Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The eigenvalue equation does not imply the matrix relation

Statement refuted

Let U,V∈SL2(C) commute and have a common invariant line on which their eigenvalues are M,L, respectively. For every primitive pair (p,q)∈Z2, if MpLq=1, then UpVq=I2.

Facts & Assumptions

Given: N=(0100), U=I2+N, V=I2, and the primitive pair (p,q)=(1,0).

[F1]

N2=0, so U is upper-triangular unipotent with determinant one and is not the identity.

[F2]

The identity matrix commutes with every matrix, and the eigenvalues of an upper-triangular matrix are its diagonal entries.

Counterexample

1.1F1F2givenalgebra

Since N2=0, U=(1101), so U∈SL2(C), has only eigenvalue 1, and is not I2. Also V=I2∈SL2(C). By [F2], the pair commutes and has common eigenline C(1,0) with eigenvalues M=L=1.

2.1F1givenstep 1.1algebra∎

For (p,q)=(1,0), the eigenvalue equation gives MpLq=1110=1, while UpVq=U≠I2. This refutes the statement. The example is only a matrix pair and does not claim to extend to any knot-group representation.

Depends on

Used by

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Sources