# Does every Lie algebra come from commutator of some associative product operation?

Suppose $\mathfrak{g}$ is an Lie algebra. Is it possible to define an associative product operation $\star$ on $\mathfrak{g}$ such that $[A,B]=A\star B - B \star A$ ? If it is not possible to do so in general, how do we know which Lie algebras come as commutator of some product operation ? (Are there any useful criteria ?)

I am aware (without proof) that every Lie algebra has an embedding $f$ into an associative algebra such that the Lie bracket $[A,B]$ corresponds to $f(A)f(B)-f(B)f(A)$,but the question that I ask above demands something stronger.

• I think the universal enveloping algebra satisfies your criteria: en.wikipedia.org/wiki/Universal_enveloping_algebra. Look at the "Direct Construction" section. – Adam Saltz Oct 28 '13 at 18:16
• I don't think it solves the problem, since the universal enveloping algebra is built on a much larger space than the Lie algebra. For sl_2, say, you want a three-dimensional associative algebra whose commutator is the Lie product. It seems wildly implausible that this is possible for general Lie algebras without being advertised along with the PBW theorem, and I would expect instead a strong No-Go theorem that it is generally not possible except in extremely special cases. – zyx Oct 28 '13 at 18:32
• @user90041 I learned about them from Kassel, Rosso, and Turaev's Quantum Groups and Knot Invariants and also Hong and Kang's Quantum Groups and Crystal Bases. – Adam Saltz Oct 28 '13 at 18:40
• Do we insist that $\star$ be bilinear? – Julian Rosen Oct 28 '13 at 18:45
• @user90041: I'm not sure myself, but I thought it might be useful. With Ado's theorem you can assume without loss of generality that your $\mathfrak g$ is subspace of matrices and operation is matrix commutator. Unfortunately the problem still remains, your subspace might not be closed under matrix multiplication. – Rafael Mrđen Oct 30 '13 at 11:07

The answer is no in general, and I suspect it's no for all simple Lie algebras. Here's an argument that works for many simple Lie algebras including $\mathfrak{sl}_2$. (Dollars to doughnuts there's a more direct elementary proof for $\mathfrak{sl}_2$.)
Suppose that $\mathfrak{g}$ is a non-commutative Lie algebra which arises as the commutators for some algebra. Then there's a map of $\mathfrak{g}$-modules $\mathfrak{g} \otimes \mathfrak{g} \rightarrow \mathfrak{g}$ given by multiplication $x \otimes y \mapsto xy$ and another map given by opposite multiplication $x \otimes y \mapsto yx$. Since $\mathfrak{g}$ is non-commutative these two maps are linearly independent, so $\dim \mathrm{Hom}_{\mathfrak{g}\text{-mod}}(\mathfrak{g} \otimes \mathfrak{g}, \mathfrak{g}) \geq 2$. So any non-commutative Lie algebra with the property that $\dim \mathrm{Hom}_{\mathfrak{g}\text{-mod}} (\mathfrak{g} \otimes \mathfrak{g}, \mathfrak{g}) = 1$ cannot come from an associative algebra. Examples include $\mathfrak{sl}_2$ and all simple Lie algebras outside the A series.
• Could you please elaborate on why ${\sf dim}({\sf Hom}({\mathfrak g} \otimes {\mathfrak g}, {\mathfrak g})=1)$ when ${\mathfrak g}$ is a simple algebra ? By e.g. giving a reference. – Ewan Delanoy Nov 9 '13 at 17:56
• @EwanDelanoy: It's just a calculation that you do seperately in every case. For $\mathfrak{sl}_2$ you should be able to check it easily by hand. For other simple Lie algebras the calculation is harder. It's certainly in Bird Tracks and for the BCD series it should be in Fulton-Harris. I know it from Vogel's uniform formula for decomposing $\mathfrak{g} \otimes \mathfrak{g}$ e.g. at the end of this paper. – Noah Snyder Nov 9 '13 at 18:32
The answer is indeed no for all semisimple Lie algebras. Such an associative product would determine a left-symmetric and a right-symmetric structure on the Lie algebra. It is known that this cannot exist for semisimple Lie algebras over a field of characteristic zero because of Whitehead's lemma. In fact, there is no left-symmetric structure. If $L$ denotes the Lie algebra, the left-multiplication defines an $L$-module $M_L$ which must satisfy $H^1(L,M_L)=0$. Since the identity $id$ is a $1$-cocylce of $L$ with values in $M$, it must be a coboundary, i.e., there exists an element $e$ such that the right-multiplication $R(e)$ by $e$ is the identity. Because of $[L,L]=L$ however, all right multiplications have zero trace. Hence the identity has zero trace - a contradiction.