1. What the declaration says in plain English The theorem FApply_GApply establishes a linear algebraic identity between two operators: the almost-product operator $F$ and the golden operator $G$. It states that applying $F$ to the output of $G(v)$ is exactly equal to scaling $F(v)$ by $1/2$ and adding the original vector $v$ scaled by $\sqrt{5}/2$. This holds as long as the underlying rank-one operator has a non-zero trace/coefficient ($\mu \neq 0$).
2. Why it matters in Recognition Science
In Recognition Science (RS), the golden ratio $\phi$ is structurally forced by self-similarity, rather than being an empirical parameter. In the multi-dimensional setting, the framework formalizes a "rank-one tensor picture" driven by an inverse metric and a covector cost gradient. From this, an idempotent projector $P$ and an involution almost-product operator $F$ ($F^2 = I$) naturally arise. The golden operator $G$ embeds the mathematics of the golden ratio into this spatial geometry. This theorem is a critical computational step used immediately after to prove GApply_square ($G^2 = G + I$), ensuring the spatial operator correctly mimics the algebraic defining polynomial of $\phi$.
3. How to read the formal statement
{n : ℕ}: The dimension of the vector space.lam : ℝ,hInv : Fin n → Fin n → ℝ,β : Vec n: The scaling, inverse metric, and covector defining the rank-one geometric update.hμ : mu lam hInv β ≠ 0: The hypothesis that the scalar trace-like coefficient $\mu$ is non-zero, ensuring the projector is well-defined.v : Vec n: An arbitrary vector being acted upon.FApply lam hInv β (GApply lam hInv β v) = ...: The conclusion evaluating the composition $F(G(v)) = \frac{1}{2} F(v) + \frac{\sqrt{5}}{2} v$.
4. Visible dependencies in the supplied source This THEOREM relies entirely on basic linear algebra formalized in the same file:
- It unfolds GApply, which is defined as $G(v) = \frac{1}{2}v + \frac{\sqrt{5}}{2}F(v)$.
- It leverages the linearity of $F$ via FApply_add and FApply_smul.
- It utilizes the involution property of $F$, proved in FApply_square, which states that applying $F$ twice yields the original vector (this is the only step requiring the
hμassumption).
5. What this declaration does not prove
- It is a purely algebraic theorem and does not prove the root Universal Forcing of $\phi$ from logic (which happens in
t6_holds/phi_forcing_complete, not present in this slice). - It works generically for any dimension $n$, and thus does not prove that spatial dimension $D = 3$ is forced.
- It does not directly derive physical constants like $G$ or $c$.