Let the set $\mathcal{g}$ be the Lie algebra of a matrix Lie group $G$. Then my book asserts that $\mathcal{g}$ is a real vector space because it's closed under real scalar multiplication. My question why is it not closed under complex scalar multiplication?
If $X \in \mathcal{g}$ and the corresponding exponentiated matrix $e^{tX}\in G$ ($t \in \mathbb{R}$), I don't see why a multiplication with a complex scalar $C$ of $X$, namely $CX$, will make $CX\notin \mathcal{g}$.
If $G$ is a real Lie group, so in particular it is a real manifold, then its tangent spaces will be real vector spaces. In particular its Lie algebra will be a real vector space. If, on the other hand, $G$ is a complex Lie group, then its Lie algebra will be a complex vector space.