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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 $$ (u_1\otimes u_2).(v_1\otimes v_2)=(u_1\otimes 1)\Big((\delta\otimes\gamma_2)\circ\Delta(u_2)[v_1]\otimes v_2\Big) $$ where $\gamma_2(m)$ is the multiplication by $m$ on the left within $\mathcal{U}(\mathfrak{g}_2)$.
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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 May 23 '18 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 May 23 '18 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 May 25 '18 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 May 25 '18 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 May 26 '18 at 9:23
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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. May 25 '18 at 22:37
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    $\begingroup$ I've specified the categories, and I hope this makes it clearer. $\endgroup$ – Thomas Poguntke May 26 '18 at 8:23
  • $\begingroup$ @ThomasPogunke Well, thank you (+1), let me check (and possibly clarify). $\endgroup$ – Duchamp Gérard H. E. May 26 '18 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 May 26 '18 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 May 26 '18 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 May 26 '18 at 22:25
  • $\begingroup$ Yes, Darij has rightly understood what I wrote (direct sum as a module). This is my concern and one of the dreams of Pr. Schützenberger : understand as much as possible left-right (non trivial) rewritings. Reference to Bourbaki added (point 2.5). $\endgroup$ – Duchamp Gérard H. E. May 27 '18 at 0:09
  • $\begingroup$ Added the case of semi-direct products as point 5. $\endgroup$ – Duchamp Gérard H. E. May 27 '18 at 6:14
  • $\begingroup$ Finally, the answer is yes $\alpha$ is one-to one in any case. I put a proof (relying on indexed rewriting) as answer (sorry :). Do not hesitate to interact and ask clarifications in here on in chat. $\endgroup$ – Duchamp Gérard H. E. Jun 10 '18 at 14:52
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The answer is yes $\alpha$ is one-to one in any case.

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 $a<b$ 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 $$ \begin{eqnarray} &&r_{w,i}(g_{w[1]}\otimes\ldots \otimes g_b\otimes g_a\otimes\ldots g_{w[n]})=\cr &&g_{w[1]}\otimes\ldots \otimes [g_b,g_a]_1\otimes\ldots g_{w[n]}+\cr &&g_{w[1]}\otimes\ldots \otimes [g_b,g_a]_2\otimes\ldots g_{w[n]}+\cr &&g_{w[1]}\otimes\ldots \otimes g_a\otimes g_b\otimes\ldots g_{w[n]}\cr && \in T_{p\,a\,s}\oplus T_{p\,b\,s}\oplus T_{p\,ab\,s} \end{eqnarray} $$ (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 $$ \beta_0(t)=\beta_0\Big(\sum_{i=1}^n t_1\otimes B(x_i,y_i)\otimes \rho(v_i\otimes t_3\otimes t_4)\Big)=0 $$

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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 Jun 9 '18 at 17:18
  • $\begingroup$ @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. $\endgroup$ – Duchamp Gérard H. E. Jun 9 '18 at 17:35
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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 Jun 9 '18 at 17:38
  • $\begingroup$ @darijgrinberg I think it is fixed, this is due to the rewriting process (I added property 4), thanks for interaction (+1) $\endgroup$ – Duchamp Gérard H. E. Jun 9 '18 at 19:09
  • $\begingroup$ 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 Jun 9 '18 at 20:39

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