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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.

h__
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2 Answers2

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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.

Pedro
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Mariano Suárez-Álvarez
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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 $U$.

Qiaochu Yuan
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  • Thanks... I just realized this and was about to delete the question. – h__ Nov 18 '12 at 05:19
  • 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. – h__ Nov 18 '12 at 05:31