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I would aim to know the complete classification of 6 dimensional non-semi simple Lie algebra (here the dimension stands for the generators; or the dimension $\leq 6$).

In this paper, in page 7, it stated that: "There is no complete classification of the six-dimensional real Lie algebras. However, all nilpotent six-dimensional Lie algebras are known." I also found this paper: J.Math.Phys. 17 (1976) 986, which lists nilpotent six-dimensional Lie algebras in Table III, p.991.

My question, again, is that: whether complete classification of 6 dimensional non-semi simple Lie algebra is known; for both real/complex non-semi simple Lie algebra of dimension 6? What is the most complete result? (In which paper/ref can I find a table of their Lie algebra?) Is that Table III a complete or incomplete for all real/complex non-semi simple Lie algebra of dimension 6?

Deep appreciation to whom concern and reply. (My question is motivated by a problem in topological field theory.)

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The table is only for nilpotent Lie algebras. Is every six-dimenisonal Lie algebra nilpotent or semisimple? (That's a rhetorical question. The answer is clearly no, right?) –  Pete L. Clark Dec 28 '13 at 7:18
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@cycles: the tag "central-simple algebras" is not relevant here (and probably neither is "representation theory"). –  YCor Dec 28 '13 at 12:33
    
This table is weird: several families depend on a parameter, but from several more modern treatments, it appears that there are finitely many 6-dimensional Lie algebras (over $\mathbf{R}$ or $\mathbf{C}$) (actually 30 over an arbitrary field of characteristic $\neq 2$, 34 over $\mathbf{R}$, see de Graff's paper: sciencedirect.com/science/article/pii/S0021869306005254, Journal of Algebra Volume 309, Issue 2, 15 March 2007, Pages 640–653) –  YCor Dec 28 '13 at 20:56
    
@Dear commenters: another refined question is posted by my colleague: mathoverflow.net/questions/153013/…; would you mind shed some light? Many thanks really. –  cycles Dec 29 '13 at 3:06
    
@Yves: In your last comment you want the word "nilpotent" somewhere, right? BTW I totally missed that some of the entries in the table have parameters! This makes my answer essentially worthless, so I have deleted it. However I amused by the fact (if it is a fact) that I was somehow actually right and there are only finitely many nilpotent Lie algebras. By the way: is it the case that there are only finitely many nilpotent Lie algebras of any fixed finite dimension (over $\mathbb{C}$, say)? –  Pete L. Clark Dec 29 '13 at 4:48
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2 Answers 2

I should add one reference, for the case of real numbers, namely the classification of all solvable real six-dimensional Lie algebras by Turkowsky:

P. Turkowski. Solvable Lie algebras of dimension six. J. Math. Phys. 31 (1990), 1344-1350.

As mentioned befiore, the classification of nilpotent Lie algebras of dimension $6$ has been done earlier by Morozow in $1958$, valid for all fields of characteristic zero. Recently this classification has been extended to arbitrary fields, by Serena Cicalo, Willem A de Graaf and Csaba Schneider here:http://arxiv.org/abs/1011.0361.

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Thanks so much Dietrich, let me read it. –  cycles Dec 28 '13 at 20:43
    
an officemate/colleague of mine, she helps me to ask another relevant question here: mathoverflow.net/questions/152997; this is what we were looking for. Perhaps you could shed some light? Many thanks. –  cycles Dec 28 '13 at 22:03
    
@Dear Dietrich: another refined focused question is posted by my colleague: mathoverflow.net/questions/153013/…; would you mind shed some light? Many thanks really. –  cycles Dec 29 '13 at 3:07
    
@Dietrich, well Cicalo, almost sister let's say... –  Nicola Ciccoli Dec 29 '13 at 15:35
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I am considering the complex case.

The list you wrote is complete for the NILPOTENT Lie algebras and was first obtained by Morozov in 1958 (paper in Russian). To obtain the calssification of ALL Lie algebras of dimension 6 one has to consider two more case:

  • non solvable

  • solvable but not nilpotent

As for the non solvable case there are only three possibilities:

1) ${\mathfrak sl}_2(\mathbb C)\oplus {\mathfrak sl}_2(\mathbb C)$

2) ${\mathfrak sl}_2(\mathbb C)\ltimes \mathbb C^3$ where the semidirect product is given by the adjoint action;

3) ${\mathfrak sl}_2(\mathbb C)\ltimes {\mathfrak n}_3(\mathbb C)$ where the semidirect product is by a sutable action on the Heisenberg -Lie algebra ${\mathfrak n}_3(\mathbb C)$ (3-dimensional, nilpotent).

Then you need to classify the solvable ones. This was achieved by Yamaguchi Mem. Fac. Sci. Kyushu Univ. Ser. A 35, pages 341-351 (1981). Some more informations are contained in a paper by Goze-Ancochea Bermudez (1985) where the variety of Lie algebras is studied as an algebraic variety for dimension up to 7. The relevant information connected to Clark's comment, is the number of irreducible components.

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There's one more non-solvable 6-dimensional Lie algebra: 4) $\mathbf{C}\times(\mathfrak{sl}_2(\mathbf{C})\ltimes\mathbf{C}^2)$. –  YCor Dec 28 '13 at 11:55
    
@Nicola Ciccoli, thanks for the reply. Please let me see. –  cycles Dec 28 '13 at 20:12
    
@Nicola and Yves, I am more interested in the real non-semi-simple Lie algebra which has non-degnerate invariant metric bilinear form $\Omega_{ab}$ (especially important if the Killing form degenerate). It is said that in 6 dimension the nilpotent one has only $A_{6,3}$ algebra admitting a non-degnerate invariant metric $\Omega_{ab}$. Also $\oplus^6 A_1$ and $A_{5,3}\oplus A_1$. Do you know more examples of 6 dim with non-degnerate invariant metric bilinear form $\Omega_{ab}$, please? –  cycles Dec 28 '13 at 20:37
    
Why not $\mathbb C^3 \times sl_2 (\mathbb C)$ –  Will Sawin Dec 28 '13 at 20:38
    
@Will: you're right: actually the missing are 4) $\mathbf{C}\times(\mathfrak{sl}_2(\mathbf{C})\ltimes\mathbf{C}^2)$; 5) $\mathbf{C}^3\times\mathfrak{sl}_2(\mathbf{C})$; 6) $\mathfrak{n}_3(\mathbf{C})\times\mathfrak{sl}_2(\mathbf{C})$, and unless I miss something, that's it. –  YCor Dec 28 '13 at 21:42
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