MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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?"

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.

share|cite|improve this question
up vote 3 down vote accepted

In this paper 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.