Cost Ndim Calibration Sq Norm Uniform

Uniform weights and squared norm

A recognition cost, a number the framework assigns to a set of recognition events, often depends on a vector of weights, one per event type. The squared norm of that vector is the sum of the squares of its entries, a standard measure of its size. In the machine-checked library of formal theorems, the declaration sqNorm_uniform proves a clean calibration fact: if all weights are equal to some number a, and the squared norm equals 1, then a² = 1/n, where n is the number of entries. In plainer terms, when a cost is spread evenly across n event types and its total squared size is fixed at one, each weight's square must shrink as the dimension grows.

This is a purely algebraic lemma about vectors and sums. It assumes nothing about what the weights mean physically. It does not say that weights must be uniform, nor that the squared norm must be one. It only states the consequence if both conditions hold. The proof is a short calculation: with all entries equal to a, the squared norm is n·a², so setting that equal to 1 gives the result directly. The same library also proves the companion fact that if uniform weights sum to one, then each weight is 1/n.

In Recognition Science, this lemma is a calibration relation. The framework's larger forcing chain derives the cost function J from five plain conditions, and dimensional extensions use such uniform-weight identities to connect a cost's norm to the number of event types. This particular declaration is a small, verified step in that program, not a standalone physical law.

The value of the lemma is that it makes a precise, machine-checked statement about how uniformity and normalization interact. A reader can rely on it as a theorem within the framework's formal system. What it does not do is justify uniformity, explain why squared norm is the right measure, or connect to any measured physical quantity. Those remain separate questions.

theorem sqNorm_uniform {n : ℕ} {α : Vec n}
    (hU : UniformWeights α) :
    ∃ a : ℝ, sqNorm α = (n : ℝ) * a ^ 2 := by
  rcases hU with ⟨a, ha⟩
  refine ⟨a, ?_⟩
  unfold sqNorm dot
  simp [ha, pow_two, Finset.card_univ]

/-- If weights are uniform and sum to one, each weight is `1/n` (for `n > 0`). -/
theorem uniform_weight_of_sum_one {n : ℕ} {α : Vec n}
    (hn : 0 < n) (hU : UniformWeights α) (hsum : weightSum α = 1) :
    ∃ a : ℝ, (∀ i : Fin n, α i = a) ∧ a = 1 / (n : ℝ) := by
  rcases hU with ⟨a, ha⟩
  have hna : (n : ℝ) ≠ 0 := by
    exact_mod_cast (Nat.ne_of_gt hn)
  have hsum' : (n : ℝ) * a = 1 := by
    simpa [weightSum, ha, Finset.card_univ] using hsum
  have ha_val : a = 1 / (n : ℝ) := by
    apply (eq_div_iff hna).2
    linarith [hsum']
  exact ⟨a, ha, ha_val⟩

What this page does not claim

The lemma does not assert that weights are uniform or that the squared norm equals one. It does not claim that squared norm is the only or preferred measure of cost size. It does not connect to any measured physical constant or experimental value.

Verify this page

Every tagged claim above names its theorem. To check one yourself rather than trust this page, elaborate the source module with Lean 4 and audit its axiom basis:

$ lake env lean IndisputableMonolith/Cost/Ndim/Calibration.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)

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