Let \(e_r\) be the \(r\)-th elementary symmetric function in the variables \(x_1, x_2, \ldots, x_n\).
Show that the determinant of \(\left(\frac{\partial e_i}{\partial x_j}\right)_{1 \leq i, j \leq n}\) is a homogeneous polynomial and calculate its degree.
Calculate the determinant.
Solution
Part (a).
1. \(e_i\) is homogeneous of degree \(i\).
\(e_i = \sum_{1 \le j_1 < \cdots < j_i \le n} x_{j_1} \cdots x_{j_i}\), a sum of monomials of degree \(i\).
2. \(\frac{\partial e_i}{\partial x_j}\) is homogeneous of degree \(i - 1\).
differentiating a homogeneous polynomial of degree \(i\) lowers the degree by \(1\).
3. The determinant \(\det\left(\frac{\partial e_i}{\partial x_j}\right)\) is homogeneous of degree \(\sum_{i=1}^n (i-1) = \frac{n(n-1)}{2}\).
the determinant is a sum of products, each product taking one entry from each row \(i\); the entry in row \(i\) has degree \(i-1\), so each term has total degree \(\sum_{i=1}^n (i-1)\).
Part (b).
1. \(\frac{\partial e_i}{\partial x_j} = e_{i-1}(x_1, \ldots, \widehat{x_j}, \ldots, x_n)\).
the terms of \(e_i\) containing \(x_j\) are exactly \(x_j\) times the elementary symmetric functions of degree \(i-1\) in the remaining variables.
2. The matrix \(\left(\frac{\partial e_i}{\partial x_j}\right)\) is the Jacobian of the map \(x \mapsto (e_1, \ldots, e_n)\).
definition.
3. \(\det\left(\frac{\partial e_i}{\partial x_j}\right) = \prod_{1 \le i < j \le n} (x_i - x_j)\) (the Vandermonde determinant).
this is the classical identity for the Jacobian of the elementary symmetric functions; it follows by noting the determinant vanishes whenever \(x_i = x_j\) (two columns equal), so it is divisible by \(\prod_{i<j}(x_i - x_j)\), and both sides have degree \(\frac{n(n-1)}{2}\) by part (a), so they agree up to a constant, which is \(1\) (checked on the leading term).
4. Q.E.D.
3 (part (a)) and 3 (part (b)).