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Credential: RUB5-PCAN5-20260813 · Status: PASS · Evidence: E0 + E1

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Canonical projector

Let $P_{\mathrm{sym}}$ be the trivial-character projector for the common $S_3$ action and $P_6$ the Casimir spectral projector at eigenvalue 6. Then

$$ P_{\mathrm{can}}=P_6P_{\mathrm{sym}}=P_{\mathrm{sym}}P_6, $$

$$ P_{\mathrm{can}}^2=P_{\mathrm{can}}=P_{\mathrm{can}}^\dagger, \qquad \operatorname{rank}P_{\mathrm{can}}=5. $$

It selects the unique $S_3$-trivial spin-2 copy inside E47:

$$ \mathbb C^{125}\supset E_{47}\supset E_{\mathrm{can}}, \qquad 125\supset47\supset5. $$

Machine certificate

• Canonical leakage: 4.163×10^-14
• Canonical S₃ displacement: 8.778×10^-15
• Occupancy drift under the declared S₃-invariant unitary: 7.078×10^-16
• Validator result: PASS 20/20
• SHA-256: 588a0ba2c095da3b3c83bfb1787522006cbe903bc5858947fc578d55d6b62cd2

Provenance

Professor’s Cube Exhibit — Shared-Carrier S₃ Symmetry Bridge

Dynamic closure certificate · E47-JSSC-20260813

The existing rank-5 projector is now the fixed projector of the positive joint generator

$A_{can}=K^2+11664(I-P_{sym})$.

The machine certificate proves $\ker A_{can}=\operatorname{im}P_{can}$ and the attained identity

$\|(I-A_{can}/99144)^n-P_{can}\|_2=(15/17)^n$.