= Solution
The <Weyl complete reducibility theorem> states that every finite-dimensional <Lie algebra representation> of a finite-dimensional complex <semisimple Lie algebra> is a <direct sum> of <Irreducible Lie algebra representations>. Equivalently, every invariant <vector subspace> has an invariant complement.
We use the permitted <Casimir operator> properties in the following precise form. There is a central quadratic operator $C$ commuting with the action on every module; it is zero on the <trivial Lie algebra representation>, and on every nontrivial finite-dimensional <Irreducible Lie algebra representation> it is a nonzero scalar $c$. This follows from the <Schur lemma> and the <Casimir eigenvalue> $c_\lambda=(\lambda,\lambda+2\rho)$, using the <Killing form> normalization. We also use the permitted one-dimensional-representation fact: a complex <semisimple Lie algebra> has only trivial one-dimensional representations. Equivalently, it is a <perfect Lie algebra>, $[\mathfrak g,\mathfrak g]=\mathfrak g$, so a character $\mathfrak g\to\mathbb C$ annihilating brackets must vanish. Neither fact assumes complete reducibility of the module being proved reducible.
First prove that every finite-dimensional <short exact sequence>
$$
0\longrightarrow W\longrightarrow E\longrightarrow\mathbb C\longrightarrow0
$$
with trivial quotient splits. If $W$ is nontrivial irreducible, the <Casimir operator> $C_E$ has image in $W$ and restricts to $cI_W$ there. Consequently $\ker C_E$ is a one-dimensional invariant complement to $W$. If $W$ is trivial irreducible, $E$ has a <basis> in which every action is $\begin{pmatrix}0&a(x)\\0&0\end{pmatrix}$. The <Lie algebra representation> identity makes $a([x,y])=0$, so the one-dimensional-representation fact gives $a=0$ and again the sequence splits.
For general $W$, induct on $\dim W$. Choose an irreducible submodule $W_1\subseteq W$. The induced sequence with kernel $W/W_1$ and middle term $E/W_1$ splits by induction. The inverse image $E_1\subseteq E$ of its invariant complement is a submodule fitting into $0\to W_1\to E_1\to\mathbb C\to0$. The irreducible-kernel case gives an invariant line in $E_1$ mapping isomorphically to the quotient. It is also an invariant complement to $W$ in $E$. The case $W=0$ starts this induction. This proves <splitting of a trivial quotient for a semisimple Lie algebra>, including kernels that are not assumed completely reducible.
Now let $U\subseteq V$ be any <invariant subspace>. On the <Hom representation> $\operatorname{Hom}(V,U)$ the action is
$$
(x\cdot T)(v)=x\cdot T(v)-T(x\cdot v).
$$
Let $E$ consist of the maps whose restriction to $U$ is a scalar multiple of $I_U$. This is a submodule, and restriction gives
$$
0\longrightarrow\operatorname{Hom}(V/U,U)\longrightarrow E\longrightarrow\mathbb C\longrightarrow0.
$$
The right-hand map is surjective because an ordinary <linear projection> $V\to U$ exists; its quotient action is trivial because a commutator with $I_U$ is zero. The splitting just proved supplies an invariant $T\in E$ with $T|_U=I_U$. Thus $T$ intertwines the actions, $T^2=T$, and
$$
\boxed{V=U\oplus\ker T\quad\text{as }\mathfrak g\text{-modules}.}
$$
The cases $U=0$ and $U=V$ are immediate. Choosing an irreducible submodule and repeating this complement construction proves the <Weyl complete reducibility theorem>. This last step is <invariant complement from an equivariant projection>.
Dropping finite dimensionality gives an example with the complex <simple Lie algebra> $\mathfrak{sl}_2$. Its <Verma module> of <highest weight> zero has a <basis> $v_0,v_1,\ldots$ with
$$
fv_k=v_{k+1},\qquad hv_k=-2kv_k,\qquad ev_k=k(1-k)v_{k-1},
$$
where $ev_0=0$. These actions obey $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The span of $v_1,v_2,\ldots$ is a proper submodule and the quotient is trivial. It has no invariant complement: such a complement would be a trivial line, while $f$ is injective on the entire module. Thus \b[a simple Lie algebra can have an infinite-dimensional representation that is not completely reducible], even over $\mathbb C$.
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