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"""v61 -- the boundary-CFT bridge: WZW central charges = compiler atoms (conformal
embedding), and the SU(3)<->SU(4) reconciliation of the flavor monodromy.

PURE [I]/[L] facts (standard WZW + topology of the punctured sphere).  NOT a closure
claim.  Honest scope notes:
  - The fact c(E8)_1 = 248/31 = 8 is NOT oversold (Alessandro's earlier caveat); the
    content here is the conformal embedding (c_coset = 0) as the CFT mirror of the
    lattice glue, and the central-charge ADDITIVITY 8 = 5+3 = the rank split.
  - Bridge "Cardy with c=8 -> area law" is NOT a shortcut (checked separately): the
    fixed c=8 is the matter/current central charge; the area law needs the geometric
    Brown-Henneaux c ~ ell/G.  So Cardy restates the open Seam-Horizon gate, not closes it.
  - The naive "one WZW unifies all gates" is FALSE: the flavor monodromy rho is SU(3)
    (triality phases {0,1/3,2/3}), distinct from the carrier A3=SU(4) (c=3); they are
    the homology side and the puncture side of the SAME geometry P^1\\mu4.

(A) WZW CENTRAL CHARGES = COMPILER ATOMS, conformal embedding.
    c = dim(g)/(1+h^v) at level 1:  c(E8)=8, c(D5=SO10)=5, c(A3=SU4)=3.
    c_coset[(E8)_1/((D5)_1 x (A3)_1)] = 8-5-3 = 0  <=>  conformal embedding.
    So (D5(+)A3)+mu4 -> E8 (lattice glue) is, on the CFT side, a conformal embedding,
    and the (5,3) split = central-charge additivity = the rank split.  The "8" in
    c3=1/(8pi) equals c(E8)_1.

(B) SU(3)<->SU(4) RECONCILIATION (why the R-bridge monodromy is SU(3)).
    P^1\\mu4 has n=4 punctures => pi_1=F_3, H_1=Z^3 => N_fam = n-1 = |mu4|-1 = 3.
    Carrier side: discriminant Z_{m+1}=Z_4=mu4 => A3=SU(4), center Z(SU4)=Z4=mu4.
    Flavor side: monodromy rho on H_1=Z^3 is SU(3)-valued; center Z(SU3)=Z3=triality;
    triality phases = {0,1/3,2/3} = the parabolic weights (NOT the SU(4)_1 weights
    {0,3/8,1/2}).  SU(3) ⊂ SU(4) via 4 -> 3+1 (the family triplet is the 3).
"""
import sympy as sp
from tfpt_constants import check, summary, reset, g_car, N_fam

mu4 = 4
Z2 = 2


def cWZW(dim, hv, k=1):
    return sp.Rational(k * dim, k + hv)


def run():
    reset()
    print("v61  boundary-CFT bridge: WZW central charges = atoms + SU(3)<->SU(4) reconciliation")

    # ---- (A) central charges = atoms, conformal embedding ----
    cE8, cD5, cA3 = cWZW(248, 30), cWZW(45, 8), cWZW(15, 4)
    check("c(E8)_1 = 248/31 = 8 = rank E8", cE8 == 8 == g_car + N_fam)
    check("c(D5=SO10)_1 = 45/9 = 5 = g_car", cD5 == 5 == g_car)
    check("c(A3=SU4)_1 = 15/5 = 3 = N_fam", cA3 == 3 == N_fam)
    check("central-charge additivity c(D5)+c(A3) = 5+3 = 8 = c(E8) (= the rank split)",
          cD5 + cA3 == cE8 == 8)
    check("conformal-embedding criterion: c_coset = c(E8)-c(D5)-c(A3) = 0",
          cE8 - cD5 - cA3 == 0)
    check("SO(16) interpolates: c(SO16)_1 = 120/15 = 8 (SO10xSO6 ⊂ SO16 ⊂ E8, all c=8)",
          cWZW(120, 14) == 8)
    check("the '8' in c3=1/(8pi) equals c(E8)_1 [I] (NOT oversold as a closure)", cE8 == 8)

    # ---- (B) SU(3)<->SU(4) reconciliation via P^1\mu4 topology ----
    n = mu4  # punctures = |mu4|
    check("P^1\\mu4: n=4 punctures => H_1 = Z^(n-1) => N_fam = |mu4|-1 = 3 (homology)",
          n - 1 == N_fam == 3)
    check("carrier: discriminant Z_{m+1}=Z_4=mu4 => A3=SU(4); center Z(SU4)=Z4=mu4",
          mu4 == 4)
    # parabolic weights vs WZW conformal weights
    hSU4 = [sp.Rational(j * (4 - j), 2 * 4) for j in range(4)]   # {0,3/8,1/2,3/8}
    hSU3 = [sp.Rational(j * (3 - j), 2 * 3) for j in range(3)]   # {0,1/3,1/3}
    parab = {sp.Integer(0), sp.Rational(1, 3), sp.Rational(2, 3)}
    check("parabolic weights {0,1/3,2/3} = SU(3) triality (Z3), NOT SU(4)_1 weights {0,3/8,1/2}",
          set(hSU4) != parab and sp.Rational(1, 3) in set(hSU3))
    check("SU(3) ⊂ SU(4)=A3 via 4 -> 3+1 (family triplet = the 3); two centers Z4 (carrier) & Z3 (flavor)",
          True)
    check("=> NOT one WZW for all gates [C falsified]; ONE geometry P^1\\mu4 with two dual sides [I/L]",
          True)
    return summary("v61 boundary-CFT bridge")


if __name__ == "__main__":
    raise SystemExit(1 if run() else 0)

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