Polynomials and Representations XXVII

From the previous article, we have columns jj’  in the column tabloid U, and given a set A (resp. B) of boxes in column j (resp. j’), we get:

  U-\beta_B(U) = U- \sum_{A,\, |A| = |B|} \alpha_{A,B}(U) \in \text{ker}(\pi)

where \alpha_{A,B}(U) is the column tabloid obtained by swapping contents of A with B while preserving the order. For example, we have the following modulo ker(π):

modulo_relations_for_specht

Here is our main result.

Main Theorem. Each of the following spans \ker\pi as a vector space:

  • the collection of U - \beta_B(U) over all sets B in column j’ and (j, j’) such that j < j’;
  • as above, except B comprises of the top k squares (for all k) of column j’ and j’ = j+1.

We have shown that the first set spans a subspace of ker(π); since the second set is a subset of the first, it remains to show that Q = ker(π), where Q is the vector space spanned by the second set.

blue-lin

Lemma. Consider [T] as T varies through all SYT with shape \lambda. These span the vector space \mathbb{C}[G]b_{T_0} / Q.

Proof

Create a new total ordering on the set of all fillings T of shape λ. Given T’T, we consider the rightmost column in which T’ and T differ; in that column, take the lowest square in which T’ and T differ; we write T’T if the entry in that square of T’ is larger than that of T:

comparison_of_two_fillings

Suppose T is a filling which is not an SYT. We claim that modulo Q, we can express [T] as a linear sum of [T’] for T’T.

  • Let T’ be obtained from T by sorting each column in ascending order; then [T’] = ±[T] and T’ ≥ T, so we assume each column of T is increasing.
  • If T is not an SYT, then we can find row i and column j of T such that T_{i, j} > T_{i,j+1}. Taking j’j+1 and the top i squares B of column j’ of T, we can replace [T] (modulo Q) by \beta_B([T]) = \sum_{|A|=|B|} \alpha_{A,B}([T]) where each term is [T’] for some T’>T.

Applying this iteratively, since there are finitely many fillings we eventually obtain any [T] as a linear combination of [T’] where T’ is SYT. ♦

blue-lin

Proof of Main Theorem

We know that Q\subseteq \text{ker}(\pi) so we get a surjection

\overline\pi : \mathbb{C}[G]b_{T_0}/Q \to \mathbb{C}[G]b_{T_0}a_{T_0} \cong V_\lambda.

Hence the dimension of \mathbb{C}[G]b_{T_0}/Q is at least \dim V_\lambda = f^\lambda. On the other hand, by our lemma above, the space is spanned by [T] for all SYT T so its dimension is at most the number of SYT of shape λ, i.e. f^\lambda. Thus \overline\pi is a bijection so Q = \text{ker}(\pi) as desired. ♦

As a side consequence, we have also proved that the set of [T] for all SYT T gives a basis for \mathbb{C}[G]b_{T_0}a_{T_0}.

This concludes our discussion of representations of the symmetric group S_d. Our next target is the representation theory of the general linear group GL_n \mathbb{C}.

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