Last update: 3 March 2013
Each of the modules constructed in Theorem 4.1 has two natural bases:
Proposition 5.3. Let be a placed skew shape with boxes and let be the defined in Theorem 4.1. Let be the column reading tableau of shape and let denote the tableau with the entries permuted according to the permutation For each standard tableau let be defined by formula (5.2). Then is a basis of
If is a standard tableau of shape let be as given in (5.1). It follows from the formulas defining the module that the basis is simply a renormalized version of the basis i.e there are constants such that
Let be a reduced word for Then, with notations as in Theorem 4.1,
where Expanding this expression yields
for some constants The second equality is a consequence of the fact that, by Proposition 2.1, the tableaux are always standard. Since is a basis it follows from the triangular relation above that is also a basis of
The construction of in Theorem 4.1 makes the notational assumption that whenever is not a standard tableau. Formula (5.1) can be used as a definition of even in the case when is not standard and, from the definition of the action in Theorem 4.1,
Theorem 5.5 proves that these relations, when expanded in terms of the basis are exactly the classical Garnir relations!!
Let be a skew shape. A pair of adjacent boxes in the same row of determines a snake in 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 be a placed skew shape and let be the defined in Theorem 4.1. Let be the basis of defined by Proposition 5.3 and let be the column reading tableau of shape
Part (a) follows immediately from the definition of in Theorem 4.1.
(b) The subgroups and have longest elements
with lengths and respectively, and is the longest element of Let where The positions of in are such that
By Proposition 3.6 and the fact that for all
Let be the permutation such that and let Then and, by (3.5),
Each element has a unique expression such that and The left factor is the minimal length representative of the coset in Then
It follows from the formulas for the action on that the element is a nonzero multiple of the basis element Furthermore, by (5.4), So
For each except the tableau is standard. In fact these are exactly the standard tableaux which are obtained by permuting entries of which are in the snake. The tableau Note that Thus we have
where the sum over all standard tableaux which are obtained from by permuting entries which are in the snake.
Proposition 5.6. Let be a placed skew shape and let be the basis of the which is defined by Proposition 5.3.
(a) It is sufficient to show that for all and all standard tableaux of shape we have
with coefficients For notational convenience let us identify each standard tableau of shape with the permutation such that where is the column reading tableau of shape
From the definitions
Let be a standard tableau such that and is not standard. The pair of boxes where the nonstandardness in occurs in either a row or a column:
Case (1): The two boxes in where the nonstandardness occurs determine a snake in the shape consisting of all the boxes above the entry and all the boxes below the entry Use the same notation as in Theorem 5.5 so that is the standard tableau which is the same as the column reading tableau except that the entries in the snake are in row reading order. Then is nonstandard and there exists a (unique) such that and The permutation is given by
and for all Thus
Let By Theorem 5.5
The permutations and are such that and so for all Furthermore, the tableaux for are exactly the standard tableaux which are obtained by permutations of the entries in which are in the snake. Thus
Case (2): Suppose that the boxes containing and in are the boxes containing k and in the column reading tableau
by Theorem 5.5 (a).
(b) Let be a standard tableau of shape and let be the permutation such that where is the column reading tableau of shape By repeatedly using the relations (1.4) we obtain
where the are polynomials in with coefficients in If acts on by integral powers of then we have
with coefficients in
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.