Question. Can you find a bijective proof of the identity $$ \operatorname{dim}(S^{\lambda} \mathbb{C}^m)\ \operatorname{dim}(S^{\lambda'} \mathbb{C}^n) \ f^{n^m} = \dim \Lambda^p (\mathbb{C}^m \otimes \mathbb{C}^n )\ f^\lambda\ f^{n^m / \lambda},$$ where $\lambda$ is a Young diagram with $p$ boxes, and with at most $m$ rows, and with at most $n$ columns?

Above, $S^\lambda \mathbb{C}^m$ is the Schur functor applied to $\mathbb{C}^m$ or, in other words, the irreducible representation of the general linear group with the highest weight $\lambda$. Thus $\operatorname{dim}(S^{\lambda} \mathbb{C}^m)$ is the number of semistandard Young tableaux with shape $\lambda$ with entries $1,\dots,m$.

Above, $f^\lambda$ is the dimension of the irreducible representation of the symmetric group which corresponds to the Young diagram $\lambda$. Thus $f^\lambda$ is equal to the number of standard Young tableaux with shape $\lambda$.

As usual, $\lambda'$ denotes the Young diagram conjugate to $\lambda$.

We denote by $n^m$ the rectangular Young diagram with $m$ rows and $n$ columns.

The skew diagram $n^m/\lambda$ is a rotation by $180$ degrees of a true Young diagram, so $f^{n^m/\lambda}$ still counts standard Young tableaux.

Note that each of the factors has a natural combinatorial interpretation as the number of (semi)standard Young tableaux of prescribed shape, except for the first factor on the right-hand side $$ \operatorname{dim} \Lambda^p (\mathbb{C}^m \otimes \mathbb{C}^n ) = \binom{m n}{ p} $$ which also has an obvious combinatorial interpretation.

The above is Problem 1.2 from a joint work with Greta Panova, "External powers of tensor products as representations of general linear groups", where two proofs if the identity are provided.

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    $\begingroup$ I think the usual agreement is that questions should not start with an imperative ("Find a bijective proof"). $\endgroup$ – LSpice May 23 '17 at 12:39

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