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I was told that a polynomial group law on (all of) $\mathbb{R}^n$ gives automatically a nilpotent (Lie, of course) group.

Is it true? Where can I find a proof?

A counterexample for open subsets of $\mathbb{R}^n$ is furnished by the halfplane with the $ax+b$ law.

flag	asked Apr 17 2010 at 17:59 Gian Maria Dall'Ara 1,408●6●19

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Torsten, make it an answer (not just a comment). – Sammy Black Apr 17 2010 at 19:13

Thank you, Torsten! – Gian Maria Dall'Ara Apr 17 2010 at 19:49

Moved a comment to an answer as per instructions. – Torsten Ekedahl Apr 17 2010 at 20:36

Torsten Ekedahl · Accepted Answer · 2010-04-17 20:35:54Z UTC

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This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"

answered Apr 17 2010 at 20:35

Torsten Ekedahl
17.9k●1●42●68

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This short paper is apparently not available online, but a version of Lazard's theorem is also written down in the 1970 book by Demazure-Gabriel, *Groupes algebriques, I*: see IV, section 4, 4.1. – Jim Humphreys Apr 17 2010 at 20:53

Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent?

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