Suppose $u(n)$ is the Lie algebra of the unitary group $U(n)$, why the dual vector space of $u(n)$ can be identified with $\sqrt{-1}u(n)$?
Joseph Malkoun
Hi Pan,
$u(n)$ is a real Lie algebra, and in particular a real vector space. Using a non-degenerate symmetric bilinear form on $u(n)$, you can identify $u(n)$ with its dual vector space. The $\sqrt{-1}$ is not that important in a sense, and probably comes from using a non-degenerate pairing between skew-hermitian and hermitian matrices (instead of a non-degenerate symmetric bilinear form). $u(n)$ is the space of skew-hermitian $n$ by $n$ matrices, and $\sqrt{-1} u(n)$ is the space of hermitian $n$ by $n$ matrices by the way.
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