Again, in this sector there is no massless solution, for the constant profile of a massless mode is incompatible with the BC (25). Also, the equation of motion and the BCs are again diagonal in the isospin index, so f 1 R n = f 2 $_{Rn}$. The right charged sector obeys the usual L 2 orthogonality property: where the factor 1 /g 2 5 has been inserted to compensate for the mass dimension of the integral, so that we can normalize: R$_{m}$ ≡ 1, again ensuring that the kinetic terms will have the canonical normalization. the equation of motion and the BCs for the vector profiles are given by: and the pseudo-scalar profiles satisfy In contrast with the charged ones, the neutral sector admits a single massless solution; we have m$_{N}$$_{0}$ = 0. Eqs. (29) and (30) imply for a massless mode that both f 3 Ln and g 3 L,Rn = 1 m$_{N n}$ f 3 $_{L,Rn}$, if m$_{N n}$ = 0 g 3 L,Rn = 0 if m$_{N n}$ = 0 (32) c 3 n = ˜ v 2 m$_{n}$ ( f 3 Ln − f 3 R n )∣ ∣ $_{πR}$. ˆ Df 3 L,Rn = − m 2 $_{N n}$f 3 $_{L,Rn}$, (29) ∂$_{y}$f 3 L,Rn = 0 at y = 0 , (30)     ( ˜ g 2 g 2 $_{5}$∂$_{y}$ − b ( πR ) m 2 N n + ˜ g $^{2}$˜ v 2 4 ) f 3 Ln − ˜ g $^{2}$˜ v 2 4 f 3 R n = 0 ˜ g ′$_{2}$ g 2 5 ∂$_{y}$ − b ( πR ) m 2 N n + ˜ g $^{′$_{2}$}$˜ v 2 4 f 3 Rn − ˜ g $^{′$_{2}$}$˜ v 2 4 f 3 Ln = 0 at y = πR, (31) W 3 $_{L µ}$( x, y ) = ∞ ∑ n =0 f 3 $_{Ln}$( y ) N ( n ) µ ( x ) , W 3 L $_{5}$( x, y ) = ∞ ∑ n =0 g 3 $_{Ln}$( y ) G ( n ) N ( x ) , W 3 $_{R µ}$( x, y ) = ∞ ∑ n =0 f 3 $_{R n}$( y ) N ( n ) µ ( x ) , W 3 R $_{5}$( x, y ) = ∞ ∑ n =0 g 3 $_{Rn}$( y ) G ( n ) N ( x ) , π $^{3}$( x ) = ∞ ∑ j =0 c 3 j G ( j $^{)}$( x ); (28) ( f 1 , 2 $_{Rn}$, f 1 , 2 Rm ) ≡ 1 g 2 5 ∫ πR 0 dy f 1 , 2 $_{R n}$f 1 , 2 R m = R 2 $_{m}$δ$_{mn}$ , (27) • Finally, a neutral sector coming from the expansion of ( W 3 $_{L}$)$_{M}$ , ( W 3 $_{R}$)$_{M}$ and π $^{3}$. The expansion has the form 12