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Let $\mathfrak{g}$ be a finite-dimensional Lie algebra with an ordered basis $x_1 < x_2 < ... < x_n$.

We define the universal enveloping algebra $U(\mathfrak{g})$ of $\mathfrak{g}$ to be the free noncommutative algebra $k\langle x_1,...,x_n\rangle$ modulo the relations $(x_ix_j - x_jx_i = [x_i,x_j])$.

The Poincaré–Birkhoff–Witt theorem states that $U(\mathfrak{g})$ has a basis consisting of lexicographically ordered monomials i.e. monomials of the form $x_1^{e_1}x_2^{e_2}...x_n^{e_n}$. Checking that this basis spans $U(\mathfrak{g})$ is trivial, so the work lies in showing that these monomials are linearly independent.

One standard proof of PBW is to construct a $\mathfrak{g}$-action on the commutative polynomial ring $k[y_1,...,y_n]$ by setting $x_1^{e_1}x_2^{e_2}...x_n^{e_n}\cdot 1 = y_1^{e_1}y_2^{e_2}...y_n^{e_n}$ and verify algebraically that this gives rise to a well-defined representation of $\mathfrak{g}$. Details can be found in Dixmier's book on enveloping algebras.

What other proofs of PBW are there out there?

Are there nice reformulations of the above proof from a different perspective, such as one that emphasizes the universal property of $U(\mathfrak{g})$?

However, I would be especially interested in learning about proofs which are not just repackaged versions of the same algebraic manipulations used in the above proof (for example, geometric proofs which appeal to some property of $U(\mathfrak{g})$ as differential operators, etc.). If we allow ourselves more tools than just plain algebra, what other proofs of PBW can we get?

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I don't know if this counts as a reformulation, but PBW can be interpreted as the statement that the associated graded of $U(\mathfrak{g})$ can be naturally identified with $S(\mathfrak{g})$; one should interpret the former as a noncommutative algebra of operators on the quantum system whose classical limit is the Poisson manifold $\mathfrak{g}^{\ast}$. – Qiaochu Yuan Feb 3 '12 at 6:03
Since you are asking for a list then I guess this should be community wiki. – DamienC Feb 3 '12 at 8:25
A small comment - the shortest and in my opinion nicest way to organise the algebraic manipulations you mention is to use Diamond Lemma, - and in this way can be generalised wonderfully, see e.g. – Vladimir Dotsenko Feb 3 '12 at 10:27
An aside: Birkhoff would sometimes remark that he didn't know what Poincaré had to do with the result. – Gerald Edgar Feb 5 '13 at 15:42
This is to answer a question with a question. Does anyone know of a published proof, nice or not, of the PBW for graded Lie algebras, with the usual sign convention on the symmetry of the graded tensor product? Ponto and I put a proof in ``More Concise algebraic Topology'', Section 22.1, generalizing the passage to associated graded version of the classical proof. – Peter May Feb 6 '13 at 2:58

The nicest one I have ever seen uses a mix of universal algebra and combinatorial algebra, and was given by P. J. Higgins in Baer Invariants and the Birkhoff-Witt Theorem, Journal of Algebra 11, pp. 469-482 (1969) (free PDF linked).

Then there is the purely computational one which works over any $\mathbb Q$-algebra as base "field" and was given in the book by Deligne-Morgan. See I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt for details.

Emanuela Petracci gave in her thesis another computational proof, which uses the language of bialgebras to make the manipulations manageable.

There is also Cohn's A remark on the Poincaré-Birkhoff-Witt Theorem, J. London Math. Soc. (1963) s1-38(1): 197-203. It also has a discussion topic on MO.

If $\mathfrak g$ is the Lie algebra of a Lie group over $\mathbb R$, then you can indeed prove PBW using geometry: see, e. g., Proposition 1.9 in PDF 1 of Chapter 2 of Helgason's Lie Groups lecture notes. However, I don't think it is realistic to use this as a general proof for PBW; Lie's Third Theorem seems to be hard and require PBW itself.

Poincaré might have proven PBW himself (at least over a field of characteristic $0$), but I don't understand his proof (at least in a modern translation, which might itself be erroneous).

I hate to say but the only of the above references that I found easily readable is Higgins's paper...

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There is also a nice homological/deformation theoretic proof by Braverman and Gaitsgory: ("The Poincare-Birkhoff-Witt theorem for quadratic algebras of Koszul type"). – DamienC Feb 3 '12 at 10:57
@Damien: I already mentioned this article in a comment to the original post ;-) – Vladimir Dotsenko Feb 3 '12 at 11:08
It seems to me that the paper "The diamond lemma for ring theory", Advances in Mathematics 29 (1978) 178-218, by George M. Bergman has not been quoted in this thread. Related link: – Pierre-Yves Gaillard Feb 3 '12 at 11:39
The book by Cartan and Eilenberg includes a nice, straightforward proof. – Mariano Suárez-Alvarez Feb 11 '12 at 4:18
Lie's Third Theorem is hard, but there is a wonderful differential geometric proof whose Lie algebraic part is really simple. I found it in a paper by Van Est, "Une demonstration de E. Cartan du troisieme theoreme de Lie". Another proof of Lie III without even less Lie algebra theory is in the book by Duistermaat-Kolk. Both proofs use that $H^2(G;\mathbb{R})=0$ for simply-connected Lie group. – Johannes Ebert Feb 5 '13 at 23:40

While the OP asked for a more geometric rather than algebraic proof, I would like to point out Bergman's very nice paper,

The diamond lemma for ring theory, Advances in Mathematics 29 (1978) 178-218.

Here's a link to the paper (thanks to Darij for pointing this out!):

Also, there is this blog post by David Speyer which summarizes it, in the context of other "diamond lemma" results:

Essentially, for algebras whose defining relations fit into a framework of "find a certain type of monomial, and replace it with a linear cominbation of simpler ones", for some suitable notion of "simple", there are obvious obstructions to linear independence of reduced expressions: one may have a monomial (AB)C=A(BC), such that both AB and BC can be reduced. In that case, the ambiguity is called resolvable if upon further simplifying to reduced expressions, the two different expressions yield the same result. Of course if they didn't, then it would mean there was a linear relation amongst reduced monomials.

Bergman's result says that if you can resolve such simple overlap ambiguities, featuring just one overlap (and also something called inclusion ambiguity - which are rarer, and can always be weeded out anyways, as he explains), then the PBW monomials indeed form a basis. In the case of Lie algebras, checking this reduces to nothing more than the Jacobi identity, as he shows.

I really like the approach via Diamond Lemma as it is very elementary to prove, and the conditions of the Lemma are easy to verify in practice. It is general enough to apply to very many "almost" commutative algebras, such as arise in quantum groups and related areas.

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All sufficiently all Adv in Math articles are in open access nowadays, thanks to , and this is no exception ( , see also corrections ). – darij grinberg Feb 6 '13 at 2:30
old, not all. Also, I am wondering how I forgot about that proof! – darij grinberg Feb 6 '13 at 2:30

$\def\fg{\mathfrak{g}}$This is an idea I've had for a bit which I both like and dislike: It comes from trying to take the very nice proof which holds when $\fg$ is the Lie algebra of a Lie group over $\mathbb{R}$ and push it to become purely algebraic. As darij writes, this nice proof can be found well explained in Helgason's notes (Prop 1.9). Of course, some may argue that I've destroyed the niceness by trying to remove the geometry.

Let our ground field $k$ have characteristics $0$. Let $k[[\fg^{\ast}]]$ be the completion of $\mathrm{Sym}(\fg^{\ast})$ at the origin, so elements of $k[[\fg^{\ast}]]$ can be thought of roughly as germs of functions on $\fg$ near $0$. For $g \in \fg$, we have the natural derivation $\delta(g)$ on $k[[\fg^{\ast}]]$, given by extending the linear pairing $\fg \times \fg^{\ast} \to k$ to a continuous derivation.

For $g \in \fg$, define a derivation $\partial(g)$ by $$\partial(g) = \sum_{n=0}^{\infty} \frac{B_n}{n!} \delta\left( (ad \ u)^n(g) \right).$$ Here $u$ is the formal variable which one can think of as "valued in (a formal nbhd of the identity in) $\fg$" and $t/(1-e^{-t}) = \sum (B_n/n!) t^n$.

It should be possible to prove that $\partial(g)$ is a Lie algebra action directly. If so, one then finishes exactly as in Helgason's notes, seeing that the PBW basis of $U(\fg)$ pairs independently against $k[[\fg^{\ast}]]$.

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This looks very similar to what Emanuela Petracci does in Chapter 2 of her thesis ( ), although her derivation is one of $S\left(\mathfrak g\right)$ and not of the completion $k\left[\left[\mathfrak g^{\ast}\right]\right]$. The computations are indeed complicated, and she gets them manageable using some tricks (some tricks I must say I don't completely understand -- can anyone explain me the proof of Lemma 2.1.1 ii) ? -- but can be substituted by lengthy calculations). – darij grinberg Feb 5 '13 at 17:02

I'm partial to Dylan Thurston's proof in his Ph.D thesis. He proves the Duflo isomorphism (which is stronger than PBW) in a graphical/knot-theoretic context. The paper is a real pleasure to read.

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It is interesting. However in my understanding Duflo is something complicated, miraculous and un-understandable :) PBW is something trivial (for Lie algebras). – Alexander Chervov Feb 11 '12 at 15:06
@Alexander: Yes I agree that Duflo is much much harder than PBW. – Jim Conant Feb 11 '12 at 20:17

Another proof is given here; Positelsky, L, Functional Analysis and Its Applications, 1993, 27:3, 197–204 :

I quote:

''The classical PBW theorem attains its natural place in this context as a particular case of the fact that every Kozsul CDG-algebra corresponds to a QLS-algebra; here a QLS=quadratic linear scalar algebra is roughly ``an algebra defined by (generators and) non-homogenious relations of degree 2.

text in russian, text in English

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$\def\fg{\mathfrak{g}}\def\cU{\mathcal{U}}\def\cT{\mathcal{T}}\def\tD{\widetilde{\Delta}}$Here is an argument Noah Snyder showed me back in grad school, which proves PBW in characteristic zero on the assumption that $\fg$ has a faithful representation. Ado's theorem says that every Lie algebra has a faithful finite dimensional representation, but here we don't need finite dimensionality -- just any vector space at all on which $\fg$ acts without kernel. For example, if $\fg$ is the Lie algebra of a group $G$, we can take the action on $C^{\infty}(G)$; if $\fg$ is given to us as a space of $n \times n$ matrices we can use the action on $\mathbb{R}^n$.

Let $\cU(\fg)$ be the universal enveloping algebra and let $\cT(\fg)$ be the tensor algebra. Let $\tD : \cT(\fg) \to \cT(\fg) \otimes \cT(\fg)$ be the unique map of rings with $\tD(v) = v \otimes 1 + 1 \otimes v$ for $v \in \fg$. (In the ring structure on $\cT$, the factors $1 \otimes \cT$ and $\cT \otimes 1$ commute.) We can check directly from the defining relations that $\tD$ descends to a map $\Delta: \cU(\fg) \to \cU(\fg) \otimes \cU(\fg)$. (This is the Hopf coproduct.)

Let $S(\fg) \subset \cT(\fg)$ be the symmetric tensors. We want to prove the composition $S(\fg) \to \cT(\fg)\to \cU(\fg)$ is injective. We show by induction on $n$ that this map is injective on $S_{\leq n}(\fg)$ -- the space of symmetric tensors of degree $\leq n$.

The inductive step is fine. Suppose (for contradiction) that $f$ is a nonzero element of the kernel, and $f \in S_{\leq n}$. Set $g:=\tD(f) - f \otimes 1 - 1 \otimes f$. Then $g \in S_{\leq n-1} \otimes S_{\leq n-1}$. And, if $n \geq 2$, then $g$ is nonzero. Checking the latter uses that we are in characteristic zero in order to make sure that certain binomial coeffcients are nonzero; for example, if the characteristic were $p$ then, for $v \in \fg$, we have $\tD(v^p) -v^p \otimes 1 - 1 \otimes v^p = (v \otimes 1 + 1 \otimes v)^p -v^p \otimes 1 - 1 \otimes v^p=0$.

But, if $f$ maps to $0$ in $\cU$, then $g=\tD(f) - f \otimes 1 - 1 \otimes f$ must map to zero in $\cU \otimes \cU$. But $g \in S_{\leq n-1} \otimes S_{\leq n-1}$ and, inductively, $S_{\leq n-1} \otimes S_{\leq n-1}$ injects into $\cU$.

The trouble is the base case, $n=1$. To do this, we assume that $\fg$ acts faithfully on some vector space $V$. Then we obtain a map $\cU(\fg) \to \mathrm{End}(V)$, such that $\fg \to \cU(\fg) \to \mathrm{End}(V)$ is injective. In particular, $\fg$ injects into $\cU(\fg)$.

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