Is a polynomial group law on \$\mathbb{R}^n\$ automatically nilpotent? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:45:21Z http://mathoverflow.net/feeds/question/21682 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21682/is-a-polynomial-group-law-on-mathbbrn-automatically-nilpotent Is a polynomial group law on \$\mathbb{R}^n\$ automatically nilpotent? Gian Maria Dall'Ara 2010-04-17T17:59:17Z 2010-04-17T20:35:54Z <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#21689 Answer by Torsten Ekedahl for Is a polynomial group law on \$\mathbb{R}^n\$ automatically nilpotent? Torsten Ekedahl 2010-04-17T20:35:54Z 2010-04-17T20:35:54Z <p>This is true and is in "Michel Lazard: Sur la nilpotence de certains groupes algébriques, Comptes Rendus, vol 241, 1955, 1687--1689"</p>