Translation functor on parabolic Verma module

Question

I want to prove that following proposition by using Theorems/propositions in Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$.

Define $ \Lambda := \{\nu \in \mathfrak{h}^* : \langle\nu,\alpha^\lor\rangle \in \mathbb{Z} \ \text{for all }\alpha \in \Phi^+\}. $

Define $ \Lambda^+ := \{\nu \in \mathfrak{h}^* : \langle\nu,\alpha^\lor\rangle \in \mathbb{Z}^{\ge 0} \ \text{for all }\alpha \in \Phi^+\}. $

Define $ \Lambda^+_I := \{\nu \in \mathfrak{h}^* : \langle\nu,\alpha^\lor\rangle \in \mathbb{Z}^{\ge 0} \ \text{for all }\alpha \in \Phi^+_I\}. $

The $\mathbb{Z}$-span $\Lambda_r$ of $\Phi$ is called the root lattice.

Recall that a weight $\eta\in \mathfrak{h}^*$ is antidominant if $\langle \eta+\rho,\alpha^{\lor}\rangle\not\in\mathbb{Z}^{>0}$ for all $\alpha\in \Phi^+$.

For $w \in W$ and $\eta\in \mathfrak{h}^*$, define a shifted action of $W$ (called the dot action) by $w \cdot \eta := w(\eta + \rho) - \rho$. If $\eta, \nu \in \mathfrak{h}^*$, then we say that $\eta$ and $\nu$ are linked if for some $w \in W$, we have $\nu = w \cdot \eta$.

The weight $\eta \in \mathfrak{h}^*$ is regular if $|W \cdot \eta| = |W|$ or, equivalently, if $\langle \eta + \rho,\alpha^\lor\rangle \neq 0$ for all $\alpha\in\Phi$

By exercise 1.13, $M\in\mathcal{O}_{\chi_\lambda}$ has a direct sum decomposition $M=\bigoplus M_i$ such that all weights of each $M_i$ are contained in a single coset of the root lattice $\Lambda_r$ in $\mathfrak{h}^*$. Therefore, the category $\mathcal{O}_{\chi_\lambda}$ decomposes as a direct sum of full subcategories, which can be indexed by the nonempty intersection of the orbit $W\cdot\lambda$ with the cosets $\mathfrak{h}^*/\Lambda_r$. We use the antidominant weight $\mu$ in the intersection to parameterize the corresponding subcategory of $\mathcal{O}_{\chi_\lambda}$. We denote this subcategory by $\mathcal{O}_{\mu}$.

We say $\eta$ and $\nu$ are compatible if $\nu - \eta \in\Lambda$.

Fix $\eta, \nu \in \mathfrak{h}^*$ such that $\eta,\nu$ are compatible and write $pr_\eta$ and $pr_\nu$ for the natural projections of the category $\mathcal{O}$ onto $\mathcal{O}_{\chi_\eta}$ and $\mathcal{O}_{\chi_\nu}$. Let $\overline{\nu-\eta}$ be the unique $W$-conjugate in $\Lambda^+$ of $\nu-\eta$. If $M \in \mathcal{O}$, then $M \mapsto pr_ \nu \left(L(\overline{\nu-\eta}) \otimes (pr_\eta M)\right)$ followed by inclusion into $\mathcal{O}$ defines an exact functor $\mathcal{O} \to \mathcal{O}$. Its restriction to $\mathcal{O}_{\chi_\eta}$ (without the inclusion) relates the two subcategories $\mathcal{O}_{\chi_\eta}$ and $\mathcal{O}_{\chi_\nu}$. Write $T^\nu_\eta$ for the resulting functor on $\mathcal{O}$ (or on $\mathcal{O}_{\chi_\eta}$). We call $T^\nu_\eta$ a translation functor.

Proposition: Let $\eta,\nu\in \Lambda_I^+$, Suppose $\eta,\nu\in \Lambda_I^+$ are integral, regular and antidominant. Then there is an equivalence of categories between $\mathcal{O}_\nu$ and $\mathcal{O}_\eta$ such that (i) $T^{\eta}_{\nu}(L(x\cdot \nu))\cong L(x\cdot \eta)$ and $T^{\eta}_{\nu}(M(x\cdot \nu))\cong M(x\cdot \eta)$ for $x\in W$. (ii) If $x\cdot\nu$ is in $\Lambda_I^+$ then $T^{\eta}_{\nu}(M_I(x\cdot \nu))\cong M_I(x\cdot \eta)$.

Proof: Since $\eta,\nu\in \Lambda_I^+$ are integral, regular and antidominant, it holds that $\eta,\nu$ are compatible. Since $\eta,\nu$ are integral, we have $\eta^\natural=\eta,\nu^\natural=\nu$ and hence $\eta^\natural,\nu^\natural\in F$ where $\Phi_F^-=\Phi^+$, $\Phi_F^0=\Phi_F^+=\emptyset$. Applying Theorem 7.6, Proposition 7.7 and Theorem 7.8, we deduce that there is an equivalence of categories $T^{\eta}_{\nu}$ between $\mathcal{O}_{\nu}$ and $\mathcal{O}_{\eta}$ satisfying (i).

For a proof of (ii) it suffices to show that $T^{\eta}_{\nu}(M_I(x\cdot \nu))\cong M_I(x\cdot \eta)$ by the properties of the full subcategory. In fact, there is an exact sequence (Theorem 9.4) \begin{equation}\label{longexact1} \bigoplus_{\alpha\in I} M(s_\alpha x \cdot \nu) \xrightarrow{f} M(x \cdot \nu) \xrightarrow{g} M_I (x \cdot \nu) \to 0. \quad\quad(1) \end{equation} Applying $T^{\eta}_{\nu}$ to (1), we get another exact sequence \begin{equation}\label{longexact2} \bigoplus_{\alpha\in I} M(s_\alpha x \cdot \eta) \xrightarrow{T^{\eta}_{\nu}(f)} M(x \cdot \eta) \xrightarrow{T^{\eta}_{\nu}(g)} T^{\eta}_{\nu}(M_I(x \cdot \nu)) \to 0. \quad\quad(2) \end{equation} On the other hand, by replacing $\nu$ in (1) with $\eta$, we can get an exact sequence in $\mathcal{O}_{\eta}$: \begin{equation}\label{longexact3} \bigoplus_{\alpha\in I} M(s_\alpha x \cdot \eta) \xrightarrow{f'} M(x \cdot \eta) \xrightarrow{g'} M_I (x \cdot \eta) \to 0. \quad\quad(3) \end{equation}

Since (2) is exact, it holds that $T^{\eta}_{\nu}(g)$ is surjective and $\ker(T^{\eta}_{\nu}(g))=\mathrm{Im}(T^{\eta}_{\nu}(f))$, we have $ T^{\eta}_{\nu}(M_I(x \cdot \nu))= \mathrm{Im}(T^{\eta}_{\nu}(g)) \cong M(x\cdot\eta)/\ker(T^{\eta}_{\nu}(g)) =M(x\cdot\eta)/\mathrm{Im}(T^{\eta}_{\nu}(f)).$

Since (3) is exact, it holds that $g'$ is surjective and $\ker(g')=\mathrm{Im}(f')$, we have $ M_I(x \cdot \eta)=\mathrm{Im}(g') \cong M(x\cdot\eta)/\ker(g') =M(x\cdot\eta)/\mathrm{Im}(f'). $

My question: How to show $\mathrm{Im}(T^{\eta}_{\nu}(f))=\mathrm{Im}(f')$?

Answer 1

From the proof of Theorem 9.4, we have $f$ and $f'$ are monomorphisms.

Since $T^{\eta}_{\nu}$ is an exact functor, it preserves monomorphism and hence $T^{\eta}_{\nu}(f)$ is an monomorphism.

Let $\pi_{M(s_\beta x\cdot\eta)}:\oplus_{\alpha\in I} M(s_\alpha x\cdot\eta)\to M(s_\beta x\cdot\eta)$ be the natural projection and let $f'|_{M(s_\beta x\cdot\eta)}:=f'\circ \pi_{M(s_\beta x\cdot\eta)}$. Note that \begin{equation*} f'=\sum_{\alpha\in I}f'|_{M(s_\alpha x\cdot\eta)}. \quad\quad (i) \end{equation*} Since $f'$ is a monomorphism, $f'|_{M(s_\alpha x\cdot\eta)}: M(s_\alpha x\cdot\eta) \to M(x\cdot\eta)$ is also a monomorphism.

By definition of monomorphism, it is clear that any monomorphism is not an zero morphism.

Since \begin{equation*} \dim \mathrm{Hom}_{\mathcal{O}}(M(s_\alpha x\cdot\eta), M(x\cdot\eta))\le 1 \end{equation*} and \begin{equation*} 0\neq f'|_{M(s_\alpha x\cdot\eta)}\in \mathrm{Hom}_{\mathcal{O}}(M(s_\alpha x\cdot\eta), M(x\cdot\eta)) \end{equation*} for any $\alpha\in I$, we have $\dim \mathrm{Hom}_{\mathcal{O}}(M(s_\alpha x\cdot\eta), M(x\cdot\eta))=1$ for any $\alpha\in I$.

Similarly, we have \begin{equation*} T^{\eta}_{\nu}(f)=\sum_{\alpha\in I} T^{\eta}_{\nu}(f)|_{M(s_\alpha x\cdot\eta)}. \quad\quad (ii) \end{equation*} Since $T^{\eta}_{\nu}(f)|_{M(s_\alpha x\cdot\eta)}\in \mathrm{Hom}_{\mathcal{O}}(M(s_\alpha x\cdot\eta), M(x\cdot\eta))$, we have $T^{\eta}_{\nu}(f)|_{M(s_\alpha x\cdot\eta)}=c_\alpha f'|_{M(s_\alpha x\cdot\eta)}$ for some $c_\alpha\neq 0$. This implies that \begin{equation*} T^{\eta}_{\nu}(f)=\sum_{\alpha\in I} c_\alpha f'|_{M(s_\alpha x\cdot\eta)}. \quad\quad (iii) \end{equation*} From (i), we have $\mathrm{Im}(f')=\bigoplus_{\alpha\in I}\mathrm{Im}(f'|_{M(s_\alpha x\cdot\eta)})$.

From (iii), we have $\mathrm{Im}(T^{\eta}_{\nu}(f))=\bigoplus_{\alpha\in I}\mathrm{Im}(c_\alpha f'|_{M(s_\alpha x\cdot\eta)})$.

Since $f'|_{M(s_\alpha x\cdot\eta)}$ is a linear map, we have $\mathrm{Im}(c_\alpha f'|_{M(s_\alpha x\cdot\eta)})=\mathrm{Im}(f'|_{M(s_\alpha x\cdot\eta)})$ for any $\alpha\in I$. Therefore, $\mathrm{Im}(T^{\eta}_{\nu}(f))=\mathrm{Im}(f')$.