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 PBW vectors $f^r{v}_m$ ($r\ge 0$) form a basis of $M(m)$. The relations in the Given imply $h{f}^r{v}_m=(m-2r)f^r{v}_m$ and, by induction from $e{f}^{r+1}{v}_m=fe{f}^r{v}_m+h{f}^r{v}_m$, imply $e{f}^r{v}_m=r(m-r+1)f^{r-1}{v}_m$ for $r\ge 1$.

For $m\in {\mathbb{Z}}_{\ge 0}$, step 1.1 makes the nonzero PBW vector $f^{m+1}{v}_m$ singular of weight $m-2(m+1)=-m-2$. The Verma universal property gives a nonzero map $M(-m-2)\to M(m)$, and the cited injectivity lemma makes it the displayed embedding, including $m=0$. For any $m\in \mathbb{C}$, the PBW weights of $M(m)$ are $m-2r$ for integers $r\ge 0$. A nonzero map $M(s\cdot m)\to M(m)$ must send its highest vector to a nonzero vector of weight $s\cdot m=-m-2$, so $r=m+1\in {\mathbb{Z}}_{\ge 0}$. If $r>0$, then $m\in {\mathbb{Z}}_{\ge 0}$; if $r=0$, then $m=-1$ and $s\cdot m=m$, so the map is a scalar identity on the cyclic module rather than a nontrivial reflection embedding. No other $m$ permits the claimed direction.

Put $p=⟨\lambda +\rho ,{\alpha }_1^{\vee }⟩$ and $q=⟨\lambda +\rho ,{\alpha }_2^{\vee }⟩$. Both are positive integers. Write $s=s_{{\alpha }_1}$, $t=s_{{\alpha }_2}$, and $w_0=sts=tst$. In the simple-root basis, the displacements $\lambda -w\cdot \lambda$ for $w=e,s,t,st,ts,w_0$ are respectively $(0,0),(p,0),(0,q),(p+q,q),(p,p+q),(p+q,p+q)\mathrm{.}$ These are six distinct points. Their coordinatewise order is exactly Bruhat comparability: $e$ is below all, $s,t$ are incomparable, $st,ts$ are incomparable, and each length-one vertex lies below each length-two vertex, with $w_0$ above all.

The simple-root singular-vector theorem gives embeddings for six of the eight Bruhat covers: $e<s$, $e<t$, $s<ts$, $t<st$, $st<w_0$, and $ts<w_0$. Their positive integral exponents are respectively $p,q,p+q,p+q,p,q$. The two remaining covers are $s<st$ and $t<ts$, each a reflection in ${\alpha }_1+{\alpha }_2$.

To construct the $s<st$ embedding directly, work in $M(s\cdot \lambda )$ with highest vector $v$. Choose normalized $A_2$ root vectors $e_i,f_i,h_i$ from the supplied finite root structure, and put $F=[f_2,f_1]$. The $A_2$ root spaces give $[f_1,F]=[f_2,F]=0$; Jacobi and $[e_i,f_j]={\delta }_{ij}{h}_i$ give $[e_1,F]=-f_2$ and $[e_2,F]=f_1$. The shifted Dynkin coordinates of $s\cdot \lambda$ are $(-p,p+q)$, so $h_{1}v=(-p-1)v$ and $h_{2}v=(p+q-1)v$. Set $c_k=(-1{)}^k\left(\genfrac{}{}{0px}{}{q}{k}\right)(p{)}_k$, where $(p{)}_0=1$ and $(p{)}_k=p(p+1)\cdots (p+k-1)$, and define $X_{p,q}={\sum }_{k=0}^q{c}_k{f}_1^{q-k}{F}^k{f}_2^{q-k}v\mathrm{.}$ The displayed PBW monomials are distinct and $c_0=1$, so $X_{p,q}\mathrm{\ne }0$. It has weight $s\cdot \lambda -q({\alpha }_1+{\alpha }_2)=st\cdot \lambda$. Using the four commutators above, the coefficient of $f_1^{q-k-1}{F}^k{f}_2^{q-k}v$ in $e_1{X}_{p,q}$ is $-(q-k)(p+k)c_k-(k+1)c_{k+1}$; the coefficient of $f_1^{q-k}{F}^k{f}_2^{q-k-1}v$ in $e_2{X}_{p,q}$ is its negative, for $0\le k<q$. Both vanish because $c_{k+1}=-(q-k)(p+k)c_k\mathrm{/}(k+1)$. Thus $X_{p,q}$ is singular, and the Verma universal property gives a nonzero map $M(st\cdot \lambda )\to M(s\cdot \lambda )$, which is injective by the cited Verma-map lemma. Interchanging the simple-root indices and $p,q$ constructs the $t<ts$ embedding.

Composing the eight cover embeddings from steps 2.1 and 2.2 gives an embedding whenever $v\le w$ in Bruhat order. Conversely, any nonzero map $M(w\cdot \lambda )\to M(v\cdot \lambda )$ sends its highest vector to a nonzero vector of weight $w\cdot \lambda$ in the target. The PBW weight model therefore requires $v\cdot \lambda -w\cdot \lambda \in Q^{+}$, which by the six displacements in step 1.1 is equivalent to $v\le w$. This proves the exact embedding poset.

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.

Put $m=⟨\lambda +\rho ,{\alpha }_1^{\vee }⟩\in {\mathbb{Z}}_{>0}$ and $\mu =s_1\cdot \lambda =\lambda -m{\alpha }_1$. The singular-vector lemma The simple-root singular vector in a Verma module gives a nonzero highest vector $f_1^m{v}_{\lambda }$ of weight $\mu$ in $M(\lambda )$. The Verma universal property The universal property of Verma modules therefore gives a nonzero map $M(\mu )\to M(\lambda )$.

Every central element acts by a scalar on each cyclic highest-weight module by The Harish-Chandra projection computes the highest-weight scalar. The nonzero map in step 1.1 intertwines these scalar actions, so ${\chi }_{\mu }={\chi }_{\lambda }$.

The PBW weight description The PBW model of a Verma module puts every weight of $M(\mu )$ in $\mu -Q^{+}$. Since $\lambda =\mu +m{\alpha }_1$ lies strictly above $\mu$, $M(\mu )$ has no vector of weight $\lambda$. A map $M(\lambda )\to M(\mu )$ must therefore kill its highest vector and is zero. Thus equal central characters do not give both embedding 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.