Restricted duality is exact and involutive on O

Given: The setting above and the hypotheses in the statement.

Fix a finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$, and a positive Borel $\mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$. Write $Q^{+}={\sum }_i{\mathbb{Z}}_{\ge 0}{\alpha }_i$, $\mu \le \lambda$ when $\lambda -\mu \in Q^{+}$, and $w\cdot \lambda =w(\lambda +\rho )-\rho$. For an $\mathfrak{h}$-semisimple module $M$ with finite-dimensional weight spaces, its restricted Chevalley dual is $D(M)={⨁}_{\mu \in {\mathfrak{h}}^{\ast }}{M}_{\mu }^{\ast },(x\phi )(m)=\phi (\tau (x)m)(x\in U(\mathfrak{g}))\mathrm{.}$ Here each functional is extended by zero on the other weight spaces and $\tau$ is the fixed anti-involution of def-chevalley-contravariant-form. In particular $\tau (h)=h$ and $D(M{)}_{\mu }=M_{\mu }^{\ast }$. A map $f:M\to N$ induces $D(f):D(N)\to D(M)$ by precomposition. The action law follows from $\tau (xy)=\tau (y)\tau (x)$; a root vector of weight $\alpha$ sends $M_{\mu }^{\ast }$ to $M_{\mu +\alpha }^{\ast }$, so the restricted sum is stable. This is a complex-linear algebraic dual, with no conjugation. Ordinary Lie-module duality has a minus sign and reverses weights; twisting that dual by the Lie automorphism $x\mapsto -\tau (x)$ gives the convention used here. (Restricted Chevalley dual)

Fix a finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$, and a positive Borel $\mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$. Write $Q^{+}={\sum }_i{\mathbb{Z}}_{\ge 0}{\alpha }_i$, $\mu \le \lambda$ when $\lambda -\mu \in Q^{+}$, and $w\cdot \lambda =w(\lambda +\rho )-\rho$. For every highest weight $\lambda$, $D(L(\lambda ))\cong L(\lambda )$ as $\mathfrak{g}$-modules. (Restricted self-duality of simple highest-weight modules)

Fix a finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$, and a positive Borel $\mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$. Write $Q^{+}={\sum }_i{\mathbb{Z}}_{\ge 0}{\alpha }_i$, $\mu \le \lambda$ when $\lambda -\mu \in Q^{+}$, and $w\cdot \lambda =w(\lambda +\rho )-\rho$. Every object of $\mathcal{O}$ has a finite composition series and is both Noetherian and Artinian. The length of zero is zero. (Every object of O has finite length)

Fix a finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$, a Cartan subalgebra $\mathfrak{h}$, and a positive Borel $\mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$. Write $Q^{+}={\sum }_i{\mathbb{Z}}_{\ge 0}{\alpha }_i$, $\mu \le \lambda$ when $\lambda -\mu \in Q^{+}$, and $w\cdot \lambda =w(\lambda +\rho )-\rho$. The category $\mathcal{O}$ is closed under submodules, quotients and finite direct sums and is an abelian category. If $0\to A\to E\to B\to 0$ is exact, $A,B\in \mathcal{O}$, and $E$ is $\mathfrak{h}$-semisimple, then $E\in \mathcal{O}$. The middle-term weight hypothesis is essential. (Category O is abelian and extension closed among weight modules)

If an object $A$ in an abelian category has two composition series, then the two series have the same length and the same composition factors up to permutation and isomorphism. (Jordan-Holder theorem in an abelian category)

The simple objects of $\mathcal{O}$ are exactly the modules $L(\lambda )$, $\lambda \in {\mathfrak{h}}^{\ast }$, and $L(\lambda )\cong L(\mu )$ if and only if $\lambda =\mu$. (The simple objects of O)