TITLE: Casimir element invariance
QUESTION [5 upvotes]: I'm sorry to bother but i'm having some problems in proving that, given a simple Lie Algebra L of finite dimension $n$ (equipped with the Killing form) and its enveloping universal algebra U(L), then the element (Casimir):
c = $\sum x_iy_i$ where $(x_i)_i$ is a basis and $(y_i)_i$ is its dual basis (with respect to the Killing form) doesn't depend on the choice of a particular basis.
The argument should be just related to linear algebra i guess. I tried to take another basis and the change-of-coordinates-matrix but i can't get to the solution.
Can someone help me? Am I missing something? Thank you

REPLY [5 votes]: The idea is that, given a finite dimensional vector space, $V$ we have a natural isomorphism $V\otimes V^*\cong \hom(V,V)$ given by $v\otimes \phi \mapsto \phi(\cdot)v$ (and extending linearly).  Thus, we have a unique element corresponding to the identity map, which we can give explicitly as follows.  Let $v_1, \ldots, v_n$ be a basis of $V$, and let $\phi_1, \ldots, \phi_n$ be the corresponding dual basis.  The identity corresponds to $\sum  v_i \otimes \phi_i$.  This is the casimir element, after we use the Killing form to establish an isomorphism $V\cong V^*$.