Yang–Mills Theory of Gravity

Article Yang–Mills Theory of Gravity

Malik Al Matwi

Department of Mathematical Science, Ritsumeikan University, 4-2-28, Kusatsu-Shi, Shiga 525-0034, Japan; [email protected]

Received: 25 June 2019; Accepted: 22 October 2019; Published: 12 November 2019

Abstract: The canonical formulation of general relativity (GR) is based on decomposition space–time manifold M into R × Σ, where R represents the time, and Σ is the three-dimensional space-like surface. This decomposition has to preserve the invariance of GR, invariance under general coordinates, and local Lorentz transformations. These symmetries are associated with conserved currents that are coupled to gravity. These symmetries are studied on a three dimensional space-like hypersurface Σ embedded in a four-dimensional space–time manifold. This implies continuous symmetries and conserved currents by i Noether’s theorem on that surface. We construct a three-form Ei ∧ DA (D represents covariant exterior a i derivative) in the phase space (Ei , Aa) on the surface Σ, and derive an equation of continuity on that surface, and search for canonical relations and a Lagrangian that correspond to the same equation of 0a i continuity according to the canonical field theory. We find that Σi is a conjugate momentum of Aa and ab i i Σi Fab is its energy density. We show that there is conserved spin current that couples to A , and show µνi i i i that we have to include the term Fµνi F in GR. Lagrangian, where F = DA , and A is complex SO(3) µνi i connection. The term Fµνi F includes one variable, A , similar to Yang–Mills gauge theory. Finally we couple the connection Ai to a left-handed spinor field ψ, and find the corresponding beta function.

Keywords: gravitational Lagrangian; complex connection Ai; left-handed spinor ﬁeld

1. Introduction Gravity can be formulated based on gauge theory by gauging the Lorentz group SO(3, 1) [1]. For this purpose, we need to fix some base space and consider that the Lorentz group SO(3, 1) acts locally on Lorentz frames which are regarded as a frame bundle over a fixed base space. We can consider this base space as an arbitrary space–time manifold M with coordinates xµ, and consider the local Lorentz frame as an element in the tangent frame bundle over M. By that we have two symmetries; invariance under continuous transformations of local Lorentz frame, SO(3, 1) group, and invariance under diffeomorphism of the space–time M, which is originally considered as a base space [2]. Since the group SO(3) is a subgroup of SO(3, 1), the Lagrangian of gravity has an internal gauge symmetry group SO(3). Thus the elements of SO(3) act locally on some spacial Lorentz orthonormal frames (e1, e2, e3), we consider these frames as a basis of the tangent vector bundle on the three-dimensional space-like hypersurface Σ. Using some local coordinate system σa, a = 1, 2, 3 on Σ, we expand these i i a i i j basis vectors into e = eadσ , so defining the gravitational field ea with metric gab = δijeaeb on Σ. Using i the isomorphism between Lie algebra of SU(2) and that of SO(3), one can regard ea also as a local su(2)-valued one-form. So we have an SU(2) vector bundle with real spin connection ωij [3]. These facts can be generated into self-dual and anti-self-dual formalism of general relativity (GR) with complex i connection A and complex conjugate one-form field Ei.

Physics 2019, 1, 339–359; doi:10.3390/physics1030025 www.mdpi.com/journal/physics Physics 2019, 1 340

This paper proceeds as follows: We start with the four-dimensional (4D) Palatini Lagrangian and perform a 3 + 1 decomposition based on the decomposition M → R × Σ, where R represents the time, and Σ is the three-dimensional (3D) space-like surface, thus we specify the Lagrangian part L1(gab) on Σ. Then we try to show that L1(gab) is independent of time on this surface, we try to prove this fact using the d fact that there are no dynamics in the space-like region of M. After that we use dt L1(gab) = 0 to get an equation of continuity on Σ, and search for canonical relations and Lagrangian that correspond to the same 0a equation of continuity according to the canonical field theory. We find that Σi is a conjugate momentum i ab i i of Aa and Σi Fab is its energy density. We obtain a Lagrangian for the connection A in 4-manifold M, then we couple it to a left-handed fermion field and find the beta function.

2. Decomposition Space–Time Manifold M into R × Σ The formulation of GR based on decomposition space–time manifold M into R × Σ is needed for expressing the metric of space–time as a solution of an equation for time evolution, such as in the Hamiltonian formulation. Thus the time evolution is the changing of the geometry of this surface. This decomposition preserves the continuous symmetries (gauge invariance and diffeomorphism invariance) of GR and its canonical quantization, so we can use it for the gauge theory of GR [4–7]. I I µ We define gravitational field as a one-form e = eµ(x)dx that is related with metric gµν(x) on an I J IJ 1 arbitrary space–time manifold M by gµν = ηIJeµeν, with spin connection ω (x) ∈ Ω (M, so(3, 1)), where so(3, 1) is Lie algebra of Lorentz group SO(3, 1). The spin connection defines covariant derivative Dµ that acts on all fields which have Lorentz indices (I, J, ...):

I I I J Dµv = ∂µv + ωµJv .

We start with the GR Lagrangian of the form

−1 µ ν IJ L(e, ω) = (16πG) eI eJ (Rµν) e, (1) where IJ IJ I KJ R = dω + ω K ∧ ω µ I µ I is the Riemannian curvature tensor and eI satisﬁes eµeJ = δJ . By the decomposition M → R × Σ, we decompose this Lagrangian into

−1 a b ij −1 a b 0J −1 0 a IJ L(e, ω) = (16πG) ei ej (Rab) e + (16πG) e0eJ (Rab) e + (16πG) eI eJ (R0a) e, (2) where i and j are Lorentz indices for I = i = 1, 2, 3, and a = 1, 2, 3. The part

−1 a b ij L1 = (16πG) ei ej (Rab) e has the gauge symmetry of the group SO(3), which is a subgroup of SO(3, 1), it also relates to the geometry a a a ij of the surface Σ(σ ) under the variation in the direction of Σ(σ ) subject to δe0 = 0, δω0 = 0, since it i j a depends only on the metric gab = δijeaeb which is defined on Σ(σ ), which is intrinsic geometry. ij k Since (Rab) is an anti-symmetric tensor, we can introduce a new one-form field Ec , the Hodge dual i j a i i a of e ∧ e in the internal spin space on the surface Σ(σ ); E = Eadσ , it is called the gravitational electric field [8,9] 1 eaeb − eaeb = εabcε Ek. 2 i j j i ijk c Physics 2019, 1 341

a In self-dual formalism of GR, Ei is complex given by [10,11]

1 Eia = εabcPi eI eJ, with Ei = Σi = Pi eI eJ (3) 2 IJ b c ab ab IJ a b

i where PIJ is a self-dual projector given by

1 i Pi = εi , for I = i, J = j, and Pi = −Pi = δi, for I = 0, J = j 6= 0. (4) IJ 2 jk 0j j0 2 j

1a 1 abc 2 3 0 1 For example E = 2 ε (ebec + iebec ).

a For the Lagrangian part L1 on Σ(σ ), we use the ﬁrst one:

−1 k abc ij L1 = (16πG) Ec ε εijk(Rab) e. (5)

The remaining part

−1 a b 0J −1 0 a IJ L2 = (16πG) e0eJ (Rab) e + (16πG) eI eJ (R0a) e, associates with the time evolution under the variation in the normal direction of Σ(σa), it is subject to a ij a δei = 0, δωa = 0, and changes the geometry of the surface Σ(σ ) during the time. Dynamics, such as 0 2 µ ν propagation gab(x) → gab(x ) in the time-like region (∆x) = gµν(x)∆x ∆x < 0, and determines how the surface Σ(σa) is embedded into the 4D manifold M, which is extrinsic geometry. But the surface Σ(σa) a is embedded in a space-like region in M; V1, V2 ∈ TpΣ(σ ), g(V1, V2) > 0, so there are no dynamics on a TpΣ(σ ). We can see this fact by noting that ∇tgab = 0 (the covariant derivative of the metric is zero), so

dL1(gab) ∂L1(gab) dt = ∇tgabdt = 0. dt ∂gab

a b a a a b We will rewrite ∇tgab = 0 as t t ∇tgab = 0, for t ∈ TpΣt(σ ). The formula t t ∇tgab = 0 is more ∗ a ∗ a a general than ∇tgab = 0 since there is ∇tgab ∈/ Γ (T Σt(σ ) × T Σt(σ )), and so its projection onto TpΣ(σ ) is zero. This case appears in the diffeomorphism maps of Σ(σa) into another space-like surface, as we will see. We can study the embedding of Σ(σa) by letting the time derivative of a position vector on its tangent a space be in the direction of the normal to this tangent space TpΣ(σ ). We can see this by considering a a a a a a a a position vector t ∈ TpΣt(σ ) that satisﬁes t˙ ta = t˙at = 0. Let t˙ ∈ NpΣt(σ ), where NpΣt(σ ) is the a a a a a normal space to Σt(σ ) = {t} × Σ(σ ), and t˙ = ∇tt is a covariant derivative of t with respect to the a b b b a a b a b time t ∈ R ⊂ R × Σ(σ ). From ta = gabt , we obtain t˙a = g˙abt + gabt˙ , so t t˙a = t g˙abt + t gabt˙ . Since a a a b a b a b t˙ ∈ NpΣt(σ ), we have t gabt˙ = 0, thus we get g˙abt t = t t ∇tgab = 0 which means that the points of a a b M do not expand nor contract covariantly in the space-like TpΣt(σ ), but t t ∂tgab 6= 0 is possible. Therefore, for a diffeomorphism map of Σ(σa) into another space-like surface (consider this as time evolution), it must be ∗ a ∗ a ∇t(gab) ∈/ Γ (T Σt(σ ) × T Σt(σ )) , and ∗ a ∗ a ∗ a ∗ a ∇t(gab) ∈ Γ(N Σt(σ ) × T Σt(σ )) ⊕ Γ(N Σt(σ ) × N Σt(σ )). Physics 2019, 1 342

0 a 0 This means for each x, x ∈ Σt(σ ), the covariant propagation gab(x) → gab(x ) ≡ gab(x) + µ ∗ a ∗ a ξ ∇µgab(x) does not occur in Tp Σt(σ ) × Tp Σt(σ ), but it occurs in

∗ a ∗ a ∗ a ∗ a Np Σt(σ ) × Tp Σt(σ ) ⊕ Np Σt(σ ) × Np Σt(σ ).

a ∗ a This is a changing in the embedding of Σ(σ ) in M. Let ta ∈ Tp Σt(σ ) satisfy (t˙a) = n, where a (·) is matrix notation and n is the normal to Σt(σ ). This normal is in the direction of the time, so 0 a a b it carries one index; n = n , thus (t˙a) = n0. It must also satisfy (t˙ ) ∈/ TpΣt(σ ), so (gab)(t˙ ) = 0. b b b b Using this in (t˙a) = (g˙ab)(t ) + (gab)(t˙ ), we obtain (t˙a) = (g˙ab)(t ), so we get n0 = (g˙ab)0bt . Thus we 0 0 b 0 0 b get n0n = n (g˙ab)0bt , and by n0n = 1, we obtain n (g˙ab)0bt = 1. This formula is for determining ∗ a ∗ a (g˙ab) = ∇t(gab) ∈ Γ(N Σt(σ ) × T Σt(σ )). ∗ a ∗ a Let us suggest a formula for determining (g˙ab) ∈ Γ(N Σt(σ ) × T Σt(σ )) like

c c c a b c ∇tgab = fabg0c, fab = fba, t t fab = 0,

∗ a ∗ a a where g0c ∈ Γ(N Σt(σ ) × T Σt(σ )), and gµν = (g00, g0a, gab) is the full metric. Let us write fbc using a a a ˜ ˜ bc a0 a0 matrix notation f , its elements are ( f )bc = fbc. Let fa satisfy ( fa) ( f )bc = δa . Thus we get the inversion

˜ ab ˜ ab c0 c0 ( fc) ∇tgab = ( fc) ( f )abg0c0 = δc g0c0 = g0c, (6) this is obtaining the metric component g0a from ∇tgab; the changing of the metric gab with respect to the time. Thus if we ﬁx the metric component g00, like g00 = −1, we obtain the map gab(t) → gµν = ∗ a ∗ a ∗ a ∗ (−1, g0a, gab). So we have an immersion Γ(T Σ(σ )) → Γ(N Σt(σ ) × T Σt(σ )) ⊂ Γ(T M). In another words, for an immersion M(n) → M(n+1), and under some hypotheses, we can construct a metric on (n+1) n a b M using the metric on M . Our hypothesis here is t t ∇tgab = 0. a ˜ a b c a > a First we ﬁnd the matrices f then fa. The matrices f are symmetric and satisfy t t ( f )bc = t ( f )t = > a 0, where the vector t = (t1, t2, t3)p is the unit vector in TpΣt(σ )), we can write them in a simple form like     0 0 0 cos(θ2) 0 sin(θ2) 1   2   f =  0 cos(θ1) sin(θ1)  , f =  0 0 0  , 0 sin(θ1) − cos(θ1) sin(θ2) 0 − cos(θ2)

  cos(θ3) sin(θ3) 0 3   f =  sin(θ3) − cos(θ3) 0  , 0 0 0

a b c the angles θ1, θ2 and θ3 can be determined to satisfy t t ( f )ab = 0, thus we obtain

2 2 2 2 2 2 t3 − t2 t3 − t1 t2 − t1 tan(θ1) = , tan(θ2) = , tan(θ3) = . 2t3t2 2t3t1 2t2t1 Physics 2019, 1 343

Therefore the matrices f˜a can be written in the form     0 c1 cos(θ3) 0 −2c1 sin(θ3) c1 cos(θ3) cos(θ3) ˜   ˜   f1 =  c1 cos(θ3) 2c1 sin(θ3) sin(θ1)  , f2 =  c1 cos(θ3) 0 c2 cos(θ1)  0 sin(θ1) 0 cos(θ3) c2 cos(θ1) 2c2 sin(θ1)

  0 cos(θ3) 0 ˜   f3 =  cos(θ3) −2c3 sin(θ1) c3 cos(θ1)  , 0 c3 cos(θ1) 0 the constants c1, c2 and c3 are determined to satisfy

˜ ab 1 ˜ ab 2 ˜ ab 2 ( f1) ( f )ab = ( f2) ( f )ab = ( f3) ( f )ab = 1.

a We can consider that as a continuous changing in the embedding of Σt(σ ) in M;which is a diffeomorphism map. Since the Lagrangian L1 depends only on gab while the term L2 depends on (g00, g0a, gab), thus by the previous discussing we have a map L1(gab) → L2(g00, g0a, gab) (time evolution), ˜ bc with g00 = −1, g0a = ( fa) ∇tgbc. As illustrated in the following Figure1.

Figure 1. Propagation of space-like surface.

0 Let ` ⊂ Σt1 which propagated to ` ⊂ Σt1

Figure 2. Changing distance between two space-like points after propagation.

p0 Z 2 02 2 a ` − ` = (−dtdt + g0adtdσ ). 0 p1 02 2 a a The limit ` − ` = 0 corresponds to dt = 0 and dσ = 0, not to −dtdt + g0adtdσ = 0 (with dt 6= 0 a a a and dσ 6= 0). For example, if two particles P1 and P2 exchange photons with wavelength λ = ` ≈ σ2 − σ1 0 in t = t1, then in t = t2 they measure a different wavelength, namely λ + ∆λ = ` . The difference is given 2 2 a by 2λ∆λ = −λ + g0aλ , where we set dt = dσ = λ. Thus the two particles P1 and P2 measure after a time dt = λ the difference 2∆λ/λ = −1 + g0a, where g0a is given in Equation (6). Thus we study the embedding of 3D surface Σt and its changing in 4D manifold M using the 3D metric gab on Σt and its derivative with respect to the time. ∗ a The Lagrangian L1(gab) is a function on the space-like space Tp Σt(σ ), therefore, it is independent of d ∗ a time, dt L1(gab) = 0 on Tp Σt(σ ). We can see this by using the fact

∗ a ∗ a ∗ a ∗ a ∇t(gab) ∈ Γ(N Σt(σ ) × T Σt(σ )) ⊕ Γ(N Σt(σ ) × N Σt(σ )), we have d ∂L1(gab) L1(gab(t + dt)) − L1(gab(t)) = L1(gab(t))dt = ∇t(gab)dt, dt ∂gab ∗ a ∗ a ∗ a ∗ a ∗ a which is a function on Np Σt(σ ) × Tp Σt(σ ) ⊕ Np Σt(σ ) × Np Σt(σ ), not on Tp Σt(σ ); which means there are no dynamics on the space-like region (∆x)2 > 0.

a 4 ∗ Let us write dt∂t L1(gab) = (Σ(σ ), dθ), where (Σ, V) is a projection of V ∈ ∧ Tp M onto a surface Σ, the inner product of V with tangent basis in TpΣ, deﬁned below in Equation (17), and θ is three-form in ij a the phase space (Ei, ω ) on Σ(σ ). Thus we write

d d Σ(σa), dt ∧ θ = dt L (g ) = 0. dt dt 1 ab

If we write θ as a −1/2 k ij θ(E, ω, Σt(σ )) = (16πG) εijkE ∧ R , (7) ij ij a b i i a with R = (Rab) dσ ∧ dσ and E = Eadσ , we obtain

d d Σ(σa), dt ∧ θ = (16πG)−1/2dt (Ekεabcε (R )ije) dt dt c ijk ab d = (16πG)1/2dt L = 0. dt 1 Physics 2019, 1 345

a a ∂ Since θ is three-form on Σ(σ ), so dσ ∧ ∂σa θ = 0, if we add it to the last formula, we get

∂ ∂ Σ(σa), dt ∧ θ + dσa ∧ θ = (Σ(σa), dθ) = 0. ∂t ∂σa

a µ This relates the Lagrangian L1 and θ with this surface. Under arbitrary transformation (t, σ ) → x , a ∂σa µ the two Lagrangian parts L1 and L2 will mix. Thus dσ = ∂xµ dx and the basis on Σ transforms as

∂xµ ∂xν ∂xρ ∂a ∧ ∂ ∧ ∂c → ∂µ ∧ ∂ν ∧ ∂ρ. (8) b ∂σa ∂σb ∂σc Therefore, the components of three-form θ transforms as

∂σa ∂σb ∂σc θ → θ = θ . abc µνρ abc ∂xµ ∂xν ∂xρ

To keep the invariance under this transformation, that is (Σ, dθ) = 0 still holds, we write the three-form ij θ in the phase space (Ei, ω ) on M, and let (Σ, dθ) be its projection onto Σ. Therefore we write

−1/2 k ij θ(E, ω) = (16πG) εijkE ∧ R , (9)

ij thus we get three-form θ in the phase space (Ei, ω ) on M, it has internal SO(3) symmetry. Its projection a onto Σt(σ ) is a −1/2 k abc ij (Σt(σ ), θ(E, ω)) = (16πG) Ec ε εijkRab. We write θ using self-dual formalism (Plebanski formalism) [11] in which the connection Ai is a three-complex one-form given by IJ i i IJ ω → A = P IJω , (10) i where PIJ is self-dual projector given by Equation (3). The curvature which associates with this connection is i i i j k F = dA + ε jk A ∧ A . a On the surface Σt(σ ), it is

1 j Fi = ∂ Ai − ∂ Ai + εi A Ak . ab 2 b b b a jk a b Also [a b] IJ [a b] ab abc eI eJ → Pi eI eJ = Ei = ε Eic. (11) Thus we have self-dual plus anti-self-dual projection:

a b IJ abc i abc ¯ ¯i eI eJ R(ω)ab → ε EiaFbc + ε EiaFbc, (12)

¯ ¯i i where Eia and Fbc are the Hermitian conjugate of Eia and Fbc. This projection relates to the decomposition of Lie algebra of the Lorentz group SO(1, 3) into two copies of Lie algebra of SL(2, R) [12].

We write the components of the curvature as

1 1 j Fi = D Ai − D Ai = ∂ Ai − ∂ Ai + εi A Ak , (13) ab 2 a b b a 2 a b b a jk a b Physics 2019, 1 346 where 1 j D Ai = ∂ Ai + εi A Ak, a b a b 2 jk a b which motivates introducing a notation of the covariant derivative like [7]

1 DVi = dVi + εi Aj ∧ Vk, 2 jk or 1 j i j j D Vi = ∂ Vi + εi A Vk = ∂ Vi − A (T )ikVk, µ ν µ ν 2 jk µ ν µ ν 2 µ A ν so i j j D = ∂ − A (T ), µ µ 2 µ A j ik jik j where the matrix elements (TA) = −iε are the elements of the generators TA in the adjoint representation of the group SU(2) [13]. The coupling constant here is g = 1. In general we write this covariant derivative as i j j D = ∂ − gA (T ). (14) µ µ 2 µ A a i Using the projection in Equation (12), we rewrite θ in the complex phase space (Ei , Aa) as

a −1/2 abc i 3 θ(E, ω, Σ(σ )) = (16πG) ε EicFab(A)ed σ, where we take in consideration only the ﬁrst part, the second is obtained by taking the Hermitian conjugate. We can write it as three-form on M as done in Equation (9), we obtain

−1/2 i θ(E, A) = (16πG) Ei ∧ DA . (15)

a Its projection onto Σt(σ ) is

a −1/2 abc i (Σt(σ ), θ) = (16πG) ε EiaFbc(A),

i where A is complex SO(3) connection, and Ei is a complex one-form as deﬁned in Equation (11).

a 3. Equation of Continuity on the Hypersurface Σt(σ ) d a We have showed that the Lagrangian L1(gab) is independent of time, dt L1(gab) = 0 on Σ(σ ) since ∗ a ∗ a ∇t(gab) ∈/ Γ(T Σt(σ ) × T Σt(σ )). This relates to the fact that there are no dynamics in the space-like a a TΣt(σ )); the points of M do not expand nor contract covariantly in this region, ∇µt 6= 0. Note that a a although ∇µt = 0, but it may be ∂µt 6= 0. We had

∗ a ∗ a ∗ a ∗ a ∇t(gab) ∈ Γ(N Σt(σ ) × T Σt(σ )) ⊕ Γ(N Σt(σ ) × N Σt(σ )), which shows that the two parts of the Lagrangian L1 and L2 mix by time evolution. Then we wrote d a ij dt dt L1(gab) = 0 as (Σ(σ ), dθ) = 0, where θ is three-form in the phase space (Ei, ω ) on M (Equation (7)), a its projection onto Σt(σ ) is Equation (17)

a −1/2 abc i jk (Σt(σ ), θ(E, ω)) = (16πG) εijkε EaRbc(ω). Physics 2019, 1 347

−1/2 i In self-dual formalism, we obtained θ(E, A) = (16πG) Ei ∧ DA , with

a −1/2 abc i (Σt(σ ), θ(E, A)) = (16πG) ε EiaFbc(A).

Our condition (Σ(σa), dθ) = 0 makes sense here because of the decomposition R × Σ and ﬁxing a coordinate system σa on the hypersurface Σ, this yields to an equation of continuity on this surface. For a this purpose, we take the inner product of the four-form dθ with a tangent basis on the surface Σt(σ ) at an arbitrary point, we get a one-form co-vector (Σ(σa), dθ) in the direction of the normal to this surface at a a that point. Then we set (Σ(σ ), dθ) = 0, we obtain an equation of continuity on Σt(σ ).

Now taking the exterior derivative of Equation (15), we obtain

1/2 i i i (16πG) dθ = d Ei ∧ DA = (DEi) ∧ DA − Ei ∧ DDA , (16)

a abc where E0i = 0. The tri-tangent basic on Σt(σ ) is ∂a ∧ ∂b ∧ ∂c, we rewrite it as (1/3!)ε ∂a∂b∂c. The projection of dθ onto this basic is

1/2 abc i abc i abc (16πG) dθ, ε ∂a∂b∂c = (DEi) ∧ DA , ε ∂a∂b∂c − Ei ∧ DDA , ε ∂a∂b∂c where (·, ·) is contraction pairing deﬁned by

µ ν ρ σ abc abc [µ ν ρ σ] Vµνρσdx ∧ dx ∧ dx ∧ dx , ε ∂a∂b∂c = ε Vµνρσdx δc δb δa , (17) where the bracket [....] is anti-symmetrization of the indices. Although DDAi is zero in 4D manifold M, i abc but the contraction pairing of Ei ∧ DDA with 3D basis ε ∂a∂b∂c is not zero as we will see, since we do not sum over the time index µ = 0 as we sum over the spatial indices µ = 1, 2, 3, because we regard dx0 as a normal to the surface Σt(σ ). µ For cotangent basis {dx } and tangent basis {∂a}, this pairing can be deﬁned simply by using inner product like [14] µ µ (dx , ∂a) = δa , a a in which we consider dx = dσ for a = 1, 2, 3, so E0i = 0 regarding to our gauge. Starting with the ﬁrst term

i abc abc i µ ν ρ σ (DEi) ∧ DA , ε ∂a∂b∂c = ε DµEνiDρ Aσ (dx ∧ dx ∧ dx ∧ dx , ∂a∂b∂c) we get

i abc 4 (DEi) ∧ DA , ε ∂a∂b∂c = abc i µ abc i µ abc i µ abc i µ − ε DaEbiDc Aµdx + ε DaEbiDµ Acdx − ε DaEµiDb Acdx + ε DµEaiDb Acdx .

The second term is

i abc abc i µ ν ρ σ Ei ∧ DDA , ε ∂a∂b∂c = ε EµiDνDρ Aσ (dx ∧ dx ∧ dx ∧ dx , ∂a∂b∂c) . Physics 2019, 1 348

Doing the same thing, we get

i abc 4 Ei ∧ DDA , ε ∂a∂b∂c = abc i µ abc i µ abc i µ abc i µ − ε EaiDbDc Aµdx + ε EaiDbDµ Acdx − ε EaiDµDb Acdx + ε EµiDaDb Acdx .

Adding the two terms, we obtain

1/2 abc 4(16πG) dθ, ε ∂a∂b∂c = abc i µ abc i µ abc i µ abc i µ −ε DaEbiDc Aµdx + ε DaEbiDµ Acdx − ε DaEµiDb Acdx + ε DµEaiDb Acdx abc i µ abc i µ abc i µ abc i µ +ε EaiDbDc Aµdx − ε EaiDbDµ Acdx + ε EaiDµDb Acdx − ε EµiDaDb Acdx .

We deﬁne the curvature by using the covariant derivative from Equation (13) as

1 1 j Fi = D Ai − D Ai = ∂ Ai − ∂ Ai + gεi A Ak , (18) µc 2 µ c c µ 2 µ c c µ jk µ c therefore 1 Fρνi = gρµgνcFi = Dρ Aνi − Dν Aρi ; Dρgµν = 0. µc 2 Its Hodge dual on the surface Σ with respect to the coordinates (σa) is

ai abc i F = ε Fbc.

Also we deﬁne the complex two-form ﬁeld from Equation (3) as

bc bc bca abc Σi = Ei = ε Eai = ε Eai.

Using them in the last formula, we get

1/2 abc abc i µ ai µ 4(16πG) dθ, ε ∂a∂b∂c = 2ε (DaEbi)Fµcdx − 2(DaEµi)F dx ai µ bc i µ ai µ ai µ + 2(DµEai)F dx + 2Ei DbFcµdx + 2EaiDµF dx − 2EµiDaF dx .

And using abc i µ acb i µ ac i µ ε (DaEbi)Fµcdx = −ε (DaEbi)Fµcdx = −(DaEi )Fµcdx , we obtain

1/2 abc 2(16πG) dθ, ε ∂a∂b∂c = ac i µ ai µ ai µ bc i µ − (DaEi )Fµcdx − (DaEµi)F dx + (DµEai)F dx − Ei DbFµcdx ai µ ai µ +EaiDµF dx − EµiDaF dx .

With ac i µ bc i µ ac i µ −(DaEi )Fµcdx − Ei DbFµcdx = −Da Ei Fµcdx , it becomes

1/2 abc ac i µ ai µ ai µ 2(16πG) dθ, ε ∂a∂b∂c = −Da(Ei Fµc)dx − Da(Eµi F )dx + Dµ(Eai F )dx . Physics 2019, 1 349

a 0 As we suggested before, we let the normal of the surface Σt(σ ) be in direction of the time dx , so abc dθ, ε ∂a∂b∂c is in direction of the time. Therefore we set µ = 0, thus we get

1/2 abc ac i 0 ai 0 ai 0 2(16πG) dθ, ε ∂a∂b∂c = −Da(Ei F0c)dx − 2Da(E0i F )dx + D0(Eai F )dx .

abc a The vector dθ, ε ∂a∂b∂c is one-form in the direction of the normal to the surface Σt(σ ). It is zero as we mentioned before, thus we get

ac i 0 ai 0 ai 0 −Da(Ei F0c)dx − Da(E0i F )dx + D0(Eai F )dx = 0, or ac i ai ai −Da(Ei F0c) − Da(E0i F ) + D0(Eai F ) = 0. We write it as ac i ai ai Da(Ei Fc0) − Da(E0i F ) + D0(Eai F ) = 0. ai ai ai ab i The term (Eai F ) is scalar, so D0(Eai F ) = ∂0(Eai F ), the vector Ei Fb0 is a usual vector ﬁeld on Σ, it ab i ab i does not carry a Lorentz index, so Da(Ei Fb0) = ∂a(Ei Fb0) and E0i = 0, thus we get

ab i ai ∂a(Ei Fb0) + ∂0(Eai F ) = 0,

ai 1 abi 1 abi using Eai F = 2 Eabi F = 2 Σabi F

1 ∂ ΣabFi + ∂ Σ Fabi = 0. (19) a i b0 0 2 abi

This equation shows that there is a relation between Σi and Fi in the space (Σi, Fi) on 3 + 1 manifold i i j i j R × Σ. Usually this relation is written as F = ψ jΣ + ψ¯ jΣ¯ . We can find that relation by regarding this equation as an equation of continuity with respect to a Lagrangian like L(F0ai, DAi) that satisfies the action principle δS(DAi) = 0 and the invariance under continuous symmetries of GR. Therefore we 1 abi 00 ab i 0a regard 2 cΣabi F as energy density T , and cΣi Fb0 as momentum density T , where c is constant for satisfying the units. Then we search for a suitable Lagrangian and Hamiltonian with canonical relations that correspond to the same continuity equation according to the quantum fields theory, we do this at the flat-space–time limit and generalize it to an arbitrary curved space–time. In scalar field φ theory, the Lagrangian is [15]

L(φ, ∂φ) = π∂0φ − H(π, φ),

0 the conjugate momentum is π = ∂0φ = −∂ φ. The conserved momentum–energy tensor is

00 0a a 0 a a T = H(π, φ) = π∂0φ − L(φ, ∂µφ) and T = P = ∂ φ∂ φ = −π∂ φ.

In the ﬂat limit of the space–time, our momentum–energy tensor is

0a a 0bi ab i ab i T = cΣ bi F = −cΣi F0b = cΣi Fb0, and 1 1 ij T00 = cΣabFi = ceaebF , (20) 2 i ab 4 i j ab Physics 2019, 1 350 comparing it with the momentum ∂0φ∂aφ = −π∂aφ, we conclude that our conjugate momentum is πbi ∼ −F0bi [16–19]. By considering a Lagrangian of the form L(F0ai, DAi) with corresponding Hamiltonian like i i H(πai, DA ) and using the action principle δS(DA ) = 0 and the diffeomorphism invariance on the a surface Σt(σ ), we obtain the momentum–energy tensor like [16–18]

00 0ai 0ai i 0a abi T = πai F − L(F , DA ) and T = cπbi F .

00 1 abi 0a a 0bi Comparing them with our momentum–energy tensor T = 2 cΣabi F and T = cΣ bi F , we conclude 0a abi a 0bi T = πbi F = cΣ bi F . (21) i In general, the curvature Fµν can be written as [20,21]

i i j 0i ¯ j Fµν = ψ jΣµν + ψ jΣµν, (22)

µν i i µν j j µν ¯ j so Σi Fµν = ψ i since Σi Σµν = δi and Σi Σµν = 0. Using Equation (22) in Equation (21), we get

0a i abj 0i abj a 0bi T = πbiψ jΣ + πbiψ jΣ¯ = cΣ bi F ,

0i i i j i −1 i j therefore we set ψ j = 0, so Fµν = ψ jΣµν, and Σµν = (ψ ) jFµν, we obtain

i abj abi bi −1 i 0bj 0bi πbiψ jΣ = −cΣ F0bi so π = c(ψ ) jF = cΣ , (23)

0bi i 00 which means that Σ is conjugate momentum of Aa. Using it in the Hamiltonian H = T :

0ai 0ai i 0ai 0ai i H = πai F − L(F , DA ) = −cΣ0ai F − L(F , DA ),

1 abi then using our energy density 2 cΣabi F , we get the Lagrangian

1 L(E, F0ai, DAi) = π F0ai − H = −cΣ F0ai − cΣ Fabi. (24) ai 0ai 2 abi Therefore 1 1 L(E, F) = − c(Σ F0ai + Σ Fa0i) − cΣ Fabi, 2 0ai 0ai 2 abi so 1 1 p L(E, A) = − cΣ Fµνi, or L(E, A) = − cΣ Fµνi −gd4x, 2 µνi 2 µνi where Σi = Ei and Fi are deﬁned in Equations (3) and (18). This Lagrangian has a symmetry of the complex group SO(3) and the self-dual of Lorentz group SO(3, 1). The contraction is deﬁned by using the metric I J gµν = ηIJeµeν. This Lagrangian corresponds to self-dual part of Equation (12). To get the total Lagrangian, we add the Hermitian conjugate, we obtain Ldual GR(E, A, E¯, A¯) = LLe f t GR(E, A) + LRight GR(E¯, A¯):

−c µν −c µν L (Σ, A, Σ¯ , A¯) = Σ Fi e + Σ¯ F¯i e, (25) dual GR 2 i µν 2 i µν ¯i i i IJ the curvature F is the Hermitian conjugate F = PIJ R (10). The Lorentz group is SO(3, 1), its Lie algebra is reducible and can be decomposed into two copies of the Lie algebra of SU(2): Physics 2019, 1 351

∼ SO(3, 1, C) = SO(3, C)Le f t × SO(3, C)Right.

i i IJ The complex connection A = PIJω takes values in Lie algebra of SO(3, C)Le f t, while its Hermitian ¯ i ¯i IJ conjugate A = PIJ R takes values in Lie algebra of SO(3, C)Right. Thus the Lorentz invariance is satisﬁed by the uni-variance under the two groups SL(2, R)Le f t and SL(2, R)Right [11,12].

To determine the constant c, we write this Lagrangian in the form

−1 µ ν IJ (16πG) eI eJ (Rµν) e,

i by using properties of the projection PIJ

1 PIJ Pi + P¯ IJ P¯i = (δI δJ − δI δJ ), and Pi P¯ IJ = 0, (26) i KL i KL 2 K L L K IJ k

µ ν IJ −1 the Lagrangian (Equation (25)) becomes (−c/2)eI eJ (Rµν) e, thus −c/2 = (16πG) . Therefore

1 µν 1 µν L (A, Σ, Σ¯ , A¯) = Σ Fi e + Σ¯ F¯i e, (27) dual GR 16πG i µν 16πG i µν this Lagrangian is similar to the Plebanisky Lagrangian, but it is not multiplied by the imaginary number i and does not include the cosmological constant term.

4. Yang–Mills Theory of Gravity By regarding the local Lorentz symmetry as a gauge symmetry with spin connection ωIJ (or Ai) as gauge ﬁelds, we recognize Yang–Mills theory in gravity. But not full gravity, since in the Yang–Mills theory, the variables are connections and conserved currents, while in the gravity the metric is also variable. The local Lorentz symmetry generates locally conserved currents, and those currents are coupled to spin connection ωIJ. This makes the local Lorentz symmetry a gauge symmetry with the Lorentz group as a gauge group. Also, these currents must be conserved and vanish in the vacuum. i i j i j i i j i j From the formula F = ψ jΣ + ψ¯ jΣ¯ , we can get the inversion Σ = ξ jF + ξ¯ jF¯ by inserting it back, we obtain i j ¯i ¯j j i ¯j ¯i ¯j ψ jξ k + ψ jξ k = δk and ψ jξ k + ψ jξ k = 0. i i j i j We can get the equation of motion Σ = ξ jF + ξ¯ jF¯ from this Lagrangian (Equation (27)) by adding terms like ¯ ¯ µν i µνi j 0 ¯ µνi j 16πGLdual GR(A, Σ, Σ, A) = Σi Fµνe − ψijΣ Σµνe − ψijΣ Σµνe + C.C, µν i i i j ¯i ¯ j i therefore the δL/Σi = 0 and δL/δFµν = 0 yields F = ψ jΣ + ψ jΣ and DΣ = 0. But by using properties µνi j µνi j ij of the self-dual projection from Equation (26), we obtain Σ¯ Σµν = 0 and Σ Σµν = δ , but this does not change the equations of motion. i But as we will see, if there is a Lorentz current, like the spin current of the spinor ﬁeld, then δL/δFµν 6= i I µνi 0, therefore to keep DΣ = 0, and to also keep De = 0, we add a term like Fµνi F e. This is done in order i i µνi j to insert back Σ ∼ F in ψijΣ Σµν into the Lagrangian. Therefore we write

¯ ¯ −1 µν i µνi j µν i Ldual GR(E, Σ, Σ, A) = (16πG) (Σi Fµνe − ψijΣ Σµν)e + kFi Fµνe + ... + C.C, where k is a constant can relate to a coupling constant of Lorentz current with the spin connection Ai. Physics 2019, 1 352

µνi i i This Lagrangian includes F Fµνe, F is given in Equation (18), similarly to Lagrangian in Yang–Mills theory with gauge group SO(3), and the self-dual of Lorentz group SO(3, 1). It depends only on the i µνi i connection A , thus it describes the changing of the local Lorentz basis. Therefore F Fµνe reads the invariance only under the local Lorentz transformations. Also, it is a topological invariance, so it allows the free propagation of spin connection Ai. But by relating Ai with the triad eI, and relating eI with the I J i metric gµν by gµν = ηIJeµeν, this breaks the free propagation of spin connection A as free waves, except in the background approximation of the metric, the result is gravitational waves. Similarly to the free electromagnetic field. By using the properties of self-dual projection, this Lagrangian can be written using the Riemannian tensor R(ω) as 1 L = Re + kR2e + .... 16πG µνi i We find the role of the term F Fµνe by including the interaction of mass-less spinor particles with gravity. Since they are massless, their energies are small so that their interaction with the gravitational field eI is weak, but their interaction with spin connection takes place. The interaction term is

IJ µ ¯ K IJ µ ωµ eKψγ SIJψe = ωµ JIJe, (28) µ µ K µ ¯ K where JIJ = eK JIJ = eKψγ SIJψ is Lorentz current. If we add this term to the Lagrangian

−1 µ ν IJ (16πG) eI eJ (Rµν) e, we get −1 µ ν IJ IJ µ (16πG) eI eJ (Rµν) e + ωµ JIJe. IJ So the equation of motion for ωµ is

−( )−1 ( µ ν ) + µ = 16πG Dµ e[I eJ]e JIJ 0.

In self-dual formalism, this equation becomes

−1 µνi νi −(16πG) DµΣ + J = 0.

These two equations say that the Lorentz current is the source for the gravitational ﬁeld eI, but this is not right since the energy is the source for it, also we choose DeI = 0. Thus, these equations do not hold. But if we use the Lagrangian

−1 µν i µν i i µ L = (16πG) Σi Fµνe + kFi Fµνe + Aµ Ji e + C.C, (29) the equation of motion for Ai becomes

1 −(16πG)−1D (Σµνie) − kD Fµνie + Jνi = 0. µ 2 µ

µνi Thus we choose Dµ(Σ e) = 0 and

1 − kD Fµνi + Jνi = 0. (30) 2 µ Physics 2019, 1 353

µνi i ν ¯ I i The term DµF includes only the spin connection A , so the Lorentz current eI ψγ S ψ contributes as a source for the spin connection Ai, and so interacts with it. This relate to the fact that we can regard the local Lorentz symmetry as a gauge group with spin connection ωIJ (or Ai) as gauge ﬁelds. The Lorentz current is conserved since

1 1 j D D Fµνi = [D , D ]Fµνi = − εi F Fµνk = 0 → D Jνi = 0. ν µ 2 ν µ 2 jk µν ν

i Furthermore, Equation (30) allows us to calculate a Lorentz current for a given curvature Fµν, although IJ i the curvatures Rµν and Fµν are calculated using only the energy–momentum tensor. Although there are Lorentz currents associated with matter, those currents relate to the local Lorentz symmetry. µ We need to prove ∇ Rµνρσ = 0 in the vacuum, where Rµνρσ is Riemann curvature tensor, it satisﬁes Rµνρσ = −Rνµρσ, Rµνρσ = −Rµνσρ and Rµνρσ = Rρσµν. In the vacuum, we have the equality

Rµν = constant × Rgµν,

µν where Rµν is a Ricci tensor and R = g Rµν. This equality is equivalent to another equality [11]:

∗ ∗ Rµν = constant × Rgµν ⇔ Rµνρσ = R µνρσ, with Hodge operator ∗ 1 µ0ν0 ∗ 1 ρ0σ0 R = e R 0 0 , R = R 0 0 e , µνρσ 2 µν µ ν ρσ µνρσ 2 µνρ σ ρσ µνρσ the anti-symmetric tensor e is the volume four-form for metric gµν. Therefore in the vacuum, we have

µ0ν0 ρ0σ0 µ0ν0 ρ0σ0 eµν Rµ0ν0ρσ = Rµνρ0σ0 e ρσ, so eµνµ0ν0 R ρσ = Rµν eρ0σ0ρσ,

µνρσ acting by ∇γ on both sides, with ∇γe = 0 we get

µ0ν0 ρ0σ0 eµνµ0ν0 ∇γR ρσ = ∇γRµν eρ0σ0ρσ.

The summing here is over µ0, ν0, ρ0 and σ0, while µ, ν, ρ, σ and γ are ﬁxed. Then multiplying both sides by eγµνα: γµνα µ0ν0 γµνα ρ0σ0 e eµνµ0ν0 ∇γR ρσ = e ∇γRµν eρ0σ0ρσ.

γµνα ρ0σ0 We note that by summing over γ, µ, ν in e ∇γRµν for each ﬁxed ρ, σ, α, we obtain the Bianchi γµνα ρ0σ0 identity e ∇γRµν = 0, therefore

γµνα µ0ν0 e eµνµ0ν0 ∇γR ρσ = 0.

It becomes

0 0 ( α γ − γ α )∇ µ ν = → ∇ αγ − ∇ γα = − ∇ γα = 2 δµ0 δν0 δµ0 δν0 γR ρσ 0 2 γR ρσ 2 γR ρσ 4 γR ρσ 0,

µ IJ µ IJ Iρ Jσ µ so ∇ Rµνρσ = 0 in the vacuum. Therefore the Lorentz current Jν = ∇ Rµν = e e ∇ Rµνρσ vanish in the vacuum. Using the self-dual projection, we ﬁnd

i µ i i µ IJ Jν = ∇ Fµν = PIJ∇ Rµν Physics 2019, 1 354 also vanish in the vacuum. Therefore this current associates only with matter. We can also prove this by i i j i j i using the formula F = ψ jΣ + ψ¯ jΣ¯ , and by setting ψ¯ j = 0 in the vacuum [11]. Using

i ν i (∗d ∗ F )µ = ∇ Fνµ,

i i j i j with ∗F = ψ j(∗Σ ) + ψ¯ j(∗Σ¯ ), and

∗ Σj = −iΣj, ∗ Σ¯ j = iΣ¯ j, (31)

Hodge operator here is with respect to the metric gµν, we get

ν i i j i ¯ j ∇ Fνµ = i(∗(−dψ jΣ + dψ¯ jΣ ))µ.

The ﬁrst term becomes

i j i j i j i j i j i i j d(ψ jΣ ) = d(ψ jΣ + ψ¯ jΣ¯ ) − d(ψ¯ jΣ¯ ), so d(ψ jΣ ) = dF − d(ψ¯ jΣ¯ ),

i i j i j and by Bianchi identity dF = 0, we obtain d(ψ jΣ ) = −d(ψ¯ jΣ¯ ). Therefore

ν i i ¯ j i ¯ j i ¯ j i j ∇ Fνµ = i(∗(d(ψ¯ jΣ ) + dψ¯ jΣ ))µ = 2i(∗d(ψ¯ jΣ ))µ = −2i(∗d(ψ jΣ ))µ. (32)

i ν i Since ψ¯ j = 0 in the vacuum, we get ∇ Fνµ = 0.

i Therefore for a Lorentz Jµ current that associates with matter, we get

ν i i ¯ j i i ¯ j i ∇ Fνµ = 2i(∗d(ψ¯ j)Σ )µ = Jµ, so − 2idψ¯ j ∧ Σ = ∗J ,

¯ j i i µ Where we used dΣ = 0 and J = Jµdx , with property of Hodge dual twice operation on p-form V in n dimensions: ∗ ∗ V = (−1)pq+tV, where q = n − p, and t is the number of negative eigenvalues of the metric tensor [14]. In our case we have n = 4, p = 3, t = 1.

i Same thing we get for ψ j, (Equation (32)):

ν i i j i i j i ∇ Fνµ = −2i(∗(dψ j ∧ Σ ))µ = Jµ, so − 2idψ j ∧ Σ = ∗J .

Therefore i j µ ν ρ iσ µ ν ρ −2i(Dµψ j)Σνρdx ∧ dx ∧ dx = J eµνρσdx ∧ dx ∧ dx /3!, so i j µνρσ iσ 2i(Dµψ j)Σνρe = J . But j νρµσ j µσ j µσ Σνρe /3! = (∗Σ ) = (−iΣ ) , (we used self-dual properties Equation (31)), we obtain

Jiν (D ψi )Σjµν = . (33) µ j 2 × 3! Physics 2019, 1 355

i jµν I IJ iν The quantity 2(Dµψ j)Σ depends on the triads e , and on spin connection ω , while J relates matter. Therefore by solving this equation for a given Lorentz current Jiσ, we get solutions for eI and IJ i ω . But this formula reads only the contributing of Lorentz current in ψ j, there is a contributing of cosmological constant and matter in symmetric part of ψ; tr(ψ) = −Λ − 2πGT, where T is trace of energy-momentum tensor Tµν [12,22]. Since Lorentz current generates local Lorentz transformation, we i iσ expect that Dµψ j in Equation (33) is anti-symmetric, this distinguishes contributing of J from those of cosmological constant and matter. So we write

ki jµν iν 2(Dµψk)ε jΣ = J , (34)

iν i ki i thus J contributes in anti-symmetric part of ψ j := ψkε j, where ψ is vector ﬁeld in local Lorentz frame.

i i IK IK IK L Let us write Dµψ = PIK Jµ , with Lorentz current Jµ = JL eµ deﬁned in Equation (28). The IK iν i Iν i i IK relationship (linear) between Jµ and J = JI e can be determined by inserting Dµψ = PIK Jµ in Equation (34). It is easier to solve i i IK Dµψ = PIK Jµ IK in region away from matter where Jµ = 0, so we can solve it in background approximation, g ≈ η, thus ψi → (ψr, ψθ, ψφ) in spherically coordinates. If we assume that the vector ﬁeld ψr depends only on the radius r, we get 1 ar ∂ (r2ψr(r)) = 0 → ψr(r) = , r2 r r2 r iν i where a is constant vector. Therefore the contributing of Lorentz current J in the curvature Fµν is

1 j Fi := a εki Σ , µν r2 k j µν therefore the contributing of Jiν in F2 is

µν 2 µν j j Fi F := a2, with Σ Σ = δ . µν i r4 i µν i √ H i = 2 = H ~ · ~ = This formula satisﬁes S Fµν dS 4π 2a constant, it is similar to electric ﬁeld S E dS Q/ε0. This is similarity between GR and Yang-Mills theory.

The Lorentz current is not associated only with spinor particles, the Lorentz symmetry for arbitrary ﬁeld produces a global conserved Lorentz current like ([15], section 22)

MIJK = xJ TIK − xKTIJ.

TIK is energy-momentum tensor in ﬂat space–time. Locally we write this as

µJK ν J µK K µJ M = a (eνT − eν T ), where aν is constant, therefore

µJK J µK µK K µν ∇µ M = 0; ∇µeν = 0, ∇µT = 0, T = eν T . Physics 2019, 1 356

i i The existence of a conserved spin current Jµ that couples to A lets us to believe in local Lorentz symmetry as a gauge group with spin connection ωIJ (or Ai) as gauge ﬁelds similarly to Yang–Mills theory.

5. Beta Function We assume that the interaction of mass-less particles with the gravity is dominated by interaction i µνi of their spin current (Lorentz current) with the connection Aµ. We can use the Lagrangian Fµνi F to i describe the interaction of left-handed fermions with the connection Aµ. We choose a representation of SL(2, C) in which we have Ki = iJi, where Ki are boost generators and Ji are rotation generators [23]. Our i i i IJ connection A is a one-form complex given by the self-dual projection Aµ = PIJωµ of the spin connection IJ ωµ according to the decomposition so(3, 1 : C) = so(3 : C) ⊕ so(3 : C) [24]. Therefore

Ai Ji = Re(Ai)Ji + iIm(Ai)Ji = Re(Ai)Ji + Im(Ai)Ki, or Ai Ji + A0iKi, where Ai and A0i are real.

For simplicity let us choose A0i = γAi, with a constant γ ∈ R. Thus the connection becomes Ai Ji + A0iKi = Ai Ji + iγAi Ji = Ai Ji + iγJi = AiTi ∈ Ω1(M, sl(2, C)), with new generators Ti = Ji + iγJi = (1 + iγ) Ji ∈ sl(2, C). i 1 i µ + I i Therefore the coupling of the connection A with left-handed spinor ﬁeld is 2 gAµeI ψ σ¯ T ψe, where g is coupling constants comes from using the covariant derivative seen in Equation (14). So we write the fermion-gravity Lagrangian as

µ 1 1 µ L(A, ψ) = ie ψ+σ¯ I ∂ ψe + F Fµνie + gAi e ψ+σ¯ I Tiψe. I µ 4 µνi 2 µ I

I J Using the metric gµν = ηIJeµeν, we obtain

µνi µ0ν0i µνi I J µ0ν0i µνi Fµνi F = gµµ0 gνν0 F F = eµeIµ0 eνeJν0 F F .

i i i I J In background space–time, we have eµ(x) = δµ + hµ(x), so gµν(x) = ηIJδµδν + ..., where η is the Minkowski metric. Therefore the gravity and spinor Lagrangian approximates to

+ µ 1 µ0ν0i µνi 1 i + µ i L(A, ψ) = iψ σ¯ ∂ ψ + η 0 η 0 F F + gA ψ σ¯ T ψ + .... µ 4 µ µ ν ν 2 µ

i The remaining term includes the interaction with the ﬂuctuated gravitational ﬁeld hµ(x), this interaction relates with local invariance under diffeomorphism of M. Let us consider only the part

+ µ 1 µ0ν0i µνi 1 i + µ i L(A, ψ) = iψ σ¯ ∂ ψ + η 0 η 0 F F + gA ψ σ¯ T ψ, µ 4 µ µ ν ν 2 µ which is invariant under local Lorentz transformation, the compatible currents take values in sl(2, C), thus Ai is coupled to Lorentz currents. By that we have included only the invariance under local Lorentz transformation and excluded the local invariance under diffeomorphism of M. Thus we describe the GR i ∗ using Lorentz frames as vector bundle over basis space M with connection A ∈ Tp M × sl(2, C). 1 i It is similar to the Lagrangian of Yang–Mills theory for spinor ﬁeld, but with the generators 2 T = 1 i i 1 i 2 (1 + iγ) J , so to get results from the usual theory, we just replace the generators J with 2 (1 + iγ) J . Physics 2019, 1 357

For example, to get the beta function β(g) = ∂g/∂ln(M) for our Lagrangian, we use the beta function of Yang–Mills theory for spinor ﬁeld with symmetry group like SU(N), it is given by [15,25]

11 4 g3 β(g) = − T(A) − n T(R) + O(g5). 3 3 f 16π2

The numbers T(A) and T(R), are given in

a b ab a b ab Tr(JR JR) = T(R)δ and Tr(JA JA) = T(A)δ ,

a a where JR are generators for the fundamental representation of SU(N) and JA are generators for the adjoint representation. To get them for our Lagrangian, we have to start with the commutation relation a b abc c 1 2 [J , J ] = i f J and note that we can multiply both sides by 4 (1 + iγ) to get

1 1 1 1 (1 + iγ) Ja, (1 + iγ) Jb = i (1 + iγ) f abc (1 + iγ) Jc, 2 2 2 2

1 abc thus we get new anti-symmetric structure constants 2 (1 + iγ) f , although the new generators are not hermitian, but this does not violates the methods of deriving the beta function, the necessary thing in deriving it is keeping T(A), T(R) and f abc constants [15,26,27]. Anyway, we will absorb the modiﬁcation 1 factor 2 (1 + iγ) into the coupling constant g, so we have SU(2) gauge group with complex coupling 1 constant like 2 (1 + iγ) g. Therefore we obtain

1 1 1 Tr[Ja Jb ] = T(R)δab → Tr (1 + iγ) Ja (1 + iγ) Jb = (1 + iγ)2 T(R)δab R R 2 R 2 R 4 and 1 1 1 Tr[Ja Jb ] = T(A)δab → Tr (1 + iγ) Ja (1 + iγ) Jb = (1 + iγ)2 T(A)δab. A A 2 A 2 A 4

1 2 Thus to get beta function for our Lagrangian, we replace T(A) with 4 (1 + iγ) T(A) and T(R) with 1 2 4 (1 + iγ) T(R), so using this in the beta function

11 4 g3 β(g) = − T(A) − n T(R) + O(g5), 3 3 f 16π2 we get 1 11 4 g3 β(g) = − (1 + iγ)2 T(A) − n T(R) + O(g5). 4 3 3 f 16π2 1 For the group SU(2), we have T(A) = 2 and T(R) = 2 , so for n f = 1, we obtain beta function like

5 g3 β(g) = − (1 + iγ)2 + O(g5). 3 16π2 Taking in consideration the ﬁrst statement we obtain

∂g 5 g3 = − (1 + iγ)2 , ∂ln(M) 3 16π2 Physics 2019, 1 358 which can be solved for energy scale M as

10 M −2( ) = ( + )2 + −2( ) g M 2 1 iγ ln g M0 . 3 · 16π M0

Usually the coupling constant is real, so for our one, we consider that the interaction strength is governed by the real part of the coupling constant g, this is

10 M −2( ) = − 2 + −2( ) gint M 2 1 γ ln gint M0 , 3 · 16π M0 thus the behavior of interaction of the left-fermions with the spin connection Ai according to our Lagrangian depends on γ, if γ > 1, then β(g) > 0, which means that this interaction becomes stronger as the energy increases, until breaking the perturbation at some energy scale. It is natural to consider γ > 1 0i i i i since we expect A = Aboost > Arotation = A in nature; the gravity effects the particles by changing their energies (like accelerating a particle via a gravitational field) not by changing their angular momentums. So from A0 = γA, we have γ > 1 for this case. i i But when does the case γ > 1; β(g) < 0 appear? it appears when Arotation > Aboost, in this case, the gravity induces a rotation of the inertial frame (the Lorentz frame moves with a particle, or the Lorentz frame in which the particle has a constant speed) more than changing the energy of that particle (accelerating). This occurs when there is the smallest distance between two particles which interact by their gravitational field. At this distance, the velocities of the two particles are constant, so the interaction by the gravity is dominated by changing their angular momentum, thus A0 > A so γ > 1. This situation appears in the back holes; which is a confinement of β(g) < 0.

6. Conclusions We have considered the spacial Lorentz orthonormal basis (e1, e2, e3) as an element in a vector bundle with real spin connection ωij, which takes values in the Lie algebra of group SO(3) or SU(2). We a considered this vector bundle as a tangent vector bundle on the 3D hypersurface of constant time Σt(σ ), i jk a ij this allowed us to define three-form θ = constant × εijkE ∧ R in the phase space (Ei , ωa ) on this surface. By arbitrary transformation of xµ, the three-form θ becomes on M, but (Σ(σa), dθ) = 0 is always satisfied. By doing the same thing, we obtained the equation (Σ(σa), dθ) = 0 using self-dual and anti-self-dual a formalism. This equation produces an equation of continuity on the hypersurface Σt(σ ). We found that 0a i ab i Σi is a conjugate momentum of Aa where Σi Fab is its energy density. We saw that we have to include the i 4 µνi 4 term L(DA )ed x = (1/4)Fµνi F ed x in the GR Lagrangian, since there is a conserved spin current that couples to Ai. This is the similarity between GR and Yang–Mills theory of gauge fields. If we can solve the GR equations using only spin current, we may consider GR as Yang–Mills theory of gauge fields on a curved space–time manifold with spin connection Ai as a gauge field.

Conﬂicts of Interest: The authors declare no conﬂict of interest.

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