1. Plain English
The theorem FApply_MetallicApply describes how the almost-product operator $F = 2P - I$ interacts with a "metallic family" operator $M_{p,q}$. When you apply $F$ to a vector that has already been transformed by $M_{p,q}$, the result is a linear combination of the original vector and its $F$-transformation. Specifically, $F(M_{p,q}(v)) = \frac{p}{2} F(v) + \frac{\sqrt{p^2 + 4q}}{2} v$.
2. Why it matters in Recognition Science
RS derives its physical structures algebraically from a parameter-free cost operation. The almost-product operator $F$ satisfies the involution property $F^2 = I$. The metallic family of operators mathematically generalizes the golden mean ($p=1, q=1$). This theorem supplies the structural operator algebra that enables RS to construct and square these metallic operators, a necessary precursor to deriving the golden ratio $\phi$ scaling operators that force RS-native fundamental constants.
3. How to read the formal statement
{n : ℕ}: The dimension of the vector space.(p q lam : ℝ): Real parameters where $p, q$ define the metallic operator and $\lambda$ scales the cost tensor.(hInv : Fin n → Fin n → ℝ)and(β : Vec n): The inverse metric kernel $h^{-1}$ and covector $\beta$ that induce the projector $P$.(hμ : mu lam hInv β ≠ 0): The strict non-zero assumption for the scalar $\mu$ (from the quadratic law $A^2 = \mu A$), required to guarantee that $F$ is an involution.(v : Vec n): The operand vector.- The conclusion translates to: $F(M_{p,q}(v)) = \frac{p}{2} F(v) + \frac{\sqrt{p^2 + 4q}}{2} v$.
4. Visible dependencies
The proof in the provided Lean source is purely algebraic. It unfolds the definition of MetallicApply and pushes $F$ through the linear combination using:
- Additive linearity: FApply_add
- Scalar multiplication linearity: FApply_smul
- The involution identity $F^2(v) = v$: FApply_square. The application of
FApply_squareis the sole step that consumes the non-zero assumptionhμ.
5. What this declaration does not prove
- Physical setting of $\mu$: It does not prove that $\mu \neq 0$ is realized in any specific physical setting; it acts purely as an algebraic hypothesis here.
- Uniqueness of the Golden Ratio: It does not prove that $p=1, q=1$ is uniquely forced (which corresponds to T6 in the wider RS forcing chain). It merely provides the generalized algebra for any metallic root.
- Physical Constants: It does not establish the RS-native physical values (like $c$ or $\hbar$); it operates entirely in the abstraction of finite-dimensional operator algebra.