The question is motivated by a more broad perspective in another MO post and here, but here we would like to understand a specific case (our question potentially connects to / is motivated b Quantum Gravity).

**Question:** We would like to know whether there **exists a finite (here six) dimensional Lie algebra which provides a symmetric invariant bilinear non-degenerate metric?**
$$
\omega_{ab}\equiv
{\begin{pmatrix}
\Omega_{h_i, h_j} & a_{i}\;\mathbb{I}\\
a_{i}\;\mathbb{I} & 0
\end{pmatrix}}
$$
With $\Omega_{h_i, h_j}\equiv
{\begin{pmatrix}
2b_1 & c_{12}& c_{13} \\
c_{12} & 2b_2 & c_{23} \\
c_{13} & c_{23} & 2b_3
\end{pmatrix}}$, and
$a_{i}\mathbb{I}\equiv
{\begin{pmatrix}
a_1 &0 &0 \\
0 &a_2 &0 \\
0& 0& a_3
\end{pmatrix}}$
(Here $a_1,a_2,a_3$ all are positive numbers. $b_1,b_2,b_3,c_{12},c_{13},c_{23}$ are either positive numbers or zeros.) if yes, **what is this Lie algebra?** **The commutation rules?** Is it semi-simple Lie algebra or non-semi-simple? nilpotent or not, solvable or non-solvable Lie algebra?

extra INFO:

**NOTE:** A specific case we had worked out (but which unfortunately whose Lie algebra does not provide the desired $\omega_{ab}$) is the doubled extension of $\mathcal{G}$ by $\mathcal{h}$. The full vector space is $\mathcal{G}\oplus \mathcal{h}\oplus \mathcal{h}^*$. Here the subalgebra is spanned by the Abelian extension of $\mathcal{G}$ by $\mathcal{h}^*$: i.e. $\mathcal{G}\oplus \mathcal{h}^*$. The full algebra is the semi-direct product of $\mathcal{h}$ by this Abelian extension, $\mathcal{G}\oplus \mathcal{h}^*$. This is the doubled extension: $\mathcal{G}\oplus \mathcal{h}\oplus \mathcal{h}^*$. Also $\mathcal{h}^*$ is the dual to $\mathcal{h}$. This Lie algebra is studied in the literature before.
Use this non-semi-simple type Lie algebra $\mathcal{G}\oplus \mathcal{h}\oplus \mathcal{h}^*$, we can find a type of bilinear form as:
$$
\Omega_{ab} = {\begin{pmatrix}
\Omega_{g_i, g_j} &0 &0 \\
0 & \Omega_{h_i, h_j} & \mathbb{I}\\
0 & \mathbb{I} & 0
\end{pmatrix}}
$$
each blocks is spanned by generators of $\mathcal{G}$, $\mathcal{h}$, $\mathcal{h}^*$ respectively. **But $\Omega_{ab}$ this does not provide us the desired form as $\omega_{ab}$.** I suppose we find the Lie algebra by ${f_{ai}}^k\omega_{jk}+{f_{aj}}^k\omega_{ik}=0$ and here $\omega_{ik}=\omega_{ki}$ symmetric for this given $\omega_{ab}$. Say here $[X_i,X_j]={f_{ij}}^k X_k+\dots$.

- Lastly pardon that our understanding is more based on the theoretical physics side. We do have PhDs and do some research works. But our math accents are unfortunately not canonical/orthodox. Please feel free to comment/reply/correct/provide Ref for us. (if you vote down, please provide useful/rational comments on why/how to improve our post.) Thank you.