Translation of "le nilradicalisé de g" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:45:40Z http://mathoverflow.net/feeds/question/13587 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13587/translation-of-le-nilradicalise-de-g Translation of "le nilradicalisé de g" Yemon Choi 2010-01-31T21:43:01Z 2010-01-31T23:49:41Z <p>I apologize for asking something that might well be found in a mathematical dictionary, but the similarity of the French word to an English one is frustrating my attempts to Google the answer (and the library is shut at time of typing). I suspect the answer should be obvious to those who, unlike me, know some basic Lie group/Lie algebra terminology.</p> <p>Some context: I am reading an old paper of Dixmier from 1969, which has the following construction/definition. Let $\mathfrak g$ be a Lie algebra (characteristic zero, finite-dimensional), let $\mathfrak n$ be its largest nilpotent ideal -- the <em>nilradical</em> -- and put ${\mathfrak h}=[{\mathfrak g},{\mathfrak g}]+{\mathfrak n}$. Dixmier calls ${\mathfrak h}$ "le nilradicalis&eacute; de ${\mathfrak g}$".</p> <p>Literal translation would surely be "the nilradicalised", but that sounds more like a mopey university indie band than a mathematical object. So what is the usual name for this object in English?</p> http://mathoverflow.net/questions/13587/translation-of-le-nilradicalise-de-g/13604#13604 Answer by Ben Webster for Translation of "le nilradicalisé de g" Ben Webster 2010-01-31T23:49:41Z 2010-01-31T23:49:41Z <p>I suspect pretty strongly that this is idiosyncratic terminology; I've never seen that subalgebra used, and the term has no google hits other than this post. </p>