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Denote by $\mathfrak{g}$ a complex semisimple Lie algebra and let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$.

Let $\Phi$ be the root system of $(\mathfrak{g},\mathfrak{h})$, write $W$ for the corresponding Weyl group of $\Phi$, and denote by $\mathfrak{g}_\alpha$ the root subspace of $\mathfrak{g}$ corresponding to a root $\alpha$.

We fix a choice of positive roots $\Phi^+$, and let $\Delta$ be the corresponding subset of simple roots in $\Phi^+$. Note that each subset $I\subseteq\Delta$ generates a root system $\Phi_I\subseteq\Phi$, with positive roots $\Phi_I^+:=\Phi_I\cap \Phi^+$.

There are a number of subalgebras of $\mathfrak{g}$ associated with the root system $\Phi_I$. Let $ \mathfrak{l}_I:=\mathfrak{h}\oplus\sum_{\alpha\in\Phi_I}\mathfrak{g}_\alpha$ be the Levi subalgebra and let $ \mathfrak{u}_I:=\sum_{\alpha\in\Phi^+\backslash\Phi_I^+}\mathfrak{g}_\alpha $ be the nilpotent radical.

Then $\mathfrak{p}_I:=\mathfrak{l}_I\oplus \mathfrak{u}_I$ is a standard parabolic subalgebra of $\mathfrak{g}$.

We note that once $I$ is fixed, there is little use for other subsets of $\Delta$. Therefore, we omit the subscript if a subalgebra is obviously associated to $I$.

Definition 1: The category $\mathcal{O}^\mathfrak{p}$ is the full subcategory of $U(\mathfrak{g})$-Mod whose objects $M$ satisfy the following conditions:

  1. $M$ is a finitely generated $U(\mathfrak{g})$-module.

  2. $M$ is a direct sum of finite-dimensional simple $U(\mathfrak{l})$-modules.

  3. $M$ is locally finite as a $U(\mathfrak{p})$-module.

Definition 2: The category $\mathcal{O}^\mathfrak{p}$ is the full subcategory of $U(\mathfrak{g})$-Mod whose objects $M$ satisfy the following conditions:

  1. $M$ is a finitely generated $U(\mathfrak{g})$-module.

  2. $M$ is a direct sum of finite-dimensional simple $U(\mathfrak{l})$-modules.

  3. $M$ is locally finite as a $U(\mathfrak{u})$-module.

My question: How to show Definition 1 is equivalent to Definition 2?

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By the PBW theorem we can write $U(\mathfrak{p}) = U(\mathfrak{l})U(\mathfrak{u}).$ By our assumption $U(\mathfrak{u})m$ is finite dimensional for any $m\in M$ and by the second point of the definition $U(\mathfrak{l})n$ is finite-dimensional for any $n \in U(\mathfrak{u})m.$

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