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

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The canonical map g→U(g) is injective

Statement

If a basis of $g$ is supplied, the canonical map $ι_{g} : g \to U (g)$ is injective. In particular, no degree-one basis vector vanishes in the enveloping algebra.

Facts & Assumptions

Given: A Lie algebra $g$ with a specified basis and any specified total order on that basis.

[L1]

Length-one ordered PBW monomials are part of a basis of $U (g)$ (Poincaré–Birkhoff–Witt theorem).

Proof

technique · direct

1.1

The images under $ι_{g}$ of the supplied basis elements are exactly the distinct length-one PBW monomials, so [L1] makes them linearly independent.

2.1

Every $x \in g$ has a unique finite basis expansion, and $ι_{g} (x) = 0$ forces all its coefficients to vanish by step 1.1. Hence $x = 0$ and $ι_{g}$ is injective; if the basis is empty, $g = 0$ and the claim is immediate.

step 1.1algebra∎

Depends on

Poincaré–Birkhoff–Witt theorem

Used by

An enveloping algebra need not be commutative False statement
Injectivity is not part of the enveloping quotient definition False statement

Dependency tree · two levels

5 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, Corollary 13.3, printed p. 75 (standard reference, not scraped)
Kirillov, An Introduction to Lie Groups and Lie Algebras, Corollary 5.13, printed p. 75 (standard reference, not scraped)