Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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Minimal van Kampen area agrees with minimal algebraic relator area

Statement

For a null word in a finite presentation, the minimal area of a van Kampen diagram equals its minimal algebraic relator area.

Facts & Assumptions

Given: A finite presentation and a word w representing the identity.

[L1]

Van Kampen diagrams exist exactly for null words in the presented group (A word is trivial in a presented group exactly when it bounds a finite van Kampen diagram).

[F1]

Algebraic relator area is the minimum number of conjugates of defining relators needed to express the word, when such a minimum exists (Algebraic relator area and the Dehn function of a finite presentation, Every null word has a minimal algebraic relator area).

Proof

technique · direct
1.1

Let D be any van Kampen diagram for w with m faces. Reading the faces one by one as in the proof of The boundary label of a van Kampen diagram is trivial in the presented group expresses w as a product of m conjugates of relators and their inverses. Hence the algebraic relator area of w is at most m.

L1F1given
1.2

Conversely, let w=k=1mukrkεkuk1 be an algebraic expression with m minimal as in [F1]. The converse construction in A word is trivial in a presented group exactly when it bounds a finite van Kampen diagram produces a van Kampen diagram with exactly m faces and boundary word w. Therefore the minimal diagram area is at most the algebraic relator area.

F1L1construct
2.1

Step 1.1 gives one inequality between the two minima and step 1.2 gives the reverse inequality. Therefore the two minimal areas are equal.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources