# About enveloping algebras of direct sums

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 $$(u_1\otimes u_2).(v_1\otimes v_2)=(u_1\otimes 1)\Big((\delta\otimes\gamma_2)\circ\Delta(u_2)[v_1]\otimes v_2\Big)$$ where $\gamma_2(m)$ is the multiplication by $m$ on the left within $\mathcal{U}(\mathfrak{g}_2)$.
• 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. May 23, 2018 at 2:45
• See mathoverflow.net/questions/88598/… for the case where $R$ is a $\mathbb{Q}$-algebra. May 23, 2018 at 6:54
• 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. May 25, 2018 at 12:15
• Yes, that is what I meant (just on the other side of the Yoneda embedding). May 25, 2018 at 18:18
• 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. May 26, 2018 at 9:23

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 $$(f_1+f_2)([x_1+x_2,y_1+y_2]) = f_1([x_1,y_1]) + f_2([x_2,y_2]) = [f_1(x_1) + f_2(x_2), f_1(y_1) + f_2(y_2)].$$

• For the hasty reader, I added the assumption "(direct sum, i.e. I suppose that they commute)". May 25, 2018 at 22:37
• I've specified the categories, and I hope this makes it clearer. May 26, 2018 at 8:23
• @ThomasPogunke Well, thank you (+1), let me check (and possibly clarify). May 26, 2018 at 8:44
• This is the Yoneda lemma: if $\alpha^*$ is a bijective for all $A \in \mathcal C$, then $\alpha$ is an isomorphism in $\mathcal C$. May 26, 2018 at 9:02
• If by one-to-one you mean isomorphism, then yes; the Yoneda lemma holds for any category, including $\mathcal C$. May 26, 2018 at 9:42

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.

• Yeah, Gérard is asking a subtler question: His direct sum is only a direct sum in the category of $R$-modules. May 26, 2018 at 22:25

Late late edit .- The answer seems to be yes $$\alpha$$ is one-to one I give below the scheme of a tentative proof (relying on indexed rewriting) as an answer (to be checked thoroughly).

We first build an arrow $$\rho\ :\ T(\mathfrak{g})\to T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2)$$ by straightening the inversions. In order not to confuse strings with numbers, we reindex $$\mathfrak{g}_1=\mathfrak{g}_a$$ and $$\mathfrak{g}_2=\mathfrak{g}_b$$ where $$a are two symbols. Using the decomposition in

Bourbaki, Algebra Chapter III § 5.5, one gets $$T(\mathfrak{g})=\oplus_{w\in \{a,b\}^*}\,T_w$$ where $$\{a,b\}^*$$ is the set of words in the symbols $$\{a,b\}$$ (the free monoid generated by $$\{a,b\}$$) i.e. mappings $$\{1,\cdots n\}\ni i\mapsto w[i]\in \{a,b\}$$ ($$n$$ is the length of the word and, if $$n=0$$, one gets the empty word i.e. the unit of the free monoid). For each word having an inversion at place "$$i$$" i.e. $$w=p\,ba\,s$$ where $$|p|=i-1$$, we have the rewrite rule $$\begin{eqnarray} &&r_{w,i}(g_{w}\otimes\ldots \otimes g_b\otimes g_a\otimes\ldots g_{w[n]})=\cr &&g_{w}\otimes\ldots \otimes [g_b,g_a]_1\otimes\ldots g_{w[n]}+\cr &&g_{w}\otimes\ldots \otimes [g_b,g_a]_2\otimes\ldots g_{w[n]}+\cr &&g_{w}\otimes\ldots \otimes g_a\otimes g_b\otimes\ldots g_{w[n]}\cr && \in T_{p\,a\,s}\oplus T_{p\,b\,s}\oplus T_{p\,ab\,s} \end{eqnarray}$$ (where $$[g,h]_i:=p_i([g,h])$$, $$p_i$$ being the projector on the summands of the decomposition $$\mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2$$) if the indexing word has no inversion at place $$i$$'', $$r_{w,i}$$ acts trivially (as identity). It is not difficult to check that this system (as usual $$r_{w,i}$$ is extended to $$T(\mathfrak{g})$$ by identity to other summands) is confluent (as, if there are two inversions, the indexing word must be of the form $$w=p\,ba\,u\,ba\,s$$) and noetherian (as the rules reduce either length or number of inversions). So, the linear map $$\rho$$ which sends any tensor to its normal form

1. is a projector
2. its image is $$\oplus_{w\in \{a,b\}^*_{irr}}\,T_w=T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2)$$ ($$\{a,b\}^*_{irr}$$ being the set of words without inversion i.e. $$a^*b^*=\{a^pb^q\}_{p,q\geq 0}$$)
3. its kernel is the submodule generated by the differences $$t-r_{w,i}(t)$$
4. due to the rewriting properties, one has $$\rho(u\otimes \rho(v)\otimes w)=\rho(u\otimes v\otimes w)$$
From now on, we return to the original labeling $$\mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2$$.

Restricting $$\rho$$ to its image, we get a surjection (still called $$\rho$$ here) $$\rho\ :\ T(\mathfrak{g}) \rightarrow T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2)$$ and the factor embedding $$j\ :\ T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2)\hookrightarrow T(\mathfrak{g})$$ is a section of it. Together with point (4) above, this suffices to prove that there exists a (unique) law of algebra (associative with unit, AAU in the sequel) on $$T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2)$$ (call it $$*$$) such that $$\rho$$ is a morphism (of AAU), this law reads $$t_1*t_2:=\rho(j(t_1)\otimes j(t_2))$$ and (we identify $$t$$ and $$j(t)$$) $$(t_1*t_2)*t_3=\rho(\rho(t_1\otimes t_2)\otimes t_3)= \rho(t_1\otimes t_2\otimes t_3)=t_1*(t_2*t_3)$$

then the kernel of $$\rho$$ is a two sided ideal which contains all the elements (with $$(g_2,h_1)\in \mathfrak{g}_2\times \mathfrak{g}_1$$) $$B(g_2,h_1)=g_2\otimes h_1 - h_1\otimes g_2 - [g_2,h_1]\ .$$ By composition, we now have a linear morphism $$\beta_0=(s_1\otimes s_2)\circ\rho$$ $$T(\mathfrak{g}) \stackrel{\rho}{\rightarrow} T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2) \stackrel{s_1\otimes s_2}{\longrightarrow} \mathcal{U}(\mathfrak{g}_1)\otimes\mathcal{U}(\mathfrak{g}_2)$$ where $$s_i$$ is the canonical surjection $$T(\mathfrak{g}_i)\to \mathcal{U}(\mathfrak{g}_i)$$. Let us show now that the kernel of $$s:T(\mathfrak{g})\to \mathcal{U}(\mathfrak{g})$$ is included in $$ker(\beta_0)$$, so that we could factorize $$\beta_0$$ through $$\mathcal{U}(\mathfrak{g})$$ as follows.
$$\require{AMScd} \begin{CD} T(\mathfrak{g}) @>{\rho}>>T(\mathfrak{g}_1)\otimes T(\mathfrak{g}_2)\\ @V{s}VV @VV{s_1\otimes s_2}V \\ \mathcal{U}(\mathfrak{g}) @>{\beta}>> \mathcal{U}(\mathfrak{g}_1)\otimes\mathcal{U}(\mathfrak{g}_2) \end{CD}$$ Let us prove that this last diagram is admissible.

We know that $$\mathcal{J}=ker(s)$$ is the two sided ideal generated by the elements $$\{B(g,h)\}_{g,h\in \mathfrak{g}}$$ where $$B(g,h)=g\otimes h - h\otimes g - [g,h]$$ As $$\mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2$$ and $$B(-,-)$$ is bilinear antisymmetric it amounts to the same to split the family in three relators $$\{B(g_1,h_1)\}_{g_1,h_1\in \mathfrak{g}_1}\cup \{B(g_2,h_2)\}_{g_2,h_2\in \mathfrak{g}_2}\cup \{B(g_2,h_1)\}_{(g_2,h_1)\in \mathfrak{g}_2\times \mathfrak{g}_1}$$ Calling $$\mathcal{J}_{ij}$$ be the corresponding (two-sided) ideals, we have already shown that $$\mathcal{J}_{21}\subset ker(\rho)\subset ker(\beta_0)$$ The fact that $$\mathcal{J}_{11}\subset ker(\beta_0)$$ is a consequence of the following identity (for $$g_2\in \mathfrak{g}_2$$ and $$g_1,h_1\in \mathfrak{g}_1$$) $$g_2\otimes B(g_1,h_1)\equiv_\rho B([g_2,g_1]_1,h_1)+B(g_1,[g_2,h_1]_1)+ B(g_1,h_1)\otimes g_2\qquad (*)$$ where $$\equiv_\rho$$ means the equivalence modulo $$ker(\rho)$$.

One shows very similarly that $$\mathcal{J}_{22}\subset ker(\beta_0)$$, hence the square diagram above. Computing on generators shows that $$\alpha$$ and $$\beta$$ are mutually inverse.

Remarks i) The same proof seems to show that, in case $$\mathfrak{g}=\mathfrak{g}_1\oplus \mathfrak{s}$$ where $$\mathfrak{s}$$ is some submodule, then the arrow $$\alpha\ :\ \mathcal{U}(\mathfrak{g_1})\otimes \mathcal{A}(\mathfrak{s})\to \mathcal{U}(\mathfrak{g})$$ (where $$\mathcal{A}(\mathfrak{s})$$ is the subalgebra generated by $$\mathfrak{s}$$) is one-to-one.

ii) Generalizing the straightening process to $$|I|<+\infty$$, and taking into account the ambiguities (squares and hexagons), it seems true that, for any decomposition $$\mathfrak{g}=\oplus_{i\in I}\, \mathfrak{g}_i$$ (equality is as $$R$$-modules but any individual $$\mathfrak{g}_i$$ is a Lie subalgebra), the ($$I$$ being linearly ordered) multiplication map (only linear) $$\stackrel{\rightarrow}{\otimes}_{i\in I}\mathcal{U}(\mathfrak{g}_i) \stackrel{\alpha}{\longrightarrow} \mathcal{U}(\mathfrak{g})$$ is one-to-one. However the construction and proofs seem considerably more difficult.

iii) If $$I$$ is infinite, one first constructs $$\stackrel{\rightarrow}{\otimes}_{i\in I}\mathcal{U}(\mathfrak{g}_i)$$ as in Bourbaki Algebra Chapter III § 4.5 using the unities but not the algebra structure. We can, again, ask the same question.

Please do not hesitate to interact if something is wrong or unclear !

Late edit i) Proof of identity $$(*)$$.

Let us use two derivations within $$T(\mathfrak{g})$$ defined on the generators $$g\in \mathfrak{g}$$ by $$ad_{g_2}^{\otimes}(g):=g_2\otimes g-g\otimes g_2=[g_2,g]_{\otimes} \ ;\ ad_{g_2}(g):=[g_2,g]$$ then, for $$g_2\in \mathfrak{g}_2$$ and because, for $$u\in\mathfrak{g}_1$$, $$ad_{g_2}^{\otimes}(u)\equiv_\rho ad_{g_2}(u)$$,
$$\begin{eqnarray} && ad_{g_2}^{\otimes}(B(g_1,h_1))=ad_{g_2}^{\otimes}([g_1,h_1]_{\otimes}-[g_1,h_1])\cr &&=[[g_2,g_1]_{\otimes},h_1]_{\otimes}+[g_1,[g_2,h_1]_{\otimes}]_{\otimes}- ad_{g_2}^{\otimes}([g_1,h_1])\cr &&\equiv_\rho [[g_2,g_1]_{\otimes},h_1]_{\otimes}+[g_1,[g_2,h_1]_{\otimes}]_{\otimes}- ad_{g_2}([g_1,h_1])\cr &&\equiv_\rho [[g_2,g_1],h_1]_{\otimes}+[g_1,[g_2,h_1]]_{\otimes}- [[g_2,g_1],h_1]-[g_1,[g_2,h_1]]\cr &&\equiv_\rho B([g_2,g_1],h_1)+B(g_1,[g_2,h_1])\cr &&\equiv_\rho B([g_2,g_1]_1,h_1)+B(g_1,[g_2,h_1]_1)\cr \end{eqnarray}$$

ii) Proof of $$\mathcal{J}_{11}\subset ker(\beta_0)$$

The two-sided ideal $$\mathcal{J}_{11}$$ is linearly generated by the elements $$s=u_1\otimes B(g_1,h_1)\otimes u_2$$ with $$g_1,h_1\in \mathfrak{g}_1$$ and $$u_i\in T_{w_i}$$

We first rewrite each $$u_i$$ with $$\rho$$ and get that the two-sided ideal $$\mathcal{J}_{11}$$ is linearly generated by elements of the form $$t\equiv_\rho t_1\otimes t_2 \otimes B(g_1,h_1)\otimes t_3\otimes t_4$$ with $$t_1\in T_{a^p},\ t_2\in T_{b^q},\ t_3\in T_{a^r},\ t_4\in T_{b^s}$$.

If $$q>0$$, we have $$t_2=t'_2\otimes g_2$$, for some $$g_2\in \mathfrak{g}_2$$

then using identity $$(*)$$ , one has $$t\equiv_\rho t_1\otimes t'_2 \otimes \Big( B([g_2,g_1]_1,h_1)+B(g_1,[g_2,h_1]_1)+B(g_1,h_1)\otimes g_2\Big)\otimes t_3\otimes t_4$$ using a recurrence on $$q$$, one gets $$t\equiv_\rho t_1\otimes \Big(\sum_{i=1}^{q} B(x_i,y_i)\otimes v_i\otimes t_3\otimes t_4\Big)$$ with $$x_i,y_i\in \mathfrak{g}_1$$ and $$v_i$$ in some $$T_{w_i}$$. Now $$t\equiv_\rho \Big(\sum_{i=1}^n t_1\otimes B(x_i,y_i)\otimes \rho(v_i\otimes t_3\otimes t_4)\Big)$$ applying $$\beta_0$$ (which is compatible with $$\equiv_\rho$$) to both members, we have $$\beta_0(t)=\beta_0\Big(\sum_{i=1}^n t_1\otimes B(x_i,y_i)\otimes \rho(v_i\otimes t_3\otimes t_4)\Big)=0$$

• Why is the multiplication $*$ associative? Also, some extraneous $\in$ signs in the definition of $r_{w,i}$. Jun 9, 2018 at 17:18
• @darijgrinberg [Why is the multiplication ∗ associative]---> Good question, let me think of it (maybe my argument is incomplete). [$\in$ signs in the definition of $r_{w,i}$]---> this is to show in which sectors the result goes. Jun 9, 2018 at 17:35
• 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). Jun 9, 2018 at 17:38
• @darijgrinberg I think it is fixed, this is due to the rewriting process (I added property 4), thanks for interaction (+1) Jun 9, 2018 at 19:09
• 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 $\mathcal{U}(\mathfrak{g}_1)$. Jun 9, 2018 at 20:39