$\DeclareMathOperator\Hom{Hom}$I'm trying to understand morphism of Verma modules and consider the following example.


**PART 1:**
Consider $\mathfrak{g}=\mathfrak{gl}\_3$ over $\mathbb{C}$ with positive roots
\begin{equation\*}\Phi\_+=\{\alpha\_1=(1,-1,0),\alpha\_2=(1,0,-1),\alpha\_3=(0,1,-1)\},\end{equation\*}
which defines a cartan decomposition $\mathfrak{g}=\mathfrak{n}^- \oplus \mathfrak{h}\oplus \mathfrak{n}$. Then to the positive roots corresponding reflections are $s\_{\alpha\_1}=(1,2)$, $s\_{\alpha\_2}=(1,3)$ and $s\_{\alpha\_3}=(2,3)\in S\_3$. Denoting by $\rho=\frac{1}{2} (\alpha\_1 +\alpha\_2 + \alpha\_3)=(1,0,-1)$ half the sum of all positive roots, we have for weights $\lambda=(0,-1,1)$ and $\mu=(-1,-1,2)$, that
\begin{equation\*}
\mu=s\_{\alpha\_2}((1,-1,0))-(1,0,-1)=s\_{\alpha\_2}(\lambda+\rho)-\rho=s\_{\alpha\_2}\cdot \lambda=\lambda -\alpha\_2<\lambda. 
\end{equation\*}


Hence by a Theorem of Verma (Theorem 4.6 in [H]: Humphrey's "[Representation of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$](https://doi.org/10.1090/gsm/094)") there exist a morphism of Verma modules $\phi:M(\mu) \rightarrow M(\lambda)$, with respective maximal vectors $v\_\mu$ and $v\_\lambda$. The morphism $\phi$ is known to be injective (Theorem 4.2 in [H]) and we have $\phi(v\_\mu)=u\cdot v\_\lambda$ for a unique $u \in U(\mathfrak{n}^-)$, which also determines $\phi$. Furthermore $\dim(\Hom(M(\mu),M(\lambda))=1$, hence up to some scalar there is only one choice for $u$, which I'm trying to find.


My thoughts so far:
The Verma modules $M(\lambda)$ and $M(\mu)$ each have a unique simple submodule $L(\mu')$, which should be isomorphic/the same and is also a Verma module (Proposition 4.1 and Theorem 4.2 in [H]). By Theorem 4.8 in [H] $\mu'$ has to be antidominant. Hence $\mu'=(-2,0,2)$. According to the proof for $\dim(\Hom(M(\mu),M(\lambda))=1$ in [H], it is enough to understand how the simple module $L(\mu')$ is mapped to itself under $\phi$. As $\mu -\mu'=\alpha\_1$ we have $\dim M(\mu)\_{\mu'}=1$, hence the maximal vector of $L(\mu') \subset M(\mu)$ is $y\_{\alpha\_1}v\_\mu$ with respect to $v\_\mu$ and fixed choosen root vectors $y\_{\alpha\_i}$ of $\mathfrak{g}\_{-\alpha\_i} \subset \mathfrak{g}$. But then I struggle as we have for the equation $\lambda-\mu'=t\_1\alpha\_1+t\_2\alpha\_2+t\_3\alpha\_3$ with $t\_i \geq 0$ two solutions, namely $(t\_1,t\_2,t\_3) \in \{(2,0,1),(1,1,0)\}$. Hence $\dim M(\lambda)\_{\mu'}=2$ and I don't know if $\phi(y\_{\alpha\_1}v\_\mu)=c\cdot y\_{a\_1}^2y\_{a\_3}v\_\lambda$ or $\phi(y\_{\alpha\_1}v\_\mu)=c\cdot y\_{a\_1}y\_{a\_2}v\_\lambda$ ($c$ some scalar). Or is this completely wrong?


**SOLUTION PART 1**: By the comments [below](https://mathoverflow.net/a/366800) it follows that $\phi(v\_\mu)=c(y\_{\alpha\_1}y\_{\alpha\_3}+2y\_{\alpha\_2})v\_\lambda$ for some $c \in \mathbb{C}$.


**ADDENDUM PART 2**: Consider then part of the (strong) BGG resolution (using the notation as in [H]) of the simple module $L((0,0,0))$
\begin{equation\*} 
C:M((-2,0,2)) \xrightarrow{\delta\_3} M((-2,1,1)) \oplus M((-1,-1,2) \xrightarrow{\delta\_{2\_1}} M((0,-1,1)) 
\end{equation\*}
with $\delta\_{2\_1}:M((-2,1,1)) \oplus M((-1,-1,2) \xrightarrow{\delta\_{2}} M((0,-1,1)) \oplus M((-1,1,0)) \xrightarrow{\pi\_1} M((0,-1,1))$.


Hence $\delta\_{2\_1}\circ \delta\_3=0$. With the same arguments as in the [comments](https://mathoverflow.net/a/366800), we have
\begin{align\*}
 \delta\_3(v\_{(-2,0,2)})&{}=(a\_1y\_{\alpha\_3}v\_{(-2,1,1)},a\_2y\_{\alpha\_1}v\_{(-1,-1,-2)}), \\ 
\delta\_{2\_1}(v\_{(-2,1,1)},v\_{(-1,-1,2)})&{}=(b\_1y\_{\alpha\_1}^2+b\_2(y\_{\alpha\_1}y\_{\alpha\_3}+2y\_{\alpha\_2}))v\_{(0,-1,1)} \end{align\*}
for some non-trivial scalars $a\_i$, $b\_i$.


So we would get
\begin{align\*} 0&{}=\delta\_{2\_1}\circ \delta\_3(v\_{(-2,0,2)})=\delta\_{2\_1}(a\_1y\_{\alpha\_3}v\_{(-2,1,1)},a\_2y\_{\alpha\_1}v\_{(-1,-1,-2)})\\&{}=(a\_1b\_1y\_{\alpha\_1}^2y\_{\alpha\_3}+a\_2b\_2(y\_{\alpha\_1}^2y\_{\alpha\_3}+y\_{\alpha\_1}y\_{\alpha\_2}))v\_{(0,-1,1)}\\
&{}=((a\_1b\_1+a\_2b\_2)y\_{\alpha\_1}^2y\_{\alpha\_3}+a\_2b\_2y\_{\alpha\_1}y\_{\alpha\_2}))v\_{(0,-1,1)}.
\end{align\*}
But why is the last term equal to zero for nontrivial $a\_i$, $b\_i$? I thought that $y\_{\alpha\_1}^2y\_{\alpha\_3}$ and $y\_{\alpha\_1}y\_{\alpha\_2}$ are linearly independent.



Morphism of Verma modules
**PART 1:**


The element $u$ must have weight $-\alpha\_2$, since $\mu = \lambda - \alpha\_2.$


In $U(\mathfrak{n^-})$ there are only two linearly independent elements that have such weight (assuming PBW basis with respect to fixed order of generators based on positive roots): $y\_{\alpha\_2}$ and $y\_{\alpha\_1}y\_{\alpha\_3}.$ Hence the sought element $u$ is a linear combination of such
$$
u = a y\_{\alpha\_2} + b y\_{\alpha\_1}y\_{\alpha\_3}.
$$


Since this has to be the image of the higest weight vector of $M(\mu)$ we must have $x\_{\alpha\_1} u = 0$ and $x\_{\alpha\_3} u = 0.$ Writing it out and using relations defining the Verma module and the Lie algebra, we end up with system of 2 linear equations for 2 unknowns. E.g. we have
$$
x\_{\alpha\_1} (ay\_{\alpha\_2}v\_\lambda) = (a[x\_{\alpha\_1}, y\_{\alpha\_2}] + ay\_{\alpha\_2} x\_{\alpha\_1})v\_\lambda
$$
where the first term on the right hand side is either zero, or some element of Cartan subalgebra acting on $v\_\lambda$, and the second term is zero from the definition of the Verma module.


**PART 2:**
I think you made a mistake in your calculations. For any $U(\mathfrak{g})$-homomorphism $\varphi$ we have $\varphi(u v) = u \varphi(v)$. Hence the composition going through $M(-2, 1, 1)$ is equal to $$
a\_1y\_{\alpha\_3} \delta\_{2\_1}(v\_{(-2, 1, 1)}) = a\_1b\_1 y\_{\alpha\_3} y\_{\alpha\_1}^2 v\_{(0, -1, 1)}.
$$


The elements $y\_{\alpha\_1}$ and $y\_{\alpha\_3}$ do not commute, in fact $[y\_{\alpha\_1}, y\_{\alpha\_3}]$ should be a multiple of $y\_{\alpha\_2}.$


Similarly, the composition going through $M(-1,-1,2)$ equals
$$
a\_2y\_{\alpha\_1} \delta\_{2\_1}(v\_{(-1,-1,2)}) = a\_2 b\_2 y\_{\alpha\_1}(2y\_{\alpha\_1} y\_{\alpha\_3} + y\_{\alpha\_2}) v\_{(0, -1, 1)}.
$$


