Let $\mathfrak{g}$ be a simple, finite-dimensional Lie algebra. We define its loop algebra $L\mathfrak{g} = \mathfrak{g} \otimes \mathbb{C}\lbrack t,t^{-1} \rbrack$ with Lie bracket $\lbrack x \otimes t^n, y \otimes t^m \rbrack = \lbrack x,y \rbrack \otimes t^{n+m}$, where the bracket to the right is the one on $\mathfrak{g}$. I would like to find a representation of $L\mathfrak{g}$ that is reducible but not decomposable. Surprisingly, I could not find any answers to this in the literature, except for a small mention here [1], though this proved far too advanced for me.
[1] https://pdfs.semanticscholar.org/8c44/102fa07290bb35516cefcdc1b7d5fada7b7d.pdf
From my old homework solutions:
Let $\mathfrak{g}$ be any simple finite-dimensional Lie algebra. Let $W$ be a nontrivial finite-dimensional irreducible representation of $\mathfrak{g}$. [Such a representation $W$ exists, because otherwise all finite-dimensional irreducible representations of $\mathfrak{g}$ would be trivial, so that the adjoint representation would have a filtration with all subquotients being trivial, and thus $\mathfrak{g}$ would be nilpotent, which contradicts to $\mathfrak{g}$ being simple.]
Consider the polynomial ring $\mathbb{C}\left[ \varepsilon\right]$ (where $\varepsilon$ is an indeterminate) and its quotient ring $\mathbb{C}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right)$ (a $2$-dimensional $\mathbb{C}$-vector space, with algebra structure similar to the dual numbers but defined over $\mathbb{C}$). Given any $u \in \mathbb{C}\left[\varepsilon\right]$, we let $\overline{u}$ be the canonical projection of $u$ onto $\mathbb{C}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right)$.
We notice that we can consider the Lie algebra $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) :=\mathfrak{g} \otimes\mathbb{C}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $ as a quotient Lie algebra of $L\mathfrak{g}$ by means of the Lie algebra homomorphism $L\mathfrak{g} =\mathfrak{g}\otimes\mathbb{C}\left[ t,t^{-1}\right] \rightarrow \mathfrak{g}\otimes\mathbb{C}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $ obtained by tensoring $\operatorname*{id} :\mathfrak{g}\rightarrow\mathfrak{g}$ with the $\mathbb{C}$-algebra homomorphism \begin{align*} \mathbb{C}\left[ t,t^{-1}\right] & \rightarrow\mathbb{C}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) ,\\ t & \mapsto\overline{1+\varepsilon}. \end{align*} Thus, in order to construct a reducible but not decomposable finite-dimensional representation of $L\mathfrak{g}$, it is enough to construct a reducible but not decomposable finite-dimensional representation of $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $.
We let $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon ^{2}\right) $ act on $W\oplus W$ as follows: \begin{align} \left( a+b\overline{\varepsilon}\right) \rightharpoonup\left( v,w\right) =\left( a\rightharpoonup v,\ b\rightharpoonup v+a\rightharpoonup w\right) . \end{align} (Here, $g \rightharpoonup w$ means the result of a Lie algebra element $g$ acting on a module element $w$.) It is clear that this representation of $\mathfrak{g}\left[ \varepsilon \right] /\left( \varepsilon^{2}\right) $ is reducible (it has $0\oplus W$ as a subrepresentation). Now, we are going to show that it is indecomposable.
In fact, it is easy to see that every nonzero $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $-subrepresentation of $W\oplus W$ contains a nonzero element of $0\oplus W$.
[Proof. Let $V$ be a nonzero $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $-subrepresentation of $W\oplus W$. Then, there exists a nonzero $\left( v,w\right) \in V$. We want to prove that $V$ contains a nonzero element of $0\oplus W$.
If $v=0$, then we are done (because then, $\left( v,w\right) $ is a nonzero element of $0\oplus W$). So assume $v\neq0$.
There exists a $b\in\mathfrak{g}$ such that $b\rightharpoonup v\neq0$ (because otherwise, $\mathfrak{g}v$ would be $0$, so that $W$ would contain a trivial subrepresentation, and thus be trivial itself (since $W$ is irreducible), contradicting the fact that $W$ is nontrivial). For this $b$, the vector $\left( b\overline{\varepsilon}\right) \rightharpoonup\left( v,w\right) =\left( 0,\ b\rightharpoonup v\right) $ is a nonzero element of $0\oplus W$ contained in $V$ (since $\left( v,w\right) \in V$). Thus, we have proven that $V$ contains a nonzero element of $0\oplus W$, qed.]
Thus, every nonzero $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $-subrepresentation of $W\oplus W$ contains $0\oplus W$.
[Proof. Let $V$ be a nonzero $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $-subrepresentation of $W\oplus W$. We have already shown that $V$ contains a nonzero element of $0\oplus W$. In other words, there exists a nonzero $w\in W$ such that $\left( 0,w\right) \in V$. But since $W$ is irreducible and $w\neq0$, we have $U\left( \mathfrak{g}\right) \cdot w=W$ (where $U\left(\mathfrak{h}\right)$ denotes the universal enveloping algebra of a Lie algebra $\mathfrak{h}$). Now, since $\left( 0,w\right) \in V$ and since $V$ is a $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $-representation, we have \begin{align} V &\supseteq\underbrace{U\left( \mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) \right) }_{\substack{\supseteq U\left( \mathfrak{g}\right) \\\text{(where we canonically include }\mathfrak{g}\text{ in }\mathfrak{g}\left[\varepsilon\right]/\left(\varepsilon^2\right) \\\text{by sending }x\text{ to } x+0\overline{\varepsilon}\text{)}}}\cdot\left( 0,w\right) \supseteq U\left( \mathfrak{g}\right) \cdot\left( 0,w\right) =\left( 0,\underbrace{U\left( \mathfrak{g}\right) \cdot w}_{=W}\right) \\ &=\left( 0,W\right) =0\oplus W. \end{align} Thus, $V$ contains $0\oplus W$, qed.]
Hence, if we were able to decompose the $\mathfrak{g}\left[ \varepsilon\right] /\left( \varepsilon^{2}\right) $-representation $W\oplus W$ into two nonzero addends, then both of these addends would contain $0\oplus W$, which is absurd. We are thus done proving the indecomposability of $W\oplus W$.