I read Nolan Wallach's Geometric Invariant Theory book. In a chapter, the author talks about the group $G$ in Aut$(\mathfrak{g})$ corresponding to ad$(\mathfrak{g})$ (here $\mathfrak{g}$ is a complex Lie algebra). I do not understand what this group is. I know that given $X\in\mathfrak{g}$, ad$(X)$ is a Lie algebra map from $\mathfrak{g}$ to itself however this cannot be an automorphism since it always has a kernel (namely, ad($X)(X)$ is always $0$).
$\textbf{Question}:$ Given a complex Lie algebra $\mathfrak{g}$, what is the subgroup of Aut$(\mathfrak{g})$ corresponding to ad$(\mathfrak{g})$?
Best Answer
As an example of a non-closed embedding, consider the abelian Lie group ${\mathbb T}^2$ with Lie algebra ${\mathbb R}^2$. Any line through the origin with irrational slope defines a Lie subalgebra of ${\mathbb R}^2$ corresponding to a non-closed embedding ${\mathbb R}\hookrightarrow {\mathbb T}^2$ densely winding ${\mathbb R}$ around ${\mathbb T}^2$.
In your context, you encounter two instances of this:
The (closed) connected subgroup $\text{Aut}^{\circ}({\mathfrak g}) \leq \text{GL}({\mathfrak g})$ corresponds to the Lie subalgebra $\text{Lie}(\text{Aut}({\mathfrak g}))\cong \text{der}({\mathfrak g})$ of $\text{Lie}(\text{GL}({\mathfrak g}))\cong {\mathfrak g}{\mathfrak l}({\mathfrak g})$.
There is a unique Lie group embedding into $\text{Aut}^{\circ}({\mathfrak g})$ corresponding to the subalgebra $\text{ad}({\mathfrak g})\subseteq\text{der}({\mathfrak g})$ of $\text{Lie}(\text{Aut}({\mathfrak g}))$, and the domain of that embedding is the group the book is talking about. Note, however, that this group is not closed in general. See e.g. Exercise 18.1.2 in Hilbert-Neeb's "Structure and Geometry of Lie Groups".