Is every three-dimensional pseudo-Riemannian group a space form? In an earlier post, I asked whether every three-dimensional Lie group equipped with a bi-invariant Riemannian metric has constant curvature. The answer is yes. Now I am curious to know what happens if we consider pseudo-Riemannian metrics, instead.
Let $M$ be a smooth manifold. A pseudo-Riemannian metric on $M$ is a smooth symmetric two-tensor field that is nondegenerate at each point of $M$.
Let $(G,g)$ be a Lie group equipped with a bi-invariant pseudo-Riemannian metric $g$. (Every compact and every semisimple Lie group admits such a metric.)
Question 1. If $\dim G =3$, does then $(G,g)$ have constant curvature?
Some comments: I know that every three-dimensional pseudo-Riemannian Einstein manifold has constant curvature. I also know that, if the Killing form $B$ of $G$ is nondegenerate, then $(G,B)$ is Einstein.
In fact, a stronger question that I would like to ask is the following.
Question 2. If $\dim G =3$, is then $(G,g)$ isometric to $(G,B)$?
 A: For simplicity, I will assume that your Lie group $G$ is simply-connected (since the question is local, the assumption is harmless). Then the biinvariant nondegenerate bilinear form on $G$ yields an ad-invariant
nondegenerate bilinear form on ${\mathfrak g}$, meaning:
$$
\langle ad_{X} Y, Z\rangle  + \langle Y, ad_{X} Z\rangle = 0$$
for all $X, Y, Z\in {\mathfrak g}$. Here is the key fact: Suppose that ${\mathfrak g}$ is a real 3-dimensional Lie algebra which admits an ad-invariant nondegenerate bilinear form. Then either  ${\mathfrak g}$ is abelian or it is simple. See for instance Proposition 2.3 in
V. del Barco, G. P. Ovando and F. Vittone, Naturally reductive pseudo-Riemannian Lie groups in low dimensions, arXiv 1211.0884.
Now:

*

*If $G$ is abelian, then it is isomorphic to ${\mathbb R}^{3}$ (with the additive group structure). Then, clearly, the invariant metric is flat (of some signature), hence, you get a space-form. I would not call such a metric the Killing form on $G$, of course.


*Suppose $G$ is simple (hence, either $SU(2)$ or the universal cover of $SL(2, {\mathbb R})$. Then the adjoint representation of $G$ is absolutely irreducible (i.e. remains irreducible after the complexification) and, hence, by the Schur Lemma, has unique, up to scaling, invariant bilinear form. Thus, it has to be the Killing form of $G$ (up to scaling). This is your stronger claim.
