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.
Surjectivity modulo n-minus for free weight-generated modules (BGG 10.5)
Statement
Assume the Axiom of Choice (The Axiom of Choice). Let . Let be a -module that is free on weight-vector generators (every element of is a finite sum with ), and let be a -module map such that each image is a weight vector of . Then is surjective if and only if the induced map is surjective.
Facts & Assumptions
Given: The Axiom of Choice, an object of the classical category of The classical BGG category O, a -module free on weight-vector generators , and a -linear map whose values on the generators are weight vectors of .
Every object of is -semisimple with finite-dimensional weight spaces, and its weight set is contained in a finite union of cones (The support description of category O with finite generation, The classical BGG category O).
, so is spanned by PBW monomials in the , and for a -module the coinvariants are (Triangular decomposition from a chosen positive root system, The PBW model of a Verma module).
, and is -stable, so is -semisimple with finite-dimensional weight spaces (The classical BGG category O).
Proof
If is surjective, then is surjective, because by -linearity, so induces a surjection of the quotients.
Conversely assume surjective; we prove that every weight vector of lies in by descending induction on the weight. Since the weight set of is contained in finitely many cones , the set of weights of with is finite for every weight (only the finitely many cones with contribute, and there the coefficients of are bounded by those of ).
Inductive step. Fix a weight and , and assume all weight vectors of of weight lie in . Since is surjective, is a linear combination of the classes , and each nonzero is a weight vector because is a weight vector by hypothesis and the quotient map is -equivariant. As is -semisimple, taking the weight- component of the relation lets us discard every generator whose class has weight different from ; hence with whenever (for the surviving indices is the original coefficient and the corresponding vectors have weight ). Therefore .
By [F3] the element has weight and lies in , so its weight- component is a sum with : indeed lowers weights by . Each has weight , so by the induction hypothesis, and then because is a -submodule. Hence .
The base of the induction is the case of a maximal weight, where the sum in step 3.1 is empty and ; the induction is well founded by step 1.2. Since is spanned by its weight vectors, , so is surjective.
Depends on
Used by
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
- Fan Zhou, The classical and the functorial BGG resolutions (Columbia thesis 2021), Part I Sec. 4.2.1 (BGG Lemma 10.5), pp. 18-20 (standard reference, not scraped)
- J. van Ekeren, Topics in representation theory (IMPA 2024), Sec. 29, p. 122 (standard reference, not scraped)