# Finite dimensional Lie algebra with non-degenerate invariant bilinear forms $\Omega_{ab}$

Firstly, my apology to MO experts that I am in a more science/physics background (a PhD). So please feel free refine/modify/comment my language if I have different math accents than yours. From reading this MO post: complete-classification-of-six-dimensional-non-semi-simple-lie-algebra, I just learned many useful comments and papers in the literature.

From what I had learned in the Reference of this post, we can (strictly?) organize the classes of finite dimensional Lie algebra by:

I. semi-simple. (Killing form is non-degenerate)

II. non-semi-simple (Killing form is degenerate):

$\bullet$ non-solvable.

$\bullet$ solvable and nilpotent.

$\bullet$ solvable but not nilpotent.

QUESTION: Here I am simply interested in focusing on:

"What are the list of 6 dimensional Lie algebra and 8 dimensional Lie algebra, which allow symmetric non-degenerate invariant bilinear forms $\Omega_{ab}$?" (hopefully the list can be as complete as possible, but a partial list is welcome.) If there is a list of corresponding $\Omega_{ab}$ metric, it will be the best.

I am mostly interested in non-semi-simple case, and in real Lie algebra more than the complex Lie algebra (of course, if providing examples of complex Lie algebra will also be nice).

eg: So far I know only one example in 6-dimension is the nilpotent Lie algebra $A_{6,3}$ with symmetric nondegenerate $\Omega_{ab}$:

The algebra is $$[e_1,e_2]=e_6,\;\;[e_1,e_3]=e_4,\;\;[e_2,e_3]=e_5,$$ with other commutators are zeros. One can find the nondegenerate $\Omega_{ab}$ to be: $$\Omega_{ab}={\begin{pmatrix} q_1 & 0&0&0 &q_2&0 \\ 0& q_3 & 0& -q_2&0 &0\\ 0& 0& q_4 & 0& 0& q_2\\ 0& -q_2 & 0& 0& 0 &0\\ q_2 &0& 0& 0& 0 &0\\ 0& 0& q_2 & 0& 0 &0 \end{pmatrix}}$$

What are other examples in 6 dimensional Lie algebra and 8 dimensional Lie algebra? (for those semi-simple Lie algebra, I suppose we can use Killing form = $\Omega_{ab}=-{f_{ak}}^l{f_{bl}}^k$. What are the complete lists of semi-simple and non-semi-simple ones of 6 and 8 dimensions?)

Papers/Ref are mostly welcome. (This question is well-motivated by constructing a type of Wess-Zumino-Witten model). Thank you for the concern.

• A comment on the first part of your question: Every finite dimensional real Lie algebra is a semi-direct product of a solvable and a semi-simple Lie algebra (Levi-Decomposition), so the classifiction of finite dimensional Lie algebras splits up in the (distinct) two cases of solvable and semi-simple lie algebras. In the class of the solvable lie algebras, you have the proper tower of classes: solvable contains nilpotent contains abelian. Dec 28, 2013 at 22:14
• @archipelago, thanks so much for the comment. I recall that learning this sometime ago reading Lie algebra. Dec 28, 2013 at 22:19
• @archipelago: the classification does not split: you still have to classify the actions of semisimple Lie algebras on solvable Lie algebras. Also if you have a given semidirect product, whether it has a nondegenerate invariant bilinear form does not reduce to looking the radical and the semisimple factor. For instance, $sl_2(C)\ltimes C^2$ has no invariant quadratic form, while $sl_2(C)\ltimes C^3$ (adjoint representation) has one.
– YCor
Dec 28, 2013 at 23:29
• @Dietrich: I think you mean indecomposable. Otherwise you also have the abelian one, as well as the direct product of the indecomposable 5-dimensional one by the 1-dimensional one.
– YCor
Dec 28, 2013 at 23:46
• @Dietrich, Thanks for comments. In 6-dimension, it should be the $A_{6,3}$ example I gave. Are there others 6-dimension (any, both semi-simple and non-semi-simple) with a symmetric non-degenerate invariant bilinear form? Dec 28, 2013 at 23:47

There are several classification lists of solvable and nilpotent quadratic Lie algebras, i.e., having a symmetric, invariant non-degenerate bilinear form. For the classification of nilpotent quadratic Lie algebras of dimension $n\le 7$ over the field of real and complex numbers, see

Piu P., Goze M., Gruppi e Algebre di Lie, appunti per un seminario, Universita degli studi di Cagliari, Dipartimento di Mathematica, 1991.

Gr. Tsagas and P. Nerantzi: Symmetric invariant non-degenerate bilinear forms on nilpotent Lie algebras - see here.

It turns out, that in dimension $6$ there is just one indecomposable nilpotent quadratic algebra, and a decomposable arising from the $5$-dimensional and $1$-dimensional quadratic algebra. In dimension $8$ there are many (two-step nilpotent) examples, but I think, no complete classification.

For the classification of solvable ones in dimension $n\le 6$ see

Tien Dat Pham, Anh vu Le, Minh thanh Duong: Solvable quadratic Lie algebras in low dimension.

In general, see double extension construction and work by Medina and Revoy, Favre and Santaroubane, and many others.

• Thanks Dietrich for the nice answer. This is what I looked for. Sorry I am a physicist with poor math. Could you fill me in: you state "solvable and nilpotent quadratic Lie algebras" <-> "having a symmetric, invariant non-degenerate bilinear form". Is this an iff statement? But how about simple Lie algebra (not solvable) which DOES provide non-degenerate bilinear form(i.e.(Killing)? Dec 31, 2013 at 23:30
• @ Dietrich, what is the equivalent statement for "having a symmetric, invariant non-degenerate bilinear form" for Lie algebra? Is that "solvable and nilpotent quadratic Lie algebras" or something more than that? Can it be real or complex Lie algebra? Many thanks. Dec 31, 2013 at 23:52
• @mystery, it is more than that. You said that you are mostly interested in the non-semisimple case, and then it is natural to consider the solvable (reps. nilpotent) case. Jan 1, 2014 at 9:53