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I think that there are two things to be motivated: one is the Casimir, and the other is the proof of semi-simplicity.

First, for the Casimir, it might help to note that there is a formula-free" construction. A symmetric bilinear form $\kappa$ defines $\mathfrak{g}\simeq \mathfrak{g}^{\vee}$, and the Casimir is the image in $U(\mathfrak{g})$ of the element corresponding to the identity under the isomorphism $\mathfrak{g}\otimes\mathfrak{g}\simeq \mathfrak{g}^{\vee}\otimes\mathfrak{g}$.

The idea of the proof is actually very simple: you construct an element of the center of the algebra which detects" the trivial representation. I.e., it acts by zero on the trivial representation and by a non-zero scalar on all simple (finite-dimensional) representations. (This is in exact analogy to what happens for finite groups in characteristic zero: then $1-\frac{1}{|G|}\sum_{g\in{G}}\delta_g$ has the same property).

Once you have such a central element, let's call it $C$, the proof that finite-dimensional representations are semi-simple is easy. For all $V,W$, note that: $$\operatorname{Ext}^i(V,W)= \operatorname{Ext}^i(\mathbb{C},\underline{\operatorname{Hom}}(V,W))$$ (where $\underline{\operatorname{Hom}}$ is internal $\operatorname{Hom}$ relative to the usual tensor product of $\mathfrak{g}$-modules, and both $\operatorname{Ext}$s are of $\mathfrak{g}$-modules) because formation of internal $Hom$ is exact in both variables. Therefore, it's enough to show that $\operatorname{Ext}^1(\mathbb{C},V)=0$ for all finite-dimensional $\mathfrak{g}$-modules $V$ (substitute $\underline{\operatorname{Hom}}(V,W)$ for $V$).

Clearly any such $V$ has finite length, so by devissage, it's enough to prove for simple modules. Either $V$ is trivial or it is not. If $V$ is non-trivial, then $C$ acts on $\operatorname{Ext}^i(\mathbb{C},V)$ by two different scalars: the one by which it acts on $V$ and by $0$ (by which it acts on $\mathbb{C}$). Therefore, this vector space must be zero. If $V=\mathbb{C}$, then $\operatorname{Ext}^1(\mathbb{C},\mathbb{C})=0$ since for any extension $E$, the homomorphism $\mathfrak{g}\to\operatorname{End}(E)$ maps to the 1-dimensional subspace sending $E$ to $\mathbb{C}$ and sending $\mathbb{C}$ to $0$, but since $\mathfrak{g}$ has no codimension 1 ideals this must be the trivial homomorphism.

Of course, basically the same proof goes through for finite groups and is essentially the same as the usual proof.

I think that there are two things to be motivated: one is the Casimir, and the other is the proof of semi-simplicity.

First, for the Casimir, it might help to note that there is a formula-free" construction. A symmetric bilinear form $\kappa$ defines $\mathfrak{g}\simeq \mathfrak{g}^{\vee}$, and the Casimir is the image in $U(\mathfrak{g})$ of the element corresponding under the isomorphism $\mathfrak{g}\otimes\mathfrak{g}\simeq \mathfrak{g}^{\vee}\otimes\mathfrak{g}$.

The idea of the proof is actually very simple: you construct an element of the center of the algebra which detects" the trivial representation. I.e., it acts by zero on the trivial representation and by a non-zero scalar on all simple (finite-dimensional) representations. (This is in exact analogy to what happens for finite groups in characteristic zero: then $1-\frac{1}{|G|}\sum_{g\in{G}}\delta_g$ has the same property).

Once you have such a central element, let's call it $C$, the proof that finite-dimensional representations are semi-simple is easy. For all $V,W$, note that: $$\operatorname{Ext}^i(V,W)= \operatorname{Ext}^i(\mathbb{C},\underline{\operatorname{Hom}}(V,W))$$ (where $\underline{\operatorname{Hom}}$ is internal $\operatorname{Hom}$ relative to the usual tensor product of $\mathfrak{g}$-modules, and both $\operatorname{Ext}$s are of $\mathfrak{g}$-modules) because formation of internal $Hom$ is exact in both variables. Therefore, it's enough to show that $\operatorname{Ext}^1(\mathbb{C},V)=0$ for all finite-dimensional $\mathfrak{g}$-modules $V$ (substitute $\underline{\operatorname{Hom}}(V,W)$ for $V$).

Clearly any such $V$ has finite length, so by devissage, it's enough to prove for simple modules. Either $V$ is trivial or it is not. If $V$ is non-trivial, then $C$ acts on $\operatorname{Ext}^i(\mathbb{C},V)$ by two different scalars: the one by which it acts on $V$ and by $0$ (by which it acts on $\mathbb{C}$). Therefore, this vector space must be zero. If $V=\mathbb{C}$, then $\operatorname{Ext}^1(\mathbb{C},\mathbb{C})=0$ since for any extension $E$, the homomorphism $\mathfrak{g}\to\operatorname{End}(E)$ maps to the 1-dimensional subspace sending $E$ to $\mathbb{C}$ and sending $\mathbb{C}$ to $0$, but since $\mathfrak{g}$ has no codimension 1 ideals this must be the trivial homomorphism.

Of course, basically the same proof goes through for finite groups and is essentially the same as the usual proof.

Post Undeleted by Moosbrugger
2 Fixed mistakes.
Post Deleted by Moosbrugger
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