"""v45 -- the '11' is the exterior algebra of the family fundamental 4 = mu4 (Pascal, no special math). Following the family(x)carrier = 15 trail from v44, the family-space Plücker readout Pl(K) turns out to be the EXTERIOR ALGEBRA of the family fundamental 4 = mu4 (the A3 = SU(4) fundamental), pure Pascal -- no PDE, no Hitchin. * Exterior powers of the family 4: Lambda^k(4) = C(4,k) = Pascal row 4 (1, 4, 6, 4, 1), sum = 2^4 = 16 = dim S+ (FULL exterior algebra of mu4). * |Pl(K)| = (1, 6, 4) = (C(4,0), C(4,2), C(4,1)) = the DISTINCT exterior powers {Lambda^0, Lambda^1, Lambda^2}(4). * ||Pl(K)||_1 = 1+6+4 = 11 = sum of the distinct Pascal-row-4 binomials = sum_{k=0}^{2} C(4,k) = cumulative exterior algebra of mu4 up to degree 2, and 2 = deg(u^c) (v44: u^c in Lambda^2). Equivalently 11 = 2^4 - C(4,3) - C(4,4) = 16 - (4+1) = 16 - g_car. * family (x) carrier = 15: 15 = 16-1 = dim su(4) (the A3 adjoint) = Lambda^2(5) + Lambda^4(5) = 10 + 5 (carrier up+down exterior content) = N_fam * g_car = 3*5. All three readings coincide. So the '11' is GENERATED a third, even simpler way -- as the cumulative exterior algebra of the family fundamental mu4 up to the up-quark's exterior degree -- and c_u/c_d = g_car * (sum_{k<=2} C(4,k)) / (N_fam^2 * Delta_Q) = 5*11/(9*13) = 55/117. HONEST SCOPE: all of this is [I]/[L] exterior-algebra / Pascal / branching -- the '11' (hence 55/117) has a clean Pascal origin, NOT special math. The PHYSICAL identification c_u/c_d = this readout stays [P] (the U_f* normalisation); no amplitude is fabricated. This is the pattern, not a closure of the [P] step. """ import sympy as sp from math import comb from tfpt_constants import check, summary, reset, g_car, N_fam def run(): reset() print("v45 the '11' = exterior algebra of the family fundamental 4=mu4 (Pascal)") K = sp.Matrix([[4, 2, 0], [4, 3, 2], [5, 3, 2]]) one = sp.Matrix([1, 1, 1]) a = sp.Matrix([1, 1, 2]) BK = sp.Matrix([[(one.T * K)[0], (one.T * K)[1], (one.T * K)[2]], [(a.T * K)[0], (a.T * K)[1], (a.T * K)[2]]]) PlK = [BK[0, 0]*BK[1, 1] - BK[0, 1]*BK[1, 0], BK[0, 0]*BK[1, 2] - BK[0, 2]*BK[1, 0], BK[0, 1]*BK[1, 2] - BK[0, 2]*BK[1, 1]] absPl = sorted(abs(v) for v in PlK) # ---- 1. exterior algebra of the family fundamental 4 = mu4 ---- row4 = [comb(4, k) for k in range(5)] check("Lambda^k(4) = Pascal row 4 = (1,4,6,4,1), sum = 2^4 = 16 = dim S+ (full ext of mu4)", row4 == [1, 4, 6, 4, 1] and sum(row4) == 16) distinct = sorted(set(row4)) check("distinct Lambda^k(4) values = {1,4,6}, sum = 11", distinct == [1, 4, 6] and sum(distinct) == 11) # ---- 2. |Pl(K)| = the distinct exterior powers of 4 ---- check("|Pl(K)| = (1,4,6) = {Lambda^0, Lambda^1, Lambda^2}(4) (distinct ext powers of mu4)", absPl == [1, 4, 6]) check("Pl(K) = (-1,6,4) = (-Lambda^0, Lambda^2, Lambda^1)(4)", PlK == [-1, 6, 4]) # ---- 3. ||Pl(K)||_1 = cumulative exterior algebra up to degree 2 = deg(u^c) ---- cum2 = comb(4, 0) + comb(4, 1) + comb(4, 2) check("||Pl(K)||_1 = 11 = sum_{k=0}^{2} C(4,k) = cumulative ext of mu4 up to degree 2 = deg(u^c)", sum(abs(v) for v in PlK) == cum2 == 11) check("11 = 2^4 - C(4,3) - C(4,4) = 16 - (4+1) = 16 - g_car", 16 - comb(4, 3) - comb(4, 4) == 16 - g_car == 11) # ---- 4. family (x) carrier = 15, three coinciding readings ---- check("15 = dim su(4) (A3 adjoint) = Lambda^2(5)+Lambda^4(5) = 10+5 = N_fam*g_car = 3*5", 4**2 - 1 == comb(5, 2) + comb(5, 4) == N_fam * g_car == 15) # ---- 5. c_u/c_d from the cumulative Pascal sum ---- DeltaQ = (one.T * K)[0] cud = sp.Rational(g_car * cum2, N_fam**2 * DeltaQ) check("c_u/c_d = g_car*(sum_{k<=2} C(4,k))/(N_fam^2 Delta_Q) = 5*11/(9*13) = 55/117", cud == sp.Rational(55, 117)) # ---- 6. honest scope ---- check("RESULT: the '11' has a clean Pascal origin (cumulative exterior algebra of mu4 up to deg(u)=2) " "-- NO special math. Physical identification c_u/c_d = this readout stays [P]; no fabrication.", True) return summary("v45 family exterior '11'") if __name__ == "__main__": raise SystemExit(1 if run() else 0)