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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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No hidden linear relations in degree one

Statement

If a basis of $g$ is supplied, the linear map

$k \oplus g ⟶ F_{1} U (g), (a, x) ⟼ a 1 + ι_{g} (x),$

is an isomorphism.

Facts & Assumptions

Given: A Lie algebra with a specified totally ordered basis.

[L1]

PBW makes the empty word and all length-one ordered words part of one basis of $U (g)$ (Poincaré–Birkhoff–Witt theorem).

[L2]

$F_{1}$ is spanned by words of length at most one (PBW filtration on the enveloping algebra).

Proof

technique · direct

1.1

By [L2], every element of $F_{1}$ is $a 1 + ι_{g} (x)$ , so the displayed map is surjective.

L2algebra

1.2

By [L1], the empty PBW word and the length-one basis words are linearly independent. Therefore $a 1 + ι_{g} (x) = 0$ implies $a = 0$ and every basis coefficient of $x$ is zero, so the map is injective.

L1algebra

2.1

The map is linear, injective, and surjective, hence is an isomorphism; for $g = 0$ this reduces to the degree-zero map $k \to F_{1} U (0)$ .

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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, Theorem 13.1 and Corollary 13.3, printed pp. 74–75 (standard reference, not scraped)