4
$\begingroup$

Let $g$ be a Lie ring (Lie algebra over $\mathbb Z$), and let $U(g)$ and $S(g)$ denote the universal enveloping and symmetric algebra of $g$. The Poincaré-Birkhoff-Witt theorem (in the form proved by Lazard, see "Sur les algebres enveloppantes universelles de certaines algebres de Lie, M Lazard - Publ. Sci. Univ. Alger. Ser. A, 1954) yields a ring isomorphism between $S(g)$ and an associated graded of $U(g)$.

I can prove that $S(g)$ and $U(g)$ are isomorphic as $\mathbb Z$-modules; this essentially follows from the proof by Lazard. Is this already known?

Improve this question
$\endgroup$
3
  • $\begingroup$ Just remark isomorphism fails for char = p mathoverflow.net/questions/99018/… $\endgroup$
    – Alexander Chervov
    Commented Aug 6, 2012 at 10:43
  • 2
    $\begingroup$ Doc, you make my head spin. Surely, PBW holds over any field. $\endgroup$
    – Bugs Bunny
    Commented Aug 7, 2012 at 12:53
  • 3
    $\begingroup$ Clarification: I consider algebras over $\mathbb Z$, in which case Lazard shows that there is a ring isomorphism between $gr(U(g))$ and $S(g)$. Indeed $\mathbb Z$ is a PID. The typical "counterexamples" to PBW involve algebras over non-domains such as $\mathbb F_p[a,b,c]/(a^p,b^p,c^p)$. My question is about isomorphism of $S(g)$ and $U(g)$ as abelian groups, not as rings. $\endgroup$
    – grok
    Commented Aug 8, 2012 at 11:29

2 Answers 2

Reset to default
4
$\begingroup$

Earlier Lazard's result (Sur les algèbres enveloppantes universelles de certaines algèbres de Lie. C. R. Acad. Sci. Paris 234, (1952). 788–791. ) does it for Lie rings over PID.

For general Lie rings the first counterexample was constructed by Shirshov (On the representation of Lie rings as associative rings. Uspehi Matem. Nauk (N.S.) 8, (1953). no. 5(57), 173–175.)

Improve this answer
$\endgroup$
0
$\begingroup$

It is known if $g$ is free as a $\mathbb{Z}$-module, and fails otherwise, see e.g. P.M.Cohn, A Remark on the Birkhoff-Witt Theorem J. London Math. Soc. (1963) s1-38(1): 197-203

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I don't think this is claimed anywhere in Cohn. $\endgroup$
    – darij grinberg
    Commented Aug 6, 2012 at 15:02
  • 1
    $\begingroup$ It fails in the category of "algebras over $\Phi$", for some rings $\Phi$ that contain zero divisors; however, this is not exactly the question I wanted to ask; I added a clarification. $\endgroup$
    – grok
    Commented Aug 8, 2012 at 11:30
  • 1
    $\begingroup$ @Dotsenko: yes indeed. Most people seek an algebra isomorphism, so replace $U(g)$ by an associated graded; I'm interested in less, but don't want to pass to an associated graded. My question only makes sense for Lie rings with $\mathbb Z$ or $\mathbb Z/q$ additive factors with $q$ non-prime. $\endgroup$
    – grok
    Commented Aug 9, 2012 at 12:03
  • 1
    $\begingroup$ Does somebody have the paper by by Lazard ? (I cannot find it) Sur les algebres enveloppantes universelles de certaines algebres de Lie, M Lazard - Publ. Sci. Univ. Alger. Ser. A, 1954 $\endgroup$
    – Duchamp Gérard H. E.
    Commented Feb 14, 2013 at 14:50
  • 1
    $\begingroup$ Paper by Cohn: justpasha.org/math/links/files/c/cohn/197.pdf $\endgroup$
    – Pasha Zusmanovich
    Commented May 10, 2013 at 15:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .