most recent 30 from http://mathoverflow.net 2013-05-21T11:10:34Z http://mathoverflow.net/feeds/question/1805 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1805/non-free-projective-modules-for-a-universal-enveloping-algebra Greg Muller 2009-10-22T04:18:51Z 2009-10-22T10:03:36Z

<p>Let g be a finite dimensional Lie algebra over k, and let U be its universal enveloping Lie algebra. Is there a left module M of U which is projective but not free? That is, is the Quillen-Suslin theorem still true for enveloping algebras?</p> <p>Quillen-Suslin says this there are no non-free projectives for S(g), the associated graded algebra of U. Thus, if the associated graded module of a projective is projective, then it is free (and so the original module was also free). Therefore, this question is equivalent to the question "Is the associated graded module of a projective U-module always projective?"</p> <p>My guess is no, because the Weyl algebra has non-free projectives, even though it's associated graded algebra is a polynomial algebra. However, the tricks I know that work for the Weyl algebra don't work for Lie algebras. I would love a simple example of a non-free projective U-module.</p>

http://mathoverflow.net/questions/1805/non-free-projective-modules-for-a-universal-enveloping-algebra/1860#1860 Simon Wadsley 2009-10-22T10:03:36Z 2009-10-22T10:03:36Z

<p>In <a href="http://www.numdam.org/numdam-bin/item?id=CM_1985__54_1_63_0" rel="nofollow">this paper</a> Stafford shows that whenever g is a finite-dimensional non-abelian Lie algebra the enveloping algebra has non-free but stably free (and therefore projective) right ideals. He also shows how to construct them. </p>