Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-Lie algebra, which is not necessarily free as a $k$-module. Let $S\left(L\right)$ denote the symmetric algebra of $L$ (over $k$), constructed as a quotient of the tensor algebra $T\left(L\right)$ of $L$. Let $U\left(L\right)$ denote the universal enveloping algebra of $L$ (over $k$), constructed as the quotient of the tensor algebra $T\left(L\right)$ modulo the two-sided ideal generated by all elements of the form $x\otimes y-y\otimes x-\left[x,y\right]$ with $x\in L$ and $y\in L$.

The canonical filtration of the tensor algebra $T\left(L\right)$ descends to a filtration of the universal enveloping algebra $U\left(L\right)$. The associated graded algebra of $U\left(L\right)$ - let us call it $GU\left(L\right)$ - is commutative (as is easily seen) and generated by the elements $\overline{\sigma\left(v\right)}\in U_1\left(L\right) / U_0\left(L\right)$ for $v\in L$ (where $\sigma$ denotes the map $L\to T\left(L\right)\to U\left(L\right)$). Thus, there exists a surjective $k$-algebra homomorphism $S\left(L\right)\to GU\left(L\right)$ which maps $v$ to $\overline{\sigma\left(v\right)}$ for every $v\in L$ (according to the universal property of the symmetric algebra).

One version of the Poincaré-Birkhoff-Witt theorem (abbreviated PBW theorem) says that under certain conditions, this homomorphism is actually an isomorphism. The Wikipedia page says that it is so if any of the following four cases holds (here I am quoting Wikipedia):

(1) $L$ is a flat $k$-module,

(2) $L$ is torsion-free as an abelian group,

(3) $L$ is a direct sum of cyclic modules (or all its localizations at prime ideals of $k$ have this property), or

(4) $k$ is a Dedekind domain.

A reference is given to a paper which I have no access to:

P.J. Higgins, Baer Invariants and the Birkhoff-Witt theorem, J. of Alg. 11, 469-482, (1969)

Most internet sources which prove PBW only prove it under the condition that $L$ is a free $k$-module. (Out of these proofs I consider Garrett's version most readable.) I am interested in a proof in case (2). I know that it is enough to consider the case when $k$ is a $\mathbb Q$-algebra.

The following two sources might give such a proof, if only I could understand them:

Source 1:

T. Ton-That, T.-D. Tran, Poincaré's proof of the so-called Birkhoff-Witt theorem Rev. Histoire Math., 5 (1999), pp. 249-284. As this is only formulated for $k$ a field, this needs some modifications, but that's not my main worry. I fail to understand this paragraph on pages 277-278:

"The first four chains are of the form

$U_1 = XH_1,\ U'_1 = H'_1Z,\ U_2 = YH_2,\ U'_2 = H'_2T$,

where each chain $H_1$, $H'_1$, $H_2$, $H'_2$ is a closed chain of degree $p - 1$; therefore by induction, each is the head of an identically zero regular sum. It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero, and therefore each of them can be considered as the head of an identically zero regular sum of degree $p$."

I don't understand the "It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero" part. This seems to be equivalent to $H_1 = H'_1 = H_2 = H'_2 = 0$, which I don't believe (the head of an identically zero regular sum isn't necessarily zero), but the authors are only using the weaker assertion that each of $U_1, U'_1, U_2, U'_2$ is the head of an identically zero regular sum of degree $p$ - which, however, is still far from being obvious to me.

Source 2:

P.M. Cohn, A remark on the Birkhoff-Witt theorem, J. London Math. Soc. 38, 197-203, (1963). This gives a rather strange argument, which doesn't really rhyme for me. Probably I don't understand it though. If anybody could write it up in modern terms I would be very thankful. (If you want to know where exactly I am stuck, it's "$1w=w\in M_n$" on page 202, but I fear that there are also some things I have not really grasped before that point.)

Is there any accessible (I'm not at a university campus right now, and I need this rather soon) source for a proof of PBW in case (2)?

UPDATE (5 JUNE 2011):

(a) I do have Higgins's paper now. It is a beautiful and well-written piece of mathematics; I can hardly believe my eyes that something that well-written has been published in a journal!

This paper does not explicitly prove PBW in the cases (1) and (2), but the things it proves (combined with Lazard's theorem that flat modules are direct limits of free modules) are enough to conclude that PBW holds in the cases (1) and (2).

(b) Emanuela Petracci has a proof of PBW in case (2), and she even claims that it is Cohn's proof. This is probably just modesty, as she shows substantially more. I am going to read the proof when I have more time.

(c) My question about Source 1 still stands, although I don't really need that proof now that I know better ones.

(d) I have read the Deligne-Morgan proof; it is beautiful (although I would hardly call it straightforward, Theo; it is algebraic acrobatics in its purest form).


I liked the Cohn paper, although it's been a while since I read it, and I read it more to understand the counterexample in the second half than the proof in the first half. A proof in the case when your $k$ is a commutative algebra over $\mathbb Q$ is available in:

  • Deligne, Pierre; Morgan, John W. Notes on supersymmetry (following Joseph Bernstein). Quantum fields and strings: a course for mathematicians, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597.

The proof is pretty much straightforward: when $k \supseteq \mathbb Q$, then the "symmetrization" map $S(L) \to T(L)$ that sends $l_1\cdot \dots \cdot l_n \mapsto \sum_{\sigma \in S_n} l_{\sigma(1)} \otimes \dots \otimes l_{\sigma(n)}$ exists, and you can check directly that the composition composition $S(L) \to T(L) \to U(L)$ is an isomorphism, is my memory is correct. Unfortunately, I don't have an electronic copy available.

(Deligne and Morgan also cite

  • Corwin, L.; Ne'eman, Y.; Sternberg, S. Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry). Rev. Modern Phys. 47 (1975), 573--603. MR0438925.

to claim that actually only you need $2,3$ to be invertible. This cann't be right, given the counterexample in the paper by Cohn.)

Actually, looking again at the Cohn paper, he argues in two steps:

  1. Reduce to the case when $k \supseteq \mathbb Q$.
  2. Prove it there.

The first step is simple enough: all you need to prove is that $L \to U(L)$ is injective, and since $L \to L\otimes_{\mathbb Z}\mathbb Q$ is injective (since $L$ is torsion-free), it's enough to check that $L\otimes_{\mathbb Z}\mathbb Q \to U(L) \otimes_{\mathbb Z}\mathbb Q$ is injective, and for this it's enough to work over $k \otimes_{\mathbb Z}\mathbb Q$.

Ok, so the second step is the one that Deligne and Morgan do. And in Cohn's paper he actually spells out the "you can check directly" in my glib write-up above. Because, of course, it takes some work. One way to do this, and Cohn does roughly this, is by describing the inverse to the symmetrization map $S(L) \to U(L)$ in explicit enough detail that he can show that the inverse map makes sense over any $k$ --- Cohn does this by studying the case when $L$ is free.

Actually, Cohn does the following. He builds the map $S(L) \to U(L)$, which is a filtered map, and hence descends to a map $GS(L) = S(L) \to GU(L)$. He supposes that on associated graded the map is not an isomorphism, so that means that there is $w \in S(L)$ homogeneous of degree $n$ so that its image in $U(L)$ is in the $(n-1)$st filtration. Then if $M_{\leq \bullet}$ is a filtered $L$ module for which $L$ acts by moving up at most one degree, $w : M_{\leq \bullet} \to M_{\leq \bullet + n - 1}$, because we wrote the image of $w$ in $U(L)$ as a product of $n-1$-and-fewer things. But on the other hand, $S(L) \to U(L)$ is an $L$-module homomorphism, and Cohn proves that the image of $w$ in $U(L)$ sends $1\in S(L)$ to $w\in S(L)$, by showing this when $L$ is free.

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    $\begingroup$ Thanks a lot! I wished Cohn would have explained this rather than leaving it to reviewers and MathOverflowers. Also, thanks for the references to the superalgebra papers, for my next question would have been how far we can go with PBW for super-Lie algebras. -- As concerning the counterexample, I don't understand how Cohn sees that "the remaining terms are unaffected by (15) and clearly also by (16)". This certainly is very suggestive, but I don't see a clear-cut proof in Cohn's paper. $\endgroup$ Apr 17 '11 at 11:12
  • $\begingroup$ I should mention that by no means do I claim to have explained here the meat of Cohn's main step, which is to show that for $w\in S(L)$, its image in $U(L)$ acts on $S(L)$ to send $1$ to $w$. He asserts this, but I didn't reread the details. $\endgroup$ Apr 18 '11 at 3:13
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    $\begingroup$ As for the counterexample, it's quite nice: define over $k=\mathbb F_p$ a Lie series $f(x,y) = (x+y)^p-x^p-y^p$ (check it's a Lie series); then check that over $R=k[a,b,c]/(a^p=b^p=c^p=0)$ the Lie algebra $\langle x,y,z|ax+by+cz=0\rangle$ has $f(ax,by)\neq 0$; it's clear that $f(ax,by)$ acts by $0$ on any module, since it acts by $(ax+by)^p-(ax)^p-(by)^p=c^pz^p-a^px^p-b^py^p=0$, and so if you can show that $f(ax,by)\neq 0$ in the Lie algebra, you have shown that $L\to U(L)$ is not injective, and hence $S(L) \to U(L)$ is not an isomorphism. Cohn cites Zassenhaus for this later fact. $\endgroup$ Apr 18 '11 at 3:17
  • $\begingroup$ But I think you can do it by hand. I'll do $p=3$: $f_3(x,y)=(x+y)^3-x^3-y^3=xxy+xyx+yxx+xyy+yxy+yyx=xxy-xyx-xyx+yxx+\dots = x[x,y]-[x,y]x + \dots = [x,[x,y]]+[y,[y,x]]$. To check that $f(ax,by)\neq 0$, you can for example give everything gradings and then work out the low-graded piece of the multiplication table; eventually nothing more can vanish, because of the gradings. $\endgroup$ Apr 18 '11 at 3:32
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    $\begingroup$ Anybody having a problem with the Cohn's counterexample is apparently in a good company: last summer I have heard from Efim Zelmanov, Leonid Bokut' and Yongshan Chen that all they have trouble with it and doubt that Cohn had a valid proof. Yongshan Chen devised a proof for some (small?) values of $p$ using the so-called Groebner-Shirshov bases technique. $\endgroup$ May 9 '11 at 5:52

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