[proofplan] We expand the Casimir element using the root space decomposition of $\mathfrak g$ and a Killing-dual choice of root vectors. After rewriting each positive-negative root contribution so that positive root vectors act on the right, we evaluate the resulting expression on a highest weight vector. The positive root terms vanish, the Cartan part gives $(\lambda,\lambda)$, and the commutator terms give $2(\lambda,\rho)$. Finally, centrality of the Casimir element makes this scalar action hold on the whole irreducible highest weight module. [/proofplan]

[step:Choose Killing-dual bases adapted to the root decomposition] Let $r:=\dim_{\mathbb C}\mathfrak h$, and choose a basis $(h_1,\dots,h_r)$ of $\mathfrak h$. Let $(k_1,\dots,k_r)$ be the $B$-[dual basis](/theorems/414) of $\mathfrak h$, so that \begin{align*} B(h_i,k_j)=\delta_{ij} \end{align*} for all $1\le i,j\le r$. Let $\Phi:=\Phi^+\cup(-\Phi^+)$ be the full root system. For each root $\alpha\in\Phi$, write \begin{align*} \mathfrak g_\alpha:=\{x\in\mathfrak g:[h,x]=\alpha(h)x\text{ for every }h\in\mathfrak h\} \end{align*} for the corresponding root space. The root-space theorem for finite-dimensional complex semisimple Lie algebras gives \begin{align*} \dim_{\mathbb C}\mathfrak g_\alpha=1 \end{align*} for every $\alpha\in\Phi$ and the direct-sum decomposition \begin{align*} \mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha. \end{align*} The Killing form is orthogonal on root spaces unless the roots sum to zero: if $x\in\mathfrak g_\alpha$ and $y\in\mathfrak g_\beta$, invariance gives \begin{align*} (\alpha+\beta)(h)B(x,y)=B([h,x],y)+B(x,[h,y])=0 \end{align*} for every $h\in\mathfrak h$, so $B(x,y)=0$ when $\alpha+\beta\ne 0$. Nondegeneracy of $B$ on $\mathfrak g$ therefore pairs each one-dimensional $\mathfrak g_\alpha$ nondegenerately with $\mathfrak g_{-\alpha}$. For each $\alpha\in\Phi^+$, choose a nonzero vector $e_\alpha\in\mathfrak g_\alpha$. Since the Killing form pairs $\mathfrak g_\alpha$ nondegenerately with $\mathfrak g_{-\alpha}$, choose $f_\alpha\in\mathfrak g_{-\alpha}$ such that \begin{align*} B(e_\alpha,f_\alpha)=1. \end{align*} Then \begin{align*} (h_1,\dots,h_r,(e_\alpha)_{\alpha\in\Phi^+},(f_\alpha)_{\alpha\in\Phi^+}) \end{align*} is a basis of $\mathfrak g$ adapted to the decomposition \begin{align*} \mathfrak g=\mathfrak h\oplus\bigoplus_{\alpha\in\Phi^+}\mathfrak g_\alpha\oplus\bigoplus_{\alpha\in\Phi^+}\mathfrak g_{-\alpha}. \end{align*} Its $B$-dual basis consists of \begin{align*} (k_1,\dots,k_r,(f_\alpha)_{\alpha\in\Phi^+},(e_\alpha)_{\alpha\in\Phi^+}). \end{align*} Therefore the Casimir element has the expansion \begin{align*} \Omega_B=\sum_{i=1}^r h_i k_i+\sum_{\alpha\in\Phi^+}(e_\alpha f_\alpha+f_\alpha e_\alpha). \end{align*} [/step]

[step:Move all positive root vectors to the right]For each $\alpha\in\Phi^+$, the defining relation in the enveloping algebra gives \begin{align*} e_\alpha f_\alpha=f_\alpha e_\alpha+[e_\alpha,f_\alpha]. \end{align*} Substituting this identity into the root contribution to $\Omega_B$ gives \begin{align*} \Omega_B=\sum_{i=1}^r h_i k_i+\sum_{\alpha\in\Phi^+}\bigl(2f_\alpha e_\alpha+[e_\alpha,f_\alpha]\bigr). \end{align*}[/step]

[guided]The point of this rewriting is to prepare the Casimir element for evaluation on a highest weight vector. Positive root vectors annihilate such a vector, but this is useful only when the positive root vector stands on the right. For a fixed positive root $\alpha\in\Phi^+$, the vectors $e_\alpha\in\mathfrak g_\alpha$ and $f_\alpha\in\mathfrak g_{-\alpha}$ are elements of $U(\mathfrak g)$. The universal enveloping algebra relation says that, for $x,y\in\mathfrak g$, \begin{align*} xy-yx=[x,y]. \end{align*} Applying this with $x=e_\alpha$ and $y=f_\alpha$, we get \begin{align*} e_\alpha f_\alpha=f_\alpha e_\alpha+[e_\alpha,f_\alpha]. \end{align*} The Casimir expansion from the previous step contains both ordered products $e_\alpha f_\alpha$ and $f_\alpha e_\alpha$. Replacing only the first one gives \begin{align*} e_\alpha f_\alpha+f_\alpha e_\alpha = f_\alpha e_\alpha+[e_\alpha,f_\alpha]+f_\alpha e_\alpha = 2f_\alpha e_\alpha+[e_\alpha,f_\alpha]. \end{align*} Summing this identity over all positive roots gives \begin{align*} \Omega_B=\sum_{i=1}^r h_i k_i+\sum_{\alpha\in\Phi^+}\bigl(2f_\alpha e_\alpha+[e_\alpha,f_\alpha]\bigr). \end{align*} This is the exact form needed: each term $f_\alpha e_\alpha$ has $e_\alpha$ acting first on a module vector.[/guided]

[step:Identify the root commutators with the Killing-form elements $h_\alpha$] For each $\alpha\in\Phi^+$, the commutator $[e_\alpha,f_\alpha]$ belongs to $\mathfrak h$. We claim that \begin{align*} [e_\alpha,f_\alpha]=h_\alpha, \end{align*} where $h_\alpha\in\mathfrak h$ is defined by $B(h_\alpha,h)=\alpha(h)$ for every $h\in\mathfrak h$. Indeed, for every $h\in\mathfrak h$, [invariance of the Killing form](/theorems/3808) gives \begin{align*} B([e_\alpha,f_\alpha],h)=B(e_\alpha,[f_\alpha,h]). \end{align*} Since $f_\alpha\in\mathfrak g_{-\alpha}$, we have $[h,f_\alpha]=-\alpha(h)f_\alpha$, hence $[f_\alpha,h]=\alpha(h)f_\alpha$. Therefore \begin{align*} B([e_\alpha,f_\alpha],h)=B(e_\alpha,\alpha(h)f_\alpha)=\alpha(h)B(e_\alpha,f_\alpha)=\alpha(h). \end{align*} By the defining uniqueness of $h_\alpha$, this proves $[e_\alpha,f_\alpha]=h_\alpha$. [/step]

[step:Evaluate the rewritten Casimir element on a highest weight vector] Let $0\ne v_\lambda\in L(\lambda)$ be a highest weight vector of weight $\lambda$. Thus \begin{align*} h v_\lambda=\lambda(h)v_\lambda \end{align*} for every $h\in\mathfrak h$, and \begin{align*} xv_\lambda=0 \end{align*} for every $x\in\mathfrak g_\alpha$ with $\alpha\in\Phi^+$. Using the rewritten expression for $\Omega_B$, we obtain \begin{align*} \Omega_B v_\lambda = \sum_{i=1}^r h_i k_i v_\lambda + \sum_{\alpha\in\Phi^+}\bigl(2f_\alpha e_\alpha+[e_\alpha,f_\alpha]\bigr)v_\lambda. \end{align*} Since $e_\alpha v_\lambda=0$, each term $f_\alpha e_\alpha v_\lambda$ is zero. Since $[e_\alpha,f_\alpha]=h_\alpha$, the root contribution becomes \begin{align*} \sum_{\alpha\in\Phi^+}[e_\alpha,f_\alpha]v_\lambda = \sum_{\alpha\in\Phi^+}h_\alpha v_\lambda = \sum_{\alpha\in\Phi^+}\lambda(h_\alpha)v_\lambda. \end{align*} By the definition of the [bilinear form](/page/Bilinear%20Form) on $\mathfrak h^*$, \begin{align*} \lambda(h_\alpha)=B(h_\lambda,h_\alpha)=(\lambda,\alpha). \end{align*} Therefore the root contribution is \begin{align*} \sum_{\alpha\in\Phi^+}(\lambda,\alpha)v_\lambda=2(\lambda,\rho)v_\lambda. \end{align*} It remains to compute the Cartan contribution. Since each $k_i$ acts on $v_\lambda$ by the scalar $\lambda(k_i)$, we have \begin{align*} \sum_{i=1}^r h_i k_i v_\lambda = \sum_{i=1}^r \lambda(k_i)\lambda(h_i)v_\lambda. \end{align*} The element $\sum_{i=1}^r \lambda(k_i)h_i\in\mathfrak h$ has the same $B$-pairing with every $h\in\mathfrak h$ as $h_\lambda$, hence equals $h_\lambda$. Thus \begin{align*} \sum_{i=1}^r \lambda(k_i)\lambda(h_i)=\lambda(h_\lambda)=B(h_\lambda,h_\lambda)=(\lambda,\lambda). \end{align*} Combining the Cartan and root contributions gives \begin{align*} \Omega_B v_\lambda=\bigl((\lambda,\lambda)+2(\lambda,\rho)\bigr)v_\lambda=(\lambda,\lambda+2\rho)v_\lambda. \end{align*} [/step]

[step:Extend the scalar action from the highest weight vector to all of $L(\lambda)$] [claim:The Killing-form Casimir element is central] For any basis $(z_a)$ of $\mathfrak g$ with $B$-dual basis $(w_a)$, the element \begin{align*} \sum_a z_a w_a\in U(\mathfrak g) \end{align*} commutes with every element of $\mathfrak g$. [/claim] [proof] Fix $y\in\mathfrak g$, and write \begin{align*} [y,z_a]=\sum_b A_{ba}z_b. \end{align*} Invariance of $B$ gives \begin{align*} B([y,w_a],z_c)=-B(w_a,[y,z_c])=-A_{ac}. \end{align*} Since $(w_c)$ is the $B$-dual basis to $(z_c)$, this says \begin{align*} [y,w_a]=-\sum_c A_{ac}w_c. \end{align*} Thus the commutator in $U(\mathfrak g)$ is \begin{align*} \left[y,\sum_a z_a w_a\right] =\sum_{a,b}A_{ba}z_bw_a-\sum_{a,c}A_{ac}z_aw_c=0, \end{align*} after relabelling the dummy indices in the two sums. Therefore the Casimir element commutes with every $y\in\mathfrak g$, and hence lies in the center of $U(\mathfrak g)$. [/proof] Applying the claim to the Killing-dual basis used to define $\Omega_B$, the element $\Omega_B$ is central in $U(\mathfrak g)$. Since $L(\lambda)$ is irreducible and $v_\lambda\ne 0$, the $U(\mathfrak g)$-submodule generated by $v_\lambda$ is all of $L(\lambda)$. Let $u\in U(\mathfrak g)$ be arbitrary. Using centrality of $\Omega_B$ and the calculation on $v_\lambda$, we get \begin{align*} \Omega_B(u v_\lambda)=u\Omega_B v_\lambda=u\bigl((\lambda,\lambda+2\rho)v_\lambda\bigr)=(\lambda,\lambda+2\rho)u v_\lambda. \end{align*} Every element of $L(\lambda)$ is a finite sum of vectors of the form $u v_\lambda$ with $u\in U(\mathfrak g)$. Hence $\Omega_B$ acts on all of $L(\lambda)$ by the scalar \begin{align*} (\lambda,\lambda+2\rho). \end{align*} This proves the formula. [/step]