Homomorphisms Between Verma Modules and Linkage — Examples

Since $\rho =1$ and $s\cdot m=-m-2$, the relevant pairing is $m+1$, which is positive integral exactly when $m\in {\mathbb{Z}}_{\ge 0}$.

The embedding proposition gives the displayed inclusion, including $m=0$; if the pairing is not positive integral, the BGG criterion rules out that reflection map.

Put $\nu =\lambda +\rho$. Regularity makes the dot orbit free, so its members are indexed by the six elements of $S_3$. For $u\in W$ and a positive root $\alpha$, $⟨u\nu ,{\alpha }^{\vee }⟩=⟨\nu ,(u^{-1}\alpha {)}^{\vee }⟩$ is positive exactly when $u^{-1}\alpha$ is positive; integrality makes every such positive value a positive integer.

Write $s=s_{{\alpha }_1}$, $t=s_{{\alpha }_2}$, and $w_0=sts=tst$. The positive-reflection pairs are $(e,s),(e,t),(e,w_0),(s,st),(s,ts),(t,st),(t,ts),(st,w_0),(ts,w_0)$. The rank inequalities defining Bruhat order give precisely these reflection comparisons (all except $(e,w_0)$ are covers). Thus checking the three positive roots of $A_2$ shows that the reflections satisfying step 1.1 are exactly the pairs $u<s_{\alpha }u$ in Bruhat order. In particular every strong-linkage edge lies in Bruhat order, while every Bruhat cover is one of these positive-integral reflection edges. Taking transitive closures and applying the BGG criterion proves the nonzero-homomorphism equivalence; the cited injectivity lemma turns every such map into an embedding. This proves both directions of the stated embedding equivalence.

The zero pairing means the shifted weight is fixed by $s_1$, hence $s_1\cdot \lambda =\lambda$.

Its stabilizer contains $\{1,s_1\}$ and is exactly this subgroup for a generic point on that wall, so orbit-stabilizer gives three distinct translates. The regular six-vertex diagram therefore cannot be used unchanged.

In a regular dominant integral $A_2$ dot orbit take $\mu =s_1\cdot \lambda$. The weights have equal central character because they are dot conjugate.

The BGG criterion gives $M(\mu )\hookrightarrow M(\lambda )$, but not the reverse map: $\lambda \uparrow ̸\mu$, since the only directed reflection lowers the dominant weight. Thus central-character equality does not determine both directions.

In ${\mathfrak{s}\mathfrak{l}}_2$, take $\lambda =-\frac{1}{2}$, so $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=\frac{1}{2}$. PBW gives vectors $f^n{v}_{\lambda }$ only for integers $n\ge 0$; $f^{1\mathrm{/}2}{v}_{\lambda }$ is not a vector of $M(\lambda )$.

Thus the asserted formal power does not even define a candidate singular vector. The integrality condition is necessary before the rank-one singular-vector calculation can begin.