2 added 4 characters in body

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 $\tilde{F}\alpha^{n\alpha}$ \tilde{F}_{\alpha}^{n_\alpha}$is just the difference in the weight multiplicities (or 0), the result follows by pigeonhole. 1 OK, I think I see the answer: by whatever character formula you like, you can see that for any fixed weight spaces (say, the$\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.