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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.
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ps. as mentioned in the maintext, if you vote down, please leave your name and provide useful/rational comments on why/how to improve our post. Thanks! –  mystery Dec 29 '13 at 20:16

1 Answer 1

up vote 4 down vote accepted

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.

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THANK you, Prof. José Figueroa-O'Farrill for the nice comments. Let me digest. ps. I enjoy reading many of your inspiring classic hep-th papers. :) –  mystery Dec 29 '13 at 7:19

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