57. The earlier disk obstruction has an exact finite matrix witness
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For the rank-three model of Section 51, put
Then \(\operatorname{tr}\mathcal J=m\) and
Its eigenvalues are \((m\pm\sqrt{x^2+y^2})/2\), so positivity is exactly the Lorentz-cone condition. The normalized base is the unit disk.
Proposition 51: Exact rational matrix witness for the rotation model
Let \(T=\operatorname{diag}(1,R)\) with the unchanged rotation
The rotate branch and the two probe branches from Proposition 46 have the identities
where \(E_{ij}\) is the corresponding \(2\times2\) matrix unit. The other rotate branch is zero. Thus this exact model has a finite rational matrix certificate without attenuation or a polygonal replacement.
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Proof
Direct multiplication gives the rotate identity. The two matrix-unit congruences retain, respectively, the upper and lower diagonal entries of \(\mathcal J(z)\) and place them in its upper diagonal entry. These are \((m+x)/2\) and \((m-x)/2\), exactly the two probe masses resetting to \((1,1,0)\). Also \(W^{\mathsf T}W=5I\) and
The actions conserve trace. The initial state embeds as \(\operatorname{diag}(1,0)\), and \((1,0,0)\) is a strictly positive interior witness. Proposition 50 applies. \(\square\)
Nothing has been changed in the observation law. Proposition 46 still rules out a finite positive-generator certificate for this model. What changes is the certificate language. A curved matrix domain and a finite set of positive coordinate states are not the same representation class.
This use of symmetric positive semidefinite matrices is mathematical. It neither requires quantum hardware nor claims a quantum description of the measured system. It borrows a well-defined operator representation rather than a physical interpretation.