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 $\Phi$.

  • $\begingroup$ 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. $\endgroup$
    – Ennar
    Nov 24, 2020 at 14:29
  • $\begingroup$ @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. $\endgroup$ Nov 24, 2020 at 14:58
  • $\begingroup$ 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). $\endgroup$
    – Ennar
    Nov 24, 2020 at 15:10
  • 2
    $\begingroup$ 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.) $\endgroup$ Apr 19, 2021 at 18:50
  • $\begingroup$ Is your question out of pure curiosity (which is, IMHO, perfectly legitimate) or is there some motivation that you could (even shortly) elaborate there ? $\endgroup$ Apr 20, 2021 at 8:53


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