6
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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 Apr 22 '16 at 21:08
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Apr 22 '16 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.