$\tilde{e}_i$ action on $e^{(n)} u$

Question

I am reading Kashiwara's paper GLOBAL CRYSTAL BASES OF QUANTUM GROUPS. On page 462, the action of $\tilde{e}_i$ on an integrable $U_q(g)$-module $M$ is defined as follows. For $u \in \ker e_i \cap M_{\lambda}$ and $0 \leq n \leq \langle h_i, \lambda \rangle$, define \begin{align} & \tilde{e}_i( f^{(n)} u ) = f^{(n-1)} u, \\ & \tilde{f}_i( f^{(n)} u ) = f^{(n+1)} u. \end{align} Can we define the action of $\tilde{e}_i$ on $M$ using by defining $\tilde{e}_i( e^{(n)} v )$, where $v \in \ker f_i \cap M_{\lambda}$? What are the formulas for $\tilde{e}_i( e^{(n)} v )$. Thank you very much.

Answer 1

The actions are defined in the Kashiwara's paper: On crystal bases of the Q-analogue of universal enveloping algebras in (2.2.5).

Let $M$ be an integrable $U_q(\mathfrak{g})$-module. Then \begin{align*} M = \oplus_{0 \leq n \leq -\langle h_i, \mu \rangle} e_i^{(n)} (\ker f_i \cap M_{\mu}), \end{align*} where $M_{\mu} = \{u \in M: h\cdot u = \langle h, \mu \rangle u, \forall \ h \in P^* \}$, $P$ is the weight lattice of $\mathfrak{g}$.

The actions of $\tilde{e}_i$ and $\tilde{f}_i$ on $M$ are given by \begin{align*} & \tilde{e}_i(e_i^{(n)} v) = e_i^{(n+1)}v, \\ & \tilde{f}_i(e_i^{(n)} v) = e_i^{(n-1)}v, \end{align*} $v \in \ker f_i \cap M_{\mu}$, $0 \leq n \leq - \langle h_i, \mu \rangle$.