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.

  • $\begingroup$ 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. $\endgroup$ Dec 28, 2013 at 22:14
  • $\begingroup$ @archipelago, thanks so much for the comment. I recall that learning this sometime ago reading Lie algebra. $\endgroup$
    – miss-tery
    Dec 28, 2013 at 22:19
  • 4
    $\begingroup$ @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. $\endgroup$
    – YCor
    Dec 28, 2013 at 23:29
  • $\begingroup$ @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. $\endgroup$
    – YCor
    Dec 28, 2013 at 23:46
  • $\begingroup$ @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? $\endgroup$
    – miss-tery
    Dec 28, 2013 at 23:47

1 Answer 1


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.

  • 1
    $\begingroup$ 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)? $\endgroup$
    – miss-tery
    Dec 31, 2013 at 23:30
  • $\begingroup$ @ 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. $\endgroup$
    – miss-tery
    Dec 31, 2013 at 23:52
  • $\begingroup$ @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. $\endgroup$ Jan 1, 2014 at 9:53

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