Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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

Statement

For every x,yg, the canonical map satisfies

ιg([x,y])=ιg(x)ιg(y)ιg(y)ιg(x).

Consequently ιg:gU(g)Lie is a Lie-algebra homomorphism into the commutator Lie algebra.

Facts & Assumptions

Given: The quotient presentation of U(g) and canonical linear map ιg from Universal enveloping algebra.

[L1]

Every generator xyyx[x,y] of the defining ideal has zero image in the quotient.

Proof

technique · direct
1.1

Applying the quotient map to the relator in [L1] gives ιg(x)ιg(y)ιg(y)ιg(x)ιg([x,y])=0, which is the displayed identity.

givenL1algebra
2.1

Since ιg is linear by construction and step 1.1 is bracket preservation, it is a Lie-algebra homomorphism. No injectivity has been used.

step 1.1

Depends on

Used by

Dependency tree · two levels

4 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