TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Linear algebraic relator area implies slim Cayley triangles

Statement

Assume the Axiom of Choice. Let a finite presentation have relator lengths at most $L \geq 0$ and satisfy $Area (w) \leq K ∣ w ∣$ for every null word, with $K \geq 0$ . Its unit-edge metric Cayley realization $X$ has uniformly slim geodesic triangles. More precisely, writing $m_{X} (P)$ for the supremum of minsize over triangles of perimeter at most $P$ , one has $m_{X} (P) \leq A (K, L) \sqrt{P + 7} + B (L) (P \geq 0),$ where $r = \max (1, L)$ , $C (L) = 20 (L + 1)$ , $A (K, L) = 2 r \sqrt{C (L) (K + 2)}$ , and $B (L) = 2 r + 10$ .

Facts & Assumptions

Given: Assume AC and fix the presentation, $K, L \geq 0$ .

[F1]

The realization is geodesic; a perimeter- $p$ triangle has a marked word of length $n \leq p + 6$ , a filling with $N \leq C (L) (Area (w) + n + 1)$ , and vertex-side Hausdorff error at most $3$ . (Algebraic relator area controls coarse filling area).

[F2]

Such a filling and error $e$ imply minsize at most $2 r \sqrt{N} + 2 r + 2 e$ . (Coarse triangle minsize is bounded by square root of area).

[F3]

Under AC a geodesic space with sublinear triangle-minsize function has uniformly slim triangles. (Sublinear triangle minsize implies hyperbolicity).

[F4]

AC is assumed, as required by the sublinear criterion. (The Axiom of Choice).

Proof

technique · direct

1.1givenF1algebra

For any chosen triangle of perimeter $p$ , its approximating word is null. The assumed area inequality and F1 give $N \leq C (L) ((K + 1) n + 1) \leq C (L) ((K + 1) (p + 6) + 1) \leq C (L) (K + 2) (p + 7) .$ The last inequality follows by subtracting the left inner expression from the right: the difference is $p + K + 7 \geq 0$ . All constants are nonnegative.

2.1step 1.1F2algebra

Apply F2 with $e = 3$ . The triangle minsize is at most $2 r \sqrt{C (L) (K + 2)} \sqrt{p + 7} + 2 r + 6 \leq A (K, L) \sqrt{p + 7} + B (L) .$ For $p \leq P$ this is at most the same expression with $P$ . Taking the supremum over those triangles proves the asserted bound for $m_{X} (P)$ , including $P = 0$ .

3.1step 2.1algebra

Set $F (P) = A (K, L) \sqrt{P + 7} + B (L)$ . This function is nonnegative and nondecreasing. For $P \geq 1$ , $0 \leq F (P) / P \leq A (K, L) \sqrt{8} / \sqrt{P} + B (L) / P .$ Given $ε > 0$ , taking $P$ larger than $1$ , $(2 A (K, L) \sqrt{8} / ε)^{2}$ , and $2 B (L) / ε$ makes the right side at most $ε$ . Thus $F (P) / P \to 0$ , and step 2.1 gives $m_{X} (P) / P \to 0$ .

4.1F1F3F4step 3.1∎

F1 establishes that $X$ is geodesic, and step 3.1 verifies precisely the sublinearity hypothesis of F3. Under the assumed AC, F3 therefore yields a finite slimness constant for all chosen geodesic triangles of $X$ . The displayed $A, B$ depend only on $K, L$ ; a common slimness constant across presentations requires the separate uniformity argument.

Depends on

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Sources

Drutu–Kapovich, Geometric Group Theory — §9.7.4, Theorem 9.100 and Proposition 9.103; §11.20, Proposition 11.167(a) (standard reference, not scraped)