Yang Mills Theory of Gravity

The canonical formulation of general relativity is based on decomposition space--time manifold $M$ into $ R\times \Sigma$, this decomposition has to preserve the invariance of general relativity, invariance under general coordinates, and local Lorentz transformations. These symmetries associate with conserved currents that are coupled to gravity. In this paper, we try to solve the equations of motion of general relativity in self-dual formalism using only the spin currents(Lorentz currents), in static case, and without needing using the Einstein's equation, that makes the general relativity similar to Yang-Mills theory of gauge fields. We give an example, matter located at a point, so we have spherical symmetric system. Then we add Yang--Mills Lagrangian to general relativity Lagrangian. Finally we use the decomposition of the space--time manifold $M=R\times \Sigma$ to find that $\Sigma^{0a}_i$ is a conjugate momentum of $A^{i}_a$ and $\Sigma^{ab}_i F_{ab}^i$ is energy density.


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 M. We can consider this base space as an arbitrary space-time manifold M with coordinates (x µ ), 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].
The local Lorentz invariance gives a conserved current, call it spin current, or Lorentz current. In flat coordinates (x I ), the spin current for arbitrary field ϕ is ( [15], section 22) the conservation law is ∂ I M IJK = 0, with conservation of energy-momentum tensor ∂ I T IJ = 0. When the field carries spinor indices, like ϕ α , this adds a term like In arbitrary coordinates (x µ ), we write (x I = e I µ x µ ), where x I becomes tangent vector, also T µν = e µ I e ν J T IJ . One can write M µIJ = e I ν x ν T µJ − e J ν x ν T µI .
Therefore D µ M µIJ = 0 when D µ e I ν = 0 and D µ T µI = 0. The current M µIJ couples to the spin connection ω IJ µ for local symmetry of Lorentz group SO (3,1). The equation D µ e I ν = 0 defines affine connection ∇ on T M × T M and spin connection ω on T M × so (3,1). In this paper, we try to solve the equation of motion of the Plebanski Lagrangian using only the spin connection, without needing using the affine connection ∇(avoiding using it).

Solutions using spin currents
By regarding the local Lorentz symmetry as a gauge symmetry with spin connection ω ∈ Ω 1 (M, so(3, 1)) (or A ∈ Ω 1 (M, so(3, C))) 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 produces locally conserved currents, and those currents are coupled to spin connection ω IJ . 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.
We try in this section to solve the equations of motion of general relativity in self-dual formalism using only the spin current(Lorentz current), in static case, and without needing solving the Einstein's equation. We give an example, matter located at a point, so we have spherical symmetric system.
In self-dual formalism of general relativity, the independent variables are B µν i and A i µ , they take values in Lie algebra so(3, C). The non-zero spin current δL matter /δA i µ implies D µ B µνi = 0(here D µ = ∇ µ + A µ ). Therefore we write B µνi = K i j Σ µνi , for some 3 × 3 complex matrix K i j of scalar functions on M in order to get D µ Σ µνi = 0 and Σ µνi = P iIJ e µ I e ν J , with gravitational field e I µ satisfying D ν e I µ = 0, here D µ = ∇ µ + ω µ , where ∇ is affine connection on M, we avoid using it. The gravitational field e I µ defines a metric g µν = e J µ e Jν , it satisfies ∇ ρ g µν = (D ρ e J µ )e Jν + e J µ (D ρ e Jν ) = 0. The field e µ I is inversion of e I µ , that is e I µ e µ J = δ I J and e I µ e ν I = δ ν µ .
We try to find the solutions of K i j and Σ µνi , so obtaining the spin connection A i µ from D µ Σ µνi = 0. By that we obtain the curvature of A i µ and the field B µνi = K i j Σ µνi .
The general relativity Lagrangian in self-dual formalism including matter is where φ is traceless matrix. The equation of motion of B µν i is The two forms F i µν and B i µν takes values in Lie algebra so(3, C), so there are always matrices ψ and ψ ′ of scalar functions, so writing The equations of motion asserts that the matrices ψ and ξ are symmetric.

The equation of motion of the spin connection
Now we try to solve the equation (4) using B µνi = K i j Σ µνi with D µ Σ µνi = 0. Therefore (3), one finds that the matrix K must also be symmetric in order to satisfy the general relativity constraints using Σ i and F i = (ψK) i j Σ j in the vacuum(ψ ′i j = 0), where we let Σ i be a solution. By SO(3, C) transformation using an invertable matrix U i j , we can get K i j = K i δ i j , for some scalar functions K i . Latter we find that K i satisfies The equation (5) means that the spin current becomes a source for D µ K i j instead of being source for D µ B µνi . The new basis Σ µνi has to be given using gravitational field e I µ with the inversion e µ I , so writing Σ µνi = P iIJ e µ I e ν J , the field e I µ satisfies D ν e I µ = 0. The matrix K i j has to be symmetric as mentioned before.
Using Σ µνi = P iIJ e µ I e ν J and J νi = P i IJ J νIJ in (5), we obtain D µ K i j P j IJ e Iµ e Jν = P i IJ J νIJ , and using J νIJ = e Iµ e ρJ J ν µρ which is possible since e Iµ e Jρ J νIJ = J ν µρ exists, we find Since K i j is real symmetric matrix, we let v i µ j be also real and symmetric in i, j. The And according to the calculations below, we find that v i µ δ i j P j IJ e Iµ = 0 implies v i µ = 0. So we drop v i µ j from now.
The matrix K i j is symmetric, so we can make SO(3) × SO(3, 1) transformation using a matrix U i j to get K i j = K i δ i j , so But D µ δ i j = 0, thus Therefore P i IJ e Jν ∂ ν ∂ µ K i = 0 for each I = 0, 1, 2, 3 and i = 1, 2, 3.
For I = 0, we have P i 0j = −iδ i j /2, it implies e iν ∂ ν ∂ µ K i = 0 for each i = 1, 2, 3. And for I = j, we have P i jk = ǫ i jk /2, it implies e iν ∂ ν ∂ µ K j = e jν ∂ ν ∂ µ K i for each i, j = 1, 2, 3. The first equation e iν ∂ ν ∂ µ K i = 0 implies the proportionality ∂ ν ∂ µ K i ∼ ǫ i jk e j ν (or ∼ ǫ νρσγ e ρi ), using it in the second equation leads to which is anti-symmetric in i, j, this is contradict with the symmetric in the second equation where we used the metric g µν = e J µ e Jν , it satisfies ∇ ρ g µν = (D ρ e J µ )e Jν + e J µ (D ρ e Jν ) = 0.
We note that we used a metric g µν defined by e Jµ e J ν , but the equation is valid for any other metric g ′ µν (at least locally). Also we let the affine connection ∇ µ be flat connection, therefore we can make coordinates transformation to obtain euclidean metric on M, and so ∇ µ = ∂ µ . Another treatment of e Iν ∂ ν ∂ µ K i = 0 is multiplying it with e Iρ and sum over I to get ∂ ρ ∂ µ K i = 0 which can be solved by letting K i depend only on one coordinate(like r in spherical coordinates), we obtain η µν ∂ µ ∂ ν K i = 0, for Lorentz metric η.
By that the functions K i do not depend on the metric e Jµ e J ν on M, but they have to depend only on one coordinate on M.
To get a solution, we first solve ∂ 2 K i = 0(by letting K i depend only on one coordinate), then using it in −(∂ µ K i )Σ µνi + J νi = 0, the equation (5), to get the solution of Σ i µν in terms of given spin current J νi , and so getting B i µν , where the spin current J iµ is assumed to be given as a function on M. Using the solution of Σ i µν , we obtain the connection In 3 + 1 decomposition of the space-time manifold M = Σ × R, let Σ t be the space-like slice of constant time t with the coordinates (x a ) a=1,2,3 (and 0 is time index). The equation in which we introduce the vector field E i and the 1-form B i , on the space-like slice Σ t (the field E ai is conjugate to the connection A i a ).
where J ai are vector currents and J 0i = Q i are charges.
In the static case J ai = 0, we can solve the first equation in terms of ∂ a K i by writing and f is scalar function on M. But when Regarding the second equation of (9), when J 0i = 0(in the vacuum), we can solve it in terms of ∂ a K i by writing for some 1-form v ∈ Γ (M; T * Σ × so(3, C), and g is scalar function on M. The equation And when J 0i = 0(existence of matter), we let E ai = g ab (∂ b K i )u i , for some vector field u ∈ Γ (M; so(3, C)). By that Thus there are two solutions of E ai (for vacuum and for matter), the total solution is as showed before.
Using the self-duality [22,11] where we used the Hodge duality theory between the forms and the tensor fields, here Σ i µν is 2-form and Σ iµν is (0,2) tensor field. Regarding the equations (7), (10), (11) and (13), one can take the solutions In order to write Σ IJ as e I ∧ e J in the solutions (14), as required to get gravity theory, we . Therefore the solutions (14) become so v a B i a = 0. By that and according to self-dual projection, there is vector fields b I and we get e J a = ∂ a K J u J 2 and e I 0 = v 0 u I 1 , for v 0 = 1. By that the solutions (15) can be written as Σ i = P i IJ Σ IJ for Σ IJ = e I ∧ e J . But we have to note that the solution (15) is a general solution and we have to find a special solution, like to choose b i = u i and let it be constant field, as we will do in the following study.
If the charges J 0i = 0 are given as a functions on M, and in order to get a solution using them, we use the solutions of K i (obtained from ∂ 2 K i = 0) in the equation (12), so obtaining the field u i = b i . And obtaining the vector v from v a ∂ a K i = 0 so obtaining the 1-form by v a = g ab v b . We obtain Σ i 0a and Σ i ab using the solutions (15), so getting the connection A i from DΣ i = 0. We note that v a E i a = 0, v a B ai = 0 and v a ∂ a K i = 0 depend on the symmetry of the system, for example, spherical symmetry, cylindrical symmetry, and so on.
Using the solutions (15), we get the connection A ∈ Ω 1 (M; so(3, C)) using DΣ i = 0, The solution of the spin connection A i is with The used metric is g µν = e I µ e J ν η IJ and Σ µν i is inversion of Σ i µν (see appendix A for more details). By using we get the inversion tensor Σ µνi , and e I µ is obtained from Σ i µν = P i IJ e I µ e J ν . We obtain the We try to give an example for obtaining that solutions. In spherical symmetry, where the matter located at a point. We use the spherically coordinates (r, θ, ϕ) on the space-like slice Σ t = Σ = R 3 . We let the solutions depend only on the radius r, and start with solution of ∂ 2 K i = 0(the equation (6)), for some constants c i ∈ R. By that the functions K i depend only on one coordinate, r, as discussed below of equation ( (6)). Therefore where in the spherical symmetry we let a 1 and a 2 do not depend on the coordinates θ and ϕ.
The values of the constants a 1 and a 2 are not significant since a i = g ij a j is Killing vector, thus we set a 1 = a 2 = 1. The used metric g ab here is the standard metric in the spherical coordinates, because a i is Killing vector of K i .
We get the solutions of the 1-form B i and the vector field E i using (15), By that we get According to self-dual map, there are at least two constant fields b I 1 and b I 2 satisfying b i = P i IJ b I 1 b J 2 . So from Σ i µν = P i IJ e I µ e J ν , we get the gravitational fields We let that matter be located at the origin (0, 0, 0) ∈ R 3 , therefore the charge (12) is Using the equation (12) And in order to get Q i (x) = Q i 0 δ 3 (x), we let So we choose c i = 1 and u i = −ǫQ i 0 /(π 2 √ 2), and as we saw before the total solution of E ai (for vacuum and for matter) is (12). Therefore in the limit ǫ → 0 + , the remaining solution is There is multiplicity in the gravitational fields (19), therefore we let the non-zero fields be We obtain the connection A i µ using (16), so getting the curvature F i (A) = dA i + ε i jk A j ∧A k . The constants b I 1 and b I 2 have to be fixed in order to satisfy Σ µν i F i µν (A) = 0 in the vacuum.

Modifying by adding Yang-Mills Lagrangian
In this section we get DΣ i = 0(De I = 0) by adding a new term to the Lagrangian in self-dual formalism. We add Yang-Mills Lagrangian of the connection A i to the Lagrangian to get the equation of motion of A i ν is (k is constant) Thus we can choose D µ Σ µνi = 0, by that we get This is same equation of motion in Yang-Mills theory of gauge fields. This relates to the fact that we can regard the local Lorentz symmetry (internal symmetry) as a gauge symmetry with spin connection ω IJ (or A i ) as a gauge field. The Lorentz current in 21 is conserved since We can get same current by using Riemann curvature tensor R µ νρσ given in terms of the Levi-Civita connection ∇. The Bianchi identity is Multiplying it by δ ρ µ , yields to the Ricci tensor R νσ = R µ νµσ vanishes in the vacuum, so ∇ µ R µ νσγ also vanishes in the vacuum, so ∇ µ R µνσγ = ∇ µ R σγµν = 0 associates only with matter. The curvature of the spin connection ω relates with the Riemann curvature by R IJ µν (ω(e)) = e Iσ e Jγ R σγµν (g(e)), with g µν (e) = η IJ e I µ e J ν , therefore ∇ µ R σγµν = 0(= 0) → D µ R IJ µν = 0(= 0), D µ e I ν = 0.
Thus D µ R IJ µν associates only with matter and vanishes in the vacuum. Using the self-dual projection, F i µν (A) = P i IJ R IJ µν (ω), we find that also vanishes in the vacuum. Therefore this current associates only with matter and the equation (21) is well defined.
In Plebaniski formalism, we have the formula in the space (Σ i , F i ), where T r(ψ) = 0,ψ i j = 0 in the vacuum without cosmological constant [10]. By using this formula in Equation (21), we get (Appendix B) This is same equation (5) We get the energy density H and we set d dt H = 0 when there is no external source. We see that d dt H = 0 is same dθ = 0 for 3-form θ on Σ. We define gravitational field as a one-form e I = e I µ (x)dx µ that is related with metric g µν (x) on an arbitrary space-time manifold M by g µν = η IJ e I µ e J ν , with spin connection ω IJ (x) ∈ Ω 1 (M, so(3, 1)), where so ( We start with the general relativity Lagrangian of the form where is the Riemannian curvature tensor and e µ I satisfies e I µ e µ J = δ I J . By the decomposition M → R × Σ, we decompose this Lagrangian into The part takes the role of the energy density on the surface Σ t of constant time, it does not include the time derivative, but it includes the spatial derivatives with respect to coordinates on Σ t . It is invariant under the SO(3, 1) transformation, thus we let d dt H = 0 when there are no external sources. The second term has the term ee 0 I e a J ∂ 0 ω IJ a with represents the variables of the phase space ee 0 I e a J , ω IJ a on the surface Σ t , where we let ee 0 I e a J = E a IJ be conjugate momentum to ω IJ a . The final term is Integrating by parts, this term becomes The variable ω IJ 0 has no time derivative, so it is Lagrangian multiplier, it gives the constraint D a ee 0 I e a J = D a E a IJ = 0.
In self-dual formalism of GR, we use the self-duality [22,11] The Lie algebra so(3, 1)(with indices I, J, ...) is replaced by the Lie algebra so(3, C)(with indices i, j, ...). The field E a IJ becomes complex given by [9,10] where P i IJ is a self-dual map given by For example E 1a = 1 2 ε abc (e 2 b e 3 c + ie 0 b e 1 c ). This map relates to the decomposition of Lie algebra of the Lorentz group SO(1, 3) into two copies of Lie algebra of SL(2, R) [11].
In the equation (27), we used the formula (26) to get Regarding the previous definition of the conjugate momentum E a IJ = ee 0 I e a J of ω IJ a , we ignore −i and just write E ia = eP i IJ e 0I e aJ which becomes conjugate momentum to complex connection A i .
The obtained connection A i is a three-complex one-form given by The curvature which associates with this connection is On the surface Σ t (σ a ), it is Thus we have self-dual plus anti-self-dual map: whereĒ ia andF i bc are the Hermitian conjugate of E ia and F i bc .
For the Lagrangian part H on Σ t , the equation (25), it becomes or where σ a is coordinates on Σ t . We can write Hd 3 σ as 3-form on Σ t , like By that the condition d dt H = 0 becomes Here (Σ, V ) is a pairing of 4-form V ∈ ∧ 4 T * p M with a surface Σ, the pairing of V with tangent basis in T p Σ, defined below in equation (40).
Since θ is three-form on Σ(σ a ), so dσ a ∧ ∂ ∂σ a θ = 0, if we add it to the last formula, we get Under arbitrary transformation (t, σ a ) → x µ , dσ a = ∂σ a ∂x µ dx µ , the basis on Σ transforms as Therefore, the components of three-form θ transforms as To keep the invariance under this transformation, we let the equation (34), (Σ, dθ) = 0 still hold. We write the components of the curvature as where which motivates introducing a notation of the covariant derivative like [12] where the matrix elements (T j A ) ik = −iε jik are the elements of the generators T j A 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 By that, we write the 3-form (32) as Its pairing with the surface Σ t (σ a ) is the energy Regarding the equations (33) and (34), we let (Σ t (σ a ), dθ) = 0, this gives a continuity equation, as we see next.
Our condition (Σ(σ a ), dθ) = 0 makes sense here because of the decomposition R × Σ and fixing a coordinate system σ a on the hypersurface Σ, this yields to an equation of continuity on this surface. For this purpose, we take the pairing of the four-form dθ with a tangent basis on the surface Σ t (σ a ) at an arbitrary point, we get a one-form co-vector (Σ(σ a ), dθ) in the direction of the normal to this surface at that point. Then we set (Σ(σ a ), dθ) = 0, we obtain an equation of continuity on Σ t (σ a ).
The pairing of dθ with this basis is where (·, ·) is contraction pairing defined by where the bracket [....] is anti-symmetrization of the indices. Although DDA i is zero when the sum is over all indices like ε abc D a D b A i c = 0, but here we do not sum over all indices of DDA i alone, since we take common functions out of the sum.
For cotangent basis {dx µ } and tangent basis {∂ a }, this pairing can be defined simply by using inner product like [14] (dx µ , ∂ a ) = δ µ a , in which we consider dx a = dσ a for a = 1, 2, 3, so E 0i = 0 regarding to our gauge.
Starting with the first term The second term is Doing the same thing, we get Adding the two terms, we obtain We define the curvature by using the covariant derivative from Equation (36) as therefore Its Hodge dual on the surface Σ with respect to the coordinates (σ a ) is Also we define the complex two-form field from Equation (27) as Using them in the last formula, we get And using we obtain it becomes As we suggested before, we let the normal of the surface Σ t (σ a ) be in direction of the time dx 0 , so dθ, ε abc ∂ a ∂ b ∂ c is in direction of the time. Therefore we set µ = 0, thus we get The vector dθ, ε abc ∂ a ∂ b ∂ c is one-form in the direction of the normal to the surface Σ t (σ a ). It is zero as we mentioned before, thus we get For De = 0 on the surface Σ, where e = det(e i a ), We write it as using F ai = ε abc F i bc , Σ bc i = E bc i = ε bca E ai and E ai = ee ai , we have We have regarded 1 4 E a i E b j F ij ab as energy density.
The equation We regard this equation as an equation of continuity with respect to a Lagrangian like L(F 0ai , DA i ) that satisfies the action principle δS(DA i ) = 0 and the invariance under continuous symmetries of general relativity. Therefore we regard 1 2 cΣ ab i F i ab as energy density T 00 , and cΣ ab i F i b0 as momentum density T 0a , where c is constant for satisfying the units.

Conclusions
We have considered the 4-manifold M as a base space and consider that the Lorentz group SO(3, 1) act locally on Lorentz frames which are regarded as a frame bundle over a fixed base space, 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. The local invariance of matter Lagrangian under SO(3, 1) transformation produces a current, called spin current J = δS matter /δA, this current couples to the spin connection, so we have gauge theory, that makes the general relativity similar to Yang-Mills theory of gauge fields. We have solved the equations of motion of general relativity in self-dual formalism using only the spin currents(Lorentz currents), in static case, and without needing using the Einstein's equation. We give an example, matter located at a point, so we have spherical symmetric system. We solved the equation δS/δA i ν = −D µ B µν i + J ν i = 0 by using B µνi = K i j Σ µνi for D µ Σ µνi = 0.
Therefore Σ µνj Σ µνi = P jIJ P iIJ = δ i j , where we used the self-dual projection property P jIJ P iIJ = δ i j . The sum is over the contracted indices.
The first term becomes D(ψ i j Σ j ) = D(ψ i j Σ j +ψ i jΣ j ) − D(ψ i jΣ j ), so D(ψ i j Σ j ) = DF i − D(ψ i jΣ j ), and by Bianchi identity DF i = 0, we obtain D(ψ i j Σ j ) = −D(ψ i jΣ j ). Therefore D ν F i νµ = i * (−D(ψ i j Σ j ) − * (Dψ i j Σ j ) µ = −2i( * D(ψ i j ) ∧ Σ j )) µ , where we used DΣ j = 0. Thus we obtain the raising and lowering is done by using the metric g µν . But so we obtain This is a linear equation in the basis {Σ i }.