Why is $h_n(1,x,\ldots,x^{m}) = h_m(1,x,\ldots,x^{n})$?

Question

Let $h_n$ denote the degree $n$ symmetric polynomial corresponding to the homogeneous basis. That is, in, e.g., $m$ variables

$$ h_n(x_1,\ldots,x_m) = \sum_{1\le i_1\le \cdots \le i_n \le m} \prod_{j\le n} x_{i_j} $$

Why is the following identity in $\mathbb{Q}[x]$ true?

$$ h_n(1,x,\ldots,x^m) = h_m(1,x,\ldots,x^n) $$

I am interested in this because its equivalent to Hermite reciprocity of representations of $\mathfrak{sl}_2$: $\text{Sym}^n(\text{Sym}^m(V)) \cong \text{Sym}^m(\text{Sym}^n(V))$ for $V$ the standard representation.

Answer 1

For a partition $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$ with $\lambda_1\geq \lambda_2\geq\cdots\geq\lambda_k$, write $x^\lambda=x_{\lambda_1}x_{\lambda_2}\cdots x_{\lambda_k}$. Then, with this notation, $$h_n(x_1,\ldots,x_m)=\sum_{\lambda\in X(n,m)} x^\lambda$$ where $X(n,m)$ is the set of partitions with $n$ parts of size $\leq m$, and $$h_n(1,x_1,\ldots,x_m)=\sum_{k=0}^{n}h_k(x_1,\ldots,x_m)=\sum_{\lambda\in\bigcup_{k=0}^nX(k,m)}x^\lambda $$ Similarly, $$ h_m(1,x_1,\ldots,x_n)=\sum_{\lambda\in \bigcup_{k=0}^m X(k,n)} x^\lambda. $$ Note that the transpose map $\lambda\to\lambda^t$ defines a bijection $\bigcup_{k=0}^mX(k,n)\to \bigcup_{k=0}^nX(k,m)$.

Now, for $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$, set $|\lambda|=\lambda_1+\cdots+\lambda_k$ and note that $|\lambda|=|\lambda^t|$. Therefore, \begin{align} h_n(1,x,x^2,\ldots,x^m)&=\sum_{\lambda\in \bigcup_{k=0}^nX(k,m)} x^{|\lambda|}\\ &=\sum_{\lambda\in \bigcup_{k=0}^mX({k,n})}x^{|\lambda^t|}\\ &=\sum_{\lambda\in \bigcup_{k=0}^mX({k,n})}x^{|\lambda|}\\ &=h_m(1,x,x^2,\ldots,x^{n}). \end{align}