Suppose that $\lambda,\mu$ are integer partitions, with conjugates $\lambda^*,\mu^*$. Could you help me to prove the following formula, please?
$\sum_{i,j}\mathrm{min}(\lambda_i,\mu_j)=\sum_k\lambda^*_k\mu^*_k$
Suppose that $\lambda,\mu$ are integer partitions, with conjugates $\lambda^*,\mu^*$. Could you help me to prove the following formula, please?
$\sum_{i,j}\mathrm{min}(\lambda_i,\mu_j)=\sum_k\lambda^*_k\mu^*_k$
The identity describes two ways of counting the cubic ‘cells’ in a plane partition:
Let $n_{i,j} = \min(\lambda_i,\mu_j)$ be a plane partition, then the $k$th ‘storey’ is rectangular with dimensions $\lambda^\star_k\times\mu^\star_k$:
The cells in the $k$th storey are the $(i,j)$ for which both $\lambda_i\geqslant k$ and $\mu_j\geqslant k$. But $\lambda^\star_k$ is the number of $i$ for which $\lambda_i\geqslant k$, and $\mu^\star_k$ is the number of $j$ for which $\mu_j\geqslant k$.
This is really the same answer as David Bevan's, but formulated a bit differently. Both sides of the equation count triples $(i,j,k)$ where $k\leq\lambda_i$ and $k\leq\mu_j$. If you fix $i$ and $j$ you have $\min(\lambda_i,\mu_j)$ different choices for $k$, while if you fix $k$ you have $\lambda^*_k$ choices for $i$ and independently $\mu^*_k$ choices for $j$.
I think induction on the integer that $\mu$ partitions works. Subtract 1 from the smallest part of $\mu$, and show that both sides decrease by the same amount?