# identifying dual of lie algebra of general linear groups

Is there any reference for the following fact? I am looking for a nice and simple proof.

Assume that $G=GL(n,C)$, the group of invertible $n\times n$ matrices with complex entries. Why can the dual of the Lie algebra of $G$ can be identified with the vector space of all $n\times n$ matrices, where $G$ acts by conjugation?

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## 1 Answer

Use $\text{trace}_{\mathbb C}(X\alpha)$ as duality pairing between $X$ in the Lie algebra and $\alpha$ in the dual; both matrices.

Edit: More details. It is a non degenerate duality, compatible with conjugation, since $\text{trace}(gXg^{-1}.\alpha)=\text{trace}(X.g^{-1}\alpha g)$. Note that $(g,\alpha)\mapsto g^{-1}\alpha g$ is the coadjoint action.

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@Peter MICHOR, Do you have any reference? –  Hassan Jolany Jun 10 '13 at 17:09
I would be happy if you give a little bit with detail –  Hassan Jolany Jun 10 '13 at 17:11
@Hassan: What Peter is emphasizing is the existence of an invariant nondegenerate form on the Lie algebra (even though the Lie algebra in this case isn't semisimple and therefore doesn't have a nondegenerate Killing form). –  Jim Humphreys Jun 10 '13 at 20:07