TITLE: Does every Lie algebra come from commutator of some associative product operation?
QUESTION [10 upvotes]: 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.

REPLY [2 votes]: 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.