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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Killing form of sl_2

Example

Let $k$ have characteristic zero and put $g = {s l}_{2} (k)$ . For

$e = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}), f = (\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}), h = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}),$

one has $K (h, h) = 8$ , $K (e, f) = K (f, e) = 4$ , and all other basis pairings are zero. Thus $K$ is nondegenerate.

Facts & Assumptions

Given: The displayed matrices over a characteristic-zero field.

[L1]

The Killing form is $K (x, y) = tr ({ad}_{x} {ad}_{y})$ (Killing form).

[L2]

A finite-dimensional characteristic-zero Lie algebra is semisimple exactly when its Killing form is nondegenerate (Cartan's semisimplicity criterion).

Verification

technique · direct matrix calculation

1.1

In the ordered basis $(e, f, h)$ , the relations $[h, e] = 2 e$ , $[h, f] = - 2 f$ , and $[e, f] = h$ give ${ad}_{e} = (\begin{array}{ccc} 0 & 0 & - 2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}), {ad}_{f} = (\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ - 1 & 0 & 0 \end{array}), {ad}_{h} = (\begin{array}{ccc} 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 0 \end{array}) .$

givenalgebra

2.1

Squaring the last matrix gives trace $8$ . Multiplying the first two matrices in either order gives trace $4$ . Their squares have trace zero, and multiplying either by ${ad}_{h}$ in either order also has trace zero. By [L1], these are precisely the claimed pairings.

L1step 1.1algebra

3.1

The Gram matrix has determinant $\det (\begin{array}{ccc} 0 & 4 & 0 \\ 4 & 0 & 0 \\ 0 & 0 & 8 \end{array}) = - 128.$ Characteristic zero makes this scalar nonzero, so $K$ is nondegenerate. In particular [L2] recovers semisimplicity. Every basis pairing was calculated, including the zero pairings, and no choice principle is used.

L2step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Kirillov, An Introduction to Lie Groups and Lie Algebras, Example 5.51 (standard reference, not scraped)