Timeline for Commutators and brackets in nilpotent Lie algebras
Current License: CC BY-SA 4.0
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| Mar 18, 2021 at 16:27 | comment | added | 57Jimmy | @YCor Thanks for your comments. Still, it seems to me that we are basically reformulating the same statement in many different ways (saying that the central series of group and Lie algebra "are the same" could probably also be deduced from the fact that the group is $d$-step nilpotent iff the Lie algebra is $d$-step nilpotent, I think), but I still do not know how to prove any of them. I could really not find the result of Malcev you mentioned. Do you have a more precise reference? I would be happy to accept it as an answer. | |
| Mar 17, 2021 at 19:37 | comment | added | YCor | OK, actually all this is true for an arbitrary nilpotent Lie $\mathbf{Q}$-algebra (and this applies to your setting, since any Lie algebra over $K$ (extension of $\mathbf{Q}$) is a Lie algebra over $\mathbf{Q}$, possibly infinite-dimensional. That the central series is the same for the bracket and for the group is part of Malcev's results. Basically the proof is by induction, using that if the Lie algebra is $d$-step nilpotent, then the $d$-fold group and Lie commutators are equal. | |
| Mar 17, 2021 at 19:36 | comment | added | 57Jimmy | @Ycor As for the first comment: as I mentioned in the question, we are over an algebraically closed field $k$. And when I say "group" I mean "algebraic group over $k$" (as in the tags) (just to clarify). But you are right, I did not write down what I intended (corrected now). | |
| Mar 17, 2021 at 19:33 | history | edited | 57Jimmy | CC BY-SA 4.0 |
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| Mar 17, 2021 at 19:26 | history | edited | 57Jimmy | CC BY-SA 4.0 |
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| Mar 17, 2021 at 19:23 | comment | added | 57Jimmy | @YCor The second comment might be what I am looking for, do you have a reference in mind? | |
| Mar 17, 2021 at 17:08 | comment | added | YCor | I'm not sure what the question is, but it's true that all $d$-iterated commutators vanish iff all $d$-iterated brackets vanish. Indeed the lower central series are the same in the algebra and the group sense. | |
| Mar 17, 2021 at 17:06 | comment | added | YCor | The statement "bracket is trivial if and only if the commutator is trivial" is correct but the argument is not correct (unipotent is not a group property, and in any case it can be supposed $G$ is the group of real points of a real group). An argument is that a subspace of $\mathfrak{g}$ is closed under bracket iff it's stable under the group law, and it follows that both conditions are equivalent to $\mathbf{R}x+\mathbf{R}y$ being a Lie subalgebra. | |
| Mar 17, 2021 at 16:42 | history | asked | 57Jimmy | CC BY-SA 4.0 |