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.


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<j. It remains to prove the first statement.

For that, note that \Delta and \prod_{i<j} (x_i - x_j) 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|.


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.


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

Theoremc_\lambda is the Schur polynomial s_\lambda in \Lambda_n.


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. ♦


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).


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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