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Using Engel's Theorem and Lie's Theorem, one can easily establish the following result:

Let $ \frak{g} $ be a finite-dimensional Lie algebra over an algebraically closed field $ \mathbb{F} $ of characteristic $ 0 $. If $ \frak{g} $ is solvable, then $ [{\frak{g}},{\frak{g}}] $ is nilpotent.

In order to apply the two theorems stated at the beginning, one must assume that (i) $ \mathbb{F} $ is algebraically closed, (ii) $ \mathbb{F} $ has characteristic $ 0 $, and (iii) $ \frak{g} $ is finite-dimensional. If we relax each of these three conditions in turn, are there certain well-known counterexamples?

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Can't one prove this for an arbitrary field of char 0 by extension of scalars? Certainly relaxing condition three will be hopeless. –  Daniel Pomerleano Dec 5 '12 at 7:50
    
Yes, I just realized that one can remove the 'algebraically closed' condition. Thank you for the comment anyway! –  Leonard Dec 5 '12 at 16:00
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@Leonard: Concerning the infinite dimensional case, I'm not sure whether examples like Salvo's exist "in nature". But there is an old book by Ian Stewart and his student Ralph Amayo which looks at many features parallel to the theory of infinite groups: Infinite-dimensional Lie algebras, Noordhoff International Publ., Leyden, The Netherlands, 1974. I wrote a review for the AMS Bulletin, freely available online at www.ams.org/journals/, but don't have the book itself handy. –  Jim Humphreys Dec 6 '12 at 16:40
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Dear Leonard, I just modified the example and put it in a "more natural" form (it is realized as a semidirect product of a Heisenberg algebra by an infinite dimensional left module). –  Salvatore Siciliano Dec 7 '12 at 15:49
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@ Leonard. What you wrote is correct. In any case, for the semidirect of a Lie algebra by a module you can look up, for instance, the book "Hilton-Stammbach: A course in homological algebra" (Chapter VII, Section 2)" –  Salvatore Siciliano Dec 8 '12 at 6:01
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2 Answers

up vote 12 down vote accepted

Condition (i) can be removed, as already observed by Daniel. In positive characteristic you can find a counterexample in the book "J. Humphreys: Introduction to Lie algebras and representation theory" (Chapter 2, Section 4, page 20, Exercise 4), so condition (ii) cannot be relaxed. Finally, let $H={\mathbb F}x+{\mathbb F}y+{\mathbb F}z$ be the Heisenberg algebra with basis $x,y,z$, where $z$ is central in $H$ and $[x,y]=z$. Consider the ring of polynomials ${\mathbb F}[t]$ as a left $H$-module with $x$ acting as $d/dt$, $y$ acting by multiplication by $t$, and $z$ acting as the identity. Now consider the split extension $L=H\ltimes {\mathbb F}[t]$. Then $L$ is solvable of derived length 3, but the derived subalgebra $[L,L]={\mathbb F}z+{\mathbb F}[t]$ is not nilpotent. Thus condition (iii) cannot be removed.

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The example is excellent. Thanks! It turns out that things are not as easy as they seemed at first sight. :) –  Leonard Dec 7 '12 at 19:41
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Update

The bug in the example in Salvo's answer above has finally been fixed! It is now entirely correct.

My original concern was that, due to the requirement that the Jacobi Identity be satisfied, an infinite-dimensional Lie algebra (or even a finite-dimensional one) cannot be defined simply by playing with relations among basis vectors (which is what Salvo and I did initially, thus leading to a number of false starts). Hence, one needs to either work with "natural" examples of infinite-dimensional Lie algebras (as Professor Humphreys has mentioned in his comment below the wording of my question) or to struggle really hard with the Jacobi Identity in order to generate "non-natural" examples.

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I didn't get around to checking the Jacobi identity in that example, but it's possible (in principle) to construct a Lie algebra by specifying brackets on a basis provided you do verify the Jacobi identity as well. This is seldom fun to do, which is why I was wondering about existence of a natural realization of such a Lie algebra. It doesn't seem to fit the standard situation in an affine Lie algebra, for instance. –  Jim Humphreys Dec 6 '12 at 23:33
    
I am wondering if it might be possible to construct a directed system of finite dimensional solvable Lie algebras such that the solvability carries on to their direct limit but where the nilpotency of their derived subalgebras does not. –  Tobias Kildetoft Dec 7 '12 at 13:47
    
I just quoted another counterexample –  Salvatore Siciliano Dec 7 '12 at 15:05
    
You can search the key word "Strongly solvable Lie algebra" +"David Tower". –  M. Shahryari Dec 7 '12 at 15:07
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@ Leonard: I do not entirely agree with your Update and, indeed, it is rather usual to present Lie algebras by structure contants (provided Jacobi identity is satified). After all, a this stage it would be an easy task to present my counterexample just specyfing the brackets on a basis... –  Salvatore Siciliano Dec 8 '12 at 15:21
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