# $U\left(\mathfrak a\right) \otimes_{U\left(\mathfrak a\cap\mathfrak b\right)} U\left(\mathfrak b\right) \cong U\left(\mathfrak a + \mathfrak b\right)$ over a ring containing $\mathbb{Q}$

While the Poincaré-Birkhoff-Witt theorem is usually proven (and sometimes even formulated) for free modules only, it is known (see also here) that it holds for arbitrary modules if the ground ring is a $\mathbb Q$-algebra. I am wondering if a similar generalization holds for the following fact, which I learnt today from Pavel Etingof:

Let $\mathfrak c$ be a Lie algebra. Let $\mathfrak a$ and $\mathfrak b$ be two Lie subalgebras of $\mathfrak c$ such that $\mathfrak a + \mathfrak b = \mathfrak c$. Clearly, $\mathfrak a \cap \mathfrak b$ is also a Lie subalgebra of $\mathfrak c$. Now, the map

$U\left(\mathfrak a\right) \otimes_{U\left(\mathfrak a\cap \mathfrak b\right)} U\left(\mathfrak b\right) \to U\left(\mathfrak c\right),$

$\alpha\otimes_{U\left(\mathfrak a\cap \mathfrak b\right)} \beta\mapsto\alpha\beta$

is an isomorphism (not of algebras, but of $\left(U\left(\mathfrak a\right),U\left(\mathfrak b\right)\right)$-bimodules).

This is proven for free modules using PBW and appropriate bases. I had no time to do any research on this.

• When I tried to right-click on the title (while in the list of recent questions) to open the question in a new tab, I was only offered to be shown the TeX source of the formula (MathJax, huh?), so maybe it is NOT a very good idea to have titles consisting of formulas only? I will fix it now, hope you don't mind. – Vladimir Dotsenko Feb 16 '12 at 8:57

Yes, the fact holds whenever the ground ring is a $\mathbb{Q}$-algebra. For the proof, see the First proof of Proposition 2.4.1 in my notes for Pavel Etingof, 18.747 Infinite-dimensional Lie Algebras, Spring term 2012 at MIT. (Of course, read "commutative $\mathbb{Q}$-algebra" for "field". In the footnote that gives two proofs for $\sigma$ being an algebra isomorphism, read the second of these two proofs; the first requires the ground ring to be a field.)
More generally, this shows that the fact holds whenever the PBW theorem holds for both $\mathfrak{a}$ and $\mathfrak{b}$ (that is, all three canonical maps $\operatorname{PBW}_{\mathfrak{a}} : \operatorname{Sym}\left(\mathfrak{a}\right) \to \operatorname{gr}\left(U \left(\mathfrak{a}\right)\right)$ and $\operatorname{PBW}_{\mathfrak{b}} : \operatorname{Sym}\left(\mathfrak{b}\right) \to \operatorname{gr}\left(U \left(\mathfrak{b}\right)\right)$ and $\operatorname{PBW}_{\mathfrak{c}} : \operatorname{Sym}\left(\mathfrak{c}\right) \to \operatorname{gr}\left(U \left(\mathfrak{c}\right)\right)$ are module isomorphisms). (The proof pretends to also require $\operatorname{PBW}_{\mathfrak{a} \cap \mathfrak{b}}$ to be an isomorphism; but if you dig a little bit deeper, you'll see that $\operatorname{PBW}_{\mathfrak{a} \cap \mathfrak{b}}$ merely has to be surjective, which it always is.)