Does the proof of the Poincare-Birkhoff-Witt theorem need the Jacobi identity? The title says it. Suppose I have a vector space $V$ equipped with a bilinear bracket such that $[x,y]=-[y,x]$, and define the universal enveloping algebra $U$ as usual: namely the tensor algebra on $V$ modulo the 2-sided ideal generated by 
$$x\otimes y-y\otimes x=[x,y]$$
Then an ordered basis $\{x_i\}_{i\in I}$ of $V$ give rise to an order bases of $U$:
$$\{\prod_{j=1}^tx_{i_j}^{k_j}\}_{i_1<i_2<\cdots<i_t, t\in\mathbb{N}}$$
Is this conclusion right?
I don't think the Jacobi identity is needed in the usual proof, but I just read one written by Paul Garrett (http://www.math.umn.edu/~garrett/m/algebra/pbw.pdf) which used it, so I'm not so sure about it.
 A: The Jacobi identity is needed. Otherwise the conclusion of the PBW theorem is false. This is discussed in Bergman's famous paper on the Diamond Lemma.
For a simple example, take your "algebra" to be generated by $x$, $y$ and $z$, with $[x,y]=y$, $[x,z]=z$ and $[y,z]=x$. Then in the "enveloping algebra", $x$ is zero.
Much later: in fact, the natural map $\mathfrak g\to\mathcal U(\mathfrak g)$ from an algebra with an antisymmetric bracket to its enveloping algebra constructed as if it $\mathfrak g$ were a Lie algebra has kernel precisely the subspace of $\mathfrak g$ you need to kill in order to get the Jacobi identity.
A: The answer to the title question is "yes" and the answer to the body question is "no." The point is that 


*

*$U$ is an algebra,

*algebras under commutator automatically satisfy the Jacobi identity, and

*PBW implies that every Lie algebra embeds into an algebra under commutator. 


More explicitly, if $V$ doesn't satisfy the Jacobi identity, then you can find $x, y, z$ such that $J(x, y, z) \neq 0$ (where $J = 0$ is the Jacobi identity). But $J = 0$ in $U$, and this leads to a nontrivial linear dependence between whatever elements you get when you evaluate $J$ in $U$. 
