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Let $k$ be a commutative ring with unit and $L$ be a Lie $k$-algebra.

Let $U(L)$ be the universal enveloping $k$-algebra of $L$ (one can define it as a quotient of the tensor algebra, as it is explained in this MO question, or one can say that $U(-)$ is left adjoint to the forgetful functor sending an associative $k$-algebra to the Lie $k$-algebra obtained by taking the same underlying $k$-module and with Lie bracket being the commutator).

The associative $k$-algebra $U(L)$ is filtered as a $k$-algebra, and there is a canonical epimorphism $S(L)\to gr\big(U(L)\big)$.

If this epimorphism is an isomorphism, then we say that $L$ has the PBW property.

All the examples of Lie $k$-algebras not satisfying the PBW property I am aware of are constructed in the following way: one first finds an example of a Lie algebra for which the map $L\to U(L)$ (the unit of the adjunction) is not injective, and then it is quite clear that the PBW property can't hold.

My question is then:

Is there any example of a Lie $k$-algebra $L$ such that the map $L\to U(L)$ is injective which does not satisfy the PBW property ?

Or is it that the PBW property is just equivalent to $L\to U(L)$ being injective (it would be great, but I have no idea why this would be true - EDIT: one might want to use that $L\to U(L)$ is injective to reduce to the cas when $k\supset\mathbb{Q}$)?

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How does $L \hookrightarrow U(L)$ reduce to the $k\supseteq \mathbb Q$ case? If $k$ is any field, then all Lie algebras over it have the PBW property. Actually, come to think of it, I think that Lie algebras always have the PBW property when $k \supseteq \mathbb Q$ — my memory is that this is proved in, for example, the article by Deligne and Morgan in QFT and Strings, but flipping through I can't find it. The idea is to prove that the canonical symmetrization map (which requires dividing by all $n!$s) is a filtered isomorphism $S(L) \to U(L)$ covering the canonical map $S(L)\to gr(U(L))$. – Theo Johnson-Freyd Nov 27 '12 at 23:18
I know that PBW holds whenever $k\supset\mathbb{Q}$. About my EDIT, this was just a dummy suggestion for a strategy. I was thinking about something like: if $L\to U(L)$ is injective and if PBW holds for $L\otimes_k\mathbb{K}$ (for some $\mathbb{K}$) then PBW holds for $L$ – DamienC Nov 28 '12 at 7:52

While I can't immediately give a precise example, it may help to focus the question a little more in order to emphasize what is actually involved.

1) In most literature on Lie algebras (related to Lie groups or algebraic groups), the emphasis is on Lie algebras over fields or perhaps over integral domains and their fraction fields. But there is another direction of study involving Lie rings (and often associated abstract groups). It's possible more generally to study Lie algebras over an arbitrary commutative ring $K$ (with 1), which is done for instance in Chapter I of the Bourbaki treatise Groupes et algebres de Lie. There the PBW theorem is developed in Section 2.7, following generalities on the universal enveloping algebra $U(L)$. But the theorem is stated and proved under the restrictive assumption that the Lie algebra is a free module over $K$. This leads directly to the embedding of $L$ in $U(L)$ and the existence of the usual PBW bases for the latter algebra. Once you assume that $L$ has a basis, what you call the PBW property is then equivalent to the embedding property.

2) Bourbaki's Exercise 9 for Section 2 provides a rather complicated construction of a Lie algebra $L$ over a commutative ring which doesn't embed in its universal enveloping algebra. But $L$ does embed in its symmetric algebra, by definition. This is certainly enough to defeat the PBW property, as you observe. [But your wording "All the examples ..." is confusing and really means "All the known examples .."] You are asking for an example of a Lie algebra which fails to be a free $K$-module and fails to have the PBW property but does embed in $U(L)$. Given the complexity of the Bourbaki exercise mentioned above, such an example (if it exists!) is probably even more complicated.

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There is also has a relatively simple example in… (see also the discussion here: ). – DamienC Nov 28 '12 at 20:13

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