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The Casimir element is dual to the Killing form. (I think. I am somewhat uncertain about this because nobody has ever said this to me, even though it seems like the right thing to say, and frankly I don't know why Lie algebra textbooks don't just say this.) That is, the nondegeneracy of the Killing form is equivalent to its providing an isomorphism $\mathfrak{g} \to \mathfrak{g}^{\ast}$, and writing this isomorphism as a tensor exhibits it as an element of $\mathfrak{g} \otimes \mathfrak{g}$ - precisely the element $\sum e_i \otimes e_i'$ where $e_i$ is a basis. This embeds into $U(\mathfrak{g}) \otimes U(\mathfrak{g})$ and the multiplication map into $U(\mathfrak{g})$ gives the Casimir element.

(Oh good, I see that this is the same definition Akhil gives in the blog post darij linked to above. That makes me feel better.)

Also, you're using the "wrong" definition of the determinant. The exterior powers are all functors, and they take linear maps $T : V \to V$ to linear maps $T : \Lambda^n V \to \Lambda^n V$. Since $\Lambda^n V$ is one-dimensional when $n = \dim V$, the linear maps from $\Lambda^n V$ to itself are canonically isomorphic to the base field $k$. There is also a slightly more transparent definition of the trace: $\text{End}(V)$ is canonically isomorphic to $V^{\ast} \otimes V$, and then one composes with the dual pairing $V^{\ast} \times V \to k$.

It seems to me the notion you're looking for is what general notion of monoidal category supports the definition you're looking at. Traces can be defined in monoidal categories with duals, although I'm not sure what the natural setting for determinants is.

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The Casimir element is dual to the Killing form. (I think. I am somewhat uncertain about this because nobody has ever said this to me, even though it seems like the right thing to say, and frankly I don't know why Lie algebra textbooks don't just say this.) That is, the nondegeneracy of the Killing form is equivalent to its providing an isomorphism $\mathfrak{g} \to \mathfrak{g}^{\ast}$, and writing this isomorphism as a tensor exhibits it as an element of $\mathfrak{g} \otimes \mathfrak{g}$ - precisely the element $\sum e_i \otimes e_i'$ where $e_i$ is a basis. This embeds into $U(\mathfrak{g}) \otimes U(\mathfrak{g})$ and the multiplication map into $U(\mathfrak{g})$ gives the Casimir element.

(Oh good, I see that this is the same definition Akhil gives in the blog post darij linked to above. That makes me feel better.)

1

The Casimir element is dual to the Killing form. (I think. I am somewhat uncertain about this because nobody has ever said this to me, even though it seems like the right thing to say, and frankly I don't know why Lie algebra textbooks don't just say this.) That is, the nondegeneracy of the Killing form is equivalent to its providing an isomorphism $\mathfrak{g} \to \mathfrak{g}^{\ast}$, and writing this isomorphism as a tensor exhibits it an element of $\mathfrak{g} \otimes \mathfrak{g}$ - precisely the element $\sum e_i \otimes e_i'$ where $e_i$ is a basis. This embeds into $U(\mathfrak{g}) \otimes U(\mathfrak{g})$ and the multiplication map into $U(\mathfrak{g})$ gives the Casimir element.

(Oh good, I see that this is the same definition Akhil gives in the blog post darij linked to above. That makes me feel better.)