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Lie $g$ be a finite dimensional complex simple Lie algebra and $U_q(g)$ the corresponding quantum group, where $q$ is not a root of unity. Every simple finite dimensional $g$-module is of the form $V(\lambda)$, where $\lambda$ is a dominant weight. Every simple finite dimensional $U_q(g)$-module is also of the form $V(\lambda)$, where $\lambda$ is a dominant weight.

My questions are:

(1) Are there some relation between the crystal basis for a $g$-module $V(\lambda)$ and a $U_q(g)$-module $V(\lambda)$?

(2) Are the crystal graphs for a $g$-module $V(\lambda)$ and a $U_q(g)$-module $V(\lambda)$ the same?

Thank you very much.

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(1) Yes, the crystal basis for $V(\lambda)$ is the specialization of the $q$-crystal basis at $q=1$. In fact, that's the only way to construct it.

(2) Again, the crystal graphs coincide by definition.

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  • $\begingroup$ Hello Knop, do you know how to give an explicit matrix realization of $V(\lambda)$ from the crystal basis? $\endgroup$
    – Reflecting_Ordinal
    Commented Dec 31, 2023 at 17:52

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