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Topological boundary state counts without universal gravity claims — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Fusion count alone leaves an entropy-to-scalar ratio undetermined

Example

For the two-label matrix $N = (\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array})$ in the finite fusion count model, $D_{4} = 2$ . Choose a positive scalar $a$ independently of $N$ and set $A_{4} : = 4 a$ . The ratio between the model entropy $\log D_{4}$ and this separately assigned scalar is $\log 2 / (4 a)$ , and it changes when $a$ is replaced by $2 a$ although $N$ and $D_{4}$ are unchanged. Thus the finite count alone does not determine a proportionality coefficient between its entropy and an independently assigned area-like scalar. The scalar $A_{4}$ is formal data in this mathematical example; no geometrical area, physical entropy, or gravitational law is asserted.

Example references

Fusion path recursion and entropy

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

Daniel Harlow, The Ryu-Takayanagi Formula from Quantum Error Correction

Robert M. Wald, The Thermodynamics of Black Holes

Facts & Assumptions

Given: The two-label matrix and return-count recursion, an independently assigned scalar $a > 0$ , and $A_{4} = 4 a$ .

[F1]

Fusion path recursion and entropy: for the two-label matrix, $D_{0} = 1$ , $D_{1} = 0$ , and $D_{n + 2} = D_{n + 1} + D_{n}$ ; when $D_{n} > 0$ , the uniform-state entropy is $\log D_{n}$ .

[F2]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\log$ is strictly increasing and $\log 1 = 0$ .

[A1]

The scalar assignment $a > 0$ is independent of the transition matrix $N$ and its counts.

Verification

1.1F1givenalgebra

Applying the recurrence in [F1] gives $D_{2} = 1$ , $D_{3} = 1$ , and $D_{4} = 2$ . Since $D_{4} > 0$ , the corresponding finite-state entropy is $s_{4} = \log 2$ .

2.1A1step 1.1algebra

With the assignment $a$ , the ratio is $c (a) : = s_{4} / A_{4} = \log 2 / (4 a)$ . Replacing $a$ by $2 a$ leaves $N$ , $D_{4}$ , and $s_{4}$ fixed while giving $c (2 a) = \log 2 / (8 a) = c (a) / 2$ .

3.1F2step 2.1algebra∎

Because $2 > 1$ , [F2] gives $\log 2 > \log 1 = 0$ ; since $a > 0$ , $c (a) > 0$ and therefore $c (2 a) \neq c (a)$ . The count model consequently does not fix this coefficient.