Then $s_i\left(s_{i}p\right)=p$ and $$s_i\mathrm{char}\left(B\right)=\sum _{p\in B}{X}^{s_i\mathrm{wt}\left(p\right)}=\sum _{p\in B}{X}^{\mathrm{wt}\left(s_{i}p\right)}=\mathrm{char}\left(B\right).$$ Hence .

Let $$\epsilon =\sum _{w\in W}\det \left(w\right)w\text{so that}{a}_{\mu }=\epsilon \left(X^{\mu }\right),\text{for}\mu \in P.$$

Since $\mathrm{char}\left(B\right)\in {\mathbb{Z}\left[P\right]}^W$, $$\mathrm{char}\left(B\right)a_{\rho }=\mathrm{char}\left(B\right)\epsilon \left(X^{\rho }\right)=\epsilon \left(\mathrm{char}\right(B\left)X^{\rho }\right)$$ and $$\mathrm{char}\left(B\right)=\frac{1}{a_{\rho }}\mathrm{char}\left(B\right)a_{\rho }=\frac{\epsilon \left(\mathrm{char}\right(B\left)X^{\rho }\right)}{a_{\rho }}$$ $$=\sum _{p\in B}\frac{\epsilon \left(X^{\mathrm{wt}\left(p\right)+\rho }\right)}{a_{\rho }}=\sum _{p\in B}\frac{a_{\mathrm{wt}\left(p\right)+\rho }}{a_{\rho }}=\sum _{p\in B}{s}_{\mathrm{wt}\left(p\right)}.$$

There is some cancellation which can occur in this sum. Assume $p\in B$ such that $p⊈C-\rho$ and let $t$ be the first time that $p$ leaves the cone $C-\rho$. In other words, let $t\in {ℝ}_{>0}$ be minimal such that there exists an $i$ with $$p\left(t\right)\in H_{{\alpha }_i,-1}\text{where}{H}_{{\alpha }_i,-1}=\{\lambda \in {𝔥}_{ℝ}^{*}|⟨\lambda ,{\alpha }_i^{\vee }⟩=-1\}.$$

Let $i$ be the minimal index such that the point $p\left(t\right)\in H_{{\alpha }_i,-1}$ and define $s_i\circ p$ to be the element of the $i$-string of $p$ such that $$\mathrm{wt}(s_i\circ p)=s_i\circ p$$

Note that since $⟨p\left(t\right),{\alpha }_i^{\vee }⟩=-1,p$ is not the head of its $i$-string and $s_i\circ p$ is well defined. If $q=s_i\circ p$ then the first time $t$ that $q$ leaves the cone $C-\rho$ is the same as the first time that $p$ leaves the cone $C-\rho$ and $p\left(t\right)=q\left(t\right)$. Thus $s_i\circ q=p$ and $s_i\circ (s_i\circ p)=p$. Since $$s_{\mathrm{wt}(s_i\circ p)}=s_{s_i\circ \mathrm{wt}\left(p\right)}=-s_{\mathrm{wt}\left(p\right)},$$ the terms $s_{\mathrm{wt}(s_i\circ p)}$ and $s_{\mathrm{wt}\left(p\right)}$ cancel in the sum in (5.29). Thus $$\mathrm{char}\left(B\right)=\sum _{\genfrac{}{}{0pt}{}{p\in B}{p\subseteq C-\rho }}{s}_{\mathrm{wt}\left(p\right)}.\square$$

(a) The path $p_{\lambda }^{+}$ is the unique ighest weight path in $B\left(\lambda \right)$. Thus, by Theorem 5.5, $\mathrm{char}\left(B\right(\lambda \left)\right)=s_{\lambda }.$

(b) By the "Leibnitz formula" for the root operators in Theorem 5.8c, the set $$B\left(\mu \right)\otimes B\left(\nu \right)=\{p\otimes q|p\in B\left(\mu \right),q\in B\left(\mu \right)\}$$ is a crystal. Since $\mathrm{wt}(p\otimes q)=\mathrm{wt}\left(p\right)+\mathrm{wt}\left(q\right),$ $$s_{\mu }{s}_{\nu }=\mathrm{char}\left(B\right(\mu \left)\right)\mathrm{char}\left(B\right(\nu \left)\right)=\mathrm{char}\left(B\right(\mu )\otimes B(\nu \left)\right)=\sum _{\genfrac{}{}{0pt}{}{p\otimes q\in B\left(\mu \right)\otimes B\left(\nu \right)}{p\otimes q\subseteq C-\rho }}{s}_{\mathrm{wt}\left(p\right)+\mathrm{wt}\left(q\right)}=\sum _{\genfrac{}{}{0pt}{}{q\in B\left(\nu \right)}{p_{\mu }^{+}\otimes q\subseteq C-\rho }}{s}_{\mu +\mathrm{wt}\left(q\right)},$$ where the third equality is from Theorem 5.5 and the last equality is because the path $p_{\mu }^{+}$ has $\mathrm{wt}\left(p_{\mu }^{+}\right)=\mu$ and is the only highest weight path in $B\left(\mu \right).$

(c) A $J$-crystal is a set of paths $B$ which is closed under the operators ${\stackrel{\sim }{e}}_i,$, for $j\in J$. Since $s_{\lambda }=\mathrm{char}\left(B\right(\lambda \left)\right)$ the statement applies by applying Theorem 5.5 to $B\left(\lambda \right)$ viewed as a $J$-crystal.$\square$