When reading Brian Hall's Lie Groups, Lie Algebras, and Representations, I found this statement confusing
Since SU(2) is simply connected, Theorem $5.6$ will tell us that the representations of $\operatorname{sl}(2 ; \mathbb{C}) \cong \operatorname{su}(2)_{\mathbb{C}}$ are in one-to-correspondence with the representations of $\mathrm{SU}(2)$. Since $\mathrm{SU}(2)$ is compact, Theorem $4.28$ then tells us that every representation of $\operatorname{sl}(2 ; \mathbb{C})$ is completely reducible. ...
and the two theorems mentioned are
Theorem $4.28 .$ If $G$ is a compact matrix Lie group, every finite dimensional representation of $G$ is completely reducible.
Theorem 5.6. Let $G$ and $H$ be matrix Lie groups with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$, respectively, and let $\phi: \mathfrak{g} \rightarrow \mathfrak{h}$ be a Lie algebra homomorphism. If $G$ is simply connected, there exists a unique Lie group homomorphism $\Phi: G \rightarrow H$ such that $\Phi\left(e^{X}\right)=e^{\phi(X)}$ for all $X \in \mathfrak{g}$.
Neither of the theorems have mentioned complexification. If $\mathfrak{g}_{\mathbb{C}}$ donate the complexification of a lie algebra $\mathfrak{g}$, it is easy to proof that a representation of $\mathfrak{g}_{\mathbb{C}}$ can induce a representation of $\mathfrak{g}$, but I don't know why the inverse is true. Moreover, why can we know a representation of $\mathfrak{g}_{\mathbb{C}}$ is completely reducible if it is true for $\mathfrak{g}$?
