The question is motivated by a more broad perspective in [another MO post](https://mathoverflow.net/questions/152997/finite-dimensional-lie-algebra-with-non-degenerate-invariant-bilinear-forms-om) and [here](https://mathoverflow.net/questions/152952/complete-classification-of-six-dimensional-non-semi-simple-lie-algebra?lq=1), but here we would like to understand a specific case (our question potentially connects to / is motivated b [Quantum Gravity](http://en.wikipedia.org/wiki/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.



The existence of a finite dimensional Lie algebra with a given symmetric invariant metric
If I understand your question correctly, then $\mathfrak{g} \ltimes \mathfrak{g}^\*\_{\textrm{abelian}}$, for $\mathfrak{g}$ any three-dimensional real Lie algebra, has an invariant inner product of this type. This is the semidirect product of $\mathfrak{g}$ with its coadjoint representation treated as an abelian Lie algebra.


In the generic case, you would have to take the matrix $\Omega = 0$, but for some particular Lie algebras you can allow nontrivial (positive-semidefinite) $\Omega$. For example, if you take $\mathfrak{g} = \mathfrak{su}(2)$, then you can let $\Omega$ be any (non-positive) multiple of the Killing form.


Since you mentioned the language of double extensions, this is just the double extension of the trivial zero-dimensional Lie algebra by $\mathfrak{g}$.


**Added**


Let me try to help with the digestion by writing the structure constants explicitly in terms of a basis. Choose a basis $(X\_i)$ for $\mathfrak{g}$ and let $(\alpha^i)$ denote the canonically dual basis for $\mathfrak{g}^\*$. Then the structure constants for $\mathfrak{g} \ltimes \mathfrak{g}^\*\_{\textrm{abelian}}$ relative to this choice of basis are given by:
$$
[X\_i, X\_j] = f\_{ij}{}^k X\_k \qquad\qquad [X\_i,\alpha^j] = - f\_{ik}{}^j \alpha^k \qquad\qquad [\alpha^i,\alpha^j] = 0
$$


This choice is such that the $a\_i$ are all equal. If you are prescribing the $a\_i$ ab initio then you don’t have to choose the canonical dual basis for $\mathfrak{g}^\*$: rescaling them should do the trick.


