Embedding of the adjoint group into $\mathrm{GL}(\mathfrak{g})$

Question

Given a connected Lie group $G$ with corresponding Lie algebra $\mathfrak{g}$, the adjoint representation/action $\mathrm{Ad} : G \to \mathrm{GL}(\mathfrak{g})$ induces a Lie group homomorphism. It's well-known that $\ker{\mathrm{Ad}} = Z(G)$, and that if $G$ is semisimple, then $\mathrm{Ad}(G)$ is the identity component of $\mathrm{Aut}(\mathfrak{g})$, and is in particular closed, which gives us that $G/Z(G)$ is an embedded submanifold.

I would like to know if this is true in general. That is, without assuming $G$ is semisimple, does $\mathrm{Ad}$ gives us the (manifold) embedding $G/Z(G) \hookrightarrow \mathrm{GL}(\mathfrak{g})$ (or, in other words, is $\mathrm{Ad}(G)$ a closed submanifold of $\mathrm{GL}(\mathfrak{g})$)?

Answer 1

I think the answer is no. Fix an irrational number $r_0\in \mathbf{R}\setminus\mathbf{Q}$. Let $G = \mathbf{R}\ltimes \mathbf{C}^2$ be the semidirect product, where $\mathbf{R}$ acts on $\mathbf{C}^2$ by $$ t\cdot(z_1, z_2) = (e^{2\pi t\sqrt{-1}}z_1, e^{2\pi r_0t\sqrt{-1}}z_2). $$

Then $\varphi:G / Z(G)\to \mathrm{Aut}(\mathfrak{g})$ is not an embedding of submanifold.

It's easy to see that $Z(G) = 1$. Suppose in contrary that $\varphi$ is an embedding. Then $\varphi$ induces an homeomorphism of $G$ onto $\mathrm{im}\; \varphi$. In particular, $\varphi(\mathbf{R})$ is a closed submanifold. However, it is easy to see that $$ \varphi(\mathbf{R})\subset \mathbf{S}^1\times \mathbf{S}^1 = \begin{pmatrix} 1 & & \\ & \mathbf{S}^1 & \\ & & \mathbf{S}^1 \\ \end{pmatrix}\subset \mathrm{GL}(\mathfrak{g}). $$ This reduces to the classical example of immersion of $\mathbf{R}$ into the flat torus, which is not a submanifold, contradiction. Hence $\varphi$ is not an embedding of submanifold.