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Simple extensions cannot cross linkage classes
Statement
Fix a finite-dimensional complex semisimple Lie algebra , a Cartan subalgebra , and a positive Borel . Write , when , and .
A short exact sequence in splits whenever and belong to distinct integral-reflection linkage classes.
Facts & Assumptions
Given: The setting above and the hypotheses in the statement.
Fix a finite-dimensional complex semisimple Lie algebra , a Cartan subalgebra , and a positive Borel . Write , when , and . Restricted Chevalley duality is an exact contravariant equivalence , with a natural isomorphism . It preserves each weight-space dimension, the formal character, and every simple composition multiplicity. (Restricted duality is exact and involutive on O)
For a -module , sending a homomorphism to is a bijection onto the vectors of weight annihilated by . Here is def-verma-module. The nonzero vectors in this target are precisely the highest-weight vectors of weight from def-highest-weight-vector-and-cyclic-highest-weight-module; the zero vector corresponds to the zero homomorphism. (The universal property of Verma modules)
If , then . (The strong linkage principle for Verma modules)
Fix a finite-dimensional complex semisimple Lie algebra , a Cartan subalgebra , and a positive Borel . Write , when , and . The set is preserved by its root reflections. If , then and . The equivalence classes generated by moves with are exactly . Every strong-linkage chain stays in one such class. (Integral reflection linkage is an equivalence relation)
The proper submodule which is the sum of all proper submodules is the unique maximal submodule of . The quotient is simple and is its unique simple quotient. (A Verma module has a unique simple quotient)
Proof
If , dualize the sequence; exact self-duality of simples reverses the two labels, and splitting of the dual implies splitting of the original by biduality. Thus it is enough to treat the orientation , which includes incomparable labels.
Lift the highest vector of to a vector . Such a weight lift exists because the sequence consists of weight modules. If a positive-root operator acted nontrivially on , its image would lie in the submodule at weight , forcing and hence . Therefore is singular, and the Verma universal property gives a map whose image surjects onto . Here the support bound for each simple follows from its being the highest-weight Verma quotient.
The intersection is either zero or all of the simple submodule. In the latter case , so is a length-two quotient of and is a composition factor of that Verma. Strong linkage would imply , and hence equality of integral-reflection classes, contrary to the hypothesis. Thus the intersection is zero, and is an isomorphism whose inverse is a section.
Depends on
Used by
Dependency tree · two levels
19 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
- §1.13 pp.30–32 and §4.9 p.83, decomposed proof route (standard reference, not scraped)