew local region coincides with the classical notion of a skew shape.

Proposition 6.4. Let (γ,J) be a local region with γ dominant and integral. Then the configuration of boxes associated to (γ,J) is a placed skew shape if and only if (γ,J) is a skew local region.

Proof.

() We shall show that if the placed configuration corresponding to the pair (γ,J) has any 2×2 blocks of the forms

a b c a b c a b Case (1) Case (2) Case (3)

then there exists a w(γ,J) such that wγ violates one of the two conditions in (c) of Section 5.1. This will show that if (γ,J) is a skew local region then the corresponding placed configuration of boxes must be a placed skew shape. In the pictures above the shaded regions indicate the absence of a box and, for reference, we have labeled the boxes with a,b,c.

Case (1). Create a standard tableau p such that the 2×2 block is filled with

i-1 i i+1

by filling the region of the configuration strictly north and weakly west of box c in row reading order (sequentially left to right across the rows starting at the top), putting the next entry in box c, and filling the remainder of the configuration in column reading order (sequentially down the columns beginning at the leftmost available column). Let w=wp be the permutation in (γ,J) which corresponds to the standard tableau p. Let p(i) denote the box containing i in p. Then, using the identity (6.2),

wγ,αi+ αi+1 = wγ,εi+1 -εi-1 =c (p(i+1))-c (p(i-1))=0,

since the boxes p(i+1) and p(i-1) are on the same diagonal. However,

wγ,αi = wγ,ei- εi-1 =c(p(i))-c (p(i-1))=1, and wγ,αi+1 = wγ,εi+1 -εi =c(p(i+1))-c (p( i))=-1,

and so condition (c)(1) of Section 5.1 is violated.

Case (2). Create a standard tableau p such that the 2×2 block is filled with

i-1 i i+1

by filling the region weakly north and strictly west of box c in column reading order, putting the next entry in box c, and filling the remainder of the configuration in row reading order. Using this standard tableau p, the remainder of the argument is the same as for Case (1).

Case (3). Create a standard tableau p such that the 2×2 block is filled with

i-1 i

by filling the region strictly north and strictly west of box b in column reading order, putting the next entry in box b, and filling the remainder of the configuration in row reading order. Let w=wp be the permutation in (γ,J) corresponding to p and let p(i) denote the box containing i in p. Then

wγ,αi= wγ,εi- εi-1 =c (p(i))- (p(i-1))=0,

since t(i) and t(i-1) are on the same diagonal. Hence, condition (c)(1) of Section 5.1 is violated.

() Let γn and λ/μ describe a placed skew shape (a skew shape placed on infinite graph paper). Let (γ,J) be the corresponding local region as defined in (5.7). We will show that every wγ is calibratable for every w(γ,J).

Let w(γ,J) and let p be the corresponding standard tableau of shape λ/μ. Consider a 2×2 block of boxes of p. If these boxes are filled with

i j k

then either i<j<k< or i<k<j<. In both cases we have i<-1 and it follows that -1 and are not on the same diagonal. Thus

wγ,α =c(p())-c (p(-1))0,

and so wγ satisfies condition (a) in the definition of calibratable.

The same argument shows that one can never get a standard tableau in which and -2 occur in adjacent boxes of the same diagonal and thus it follows that wγ satisfies condition (b) in the definition of calibratable. Thus (γ,J) is a skew local region.

6.5. Ribbon shapes

Classically, a border strip (or ribbon) is a skew shape which contains at most one box in each diagonal. Although the convention, [Mac1995, I, Section 1 p. 5], is to assume that border strips are connected skew shapes we shall not assume this.

Recall from (b) of Section 5.1 that a placed shape (γ,J) is a placed ribbon shape if γ is regular, i.e., γ,α0 for all αR.

Proposition 6.6. Let (γ,J) be a placed ribbon shape such that γ is dominant and integral. Then the configuration of boxes corresponding to (γ,J) is a placed border strip.

Proof.

Let (γ,J) be a placed ribbon shape with γ dominant and regular. Since γ=(γ1,,γn) is regular, γiγj for all ij. In terms of the placed configuration γi=c(boxi) is the diagonal that boxi is on. Thus the configuration of boxes corresponding to (γ,J) contains at most one box in each diagonal.

Example. If γ= ( -6,-5,-4,0,1, 3,4,5,6,7 ) and J= { ε2-ε1, ε5-ε4, ε7-ε6, ε9-ε8, ε10-ε9 } then the placed configuration of boxes corresponding to (γ,J) is the placed border strip

7 6 4 5 3 1 0 -5 -4 -6

where the boxes are labeled with their contents.

6.7. Conjugation of shapes

Let (γ,J) be a placed shape with γ dominant and integral (i.e., γ=(γ1,,γn) with γ1γn and γi) and view (γ,J) as a placed configuration of boxes. In terms of placed configurations, conjugation of shapes is equivalent to transposing the placed configuration across the diagonal of boxes of content 0. The following example illustrates this.

Example. Suppose γ=(-1,-1,-1,0,0,1,1) and J= ( ε4-ε2, ε4-ε3, ε6-ε5, ε7-ε5 ) . Then the placed configuration of boxes corresponding to (γ,J) is

-1 0 1 -1 1 -1 0

in which the shaded box is not a box in the configuration.

The minimal length representative of the coset w0Wγ is the permutation

u= ( 1234567 5673412 ) .

We have -uγ=-w0γ= ( -1,-1,0,0, 1,1,1 ) and

-u(p(γ)\J) = -u { ε4-ε1, ε5-ε1, ε5-ε2, ε5-ε3, ε6-ε4, ε7-ε4 } = - { ε3-ε5, ε4-ε5, ε4-ε6, ε4-ε7, ε1-ε3, ε2-ε3 } = { ε5-ε3, ε5-ε4, ε6-ε4, ε7-ε4, ε3-ε1, ε3-ε2 } .

Thus the configuration of boxes corresponding to the placed shape (γ,J) is

1 0 1 -1 1 -1 0

6.8. Row reading and column reading tableaux

Let (γ,J) be a placed shape such that γ is dominant and integral and consider the placed configuration of boxes corresponding to (γ,J). The minimal box of the configuration is the box such that

  1. (m1) there is no box immediately above,
  2. (m2) there is no box immediately to the left,
  3. (m3) there is no box northwest in the same diagonal, and
  4. (m4) it has the minimal content of the boxes satisfying (m1)-(m3).

There is at most one box in each diagonal satisfying (m1)-(m3). Thus, (m4) guarantees that the minimal box is unique. It is clear that the minimal box of the configuration always exists.

The column reading tableaux of shape (γ,J) is the filling pmin which is created inductively by

  1. filling the minimal box of the configuration with 1, and
  2. if 1,2,,i have been filled in then fill the minimal box of the configuration formed by the unfilled boxes with i+1.

The row reading tableau of shape (γ,J) is the standard tableau pmax whose conjugate pmax is the column reading tableaux for the shape (γ,J) (the conjugate shape to (γ,J)).

Recall the definitions of the weak Bruhat order and closed subsets of roots given after Eq. (4.5).

Theorem 6.9. Let (γ,J) be a placed shape such that γ is dominant and integral (i.e., γ=(γ1,,γn) with γ1γn and γi). Let pmin and pmax be the column reading and row reading tableaux of shape (γ,J), respectively, and let wmin and wmax be the corresponding permutations in (γ,J). Then

R(wmin)=J, R(wmax)= (P(γ)\J) Z(γ)c ,and (γ,J)= [wmin,wmax],

where Kc denotes the complement of K in R+ and [wmin,wmax] denotes the interval between wmin and wmax in the weak Bruhat order.

Proof.

(a) Consider the configuration of boxes corresponding to (γ,J). If k>i then either c(boxk)> c(boxi), or boxk is in the same diagonal and southeast of boxi. Thus when we create pmin we have that

ifk>ithen boxkgets filled before boxiboxk isnorthwestof boxi,

where the northwest is in a very strong sense: There is a sequence of boxes

boxi=boxio, boxi1,, boxir=boxk

such that boxim is either directly above boxim-1 or in the same diagonal and directly northwest of boxim-1. In other words,

ifk>ithen pmin(boxk)< pmin(boxi) boxkisnorthwest ofboxi.

So, from the formula for wp in (5.4) we get

ifk>ithen wmin(k)< wmin(i) εk-εiJ,

where wmin is the permutation in (γ,J) which corresponds to the filling tmin and J is the closure of J in R. It follows that

R(wmin)=J.

(b) There are at least two ways to prove that R(wmax)= (P(γ)\J) Z(γ)c . One can mimic the proof of part (a) by defining the maximal box of a configuration and a corresponding filling. Alternatively one can use the definition of conjugation and the fact that R(w0w)= R(w)c. The permutation wmin is the unique minimal element of (γ,J) and the conjugate of wmax is the unique minimal element of (γ,J). We shall leave the details to the reader.

(c) An element wW is an element of (γ,J) if and only if R(w)P(γ)=J and R(w)Z(γ)=. Thus (γ,J) consists of those permutations wW such that

JR(w) (P(γ)\J) Z(γ)c .

Since the weak Bruhat order is the ordering determined by inclusions of R(w), it follows that (γ,J) is the interval between wmin and wmax.

Example. Suppose γ= ( -1,-1,-1,0,0,1,1 ) and J= { ε4-ε2, ε4-ε3, ε6-ε5, ε7-ε5 } . The minimal and maximal elements in (γ,J) are the permutations

wmin= ( 1234567 1342756 ) andwmax= ( 1234567 1662734 ) .

The permutations correspond to the standard tableaux

1 2 5 3 6 4 7 and 1 2 3 5 4 6 7

Notes and References

This is an excerpt of the paper entitled Affine Hecke algebras and generalized standard Young tableaux written by Arun Ram in 2002, published in the Academic Press Journal of Algebra 260 (2003) 367-415. The paper was dedicated to Robert Steinberg.

Research supported in part by the National Science Foundation (DMS-0097977) and the National Security Agency (MDA904-01-1-0032).

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