Paper I proved |G(d,d')| ≤ 2 min(G(d,d), G(d',d')) for the Heegner operator set at level 163. Does this inequality hold across the entire class-number-one Heegner-prime family H' = {11, 19, 43, 67, 163}?
Let H = {2, 3, 7, 11, 19, 43, 67, 163} be the class-number-one Heegner primes. At each level N in H', the gate set is SN = {p in H \ {2} : p ≤ N}, with TN = UN = −wN at the prime level:
| Level N | Gate set SN | dim S₂(Γ₀(N)) |
|---|---|---|
| 11 | {3, 7, 11} | 1 |
| 19 | {3, 7, 11, 19} | 1 |
| 43 | {3, 7, 11, 19, 43} | 3 |
| 67 | {3, 7, 11, 19, 43, 67} | 5 |
| 163 | {3, 7, 11, 19, 43, 67, 163} | 13 |
N = 19, S₁₉ = {3, 7, 11, 19} — the 4 x 4 trace Gram matrix G₁₉(d, d') = Tr(TdTd' | S₂(Γ₀(19))):
| 3 | 7 | 11 | 19 | |
|---|---|---|---|---|
| 3 | 4 | 2 | −6 | −2 |
| 7 | 2 | 1 | −3 | −1 |
| 11 | −6 | −3 | 9 | 3 |
| 19 | −2 | −1 | 3 | 1 |
Violation: |G₁₉(7, 11)| = 3, but 2 min(G₁₉(7,7), G₁₉(11,11)) = 2 min(1, 9) = 2. Ratio: 3/2.
N = 67, S₆₇ = {3, 7, 11, 19, 43, 67} — the 6 x 6 Gram matrix:
| 3 | 7 | 11 | 19 | 43 | 67 | |
|---|---|---|---|---|---|---|
| 3 | 14 | −4 | 14 | −26 | −10 | 2 |
| 7 | −4 | 30 | −6 | 0 | −12 | 0 |
| 11 | 14 | −6 | 28 | −54 | 14 | −2 |
| 19 | −26 | 0 | −54 | 135 | 2 | −5 |
| 43 | −10 | −12 | 14 | 2 | 134 | −14 |
| 67 | 2 | 0 | −2 | −5 | −14 | 5 |
Violation: |G₆₇(43, 67)| = 14, but 2 min(G₆₇(43,43), G₆₇(67,67)) = 2 min(134, 5) = 10. Ratio: 14/10 = 7/5.
| Level N | Pairs tested | Status | Violating pair | Violation ratio |
|---|---|---|---|---|
| 11 | 3 | Holds | -- | -- |
| 19 | 6 | Fails | (7, 11) and (11, 19) | 3/2 |
| 43 | 10 | Holds | -- | -- |
| 67 | 15 | Fails | (43, 67) | 7/5 |
| 163 | 21 | Holds | -- | -- |
At N in {11, 19}, the cuspidal space is one-dimensional, so GN has rank one and factors as GN(d, d') = ad(fN) · ad'(fN). In this case the inequality reduces to:
For N = 11 (11a1): (a₃, a₇, a₁₁) = (−1, −2, 1), absolute values (1, 2, 1), ratio 2 — holds at the boundary. For N = 19 (19a1): (a₃, a₇, a₁₁, a₁₉) = (−2, −1, 3, 1), absolute values (2, 1, 3, 1), ratio 3 > 2 — fails.