LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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The boundary label of a van Kampen diagram is trivial in the presented group

Statement

Let $D$ be a van Kampen diagram over a presentation $⟨ X ∣ R ⟩$ . Then the boundary label of $D$ represents the identity in the presented group.

Facts & Assumptions

Given: A van Kampen diagram $D$ over $⟨ X ∣ R ⟩$ .

[F1]

The presented group is the quotient of the free group on $X$ by the normal closure of $R$ (Group presentation by generators and relations, The normal closure of a subset of a group).

[L1]

A van Kampen diagram is either the degenerate one-vertex diagram or a finite planar disc complex whose 2-cells are labelled by cyclic conjugates of relators and their inverses (Van Kampen diagrams, boundary labels, and diagram area for a presentation).

Proof

technique · direct

1.1

If $D$ has area $0$ , then [L1] forces $D$ to be the degenerate one-vertex diagram. Its boundary label is the empty word, so it represents the identity in the free group and therefore in the quotient group of [F1].

F1L1given

1.2

Assume that $D$ has positive area. Choose a base vertex on the outer boundary, a spanning tree in the $1$ -skeleton, and a spanning tree in the dual graph rooted at the exterior face. Reading the 2-cells in an order compatible with the rooted dual tree gives the standard disc-shelling identity $Lab (\partial D) = \prod_{k = 1}^{m} u_{k} r_{k}^{ε_{k}} u_{k}^{- 1}$ in the free group, where each $r_{k} \in R$ , each $ε_{k} \in {\pm 1}$ , and the words $u_{k}$ are labels of paths from the base vertex to the corresponding cells. Interior edges cancel in opposite orientations, leaving exactly the outer boundary label.

L1givenconstruct

2.1

Every factor in step 1.2 lies in the normal closure of $R$ . Hence the boundary label lies in that normal closure and represents the identity in the quotient group of [F1]. Together with step 1.1 this proves the claim in all cases.

F1step 1.1step 1.2∎

Depends on

Used by

A word is trivial in a presented group exactly when it bounds a finite van Kampen diagram Theorem

Dependency tree · two levels

8 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

GAP SmallCancellation manual, Chapter 1: Small Cancellation Theory — the classical conditions (standard reference, not scraped)
Jay Williams, Universal Countable Borel Quasi-Orders (standard reference, not scraped)
Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, Section 3.5 (standard reference, not scraped)
Clara Löh, Geometric Group Theory: An Introduction, Section 7.4.1 (standard reference, not scraped)