About Hom and weight space of nilpotent Lie algebra cohomology

Question

Let $\mathfrak{g}$ be a complex semisimple Lie algebra. Denote by $\Phi$ the root system of $(\mathfrak{g},\mathfrak{h})$ and denote by $\mathfrak{g}_\alpha$ the root subspace of $\mathfrak{g}$ corresponding to a root $\alpha$.

We fix a choice of the set 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.

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 $I$.

Denote $N_\mu$ the $\mu$-weight space of $\mathfrak{l}$-module $N$. Let $F(\lambda)$ is the finite dimensional simple $\mathfrak{l}$-module with highest weight $\lambda$.

1. Does $\text{Hom}_\mathfrak{l}(F(\lambda), H^i(\mathfrak{u},V))\cong H^i(\mathfrak{u},V)_{\lambda}$?

Answer 1

This is true for general modules, you don't need anything specific about nilpotent cohomology. Every homomorphism from a $F(\lambda)$ is uniquely determined by it's highest weight vector.