Particular boundary condition ensures that a fermion in d=1+5, compactified on a finite disk, manifests in d=1+3 as massless spinor with a charge 1/2, mass protected and chirally coupled to the gauge field

The genuine Kaluza-Klein-like theories--with no fields in addition to gravity--have difficulties with the existence of massless spinors after the compactification of some space dimensions \cite{witten}. We proposed in previous paper a boundary condition for spinors in d=(1+5) compactified on a flat disk that ensures masslessness of spinors (with all positive half integer charges) in d=(1+3) as well as their chiral coupling to the corresponding background gauge gravitational field. In this paper we study the same toy model, proposing a boundary condition allowing a massless spinor of one handedness and only one charge (1/2) and infinitely many massive spinors of the same charge, allowing disc to be curved. We define the operator of momentum to be Hermitean on the vector space of spinor states--the solutions on a disc with the boundary.


I. INTRODUCTION
The major problem of the compactification procedure in all Kaluza-Klein-like theories with only gravity and no additional gauge fields is how to ensure that massless spinors be mass protected after the compactification. Namely, even if we start with only one Weyl spinor in some even dimensional space of d = 2 modulo 4 dimensions (i.e. in d = 2(2n + 1), n = 0, 1, 2, · · ·) so that there appear no Majorana mass if no conserved charges exist and families are allowed, as we have proven in ref. [3], and accordingly with the mass protection from the very beginning, a compactification of m dimensions gives rise to a spinor of one handedness in d with both handedness in (d − m) and is accordingly not mass protected any longer.
And besides, since the spin (or the total conserved angular momentum) in the compactified part of space will in space of (d − m) dimensions manifest accordingly a charge of both signs while in the second quantization procedure antiparticles of opposite charges appear anyhow, doubling the number of massless spinors when coming from d(= 2(2n + 1))dimensional space down to d = 4 (after the second quantized procedure) is not in agreement with what we observe. Therefore there must be some requirements, some boundary conditions [2], which ensure in a compactification procedure that only spinors of one handedness and one charge survive if Kaluza-Klein-like theories have some meaning, what due to the beautifulness of the idea of the gravity as the only gauge field we would hope for.
One of us [4,5,6,7,8,9] has for long tried to unify the spin and all the charges to only the spin, so that spinors would in d ≥ 4 carry nothing but (two kinds of [13]) a spin and interact accordingly with only the gauge fields of the corresponding generators of the infinitesimal transformations (of translations and two kinds of the Lorentz transformations in the space of spinors), that is with vielbeins f α a [14] and (two kinds of) spin connections (ω abα , which are the gauge fields of S ab = i 4 (γ a γ b − γ b γ a ) andω abα , which are the gauge fields ofS ab = i 4 (γ aγb −γ bγa ), with {γ a ,γ b } + = 0). In this paper we take (as we did in ref. [2]) for the covariant momentum of a spinor The corresponding Lagrange density L for a Weyl spinor has the form L = E 1 2 [(ψ † γ 0 γ a p 0a ψ) + (ψ † γ 0 γ a p 0a ψ) † ] and leads to with E = det(e a α ). If we have no gravity in d = (1 + 3), and for ω abc = 0, while f σ s = δ σ s f (ρ), the equations of motion follow (We use s, t, or σ, τ, to denote the two compactified dimensions (x 5 and x 6 for the flat and the Einstein index, respectively), m, n, to denote the flat experimentally observed 1 + 3 = d − m dimensions, and a, b, and α, β, to denote all the flat and Einstein indices, respectively.) The authors of this work found a way out of the "Witten's no go theorem" for a toy model of M (1+3) × a flat finite disk in (1 + 5)-dimensional space [2] by postulating a particular boundary condition, which allows a spinor to carry only one handedness after the compactification-but many charges (all positive integers). Massless spinors then chirally couple to the corresponding background gauge gravitational field, which solves equations of motion for a free field, linear in the Riemann curvature, while the current through the boundary for the massless and all the massive solutions is equal to zero. In ref. [2] the boundary condition was written in a covariant way aŝ with n (ρ) = (0, 0, 0, 0, cos φ, sin φ), n (φ) = (0, 0, 0, 0, − sin φ, cos φ), which are the two unit vectors perpendicular and tangential to the boundary of the disk at ρ 0 , respectively. The operatorR is a projector. It can for the above choice of the two vectors n (ρ) and n (φ) be written asR In Appendix A properties of the Clifford algebra objects In this paper: i. We formulate the boundary condition allowing one massless state of one handedness and only one charge and infinitely many massive states with the same charge so that to each mass only one state corresponds.
ii. We define the operator for momentum p a so that it becomes Hermitean on the vector space of states fulfilling the boundary conditions and we comment on the orthogonality relations of these states.
iii. We study the properties of states on a curved disc with the boundary.
unitary. We shall see that when requiring that solutions of equations of motion obey the conditionR ′ k ψ| ρ=ρ 0 = 0, for k = 1, the same states are projected out for any k. In this sense alsoÔ ′ manifests unitarity on the states of solutions (obeying boundary conditions on the boundary, where both operators only apply).

III. EQUATIONS OF MOTION AND SOLUTIONS
We study two cases: i. We assume that the two dimensional space of coordinates x 5 and x 6 is a Euclidean plane M (2) (with no gravity) so that f σ s = δ σ s , ω 56s = 0, with the rotational symmetry around the origin, ii. We assume that space of x 5 and x 6 is curved with the zweibein f σ s = δ σ s f (ρ), with ρ, defined by x 5 = ρ cos φ, x 6 = ρ sin φ, so that the rotational symmetry around the axis perpendicular to the plane of x 5 and x 6 is preserved.
If the Euclidean plane is curved on S 2 with the radius ρ 0 and the rotational symmetry around the axis perpendicular to the plane of x 5 and x 6 , then where ψ 0 is a vacuum state. If we write the operators of handedness in d = (1 + 5) as and in the two dimensional space as Γ (2) = iγ 5 γ 6 (= 2S 56 ), we find that all four states are left handed with respect to Γ (1+5) , with the eigenvalue −1, the first two states are right handed and the second two states are left handed with respect to Γ (2) , with the eigenvalues 1 and −1, respectively, while the first two are left handed and the second two right handed with respect to Γ (1+3) with the eigenvalues −1 and 1, respectively. Taking into account Eq. (8) we may write the most general wave function ψ (6) obeying Eq.(3) in d = (1 + 5) as where A and B depend on x 5 and x 6 , while ψ (+) and ψ (4) [−] determine the spin and the coordinate dependent parts of the wave function ψ (6) Using ψ (6) in Eq. (3) we recognize the following expressions as the mass terms: where z := x 5 + ix 6 = ρ e iφ ,z := . We can rewrite Eq.(11) in a more compact form as follows Having the rotational symmetry around the axis perpendicular on the fifth and the sixth dimension we require that ψ (6) is the eigenfunction of the total angular momentum operator [−]0 .
There is a solution for any positive integer n and there is obviously no mass protection, since the solution for any chosen n is the superposition of the left and the right handed components in d = (1 + 3). Taking into account the boundary condition of Eq.(4) one sees that β n must be zero for all n, accordingly ψ (+)0 for any f (ρ), which means that only solutions of one handedness in d = (1 + 3) are allowed, assuring the mass protection mechanism in d = (1 + 3). However, all the positive integers n are allowed (we require n ≥ 0 to ensure the integrability of solutions at the origin).
Taking into account the boundary condition of Eq. (7) we see that β n must still be zero for any n, while now the conditionR ′ ψ (6) n+1/2 0 | ρ=ρ 0 = 0 leads to the condition In the case that a disc is flat with the boundary at ρ = ρ 0 we get that only n = 0 fulfils the boundary condition of Eqs. (7,15). If a disk is curved on a sphere with radius ρ 0 and we put a boundary at ρ = ρ 0 , then f = (1 + ( ρ ρ 0 ) 2 ) and the boundary condition requires n(n + 1 2 ) = 0. Again the only solution is n = 0, since n is an integer. More general f would lead to rational or irrational numbers (and so would a boundary at some other ρ 1 = ρ 0 ), so that we can conclude that n = 0 is the only solution even if the disk is not flat.
Therefore for m = 0 we get as the only solution for any curvature f (f σ s = f δ σ s , ω 56s = 0, s = 5, 6; σ = 5, 6,) where J n are the Bessel's functions of the first order, which depend on ρ, while N n determines the normalization [3].
If we require that the boundary condition of Eq.(5) should be fulfilled, then J n+1 | ρ=ρ 0 = 0 and the zeros of J n+1 determine for each n and each zero of J n a (different) mass m.
If we require that the boundary condition of Eq. (7) is fulfilled, then only n = 0 is the solution. In this case we namely have J n+1 | ρ=ρ 0 = 0 and n ( 1 ρ ∂Jn ∂ρ )| ρ=ρ 0 = 0. It turns out that n = 0 is the only possibility, since J n+1 | ρ=ρ 0 = 0 = ∂Jn ∂ρ | ρ=ρ 0 is true only for n = 0. This relation is fulfilled for infinitely many masses m i = α 1i /ρ 0 , i = 1, · · ·, where index i determines the successive number of a zero of J 1 at ρ = ρ 0 . There are accordingly infinitely many massive solutions, which obey the equations of motion (Eq.(12)) for f = 1 and the boundary condition of Eq. (7), all having eigenvalue of M 56 equal to 1/2 (18) For f = (1 + ( ρ ρ 0 ) 2 ) the solutions of Eq.(12) obeying the boundary condition of Eq. (7) can not be found among the known functions, but we still know that they have the eigenvalue of M 56 equal to 1/2 and we also guess that they behave pretty like the two Bessel's functions in Eq.(18).

IV. CURRENT THROUGH THE WALL
The current perpendicular to the wall can be written as For physically acceptable cases when spinors are localized inside the disk the current through the wall must be equal to zero One easily checks that-since the current operatorĵ ⊥ changes the handedness of a state-in the massless case (since massless states of only one handedness exist (Eq.(16)) and all the massive cases (the product of A and B appears in the current and B is zero, in particular in Eq.(18) J 1i | ρ=ρ 0 = 0) the current through the wall is for both types of the boundary conditions equal to zero.

V. HERMITICITY OF THE OPERATORS AND ORTHOGONALITY OF SOLU-TIONS
The operators p s (and consequently also (γ s p s ) 2 ) are not Hermitean on the space of solutions which have nonzero values on the boundary ρ = ρ 0 , since then d 2 xp s (ψ i † ψ j ) = 0.
We define therefore a new operatorp s . We take care of a flat disc with the boundary.

Statement:
The operatorsp s , are Hermitean on the vector space of solutions presented in Eqs.(16,18).
Proof: Since the expectation value ofp s is zero between the massless states (Eq.(16)) and so is also between the massless and all the massive states (Eq.(18)), we check the Hermiticity of the operator (γ sp . We ought to check only the x 5 and x 6 part, with the corresponding spin components included. We obtain for the expectation values of (γ sp s ) 2 between the massless and a massive state the values: ∂ρ one finds that for i = k it follows since J 1k (α 1k ) = 0. We checked accordingly the Hermiticity of the operator ρ(γ sp s ) 2 on the vector space of the massive states and correspondingly also the orthogonality of these states.
Let us add the normalization property of the massive states since m i = α 0 i ρ 0 . We conclude that on the space of solutions (Eqs. (16,18)) the operatorsp s (Eq.(21)) are Hermitean and the solutions are orthogonal. Since we do not know the explicit expressions for solutions on the curved disc (f = 1), we do not comment orthogonality properties of these functions.
with f σ m = A µ δ µ m ε σ τ x τ and the spin connection field The U(1) gauge field A µ depends only on x µ . All the other components of the spin connection fields are zero, since for simplicity we allow no gravity in (1 + 3) dimensional space.
To determine the current, coupled to the Kaluza-Klein gauge fields A µ , we analyze the spinor action ψ (6) are solutions of the Weyl equation in d = (1 + 5) . E is for f α a from equal to f −2 . The first term on the right hand side of Eq.(26) is the kinetic term (together with the last term defines the covariant derivative p 0µ in d = (1 + 3)). The second term on the right hand side contributes nothing when the integration over the disk is performed, since it is proportional to x σ (ω smµ = − 1 2 F µν δ ν m ε sσ x σ ).
We end up with couples to the corresponding Kaluza-Klein gauge field (so that after the second quantization procedure a particle and an antiparticle of only that particular charge and the opposite one appear, respectively).
We propose the boundary condition which is in Eq.(6) presented in a Lorentz invariant way, with θ which is an arbitrary parameter = (2k + 1)π/2, and k is any integer. This boundary condition allows in the massless case only the right handed spinor to live on the disk and accordingly manifests left handedness in M (1+3) . The massless and the massive solutions have the eigen value of the total angular momentum in the fifth and the sixth dimension equal to 1/2, which then manifests as a charge in d = (1 + 3). The massless solution is mass protected. When a disk is flat, the massless solution is independent of x 5 and x 6 , while the massive solutions are expressible in terms of the Bessel's functions J 0 (α i ρ/ρ 0 ) and J 1 (α i ρ/ρ 0 ), defining masses m i = α i /ρ 0 through the requirement that the i − th zero of J 1 is zero at ρ = ρ 0 .
We define a generalized momentum which is Hermitean on the vector space of states obeying equations of motion (Eq. (3)) for f σ s = δ σ s and our boundary condition (Eq.(28)). Accordingly also the operator γ sp s γ tp t is Hermitean on the same vector space, and the states are accordingly orthogonal, with the eigen values of this operator which demonstrate the masses of states.
The negative −1/2 charge states appear only after the second quantization procedure in agreement with what we observe.
If the disc is curved, so that f σ s = δ σ s f , with f = (1 + ( ρ ρ 0 ) 2 ) if it is on S 2 with a radius ρ 0 , the solutions obeying Eqs. (3,28) have similar properties as for f = 1, but in this case we present only the explicit expression for the massless state, while the massive ones stayed to be determined.

03
[±i], and to ± 1 2 for 12 (±) and 12 [±], as well as (+) ψ 0 , where ψ 0 is a vacuum state (any, which is not annihilated by the operator in front of the state) has the eigenvalues of S 03 , S 12 and S 56 equal to i 2 , 1 2 and 1 2 , correspondingly. All the other states of one representation of SO(1, 5) follow from this one by just the application of all possible S ( ab), which do not belong to the Cartan subalgebra.