Let $\lambda = \sum_{i = 1}^{n - 1} m_i \omega_i$ be the highest weight of irreducible representation $V(\lambda)$ of Lie algebra $\mathfrak{sl}_n$. As we know from the Weyl formula, $$\dim V(\lambda) = \prod\limits_{i \leq j} \frac{m_i + ... + m_j + j - i + 1}{j - i + 1}.$$ Consider the multivariate generating function in variables $z_1, ..., z_{n - 1}$ $$g_n(z) = \sum\limits_{\lambda} \dim V(\lambda)z^{\lambda}.$$ For example, $$g_2(z_1) = \frac{1}{1 - z_1},\hspace{1cm} g_3(z_1, z_2) = \frac{1 - z_1z_2}{(1 - z_1)^2(1 - z_2)^2},$$ $g_4(z_1, z_2, z_3)$ is less compact, but i can provide it.

The denomenator is always given by $\prod_{i=1}^{n - 1} (1 - z_i)^{i(n - i - 1) + 1}$. I computed the number of terms in the numerator (weighted by coefficients), it is given by the sequence $1, 2, 36, 9836, ...$ which does not appear on OEIS.

Question: Is there a combinatorial interpretation for numerator of this generating function? Even if someone presents combinatorial objects counted by the sequence $1, 2, 36, 9836, ...$, then it would already be interesting.

Motivation: If we restrict to representations of the form $V(m_i\omega_i)$ (corresponding to rectangular diagrams), or equivalenty, if we set all variables except for $z_i$ to zero, then the numerator is given by d-dimensional Narayana polynomial $\mathcal{N}_{i, n - i}(z_i)$ which has many good interpretations.

  • 3
    $\begingroup$ See math.mit.edu/~rstan/pubs/pubfiles/44.pdf for a combinatorial formula for the "character generator" as a sum of rational functions. $\endgroup$ Aug 7, 2021 at 14:39
  • $\begingroup$ @RichardStanley It is indeed very close to what i want. Getting a sum with a common denominator would be the best. Also, in this work you mention that no analogue is known for $\mathfrak{sp}$ and $\mathfrak{so}$. Has that changed? $\endgroup$ Aug 8, 2021 at 13:24
  • $\begingroup$ To my knowledge that has not changed. $\endgroup$ Aug 9, 2021 at 22:31
  • $\begingroup$ @RichardStanley Okay, thank you! $\endgroup$ Aug 14, 2021 at 11:29


Your Answer

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