[proofplan] The proof identifies the Casimir operator as the image under $\rho \otimes \rho$ of the canonical tensor $\Omega=\sum_i x_i \otimes y_i$ determined by the Killing form. We first prove that $\Omega$ is invariant under the diagonal adjoint action of $\mathfrak g$, using the invariance and nondegeneracy of the Killing form. Applying the representation $\rho$ to this tensor identity gives exactly the commutator identity $[\rho(z),C_V]=0$. [/proofplan] [step:Define the canonical tensor determined by the Killing form] Let $n := \dim_K \mathfrak g$. Define the tensor \begin{align*} \Omega := \sum_{i=1}^n x_i \otimes y_i \in \mathfrak g \otimes_K \mathfrak g. \end{align*} Since $(x_i)_{i=1}^n$ and $(y_i)_{i=1}^n$ are $\kappa$-dual bases, every $u \in \mathfrak g$ has the expansions \begin{align*} u = \sum_{i=1}^n \kappa(u,y_i)x_i \qquad\text{and}\qquad u = \sum_{i=1}^n \kappa(x_i,u)y_i. \end{align*} [/step] [step:Prove the canonical tensor is invariant under the diagonal adjoint action] Fix $z \in \mathfrak g$. We prove \begin{align*} \sum_{i=1}^n [z,x_i]\otimes y_i+\sum_{i=1}^n x_i\otimes [z,y_i]=0 \end{align*} in $\mathfrak g \otimes_K \mathfrak g$. Let $a,b \in \mathfrak g$. Pair the displayed tensor with $a \otimes b$ through the nondegenerate [bilinear form](/page/Bilinear%20Form) \begin{align*} (\mathfrak g \otimes_K \mathfrak g)\times(\mathfrak g \otimes_K \mathfrak g) &\to K,\\ (u_1\otimes u_2,\ v_1\otimes v_2) &\mapsto \kappa(u_1,v_1)\kappa(u_2,v_2). \end{align*} The pairing gives \begin{align*} &\sum_{i=1}^n \kappa([z,x_i],a)\kappa(y_i,b) +\sum_{i=1}^n \kappa(x_i,a)\kappa([z,y_i],b). \end{align*} The Killing form is invariant, so for all $r,s,t \in \mathfrak g$, \begin{align*} \kappa([r,s],t)+\kappa(s,[r,t])=0. \end{align*} Using this first with $(r,s,t)=(z,x_i,a)$ and then with $(r,s,t)=(z,y_i,b)$, the pairing becomes \begin{align*} &-\sum_{i=1}^n \kappa(x_i,[z,a])\kappa(y_i,b) -\sum_{i=1}^n \kappa(x_i,a)\kappa(y_i,[z,b]). \end{align*} By the dual-basis expansions, the first sum is $\kappa([z,a],b)$ and the second sum is $\kappa(a,[z,b])$. Hence the pairing equals \begin{align*} -\kappa([z,a],b)-\kappa(a,[z,b])=0 \end{align*} again by invariance of $\kappa$. Since $a,b \in \mathfrak g$ were arbitrary and the induced tensor pairing is nondegenerate, the tensor identity follows. [guided] Fix $z \in \mathfrak g$. The tensor $\Omega=\sum_i x_i\otimes y_i$ should be thought of as the inverse of the Killing form. To prove that it is invariant, we must show that applying $\operatorname{ad}z$ to the first tensor factor cancels applying $\operatorname{ad}z$ to the second tensor factor: \begin{align*} \sum_{i=1}^n [z,x_i]\otimes y_i+\sum_{i=1}^n x_i\otimes [z,y_i]=0. \end{align*} Because $\kappa$ is nondegenerate, it induces a nondegenerate pairing on $\mathfrak g\otimes_K\mathfrak g$ by \begin{align*} (u_1\otimes u_2,\ v_1\otimes v_2) \mapsto \kappa(u_1,v_1)\kappa(u_2,v_2). \end{align*} Thus it is enough to pair the tensor above against an arbitrary elementary tensor $a\otimes b$, where $a,b\in\mathfrak g$, and prove that the result is zero. The pairing is \begin{align*} &\sum_{i=1}^n \kappa([z,x_i],a)\kappa(y_i,b) +\sum_{i=1}^n \kappa(x_i,a)\kappa([z,y_i],b). \end{align*} The key input is [invariance of the Killing form](/theorems/3808): \begin{align*} \kappa([r,s],t)+\kappa(s,[r,t])=0 \end{align*} for all $r,s,t\in\mathfrak g$. Applying this with $(r,s,t)=(z,x_i,a)$ gives \begin{align*} \kappa([z,x_i],a)=-\kappa(x_i,[z,a]), \end{align*} and applying it with $(r,s,t)=(z,y_i,b)$ gives \begin{align*} \kappa([z,y_i],b)=-\kappa(y_i,[z,b]). \end{align*} Therefore the pairing becomes \begin{align*} &-\sum_{i=1}^n \kappa(x_i,[z,a])\kappa(y_i,b) -\sum_{i=1}^n \kappa(x_i,a)\kappa(y_i,[z,b]). \end{align*} Now we use the dual-basis identities. Since $(x_i)$ and $(y_i)$ are dual with respect to $\kappa$, \begin{align*} [z,a]=\sum_{i=1}^n \kappa(x_i,[z,a])y_i, \qquad a=\sum_{i=1}^n \kappa(x_i,a)y_i. \end{align*} Pairing these expansions with $b$ and $[z,b]$ respectively gives \begin{align*} \sum_{i=1}^n \kappa(x_i,[z,a])\kappa(y_i,b)=\kappa([z,a],b), \end{align*} and \begin{align*} \sum_{i=1}^n \kappa(x_i,a)\kappa(y_i,[z,b])=\kappa(a,[z,b]). \end{align*} Thus the original pairing equals \begin{align*} -\kappa([z,a],b)-\kappa(a,[z,b]). \end{align*} This is zero by the same invariance identity for $\kappa$. Since every pairing against $a\otimes b$ is zero and the tensor pairing is nondegenerate, the tensor itself is zero. [/guided] [/step] [step:Expand the commutator of the representation with the Casimir operator] For the fixed $z \in \mathfrak g$, compute the commutator in $\operatorname{End}_K(V)$: \begin{align*} [\rho(z),C_V] &= \rho(z)\sum_{i=1}^n \rho(x_i)\rho(y_i)-\sum_{i=1}^n \rho(x_i)\rho(y_i)\rho(z)\\ &= \sum_{i=1}^n \bigl(\rho(z)\rho(x_i)-\rho(x_i)\rho(z)\bigr)\rho(y_i) +\sum_{i=1}^n \rho(x_i)\bigl(\rho(z)\rho(y_i)-\rho(y_i)\rho(z)\bigr). \end{align*} Since $\rho: \mathfrak g \to \mathfrak{gl}(V)$ is a Lie algebra representation, \begin{align*} \rho([u,v])=\rho(u)\rho(v)-\rho(v)\rho(u) \end{align*} for all $u,v \in \mathfrak g$. Hence \begin{align*} [\rho(z),C_V] = \sum_{i=1}^n \rho([z,x_i])\rho(y_i) +\sum_{i=1}^n \rho(x_i)\rho([z,y_i]). \end{align*} [/step] [step:Apply the invariant tensor identity to obtain centrality] Define the bilinear map \begin{align*} m_\rho: \mathfrak g \otimes_K \mathfrak g &\to \operatorname{End}_K(V),\\ u\otimes v &\mapsto \rho(u)\rho(v). \end{align*} Applying $m_\rho$ to the tensor identity \begin{align*} \sum_{i=1}^n [z,x_i]\otimes y_i+\sum_{i=1}^n x_i\otimes [z,y_i]=0 \end{align*} gives \begin{align*} \sum_{i=1}^n \rho([z,x_i])\rho(y_i) +\sum_{i=1}^n \rho(x_i)\rho([z,y_i]) =0. \end{align*} By the commutator expansion from the previous step, this says \begin{align*} [\rho(z),C_V]=0. \end{align*} Equivalently, \begin{align*} \rho(z)C_V=C_V\rho(z). \end{align*} Since $z \in \mathfrak g$ was arbitrary, $C_V$ commutes with the action of every element of $\mathfrak g$. [/step]