Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over a field $k$, 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 (PBW) 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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    $\begingroup$ 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 $\mathfrak{g}^{\ast}$. $\endgroup$ – Qiaochu Yuan Feb 3 '12 at 6:03
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    $\begingroup$ Since you are asking for a list then I guess this should be community wiki. $\endgroup$ – DamienC Feb 3 '12 at 8:25
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    $\begingroup$ 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. arxiv.org/abs/hep-th/9411113. $\endgroup$ – Vladimir Dotsenko Feb 3 '12 at 10:27
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    $\begingroup$ An aside: Birkhoff would sometimes remark that he didn't know what Poincaré had to do with the result. $\endgroup$ – Gerald Edgar Feb 5 '13 at 15:42
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    $\begingroup$ 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. $\endgroup$ – 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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    $\begingroup$ There is also a nice homological/deformation theoretic proof by Braverman and Gaitsgory: arxiv.org/abs/hep-th/9411113 ("The Poincare-Birkhoff-Witt theorem for quadratic algebras of Koszul type"). $\endgroup$ – DamienC Feb 3 '12 at 10:57
  • $\begingroup$ @Damien: I already mentioned this article in a comment to the original post ;-) $\endgroup$ – Vladimir Dotsenko Feb 3 '12 at 11:08
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    $\begingroup$ 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: math.berkeley.edu/~gbergman/papers/updates/diamond.html $\endgroup$ – Pierre-Yves Gaillard Feb 3 '12 at 11:39
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    $\begingroup$ The book by Cartan and Eilenberg includes a nice, straightforward proof. $\endgroup$ – Mariano Suárez-Álvarez Feb 11 '12 at 4:18
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    $\begingroup$ 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. $\endgroup$ – Johannes Ebert Feb 5 '13 at 23:40

Note: the paper https://arxiv.org/abs/hep-th/9411113 by Braverman and Gaitsgory was already mentionned in the comments by Vladimir Dotsenko and DamienC but I wanted to write down a sketch of the argument (at least for my own understanding).

Let $\mathfrak{g}$ be a Lie algebra over a field $k$, let $S(\mathfrak{g})$ be the symmetric algebra (it is a graded commutative associative algebra) and let $U(\mathfrak{g})$ be the universal enveloping algebra (it is a filtered associative algebra). There is a natural surjective morphism of algebras $S(\mathfrak{g}) \rightarrow \text{gr} U(\mathfrak{g})$ and PBW is the statement that it is an isomorphism.

Consider $A$ the quotient of the tensor algebra $T(\mathfrak{g}[t])$ by the 2-sided ideal generated by the elements of the form $x \otimes y -y \otimes x - t[x,y]$. Considering $t$ as a variable of degree 1, $A$ is a graded $k[t]$-algebra, which specializes to $S(\mathfrak{g})$ for $t=0$ and $U(\mathfrak{g})$ for $t=1$. The introduction of the parameter $t$ makes precise the idea that $U(\mathfrak{g})$ is a deformation of $S(\mathfrak{g})$ by $[-,-]$, and PBW is equivalent to the fact that $A$ is free as $k[t]$-module, i.e. that there is no jump in the dimensions of the graded pieces when going from $t \neq 0$ to $t=0$.

From this point of view, it is clear that the natural setting for PBW is the deformation theory of $S(\mathfrak{g})$ as a graded associative algebra, and the key point is that such deformation problem has a cohomological description. First order (i.e. over $k[t]/(t^2)$) flat deformations of $S(\mathfrak{g})$ as associative algebra are parametrized by the Hochschild cohomology group $HH^2(S(\mathfrak{g}), S(\mathfrak{g}))=\wedge^2 \mathfrak{g}^* \otimes S(\mathfrak{g})$ (computing this $HH^\bullet$ is easy: it is equivalent to writing down a resolution of the diagonal of the affine space $\mathfrak{g}^*$, i.e. a Koszul resolution). For graded first order flat deformations, with the deformation parameter $t$ of degree $1$, we want the degree $-1$ part of $HH^2$, i.e. $\wedge^2 \mathfrak{g}^* \otimes \mathfrak{g}$, i.e. some $[-,-] \colon \wedge^2 \mathfrak{g} \rightarrow \mathfrak{g}$.

The obstruction to lift such flat deformation to the second order (over $k[t]/(t^3)$) lives in the degree $-2$ part of $HH^3(S(\mathfrak{g}), S(\mathfrak{g}))=\wedge^3 \mathfrak{g}^* \otimes S(\mathfrak{g})$, i.e. in $\wedge^3 \mathfrak{g}^* \otimes \mathfrak{g}$ and the vanishing of this obstruction is exactly the Jacobi identity for $[-,-]$. The set of possible lifts is then parametrized by the degree $-2$ part of $HH^2$, i.e. $\wedge^2 \mathfrak{g}^*$. Choosing the trivial lift (a non-trivial lift would correspond to a further deformation $x \otimes y -y \otimes x - t [x,y] -t^2 \phi(x,y)$ for some $\phi \in \wedge^2 \mathfrak{g}^*$), the obstruction to lift to the third order, living in the degree $-3$ part of $HH^3$, i.e. $\wedge^3 \mathfrak{g}^*$, automatically vanishes.

Now the key point is that the degree $i$ part of $HH^2$ vanishes for $i<-2$ and the degree $i$ part of $HH^3$ vanishes for $i<-3$ so our flat deformation automatically lifts to all orders in a unique way. We get some graded $k[t]$-algebra $B$, locally free as $k[t]$-module, which specializes to $S(\mathfrak{g})$ at $t=0$. Because the relations defining $U(\mathfrak{g})$ are satisfied in the specialization $B_1$ of $B$ at $t=1$, we have a natural surjective map $U(\mathfrak{g}) \rightarrow B_1$ of filtered associative algebras. So, we have obtained natural surjective maps $S(\mathfrak{g}) \rightarrow \text{gr} U(\mathfrak{g}) \rightarrow \text{gr} B_1=S(\mathfrak{g})$ whose composition is easily seen to be the identity, and so are isomorphisms, proving PBW.

This proof is not simple in the sense of elementary but is simple in the sense of conceptual. The thing to realize is that it is a deformation question and then one can apply the cohomological machinery without thinking. In particular, there is no computational trick and it is clear from this proof that the Jacobi identity is exactly the right hypothesis for a result such as PBW because it is the vanishing of some cohomological obstruction. The same approach gives PBW for Lie superalgebras, Weyl algebras or Clifford algebras, and again the cohomological machinery will tell us directly what to do, without having to invent a new trick for each special case.

  • $\begingroup$ can you get the coalgebra structure from this argument as well? $\endgroup$ – Saal Hardali Nov 28 '18 at 15:14

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.


$\def\fg{\mathfrak{g}}\def\cU{\mathcal{U}}\def\cT{\mathcal{T}}\def\tD{\widetilde{\Delta}}$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 $\cU \otimes \cU$.

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 $\cU(\fg)$.


$\def\fg{\mathfrak{g}}$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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    $\begingroup$ This looks very similar to what Emanuela Petracci does in Chapter 2 of her thesis ( iecn.u-nancy.fr/~petracci/tesi.pdf ), 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). $\endgroup$ – darij grinberg Feb 5 '13 at 17:02

There is a recent, geometric proof of PBW by Michael Eastwood based on the fact that the $n$-sphere is simply-connected for $n\geq2$:


  • $\begingroup$ full ref info: M. Eastwood. A geometric proof of the Poincaré-Birkhoff-Witt Theorem. São Paulo Journal of Mathematical Sciences (2018) 12:246-251, doi.org/10.1007/s40863-018-0095-y $\endgroup$ – YCor Mar 6 '19 at 18:04

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


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

Recently, me and Vladimir Dotsenko formulated PBW in terms of maps of monads in https://arxiv.org/abs/1804.06485. As a by product, one gets the classical PBW theorem is the manifestation that, in characteristic zero, the associative operad is free as a right Lie module through the map $\phi:\mathsf{Lie} \to\mathsf{Ass}$ that sends the bracket to the commutator.

The left adjoint to the forgetful map from associative algebras to Lie algebras is the enveloping functor, and since $\mathsf{Ass} = \mathsf{Com}\circ\mathsf{Lie}$, one gets the desired natural isomorphism between $\phi_!(\mathfrak g) = U(\mathfrak g)$ and $\mathsf{Com}(\mathfrak g) = S(g)$.

To prove that $\phi$ is free, one observes that we're dealing with weight graded connected operads $P$, so we can prove a weight graded module $M$ is free by showing that the bar complex $B(M,P,1)$ is acyclic (i.e. concentrated in degree 0).

Since freeness is an homological statement, one can use homological methods to prove this: we filter $\mathsf{Ass}$ by the powers of the ideal generated by the Lie bracket, and note that the associated graded object that results is the Poisson operad, which is $\mathsf{Com}\circ\mathsf{Lie}$, giving the result.


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