I'm trying to prove that $\mathfrak{so}_6(\Bbb C)$ is semisimple.
There exists a criterion which says that, given a Lie algebra $L\le\mathfrak{gl}(V)$, where $V$ is an irreducible $L$-module, then $Z(L)=\operatorname{Rad}(L)$ and $\dim Z(L)\le1$.
From this, we know that if $L\le\mathfrak{sl}(V)$ then $L$ is semisimple.
My problem is: how can I prove that the action of $L=\mathfrak{so}_6(\Bbb C)$ on $V=\Bbb C^6$ is irreducible?
$\mathfrak{so}_6(\Bbb C)\hookrightarrow\mathfrak{sl}_6(\Bbb C)$, so the action is, given $x\in\mathfrak{so}_6(\Bbb C)$, $\alpha_x:\Bbb C^6\to\Bbb C^6$, $v\mapsto xv$.
My teacher said that it's enough to find a vector $v\in\Bbb C^6$ such that $L.v=\Bbb C^6$. Is this right? If yes, is brute force (hand computing) the right way to find $v$?
How can we prove that this action is irreducible?
Many thanks
The proof is given in Humphreys book on Lie algebras, in section $19.2$, "the classical algebras". He writes: "Therefore, to prove that $\mathfrak{so}_n$ acts irreducibly in its natural representation (i.e., $v\mapsto xv$), it will suffice to prove that all endomorphisms of $V$ are obtainable from $1$ and $\mathfrak{so}_n$ using addition, scalar multiplication and ordinary multiplication. From $1$ we get all scalars. From the diagonal matrices we get all possible diagonal matrices. Then multiplying various other basis elements, such as $E_{ij}-E_{ji}$ for $i\neq j$, by suitable diagonal matrices $diag(0,\ldots ,1,\ldots ,0)$ yields all off-diagonal matrix units $E_{ij}$, as the reader can quickly verify."