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.


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.



  • 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 \mathbb{C}[G]b_{T_0}. For any column tabloid U = [T], where T is a filling, write T = g(T_0) for a unique g\in S_d. The corresponding element is gb_{T_0} \in \mathbb{C}[G]b_{T_0}.
    • 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}.


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.


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.


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).


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. ♦


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.


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 jj’). 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:

(\chi(g')g' + \chi(g'h)g'h) \chi(g'') \{g''T\} = \chi(g')\chi(g'')g'(e - h) \{g''T\} = 0.

Summing over all cosets of \{e,h\} \le H and g'' \in C(T) we see that (*) sums to zero. ♦

From lemmas 1 and 2, we have:

Corollary. We have:

\beta_{Y'}(U) - \gamma_\emptyset(U) = \beta_{Y'}(U) - U\in \text{ker}(\pi).

E.g. modulo ker(π), we have:


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