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4 added 11 characters in body

Concerning your first question: it's really an issue about finite dimensional vector spaces.

(1). If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map

$$e: V^\ast \otimes V \to k$$

defined by $f\otimes v \mapsto f(v)$. Then $$e \in (V^{\ast} \otimes V)^\ast .$$ There is a canonical isomorphism $V \otimes V^\ast\to (V^\ast \otimes V)^\ast$ given by $$(v,f) \mapsto ((g,w) \mapsto g(v)f(w)) .$$ Using this isomorphism, we can regard $e$ corresponds to element $e'\in V\otimes V^\ast$. This is the Casimir element.

Remark: $e'$ is characterized by the following property: it is the unique vector in $V\otimes V^{\ast}$ such that for all vector spaces $W$, the map $$\hom(V,W) \to W\otimes V^\ast$$ given by $f\mapsto (f\otimes 1_{V^\ast})(e')$ is an isomorphism.

(2). More generally, if one is given a perfect pairing: $$e : U \otimes V \to k$$ (meaning that the adjoint $U \to V^\ast$ is an isomorphism), one gets by the same procedure an element $e'\in V\otimes U$.

(3). $V = \frak g$ is a semi-simple Lie algebra, then the killing form $V\otimes V \to k$ is perfect. So we get by (2) an the (Casimir) element $e' \in V\otimes V$.

3 added 379 characters in body

Concerning your first question: it's really an issue about finite dimensional vector spaces.

(1). If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map

$$e: V^\ast \otimes V \to k$$

defined by $f\otimes v \mapsto f(v)$. Then $$e \in (V^{\ast} \otimes V)^\ast .$$ There is a canonical isomorphism $V \otimes V^\ast\to (V^\ast \otimes V)^\ast$ given by $$(v,f) \mapsto ((g,w) \mapsto g(v)f(w)) .$$ Using this isomorphism, we can regard $e$ corresponds to element $e'\in V\otimes V^\ast$. This is the Casimir element.

Remark: $e'$ is characterized by the following property: it is the unique vector in $V\otimes V^{\ast}$ such that for all vector spaces $W$, the map $$\hom(V,W) \to W\otimes V^\ast$$ given by $f\mapsto (f\otimes 1_{V^\ast})(e')$ is an isomorphism.

(2). More generally, if one is given a perfect pairing: $$e : U \otimes V \to k$$ (meaning that the adjoint $U \to V^\ast$ is an isomorphism), one gets by the same procedure an element $e'\in V\otimes U$.

(3). $V = \frak g$ is a semi-simple Lie algebra, then the killing form $V\otimes V \to k$ is perfect. So we get by (2) an element $e' \in V\otimes V$.

2 added 182 characters in body; deleted 4 characters in body; deleted 8 characters in body

Concerning your first question: it's really an issue about finite dimensional vector spaces. If $V$ is a finite dimensional vector space over a field $k$, then one has an evaluation map

$$e: V^\ast \otimes V \to k$$

defined by $f\otimes v \mapsto f(v)$. Taking the linear dual of $e$, you get a mapThen $$e^\ast: k e \to in (V^\ast V^{\ast} \otimes V)^\ast .$$ There is a canonical isomorphism $V \otimes V^\ast\to (V^\ast \otimes V)^\ast$ given by $$(v,f) \mapsto ((g,w) \mapsto g(v)f(w)) .$$ If we combine Using this isomorphismwith the evaluation map, we obtain a can regard $e$ corresponds to element $e'\in V\otimes V^\ast$. This is the Casimir element.

Remark: $e'$ is characterized by the following property: it is the unique vector in $V\otimes V^{\ast}$ such that for all vector spaces $W$, the map $$k \hom(V,W) \to V\otimes V^* W\otimes V^\ast$$ which defines the Casimir element given by evaluating at $1\in k$f\mapsto (f\otimes 1_{V^\ast})(e')\$ is an isomorphism.

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