Morphism of Verma modules

Question

$\DeclareMathOperator\Hom{Hom}$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 $\mathcal{O}$") 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 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, 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.

Answer 1

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 $$ a_2y_{\alpha_1} \delta_{2_1}(v_{(-1,-1,2)}) = a_2 b_2 y_{\alpha_1}(2y_{\alpha_1} y_{\alpha_3} + y_{\alpha_2}) v_{(0, -1, 1)}. $$

Answer 2

For part 2:

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(2)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(2)$ again and additionally $(1)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+2a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}. \end{align*}

and $a_1b_1=-a_2b_2$ is enough. As in [H| mentioned it is possible to choose $a_i,b_i \in \{-1,1\}$.