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.
A root-graded Manin triple gives dual Lie bialgebras
Statement
Let be a field of characteristic , and let be a locally finite root-graded Manin triple in the sense of Lie bialgebras, degreewise duality, and root-graded Manin triples. Define and by transposing the opposite brackets under the degreewise perfect pairing induced by :
Then these maps are well defined and make and dual Lie bialgebras.
Facts & Assumptions
Given: A field of characteristic and a locally finite root-graded Manin triple.
The two complementary isotropic subalgebras have finite-dimensional graded pieces, a degreewise perfect cross pairing, and only finitely many degree decompositions within each subalgebra in any fixed degree (Lie bialgebras, degreewise duality, and root-graded Manin triples).
The wedge pairing is the determinant pairing on exterior powers, and a Lie bialgebra cobracket is a cocycle satisfying co-Jacobi (The graded exterior algebra , Lie bialgebras, degreewise duality, and root-graded Manin triples).
Proof
By [F1], for each homogeneous only finitely many opposite-degree pairs of components of can bracket to a degree paired with . The perfect pairings therefore give a unique finite sum satisfying the first transpose identity; the same construction defines , and both maps are linear and preserve total root degree.
Choose dual bases and in the finitely many homogeneous pieces involved in a fixed calculation, and write and . Invariance gives and ; since the two subalgebras are isotropic and pair perfectly, these identities determine . For each fixed input pair, grading restricts these sums to two fixed finite-dimensional pieces, so they are finite by [F1].
Pair with in the opposite subalgebra. By the defining transpose identity, the result is by Jacobi in . Degreewise perfectness makes the co-Jacobi expression zero; the identical argument with signs exchanged proves co-Jacobi for .
Jacobi in for , paired with , gives . By the transpose definition, this is exactly the coefficient identity for . Interchanging and and using Jacobi for proves the cocycle identity for . Since the homogeneous bases were arbitrary and the pairings are perfect, both cocycle identities hold for all elements. Together with step 2.1, this proves that the two transposed maps are dual Lie bialgebra structures.
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.