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

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Universal property of the enveloping algebra

Statement

Let $A$ be a unital associative $k$ -algebra, equipped with its commutator Lie bracket. Every Lie-algebra homomorphism $f : g \to A_{L i e}$ extends uniquely to a unital algebra homomorphism $\overline{f} : U (g) \to A$ satisfying $\overline{f} ι_{g} = f$ .

Facts & Assumptions

Given: A Lie-algebra map $f : g \to A_{L i e}$ into a unital associative $k$ -algebra.

[L1]

A linear map from $g$ extends uniquely to an algebra map from $T (g)$ (Universal property of the tensor algebra).

[L2]

A map killing an ideal factors uniquely through the quotient (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring).

[L3]

The defining relators and canonical map are those of Universal enveloping algebra.

Proof

technique · direct

1.1

By [L1], $f$ extends uniquely to a unital algebra homomorphism $F : T (g) \to A$ . Since $f$ preserves Lie brackets, $F (x \otimes y - y \otimes x - [x, y]) = f (x) f (y) - f (y) f (x) - f ([x, y]) = 0$ for every defining relator.

L1algebra

2.1

The kernel of $F$ is a two-sided ideal containing all defining relators, hence contains their generated ideal $I$ . By [L2], $F$ factors uniquely as $T (g) ↠ U (g) \overset{\overline{f}}{\to} A$ , and the factor satisfies $\overline{f} ι_{g} = f$ .

step 1.1L2L3

3.1

If $G : U (g) \to A$ is another unital algebra map with $G ι_{g} = f$ , its composite with $T (g) ↠ U (g)$ is an algebra extension of $f$ . It equals $F$ by [L1], and quotient-map surjectivity gives $G = \overline{f}$ .

step 2.1L1L3algebra

4.1

Thus the required extension exists uniquely; the argument uses only the quotient presentation and not injectivity of $ι_{g}$ .

step 2.1step 3.1∎

Depends on

Used by

PBW linear independence via the ordered-monomial model Lemma
Enveloping algebra of a direct sum Proposition
Functoriality of the enveloping algebra Proposition
Hopf-algebra structure on U(g) Remark
Lie representations are U(g)-modules Theorem

Dependency tree · two levels

16 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, §12.1, printed pp. 69–70 (standard reference, not scraped)
Kirillov, An Introduction to Lie Groups and Lie Algebras, §5.1, printed pp. 71–72 (standard reference, not scraped)