5
$\begingroup$

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.

$\endgroup$
2
2
$\begingroup$

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}

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.