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This question is imported from MSE. It is linked to this one in the case of semi-direct products.

My question Let us consider a Lie $R$-algebra ($R$ is a commutative ring) written as a (module) direct sum of two of its subalgebras $$ \mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2\ (\oplus=\oplus_{R-mod}) $$ and the natural mapping $$ \alpha : \mathcal{U}(\mathfrak{g}_1)\otimes_R\mathcal{U}(\mathfrak{g}_2)\to\mathcal{U}(\mathfrak{g}) $$ ($\mathfrak{g}_i$ are not necessarily ideals).

On can check, using generators, that $\alpha$ is onto (and, in some usual cases - see below - one-to-one).

What is true/known in the general case ?

I put here the explicit construction in case one of the $\mathfrak{g}_i$ is an ideal. The proof goes as follows :

Take it that $\mathfrak{g}_1$ is such.

  1. Consider the action $\delta : \mathfrak{g}_2\to \mathfrak{Der} (\mathfrak{g}_1)$ by derivations (adjoint representation)
  2. Extend $\delta$ to $\mathfrak{Der}(\mathcal{U}(\mathfrak{g}_1))$ as in Bourbaki Lie ch 1 paragraph 2.8 prop 7.
  3. Extend $\delta$ as a morphism of $R$-algebras $\mathcal{U}(\mathfrak{g}_2)\to\mathrm{End}(\mathcal{U}(\mathfrak{g}_1))$ by universal property
  4. Set a law of $R$-unital associative algebra on $\mathcal{U}(\mathfrak{g}_1)\otimes_R\mathcal{U}(\mathfrak{g}_2)$ by $$ (u_1\otimes u_2).(v_1\otimes v_2)=(u_1\otimes 1)\Big((\delta\otimes\gamma_2)\circ\Delta(u_2)\Big)[v_1\otimes v_2] $$ where $\gamma_2(m)$ is the multiplication by $m$ on the left within $\mathcal{U}(\mathfrak{g}_2)$.

Late edit For those who know the smash product, formula of point 4 says that $$ \mathcal{U}(\mathfrak{g})=\mathcal{U}(\mathfrak{g}_1)\,\sharp\, \mathcal{U}(\mathfrak{g}_2) $$ where $\sharp$ stands for the smash product defined by the action of $\mathfrak{g}_2$ by derivations on $\mathcal{U}(\mathfrak{g}_1)$.
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    $\begingroup$ Another fairly trivial case is when both Lie algebras are abelian: the universal enveloping algebra is the symmetric algebra and it follows from the functorial properties of the symmetric algebras that $\alpha$ is an isomorphism. $\endgroup$
    – Victor Protsak
    Commented May 23, 2018 at 2:45
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    $\begingroup$ See mathoverflow.net/questions/88598/… for the case where $R $ is a $\mathbb{Q}$-algebra. $\endgroup$
    – darij grinberg
    Commented May 23, 2018 at 6:54
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    $\begingroup$ Ah, sorry, I thought $\alpha$ was supposed to be an isomorphism of algebras, which only makes sense if $\mathfrak g_1$ and $\mathfrak g_2$ commute. $\endgroup$
    – Thomas Poguntke
    Commented May 25, 2018 at 12:15
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    $\begingroup$ Yes, that is what I meant (just on the other side of the Yoneda embedding). $\endgroup$
    – Thomas Poguntke
    Commented May 25, 2018 at 18:18
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    $\begingroup$ If one of them (say $\mathfrak g_1$) is an ideal, $\alpha$ must be one-to-one: in $\mathcal U(\mathfrak g)$ there is a monomial basis of the form $m_1m_2$ where $m_i$ are monomials from $\mathfrak g_i$, since $x_2x_1=x_1x_2+y_1$ with $x_i,y_i\in\mathfrak g_i$, so that any monomial not of the above form is a sum of a monomial "closer to" the above form and a shorter monomial. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 26, 2018 at 9:23

3 Answers 3

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I'm adding my comment as a partial answer, as discussed there; this is a reformulation of point 2 that has been added to the question.

Let $\mathcal L$ be the category of Lie $R$-algebras. Assume that $\mathfrak g = \mathfrak g_1 \oplus \mathfrak g_2$ in $\mathcal L$ (direct sum, i.e. I suppose that they commute). Let $\mathcal C$ be the category of unital $R$-algebras. Then $\alpha$ is an isomorphism in $\mathcal C$, since for every unital $R$-algebra $A$, the map $$\text{Hom}_{\mathcal L}(\mathfrak g,A) \cong \text{Hom}_{\mathcal C}(U(\mathfrak g),A) \stackrel{\alpha^*}{\to} \text{Hom}_{\mathcal C}(U(\mathfrak g_1) \otimes_R U(\mathfrak g_2),A)$$ is bijective (by the Yoneda lemma). This is because the right-hand side is given by all $$(f_1,f_2) \in \text{Hom}_{\mathcal C}(U(\mathfrak g_1),A) \times \text{Hom}_{\mathcal C}(U(\mathfrak g_2),A) \mbox{ with } [f_1(x_1),f_2(x_2)]=0 \mbox{ for } x_i \in \mathfrak g_i,$$ that is, all pairs of maps $(f_1,f_2)\!: \mathfrak g_1 \oplus \mathfrak g_2 \to A$ in $\mathcal L$ with $[f_1(x_1),f_2(x_2)]=0$ for all $x_i \in \mathfrak g_i$. Then the inverse of $\alpha^*$ is defined by $(f_1,f_2) \mapsto f_1+f_2$, which is well-defined, since $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

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  • $\begingroup$ For the hasty reader, I added the assumption "(direct sum, i.e. I suppose that they commute)". $\endgroup$
    – Duchamp Gérard H. E.
    Commented May 25, 2018 at 22:37
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    $\begingroup$ I've specified the categories, and I hope this makes it clearer. $\endgroup$
    – Thomas Poguntke
    Commented May 26, 2018 at 8:23
  • $\begingroup$ @ThomasPogunke Well, thank you (+1), let me check (and possibly clarify). $\endgroup$
    – Duchamp Gérard H. E.
    Commented May 26, 2018 at 8:44
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    $\begingroup$ This is the Yoneda lemma: if $\alpha^*$ is a bijective for all $A \in \mathcal C$, then $\alpha$ is an isomorphism in $\mathcal C$. $\endgroup$
    – Thomas Poguntke
    Commented May 26, 2018 at 9:02
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    $\begingroup$ If by one-to-one you mean isomorphism, then yes; the Yoneda lemma holds for any category, including $\mathcal C$. $\endgroup$
    – Thomas Poguntke
    Commented May 26, 2018 at 9:42
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There are probably not many textbook references for these generalities, but I'd suggest looking at the first chapter of the Bourbaki treatise Groupes et algebres de Lie (whose chapters I-IX have been translated into English, possibly from later editions than the 1960 Hermann edition of Chapter I which is at hand).

Keep in mind that the first three sections of this chapter treat very general Lie algebras over a commutative ring (with 1) called $K$. The universal enveloping algebra of a "product" of two Lie algebras over $K$ is considered in $\S2.2$. I think this is essentially the same set-up you start with, in which the two Lie subalgebras commute with each other; otherwise I'm not sure what you mean by "direct sum". Anyway, in this framework it's fairly easy to prove that the universal enveloping algebra of such a product is isomorphic to the product of the two universal enveloping algebras: see their Proposition 2.

Are you asking a more subtle question here? Semi-direct products are of course more difficult to deal with.

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    $\begingroup$ Yeah, Gérard is asking a subtler question: His direct sum is only a direct sum in the category of $R$-modules. $\endgroup$
    – darij grinberg
    Commented May 26, 2018 at 22:25
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The answer is YES in all cases ($k$ a commutative ring and $\mathfrak{g}=\mathfrak{g}_1\oplus \mathfrak{g}_2$ a direct sum of two subslgebras with no other condition).

A complete and detailed proof can be found there.

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    $\begingroup$ Why is the multiplication $*$ associative? Also, some extraneous $\in$ signs in the definition of $r_{w,i}$. $\endgroup$
    – darij grinberg
    Commented Jun 9, 2018 at 17:18
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    $\begingroup$ Let me know when you've fixed the associativity argument (or maybe you don't need it?). BTW: If you can show the analogous fact for an arbitrary direct sum (not just of $2$ submodules), then you'll have gotten a new proof of Poincaré-Birkhoff-Witt (even one of the more general versions: the one where $\mathfrak{g}$ is assumed to be a direct sum of cyclic modules). $\endgroup$
    – darij grinberg
    Commented Jun 9, 2018 at 17:38
  • $\begingroup$ This is looking better and better! But I don't quite see how you get $\mathcal{J}_{11} \subseteq \ker\beta_0$. Also, minor typo: one of the $\mathcal{U}(\mathfrak{g})$s in the commutative diagram should be a $\mathcal{U}(\mathfrak{g}_1)$. $\endgroup$
    – darij grinberg
    Commented Jun 9, 2018 at 20:39
  • $\begingroup$ Also, it might be that you don't need the AAU structure any more, since the fact that $\ker \rho$ is an ideal follows straight from property 4. Unless you tacitly use it in the proof of $\mathcal{J}_{11} \subseteq \ker\beta_0$ (is $s_1 \otimes s_2$ an algebra homomorphism?). $\endgroup$
    – darij grinberg
    Commented Jun 9, 2018 at 20:41
  • $\begingroup$ @darijgrinberg [But I don't quite see how you get ...]---> One has $\ker(\rho)\subset \ker(\beta_0)$ then, due to the last identity $$\beta_0(u\otimes B(g_1,h_1)\otimes v)=\beta_0(u_1\otimes B(g_1,h_1)\otimes u_2)=0$$ where $u_i\in T(\mathfrak{g}_i)$ $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 9, 2018 at 21:21

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