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 $.

  • $\begingroup$ 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. $\endgroup$
    – Berci
    Jan 6, 2021 at 23:40
  • $\begingroup$ @Berci They are the algebras generated by the negatives and positive roots in the cartan chevalley decomposition. $\endgroup$ Jan 7, 2021 at 0:05
  • $\begingroup$ @Berci I just added clarification for $\mathfrak{n}$ and $\mathfrak{n}^-$. $\endgroup$ Jan 7, 2021 at 0:10
  • $\begingroup$ 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.) $\endgroup$ Jan 7, 2021 at 1:53
  • 2
    $\begingroup$ 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.) $\endgroup$ Jan 7, 2021 at 2:01

1 Answer 1


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 $x \neq 0 \neq y \in \mathfrak g$.

  • $\begingroup$ Thank you very much, this is very nice. You are right about it being $\mathfrak{n}^-$ I just edited it. $\endgroup$ Jan 8, 2021 at 0:34

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