14 A. The expanded Lagrangian After the expansion in mass eigenstates, the gauge Lagrangian, taking into account con- tributions from both brane and bulk terms, is reduced to the form: L gauge = − 1 2 W + ( n ) L µν $_{W}$− ( n ) µν L − 1 2 W + ( n ) R µν $_{W}$− ( n ) µν R − 1 4 N ( n ) µν N ( n ) µν − ∣ ∂$_{µ}$G + ( n ) L − m$_{L n}$W + ( n ) L µ ∣ ∣$^{2}$ − ∣ ∂$_{µ}$G + ( n ) R − m$_{R n}$W + ( n ) L µ ∣ ∣$^{2}$ − 1 2 ( ∂$_{µ}$G ( n ) N − m$_{N n}$N ( n ) µ ) 2 + { i g L klm [ N ( m ) µν W + ( k ) µ L $_{W}$− ( l ) ν L + N ( m ) µ ( $_{W}$− ( l ) µν L ) W + ( k ) L ν − h.c. ) ] + g 2 LL klmn W + ( k ) µ L $_{W}$− ( l ) ν L W + ( m ) ρ L $_{W}$− ( n ) σ L ( η$_{µρ}$η$_{νσ}$ − η$_{µν}$η$_{ρσ}$ ) ] + g 2 LN klmn [ W + ( k ) µ L $_{W}$− ( l ) ν L N ( m ) $^{ρ}$N ( n ) $^{σ}$( η$_{µρ}$η$_{νσ}$ − η$_{µν}$η$_{ρσ}$ ) ] + ( L ↔ R ) } . (38) The bilinear part of the Lagrangian is, as announced, diagonal. The trilinear and quadrilin- ear coupling constants g L $_{klm}$, g 2 LL $_{klmn}$, g 2 LN klmn are defined in terms of the gauge profiles: g L klm = 1 g 2 5 ∫ πR 0 dyf 1 $_{L k}$f 1 $_{L l}$f 3 L m + 1 ˜ g $_{2}$f 1 $_{L k}$f 1 $_{L l}$f 3 L m | $_{πR}$, (39) g 2 LL klmn = 1 g 2 πR dyf 1 $_{L k}$f 1 $_{L l}$f 1 $_{L m}$f 1 L n + 1 ˜ g $_{2}$f 1 $_{L k}$f 1 $_{L l}$f 1 $_{L m}$f 1 L n | $_{πR}$, (40) 5 0 g 2 LN klmn = 1 g 2 5 ∫ πR 0 dyf 1 $_{L k}$f 1 $_{L l}$f 3 $_{L m}$f 3 L n + 1 ˜ g $_{2}$f 1 $_{L k}$f 1 $_{L l}$f 3 $_{L m}$f 3 L n | πR (41) (remember that f 1 L ( R ) n ≡ f 2 L ( R ) $_{n}$); similar definitions hold for the coupling constants g R $_{klm}$, g 2 RR $_{klmn}$, g 2 RN klmn of the right sector, but without any contribution from boundary terms due to eq. (25). An important observation concerns the couplings g L ( R ) kl 0 . These give the coupling of N (0) , which we identified with the photon, with the charged fields; as a consequence, they should all be equal to the electric charge, for any value of k, l . By the definition (39) and eq. (33), we immediately get: g L kl 0 = g R kl 0 ≡ f$_{0}$δ$_{kl}$, (42) thanks to the fact that the wave-functions f 1 L k and f 1 R k form an orthonormal basis. Then we conclude that f$_{0}$ = e. (43)