Polynomials and Representations XXVI

Let us fix a filling $T_0$ of shape $\lambda$ and consider the surjective homomorphism of $\mathbb{C}[G]$ -modules

$\pi : \mathbb{C}[G]b_{T_0} \to \mathbb{C}[G]b_{T_0} a_{T_0} \subseteq \mathbb{C}[G]a_{T_0},$

given by right-multiplying by $a_{T_0}.$ Specifically, we will describe its kernel, which will have interesting consequences when we examine representations of $GL_n\mathbb{C}$ later.

Row and Column Tabloids

By the lemma here, $\mathbb{C}[G]a_{T_0}$ and $\mathbb{C}[G]b_{T_0}$ can be described as follows.

For the former, take a basis comprising of “row tabloids“, i.e. for each filling T, take each row as a set, thus giving a partitioning $[d] = A_1 \coprod A_2 \coprod \ldots.$ E.g. in the example below, the first two row tabloids are identical since they give rise to the same partitioning.

row_tabloids_example1

For $\mathbb{C}[G]b_{T_0}$ , take a basis of “column tabloids“, by taking a filling T and taking its columns as sets, with the condition that swapping two elements in the same column flips the sign. E.g.

column_tabloid

Note

Given a filling T, the corresponding row tabloid is denoted by {T} and the row tabloid is [T].

If $g\in S_d$ we have $g(\{T\}) = \{g(T)\}$ and $g([T]) = [g(T)]$ . E.g. (1,2) takes the above column tabloid to its negative while (1,2)(3,4) leaves it invariant.

Let us describe explicitly the correspondence between a column tabloid and an element of For any column tabloid U = [T], where T is a filling, write for a unique The corresponding element is
- The reader may worry that we left out the twist $\chi(g),$ but you may check that this is consistent when we swap two elements in the same column.

For a row tabloid, we take $\{g(T_0)\}$ to the element $ga_{T_0} \in \mathbb{C}[G]a_{T_0}$ .

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Column Swaps

Let us define some operations on column tabloids. Consider a column tabloid U := [T].

Suppose, in U, we have a (possibly empty) set B of boxes in column j’ and a set A of squares in column j such that |A| = |B|, with j’ > j. Let $\alpha_{A,B}(U)$ denote the resulting column tabloid obtained by swapping squares A and B in order.

column_tabloid_swap

Now given a set B of squares in column j’ of U, define a linear map

$\displaystyle \beta_B : \mathbb{C}[G]b_{T_0} \to \mathbb{C}[G]b_{T_0},\qquad \beta_B(U) := \sum_A \alpha_{A,B}(U),$

where A runs through all sets of squares in column j of U such that |A| = |B|. Next, given a set Y of squares in column j’ of U, let:

$\displaystyle\gamma_Y(U) := \sum_{B\subseteq Y} (-1)^{|B|}\beta_B(U) = \sum_{\substack{A,B\\ B\subseteq Y,\, |A| = |B|}}(-1)^{|B|}\alpha_{A,B}(U).$

Clearly $\gamma_Y(U)$ is an integer linear combination of the $\beta_B(U)$ ; the following shows that the converse is true.

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Lemma 1. For each set $Y'$ of squares in column $j'$ , we have:

$\displaystyle\sum_{Y \subseteq Y'} (-1)^{|Y|}\gamma_{Y}(U) = \beta_{Y'}(U).$

Proof

Fix A, B satisfying $B\subseteq Y'$ and |A| = |B|; we keep track of the coefficient of $\alpha_{A,B}(U)$ in the expansion of LHS. The term is included in $\gamma_Y(U)$ for all Y satisfying $B\subseteq Y \subseteq Y'.$ Its coefficient is thus:

$\displaystyle(-1)^{|B|}\cdot \sum_{Y,\, B\subseteq Y\subseteq Y'} (-1)^{|Y|}.$

If B = Y’, this is 1; otherwise, fixing some $y\in Y' - B$ , there is a 1-1 correspondence between Y containing y and Y not containing y; these cancel each other out so the sum is 0. Hence the overall sum is:

$\displaystyle \sum_{A,\, |A| = |Y'|} \alpha_{A, Y'}(U) = \beta_{Y'}(U)$

as desired. ♦

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Lemma 2. We have $\gamma_Y(U) \in \text{ker}\pi$ if Y is a non-empty set of squares in the j’-th column of U.

Proof

Let $K\le H\le S_d$ be subgroups defined as follows:

H is the set of $g \in S_d$ acting as the identity outside Y or the j-th column of U;
K is the set of $g\in H$ taking $Y\to Y$ and the j-th column of U back to itself.

For each $g \in K$ we have $\chi(g) g(U) = U$ so $\sum_{g\in K} \chi(g) g(U) = |K|U.$

Let $h\in S_d$ run through all swaps of a subset A of the j-th column with $B\subseteq Y$ preserving the order; this gives a complete set of coset representatives for $H/K.$ On the other hand, the sum of these $h(U)$ gives us $\gamma_Y(U)$ so:

$\displaystyle \sum_{g'\in H} \chi(g')g'(U) = \text{positive multiple of } \gamma_Y(U).$

It remains to show that the LHS is in ker(π).

Fix filling T such that U = [T]. Unwinding the definition for map π, we first take $g\in S_d$ such that $T = g(T_0)$ then take the row tabloid for $gb_{T_0}$ . Now, $gb_{T_0} = gb_{T_0} g^{-1} g = b_T g$ so

$\pi:[T] \ \mapsto\ b_{T}\{T\} = \sum_{g''\in C(T)} \chi(g'') g''\{T\}.$

Since $\pi$ is $\mathbb{C}[G]$ -linear, we now need to show:

$\displaystyle \sum_{g'\in H} \chi(g') g' \sum_{g''\in C(T)} \chi(g'')g''\{T\} \overset{?}=0.\quad (*)$

Since $Y\ne\emptyset,$ pick a square in it and a square to its left in the j-th column (since j < j’). Fix the element $g''\in C(T)$ and let $h\in H$ be the transposition that swaps those two squares in $g''T$ so $h(g''\{T\}) = \{g''T\}$ . Taking $g', g'h$ only in the outer sum of (*) gives: