Proof that the Casimir invariant of a representation commutes with everything We have some representation $\rho:L\longrightarrow End(V)$ of a $d$-dimensional complex Lie algebra $L$. Let $\{X_1,...,X_d\}$ be any basis of $L$, and let $\{\xi_1,...,\xi_d\}$ be another basis such that $\kappa_\rho(X_i,\xi_j):=tr(\rho(X_i)\circ\rho(\xi_j))=\delta_{ij}$ where $\kappa_\rho$ is the Killing form.
Define the Casimir invariant by $$\Omega_{\rho}:=\sum\limits_{i=1}^{d}\rho(X_i)\circ\rho(\xi_i).$$
I want to prove that $\forall a\in L:$ $$[\Omega_{\rho},\rho(a)]=0.$$ By taking [] to be the commutator and thinking of $\circ$ as a multiplication (matrix multiplication? I think this is justified because a basis has been chosen for $L$), I've reached
$$[\Omega_{\rho},\rho(a)]=\sum\limits_{i=1}^{d}\rho(X_i)\rho(\xi_i)\rho(a)-\rho(a)\rho(X_i)\rho(\xi_i)=...$$ $$...=\sum\limits_{i=1}^{d}\rho(X_i)[\rho(\xi_i),\rho(a)]-
[\rho(a),\rho(X_i)]\rho(\xi_i)$$
but I don't know where to go from here. Any push in the right direction would be much appreciated.
 A: For the record, there's a basis-free approach to the construction of the Casimir operator which (I think) is much clearer than the one depending on a choice of basis. The Casimir operator on $V$ is the image of the ${\mathfrak g}$-invariant element $1\in k$ under the ${\mathfrak g}$-homomorphism
$$(\ast)\qquad k\to\text{Hom}_k({\mathfrak g},{\mathfrak g})\cong {\mathfrak g}\otimes_k {\mathfrak g}^{\ast}\cong {\mathfrak g}\otimes_k {\mathfrak g}\to {\mathscr U}({\mathfrak g})\xrightarrow{\rho}\text{gl}(V)$$
and as such ${\mathfrak g}$-invariant, too. But the ${\mathfrak g}$-invariant elements in $\text{gl}(V)$ are precisely the endomorphisms $\varphi: V\to V$ which commute with all $\rho(X), X\in{\mathfrak g}$.
The decisive part in $(\ast)$ is the isomorphism of ${\mathfrak g}$-modules ${\mathfrak g}\to {\mathfrak g}^{\ast}, X\mapsto \kappa(X,-)$ given by the Killing form. The associativity of the Killing form is equivalent to this map being ${\mathfrak g}$-linear.
A: Note that (with different notation $x^j,x_j$)
\begin{align*}
\rho(x^j)\rho(x_j)\rho(z)-\rho(z)\rho(x^j)\rho(x_j) & = \rho(x^j)\Bigl( \rho(x_j)\rho(z)-\rho(z)\rho(x_j) \Bigr) \\
 & - \Bigl( \rho(z)\rho(x^j)-\rho(x^j)\rho(z) \Bigr) \rho(x_j).
\end{align*}
Then it follows, with $\Omega=\Omega (\beta, \rho)$,
\begin{align*}
\Omega \rho(z)-\rho(z)\Omega & = \sum_{j=1}^n \rho(x^j)\rho(x_j)\rho(z)-\rho(z)\rho(x^j)\rho(x_j)\\
 & = \sum_{j=1}^n \rho(x^j)[\rho(x_j),\rho(z)] - [\rho(z),\rho(x^j)]\rho(x_j) \\
 & = \sum_{j=1}^n \rho(x^j)\rho([x_j,z]) - \rho([z,x^j])\rho(x_j) \\
 & = \sum_{j,k=1}^n -a_{kj}\rho(x^j)\rho(x_k)-a^{jk}\rho(x^k)\rho(x_j) \\
 & = \sum_{j,k=1}^n a^{kj}\rho(x^j)\rho(x_k)-a^{jk}\rho(x^k)\rho(x_j) \\
 & = \sum_{j,k=1}^n a^{kj}\rho(x^j)\rho(x_k) - \sum_{j,k=1}^n a^{kj}\rho(x^j)\rho(x_k)\\
 & = 0.
\end{align*}
For details see Lemma $2.3.1$ here.
