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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$(\mathfrak{g})$?

Levent
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  • For semisimple $\mathfrak{g}$ (over the complex numbers), these are exactly the inner automorphisms. Compare https://math.stackexchange.com/q/3178915/96384 – Torsten Schoeneberg Jun 23 '19 at 21:02
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    For a finite-dimensional real (or complex) Lie algebra, this inner automorphism group of $\mathfrak{g}$ is the subgroup generated by the $\exp\mathrm{ad}(x)$ for $x\in\mathfrak{g}$. – YCor Jun 24 '19 at 08:27

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Theorem: For any Lie group $G$, the assignment $H\mapsto\text{Lie}(H)$ is a bijection between (equivalence classes of) embeddings (not necessarily closed) $H\hookrightarrow G$ of connected Lie groups $H$ into $G$ and Lie subalgebras ${\mathfrak h}\subseteq {\mathfrak g}:=\text{Lie}(G)$.

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:

  1. 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 $\text{Lie}(\text{GL}({\mathfrak g}))\cong {\mathfrak g}{\mathfrak l}({\mathfrak g})$.

  2. 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".

Hanno
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  • Okay, now I understand :) Thank you very much. I assume that there is no explicit description of $G$ in this case? – Levent Jun 23 '19 at 20:33
  • To clarify for myself : Basically $G$ is a subgroup of Aut$(\mathfrak{g})$ such that Lie$(G)$ is ad$(\mathfrak{g})$, right? – Levent Jun 23 '19 at 20:34
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    @Levent: Yes that's right, but one has to be a bit careful about what 'subgroup' means here, because it need not be a closed one. – Hanno Jul 01 '19 at 07:01