DefinitionDefinition: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

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.

Low-degree transgression for a group extension

Definition

Assume AC. For $1 \to N \to G \overset{π}{\to} Q \to 1$ , a left G-module A and $[d] \in H^{1} (N, A)^{Q}$ , put $D_{d} = {(d (n), n) : n \in N} \leq A ⋊ G$ and $L_{d} = N_{A ⋊ G} (D_{d})$ . Define $Tra [d]$ to be the class of $0 ⟶ A^{N} ⟶ L_{d} / D_{d} ⟶ Q ⟶ 1.$ It is a well-defined homomorphism to normalized bar $H^{2} (Q, A^{N})$ . Explicitly choose normalized $α : Q \to G$ and $η : Q \to A$ with $(α (q) \cdot d - d) (n) = n η (q) - η (q)$ ; put $f (q, r) = α (q) α (r) α (q r)^{- 1}$ . Then Tra is represented by $F (q, r) = η (q) + α (q) η (r) - f (q, r) η (q r) - d (f (q, r)) .$ This fixes the DHW sign convention. The derived interpretation of bar cohomology retains its comparison convention.

Facts & Assumptions

Given: AC, the extension, A, and a Q-invariant crossed-map class [d].

[F1]

The conjugation action on crossed-map classes factors through Q (Degree-one maps and the quotient action are well-defined).

[F2]

Normalized factor sets classify the extensions with fixed action (Bar two-cocycles classify abelian-kernel extensions).

[F3]

AC chooses elements of arbitrary nonempty indexed families (The Axiom of Choice).

Proof

1.1

The crossed identity makes $D_{d}$ a subgroup mapping isomorphically onto N. In $A ⋊ G$ , conjugating $(d (g^{- 1} n g), g^{- 1} n g)$ by (a,g) gives $(a + g d (g^{- 1} n g) - n a, n)$ . Thus $(a, g) \in L_{d}$ exactly when $(g \cdot d - d) (n) = n a - a$ for every n. By invariance in F1 there exists such an a for each g, so $L_{d} \to G$ is onto. Setting g=1 shows $L_{d} \cap A = A^{N}$ . Its elements over N are precisely $A^{N} D_{d}$ , since division by the unique D_d element over the same n lies in this intersection. Consequently the quotient extension is exact; its action on $A^{N}$ is $a \mapsto g a$ , the quotient Q-action.

F1givenalgebra

2.1

Apply AC to the fibers of $G \to Q$ , normalizing $α (1) = 1$ . For each q the set of a solving the normalizer equation for $α (q)$ is nonempty by step 1.1. Apply AC to these Q-indexed sets and normalize $η (1) = 0$ , which is a solution. Then $l_{q} = (η (q), α (q)) \in L_{d}$ . For $f = f (q, r)$ , direct multiplication gives $l_{q} l_{r} = (F (q, r), 1) (d (f), f) l_{q r}$ with exactly the F printed in the Definition. All factors except possibly $(F, 1)$ are in $L_{d}$ , so it too is, and step 1.1 gives $F \in A^{N}$ .

F3step 1.1algebra

2.2

If $d_{b} = d + δ b$ for a fixed b in A, then $D_{d_{b}} = (- b, 1) D_{d} (- b, 1)^{- 1}$ by the semidirect multiplication. Conjugation by (-b,1) therefore identifies their normalizers and quotients, fixes every element of $A^{N}$ , and is the identity on Q. Hence the extension class depends only on [d].

F2step 1.1algebra

3.1

In the quotient $L_{d} / D_{d}$ , the elements ${\overset{ˉ}{l}}_{q}$ form a normalized section and satisfy ${\overset{ˉ}{l}}_{q} {\overset{ˉ}{l}}_{r} = i (F (q, r)) {\overset{ˉ}{l}}_{q r}$ . Associativity yields $F (q, r) + F (q r, s) = q F (r, s) + F (q, r s)$ by comparing the two triple products. Also F(1,q)=F(q,1)=0. Thus F is a normalized $A^{N}$ -valued cocycle representing the extension by F2. Different alpha or eta give another normalized section of the same extension; their unique difference $b : Q \to A^{N}$ changes F by $δ b$ , as in F2.

F2step 2.1algebra

4.1

For d and e choose the same alpha and compatible eta_d,eta_e; their sum solves the normalizer equation for d+e. The displayed formula then gives $F_{d + e} = F_{d} + F_{e}$ term by term. For d=0 choose eta=0, giving F=0. Independence in steps 2.2 and 3.1 makes these choices irrelevant, proving additivity on cohomology. In particular if N acts trivially on A, invariance is literal equality of crossed maps, so eta=0 is allowed and $F = - d (f)$ . If N=1 or Q=1, the normalized formulas give the zero transgression.

step 2.1step 3.1step 2.2algebra∎

Depends on

Used by

Dependency tree · two levels

9 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

Dekimpe–Hartl–Wauters, A seven-term exact sequence for the cohomology of a group extension, Sections 2–5 pp.2–11 and Section 10.2 p.21 (standard reference, not scraped)