A formula on Kronecker coefficients

Question

Accidentally, I proved the following formula for the Kronecker coefficients using some obscure method. $$g\bigl(\ell^{mn}, m^{\ell n},n^{\ell m}\bigr)=1,\ \forall l,m,n\in\mathbb{N},$$ where $n^m$ is the rectangle partition. By definition, the Kronecker coefficients $g_{\mu,\nu}^\lambda$ is the structure coefficient in the tensor product $$S_\mu\otimes S_\nu = \sum_{\lambda} g_{\mu,\nu}^\lambda S_\lambda,$$ where $S_\mu$ is the irreducible representation of the symmetric group defined by the partition $\mu$. Is this a known result or maybe just some exercise?

Answer 1

I have not seen this particular statement appearing in the literature, but here is an easy way to see why it is true:

The Kronecker coefficients $g(\lambda,\mu,\nu)$ (note that such notation makes sense since these coefficients are actually invariant under any permutation of $\lambda,\mu,\nu$) can be interpreted as the coefficient of $s_{\lambda}(x)s_{\mu}(y)$ in $s_{\nu}(xy)$: $$(*) \qquad s_{\nu}(x.y) = \sum_{\lambda,\mu} g(\lambda,\mu,\nu)s_{\lambda}(x)s_{\mu}(y).$$ Here $s$ are the Schur functions, $x=(x_1,x_2,\ldots)$ and $y=(y_1,y_2,\ldots)$ are variables and $x.y = (x_1y_1,x_1y_2,\ldots,x_2y_1,x_2y_2,\ldots)$.

In this particular case, let $\nu = m^{n\ell}$, $\lambda = (\ell m)^n$ and $\mu = (mn)^\ell$, and $x=(x_1,\ldots,x_n)$, $y=(y_1,\ldots,y_\ell)$. It is easy to see that for any $a,b$, $$s_{a^b}(z_1,\ldots,z_b) =(z_1\cdots z_b)^a.$$ (there is only one SSYT of shape $a^b$ filled with at most $b$ distinct numbers)

Then $x.y$ consists of $n\ell$ variables, so $$s_{m^{n\ell}}(x.y) = (\prod_{i=1}^n \prod_{j=1}^\ell (x_i y_j) )^m = \prod_i x_i^{m\ell} \cdot \prod_j y_j^{mn} =s_{ (m\ell)^n} (x) s_{(mn)^\ell}(y)$$ and comparing with (*) we get the Kronecker coefficient in question.