Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-26T00:45:21Zhttp://mathoverflow.net/feeds/question/21682http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/21682/is-a-polynomial-group-law-on-mathbbrn-automatically-nilpotentIs a polynomial group law on $\mathbb{R}^n$ automatically nilpotent?Gian Maria Dall'Ara2010-04-17T17:59:17Z2010-04-17T20:35:54Z
<p>I was told that a polynomial group law on (all of) $\mathbb{R}^n$ gives automatically a nilpotent (Lie, of course) group. </p>
<p>Is it true? Where can I find a proof?</p>
<p>A counterexample for open subsets of $\mathbb{R}^n$ is furnished by the halfplane with the $ax+b$ law. </p>
http://mathoverflow.net/questions/21682/is-a-polynomial-group-law-on-mathbbrn-automatically-nilpotent/21689#21689Answer by Torsten Ekedahl for Is a polynomial group law on $\mathbb{R}^n$ automatically nilpotent?Torsten Ekedahl2010-04-17T20:35:54Z2010-04-17T20:35:54Z<p>This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"</p>