ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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.

Root-space brackets for matrix units

Example

In ${s l}_{n} (C)$ with the diagonal Cartan subalgebra $h$ and roots $ε_{i} - ε_{j}$ of Diagonal Cartan subalgebra and roots of ${s l}_{n}$ , the matrix units satisfy $[E_{i j}, E_{k l}] = δ_{j k} E_{i l} - δ_{l i} E_{k j},$ and the bracket of the root lines $g_{ε_{i} - ε_{j}} = C E_{i j}$ and $g_{ε_{k} - ε_{l}} = C E_{k l}$ lies in the root space of the sum of the two functionals, as required by Brackets of root spaces: $[g_{ε_{i} - ε_{j}}, g_{ε_{k} - ε_{l}}] \subseteq g_{(ε_{i} - ε_{j}) + (ε_{k} - ε_{l})} .$ For $k \neq j$ and $l \neq i$ both sides vanish; for $k = j$ and $l \neq i$ the bracket is $C E_{i l}$ , whose root is $ε_{i} - ε_{l} = (ε_{i} - ε_{j}) + (ε_{j} - ε_{l})$ .

Facts & Assumptions

Given: The algebra ${s l}_{n} (C)$ with its diagonal Cartan subalgebra and roots $ε_{i} - ε_{j}$ as in Diagonal Cartan subalgebra and roots of ${s l}_{n}$ , the root lines $g_{ε_{i} - ε_{j}} = C E_{i j}$ , and the inclusion $[g_{α}, g_{β}] \subseteq g_{α + β}$ of Brackets of root spaces; the root-space convention is Root and root space.

Verification

technique · direct

1.1givenalgebra

The product formula $E_{i j} E_{k l} = δ_{j k} E_{i l}$ is immediate from matrix multiplication: the product has the single nonzero entry $1$ in position $(i, l)$ exactly when the middle indices match. Hence $[E_{i j}, E_{k l}] = δ_{j k} E_{i l} - δ_{l i} E_{k j}$ .

2.1givenstep 1.1algebra∎

There are four cases. If $k = j$ and $l \neq i$ , then $[E_{i j}, E_{j l}] = E_{i l}$ and the root is $(ε_{i} - ε_{j}) + (ε_{j} - ε_{l}) = ε_{i} - ε_{l}$ . If $l = i$ and $k \neq j$ , then $[E_{i j}, E_{k i}] = - E_{k j}$ and the root is $(ε_{i} - ε_{j}) + (ε_{k} - ε_{i}) = ε_{k} - ε_{j}$ . If $k = j$ and $l = i$ , then $[E_{i j}, E_{j i}] = E_{i i} - E_{j j} \in h = g_{0}$ and the functional sum is zero. Finally, if $k \neq j$ and $l \neq i$ , both Kronecker terms vanish; the functional sum has no cancellation producing a root or zero, so its root space is zero. Thus every case has the asserted bracket inclusion.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I, Lectures 19–24 (standard reference, not scraped)