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I am reading through "Crystal Base and a Generalization of the Littlewood-Richardson Rule for the Classical Lie Algebras" by Nakashima, and there is something I am not understanding correctly.

Line (2.3.5) gives a description of $B(\omega_M+\omega_N)$. If $M=N=1$, then it seems like $w$ can never be in an $(a,b)$-configuration, since there cannot exist two distinct integers between $1$ and $M$. Should we then interpret condition (M.N.2) as being vacuous?

If so, then there is something else I am getting wrong. If (M.N.2) is vacuous when $M=N=1$, then $B(\omega_1+\omega_1)$ has fourteen elements. But then this means the corresponding representation has dimension 14. However, the dimension formula for the representation parameterized by $(\lambda_1\geq\ldots\lambda_n\geq 0)$ is

$\prod_{1\leq i\le j\leq n}\frac{l_i^2-l_j^2}{m_i^2-m_j^2}\prod_{1\leq i\leq n} \frac{l_i}{m_i}, $

where $l_i=\lambda_i+n-i+1/2$ and $m_i=n-i+1/2$. (I got this from Fulton and Harris). Therefore the representation corresponding to $(1,1)$ should have dimension 10, not 14.

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    $\begingroup$ I compute that the irrep of highest weight 2\omega_1 has dimension 14, not 10 $\endgroup$
    – Peter McNamara
    Jun 14, 2012 at 9:25
  • $\begingroup$ I was indeed being very, very, silly. It's the representation corresponding to (2,0), not (1,1), and this has dimension 14. $\endgroup$
    – Jeffrey Kuan
    Nov 27, 2012 at 22:11

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