PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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The enveloping algebra of an abelian Lie algebra is symmetric

Statement

If $g$ is abelian, the canonical algebra map $S (g) \to U (g)$ is an isomorphism.

Facts & Assumptions

Given: An abelian Lie algebra $g$ over $k$ .

[L1]

$U (g)$ is the quotient of $T (g)$ by relators $x \otimes y - y \otimes x - [x, y]$ (Universal enveloping algebra).

[L2]

$S (g)$ is the quotient by relators $x \otimes y - y \otimes x$ (Symmetric algebra of a vector space).

Proof

technique · direct comparison of quotient presentations

1.1

Since $g$ is abelian, $[x, y] = 0$ for all $x, y$ , so every defining enveloping relator in [L1] is exactly the corresponding symmetric relator in [L2]. The two generated two-sided ideals are equal.

givenL1L2algebra

2.1

Quotienting the same tensor algebra by the same ideal gives a canonical unital algebra isomorphism $S (g) ≅ U (g)$ fixing the image of $g$ . This includes $g = 0$ and uses no choice of basis; PBW is consistent with, but unnecessary for, this presentation argument.

step 1.1algebra∎

Depends on

Used by

The enveloping algebra of a one-dimensional abelian Lie algebra Example

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Etingof, MIT 18.745 notes, Example 12.2, printed p. 69 (standard reference, not scraped)
Kirillov, An Introduction to Lie Groups and Lie Algebras, §5.1, printed pp. 71–72 (standard reference, not scraped)