ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A diagonal tensor has large minimal TT ranks

Example

For integers $d \geq 3$ and $n \geq 2$ , let $A = \sum_{j = 1}^{n} e_{j}^{(1)} \otimes e_{j}^{(2)} \otimes \dots \otimes e_{j}^{(d)} \in R^{n \times \dots \times n} .$

Verification

Given: The tensor above with $d \geq 3$ and $n \geq 2$ .

[F1] Finite tensor arrays and the Frobenius inner product: a cut unfolding has row tuple $(i_{1}, \dots, i_{k})$ and column tuple $(i_{k + 1}, \dots, i_{d})$ .

[F2] Tensor train cores, bond ranks, and canonical gauges: TT cores contract along the bond indices.

[F3] Exact TT ranks and root-sum-square truncation error: minimal exact TT bond size at a cut equals the rank of that cut unfolding.

1.1F1givenalgebra

For each cut $k$ , only the $n$ row tuples $(j, \dots, j)$ and the matching $n$ column tuples $(j, \dots, j)$ contain nonzero entries of $A^{⟨ k ⟩}$ . The resulting matrix has $n$ ones in distinct rows and columns and zeros elsewhere, so its rank is exactly $n$ .

2.1F2F3step 1.1constructalgebra∎

By [F3], every exact TT representation has bond size at least $n$ at every cut. Such a representation is explicit: the first core enforces $α_{1} = i_{1}$ , each interior core enforces $α_{k - 1} = i_{k} = α_{k}$ , and the last core enforces $α_{d - 1} = i_{d}$ . Contracting these Kronecker-delta entries yields $1$ exactly when all mode indices are the same and $0$ otherwise, which is the stated tensor.

Depends on

Used by

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Dependency tree · two levels

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Sources

Mi-Song Dupuy, Tensor trains for high-dimensional problems, full lecture notes, Chapter 1, §1.2.3, Theorem 1.2.12 (2026) (standard reference, not scraped)
Ivan V. Oseledets, Tensor-Train Decomposition, SIAM Journal on Scientific Computing 33(5) (2011), §2 (standard reference, not scraped)