Let $\lambda$ be a partition with at most $p$ parts, let $\mu$ be a partition with at most $q$ parts, and let $\nu$ be a partition with at most $p+q$ parts. Let $m\geq \nu\_1$ be an integer. We denote by $\lambda^\ast$, $\mu^\ast$, $\nu^\ast$ the conjugate partitions, and pad them with $0$'s so the conjugates have $m$ parts. Consider the Littlewood-Richardson coefficients
$$s\_\lambda s\_\mu=\sum c\_{\lambda\mu}^\nu s\_\nu$$
Then I won't bet my life on it, but I think I found the following symmetry:


$$c\_{(p-\lambda^\ast\_{m+1-i})\_{i=1}^m,(q-\mu^\ast\_{m+1-i})\_{i=1}^m}^{(p+q-\nu^\ast\_{m+1-i})\_{i=1}^m}=c\_{\lambda\mu}^\nu$$


Provided this is true, is it already known? I have a proof and I've checked it in several cases with coefficients with unusual values, and it checks out (no sort of comprehensive test yet), but errors are always possible, and it seems like something simple enough that if it's true it would be known.



Is this simple symmetry of Littlewood-Richardson coefficients known?
I think you can understand this if you think of them as characters of $GL\_m$-representations, i.e., $s\_\lambda$ is the character of the irreducible representation which I'll denote by $V\_\lambda$. The convention is that the highest weight is $\lambda$.


First, conjugating (I assume this just means the usual transpose operation of swapping rows and columns of the Young diagram) is a bit of a red herring here since $c\_{\lambda,\mu}^\nu = c\_{\lambda^\*,\mu^\*}^{\nu^\*}$.


Second, the character of $V\_\lambda^\vee$ (dual representation) has highest weight $(-\lambda\_m, -\lambda\_{m-1},\dots, -\lambda\_1)$, which is your reversing operation. To make that into a partition, we can add $p$ to all components (for large enough $p$) and this corresponds to taking the representation $(\det V)^{\otimes p} \otimes V\_{\lambda^\vee}$.


So putting it all together, we start with


$$V\_{\lambda^\*} \otimes V\_{\mu^\*} \cong \bigoplus\_\nu V\_{\nu^\*}^{\oplus c^\nu\_{\lambda, \mu}}$$


Apply duals:


$$V\_{\lambda^\*}^\vee \otimes V\_{\mu^\*}^\vee \cong \bigoplus\_\nu (V\_{\nu^\*}^\vee)^{\oplus c^\nu\_{\lambda, \mu}}$$


Now tensor with $(\det V)^{\otimes (p+q)}$:


$$(V^{\otimes p} \otimes V\_{\lambda^\*}^\vee) \otimes (V^{\otimes q} \otimes V\_{\mu^\*}^\vee) \cong \bigoplus\_\nu (V^{\otimes (p+q)} \otimes V\_{\nu^\*}^\vee)^{\oplus c^\nu\_{\lambda, \mu}}$$


and take characters to get the identity you mentioned.


