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.


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.

  • $\begingroup$ (If the basefield is of characteristic not two...) $\endgroup$ – Mariano Suárez-Álvarez Nov 18 '12 at 5:29
  • 2
    $\begingroup$ It is a result of A. Braverman and D. Gaitsgory (J. Algebra, 181 (1996), pp. 315–328) that the PBW property is essentially the same as the Jacobi identity, in the general context of Koszul quadratic algebras. $\endgroup$ – Mariano Suárez-Álvarez Nov 18 '12 at 5:32

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 $U$.

  • $\begingroup$ Thanks... I just realized this and was about to delete the question. $\endgroup$ – h__ Nov 18 '12 at 5:19
  • $\begingroup$ In other words even $V\to U$ is not injective without assuming Jacobi's identity. So Garrett's remark "It is not clear a priori that the Jacobi identity $[x; [y; z]] + [y; [x; z]] = [[x; y]; z]$ plays a role in the argument" is not that interesting. $\endgroup$ – h__ Nov 18 '12 at 5:31

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