Polynomials and Representations XI

Here, we will give a different interpretation of the Schur polynomial, however this definition only makes sense in the ring $\Lambda_n.$

For a given vector $\alpha = (\alpha_1, \ldots, \alpha_n)$ of non-negative integers, define the following determinant, a polynomial in $x_1, \ldots, x_n$:

$\Delta_\alpha := \det\left( x_j^{\alpha_i} \right)_{1\le i, j \le n} =\det \begin{pmatrix}x_1^{\alpha_1 } & x_2^{\alpha_1 } & \ldots & x_n^{\alpha_1}\\x_1^{\alpha_2 } & x_2^{\alpha_2 } & \ldots & x_n^{\alpha_2}\\\vdots & \vdots & \ddots & \vdots \\x_1^{\alpha_n} & x_2^{\alpha_n} & \ldots & x_n^{\alpha_n}\end{pmatrix}.$

For the case where $\delta = (n-1, n-2, \ldots, 0)$, we also denote $\Delta := \Delta_\delta = \Delta_{n-1, n-2, \ldots, 0}.$ The following result is classical.

Lemma. We have $\Delta = \prod_{1\le i < j \le n} (x_i - x_j).$ Also, for any $\alpha,$ the polynomial $\Delta_\alpha$ is divisible by $\Delta.$

Proof

In the above matrix, swapping any distinct $x_i, x_j$ results in an exchange of two columns, which flips the sign of the determinant. Thus when $x_i = x_j$, $\Delta_\alpha$ vanishes. Hence $\Delta_\alpha$ is divisible by $x_i - x_j$ for all $i It remains to prove the first statement.

For that, note that $\Delta$ and $\prod_{i are both homogeneous of degree $\frac {n(n-1)} 2$ so they are constant multiples of each other. Since the largest term (in lexicographical order) is $x_1^{n-1} x_2^{n-2} \ldots x_{n-1}$ on both sides, equality holds. ♦

Definition. Suppose $\lambda$ is a partition with $l(\lambda) \le n$; append zeros to it so that $\lambda$ has $n$ elements. Now define:

$\displaystyle c_\lambda := \frac{\Delta_{\lambda + \delta}}{\Delta_\delta}.$

Being a quotient of two alternating polynomials, $c_\lambda(x_1, \ldots, x_n)$ is symmetric; note that it is homogeneous of degree $|\lambda|.$

Example

Suppose $\lambda = (2, 1)$. We have:

\begin{aligned} n=2 &\implies c_\lambda(x_1, x_2) = x_1^2 x_2 + x_1 x_2^2,\\ n=3 &\implies c_\lambda(x_1, x_2, x_3) = (x_1^2 x_2 + x_1 x_2^2 + x_2^2 x_3 + x_2 x_3^2 + x_3^2 x_1 + x_3 x_1^2) + 2x_1 x_2 x_3.\end{aligned}

It turns out $c_\lambda$ is precisely the Schur polynomial $s_\lambda \in \Lambda_n$; this will be proven through the course of the article.

Pieri’s Formula

Theorem. Take a partition $\mu$ with $l(\mu) \le n$ and let $r \ge 0$; we have:

$\displaystyle c_\mu e_r = \sum_\lambda c_\lambda,$

where $\lambda$ is taken over all partitions with $l(\lambda) \le n$ obtained by adding $r$ squares to $\mu$ such that no two of them lie in the same row.

Proof

We need to show

$\displaystyle\Delta_{\mu + \delta} e_r = \sum_\lambda \Delta_{\lambda + \delta}, \qquad \delta := (n-1, n-2, \ldots, 0).$

Note that both sides are homogeneous of degree $|\mu| + \frac {n(n-1)} 2 + r$; hence let us compare their coefficients of $x^\alpha$ where $x^\alpha$ is short for $\prod_i x_i^{\alpha_i}.$ Since both sides are alternating polynomials, we may assume $\alpha_1 > \alpha_2 > \dots.$ Now expand the LHS:

$\displaystyle \Delta_{\mu+\delta} e_r = \sum_{i_1 < \ldots < i_r} \Delta_{\mu + \delta} x_{i_1} \ldots x_{i_r}.$

In $\Delta_{\mu + \delta}$, each term is of the form $\pm x^{ \sigma (\mu+\delta)}$ so the exponents of $x_1, \ldots, x_n$ are all distinct. Now, multiplying by $x_{i_1} \ldots x_{i_r}$ increases each exponent by at most 1. Thus to obtain $x^\alpha$ with strictly decreasing exponents, we must begin with strictly decreasing exponents in the first place (i.e. $x^{\mu + \delta}$), then pick $x_{i_1} \ldots x_{i_r}$ such that the exponents remain strictly decreasing.

Thus each $(i_1, \ldots, i_r)$ corresponds to a binary n-vector $\gamma$ of weight r, such that $\mu + \gamma + \delta$ is strictly decreasing. And the latter holds if and only if $\mu + \gamma$ is a partition.

E.g. suppose $\mu = (5, 3, 3, 1, 0), n=5, r=3;$ we get:

$\mu + \delta = (9, 6, 5, 2, 1) \implies \gamma =\begin{cases}(1, 1, 1, 0, 0),\ (1, 1, 0, 1, 0), \ (1, 1, 0, 0, 1), \\(1, 0, 0, 1, 1),\ (0, 1, 1, 1, 0),\ (0, 1, 0, 1, 1).\end{cases}$

In diagrams, this corresponds to:

i.e. partitions obtained by adding r boxes to $\mu$ such that no two lie in a row. ♦

Main Result

Theorem$c_\lambda$ is the Schur polynomial $s_\lambda$ in $\Lambda_n.$

Proof

We wish to show $\mathbf c = \mathbf K\mathbf m.$ Successively applying Pieri’s formula gives:

$e_\lambda = 1\cdot e_{\lambda_1} \ldots e_{\lambda_l} = \sum_{\mu_l} \ldots \sum_{\mu_1} c_{\mu_l}$

where $\mu_0 = \emptyset$ and the Young diagram for $\mu_{i+1}$ is obtained from $\mu_i$ by attaching $\lambda_i$ boxes such that no two lie in the same row; the sum is over the set of all such $(\mu_0, \ldots, \mu_l).$ Label the additional boxes in $\mu_i \to \mu_{i+1}$ by $i+1$ and take the transpose; we obtain an SSYT of shape $\overline{\mu_l}$ and type $\lambda$.

For example, suppose $\lambda = (5, 4, 3, 1)$; here is one way we can successively add squares:

which gives us:

Hence for each $\mu$, the number of occurrences of $c_\mu$ in the above nested sum is $K_{\overline \mu\lambda}.$ This gives:

$\displaystyle e_\lambda = \sum_\mu K_{\overline \mu \lambda} c_\mu \implies \mathbf e = \mathbf K^t\mathbf J \cdot \mathbf c.$

From $\mathbf e = \mathbf K^t \mathbf J \mathbf K \mathbf m$, we get $\mathbf c = \mathbf K\mathbf m= \mathbf s$ as desired. ♦

Corollary

Pieri’s Formulae. Take a partition $\mu$ and let $r \ge 0$; we have, in the formal ring $\Lambda$:

$\displaystyle s_\mu e_r = \sum_\lambda s_\lambda,\quad s_\mu h_r = \sum_{\lambda'} s_{\lambda'},$

where $\lambda$ (resp. $\lambda'$) is taken over all partitions obtained by adding $r$ squares to $\mu$, such that no two of them lie in the same row (resp. column).

Proof

By the earlier Pieri’s formula, the first formula holds in all $\Lambda_n$. Hence it holds in $\Lambda$ too. Applying $\omega$ to it and using $\omega(s_\lambda) = s_{\overline\lambda},$ we get the second formula. ♦

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