# Translation functor on parabolic Verma module

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) $$$$\label{longexact1} \bigoplus_{\alpha\in I} M(s_\alpha x \cdot \nu) \xrightarrow{f} M(x \cdot \nu) \xrightarrow{g} M_I (x \cdot \nu) \to 0. \quad\quad(1)$$$$ Applying $$T^{\eta}_{\nu}$$ to (1), we get another exact sequence $$$$\label{longexact2} \bigoplus_{\alpha\in I} M(s_\alpha x \cdot \eta) \xrightarrow{T^{\eta}_{\nu}(f)} M(x \cdot \eta) \xrightarrow{T^{\eta}_{\nu}(g)} T^{\eta}_{\nu}(M_I(x \cdot \nu)) \to 0. \quad\quad(2)$$$$ On the other hand, by replacing $$\nu$$ in (1) with $$\eta$$, we can get an exact sequence in $$\mathcal{O}_{\eta}$$: $$$$\label{longexact3} \bigoplus_{\alpha\in I} M(s_\alpha x \cdot \eta) \xrightarrow{f'} M(x \cdot \eta) \xrightarrow{g'} M_I (x \cdot \eta) \to 0. \quad\quad(3)$$$$

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 $$\mathrm{Im}(T^{\eta}_{\nu}(f))=\mathrm{Im}(f')$$?

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, $$\mathrm{Im}(T^{\eta}_{\nu}(f))=\mathrm{Im}(f')$$.