Lie algebra of the invertible morphisms of a Lie algebra.

Question

I am confused with some facts of Lie groups and Lie algrebras.

1. If a have a Lie algebra $\mathfrak{g}$ and take its set of invertible morphisms $GL(\mathfrak{g})$ it is clear to me that this is a group. However I don't see why it is a Lie group. What is the differentiable structure on $GL(\mathfrak{g})$?

2. My next confusion is: why can the Lie algebra of $GL(\mathfrak{g})$ be identified with the set of ALL morphisms $\mathfrak{g}\to \mathfrak{g}$?

Answer 1

I suppose the dimension of ${\cal g}$ is finite. Let $GLin({\cal g})$ be the group of linear maps of ${\cal g}$ it is a Lie group. It is an open subset of $M({\cal g})$ the set of linear endomorphisms of ${\cal g}$. $GL({\cal g})$ is a closed subgroup of $GLin({\cal g})$ thus it is a Lie group.

To determine the Lie algebra of $GL({\cal g})$, you have to find matrices of $M({\cal g})$ whose exponential are in $GL({\cal g})$. If $A$ is such a matrix, by writing the fact that ${d\over{dt}}(exp(tA)[x,y]-[exp(tA)x,exp(tA)y])=0$ you obtain that the Lie algebra of $GL({\cal g})$ is the set of derivations of ${\cal g}$.