Computing kernel in the category $\mathcal{O}$

Question

Let $\mathfrak{g}=\mathfrak{gl}_3$ 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*} Consider the morphism \begin{align} M((-2,1,1)) \oplus M((-1,-1,2) &\xrightarrow\phi M((0,-1,1))\\ (v_{(-2,1,1)},v_{(-1,-1,2)}) &\mapsto(y_{\alpha_1}^2-(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)} \end{align} with respect to $\mathfrak{g}$ inside the BGG category $\mathcal{O}$. Notice that $\phi$ is the sum of the embeddings
\begin{align} M((-2,1,1)) &\xrightarrow{g} M((0,-1,1))\,,\,v_{(-2,1,1)} \mapsto y_{\alpha_1}^2 v_{(0,-1,1)}, \\ M((-1,-1,2)&\xrightarrow{h} M((0,-1,1))\,,\,v_{(-1,-1,2)} \mapsto -(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)}. \end{align}

Here $M(\lambda)$ is the Verma module with the highest weight $\lambda$, $v_\lambda \in M(\lambda)$ is the corresponding highest weight vector and $y_{\alpha_i}$ is the standard generator in the weight space $\mathfrak{g}_{-\alpha_i}$.

Then I'm struggling to compute $\ker(\phi)$, as we have relations inside $U(\mathfrak{g})$. What is a way to tackle this problem?

Answer 1

Here is a less direct, but shorter, proof using some non-trivial machinery. Denote by $s$, $t$ the simple reflections, and $M_w := M(w \cdot 0)$ where $w \in W$, and $\cdot$ is the "shifted" action $w \cdot \lambda := w(\lambda+\rho)-\rho$, where $\rho:=(1,0,-1)$, and analogously the simple modules $L_w$.

By the construction in [1, 4.2] we have a chain complex (called generalized BGG complex): $$0 \to M_{sts} \to M_{st} \oplus M_{ts} \to M_{t} \to L_t \to 0, $$ where the non-trivial maps are up to scalar your $(e,f)$ and $\phi$. Since $L_t$ is a Kostant module (as is any simple module in $\mathcal{O}$ in rank $2$; you can check this by calculating the Kazhdan–Lusztig polynomials and using the last sentence in [1, 3.4]), by [1, Theorem 4.3.] the above complex is exact. Your claim follows.

[1]. Brian D. Boe & Markus Hunziker (2009) Kostant Modules in Blocks of Category $\mathcal O_S$, Communications in Algebra, 37:1, 323-356.

Answer 2

So far I only found the following messy proof. In case I did a mistake or you know a more elegant way it would be a pleasure to know. It seems to me that can be generalised to any "BGG square". Therefore I'm wondering that I couldn't find anything in the standard literature.

Claim: $\ker(\phi) \cong M((-2,0,2))$. Consider at first the embeddings \begin{align} M((-2,0,2)) \stackrel{e}\rightarrow M((-2,1,1))\,,\,&v_{(-2,0,2)} \mapsto y_{\alpha_3}v_{(-2,1,1)}\\ M((-2,0,2)) \stackrel{f}\rightarrow M((-1,-1,2))\,,\,&v_{(-2,0,2)} \mapsto y_{\alpha_1}v_{(-1,-1,2)},\\ \end{align} which are choosen in a way such that $ge+hf=0$. Notice that \begin{equation*} (e,f):M((-2,0,2)) \rightarrow M((-2,1,1)) \oplus M((-1,-1,2)) \end{equation*} is injective by definition of $e$ and $f$ and the latter equation implies immediately that $(e,f)(M((-2,0,2)) \subset \ker(\phi)$.

Let $(v,w) \in \ker(\phi) \subset M((−2,1,1)) \oplus M((−1,−1,2)$. As all objects are weight spaces we can assume that $v \in M((−2,1,1))_\mu$ and $w \in M((−1,−1,2)_{\mu}$ for occurring weight $\mu$.

Observe \begin{align*} (-2,0,2)-\lambda_1\alpha_1-\lambda_2\alpha_2-\lambda_1\alpha_2&=(0,-1,1)-(2+\lambda_1+\lambda_2)\alpha_1-(1+\lambda_2+\lambda_3)\alpha_3, \\ (-1,-1,2)-\lambda_1\alpha_1-\lambda_2\alpha_2-\lambda_1\alpha_2&=(0,-1,1)-(1+\lambda_1+\lambda_2)\alpha_1-(1+\lambda_2+\lambda_3)\alpha_3, \\ (-2,1,1)-\lambda_1\alpha_1-\lambda_2\alpha_2-\lambda_1\alpha_2&=(0,-1,1)-(2+\lambda_1+\lambda_2)\alpha_1-(\lambda_2+\lambda_3)\alpha_3, \\ (0,-1,1)-\lambda_1\alpha_1-\lambda_2\alpha_2-\lambda_1\alpha_2&=(0,-1,1)-(\lambda_1+\lambda_2)\alpha_1-(\lambda_2+\lambda_3)\alpha_3. \\ \end{align*} Every possible occurring weight is of the form $\mu =(0,-1,1)-n\alpha_1-m\alpha_3$ with $n,m\geq0$ and we see that \begin{align*} \dim M((-2,0,2))_\mu&=\max\{0,\min\{n-1,m\}\}, \\ \dim M((-1,-1,2))_\mu&=\min\{n,m\}, \\ \dim M((-2,1,1))_\mu&=\max\{0,\min\{n-1,m+1\}\}, \\ \dim M((0,-1,1))_\mu&=\min\{n+1,m+1\}. \end{align*}

Thus for $(v,w) \in \ker(\phi)$ with $v \in M((−2,1,1))_\mu$ and $w \in M((−1,−1,2)_{\mu}$ we have:

Case 1: $n-1 \geq m+1$. Then

\begin{align*} \dim M((-2,0,2))_\mu&=m, \\ \dim M((-1,-1,2))_\mu&=m, \\ \dim M((-2,1,1))_\mu&=m+1, \\ \dim M((0,-1,1))_\mu&=m+1 \end{align*}

and

\begin{align} g(v)=-h(w) &\in g(M((-2,1,1))_\mu)\cap h(M((-1,-1,2))_\mu) \\ &=M((0,-1,1))_\mu \cap h(M((-1,-1,2))_\mu) \\ &=h(M((-1,-1,2))_\mu)=hf(M((-2,0,2))_\mu). \end{align}

Hence we find $z\in M((-2,0,2))_\mu$ such that $ge(z)=-hf(z)=-h(w)=g(v)$. By injectivity of $g$ and $h$, we see that $(v,w)=(e(z),f(z)) \in (e,f)(M((-2,0,2))$.

Case 2: $n-1 \leq m$

Case 2.1: $n-1=m$. Then

\begin{align*} \dim M((-2,0,2))_\mu&=n-1, \\ \dim M((-1,-1,2))_\mu&=n-1, \\ \dim M((-2,1,1))_\mu&=n-1, \\ \dim M((0,-1,1))_\mu&=n. \end{align*}

and

\begin{align} g(v)=-h(w) &\in g(M((-2,1,1))_\mu)\cap h(M((-1,-1,2))_\mu) \\ &=ge(M((-2,0,2))_\mu) \cap hf(M((-2,0,2))_\mu)\\ &=hf(M((-2,0,2))_\mu)\\ \end{align}

and we can proceed as in case 1.

Case 2.2: $n-1<m$. Then

\begin{align*} \dim M((-2,0,2))_\mu&=n-1, \\ \dim M((-1,-1,2))_\mu&=n, \\ \dim M((-2,1,1))_\mu&=n-1, \\ \dim M((0,-1,1))_\mu&=n. \end{align*}

Thus it is analog to case 1.

Hence we see that $\ker(\phi)=(e,f)(M((-2,0,2))\cong M((-2,0,2))$.