17 Taking into account the boundary conditions (50), (51), and defining | p $^{2}$|  ≡ ω $^{2}$, eqs. (52) are simply solved; the solutions are given by with As it can be seen, the first two components of the right sector drop out from the low-energy effective Lagrangian altogether; this corresponds to the fact that in general the right charged sector does not contain any light mode, in contrast with the left charged and neutral ones. Before substituting the solutions, note that the bulk contribution to the Lagrangian in Eq. (8) can be reduced - through an integration by parts - to a surface term plus a term proportional to the equations of motion, After the substitution, most of the terms vanish due to the BCs; we are left with taking into account the definition of the ˜ C and ˜ S functions (54), eq. (56) reduces to Eq. (57) has a complicated dependence on ω hidden in the functions ˜ S $^{′}$| $_{0}$, ˜ C $^{′}$| $_{0}$. In order to extract the low-energy behaviour of the theory, let’s expand in ω . This can be done in L (2) bulk = ω b ( πR ) 2 g 2 5 ˜ C ′ ˜ S ′ ∣ 0 ( ˜ W a, tr µ ˜ W a, tr µ + ˜ B tr µ ˜ B tr µ ) . (57) L (2) bulk = − 1 2 g 2 5 b ( y ) W a L µ ∂$_{y}$W a µ L | πR + ( L → R, a → 3); (56) L (2) bulk = − 1 2 g 2 5 ( ∂$_{y}$ ( b ( y ) W a, tr L µ W a, tr µ L ) − W a, tr L µ (( ˆ D − p $^{2}$) δ µ ν + p $^{µ}$p$_{ν}$ ) W a, tr ν L ) + ( L → R ) . (55) ˆ D ( ˜ S, ˜ C ) = − ω $^{2}$( ˜ S, ˜ C ); ˜ S ( πR, ω ) = 0 , ˜ $_{S}$′$_{(}$$_{πR, ω}$$_{) =}$ ω ; ˜ C ( πR, ω ) = 1 , ˜ $_{C}$′$_{(}$$_{πR, ω}$$_{) = 0}$$_{.}$ (54)      $^{W}$$^{a, tr}$ L µ = ( ˜ C ( y, ω ) − ˜ C ′ (0 , ω ) ˜ S ′ (0 , ω ) ˜ S ( y, ω )) ˜ W a, tr µ W a, long L µ = ˜ W a, long µ W 3 , tr R µ = ( ˜ C ( y, ω ) − ˜ C ′ (0 , ω ) ˜ S ′ (0 , ω ) ˜ S ( y, ω )) ˜ B tr µ W 3 , long R µ = ˜ B long µ W 1 , 2 R µ = 0 (53)