$\DeclareMathOperator\Hom{Hom}$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.

  • $\begingroup$ You referred to "the comments below", but there were (prior to this one) no comments. I assumed you meant @VitTuček's answer, so I edited in a link to it, as well as to Humphreys's book. (Also, a TeX comment: note that $\mathfrak g$=$\mathfrak{gl}_3$ $\mathfrak g$=$\mathfrak{gl}_3$ spaces poorly compared to $\mathfrak g=\mathfrak{gl}_3$ $\mathfrak g=\mathfrak{gl}_3$. I adjusted to the latter.) $\endgroup$
    – LSpice
    Nov 13, 2020 at 20:18

2 Answers 2



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

  • $\begingroup$ Maybe this questions is dumb, but why you are no considering "$x_{\alpha_2}u=0$" too? $\endgroup$
    – KKD
    Jul 28, 2020 at 16:20
  • 2
    $\begingroup$ @CJS Because $x_{\alpha_2} = [x_{\alpha_1}, x_{\alpha_3}].$ In general, it is sufficient to consider only action by elements corresponding to simple roots. Alas, this straightforward strategy of finding $u$ gets unfeasible pretty quickly because dimension of graded-homogeneous components of $U(\mathfrak{n}^-)$ grows very fast. $\endgroup$ Jul 28, 2020 at 17:29
  • $\begingroup$ Thanks for all these answers. At least I understood what's going on. $\endgroup$
    – KKD
    Jul 28, 2020 at 17:47
  • 1
    $\begingroup$ So after doing the computation it turns out, that $a=2$ and $b=1$ is a solution (and any multiple of (2,1)) $\endgroup$
    – KKD
    Jul 28, 2020 at 18:11
  • 3
    $\begingroup$ @CJS This is generally discouraged on this site as it can lead to Questions and Answers which are hard to parse for people coming late to the party. I've amended my answe but since it's midnight for me, I will leave the rest of the calculations to you. If you do them, please share the result with us. Also, I vaguely remember that the choice of constants in these "diamonds" is dealt with in the original BGG article. $\endgroup$ Jul 30, 2020 at 22:03

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 $a_i,b_i \in \{-1,1\}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.