13 f 3 Rn must be constant; then, using also eq. (31) we get: f 3 L 0 = f 3 R 0 ≡ f$_{0}$, (33) where f$_{0}$ is a constant. The massless mode has to be identified with the photon ⇒ N (0) µ ≡ A$_{µ}$ ; since it is the only massless mode in the spectrum we have that the symmetry of the vacuum is, correctly, just U (1)$_{e.m.}$. The “charged” and “neutral” labels we have given to the three sectors refer to their transformation properties with respect to this unbroken symmetry. As in the case of the left charged sector, the BCs at y = πR in this case explicitly contain the mass of the n th mode, so that again the basis wave-functions f 3 Ln and f 3 Rn have nonstandard orthogonality properties. The correct relations are: f 3 Ln = K$_{n}$f 3 $_{Rn}$, (35) ( N L $_{n}$) 2 + ( N R $_{m}$) 2 = 1; (36) and the constants K$_{n}$ are fixed by the BCs at y = πR (31). To get, also in this case, canonically normalized kinetic terms we have to set: the ratio ( N L $_{n}$) / ( N R n ) will be fixed by the value of K$_{n}$ and by eq. (34). In particular, for the massless mode it is easy to get 1 f 2 0 = 2 πR g 2 5 + 1 ˜ g 2 + 1 ˜ g ′ $_{2}$. (37)