6
$\begingroup$

I'm trying to follow Example 9 in Section 1.5 of the 2nd edition of Macdonald's book "Symmetric Functions and Hall Polynomials". I have trouble with understanding some points.

Before stating my question, I will first explain the example briefly. If $R$ is a root system, then the Weyl denominator formula for $R$ is $$\sum_{w\in W}\varepsilon(w)e^{w\rho}=\prod_{\alpha\in R^{+}}(e^{\alpha/2}-e^{-\alpha/2}),$$ where $W$ is the Weyl group of $R$, $R^{+}$ is the set of positive roots (relative to some base), $\rho$ is the half-sum of the positive roots and $\varepsilon(w)$ is the sign of $w\in W$. The goal of the example is to express the products $$\prod_{i<j} (1-x_{i}x_{j}),\quad\prod_{i}(1-x_{i})\prod_{i<j}(1-x_{i}x_{j}),\quad\prod_{i}(1-x_{i}^{2})\prod_{i<j}(1-x_{i}x_{j})$$ as linear combinations of Schur functions using the Weyl denominator formula for root systems of types $D_{n}, B_{n}, C_{n}$, respectively. More precisely, \begin{align*} \prod_{i<j}(1-x_{i}x_{j})&=\sum_{\pi}(-1)^{|\pi|/2}s_{\pi}(x_{1},\ldots,x_{n})\\ \prod_{i}(1-x_{i}^{2})\prod_{i<j}(1-x_{i}x_{j})&=\sum_{\mu}(-1)^{|\mu|/2}s_{\mu}(x_{1},\ldots,x_{n})\\ \prod_{i}(1-x_{i})\prod_{i<j}(1-x_{i}x_{j})&=\sum_{\nu}(-1)^{(|\nu|+p(\nu))/2}s_{\nu}(x_{1},\ldots,x_{n}), \end{align*} where the first sum ranges over all partitions $\pi=(\alpha_{1}-1,\ldots,\alpha_{p}-1\mid\alpha_{1},\ldots,\alpha_{p})$ with $\alpha_{1}\leq n-1$, the second sum ranges over all partitions $\mu=(\alpha_{1}+1,\ldots,\alpha_{p}+1\mid\alpha_{1},\ldots,\alpha_{p})$ with $\alpha_{1}\leq n-1$, the third sum ranges over all self-conjugate partitions $\nu=(\alpha_{1},\ldots,\alpha_{p}\mid\alpha_{1},\ldots,\alpha_{p})$ with $\alpha_{1}\leq n-1$ and $p(\nu)=p$.

Now I'm ready to state my question. Here I will just consider the first product. Let $\epsilon_{1},\ldots,\epsilon_{n}$ be the standard basis for $\mathbb{R}^{n}$. The root system $R=D_{n}$ is given by $$\{\pm(\epsilon_{i}\pm\epsilon_{j}):i\neq j\},$$ and the subset $$\Delta=\{\epsilon_{1}-\epsilon_{2},\epsilon_{2}-\epsilon_{3},\ldots,\epsilon_{n-1}-\epsilon_{n},\epsilon_{n-1}+\epsilon_{n}\}$$ is a base. Then \begin{align*} R^{+}&=\{\epsilon_{i}\pm\epsilon_{j}:1\leq i<j\leq n\},\\ \rho&=(n-1)\epsilon_{1}+(n-2)\epsilon_{2}+\cdots+\epsilon_{n-1}. \end{align*} On the other hand, the reflection $\sigma_{\epsilon_{i}-\epsilon_{i+1}}$ in the orthogonal complement of $\epsilon_{i}-\epsilon_{i+1}$ switches $\epsilon_{i}$ with $\epsilon_{i+1}$, and $\sigma_{\epsilon_{i}+\epsilon_{i+1}}$ switches $\epsilon_{i}$ to $-\epsilon_{j}$ and $\epsilon_{j}$ to $-\epsilon_{i}$. Thus the Weyl group $W$ can be identified with the group of all elements that permute $n$ elements as well as switching an even number of their signs, hence isomorphic to $S_{n}\ltimes\mathbb{Z}_{2}^{n-1}$. If we replace $e^{-\epsilon_{i}}$ by $x_{i}$, the right-hand side of the denominator formula becomes: \begin{align*} \prod_{\alpha\in R^{+}}(e^{\alpha/2}-e^{-\alpha/2})&=e^{\rho}\prod_{\alpha\in R^{+}}(1-e^{-\alpha})\\ &=x_{1}^{-(n-1)}x_{2}^{-(n-2)}\cdots x_{n-1}^{-1}\prod_{i<j}(1-x_{i}x_{j})\left(1-\frac{x_{i}}{x_{j}}\right)\\ &=(x_{1}\cdots x_{n})^{-(n-1)}\prod_{i<j}(1-x_{i}x_{j})\left(x_{j}-x_{i}\right). \end{align*} The product $\prod_{i<j}(1-x_{i}x_{j})$ is the product we are interested in and the product $\prod_{i<j}(x_{j}-x_{i})$ is involved in the definition of Schur function. So it remains to relate the left-hand side of the denominator formula to the partitions $\pi=(\alpha_{1}-1,\ldots,\alpha_{p}-1\mid\alpha_{1},\ldots,\alpha_{p})$ with $\alpha_{1}\leq n-1$. But I have no idea. If someone has any idea, please let me know. Thank you for reading.

This question was posted in Math Stack Exchange a few days ago by myself (https://math.stackexchange.com/questions/2967011/macdonalds-symmetric-functions-and-hall-polynomials-section-1-5-example-9). I didn't get an answer yet, so I repost the question here.

$\endgroup$
2
  • 1
    $\begingroup$ It's often relevant Ithough much less so here) to notie that there are two editions for Macdonald's book, the first in 1979 and the second in 1995. $\endgroup$ Nov 2, 2018 at 19:03
  • 2
    $\begingroup$ The magic spell is SSYT $\endgroup$ Nov 3, 2018 at 7:26

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.