reproduced certificate script asserted: CERTIFIED chi(T3^3) >= 6
Pairs with the rank-8 upper bound to put this cell at 6 ≤ χ ≤ 8. Every rank-5 decomposition would put five stabilizer states on one projective line off the Galois space; the 45 candidate groups are enumerated and each candidate is rejected twice, by least squares and then in exact arithmetic over Q(ω9), where adjoining the target must raise the rank of the five states. The harvest of candidate lines is numerical and rests on the parallelism margin the certificate documents; the rejection of what it harvests is exact. The bound therefore still rests on the harvest margin, which is why it stays at the reproduced tier. The groups do exist, so the argument cannot be closed by observing that none do. Grouping is done on a random projection to C^6, which can only propose extra candidates, never lose one, since a linear map preserves collinearity.
this repository · stabrank PR #16 · Galois descent to cut the search to finitely many candidate lines, then exhaustion
{
"schema_version": "0.1",
"orbit": "T3",
"m": 3,
"direction": "lower",
"rank": 6,
"certificate": {
"script": "verify_challenge/cert_t3m3_rank6.py",
"expect": "CERTIFIED chi(T3^3) >= 6"
},
"provenance": {
"author": "this repository",
"reference": "stabrank PR #16",
"method": "Galois descent to cut the search to finitely many candidate lines, then exhaustion",
"date": "2026-09-07",
"github": [
"vprusso"
]
},
"notes": "Pairs with the rank-8 upper bound to put this cell at 6 <= chi <= 8. Every rank-5 decomposition would put five stabilizer states on one projective line off the Galois space; the 45 candidate groups are enumerated and each candidate is rejected twice, by least squares and then in exact arithmetic over Q(w9), where adjoining the target must raise the rank of the five states. The harvest of candidate lines is numerical and rests on the parallelism margin the certificate documents; the rejection of what it harvests is exact. The bound therefore still rests on the harvest margin, which is why it stays at the reproduced tier. The groups do exist, so the argument cannot be closed by observing that none do. Grouping is done on a random projection to C^6, which can only propose extra candidates, never lose one, since a linear map preserves collinearity."
}