Proving N2: why n_solar = 4 (not a free +1 guess)

Tier-B EXACT proof that the solar numerator is 4 via three equivalent routes: Q₃/2, n_Y−1, n_W+1 — algebraically identical under Direction-1 integers. Plus light-sector 2×2 count. Easy language. Script: spt_master_map_n2_solar.py.

Created 07/19/2026, 22:41 GMT+7Updated 07/19/2026, 22:41 GMT+7

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Problem. Law N2 used to sound soft: “adjacent role = n_W+1 = 4.” Skeptics hear: you added one to fit 4/13. Result. Under Direction-1 integers (n_W=3, n_Y=5, Q₃=8), three independent-looking counts are the same number 4, and “+1” is just a rewrite of that identity — Tier-B EXACT.

Parent pages: What is Map F? · F from Action. Glossary: /glossary?q=N2 · /glossary#spt-master-map.

§1 Plain picture

On lock 13, the weak tooth is 3. The solar tooth is 4. Why not 5 or 2?

Because three natural counts — half of the 8 Bagua cells, hypercharge-budget minus vacuum, and “one step after weak” — all give 4 at once. You cannot change one without breaking the Direction-1 integers already fixed by gauge content.

§2 Four routes to 4

Route	Formula	Plain reading
R-A Half-cell	Q₃/2 = 8/2 = 4	Solar role = half the 8 trigrams
R-B Y − vacuum	n_Y − 1 = 5 − 1 = 4	Hypercharge budget minus vacuum pole
R-C Adjacent	n_W + 1 = 3 + 1 = 4	Next tooth after weak on shell W
R-D Light ν	2 × 2 = 4	m₁=0 ⇒ 2 massive ν × 2 real slots

§3 The theorem (why +1 is not free)

From Direction 1: n_Y = Q₃ − n_W. Then:

text

n_Y − 1 = Q_3 − n_W − 1
n_W + 1 = n_W + 1

Equal  ⇔  Q_3 − 2 = 2·n_W  ⇔  n_W = (Q_3−2)/2

With Q_3=8: n_W = 3  (already true from SU(2)).
Hence:   Q_3/2 = n_W+1 = n_Y−1 = 4.

Punchline. “Adjacent +1” is not an extra assumption on top of Direction 1. Once n_W=3 and Q₃=8, it is the same statement as half-cell and (n_Y−1). N2 is upgraded from slogan → identity.

§4 Consequence for Map F

sin²θ₁₂ = 4/13 forced; gap to tree θ_W is exactly 1/13 (ID-B).
Unique integer in 1..12 equal to all three of Q₃/2, n_W+1, n_Y−1 is 4.
If DUNE forces sin²θ₁₂ far from 4/13, the triple identity dies — not one soft rule.

PDG check (OUTPUT only): 4/13 ≈ 0.3077 vs 0.307 → Δ ≈ 0.23%.

§5 Importance

Axis	Score	Why
Closing softest tooth on W	9/10	Solar 4 was the main residual hand-wave
Anti-numerology	9/10	Three routes + uniqueness, not one slogan
Path-integral depth	5/10	Algebra + structural count; not Feynman yet

§6 Honest remaining

Still open: a dynamical path-integral story of why the solar role is the half-Q₃ projection (beyond the algebraic identity). The integer 4 and the triple equivalence are closed.

§7 SymPy

SymPy verify — download for offline testSYMPY ✓

N2 proof script

scripts/spt_master_map_n2_solar.py

spt_master_map_n2_solar.py — R-A/B/C/D = 4; equivalence theorem; uniqueness; ID-B

180 LOCDownload

scripts/spt_master_map_numerators.py

Full N1–N6 package — All numerators + identities A,B,C

240 LOCDownload

Reproduce in 30 seconds

pip install sympy numpy && python3 scripts/spt_master_map_n2_solar.py && python3 scripts/spt_master_map_numerators.py

Or quick-verify with AI (Grok / Claude / ChatGPT)

Don't want to install Python? Paste the prompt straight into Grok / Claude / ChatGPT / Gemini — the AI fetches the public script URL below and independently verifies each assertion in ~30 s. Open grok.com or claude.ai , paste, send.

⚠️ AI can be wrong — running the Python above is the only 100% certain check. Full AI guide →

Inputs: Bagua integers + π/√ only — no CODATA, no PDG, no calibration (Tier B). SymPy-verified as exact fractions (not floating-point). See full context at /theory/sympy-breakthrough-2026.

Bottom line for humans. N2 is no longer “we added one.” It is: half of 8 = one more than 3 = one less than 5 = 4, and that forces solar 4/13 and the 1/13 gap. That is the accurate law of the solar tooth.