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Let $\mathfrak{g}$ be a finite-dimensional abelian Lie algebra over a field $k$ of characteristic zero. I was wondering how many constructions do we have to produce a non-abelian Lie algebra $\overline{\mathfrak{g}}$ out of $\mathfrak{g}$ i.e. using just Lie algebra $\mathfrak{g}$ itself, no central extension/semi-direct product.

So far, I have just two constructions:

  1. Consider the Lie algebra of derivations $Der(\mathfrak{g})$. In this case, we are going to get the set of linear maps on $\mathfrak{g}$.

  2. Consider $\overline{\mathfrak{g}}=\mathfrak{g}\oplus\wedge^2\mathfrak{g}$ with a bracket $[x,y]=x\wedge y$ if $x,y\in \mathfrak{g}$ and $[x,y\wedge z]=[x\wedge y,z]=[x\wedge y, z\wedge w]=0$. Is there something specific about this particular Lie algebra?

Are there any other way to produce a non-abelian Lie algebra?

eightc
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    Of course, an abelian Lie algebra is indistinguishable from a vector space, except for choice of font, maybe!?!?! :), and vice-versa, so you're really asking about making non-abelian Lie algebras from vector spaces? – paul garrett Dec 13 '21 at 20:25
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    @paulgarrett, yeah, that is correct. Do you know anything about this? Then, I would be probably interested what is going on geometrically (if that's possible) as I know that I can produce a Lie group for a given Lie algebra. – eightc Dec 13 '21 at 20:31
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    Just so we're clear, you are basically asking for a full classification of all finite dimensional Lie algebras over a characteristic 0 field. A little search should show you that this is easy for dimension $\le 2$, but beyond that it's basically hopeless. If you restrict to semisimple Lie algebras, classifications over global and local fields are available but intricate. I don't think there's a full classification even in this case for more general fields. – Torsten Schoeneberg Dec 13 '21 at 22:02
  • @TorstenSchoeneberg dimension 3,4 is not that hopeless. In dimension 3 for instance we have the simple ones (classified by 3-dimensional quadratic forms modulo scalar multiplication) and the non-simple ones are semidirect products $K\ltimes K^2$, classified by $2\times 2$-matrices modulo conjugation and scalar multiplication. – YCor Dec 14 '21 at 16:41
  • @TorstenSchoeneberg In characteristic zero, for classical Lie algebra (except maybe $\mathfrak{so}_8$) there's a classification of forms over all fields, essentially modulo the classification of something else: extensions of the field, central division algebra over these extensions, quadratic forms over these extensions, and hermitian forms over quadratic extensions over these extensions. – YCor Dec 14 '21 at 16:43
  • To OP: Construction (2) is a particular case of a central extension, so I'm not sure what you mean with imposing "no central extension". – YCor Dec 14 '21 at 16:45
  • @YCor, oh, that was a lack of understanding. Now, I can see how we can obtain all varieties of Lie algebras starting just with a vector space $V$. Also, do you have any good papers/books in mind that you would recommend to read? – eightc Dec 14 '21 at 19:20
  • @YCor (re classical Lie algebras) I am kind of aware of that, a whole chapter in my thesis is about that. I still claim that these things are not completely clear over just any field of characteristic $0$. In particular, I am not aware that there is a comprehensive classification of anisotropic quadratic / hermitian forms over just any characteristic zero field. I for my part would be unable to compute the Brauer group of an arbitrary characteristic $0$ field either. In any case, my main point was that OP seemed unaware how vast the theory is even in this very restricted case. – Torsten Schoeneberg Dec 14 '21 at 23:19
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    @TorstenSchoeneberg so, well, I think we agree. At the Lie algebra level things are solved and boil down to classification of arithmetic nature, for which the level of knowledge is quite sensitive on the field. (In a more extreme way: we believe that abelian simple groups are classified, because these are cyclic groups of prime order. But we don't entirely understand primes...) – YCor Dec 15 '21 at 09:24

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As other people have said, you might as well say how can you build Lie algebras from a vector space $V$. There are bewilderingly many in general. If we restrict our attention to semisimple ones then we can easily construct $\mathfrak{sl}(V)$ as the tracefree linear maps and $\mathfrak{so}(V)$, $\mathfrak{sp}(V)$ (the second only if $V$ is even dimensional) as subspaces of $\mathfrak{sl}(V)$ which are skew for an appropriate symmetric or symplectic form on $V$. Note these two are isomorphic to $\bigwedge^2 V $, $S^2 V$.

We can also make many more by doing the same to $V \oplus V$, $V \otimes V$ and so on, or by sums and products of our resulting Lie algebras. You can do tonnes of different things here, especially if we're allowed to split $V$ up etc. If $V$ is 1-dimensional we can reconstruct the entirety of the classification of semisimple Lie algebras in this way.

Callum
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  • Yes, now I see. So, we can basically can start let's say with any finite dimensional vector space $V$ over $\mathbb{C}$. Consider an element $B$ from its $\wedge^2 V$ and take a Lie algebra $\mathfrak{g}$ that consists of linear transformation that preserves $B$. Got it. Thank you! – eightc Dec 14 '21 at 19:23
  • That's not quite what I'm suggesting in general. Your construction, assuming $V$ has even dimension, $B$ is nondegenerate and by "preserves" you mean "is skew for" produces the symplectic Lie algebra on $\mathfrak{g}$. Note we should also really choose $B$ in the dual space although it doesn't matter too much. But the orthogonal Lie algebra comes from a chose of symmetric bilinear form instead and the exceptional ones require a bit more thought – Callum Dec 17 '21 at 10:52