This is an argument in the proof of complete reducibility of representation of semisimple Lie Algebras.
We have the following SES of a complex Lie Algebra representation,
$$0\to \mathbb{C} \to W \to \mathbb{C} \to 0$$ Now the easy claim would be: $(\rho,W)$ is a two dimension representation and $\rho(x)$ is strictly upper triangular for all $x \in \mathfrak{g}$. It's not obvious to me why $\rho(x)$ is strictly upper triangular?
Edit: As mentioned by @krm2233, it should be clear once I noticed $\mathbb{C}$ is seen as a trivial rep: Since any representation of semisimple Lie Algebra $L$ has vanishing trace: $\rho(L) \subseteq sl(V)$ by using the fact that $[L,L]=L$. So any $1$-dimensional representation of $L$ acts as $0$.
This should probably be just a comment, but the two outer terms in the short exact sequence, are denoted "$\mathbb C$" without any further comment. Does this mean that they should be taken, as representations of the Lie algebra $\mathfrak g$, to be copies of the trivial representation?
If that is what is intended, then the "easy claim" is indeed straight-forward: if $i\colon \mathbb C\to W$ denotes the injective map with target $W$ from the short exact sequence, and $q\colon W \to \mathbb C$ the surjective map with source $W$, that is: $$ 0\rightarrow \mathbb C \rightarrow^iW\rightarrow^q \mathbb C \rightarrow 0 $$ is you pick a basis of $W$ consisting of $e_1 =i(1)$ and $e_2 \in W$ any vector such that $q(e_2)=1 \in \mathbb C$, then since $\mathbb C$ is the trivial representation and $i$ is a $\mathfrak g$-homomorphism, for all $x \in \mathfrak g$ we have $$ \rho(x)(e_1) = \rho(x)i(1) = i(x(1)) = i(0)=0. $$ Similarly $q(\rho(x)(e_2)) = x(q(e_2)) = x(1)=0$, so that $\rho(x)(e_2) \in \ker(q) = \mathbb C.e_1$, and hence the matrix of $\rho(x)$ with respect to the basis $\{e_1,e_2\}$ has the form $\left(\begin{array}{cc} 0 & * \\ 0 & 0 \end{array} \right)$, that is, $\rho(x)$ is strictly upper triangular for all $x \in \mathfrak g$.