CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck pass

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 collinear labeling admits a nontrivial rotation

Statement refuted

Every nonempty labeling $S \subseteq R^{3}$ determines the rotation: if $R \in S O (3)$ satisfies $R s = s$ for every $s \in S$ , then $R = I_{3}$ .

Facts & Assumptions

Given: Write $e_{1} = (1, 0, 0)$ and $e_{3} = (0, 0, 1)$ , where the tuple lists the values at the imported indices $0, 1, 2$ . Matrix-vector action is $(A v) (k) = \sum_{j < 3} A (k, j) v (j)$ , the sole column of the rectangular product with $C_{v} (j, 0) = v (j)$ . The set $S = {0, e_{3}} \subseteq R^{3}$ and the matrix $R = diag (- 1, - 1, 1)$ .

[L1]

$R \in S O (3)$ means $R^{T} R = I_{3}$ and $\det R = 1$ (The orthogonal group $O (3)$ and the rotation group $S O (3)$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[L2]

Transposition is involutive and reverses matrix products (Transpose is linear and involutive, and $(A B)^{T} = B^{T} A^{T}$ ). The determinant is the finite Leibniz sum $\det A = \sum_{π \in S_{3}} sgn (π) \prod_{i < 3} A (π (i), i)$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

Counterexample

technique · direct; exhibit a nontrivial rotation fixing a nonempty labeling

1.1givenalgebra

Take $S = {0, e_{3}} \subseteq R^{3}$ and $R = (\begin{array}{ccc} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{array})$ . Then $R e_{3} = e_{3}$ and $R 0 = 0$ , while $R \neq I_{3}$ because $R e_{1} = - e_{1}$ .

2.1step 1.1L1L2algebra∎

The matrix $R$ lies in $S O (3)$ : its transpose equals itself, so $R^{T} R = R^{2} = diag (1, 1, 1) = I_{3}$ , and in the Leibniz sum of [L2] every nonidentity permutation has an index $i$ with $π (i) \neq i$ , giving a zero off-diagonal factor. Only the identity permutation contributes, with sign $1$ , so $\det R = (- 1) \cdot (- 1) \cdot 1 = 1$ .

Remarks

The set $S = {0, e_{3}}$ is collinear: all its points lie on the line $R e_{3}$ . Its pointwise stabilizer is defined in A labeling containing two independent vectors has trivial $S O (3)$ -stabilizer. The witness includes $0$ because every linear map fixes $0$ ; that lemma requires only two independent vectors in $S$ and makes no separate origin assumption.
The displayed matrix fixes the axis $R e_{3}$ and sends $e_{1}$ to $- e_{1}$ ; this verifies its nonidentity and its action on $S$ directly. No physical-body claim is made.

Depends on

Dependency tree · 2 levels

Sources

D. Tong, Classical Dynamics (Cambridge lecture notes, 2015), Section 3.1, printed pages 46-48