Spin current in BF theory

In this paper, we introduce a current which we call spin current corresponding to the variation of the matter action in BF theory with respect to the spin connection $A$ which takes values in Lie algebra $\mathfrak{so}(3,\mathbb{C})$ in self-dual formalism. For keeping the constraint $DB^i=0$ satisfied, we suggest adding a new term to the BF Lagrangian using a new field $\psi^i$ which can be used for calculating the spin current. We derive the equations of motion and discuss the solutions. We will see that the solutions of that equations do not require a specific metric on the manifold $M$, we just need to know the symmetry of the system and the information about the spin current. Finally we find the solutions in a spherical and cylindrical symmetric systems.


Introduction
The BF theory on 4-manifold M is a topological theory which when includes constraints terms turns to gravity theory. The fundamental variables are 2-form B ∈ Ω 2 (M; so(3, 1)) and spin connection ω which takes values in so (3,1), all derivatives are linear and applied only on ω, which makes it easy for canonical formalism; finding the phase space, Hamiltonian equations, quantization,... [1]. This theory does not require a metric to be formulated, the metric is a derived quantity from the solutions of B. That gives motivation to formulate Einstein's gravity as a theory of 2-forms rather than the metric tensors and so no pre-exiting geometrical structure is needed to obtain the gravity. Let F (ω) ∈ Ω 2 (M; so(3, 1)) be the curvature of ω. The pure BF theory action is There are no local degrees of freedom because the system has so much symmetry, that all solutions are locally equivalent under gauge transformation of the group SO(3, 1) and under diffeomorphisms of M. Hence the pure BF theory is a topological theory [2,3]. In QFT, using fields(not forms), the action changes(under f ) as δS = ∂ µ (g νρ δx ρ δS/δg µν ) when the equation of motion are satisfied. Therefore in order to get TQFT, it must be δS/δg µν = 0, so independent of any metric(standard or other).
In constrained BF, the Lagrangian includes the constraint term ϕ IJKL B IJ ∧ B KL . The traceless matrix ϕ plays the role of a Lagrangian multiplier and that imposes the constraint on the 2-form B IJ , so that its solutions are given in terms of 1-forms e I = e I µ dx µ , that is B IJ = e I ∧ e J , where I, J, ... = 0, 1, 2, 3 are Lorentz indices and µ, ν, ... = 0, 1, 2, 3 are spacetime tangent indices, we regard the frame fields e I µ dx µ as gravitational fields, therefore the constrained BF theory turns to general relativity theory, the reason is that when ϕ IJKL is not zero, the term ϕ IJKL B IJ ∧ B KL breaks the diffeomorphisms invariance of BF action, thus there are non-equivalent local solutions and so local degrees of freedom exist as known in general relativity in the vacuum. The problem with constrained BF theory is that the equation of motion δS/δB = 0 contains the non-physical variable ϕ IJKL , but we can remove it by taking the trace of the equations(ϕ is traceless matrix), but also there is a problem with the trace operation, it reduces the equations to one equation which is not enough for getting a solution. For that reason we search for solutions of BF theory by using the equation δS/δω = 0(as done in this paper). In general, the equations of motion of constrained BF theory including matter give a relation between the curvature F IJ (ω) and the frame fields Σ IJ = e I ∧ e J (the Plebanski 2-form), in matrix notation, that is F = χΣ + ξΣ, where χ and ξ are symmetric matrices of scalar fields [4]. Therefore the problem turns to finding χ and ξ. Since the field ϕ IJKL is not a physical variable, the equations of motion of general relativity have not to include it(appendix A).
In this paper, we start with definition of the spin current J and discuss its conservation in BF theory including matter(for a general we do not specify a matter Lagrangian). The spin current J appears in the equations of motion as a source for d ω B by the equation * (d ω B) + J = 0, where ' * ' is Hodge star operator. And in order to get d ω B = 0 in this paper, we add a new term to BF Lagrangian, like tr (ψB ∧ F (ω)), using a new field ψ.
That can be seen as a redefinition B → B + ψB, by which the equation * (d ω B) + J = 0 becomes * (d ω B) + * (d ω (ψB)) + J = 0, so we choose d ω B = 0 and get * ((d ω ψ)B) + J = 0. Therefore the spin current becomes a source for the field ψ instead of B and we get a new formula(definition) for the spin current using ψ, and since the spin current regards symmetry of the system, the field ψ also regards that symmetry. In constraint BF theory, solving the equation * ((d ω ψ)B) + J = 0 is easy as we will see. We find that the equation of motion of ψ is same conservation equation D µ J µ = 0 of the spin current vector field J. We see that we can solve the equations of BF theory only by solving the spin current equation δS/δω = 0, J = 0 with d ω B = 0 and without needing solving the equation δS/δB = 0 which includes the Lagrangian multiplier ϕ IJKL (a non-physical variable), and without using a gravitational metric on M, we just need using the spin current and knowing the symmetry of the system.
That means that we can solve the BF equations only by using the coupling term M ω IJ µ J µ IJ which makes them easy to solve, and makes the theory similar to the gauge theory. And since ω IJ is 1-form and J IJ is vector field, the term ω IJ µ J µ IJ is naturally defined on M without needing using additional structure(like a metric), thus solving the system equation using only that coupling term gives a topological theory, i.e, the theory turns to finding 1-forms and vector fields, and does not need to use a gravitational metric, similarly to Chern-Simons theory which includes the Wilson loops as a source for the gauge field. That makes it easy to solve the equations in different cases of the spin current, e.g, point charge, straight line current, circular current,... . The lines of the spin current can be described using any coordinates system, e.g, Euclidean coordinates,..., so the BF theory can be studied in any coordinates system, but in order to avoid an effect of the coordinates on the lines of spin currents, we let that coordinates be flat(not curved). And since the spin current is source for the field ψ, this field has singularities on the lines of that spin current. We see that our solution of B IJ can be always written as e I ∧ e J , so we get the gravity theory. Finally we give an example of explicit solution of the equations in spherical and cylindrical symmetric systems in static case just by finding the field ψ using the spin charges J 0 IJ .

Spin current in BF theory
Let M be connected oriented smooth 4-manifold and P → M be an SO(3, 1)-principal bundle(Appendix A) with a spin connection ω which is locally a 1-form with values in so(3, 1) and F ∈ Ω 2 (M; so(3, 1) P ) is its curvature. The BF theory action is invariant under global and local Lorentz transformation, that gives a conserved current, we call it spin current (Appendix C). Before discussion the conservation of the spin current, we introduce the self-dual formalism.
That self-dual projection relates to the fact that the complexified Lie algebra of SO (3,1) has the decomposition so(3, 1) C = so(3, C) ⊕ so(3, C) [6]. The new connection is locally an so(3, C)-valued 1-form A on M whose components are and its curvature is The two form B IJ is mapped to where ∇ µ is the affine connection on T M.
Using the new variables we can write the Lagrangian of matter(without specifying matter fields) L matter (e I , ω IJ ) as L matter (B i , A i ,B i ,Ā i ), whereB i andĀ i (anti-selfdual representation) are the complex conjugation of B i and A i . The Urbantke formula (equation 49, appendix B) writes the metric g µν using only the constrained B i without using the con- or just writing L matter (B i , A i ).

Definition 2. Let
A be be the self-dual connection on the so(3, C)-bundle so(3, C) P → M.
Let L matter be the Lagrangian of matter fields on M. Then the spin current J µ i is defined to be The matter action S matter is required to be invariant under any infinitesimal local Lorentz transformation ω IJ µ → ω IJ µ + D µ Λ IJ for infinitesimal transformation parameter Λ IJ ∈ Ω 0 (M; so(3, 1) P ). Now we assume S matter has this property. Then we have the following.
Lemma 1. The spin current J µ i given by J µ i = δ δA i µ L matter in gravity theory is conserved [7]. Proof. Since S matter is invariant under infinitesimal gauge transformation Λ IJ , it is invariant under (we may suggest the condition (4)) The variation vanishes for arbitrary Λ i only when D µ δ δA i µ L matter = 0, where we let Λ i vanish on the boundary ∂M. Thus the current J µ i = δ δA i µ L matter is conserved. Actually the previous calculation based on the idea that A i andĀ i transform independently under infinitesimal local Lorentz transformation ω IJ µ → ω IJ µ + D µ Λ IJ , therefore there is another current that associates with the connectionĀ i when the matter Lagrangian depends also onĀ i . Same thing we find for the general relativity action, by using the variables (Σ i , A i ), we obtain the equation In 3 + 1 decomposition of the space-time manifold M = Σ × R, let Σ t be space-like slice of constant time t, with the coordinates (x a ) a=1,2,3 , let 0 be time index. In Hamilton-Jacobi system, by using the variables (E a i , A i a , A i 0 ) on the slice of constant time Σ t , that equation becomes which is satisfied when D a E a i = 0, where E a i is conjugate momentum to A i a , and we used the [5,8,9]. The equation D a E a i = 0 is satisfied in BF theory(appendix A).
Remark 1. We note that the current J µ i is similar to the currents in Yang-Mills theory of the gauge fields, we see this clearly when we regard the connection A i µ as a gauge field, by that the current J µ i relates to the local Lorentz invariance (local symmetry). The metric g µν = e I µ e J ν η IJ is invariant under arbitrary local Lorentz transformations, like e I µ (x) → U I J (x)e J µ (x), for U(x) ∈ SO(3, 1), therefore the local Lorentz symmetry is an internal degree of freedom.
where ϕ ∈ Γ (M; End(so(3, C) P )) is traceless matrix of scalar fields ϕ ij , actually we do not require it to be symmetric since we will add a new term to BF Lagrangian(see the discussion below the equation (25)). The connection A on the Lie algebra bundle so(3, C) P which is locally a 1-form with values in so(3, C) and its curvature F (A) ∈ Ω 2 (M; so(3, C) P ) are defined in the equations (2) and (3). The index contraction is done by using δ ij , the Killing form on so(3, C).
Since the matrix ϕ is traceless, we can write ϕ ij = m ij −(m 11 +m 22 +m 33 )δ ij /3, for some not traceless matrix (m ij ). The variation of the action with respect to m ij produces a quadratic equation in B i whose solution turns the theory into general relativity. These are The solutions to this are all of the following form B i = P i IJ e I ∧ e J , in which the gravitational fields e I µ are considered as frame fields [11]. Using the self-dual formula (1), the constrained 2-form B i is written as this is B i constrained = Σ i , with using the notation Σ i = P i IJ e I ∧ e J .
The equation of motion with respect to B i is Since F i (A) ∈ Ω 2 (M; so(3, C) P ) is 2-form with values in so(3, C), the δ δBi S matter is also 2-form with values in so(3, C).

Lemma 2.
In constrained B i , the variation δ δBi S matter ∈ Ω 2 (M; so(3, C) P ) has the form for some matrices T i j , ξ i j ∈ Γ (M; End(so(3, C) P )), with T i j = T j i and ξ i j =ξ j i (See appendix B, for more details).
Since tr(ϕ) = 0, so tr(ψ) = −tr(T ), the equation (8) yields Thus in the vacuum T i j = 0, we have Σ µν i F i µν = 0. We call Σ µν i F i µν the trF . The equation (9) does not contain the non-physical variable ϕ, but the problem with it is that the trace process decreases the number of equations. Therefore Σ µν i F i µν = −tr(T ) is a condition on the solutions. The (0,2) tensor Σ µν j is inverse of the 2-form Σ i µν (Appendix C).
The equation of motion with respect to the connection A i is We see that we can not choose ε µνρσ D ν B i ρσ = 0 when J µi = 0. But the condition ε µνρσ D ν B i ρσ = 0 leads to the equations D a E ai = 0(appendix A) and De I = 0, which makes the spin connection ω IJ compatible with the gravitational field e I , that makes the equations of motion easy to solve as we will see. We can get DB i = 0 by a redefinition of B i , like We get this redefinition by adding a new term to the BF action (6), as done below. We see that the equations of motion of general relativity are still satisfied in BF theory(appendix A) after adding the new term, therefore there is no problem with it. In constraint BF theory, solving the equation (11) is easy, because the field ψ i satisfies the equation D 2 ψ i = 0 (equation (34)) with respect to a metric that satisfies ∇ ρ g µν = 0.
By acting by D µ on the equation (10), we get and using D µ J µi = 0, we obtain We can regard the equation (12) as an equation of motion with respect to a new field ψ i , with the possibility of choosing DB i = 0 with J µi = 0.
In order to include the equations DB i = 0, J µi = 0 and D µ J µi = 0 in BF theory, we suggest the following action.
Definition 4. We add a new term to the BF action (6) to get in which we have added We can get the new term by a redefinition There is no problem with the redefinition of ϕ ij since they are just Lagrangian multipliers.
The equation of motion of this action with respect to the field ψ i is which is same equation (12), therefore the field ψ i does not change the equations of motion.
as we suggested before, thus δ δψ i S matter = 0 is satisfied. Also the equation (14) is satisfied in constraint BF theory by letting the matrix The equation of motion of this action with respect to the connection A k is or 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. The field Σ µν i is inverse of Σ i µν , that is Σ µν j Σ i µν = δ i j (see the appendix C for more details). By re-scaling Σ µνi by e = det(e I µ ), we write The remaining equation of (15) in which we used the spin current J µ i = δS matter /δA i µ and the condition D ν Σ µνj = 0, also re-scaled Σ µνj by e = det(e I µ ). Later we will discuss D µ J µk = 0. Here both Σ µνj and J µk are tensor fields. The equation (18) is same equation obtained in (11).
Remark 2. We note that the equation (18) is similar to the current J µn = ∂ µ ϕ i T n ij ϕ j in scalar field theory with symmetry and generators T n ij , we have J 0n = ∂ 0 ϕ i T n ij ϕ j = π i T n ij ϕ j , where π i is conjugate momentum to ϕ i . Like that, the equation (18) gives J 0i = 2i(D µ ψ k )ε ki j Σ µ0j (for µ = a = 1, 2, 3), so here D a ψ i is conjugate momentum to Σ a0i with noting that the indices raising in Σ µνj is done by using a metric g µν .
The equation of motion of the action (13) with respect to B i in constrained BF (like deriving the equation (8)) is Multiplying with Σ µν i , summing over the indices and using Σ µν iΣ j µν = 0, we obtain Since tr(ϕ) = 0 and ε i jk Σ µν i F k µν = 0(by the equations (14) and (16)), we obtain The equation (19) leads us to write F i (A) in terms of Σ i andΣ i , and since ε i jk Σ µν i F k µν = 0, we can write for some symmetric matrix (χ ij ) and skew-hermitian matrix (χ ′ ij ). Using this equation in the equation (21), we obtain In addition to this relation, there is another relation between the vector field ψ i and the symmetric matrix χ i j when T i j = 0 and J µi = 0, from the conservation of the current (18), and using the equation (22), we get This is another relation between the vector field ψ i and the symmetric matrix χ i j in existence of matter T i j = 0 with J µi = 0. In this case, the matrix χ i j has to satisfy det χ i j − tr(χ i j ) = 0 in order to get ψ i = 0, of course we do not need this condition in the vacuum T i j = 0, J µi = 0.
Using the equation (22) in (19), we obtain That yields This equation relates to the equation of motion δS/δB = 0, it includes the Lagrangian multiplier ϕ ij which is a non-physical variable which increasing the arbitrary solutions of (25), so increasing the local degrees of freedom. Therefore we need to find χ and ψ using the other equations of motion we got before. We see that we do not require the traceless matrix ϕ to be symmetric, since the third term in (25) is not symmetric in general. The symmetric matrix T ij is assumed to be given using the matter Lagrangian (Appendix B), thus the total unknown variables are 3 + 5 + 8 = 16 of the vector ψ, the symmetric matrix χ(with (23)) and the traceless matrix ϕ. The formula (25) gives 9 equations, therefore we have 16 − 9 = 7 unknown variables, but when J µi = 0, they reduce to 6 unknown variables(regarding the equation (24)). But if we choose a solution for which the symmetric matrix χ ij becomes diagonal, like for some scalar functions K 1 , K 2 and K 3 on M, the unknown variables reduce to 4 variables and to 3 variables when J µi = 0.
is another solution, and that makes the components ψ 1 , ψ 2 and ψ 3 of the vector field ψ i independent variables, therefore we can regard them as the degrees of freedom of the system and solve the equations of motions in terms of them. We note that The Bianchi identity DF i = 0 implies Dχ i j ∧ Σ j = 0 (for DΣ i = 0), hence (dK i )δ i j ∧ Σ j = 0, where we used Dδ ij = 0, with using the covariant derivative Dv i = dv i + ε i jk A j v k .
Therefore we obtain 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 ). The covariant The equation (27) We can solve them by writing (for non-zero curvature F i (A)) for some r i ∈ Ω 1 (M; so(3, C) P ) and u i ∈ Γ (M; so(3, C) P ). The functions K i are scalars, the indices are just for distinguishing each from the others. Thus we get the solutions The equation (28) implies Dr i = 0 and Du i = 0.
In the static case J a i = 0(zero current) with J 0 i = 0(non-zero charge), the spin current formula J ν k = 2iε kij D µ ψ i Σ µνj , equation (18), decomposes to two equations We can solve the first equation in terms of D a ψ i by writing for some vector v ∈ Γ (M; T Σ) that satisfies v a D a ψ i = 0, and f is scalar function on M. We without needing specifying the used metric g ab (the solution is satisfied by using an arbitrary Regarding the second equation of (32), when J 0i = 0, we get the solution Σ 0ai = −Σ a0i = f g ab D b ψ i . And when J 0i = 0, we let Σ 0ai = f g ab D b ξ i for some vector field ξ i = ψ i , and without needing specifying the used metric g ab because the solution is satisfied by using an arbitrary metric. Using the equation (17), D µ Σ µνi = 0, and for ν = 0, we have D a Σ a0i = 0.
for constant f and with using ∇ a g bc = 0 which defines affine connection on Σ t in terms of the arbitrary metric g ab . Therefore in the vacuum, where ξ i = ψ i , we obtain g ab D a D b ψ i = 0.
under the immersion map i : Σ t → M gives g ab D a D b ψ i = 0, therefore g µν is also arbitrary metric on M. Regarding the discussion in the introduction, we let the metric in g µν D µ D ν ψ i = 0 be flat, at least locally, in order to avoid the influences of geometry of M on the field ψ i which relates with the spin current by the formula (18).  (31), and in order to get a correspondence between that solutions, e.g, by using a metric, we find that for some vector fields b i , u i ∈ Γ (M; so(3, C) P ).
By that we obtain the solutions without needing using a specific metric. Thus the solutions are described by three complex scalar functions K i , a Killing vector v a = g ab v b satisfying v a ∂ a K i =0, and two vector fields b i and u i . In the vacuum, it must be b i = u i .
Remark 4. Regarding the solutions of the equations (31) and (32), we note that for every two solutions of Σ i ab and Σ abi , we get a metric g ab satisfying Σ i ab = g aa ′ g bb ′ Σ a ′ b ′ i . Also we note that the metric used in Σ i ab = g aa ′ g bb ′ Σ a ′ b ′ i is not necessary the same metric used in D a ψ i = g ab D b ψ i for getting the solutions of (32). Therefore the metric in BF theory is a derived quantity from the solutions of the 2-form B ∈ Ω 2 (M; so(3, C)).
Remark 5. In the solution (36), we see that Σ IJ can be written as e I ∧ e J , as required in constraint BF theory to get gravity theory, that is, according to self-dual projection, there is vector fields b I and K I satisfying we get e J a = ∂ a K J u J and e I 0 = v 0 u I , for v 0 = 1. A more general case is to find three vector fields b I 1 , b I 2 and K I satisfying 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 (36) 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 (36) is a general solution and we have to find a special solution, like to let b i be constant field and write u i in terms of it, as we will do in the following study.
Using the solution Σ 0ai = −Σ a0i = g ab D b ξ i = g ab (∂ b K i )u i in the second equation of (32), We see that J 0 k = 0 takes place only when b = u. Therefore in the vacuum it must be b = u. Regarding the equation (34), the field ψ i = K i b i satisfies D 2 ψ i = 0, so Db i = 0 implies If the charges J 0i = 0 are given as a functions on M, and in order to get a solution using them, we let b i ∈ Γ (M; so(3, C) P ) be constant field on M, so we can determine the scalar functions K i using the equation (37), and so obtaining the vector v ∈ Γ (M; T Σ) using v a ∂ a K i = 0. But to satisfy Db i = db i +ε i jk A j b k = 0 for a constant vector field b i , the connection A i µ must be written as Another way of getting the solutions using the charges J 0i = 0 is by using the solutions of K i (obtained from ∇ 2 K i = 0) in the equation (37), and by using u i = b i + f (x)b i + a i (for constant a i ), one gets the field b i , so getting the connection A i from Db i = 0. We obtain B ai and E ai using the equations (36), and obtain the matrix χ using the equation (26), so obtaining the curvature F = χΣ. 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. Thus we have seen that we can solve the equations of motion of BF theory without need using a specific metric on the manifold M, and the metric can be obtained from the solutions of (Σ 0ai , Σ abi ) and (Σ i 0a , Σ i ab ) according the remark (4).

Solutions in spherical symmetric system
We have seen that we can solve the equations of motion in BF theory by using a complex vector field ψ i = K i b i ∈ Γ (M; so(3, C) P ) which allows us to obtain v, E i , B i and J 0i according to the equations (35), (36) and (37). We try to find the solutions in spherical symmetric system in the vacuum(u = b) and then apply it for matter located at a point. As we have seen that the solution of the system regards the symmetry of that system since we search for a vector v ∈ Γ (M; T Σ) that satisfies v a D a ψ i = 0, v a E i a = 0 and v a B i a = 0. For example, in spherical symmetry, we use the spherically coordinates (r, θ, ϕ) on the space-like slice Σ t = Σ = R 3 . And according to the equation (33), we have D 2 ψ i = 0 on Σ t without specifying the used metric, therefore we let it be the standard metric in the spherical coordinates.
In spherical symmetry and static case, we let the vector field ψ i depend only on the radius r, we get (for Db i = 0) so ψ i = c i b i /r, for some constants c i ∈ R. Actually we can include c i in b i and just write 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 we do not define any other metric.
Using the equation (36), we get the solutions of the 1-form B i and the vector field E i , By that we get while the other components like Σ i 0θ , Σ i 0ϕ , ..., can be obtained by using gravitational fields e I µ derived from the solutions (39). That is according to self-dual map, there are at least two constant fields b I 1 and b I 2 So from Σ i µν = P i IJ e I µ e J ν (regarding remark (5)), we get the gravitational fields .
We obtain the matrix χ using the equation (26) with the solution (38), where the constants c i have to be determined in order to satisfy the condition trχ = 0(in the vacuum), so 3 i=1 c i = 0. Thus we get the curvature F = χΣ + χ ′Σ (with setting χ ′ = 0 in the vacuum [15]), Now we calculate the connection A i and the field b i which satisfies Db i = 0. Using Since we study a spherical symmetric system, we let the connection A i depend only on r. If we choose the gauge A i r = 0, i = 1, 2, 3, we get therefore by using the solution (40), we obtain Therefore the field b i ∈ Γ (M; so(3, C) P ) is constant and we get the solution where we used ε i jk b j b k = 0. Thus in this solution the field b i ∈ Γ (M; so(3, C) P ) is constant on M = Σ × R. Next we try to find b i in the case of matter located at a point.

Solutions for matter located at a point
In spherical symmetry, the functions K i are given by Therefore in order to get Q i = Q i 0 δ 3 (x), we replace 1/r 4 with 1/(r 4 + ǫ 4 ), for some infinitesimal parameter ǫ → 0 + , and we choose a solution for the field u j like for some function f on M that is needed for satisfying Du i = 0 and a constant vector field a i ∈ Γ (M; so(3, C) P ). With that we obtain (for i, j = k) By that, for r > 0, ǫ → 0 + , we obtain the same solutions as in the equations (39) and it is not sufficient to let the constants c i take arbitrary values, so we choose them to be By that we have given an example for the possibility of solving the equations of motion in BF theory without need using a gravitational metric on M, we just need using a vector field ψ i ∈ Γ (M; so(3, C) P ) which is defined in the spin current J µ k = 2iε kij D ν ψ i Σ µνj of matter, the equation (18). Also we saw that the solutions depend on the symmetry of the system, since we need for obtaining the solutions some vector v that satisfies v a ∂ a K i = 0, v a E i a = 0, v a B ai = 0.

Solutions in cylindrical symmetric system
In cylindrical symmetric system, we let matter be homogeneously located along the Z-axis.
As we did in spherically symmetry, we search for the field ψ i = K i b i ∈ Γ (M; so(3, C) P ), then we calculate v, E i , B i and J 0i according to the equations (35), (36) and (37). The vector v ∈ Γ (M; T Σ) satisfies v a D a ψ i = 0, v a E i a = 0 and v a B i a = 0, thus it is Killing vector. We try to find the solution in the vacuum(u = b) and then apply it for matter located homogeneously along the Z-axis. The needed information for solving the equations of motion is only the spin charge Q i (x), the equation (37). As we mentioned before, we do not need to use a gravitational metric, we just use a standard metric. In cylindrical symmetry, we use the cylindrical coordinates (ρ, ϕ, z) on the space-like slice Σ t = Σ = R 3 of constant time t. We let the vector field ψ i depend only on the radius ρ, we get (for D 2 ψ i = 0 and Db i = 0) so ψ i = c i b i log(ρ), for some constants c i ∈ R. Therefore where in the cylindrical symmetry we let a 1 and a 2 do not depend on the coordinates z 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 cylindrical coordinates, because we do not define any other metric.
Using the equation (36), we get the solutions of the 1-form B i and the vector field E i , By that we get while the other components like Σ i 0z , Σ i 0ϕ , ... are zeros.
We obtain the matrix χ using the equation (26) with the solution (42), where the constants c i have to be determined in order to satisfy the condition trχ = 0(in the vacuum), so 3 i=1 c i = 0. Thus we get the curvature F = χΣ + χ ′Σ (with setting χ ′ = 0 in the vacuum [15]), Using the gauge A i ρ = 0, i = 1, 2, 3, with letting A i depend only on ρ, we obtain As we did in spherical symmetry, we find b i by using the spin charge Q i (x) which is given by the equation (37). Since the system is static and the matter homogeneously located along the Z-axis, the spin charge Q i (ρ, ϕ, z) is given by Here Q i 0 is point charge located at each point of Z-axis.
In order to get a same solution, as in the equations (43) and (44), we keep the field b i be constant, and in the formula (37) we use Q i (x) = Q i 0 δ(ρ)/2πρ.
In cylindrical symmetry, the functions K i are given by K i = c i log(ρ), the equation (42), Therefore in order to get Q i = Q i 0 δ(ρ)/2πρ, we replace 1/ρ 2 with 1/ρρ 1−ǫ , for some infinitesimal parameter ǫ → 0 + , and we choose a solution for the field u j like for some scalar function f on M that is needed for satisfying Du i = 0, with a constant vector field a i ∈ Γ (M; so(3, C) P ). By that we obtain (for i, j = k) By that, for ρ > 0, ǫ → 0 + , we obtain the same solutions as in the equations (43) and it is not sufficient to let the constants c i take arbitrary values, so we choose them to be (c i ) = (1, 1, −2).

Conclusions
We have studied the BF theory including matter by redefinition the 2-form B i , as B i + ε i jk ψ j B k , or redefinition the Lagrangian multipliers ϕ ij as ϕ ij + ε ℓik ψ ℓ χ k j , so that we can get DB i = 0, in the case of non-zero spin current of matter fields. The new field ψ i is defined using the spin current vector J νi = (D µ ψ k )ε ki j Σ µνj . We saw that we can solve the BF equations by using only the spin current of matter, that is it is enough to solve the equations δS/δA i = 0, DB i = 0 and J νi = (D µ ψ k )ε ki j Σ µνj without using a gravitational metric on M and without needing solving the equation δS/δB i = 0 which includes the Lagrangian multiplier ϕ ij (a non-physical variable), so we get ϕ ij by using the solutions in δS/δB i = 0. We found that to obtain the solutions of BF theory, it is enough to use(find) the field ψ i and the Killing vector v(satisfies v a D a ψ i = 0) in euclidean coordinates, where it is convenient to describe the spin currents and their lines in euclidean coordinates and not need to describe them in curved coordinates. Also it is possible to obtain the solutions of BF theory using only the charges J 0i = 0 when they are given as a functions on M in the static case (discussion below the equation (37)). We saw that the singularities appear in solution of ψ i , that relates to the idea that the spin current J µi is the source for ψ i , therefore ψ i has singularities on line of that spin current, and by that the singularities appear and not by using a gravitational metric. We found that the solutions of BF theory equations depend on the symmetry of the system and every two solutions of (Σ i ab , Σ i 0a ) and (Σ abi , Σ 0ai ) determine a metric(remark 4), so there is no specific metric needed in BF theory because it is a topological theory, and those solutions are able to be written as e I ∧ e J (remark 5). Finally we applied the solutions of BF theory in a spherical and cylindrical symmetric systems in static case of matter.

Acknowledgements
I am grateful to Professor Yoshihiro Fukumoto at Ritsumeikan University for many useful notes on the manuscript.
These constraints are generators of gauge symmetry and diffeomorphism invariance up to boundary terms(ignoring the boundary terms). We get D a E ai = 0 from ε µνρσ D ν B i ρσ = 0 by setting µ = 0 to get ε 0abc D a B i bc = 0, and using ε 0abc ≡ ε abc to get ε abc D a B i bc = 2D a E ai = 0, where E ai = ε abc B i bc /2 is conjugate to the connection A i a on space-like slice of constant time on which we use the coordinates (x a ) a=1,2,3 . Thus the constraint D a E ai = 0 is satisfied in BF theory.
Using the equation (22), in the vacuum χ ′ i j = 0, it reduces to F i (A) = χ i j Σ j , on the spacelike surface of constant time Σ t , it becomes F i (A) ab = χ i j Σ j ab . Using E ai = ε abc Σ i bc , we obtain Multiplying it by E a i and summing over the contracted indices, we get where we used the fact that the matrix χ ij is symmetric. Therefore the second constraint of (46) is satisfied.
Then using E a i = ee a i , where e a i is the inverse of the gravitational field e i a and e = det(e i a ), we obtain But in the vacuum trχ = 0(the equation (23)), thus C = ε ijk E ai E bj F k ab = 0 is satisfied. By that we find that the general relativity constraints are satisfied in the vacuum using the equations of motion of BF theory. As required, the general relativity equations do not include the matrices χ and ϕ and do not include the new field ψ i . II-A principal G-bundle consists of the following data: 1-a manifold P , called the total space, 2-a Lie group G acting freely on P on the right: The free action means that the stabilizer of every point is trivial, that every element of G (except the identity) moves every point in P. We assume that the space of orbits M = P/G is a manifold (the base space). With projection π : P → M and for every p ∈ M, the submanifold π −1 (p) ⊂ P is fibre over M. Let {U α } be open cover of M, the local trivialization is G-equivariant diffeomorphisms given by ψ α (p) = (π(p), g α (p)) for some G-equivariant map g α : π −1 (U α ) → G. Equivariance means that g α (pg) = g α (p)g. We say that the bundle is trivial if there exists a diffeomorphism ψ : P → M × G such that ψ(p) = (π(p), ψ(p)) and such that ψ(pg) = ψ(p)g. This last condition is simply the G-equivariance of ψ.
We separate T p P to vertical and horizontal vector spaces at each point p ∈ P , we get the vertical vector fields by acting of group G on P by for every vector X ∈ g (where g is Lie algebra of G), this satisfies π * σ p (X) = d dt π pe tX t=0 = d dt (π (p))| t=0 = 0, thus σ p (X) is vertical vector field at p ∈ P . In this bundle the connection is defined as a map D : Γ(T P ) → Γ(T e G), T e G = lie(G), as usual, the connection is a map from the tangent space to itself. And since P is locally We can get this map by letting D = ∇ + A, for A ∈ Γ(T * U ⊗ T e G), and ∇ is connection on T U.
where T µν is energy-momentum tensor, and S m is matter action without specifying. Using We use the Urbantke formula [12] from which we get We use it for calculating I 1 in (48), To calculate I 2 , we use By this the equation (47) becomes Using T µ1ν1 = T IJ e µ1 I e ν1 J , the first term of (51) becomes We use the property of the self-dual projection − 1 2 ε ijℓ P i IJ P j KL = 1 4 (η JK P ℓIL − η JL P ℓIK ) − 1 4 (I ↔ J) , which can be easily checked when I = 0 and J, K, L are spatial indices, and when I = K = 0 and J, L are spatial indices, so the SO(3, 1) × SO(3, C) invariance asserts that this property is also satisfied when I, J, K, L are all spatial indices. By using this property, we obtain 2I 1 = T IJ ε µνσ1ρ1 −1 2 (η LK1 P ℓIJ − η LJ P ℓIK1 − η IK1 P ℓLJ + η IJ P ℓLK1 ) e L µ e K1 ν = T IJ ε µνσ1ρ1 −1 2 P ℓIJ g µν − η LJ P ℓIK1 e L µ e K1 ν − η IK1 P ℓLJ e L µ e K1 ν + η IJ Σ ℓµν Finally We use the selfdual property P i IJ ε IJKL = −2iP iKL , to get And using 1 2! e −1 ε µνρσ Σ i ρσ = * Σ i µν = −iΣ i µν = −iΣ µνi , e = √ −g, we obtain We get the third term of (51) by the replacing σ 1 ↔ ρ 1 in I 2 with reversing its sign, we obtain The fourth term of (51) is Using T µν g µν = T IJ η IJ = T , the equation (51) the complex matrices T and ξ are given by and ξ ij = 1 4 T IJ (η LJ P iIK + η IK P iLJ )P LK j + 1 2 The self-dual projection matrices P i IJ are given in the equation (1), andP i IJ are their complex conjugate. We get P iIK P K jJ T IJ = P iI0 P 0 jJ + P iIℓ P ℓ jJ T IJ = P iℓ0 P 0 jm T ℓm + P i0ℓ P ℓ j0 T 00 + P imℓ P ℓ j0 T m0 + P i0ℓ P ℓ jm T 0m + P inℓ P ℓ jm T nm = −P iℓ0 P jm0 T ℓm + P i0ℓ P ℓ j0 T 00 + P imℓ P ℓ j0 T m0 + P i0ℓ P ℓ jm T 0m + P inℓ P ℓ jm T nm Then using Σ i µν = P i IJ e I µ e J ν , implies −iΣ µνj Σ µνi = 1 2 e −1 ε µνρσ P jIJ P i KL e I µ e J ν e K ρ e L σ .
Therefore D µ M µIJ = 0 is satisfied when D µ e I ν = 0 and D µ T µI = 0. Here e I µ and T µI are functions of the coordinates (x µ ), so also M µIJ are functions of (x µ ). 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). When the coordinates (x µ ) are flat, so e I µ = δ I µ , by that the equation (58) becomes (57).