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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?

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2 Answers 2

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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.

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    $\begingroup$ Said another way, the existence of the basis would imply the representation ring decomposes as an infinite direct sum of rings, which cannot be unital. $\endgroup$
    – Julian Rosen
    Commented Apr 22, 2016 at 21:08
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The vector space spanned by the irreps of $G=GL_n$ can be identified, by means of the character, with the vector space of $G$-invariant algebraic functions on $G$, for the adjoint action.

If you disregard the distinction between various kinds of functions (algebraic functions, smooth functions, distributions,...), then the Dirac delta functions at the various points of $G/G_{ad}$ can be thought of as a basis of this vector space.

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  • $\begingroup$ Even more explicitly, the ring of $GL_n$-invariant algebraic functions on $GL_n$ is the free ring on the coefficients of the characteristic polynomial, plus the inverse on the determinant, and is so the ring of algebraic functions on $\mathbb A^n - \mathbb A^{n-1}$. $\endgroup$
    – Will Sawin
    Commented Apr 22, 2016 at 21:50

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