Irreducible representation and reductive Lie algebra

Question

If we have complex reductive Lie algebra L and her finite dimensional representation $\phi$. How can we show that $\phi$ is irreducible iff restriction $\phi|_{[L,L]}$ is irreducible?

Answer 1

If $\phi|_{[L,L]}$ is irreducible, it's immediate that $\phi$ is irreducible (because any invariant subspace for whole $L$ would be also invariant for $[L,L] \subseteq L$).

Conversly, suppose $\phi \colon L \to \mathfrak{gl}(V)$ is irreducible, V finite-dimensional. It can be shown that every reductive Lie algebra is direct sum of it's two ideals: $L=Z(L) \dot{+} [L,L]$, where Z(L) is center of L (hence commutative), and [L,L] is semisimple.

First we show that any element $x \in Z(L)$ acts on $V$ by scalar, i.e. there is $\lambda(x) \in \mathbb C$ such that $\phi(x) v=\lambda(x) \cdot v$. To see this, note that for any $y \in L$ we have $[x,y]=0$, so $\phi(x)\circ\phi(y)=\phi(y)\circ\phi(x)$. From this, we can see that endomorphism $\phi(x) \colon V \to V$ respects $L$-action on V: $$ \phi(x) (y.v)= \phi(x) (\phi(y) v) = \phi(y) (\phi(x) v) = y. (\phi(x) v), \quad y \in L, \ v \in V, $$ and Schur's lemma does the rest. We get a linear functional $\lambda \colon Z(L) \to \mathbb C$, such that $$ \phi(x+y)v = \lambda(x)\cdot v + \phi(y)v, \quad x \in Z(L), \ y \in [L,L], \ v \in V. $$

From this we see that a subspace $W \leq V$ is invariant for all $L=Z(L) \dot{+} [L,L]$ if and only if it is invariant for all $[L,L]$, and this proves your claim.