# An identity on symmetric polynomial

In the polynomial algebra $\mathbb{C}[X_1, X_2,\ldots, X_n]$, we define a set of symmetric polynomials as follows $h_i(X_k, X_{k+1}, \ldots, X_n)$ = sum of all monomials of total degree $i$ in the set of variables $X_k, X_{k+1}, \ldots X_n$; $\forall i,k = 1,2, \ldots n$. Let $\sigma_i$ be the elementary symmetric polynomial of degree $i$ in the variables $X_i$'s, i.e., $\sigma_i = \Sigma_{t_1 < t_2< \cdots < t_i} X_{t_1}X_{t_2}\ldots X_{t_i}$. Then it is claimed in the book Algorithms of invariant theory of Bernd Sturmfels, on page 12, that $$h_k(X_k, X_{k + 1}, \ldots, X_n) + \sum_{i = 1}^k (-1)^i h_{k-i}(X_k, X_{k + 1}, \ldots, X_n) \sigma_i = 0.$$ I am not able to see its proof. Please help me.

(Edit:) This follows from the generating function identities

$$E(t) = \sum_{i \ge 0} \sigma_i t^i = (1 + t X_1)(1 + t X_2)\cdots(1 + t X_n)$$

$$H(t) = \sum_{i \ge 0} h_i t^i = \frac{1}{(1 - t X_k)(1 - t X_{k+1})\cdots(1 - t X_n)}.$$

The desired identity follows from comparing the coefficient of $t^k$ of both sides of the identity

$$H(t) E(-t) = (1 - t X_1)\cdots(1 - t X_{k-1}).$$

• In the identity $h_k$ involves only the variables $X_k, X_{k+1}, \ldots X_n$. but $\sigma_i$ involves all the variables $X_1, X_2, \ldots, X_n$. Also the identity is equal to zero. Can you clarify more. – A.G Mar 15 '11 at 19:31
• @Anjan: sorry, I misread the question. But it is not hard to get your identity from the general identity. – Qiaochu Yuan Mar 15 '11 at 19:40
• Thanks Yuan for your help. – A.G Mar 15 '11 at 21:17