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.



Crystal basis for quantum groups and Lie algebras
(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.


