# Is $\mathrm{U}(\mathfrak{a} \oplus \mathfrak{b}) \cong \mathrm{U}(\mathfrak{a}) \otimes \mathrm{U}(\mathfrak{b})$ over a commutative ring?

Let $$\mathbb{k}$$ be a commutative ring and let $$\mathfrak{g}$$ be a Lie-Algebra over $$\mathbb{k}$$. Suppose that $$\mathfrak{a}$$ and $$\mathfrak{b}$$ are two Lie subalgebras of $$\mathfrak{g}$$ such that $$\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{b}$$ as $$\mathbb{k}$$-modules. We then have a homomorphism of $$\operatorname{U}(\mathfrak{a})$$-$$\operatorname{U}(\mathfrak{b})$$-bimodules $$\Phi \colon \operatorname{U}(\mathfrak{a}) \otimes_{\mathbb{k}} \operatorname{U}(\mathfrak{b}) \to \operatorname{U}(\mathfrak{g}) \,, \quad x \otimes y \mapsto xy \,.$$

Question. Is the homomorphism $$\Phi$$ an isomorphism of bimodules?

My thoughts so far:

• If both $$\mathfrak{a}$$ and $$\mathfrak{b}$$ are free as $$\mathbb{k}$$-modules (e.g. if $$\mathbb{k}$$ is a field), then $$\mathfrak{g}$$ is also free as a $$\mathbb{k}$$-module. One can then use the PBW-theorem to see that $$\Phi$$ is a bijection on the induced bases. It is then an isomorphism of $$\mathbb{k}$$-modules, and thus an isomorphism of bimodules.

• If $$\mathfrak{a}$$ and $$\mathfrak{b}$$ is are ideals of $$\mathfrak{g}$$, then the decomposition $$\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{b}$$ is one of Lie algebras. One can then see that both sides of $$\Phi$$ satisfy the same universal property as $$\mathbb{k}$$-algebras, and that the above map $$\Phi$$ is even an isomorphism of $$\mathbb{k}$$-algebras.

• I think that $$\Phi$$ is in general still surjective. One should still be able to get module generating sets by PBW-monomials for the universal enveloping algebras, and then see that $$\Phi$$ is surjective on the induced generators. But I’m not sure what happens with the injectivity of $$\Phi$$.

• If you can prove that $U({\frak a})\otimes U({\frak b})$ is the coproduct of $U({\frak a})$ and $U({\frak b})$, the isomorphism follows by $U$ being a left adjoint. This is certainly true in the case of a field, and I assume is true in this case as well, but didn't check if anything goes wrong without field assumption. Nov 24 '20 at 14:29
• @Ennar What kind of coproduct do you mean? I think your argument only works if $\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{b}$ as Lie algebras, i.e. if both $\mathfrak{a}$ and $\mathfrak{b}$ are ideals in $\mathfrak{g}$. (Then $\mathrm{U}(\mathfrak{a}) \otimes \mathrm{U}(\mathfrak{b})$ is the “commutative coproduct” of $\mathrm{U}(\mathfrak{a})$ and $\mathrm{U}(\mathfrak{b})$ in the category of $\mathbb{k}$-algebras.) This is what I’ve tried to express in the second item of the list. Nov 24 '20 at 14:58
• Ah, ok, sorry, I didn't read your question carefully enough, I interpreted $\frak a\oplus \frak b$ as coproduct of Lie algebras, but now I see you didn't mean that. Anyway, the map in the other direction should be $x+y\mapsto x\otimes 1 + 1\otimes y$, so maybe you can work with that directly (elements of Lie algebras still generate the universal enveloping algebras). Nov 24 '20 at 15:10
• This has been done by Gérard Duchamp in mathoverflow.net/questions/300851/… . (It might grow into a paper as there appears to be a generalization to $k$ Lie subalgebras.) Apr 19 at 18:50
• Is your question out of pure curiosity (which is, IMHO, perfectly legitimate) or is there some motivation that you could (even shortly) elaborate there ? Apr 20 at 8:53