Exact Cubic Dictionary Capacity
Rank-One Symmetric Cubics over Dyadic Rings

Muhammad Arshad · September 2026 · Free rank, parity collapse, Boolean support geometry, and a sharp 92-atom construction

Canonical paper (v1, CC BY 4.0): 10.5281/zenodo.22850810. All five papers and DOI links →

Abstract

A Sharp Capacity Theorem

Let Rₘ=Z/(2ᵐ) and consider symmetric rank-one cubic atoms v^{⊗3} with v∈Rₘⁿ. This chapter determines the maximum size of a free Rₘ-linearly independent dictionary made entirely from such atoms:

C₃(n) = C(n,1)+C(n,2)+C(n,3).

The upper bound comes from reduction modulo two: every cubic evaluation on a binary vector collapses to a square-free monomial whose support has size one, two, or three. The lower bound is attained by an explicit family indexed by those same supports. For n=8 the exact capacity is

C₃(8)=8+28+56=92.

This is a dictionary-capacity theorem, not a claim that every cubic tensor has rank 92 or that arbitrary 92-component decompositions are identifiable.


01 · Free Dictionary Capacity

The Question

Because Rₘ is not a field, ordinary vector-space dimension is not the right language for a family of atoms. We call atoms C₁,…,C_r free independent if

Σ_q λ_q C_q = 0, λ_q∈Rₘ ⇒ λ_q=0 for every q.

Define C₃(n,m) as the largest possible cardinality of a free independent set of atoms v^{⊗3}. The theorem below shows that the answer is independent of m.


02 · Parity Collapse

Cubics Become Square-Free Supports

Reduce a rank-one atom modulo two. For x_i∈F₂,

x_i²=x_i,   x_i³=x_i.

Therefore a symmetric cubic coordinate depends only on the set of distinct indices appearing in it. A degree-three monomial can have support size one, two, or three. The entire parity image of all cubic atoms is contained in a binary space of dimension at most

N₃(n)=C(n,1)+C(n,2)+C(n,3).

03 · Upper Bound

Why N₃(n)+1 Cannot Survive

Take more than N₃(n) cubic atoms. Their reductions modulo two are linearly dependent, so there is a nonzero α∈F₂ʳ with

Σ_q α_q C_q ≡ 0 (mod 2).

The sum on the left has only even entries. Multiply it by 2^{m−1}:

Σ_q 2^{m−1} α_q C_q = 0 (mod 2ᵐ).

The coefficient vector is nonzero in Rₘ. Hence the original atoms are not free independent. Therefore

C₃(n,m) ≤ N₃(n).

04 · Explicit Attaining Family

The Bound Is Sharp

Index atoms by every nonempty subset S⊆{1,…,n} of size at most three. For m≥2 define the dense vector

(v_S)_i = 1 if i∈S,   2 if i∉S.

For each support R of size one, two, or three, inspect one cubic coordinate carrying exactly support R. Modulo two the resulting square matrix is

Z_{R,S}=1 if R⊆S, otherwise 0.

Order the subsets by size. Then R⊆S implies |R|≤|S|, and equal sizes force R=S. Thus Z is triangular with diagonal one. Its determinant is odd, so the corresponding minor is a unit in every Z/(2ᵐ). The cubic atoms are therefore free independent.


05 · Main Theorem

Exact Cubic Dictionary Capacity

Theorem. For every n≥1 and m≥1, the maximum free rank of a dictionary of symmetric rank-one cubic tensors v^{⊗3} over Rₘ=Z/(2ᵐ) is

C₃(n,m)=Σ_{k=1}^{min(3,n)} C(n,k).

For m≥2 the bound is attained by a dense explicit family whose entries are only 1 and 2.


06 · Quantinion Instance

The 92 Result

n=8
C(8,1)=8
C(8,2)=28
C(8,3)=56
C₃(8)=8+28+56=92.

The result is an 11.5× free overcomplete dictionary relative to the base vector dimension:

92 / 8 = 11.5.

This number is a capacity of a chosen rank-one dictionary, not the symmetric rank of a generic cubic.


07 · Explicit Failure Beyond Capacity

Dependence Is 2-Adic

The upper bound gives a dependency for every set of 93 atoms. There are also compact explicit witnesses. Fix four coordinates Q. Consider the 15 nonempty subsets S⊆Q and their binary support atoms. Every square-free monomial of support at most three is contained in an even number of those subsets, so

Σ_{∅≠S⊆Q} C_S ≡ 0 (mod 2).
2^{m−1} Σ_{∅≠S⊆Q} C_S = 0 (mod 2ᵐ).

Thus torsion produces exact ring dependencies that ordinary real-rank intuition would miss.


08 · Relation to the Simplex Cubic Theorem

Capacity and Identifiability Are Different

Chapter 23 proves global recovery for a highly structured r=n+1 simplex family. This chapter asks only whether a fixed dictionary of rank-one cubics can be free independent. The capacity can be much larger than an identifiable unknown-atom family:

n=8: simplex recovery r=9
balanced residual-simplex recovery r=10 within family
free known-dictionary capacity r=92.

09 · Verification

Exact Integer Regression

The accompanying verifier constructs the explicit subset family, checks the binary zeta minor for n=1,…,8, verifies its full rank, and confirms the n=8 capacity 92. It also constructs exact parity dependencies that lift to Z₂₅₆ by multiplication by 128.


10 · Status Register

What Is Earned

StatementStatus
Parity image uses supports of sizes 1,2,3 onlyProven
Upper bound Σ_{k≤3} C(n,k)Proven by mod-2 dependence and 2-adic lift
Explicit attaining familyProven by odd determinant of subset-incidence minor
n=8 capacity =92Immediate corollary
All arbitrary 92-term decompositions identifiableNot claimed
Generic symmetric rank =92Not claimed