## 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. 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 $\mathbb{C}[z_{i,1}, \ldots, z_{i,n}]$. 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: 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 $\sum_{j_1, \ldots, j_d} g_{j_1, i_1} \ldots g_{j_d, i_d} D_{T'}$. ♦ 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} This entry was posted in Uncategorized and tagged , , , , , , . Bookmark the permalink.