Let's consider the Littlewood-Richardson coefficients $c^{\lambda}\_{\mu \nu}$ so that
\begin{equation}
V\_\mu \otimes V\_\nu = \bigoplus\_\lambda V\_\lambda^{\oplus c^{\lambda}\_{\mu \nu}}
\end{equation}
where $V\_\mu$ are representations of $GL\_n$.


Usually the basis elements of the (infinite dimensional) vector space of irreducible representations of $GL\_n$ are labelled by partitions. I have two questions:


1. Is there a basis (with basis elements labelled by $i, j, k, \cdots$) that "diagonalizes" the Littlewood-Richardson coefficients? In other words, $c^{i}\_{jk} = 0$ unless $i=j=k$?
2. If so, is there an elegant way of relating such a basis to the basis labelled by partitions?



"Diagonalizing" Littlewood-Richardson coefficients
If a diagonal basis existed, tensoring with a fixed representation would kill all but finitely many basis elements. This is not the case because e.g. tensoring with the $1$-dimensional trivial representation doesn't kill anything.


