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I've come across an annoying lemma trying to finish up an argument, and I was hoping one of you guys knew about it.

Question: Given

  1. a weight $\lambda$ of a simple Lie algebra $\mathfrak g$, and
  2. integers $n_\alpha$ for each simple root $\alpha$,

Is there a highest weight $\nu$, such that in the crystal of with highest weight $\nu$ there is an element $x$ of weight $\lambda$ such that $\tilde{F}_\alpha^{n_\alpha}x\neq 0$?

This is true in $\mathfrak{sl}_2$, which makes me hopeful about other Lie algebras, but the argument isn't coming together for me.

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1 Answer 1

OK, I think I see the answer: by whatever character formula you like, you can see that for any fixed weight spaces (say, the $\lambda$ and $\lambda-n_\alpha\alpha$ for all $\alpha$), you can choose $\nu$ so that the multiplicities at all these points are very large compared to the differences between the those multiplicities. Since the number of elements in the crystal killed by $\tilde{F}_{\alpha}^{n_\alpha}$ is just the difference in the weight multiplicities (or 0), the result follows by pigeonhole.

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