THEORY-271: In a toy subject–verb–object corpus whose eight subjects carry every combination of three binary aspects, the subjects' raw co-occurrence rows form a parallelepiped if and only if, for each object, its four sentences are equally probable; any other sentence distribution breaks it

Source

Torii, Maeda and Hidaka (2024), “Distributional hypothesis as isomorphism between word-word co-occurrence and analogical parallelograms”, PLOS ONE 19(10): e0312151, LIT-977. Read in full in NOTE-781: §5.3 (the toy corpus, Eq. 3, Figs. 5–7), §5.5 (Definition 1, Theorem 1 and its proof, Eqs. 4–18), §5.6.

What was actually shown

• The corpus.
  • 8 subjects, 3 verbs and 6 objects form 24 three-word sentences.
  • Each verb expresses one aspect of its subject through two hidden states: Sex, Royalty or Number. Its object is fixed by the subject’s value on that aspect.
  • Every subject occurs with every verb, and the subjects are all 2³ aspect combinations.
  • Every pair of words in a sentence co-occurs.
• Definition 1. The eight subject rows form a parallelepiped if, under some labelling, each of three edge vectors is repeated four times (Eq. 4).
• Theorem 1. The rows form a parallelepiped if and only if the sentence probabilities are equal within each of six groups of four (Eq. 5). Each group is the four sentences that end in one object.
• Necessity.
  • The edge equalities (Eq. 6) say N C = 0. The kernel of N is spanned by the constant vector and the three aspect indicators (Eq. 10), so every column of C is affine in the aspects.
  • An object’s column is zero on the four subjects that do not take it. An affine function on the cube that vanishes on a face is a multiple of the opposite face’s indicator, so the four non-zero entries are equal. This is the reader’s restatement of the step.
  • The paper extends the argument over labellings by a case analysis.
• Sufficiency. The paper asserts it from the uniform corpus (p_i = 1/24), whose rows lie in a 3D affine space and form a parallelepiped with axes Sex, Royalty and Number (Fig. 7a–b).
• A numerical check. One draw of p_i uniform on [0, 1] gives neither a parallelepiped nor parallelograms (Fig. 7c).
• What could have come out otherwise. The shape might have survived some non-uniform distributions beyond Eq. 5. The theorem says it does not.

What this does not say

• Not about PMI or log counts. The rows are raw counts. Here aspects combine as a mixture: each sentence carries one aspect, so counts add. The logarithm would break the shape, and zero cells have no logarithm. The paper’s real-text analogies use log counts and an SVD, which this theorem does not cover.
• Not about approximate shapes. It is all or nothing. The authors say Eq. 6 has no solution for noisy vectors, and propose without doing it a least-squares version for real corpora.
• Not about natural text. The only natural-text evidence is one “parallelepiped-like” set of Family words in a 2D PCA, which “merely demonstrates the ‘existence’” of one.
• Not about hierarchy or exclusion. All eight combinations occur and the aspects are independent, so the concept lattice of the incidence is Boolean. What happens when some combinations are missing is not asked.
• Not a group-theoretic result. The 48 symmetries the paper calls “the dihedral group” are the cube’s symmetry group. The theorem does not depend on the name.

Connections

• THEORY-185 (from LIT-863). A rival account for a different generative structure, not a second derivation of the same one.
  • There, attributes act multiplicatively, so the PMI is affine in them for any strengths, and the raw ratio is not.
  • Here, aspects act additively through a mixture over sentences, so raw counts are affine, and only under the local-uniformity condition.
  • Both locate the parallelogram in the data. They disagree on which transform reveals it, because they assume different ways for attributes to combine.
  • Which assumption fits natural co-occurrence is not tested by either.
• QUESTION-025. No answers link.
  • The toy corpus is the independent, full-hypercube case the question says does not settle it.
  • Its one bearing is on the question’s third case, attributes that “interact in co-occurrence”. With the incidence held Boolean, any departure from local uniformity breaks the exact shape in raw counts.
  • The proof’s method, the kernel of the constraint matrix for a target shape, would apply to a corpus with implications. The paper names that generality but does not use it.
• THEORY-183. That account gets geometry from a translation symmetry of co-occurrence. This one gets it from a symmetry of the sentence distribution. Both read word geometry as a symmetry of the statistics.

Note of 2026-10-20: Theorem 1 for every labelling and any number of aspects

Promotion audit, batch 050.

Setting.

• Subjects are s ∈ {0,1}^d, all 2^d of them. There is one verb per aspect v, and two objects o(v,b) per verb, b ∈ {0,1}.
• Sentence (s, v, o(v, s_v)) has probability p(s,v).
• The raw row of subject s has entry p(s,v) at verb v and p(s,v)·1[s_v = b] at object o(v,b). Subjects never co-occur.
• The rows form a parallelepiped (Definition 1) if, for some bijection σ from the cube’s vertices to the subjects, x ↦ row(σ(x)) is affine in x. On the cube, equal edge vectors along each axis are the same thing.

Sufficiency. Suppose p(s,v) = q(v, s_v) for every object (Eq. 5). Take σ to be the identity.

• The verb entry is q(v,0)(1 − s_v) + q(v,1)s_v, and the object entries are q(v,1)s_v and q(v,0)(1 − s_v). All are affine in s.
• Aspect k’s edge vector is supported on verb k and its two objects. So the d edge vectors have disjoint supports, and they are independent whenever every verb has positive probability.

Necessity, for any σ.

1. Fix an object o(v,b) and let f(x) be its column, read through σ. Then f is affine and nonnegative on the vertices, hence on the whole cube.
2. If f is not identically zero, its zero set on the cube is the cube’s intersection with a supporting hyperplane. That is a face, and the zero vertices are the face’s vertices.
3. f vanishes on at least the 2^(d−1) subjects with s_v ≠ b. A face with at least 2^(d−1) vertices that is not the whole cube is a facet, so the face is a facet {x_k = c} and holds exactly those subjects.
4. An affine function vanishing on {x_k = c} is a multiple of x_k − c, so it is constant on the opposite facet. So the 2^(d−1) sentences ending in o(v,b) are equally probable.
5. If f is identically zero, they are all 0.

The verb columns are sums of two object columns, so they add nothing. This one argument replaces the paper’s case analysis over permutations and holds for every d.

A numerical check (d = 3). Build the rows as above. For each of the 8! = 40,320 permutations, test whether the rows equal their projection onto span{1, x₁, x₂, x₃} (tolerance 10⁻⁹).

• A distribution meeting Eq. 5 forms a parallelepiped, also with one object’s group at zero.
• One sentence perturbed by 0.1% forms none.
• A uniform random p forms none.