# Prove isomorphism between universal enveloping algebras

I have the following problem I can't solve. I am following a couse in Lie Algebras, and one of the exercises is prove the following isomorphism: Given a semisimple lie algebra $$\mathfrak{g}$$ over a field $$\mathfrak{k}$$ of characateristic 0, and let $$\mathfrak{h}$$ be a toral maximal subalgebra, then by the cartan-chevalley decomposition, $$\mathfrak{g}$$ splits as $$\mathfrak{g}=\mathfrak{n}^- \oplus \mathfrak{h} \oplus \mathfrak{n}$$. Let $$\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}$$. Then one has that the map:

$$f:\mathfrak{U}(\mathfrak{n}^-)\otimes\mathfrak{U}(\mathfrak{b}) \rightarrow \mathfrak{U}(\mathfrak{g})$$

$$x\otimes y\longrightarrow xy$$

The exercise is to prove that this map is in fact an isomorphism. Where $$\mathfrak{U}(\mathfrak{g})$$ is the universal enveloping algebra of the $$\mathfrak{g}$$ and respectively for the rest.

I am very confused, because if you remove the $$\mathfrak{U}$$ part this map is no longer injective, since it's easy to find $$x\in\mathfrak{n},y\in\mathfrak{b}$$ so that $$xy=0$$, but I don't understand how that is possible without violating the injectivity of $$f$$.

Any help is appreciated. Thanks to everyone.

P.D: For clarification, the Cartan-Chevalley decomposition establishes than $$\mathfrak{g}= \mathfrak{h} \oplus \bigoplus_{\alpha\in\Phi} \mathfrak{g}_\alpha$$ where $$\mathfrak{h}$$ is a maximal abelian subalgebra of ad-diagonalizable elements. (This just means that the linear map $$ad(x)$$ is diagonalizable). Furthermore $$\mathfrak{g}_\alpha=\{ x\in\mathfrak{g} : [h,x]=\alpha(h)x \}$$. Where $$\Phi\subset\mathfrak{h}^*$$ and $$\Phi$$ can be split into $$\Phi=\Phi^+\cup\Phi^-$$. Then we define $$n = \bigoplus_{\alpha\in\Phi^+} \mathfrak{g}_\alpha$$ and $$n^- = \bigoplus_{\alpha\in\Phi^-} \mathfrak{g}_\alpha$$.

• What are $\mathfrak n$ and $\mathfrak n^-$? Can you tell an explicit example of $x,y$ as in your last paragraph? Note that $[x,y]=0$ doesn't imply $xy=0$ (but only $xy=yx$) in the enveloping algebra. Jan 6 at 23:40
• @Berci They are the algebras generated by the negatives and positive roots in the cartan chevalley decomposition. Jan 7 at 0:05
• @Berci I just added clarification for $\mathfrak{n}$ and $\mathfrak{n}^-$. Jan 7 at 0:10
• Shouldn't the first factor be $\mathfrak{U}(\mathfrak n^-)$? (Actually I guess $\mathfrak{U}(\mathfrak n^-) \simeq \mathfrak{U}(\mathfrak n)$, but it looks more plausible that way.) Jan 7 at 1:53
• The "I am very confused paragraph" indeed seems confused: When you say it's easy to find $x,y$ such that $xy=0$, do you mean "so that $[x,y]=0$"? Then that's true but rather irrelevant, because the product $xy$ in $\mathfrak{U}(\mathfrak g)$ on the RHS of the map $f$ is the associative product in the universal enveloping algebra, which has little to do with the Lie bracket. (Well actually, the Lie bracket is recovered from it via $[x,y]= xy-yx$, but the point is it's not the same, and maybe that is the basis of your confusion.) Jan 7 at 2:01

## 1 Answer

In fact, if $$L$$ is any Lie algebra over any field $$k$$, and $$L_1, L_2$$ are subalgebras such that $$L = L_1 \oplus L_2$$ as $$k$$-vector spaces, then the $$k$$-linear map $$f$$ given by

$$U(L_1) \otimes_k U(L_2) \rightarrow U(L)$$ $$x \otimes y \mapsto xy$$

is an isomorphism of $$k$$-vector spaces (and hence of $$(U(L_1),U(L_2))$$-bimodules).

As noticed in the first bullet point in this question, this follows from the Poincaré-Birkhoff-Witt theorem, compare e.g. Bourbaki, Lie Groups and Algebras ch. 1 §7. Indeed, it is a corollary of PBW (in Bourbaki, Corollary 3) that if $$(e_1, ..., e_n)$$ is an ordered $$k$$-basis of a Lie algebra $$\mathfrak g$$, then the monomials $$e_1^{r_1} \cdot ... \cdot e_n^{r_n}$$ are a $$k$$-basis of $$U(\mathfrak g)$$. From this the assertion (which is stated in greater generality in Bourbaki's Corollary 6) follows. Namely, if now $$(x_1, ..., x_m)$$ is an ordered $$k$$-basis of $$L_1$$, and $$y_1, ..., y_n$$ one of $$L_2$$, then on the one hand PBW and tensor products over fields tell us that the set of all $$x_1^{r_1} \cdot ... \cdot x_m^{r_m} \otimes y_1^{l_1} \cdot ... \cdot y_n^{l_n}$$ is a $$k$$-basis of $$U(L_1) \otimes U(L_2)$$, on the other hand the images of these elements under $$f$$, namely, all $$x_1^{r_1} \cdot ... \cdot x_m^{r_m} \cdot y_1^{l_1} \cdot ... \cdot y_n^{l_n}$$, again by PBW are a $$k$$-basis of $$U(L)$$, because by assumption $$(x_1, ..., x_m, y_1, ..., y_m)$$ is an ordered $$k$$-basis of $$L$$.

Now apply this to $$L=\mathfrak g, L_1 = \mathfrak{n}^-, L_2= \mathfrak b$$.

To see where your supposed counterargument goes wrong (copied from comments and discussion):

Your computation of $$xy=0$$ for certain $$x, y \neq 0$$ seems to be happening in $$M_3(\mathfrak k)$$ (or $$End_{\mathfrak k} ({\mathfrak k}^3)$$). That matrix ring is an enveloping associative algebra of $$\mathfrak{sl}_3$$, and maybe likewise would be certain associative subalgebras $$B:= \pmatrix{*&*&*\\0&*&*\\0&0&*}$$, $$N^-:=\pmatrix{0&0&0\\*&0&0\\*&*&0}$$ corresponding to $$\mathfrak b, \mathfrak n^-$$. These algebras are quotients of the respective universal enveloping algebras (by the latter's universal property, although surjectivity might be more subtle), but of course a product being zero in a quotient tells us little about the product being zero in the original ring.

What this computation shows, then, is that even though there are canonical surjections of associative algebras $$U(\mathfrak n^-) \twoheadrightarrow N^-, U(\mathfrak b) \twoheadrightarrow B, U(\mathfrak g) \twoheadrightarrow M_3(k)$$ the map $$N^- \otimes_{\mathfrak k} B \rightarrow M_3(k)$$ induced via those projections from the given map $$U(n^-) \otimes U(b) \simeq U(g)$$, is no longer an isomorphism. But viewed this way, that is maybe not as surprising.

Full disclosure, I have made the mistake of mixing up standard enveloping matrix algebras with the universal enveloping algebra before, and I still sometimes do calculations in those matrix rings to get a first feeling for the universal enveloping one, but it's important to remember that the matrix algebras (like the above $$M_3(\mathfrak k), B, N^-$$) are just quotients -- and for that matter, with a lot factored out -- of the "much, much bigger" universal enveloping algebras. In particular, note that the above PBW corollaries also imply that $$x^my^n \neq 0$$ in $$U(\mathfrak g)$$ for all $$m,n \in \mathbb N$$ and any $$x \neq 0 \neq y \in \mathfrak g$$.

• Thank you very much, this is very nice. You are right about it being $\mathfrak{n}^-$ I just edited it. Jan 8 at 0:34