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

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The canonical map to U(g) is a Lie homomorphism

Statement

For every $x, y \in g$ , the canonical map satisfies

$ι_{g} ([x, y]) = ι_{g} (x) ι_{g} (y) - ι_{g} (y) ι_{g} (x) .$

Consequently $ι_{g} : g \to U (g)_{L i e}$ is a Lie-algebra homomorphism into the commutator Lie algebra.

Facts & Assumptions

Given: The quotient presentation of $U (g)$ and canonical linear map $ι_{g}$ from Universal enveloping algebra.

[L1]

Every generator $x \otimes y - y \otimes x - [x, y]$ of the defining ideal has zero image in the quotient.

Proof

technique · direct

1.1givenL1algebra

Applying the quotient map to the relator in [L1] gives $ι_{g} (x) ι_{g} (y) - ι_{g} (y) ι_{g} (x) - ι_{g} ([x, y]) = 0$ , which is the displayed identity.

2.1step 1.1∎

Since $ι_{g}$ is linear by construction and step 1.1 is bracket preservation, it is a Lie-algebra homomorphism. No injectivity has been used.

Depends on

Universal enveloping algebra

The canonical map to U(g) is a Lie homomorphism

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources