TheoremDB

Problem packetResearch packetR1305

R1305Sourced evidence

Current checked status and unresolved remainder

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Authored summary

UNKNOWN as of 2026-07-27. The source thread verifies the property through dimension four using classifications of integral torsion conjugacy classes. The dated search found no general theorem or higher-dimensional counterexample.

The record cites sources for its explanation.

Recorded status: reported

Recorded scope: No scope is recorded.

Originating problem: Ternary representatives of finite-order integral matrices

Authored record and scope
Authored title
Current checked status and unresolved remainder
Record type
claim
Stored status
reported
Evidence grade
sourced

2Authored explanation

A dated independent review on 2026-08-01 checked the structured sources below, the complete visible source discussion, exact-title and equivalent-formulation searches, and the current TheoremDB corpus. On 2026-07-27 both MathOverflow answers and all their comments were checked. They reduce dimensions at most four to published lists and report ternary representatives in every listed class. Tahara classifies the relevant low-dimensional finite subgroups, and Yang's 2015 Electronic Journal of Linear Algebra paper lists 45 torsion conjugacy classes in \(\operatorname{GL}_4(\mathbb Z)\). These checks set the first possible counterexample dimension at five. Rational canonical form is insufficient because rational conjugacy can split into several integral conjugacy classes. A search must enumerate integral lattices or integral conjugacy classes. A matrix outside the ternary alphabet is not itself a counterexample. One must certify that its entire \(\operatorname{GL}_n(\mathbb Z)\)-conjugacy class contains no ternary matrix. Trap: allowing conjugation by \(\operatorname{GL}_n(\mathbb Q)\), changing the lattice, or using a larger bounded entry set answers a weaker question.

A complete resolution must satisfy: For a positive answer, prove that every finite-order element of every \(\operatorname{GL}_n(\mathbb Z)\) has an integrally conjugate ternary representative. For a negative answer, give a finite-order integral matrix in the least possible dimension and prove that no integral conjugate is ternary; also certify all smaller dimensions.

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3Evidence

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: mathoverflow.net ↗, Dataset references and independent 2026-08-01 status search.

4How it connects

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  "slug": "finite-order-integer-matrix-ternary-conjugate-status-20260801",
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  "title": "Current checked status and unresolved remainder",
  "summary": "UNKNOWN as of 2026-07-27. The source thread verifies the property through dimension four using classifications of integral torsion conjugacy classes. The dated search found no general theorem or higher-dimensional counterexample.",
  "relevance": "Records the strongest checked neighboring results and the exact remainder future work must settle.",
  "relevance_source": "recorded",
  "body": "A dated independent review on 2026-08-01 checked the structured sources below, the complete visible source discussion, exact-title and equivalent-formulation searches, and the current TheoremDB corpus. On 2026-07-27 both MathOverflow answers and all their comments were checked. They reduce dimensions at most four to published lists and report ternary representatives in every listed class. Tahara classifies the relevant low-dimensional finite subgroups, and Yang's 2015 Electronic Journal of Linear Algebra paper lists 45 torsion conjugacy classes in \\(\\operatorname{GL}_4(\\mathbb Z)\\). These checks set the first possible counterexample dimension at five. Rational canonical form is insufficient because rational conjugacy can split into several integral conjugacy classes. A search must enumerate integral lattices or integral conjugacy classes. A matrix outside the ternary alphabet is not itself a counterexample. One must certify that its entire \\(\\operatorname{GL}_n(\\mathbb Z)\\)-conjugacy class contains no ternary matrix. Trap: allowing conjugation by \\(\\operatorname{GL}_n(\\mathbb Q)\\), changing the lattice, or using a larger bounded entry set answers a weaker question.\n\nA complete resolution must satisfy: For a positive answer, prove that every finite-order element of every \\(\\operatorname{GL}_n(\\mathbb Z)\\) has an integrally conjugate ternary representative. For a negative answer, give a finite-order integral matrix in the least possible dimension and prove that no integral conjugate is ternary; also certify all smaller dimensions.",
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    "locator": "Dataset references and independent 2026-08-01 status search."
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      "title": "Complete the stated acceptance conditions",
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      "slug": "finite-order-integer-matrix-ternary-conjugate",
      "title": "finite order integer matrix ternary conjugate",
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6Provenance

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A statement this project treats as settled at the recorded evidence grade, with the work that backs it.

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