Linear complex extension of a representation Consider the following matrices,
$$H=\left(\begin{matrix}
1 & 0 \\
0 & -1 
\end{matrix}\right) 
\qquad X=\left(\begin{matrix}
0 & 1 \\
0 & 0 
\end{matrix}\right)
\qquad Y=\left(\begin{matrix}
0 & 0 \\
1 & 0 
\end{matrix}\right)
$$
Let $\rho:\mathfrak{su}(2)\rightarrow \text{End}(V)$ be a lie algebra homomorphism such that $\rho(X)^*=-\rho(X)$. Does this imply that $\rho(Y)^*=-\rho(Y)$?
I have been trying to use the commutation relations of $H,X,Y$ and the fact that $\rho$ is a lie algebra homomorphism to show this but I can't see to do it. 
Am I overlooking something obvious here?
 A: The assertion is true, but for a striking reason: If $\rho: \mathfrak{su}(2)_{\mathbb C} (\simeq \mathfrak{sl}_2(\mathbb C)) \rightarrow End_{\mathbb C}(V)$ is a (finite dimensional, complex) Lie algebra representation with $\rho(X)^\ast =-\rho(X)$, then $\rho$ is the trivial representation $\rho =0$.
Namely, it is well-known or easily proven that $\rho(X)$ must be nilpotent. But:
Lemma: Let $V$ be a finite-dimensional $\mathbb C$-vector space with hermitian inner product $\langle \cdot, \cdot\rangle$, and let $A \in End_{\mathbb C}(V)$ be nilpotent. Then $A^\ast =-A \Rightarrow A=0$.
Proof: Assume $A \neq 0$, then there is $0 \neq v_1 \in V$ with $v_0 := Av_1 \neq 0$ and $Av_0 =0$. So $$ 0 = \langle 0, v_1 \rangle =\langle Av_0, v_1 \rangle = \langle v_0, A^\ast v_1\rangle = \langle v_0, -Av_1 \rangle = -\langle v_0, v_0 \rangle \neq 0,$$
contradiction. (For different proofs and generalisations, compare e.g. $A$ is normal and nilpotent, show $A=0$.)
Now from $\rho(X)=0$ it follows easily (if one does not know it from the representation theory of $\mathfrak{sl}_2(\mathbb C)$) that $\rho(H)$, and a fortiori $\rho(Y)$, are $=0$.
