Alphabeta MathHow 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.
Example
For the symmetric group ,
Facts & Assumptions
Given: The symmetric group .
Over an algebraically closed field of characteristic prime to , the number of irreducible representations equals the number of conjugacy classes (If is algebraically closed and , the number of irreducible representations of equals the number of conjugacy classes).
Under the same hypotheses, the irreducible degrees satisfy the sum-of-squares formula (If is algebraically closed and , then ).
Under the same hypotheses, the group algebra is a product of full matrix algebras over the base field (If is algebraically closed and , then ).
Verification
The conjugacy classes of are Hence [L1] gives exactly three irreducible complex representations, with degrees , and [L2] gives
Each is a positive integer. Since and any larger square would already exceed , the only possible degree multiset is . Applying [L3], the Wedderburn factors are therefore , , and , so
Depends on
- If $k$ is algebraically closed and $\operatorname{char} k \nmid |G|$, then $\sum_i (\dim_k V_i)^2=|G|$
- If $k$ is algebraically closed and $\operatorname{char} k \nmid |G|$, the number of irreducible representations of $G$ equals the number of conjugacy classes
- If $k$ is algebraically closed and $\operatorname{char} k \nmid |G|$, then $k[G]\cong\prod_{i=1}^r M_{n_i}(k)$
Used by
- False statement: the sum of the irreducible degrees equals |G| False statement
Dependency tree · two levels
13 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 et al., Introduction to Representation Theory, Section 3.3 (standard reference, not scraped)