Polynomials and Representations XXXVIII

Determinant Modules

We will describe another construction for the Schur module.

Introduce variables z_{i,j} for i\ge 1, j\ge 1. For each sequence i_1, \ldots, i_p\ge 1 we define the following polynomials in z_{i,j}:

D_{i_1, \ldots, i_p} := \det{\small \begin{pmatrix} z_{1, i_1} & z_{1, i_2} & \ldots & z_{1, i_p} \\ z_{2, i_1} & z_{2, i_2} & \ldots & z_{2, i_p} \\ \vdots & \vdots & \ddots & \vdots \\ z_{p, i_1} & z_{p, i_2} & \ldots & z_{p, i_p}\end{pmatrix}}.

Now given a filling T of shape λ, we define:

D_T := D_{\text{col}(T, 1)} D_{\text{col}(T, 2)} \ldots

where \text{col}(T, i) is the sequence of entries from the i-th column of T. E.g.

determinant_products_and_filling

Let \mathbb{C}[z_{i,j}] be the ring of polynomials in z_{ij} with complex coefficients. Since we usually take entries of T from [n], we only need to consider the subring \mathbb{C}[z_{i,1}, \ldots, z_{i,n}].

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Let \mu = \overline \lambda. Recall from earlier that any non-zero GL_n\mathbb{C}-equivariant map

\displaystyle \bigotimes_j \left(\text{Alt}^{\mu_j} \mathbb{C}^n\right) \longrightarrow \bigotimes_i \left(\text{Sym}^{\lambda_i} \mathbb{C}^n \right)

must induce an isomorphism between the unique copies of V(\lambda) in the source and target spaces. Given any filling T of shape \lambda, we let e^\circ_T be the element of \otimes_j \text{Alt}^{\mu_j} \mathbb{C}^n obtained by replacing each entry k in T by e_k, then taking the wedge of elements in each column, followed by the tensor product across columns:

basis_element_of_alt_tensor_module

Note that the image of e^\circ_T in F(V) is precisely e_T as defined in the last article.

Definition. We take the map

\displaystyle \bigotimes_j \left(\text{Alt}^{\mu_j} \mathbb{C}^n\right) \longrightarrow \bigotimes_i \left(\text{Sym}^{\lambda_i} \mathbb{C}^n \right), \quad e_T^\circ \mapsto D_T

where z_{i,j} belongs to component \text{Sym}^{\lambda_i}.

E.g. in our example above, D_T is homogeneous in z_{1,j} of degree 5, z_{2,j} of degree 4 and z_{3,j} of degree 3. We let g \in GL_n\mathbb{C} act on \mathbb{C}[z_{i,j}] via:

g = (g_{i,j}) : z_{i,j} \mapsto \sum_k z_{i,k}g_{k,j}.

Thus if we fix i and consider the variables \mathbf z_i := \{z_{i,j}\}_j as a row vector, then g: \mathbf z_i \mapsto \mathbf z_i g^t. From another point of view, if we take z_{i,1}, z_{i,2},\ldots as a basis, then the action is represented by matrix g since it takes the standard basis to the column vectors of g.

Proposition. The map is GL_n\mathbb{C}-equivariant.

Proof

The element g = (g_{i,j}) takes e_i \mapsto \sum_j g_{j,i} e_j by taking the column vectors of g; so

\displaystyle e_T \mapsto \sum_{j_1, \ldots, j_d} g_{j_1, i_1} g_{j_2, i_2} \ldots g_{j_d, i_d} e_{T'}

where T’ is the filling obtained from T by replacing its entries i_1, \ldots, i_d with j_1, \ldots, j_d correspondingly.

On the other hand, the determinant D_{i_1, \ldots, i_p} gets mapped to:

\det{\small \begin{pmatrix} z_{1, i_1} & z_{1, i_2} & \ldots & z_{1, i_p} \\ z_{2, i_1} & z_{2, i_2} & \ldots & z_{2, i_p} \\ \vdots & \vdots & \ddots & \vdots \\ z_{p, i_1} & z_{p, i_2} & \ldots & z_{p, i_p}\end{pmatrix}} \mapsto \det{\small \begin{pmatrix} \sum_{j_1} z_{1,j_1} g_{j_1, i_1} & \ldots & \sum_{j_p} z_{1, j_p}g_{j_p i_p}\\ \sum_{j_1}z_{2, j_1} g_{j_1, i_1} & \ldots & \sum_{j_p} z_{2, j_p}g_{j_p, i_p}  \\ \vdots & \ddots & \vdots \\ \sum_{j_1} z_{p, j_1} g_{j_1, i_1} & \ldots & \sum_{j_p} z_{p, j_p} g_{j_p, i_p} \end{pmatrix}}

which is \sum_{j_1, \ldots, j_d} g_{j_1, i_1} \ldots g_{j_d, i_d} D_{T'}. ♦

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Since \otimes_j \text{Alt}^{\mu_j} \mathbb{C}^n contains exactly one copy of V(\lambda), it has a unique GL_n\mathbb{C}-submodule Q such that the quotient is isomorphic to V(\lambda). The resulting quotient is thus identical to the Schur module F(V), and the above map factors through

F(V) \to \otimes_i \text{Sym}^{\lambda_i} \mathbb{C}^n, \quad e_T \mapsto D_T.

Now we can apply results from the last article:

Corollary 1. The polynomials D_T satisfy the following:

  • D_T = 0 if T has two identical entries in the same column.
  • D_T + D_{T'} = 0 if T’ is obtained from T by swapping two entries in the same column.
  • D_T = \sum_S D_S, where S takes the set of all fillings obtained from T by swapping a fixed set of k entries in column j’ with arbitrary sets of k entries in column j (for fixed j < j’) while preserving the order.

Proof

Indeed, the above hold when we replace D_T by e_T. Now apply the above linear map. ♦

Corollary 2. The set of D_T, for all SSYT T with shape λ and entries in [n], is linearly independent over \mathbb{C}.

Proof

Indeed, the set of these e_T is linearly independent over \mathbb{C} and the above map is injective. ♦

Example 1.

Consider any bijective filling T for \lambda = (2, 1). Writing out the third relation in corollary 1 gives:

\left[\det\begin{pmatrix} a & c \\ b & d\end{pmatrix}\right] x = \left[\det\begin{pmatrix} x & c \\ y & d\end{pmatrix}\right] a + \left[ \det\begin{pmatrix} a & x \\ b & y\end{pmatrix}\right] c.

More generally, if \lambda satisfies \lambda_j = 2 and \lambda_{j'} = 1, the corresponding third relation is obtained by multiplying the above by a polynomial on both sides.

Example 2: Sylvester’s Identity

Take the 2 \times n SYT by writing 1,\ldots, n in the left column and n+1, \ldots, 2n in the right. Now D_T = D_{1,\ldots, n}D_{n+1, \ldots, 2n} is the product:

\det \overbrace{\begin{pmatrix} z_{1,1} & z_{1,2} & \ldots & z_{1,n} \\ z_{2,1} & z_{2,2} &\ldots & z_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ z_{n,1} & z_{n,2} & \ldots & z_{n,n} \end{pmatrix}}^M \det \overbrace{\begin{pmatrix} z_{1,n+1} & z_{1,n+2} & \ldots & z_{1,2n} \\ z_{2,n+1} & z_{2,n+2} &\ldots & z_{2,2n} \\ \vdots & \vdots & \ddots & \vdots \\ z_{n,n+1} & z_{n,n+2} & \ldots & z_{n,2n} \end{pmatrix}}^N.

In the sum D_T = \sum_S D_S, each summand is of the form D_S= \det M' \det N', where matrices M’N’ are obtained from MN respectively by swapping a fixed set of k columns in N with arbitrary sets of k columns in M while preserving the column order. E.g. for n=3 and k=2, picking the first two columns of N gives:

\begin{aligned} \det ( M_1 | M_2 | M_3) \det(N_1 | N_2| N_3) &= \det(N_1 | N_2 | M_3) \det(M_1 | M_2 | N_3) \\ +\det(N_1 | M_2 | N_2) \det(M_1 | M_3 | N_3) &+ \det(M_1 | N_1 | N_2) \det(M_2 | M_3 | N_3).\end{aligned}

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