# Casimir element invariance

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

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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 $V\cong V^*$.
Thank you very much. I'm sorry to ask a (probably stupid) question, but formally (once we've identified $V$ and $V^*$ the element $\sum v_i \times \phi_i$ lies in the tensor algebra of $L$, correct? And it's a homogeneus tensor of order 2.. that's why it's ok to say $\sum v_i \phi_i \in U(L)$? – claudia Jun 8 '11 at 18:42
@claudia It does lie in the tensor algebra. While it is homogenous of order $2$, we do not need this: there is a surjective map of rings $T(L)\to \mathcal{U}(L)$, and we simply identify the element with its image. – Aaron Jun 8 '11 at 18:50
The isomorphism map should be: $v\otimes \phi \mapsto \phi(\cdot)v$. – AlbertH Jun 8 '11 at 19:04