The single-wall tensor-weight exclusion lemma

Let $w,w^{\mathrm{\prime }}\in W$ and let $\gamma$ be a weight of $E$ with $w^{\mathrm{\prime }}\cdot \mu =w\cdot \lambda +\gamma$. Since $w^{\mathrm{\prime }}\cdot \mu -[w\cdot \lambda ]=w^{\mathrm{\prime }}({\mu }_{\bullet })-w({\lambda }_{\bullet })$, setting $x=(w^{\mathrm{\prime }}{)}^{-1}w$ gives $x{\lambda }_{\bullet }={\mu }_{\bullet }-(w^{\mathrm{\prime }}{)}^{-1}\gamma$, that is, $(w^{\mathrm{\prime }}{)}^{-1}\gamma ={\mu }_{\bullet }-x{\lambda }_{\bullet }$.

By [F2] one has $\mid \gamma \mid \le \mid \nu \mid =\mid {\mu }_{\bullet }-{\lambda }_{\bullet }\mid$, and $W$-invariance of the form gives $\mid (w^{\mathrm{\prime }}{)}^{-1}\gamma \mid =\mid \gamma \mid$, so the identity of step 1.1 yields $\mid {\mu }_{\bullet }-x{\lambda }_{\bullet }\mid \le \mid {\mu }_{\bullet }-{\lambda }_{\bullet }\mid$. On the other hand [F3] applied to the dominant vectors $\xi =-{\mu }_{\bullet }$ and $\eta =-{\lambda }_{\bullet }$ (dominant and regular by [F1]) gives $\mid {\mu }_{\bullet }-{\lambda }_{\bullet }\mid =\mid \xi -\eta \mid \le \mid \xi -x\eta \mid =\mid {\mu }_{\bullet }-x{\lambda }_{\bullet }\mid$.

The two inequalities of step 2.1 are equalities, so the equality case of [F3] applies: $x\eta \in {\mathrm{Stab}}_W(\xi )\eta$ with ${\mathrm{Stab}}_W(\xi )=\{1,s\}$ by [F1], that is, $x(-{\lambda }_{\bullet })\in \{-{\lambda }_{\bullet },-s{\lambda }_{\bullet }\}$, so $x{\lambda }_{\bullet }\in \{{\lambda }_{\bullet },s{\lambda }_{\bullet }\}$. Regularity of ${\lambda }_{\bullet }$ then forces $x\in \{1,s\}$: from $x{\lambda }_{\bullet }=s{\lambda }_{\bullet }$ we get $s^{-1}x\in {\mathrm{Stab}}_W({\lambda }_{\bullet })=\{1\}$, and from $x{\lambda }_{\bullet }={\lambda }_{\bullet }$ directly $x=1$. Both $1$ and $s$ fix ${\mu }_{\bullet }$, hence $x\cdot \mu =\mu$ and $w^{\mathrm{\prime }}\cdot \mu =w\cdot (x\cdot \mu )=w\cdot \mu$. Moreover $\gamma =w^{\mathrm{\prime }}\cdot \mu -w\cdot \lambda =w\cdot \mu -w\cdot \lambda =w(\mu -\lambda )$. Finally $\mid \gamma \mid =\mid {\mu }_{\bullet }-{\lambda }_{\bullet }\mid =\mid \nu \mid$ and $\gamma =w(\mu -\lambda )\in W(\mu -\lambda )=W\nu$, so by [F2] the weight $\gamma$ occurs in $E$ with multiplicity one.

For the reformulation, fix $w$ and let $\eta$ be a weight in the dot orbit of $\mu$, so $\eta =w^{\mathrm{\prime }}\cdot \mu$ for some $w^{\mathrm{\prime }}$ and $\eta$ has the central character of $\mu$ by [F5]. If $\dim E_{\eta -w\cdot \lambda }\mathrm{\ne }0$, then $\gamma :=\eta -w\cdot \lambda$ is a weight of $E$ with $w^{\mathrm{\prime }}\cdot \mu =w\cdot \lambda +\gamma$, so step 3.1 gives $\eta =w^{\mathrm{\prime }}\cdot \mu =w\cdot \mu$ and $\gamma =w(\mu -\lambda )$; conversely $\dim E_{w(\mu -\lambda )}=1$. Hence among labels in the dot orbit of $\mu$ only $\mathrm{\Delta }(w\cdot \mu )$ occurs, with multiplicity one, in the Verma flag of $E\otimes \mathrm{\Delta }(w\cdot \lambda )$ supplied by [F4].

Steps 3.1 and 4.1 prove both formulations: the tensor-weight identity forces $w^{\mathrm{\prime }}\cdot \mu =w\cdot \mu$ and $\gamma =w(\mu -\lambda )$ with multiplicity one, and the only standard factor of $E\otimes \mathrm{\Delta }(w\cdot \lambda )$ with central character ${\chi }_{\mu }$ is $\mathrm{\Delta }(w\cdot \mu )$, once.