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Everything you wanted to know about symmetric polynomials, part IV

December 3, 2013 6:28 am / 3 Comments on Everything you wanted to know about symmetric polynomials, part IV

Alternating and Symmetric Polynomials

Consider the following recipe for building symmetric polynomials, using alternating polynomials. Consider the Vandermonde determinant

$\displaystyle{ \Delta(x_1, \ldots, x_n) = \left|\begin{matrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ & \vdots & & \vdots & \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{matrix}\right| = \prod_{i < j} (x_j - x_i). }$

To see the last equality, note that the determinant is zero if $x_i = x_j$ for any $i \ne j$ , so it is divisible by $(x_j-x_i)$ , and all these factors are distinct. By degree-counting (the polynomial is evidently homogeneous of degree $n \choose 2$ ), this is the whole thing, up to a scalar. Finally, to get the scalar, we do some computation (e.g. plugging in convenient values like $x_i = i$ ).

If, instead of the row $(1 \ x_i\ x_i^2\ \cdots\ x_i^{n-1})$ , we used some other sequence of polynomials, like $(f_1(x_i)\ f_2(x_i)\ \cdots \ f_{n-1}(x_i))$ , the result would still be alternating in the $x_i$ ‘s, so it would still be divisible by the product above. However, the degree-counting argument might no longer be relevant. (For example, if we use $(x_i\ x_i^2\ \cdots\ x_i^n)$ , then the result is the original Vandermonde determinant times $x_1 \cdots x_n$ .)

Still, we can divide out the Vandermonde determinant, and (surprise!) the result will be a symmetric polynomial, since the sign-change is “divided out” as well.

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