by noncentral | 2026-02-05 03:02:30+00:00
Great questions! let me answer each directly in a way that keeps RCC falsifiable, concrete, and mathematically grounded.

1. Mathematical formalization

Yes — RCC is formalized at the level required for a boundary theory.

There are two layers:

(A) Conceptual geometric axioms

A1. Internal State Inaccessibility
The system cannot observe its full internal state; only lossy projections.

A2. Container Opacity
The system cannot access the manifold that contains it (training distribution, upstream causal structure, global structure).

A3. Absence of a Global Reference Frame
All inference is local; no operator enforces global consistency.

A4. Forced Local Optimization
Even under uncertainty, the system must still produce the next update using only local information.

From A1–A4:

Any embedded inference system satisfying these axioms cannot maintain globally-stable, non-drifting long-horizon inference.
This boundary statement is the formalization.

Ongoing work focuses on extensions (curvature mappings, collapse curves), not the axioms themselves — those are already minimal and falsifiable.

(B) Symbolic formalization

(Some people prefer mathematical notation, so here is the same content expressed formally.)
A1. (Internal State Inaccessibility)
Let Ω be the full internal state.
The observer sees only π(Ω), where:
π : Ω → Ω'
|Ω'| < |Ω|
All inference is based on Ω'.

A2. (Container Opacity)
Let M be the containing manifold.
Visibility(M) = 0
⇒ ∂M and curvature(M) are unobservable.

A3. (No Global Reference Frame)
No global frame Γ exists such that:
Γ : Ω' → globally consistent coordinates
Inference occurs in local frames φ_i with:
φ_i ↛ φ_j (non-invertible over long distances)

A4. (Forced Local Optimization)
At each step t:
x_(t+1) = argmin L_local(φ_t, π(Ω))
even under ∂information/∂M = 0.

From A1–A4:
No embedded system can maintain stable, non-drifting long-horizon inference
when ∂Ω > 0, ∂M > 0, and Γ does not exist.

This is the boundary condition RCC asserts.

2. What counts as proof or disproof?

RCC is falsified immediately if someone presents a system that:

• lacks global internal state access,
• lacks visibility of its container manifold,
• lacks a global reference frame,

and still performs stable, non-drifting long-horizon inference.

A single counterexample disproves RCC.

Conversely, RCC is supported where we observe:

• horizon-dependent drift,
• inconsistencies under partial visibility,
• corrections that fail to converge globally,
• collapse around 8–12 reasoning steps.

These signatures follow directly from the axioms.

3. Probabilistic programming

Probabilistic programming assumes a coherent global probability space.
RCC’s point is that collapse arises because the observer cannot construct or access such a global frame.

PPP models inference inside a slice of the manifold,
but cannot remove A1–A4 or the geometric limits they imply.

PPP fits inside RCC, not vice versa.

4. Probabilistic concolic execution & formal verification

These approaches still require:

• a symbolic state graph,
• a coherent environment model,
• or globally evaluable correctness conditions.

RCC applies exactly where these assumptions fail.

You can verify correctness inside the frame,
but you cannot verify the frame from within the system.

That geometric asymmetry is the core of RCC.

Happy to go deeper into collapse-operators, curvature terms, or empirical predictions if you’d like.
The goal is to keep RCC falsifiable and mathematically clean.