Relation of complete homogeneous symmetric polynomials and the elementary symmetric polynomials

Question

I was reading about the symmetric polynomials and saw the following relation:

$$\sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},\ldots ,X_{n})h_{{m-i}}(X_{1},\ldots ,X_{n})=0\text{ for } m>0$$

The proof is constructed by using a generating function with respect to the variable $t$, in Symmetric Functions and Hal Polynomials, SECOND EDITION I. G. MACDONALD, page 21.

(Question) I was wondering if this relation still holds for any length $k$, i.e.

$$\sum _{{i=0}}^{m}(-1)^{i}e_{i}(X_{1},\ldots ,X_{n})h_{{m-i}}(X_{1},\ldots ,X_{k})=0\text{ for }k \in \{1,...,n\}$$

P.S. I tried cases and it holds for any case.

Answer 1

We have $$h_p(X_1,\ldots,X_n)=\sum h_r(X_1,\ldots,X_k)h_{p-r}(X_{k+1},\ldots,X_n)$$ $$e_p(X_1,\ldots,X_n)=\sum e_r(X_1,\ldots,X_k)e_{p-r}(X_{k+1},\ldots,X_n)$$ $$\sum_{i=0}^m\sum_{p=0}^i\sum_{q=0}^{m-i} (-1)^ie_{i-p}(X_1,\ldots,X_k)h_{(m-p-q) -(i-p) }(X_1,\ldots,X_k)e_p(X_{k+1},\ldots,X_n)h_q(X_{k+1},\ldots,X_n)$$ If we fix $p$ and $q$, this still sums to $0$ since it is just an instance of the same formula in fewer variables, replacing $i$ with $i-p$ and $m$ with $m-p-q$. Thus if we take $q=0$ and sum over all $p$ we get the formula you want.

Answer 2

The desired identity $$\sum_{i=0}^{m} (-1)^i e_i(X_1,\ldots,X_n) h_{m-i}(X_1,\ldots,X_k) =0$$ is only valid for $m > n-k$.

For example, with $n=2$, $k=1$, and $m=1$, we get $$\sum_{i=0}^{1} (-1)^i e_i(X_1,X_2) h_{1-i}(X_1) = e_0(X_1, X_2) h_1 (X_1) - e_1(X_1,X_2) h_0(X_1) = - X_2 \neq 0.$$

To see that this identity does hold for $m > n-k$, however, just multiply the generating functions $$\sum_{j=0}^{\infty} h_j(X_1,\ldots,X_k) z^j = \prod_{i=1}^{k} \frac{1}{1-X_iz}$$ and $$\sum_{j=0}^{\infty} (-1)^je_j(X_1,\ldots,X_n) z^j = \prod_{i=1}^{n} (1-X_iz)$$ and extract the coefficient of $z^m$: we get $0$ since this produces a polynomial in $z$ of degree $n-k$.