# Spin current in BF theory

Malik Almatwi

(Department of Mathematical Science, Ritsumeikan University, 6-1-1 Matsugaoka,
Otsu, Shiga 520-2102, Japan
email<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
  2. 2 Spin current in BF theory
  3. 3 Solutions in spherical symmetric system
    1. 3.1 Solutions for matter located at a point
  4. 4 Solutions in cylindrical symmetric system
  5. 5 Conclusions
  6. 6 Acknowledgements
  7. 7 Appendix A
  8. 8 Appendix B: Calculating $\frac{{\delta S_{m}}}{{\delta\Sigma^{i}}}$
  9. 9 Appendix C

## Abstract

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.

Keywords: BF theory; local Lorentz symmetry; local Lorentz currents.

##  1 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 $B\in\Omega^{2}(M;\mathfrak{so}(3,1))$ and spin connection $\omega$
which takes values in $\mathfrak{so}(3,1)$, all derivatives are linear and
applied only on $\omega$, 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
$F(\omega)\in\Omega^{2}(M;\mathfrak{so}(3,1))$ be the curvature of $\omega$.
The pure BF theory action is $\int\limits_{M}tr({B\wedge F(\omega)})$ which is
invariant(symmetric) under local Lorentz transformation(regarded as gauge
group) and under arbitrary diffeomorphisms of $M$ and does not need using a
metric. The equations of motion are $F(\omega)=0$ and $d_{\omega}B=0$, thus
$B$ defines a twisted de Rham cohomology class $[B]\in
H^{2}_{DR}(M,\mathfrak{so}(3,1))$, and the solution of $F(\omega)=0$ is unique
up to gauge and diffeomorphism transformations. 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].

For example, let $f:M\to M$ be an infinitesimal diffeomorphism generated by a
vector field $v$, then $B\mapsto B+d_{\omega}(i_{v}(B))$ for $d_{\omega}B=0$.
Therefore $d_{\omega}B\mapsto d_{\omega}B+d^{2}_{\omega}(i_{v}(B))$, but
$F(\omega)=0$ implies $d^{2}_{\omega}=0$, thus the equation of motion
$d_{\omega}B=0$ maps to the equation of motion $d_{\omega}B=0$, therefore all
equations of motion are equivalent under diffeomorphisms of $M$. But when
$F(\omega)\neq 0$, then $d^{2}_{\omega}\neq 0$, therefore the equation of
motion $d_{\omega}B=0$ maps to the equation of motion $d_{\omega}B\neq 0$, so
there are local degrees of freedom. That relates to the fact that $d_{\omega}$
acts locally when $F(\omega)\neq 0$, in other words $f^{*}(d_{\omega})\neq
d_{\omega}$ when $F(\omega)\neq 0$, but when $F(\omega)=0$, then
$f^{*}(d_{\omega})=d_{\omega}$. In QFT, using fields(not forms), the action
changes(under $f$) as $\delta S=\int\partial_{\mu}(g_{\nu\rho}\delta
x^{\rho}\delta S/\delta g_{\mu\nu})$ when the equation of motion are
satisfied. Therefore in order to get TQFT, it must be $\delta S/\delta
g_{\mu\nu}=0$, so independent of any metric(standard or other).

In constrained BF, the Lagrangian includes the constraint term
$\varphi_{IJKL}B^{IJ}\wedge B^{KL}$. The traceless matrix $\varphi$ 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
$e^{I}=e^{I}_{\mu}dx^{\mu}$, that is $B^{IJ}=e^{I}\wedge e^{J}$, where
$I,J,...=0,1,2,3$ are Lorentz indices and $\mu,\nu,...=0,1,2,3$ are spacetime
tangent indices, we regard the frame fields $e^{I}_{\mu}dx^{\mu}$ as
gravitational fields, therefore the constrained BF theory turns to general
relativity theory, the reason is that when $\varphi_{IJKL}$ is not zero, the
term $\varphi_{IJKL}B^{IJ}\wedge B^{KL}$ 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 $\delta
S/\delta B=0$ contains the non-physical variable $\varphi_{IJKL}$, but we can
remove it by taking the trace of the equations($\varphi$ 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 $\delta
S/\delta\omega=0$(as done in this paper). In general, the equations of motion
of constrained BF theory including matter give a relation between the
curvature $F^{IJ}(\omega)$ and the frame fields $\Sigma^{IJ}=e^{I}\wedge
e^{J}$(the Plebanski 2-form), in matrix notation, that is
$F=\chi\Sigma+\xi\bar{\Sigma}$, where $\chi$ and $\xi$ are symmetric matrices
of scalar fields [4]. Therefore the problem turns to finding $\chi$ and $\xi$.
Since the field $\varphi_{IJKL}$ 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 $d_{\omega}B$ by the equation $*(d_{\omega}B)+J=0$, where
’$*$’ is Hodge star operator. And in order to get $d_{\omega}B=0$ in this
paper, we add a new term to BF Lagrangian, like $tr\left(\psi B\wedge
F(\omega)\right)$, using a new field $\psi$. That can be seen as a
redefinition $B\to B+\psi B$, by which the equation $*(d_{\omega}B)+J=0$
becomes $*(d_{\omega}B)+*(d_{\omega}(\psi B))+J=0$, so we choose
$d_{\omega}B=0$ and get $*((d_{\omega}\psi)B)+J=0$. Therefore the spin current
becomes a source for the field $\psi$ instead of $B$ and we get a new
formula(definition) for the spin current using $\psi$, and since the spin
current regards symmetry of the system, the field $\psi$ also regards that
symmetry. In constraint BF theory, solving the equation
$*((d_{\omega}\psi)B)+J=0$ is easy as we will see. We find that the equation
of motion of $\psi$ is same conservation equation $D_{\mu}J^{\mu}=0$ 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 $\delta S/\delta\omega=0$,
$J\neq 0$ with $d_{\omega}B=0$ and without needing solving the equation
$\delta S/\delta B=0$ which includes the Lagrangian multiplier
$\varphi_{IJKL}$(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 $\int\limits_{M}\omega^{IJ}_{\mu}J^{\mu}_{IJ}$ which makes them
easy to solve, and makes the theory similar to the gauge theory. And since
$\omega^{IJ}$ is 1-form and $J_{IJ}$ is vector field, the term
$\omega^{IJ}_{\mu}J^{\mu}_{IJ}$ 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 $\psi$, this field has singularities on the lines of that spin current.
We see that our solution of $B^{IJ}$ can be always written as $e^{I}\wedge
e^{J}$, 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 $\psi$ using the spin charges
$J^{0}_{IJ}$.

##  2 Spin current in BF theory

Let $M$ be connected oriented smooth 4-manifold and $P\to M$ be an
$SO(3,1)$-principal bundle(Appendix A) with a spin connection $\omega$ which
is locally a 1-form with values in $\mathfrak{so}(3,1)$ and
$F\in\Omega^{2}(M;\mathfrak{so}(3,1)_{P})$ 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.

######  Definition 1.

The self-dual projection is a homomorphism

$\mathfrak{so}(3,1)_{P}=P\times_{SO(3,1)}\mathfrak{so}(3,1)\to\mathfrak{so}(3,\mathbb{C})_{P}=P\times_{SO(3,1)}\mathfrak{so}(3,\mathbb{C})$

defined by

$\left({P_{IJ}^{i}}\right):\mathfrak{so}(3,1){\hookrightarrow}\mathfrak{so}(3,1)_{\mathbb{C}}=\mathfrak{so}(3,\mathbb{C})\oplus\mathfrak{so}(3,\mathbb{C})\to\mathfrak{so}(3,\mathbb{C}),$

with using the matrices [5]

$P^{i}_{IJ}=\frac{1}{2}\varepsilon^{i}{{}_{jk}},\text{ for }I=i,J=j,\text{ and
}P^{i}_{0j}=-P^{i}_{j0}=-\frac{i}{2}\delta^{i}_{j},\text{ for }I=0,J=j\neq 0.$
(1)

We used
$P\times_{SO(3,1)}\mathfrak{so}(3,1)={{\left({P\times\mathfrak{so}(3,1)}\right)}\mathord{\left/{\vphantom{{\left({P\times\mathfrak{so}(3,1)}\right)}{SO(3,1)}}}\right.\kern-1.2pt}{SO(3,1)}}$,
which is locally isomorphic to $U_{\alpha}\times\mathfrak{so}(3,1)$ for open
sets $\\{U_{\alpha}\\}$ of $M$.

That self-dual projection relates to the fact that the complexified Lie
algebra of $SO(3,1)$ has the decomposition
$\mathfrak{so}(3,1)_{\mathbb{C}}=\mathfrak{so}(3,{\mathbb{C}})\oplus\mathfrak{so}(3,{\mathbb{C}})$
[6]. The new connection is locally an $\mathfrak{so}(3,\mathbb{C})$-valued
1-form $A$ on $M$ whose components are

$A^{i}_{\mu}=P^{i}_{IJ}\omega^{IJ}_{\mu}=\frac{1}{2}\varepsilon{{}^{i}}_{jk}\omega^{jk}_{\mu}-i\omega^{0i}_{\mu},\quad
i=1,2,3,$ (2)

and its curvature is

$F^{i}(A)=P^{i}_{IJ}F^{IJ}(\omega)=dA^{i}+\varepsilon{{}^{i}}_{jk}A^{j}\wedge
A^{k}.$ (3)

The two form $B^{IJ}$ is mapped to $B^{i}=P^{i}_{IJ}B^{IJ}$. The covariant
derivative $D_{\mu}=\nabla_{\mu}+\omega_{\mu}^{IJ}$ acting on sections of
$TM\otimes\mathfrak{so}(3,1,\mathbb{C})_{P}$ becomes
$D_{\mu}=\nabla_{\mu}+A_{\mu}$, with
$A_{\mu}^{ij}=\varepsilon{{}^{ij}}_{k}A^{k}_{\mu}$, where $\nabla_{\mu}$ is
the affine connection on $TM$.

Using the new variables we can write the Lagrangian of matter(without
specifying matter fields) $L_{matter}(e^{I},\omega^{IJ})$ as
$L_{matter}(B^{i},A^{i},\bar{B}^{i},\bar{A}^{i})$, where $\bar{B}^{i}$ and
$\bar{A}^{i}$(anti-selfdual representation) are the complex conjugation of
$B^{i}$ and $A^{i}$. The Urbantke formula (equation 49, appendix B) writes the
metric $g_{\mu\nu}$ using only the constrained $B^{i}$ without using the
constrained $\bar{B}^{i}$. And the self-dual connection ${A}^{i}$ is
compatible with $B^{i}$ via $d_{A}B^{i}=0$, while the anti-self-dual
connection ${\bar{A}}^{i}$ is compatible with $\bar{B}^{i}$ via
$d_{\bar{A}}\bar{B}^{i}=0$. By that we may suppose

$\frac{\delta}{{\delta\bar{B}^{i}}}L_{matter}=0,\quad\frac{\delta}{{\delta\bar{A}^{i}}}L_{matter}=0,$
(4)

or just writing $L_{matter}(B^{i},A^{i})$.

######  Definition 2.

Let $A$ be be the self-dual connection on the
$\mathfrak{so}(3,\mathbb{C})$-bundle $\mathfrak{so}(3,\mathbb{C})_{P}\to M$.
Let $L_{matter}$ be the Lagrangian of matter fields on $M$. Then the spin
current $J_{i}^{\mu}$ is defined to be

$J_{i}^{\mu}=\frac{\delta}{{\delta A_{\mu}^{i}}}L_{matter}.$

The matter action $S_{matter}$ is required to be invariant under any
infinitesimal local Lorentz transformation
$\omega_{\mu}^{IJ}\mapsto\omega_{\mu}^{IJ}+D_{\mu}\Lambda^{IJ}$ for
infinitesimal transformation parameter
$\Lambda^{IJ}\in\Omega^{0}(M;\mathfrak{so}(3,1)_{P})$. Now we assume
$S_{matter}$ has this property. Then we have the following.

######  Lemma 1.

The spin current $J_{i}^{\mu}$ given by $J_{i}^{\mu}=\frac{\delta}{{\delta
A_{\mu}^{i}}}L_{matter}$ in gravity theory is conserved [7].

###### Proof.

Since $S_{matter}$ is invariant under infinitesimal gauge transformation
$\Lambda^{IJ}$, it is invariant under (we may suggest the condition (4))

$A_{\mu}^{i}\mapsto A_{\mu}^{i}+D_{\mu}\Lambda^{i}\quad
for\quad\Lambda^{i}=P^{i}_{IJ}\Lambda^{IJ}\in\Omega^{0}(M;\mathfrak{so}(3,\mathbb{C})_{P}).$

The variation

$\begin{split}S_{matter}&\left({A^{i}+D\Lambda^{i}}\right)-S_{matter}\left({A^{i}}\right)=\int\limits_{M}d^{4}x{\left({D_{\mu}\Lambda^{i}}\right)\frac{\delta}{{\delta
A_{\mu}^{i}}}}L_{matter}\\\
&=-\int\limits_{M}d^{4}x{\Lambda^{i}D_{\mu}\left({\frac{\delta}{{\delta
A_{\mu}^{i}}}L_{matter}}\right)}+\int\limits_{\partial
M}{d^{3}x^{\mu}\Lambda^{i}\frac{\delta}{{\delta
A_{\mu}^{i}}}L_{matter}}\end{split}$

vanishes for arbitrary $\Lambda^{i}$ only when
$D_{\mu}\left(\frac{\delta}{{\delta A_{\mu}^{i}}}L_{matter}\right)=0$, where
we let $\Lambda^{i}$ vanish on the boundary $\partial M$. Thus the current
$J_{i}^{\mu}=\frac{\delta}{{\delta A_{\mu}^{i}}}L_{matter}$ is conserved.
Actually the previous calculation based on the idea that ${A}^{i}$ and
${\bar{A}}^{i}$ transform independently under infinitesimal local Lorentz
transformation
$\omega_{\mu}^{IJ}\mapsto\omega_{\mu}^{IJ}+D_{\mu}\Lambda^{IJ}$, therefore
there is another current that associates with the connection ${\bar{A}}^{i}$
when the matter Lagrangian depends also on ${\bar{A}}^{i}$. ∎

Same thing we find for the general relativity action, by using the variables
$(\Sigma^{i},A^{i})$, we obtain the equation

$\int\limits_{M}d^{4}x{\left({D_{\mu}\Lambda^{i}}\right)\frac{\delta}{{\delta
A_{\mu}^{i}}}}S_{GR}=0\Rightarrow-\int\limits_{M}d^{4}x{\Lambda^{i}D_{\mu}\frac{\delta}{{\delta
A_{\mu}^{i}}}}S_{GR}=0.$

In $3+1$ decomposition of the space-time manifold $M=\Sigma\times\mathbb{R}$,
let $\Sigma_{t}$ 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 $(E_{i}^{a},A_{a}^{i},A_{0}^{i})$ on the slice
of constant time $\Sigma_{t}$, that equation becomes

$\begin{split}&-\int\limits_{\Sigma\times\mathbb{R}}d^{4}x\Lambda^{i}D_{a}\left(\frac{\delta}{{\delta
A_{a}^{i}}}S_{GR}\right)-\int\limits_{\Sigma\times\mathbb{R}}d^{4}x\Lambda^{i}D_{0}\left(\frac{\delta}{{\delta
A_{0}^{i}}}S_{GR}\right)\\\
&=-constant\times\int\limits_{\Sigma\times\mathbb{R}}d^{4}x\Lambda^{i}\left(D_{a}E^{a}_{i}+D_{0}(D_{a}E^{a}_{i})\right)=0,\end{split}$
(5)

which is satisfied when $D_{a}E^{a}_{i}=0$, where $E^{a}_{i}$ is conjugate
momentum to $A_{a}^{i}$, and we used the relations
$E^{a}_{i}=constant\times\frac{\delta}{{\delta A_{a}^{i}}}S_{GR}$ and
$D_{a}E^{a}_{i}=constant\times\frac{\delta}{{\delta A_{0}^{i}}}S_{GR}$ [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 $J_{i}^{\mu}$ is similar to the currents in Yang-
Mills theory of the gauge fields, we see this clearly when we regard the
connection $A_{\mu}^{i}$ as a gauge field, by that the current $J_{i}^{\mu}$
relates to the local Lorentz invariance (local symmetry). The metric
$g_{\mu\nu}=e^{I}_{\mu}e^{J}_{\nu}\eta_{IJ}$ is invariant under arbitrary
local Lorentz transformations, like $e^{I}_{\mu}(x)\mapsto
U^{I}{{}_{J}}(x)e^{J}_{\mu}(x)$, for $U(x)\in SO(3,1)$, therefore the local
Lorentz symmetry is an internal degree of freedom.

######  Definition 3.

The action of BF theory including matter(without cosmological constant) on
$SO(3,1)$-principal bundle $P\to M$ is defined to be [10]

$S=S_{topological}+S_{constraints}+S_{matter},$

with

$S_{topological}=\int\limits_{M}{B_{i}\wedge F^{i}(A)},\quad\text{and}\quad
S_{constraints}=\frac{1}{2}\int\limits_{M}{\varphi_{ij}B^{i}\wedge B^{j}},$

where
$\varphi\in\Gamma\left(M;\rm{End}(\mathfrak{so}(3,\mathbb{C})_{P})\right)$ is
traceless matrix of scalar fields $\varphi_{ij}$, 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 $\mathfrak{so}(3,\mathbb{C})_{P}$ which is locally a 1-form with values
in $\mathfrak{so}(3,\mathbb{C})$ and its curvature
$F(A)\in\Omega^{2}\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$ are defined
in the equations (2) and (3). The index contraction is done by using
$\delta_{ij}$, the Killing form on $\mathfrak{so}(3,\mathbb{C})$.

Hence

$S=\int\limits_{M}{\left({B_{i}\wedge
F^{i}(A)+\frac{1}{2}\varphi_{ij}B^{i}\wedge B^{j}}\right)}+S_{matter}.$ (6)

Since the matrix $\varphi$ is traceless, we can write
$\varphi_{ij}=m_{ij}-(m_{11}+m_{22}+m_{33})\delta_{ij}/3$, 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

$B^{i}\wedge B^{j}=\frac{1}{3}\delta^{ij}B_{i}\wedge B^{j}.$

The solutions to this are all of the following form
$B^{i}=P^{i}_{IJ}e^{I}\wedge e^{J}$, in which the gravitational fields
$e^{I}_{\mu}$ are considered as frame fields [11]. Using the self-dual formula
(1), the constrained 2-form $B^{i}$ is written as

$B^{i}=\frac{1}{2}\varepsilon^{i}{{}_{jk}}e^{j}\wedge e^{j}-ie^{0}\wedge
e^{i}=\Sigma^{i},$

this is $\left.{B^{i}}\right|_{constrained}=\Sigma^{i}$, with using the
notation $\Sigma^{i}=P^{i}_{IJ}e^{I}\wedge e^{J}$.

The equation of motion with respect to $B^{i}$ is

$F^{i}(A)+\varphi^{i}{{}_{j}}B^{j}+\frac{{\delta S_{matter}}}{{\delta
B_{i}}}=0,$

or

$F^{i}(A)=-\varphi^{i}{{}_{j}}B^{j}-\frac{{\delta S_{matter}}}{{\delta
B_{i}}}.$ (7)

Since $F^{i}(A)\in\Omega^{2}\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$ is
2-form with values in $\mathfrak{so}(3,\mathbb{C})$, the
$\frac{{\delta}}{{\delta B_{i}}}S_{matter}$ is also 2-form with values in
$\mathfrak{so}(3,\mathbb{C})$.

######  Lemma 2.

In constrained $B^{i}$, the variation $\frac{{\delta}}{{\delta
B_{i}}}S_{matter}\in\Omega^{2}\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$
has the form

$\left.{\frac{{\delta S_{matter}}}{{\delta
B_{i}}}}\right|_{constraint}=T^{i}{{}_{j}}\Sigma^{j}+\xi^{i}{{}_{j}}\bar{\Sigma}^{j},$

for some matrices
$T^{i}{{}_{j}},\xi^{i}{{}_{j}}\in\Gamma\left(M;\rm{End}(\mathfrak{so}(3,\mathbb{C})_{P})\right)$,
with $T^{i}{{}_{j}}=T{{}_{j}}^{i}$ and $\xi^{i}{{}_{j}}=\bar{\xi}{{}_{j}}^{i}$
(See appendix B, for more details).

Therefore in the vacuum we set $T^{i}{{}_{j}}=0$. Using this formula in the
equation (7) implies

$F^{i}(A)=-\varphi^{i}{{}_{j}}\Sigma^{j}-T^{i}{{}_{j}}\Sigma^{j}-\xi^{i}{{}_{j}}\bar{\Sigma}^{j},$

or

$F^{i}(A)=\psi^{i}{{}_{j}}\Sigma^{j}-\xi^{i}{{}_{j}}\bar{\Sigma}^{j},$ (8)

for some matrix
$\psi=-\varphi-T\in\Gamma\left(M;\rm{End}(\mathfrak{so}(3,\mathbb{C})_{P})\right)$
[12].

Since $tr(\varphi)=0$, so $tr(\psi)=-tr(T)$, the equation (8) yields

$\Sigma_{i}^{\mu\nu}F^{i}_{\mu\nu}=-tr(T),\quad\text{ for
}\quad\Sigma^{i}_{\mu\nu}\Sigma_{j}^{\mu\nu}=\delta^{i}_{j},\text{ and
}\Sigma^{i}_{\mu\nu}\overline{\Sigma}_{j}^{\mu\nu}=0.$ (9)

Thus in the vacuum $T^{i}{{}_{j}}=0$, we have
$\Sigma_{i}^{\mu\nu}F^{i}_{\mu\nu}=0$. We call
$\Sigma_{i}^{\mu\nu}F^{i}_{\mu\nu}$ the $trF$. The equation (9) does not
contain the non-physical variable $\varphi$, but the problem with it is that
the trace process decreases the number of equations. Therefore
$\Sigma_{i}^{\mu\nu}F^{i}_{\mu\nu}=-tr(T)$ is a condition on the solutions.
The (0,2) tensor $\Sigma_{j}^{\mu\nu}$ is inverse of the 2-form
$\Sigma^{i}_{\mu\nu}$ (Appendix C).

The equation of motion with respect to the connection $A^{i}$ is

$DB^{i}+\frac{\delta}{{\delta A^{i}}}S_{matter}=0,$

or

$\varepsilon^{\mu\nu\rho\sigma}D_{\nu}B_{\rho\sigma}^{i}+J^{\mu i}=0.$ (10)

We see that we can not choose
$\varepsilon^{\mu\nu\rho\sigma}D_{\nu}B_{\rho\sigma}^{i}=0$ when $J^{\mu
i}\neq 0$. But the condition
$\varepsilon^{\mu\nu\rho\sigma}D_{\nu}B_{\rho\sigma}^{i}=0$ leads to the
equations $D_{a}E^{ai}=0$(appendix A) and $De^{I}=0$, which makes the spin
connection $\omega^{IJ}$ 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 $B^{i}\to
B^{i}+\varepsilon^{i}{{}_{jk}}\psi^{j}B^{k}$, so the equation (10) yields

$\varepsilon^{\mu\nu\rho\sigma}\varepsilon^{i}{{}_{jk}}(D_{\nu}\psi^{j})B^{k}_{\rho\sigma}+J^{\mu
i}=0,\quad DB^{i}=0.$ (11)

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 $\psi^{i}$ satisfies the equation $D^{2}\psi^{i}=0$
(equation (34)) with respect to a metric that satisfies
$\nabla_{\rho}g_{\mu\nu}=0$.

By acting by $D_{\mu}$ on the equation (10), we get

$\varepsilon^{\mu\nu\rho\sigma}D_{\mu}D_{\nu}B_{\rho\sigma}^{i}+D_{\mu}J^{\mu
i}=0,$

and using $D_{\mu}J^{\mu i}=0$, we obtain

$\varepsilon^{\mu\nu\rho\sigma}D_{\mu}D_{\nu}B_{\rho\sigma}^{i}=\varepsilon^{\mu\nu\rho\sigma}[D_{\mu},D_{\nu}]B_{\rho\sigma}^{i}/2=0,$

but $[D_{\mu},D_{\nu}]=F_{\mu\nu}(A)$, therefore

$\varepsilon^{\mu\nu\rho\sigma}(F_{\mu\nu}(A))^{i}{{}_{j}}B_{\rho\sigma}^{j}=0.$

Then using $(F_{\mu\nu}(A))_{ij}=\varepsilon_{ijk}F^{k}_{\mu\nu}(A)$, implies

$\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}F^{j}_{\mu\nu}(A)B_{\rho\sigma}^{k}=0.$
(12)

We can regard the equation (12) as an equation of motion with respect to a new
field $\psi^{i}$, with the possibility of choosing $DB^{i}=0$ with $J^{\mu
i}\neq 0$.

In order to include the equations $DB^{i}=0$, $J^{\mu i}\neq 0$ and
$D_{\mu}J^{\mu i}=0$ in BF theory, we suggest the following action.

######  Definition 4.

We add a new term to the BF action (6) to get

$S=\int\limits_{M}{\left({B_{i}\wedge
F^{i}(A)+\frac{1}{2}\varphi_{ij}B^{i}\wedge
B^{j}}\right)}+\int\limits_{M}{\varepsilon_{ijk}\psi^{i}B^{j}\wedge
F^{k}(A)}+S_{matter},$ (13)

in which we have added $\int\limits_{M}{\varepsilon_{ijk}\psi^{i}B^{j}\wedge
F^{k}(A)}$, for some vector field
$\psi^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$. We can get
the new term by a redefinition $B^{i}\to
B^{i}+\varepsilon^{i}{{}_{jk}}\psi^{j}B^{k}$ in pure BF Lagrangian
$B_{i}\wedge F^{i}(A)$. Also we can get it by a redefinition of $\varphi_{ij}$
as
${\varphi^{\prime}}_{ij}B^{i}=\varphi_{ij}B^{i}+\varepsilon_{ijk}\psi^{i}F^{k}(A)$,
and by using $F^{i}(A)=\chi^{i}{{}_{j}}B^{j}+\xi^{i}{{}_{j}}\bar{B}^{j}$ with
$B^{i}\wedge\bar{B}^{j}=0$, we obtain
${\varphi^{\prime}}_{ij}=\varphi_{ij}+\varepsilon_{\ell
ik}\psi^{\ell}\chi^{k}{{}_{j}}$. There is no problem with the redefinition of
$\varphi_{ij}$ since they are just Lagrangian multipliers.

The equation of motion of this action with respect to the field $\psi^{i}$ is

$\varepsilon_{ijk}\varepsilon^{\mu\nu\rho\sigma}B^{j}_{\mu\nu}F^{k}_{\rho\sigma}(A)=0,$
(14)

which is same equation (12), therefore the field $\psi^{i}$ does not change
the equations of motion. By using
$(F_{\mu\nu}(A))_{ij}=\varepsilon_{ijk}F^{k}_{\mu\nu}(A)$, we get

$\varepsilon^{\mu\nu\rho\sigma}B^{j}_{\mu\nu}(F_{\rho\sigma}(A))^{i}{{}_{j}}=0,$

but $[D_{\mu},D_{\nu}]=F_{\mu\nu}(A)$. Therefore

$\varepsilon^{\mu\nu\rho\sigma}[D_{\rho},D_{\sigma}]B^{i}_{\mu\nu}=0,$

but we choose
$\varepsilon^{\mu\nu\rho\sigma}D_{\rho}B^{i}_{\mu\nu}=0$$(DB^{i}=0)$ as we
suggested before, thus

$\frac{{\delta}}{{\delta\psi^{i}}}S_{matter}=0$

is satisfied. Also the equation (14) is satisfied in constraint BF theory by
letting the matrix $\chi_{ij}$ in
$F^{i}(A)=\chi^{i}{{}_{j}}B^{j}+\xi^{i}{{}_{j}}\bar{B}^{j}$ be symmetric.

The equation of motion of this action with respect to the connection $A^{k}$
is

$DB^{k}+D\left({\varepsilon^{k}{{}_{ij}}\psi^{i}B^{j}}\right)+\frac{\delta}{{\delta
A^{k}}}S_{matter}=0,$ (15)

or

$\varepsilon^{\mu\nu\rho\sigma}D_{\nu}B_{\rho\sigma}^{k}+\varepsilon^{\mu\nu\rho\sigma}D_{\nu}\left({\varepsilon^{k}{{}_{ij}}\psi^{i}B_{\rho\sigma}^{j}}\right)+\frac{\delta}{{\delta
A_{\mu}^{k}}}S_{matter}=0.$

We choose $DB^{i}=0$, which implies $De^{I}=0$ in constrained BF theory, since
$\left.{B^{i}}\right|_{constraint}=\Sigma^{i}=P^{i}_{IJ}e^{I}\wedge e^{J}$.
Therefore $\varepsilon^{\mu\nu\rho\sigma}D_{\nu}B_{\rho\sigma}^{k}=0$, which
is equivalent to $D_{\mu}\left(e\Sigma^{\mu\nu i}\right)=0$ in constrained
$B^{i}$, because of the self-duality [13, 14]

$\frac{1}{{2!}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\rho\sigma}^{i}=\left({*\Sigma^{i}}\right)^{\mu\nu}=\left({-i\Sigma^{i}}\right)^{\mu\nu}=-i\Sigma^{\mu\nu
i},\quad e=\det(e_{\mu}^{I}),$ (16)

where we used the Hodge duality theory between the forms and the tensor
fields, here $\Sigma^{i}_{\mu\nu}$ is 2-form and $\Sigma^{i\mu\nu}$ is (0,2)
tensor field. The field $\Sigma^{\mu\nu}_{i}$ is inverse of
$\Sigma^{i}_{\mu\nu}$, that is
$\Sigma^{\mu\nu}_{j}\Sigma^{i}_{\mu\nu}=\delta^{i}_{j}$ (see the appendix C
for more details). By re-scaling $\Sigma^{\mu\nu i}$ by $e=\det(e_{\mu}^{I})$,
we write

$D_{\mu}\Sigma^{\mu\nu i}=0.$ (17)

The remaining equation of (15) in constrained $B^{i}$ is

$D\left({\varepsilon^{k}{{}_{ij}}\psi^{i}\Sigma^{j}}\right)+\frac{\delta}{{\delta
A^{k}}}S_{matter}=0,$

or

$\varepsilon^{\mu\nu\rho\sigma}\varepsilon^{k}{{}_{ij}}\left({D_{\nu}\psi^{i}}\right)\Sigma_{\rho\sigma}^{j}+\frac{\delta}{{\delta
A_{\mu}^{k}}}S_{matter}=0,$

hence

$-2i\varepsilon^{k}{{}_{ij}}\left(D_{\nu}\psi^{i}\right)\Sigma^{\mu\nu
j}+J^{\mu k}=0,$ (18)

in which we used the spin current $J^{\mu}_{i}=\delta S_{matter}/\delta
A_{\mu}^{i}$ and the condition $D_{\nu}\Sigma^{\mu\nu j}=0$, also re-scaled
$\Sigma^{\mu\nu j}$ by $e=\det(e_{\mu}^{I})$. Later we will discuss
$D_{\mu}J^{\mu k}=0$. Here both $\Sigma^{\mu\nu j}$ and $J^{\mu k}$ 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 $J^{\mu
n}=\left({\partial^{\mu}\varphi^{i}}\right)T_{ij}^{n}\varphi^{j}$ in scalar
field theory with symmetry and generators $T_{ij}^{n}$, we have
$J^{0n}=\left({\partial^{0}\varphi^{i}}\right)T_{ij}^{n}\varphi^{j}=\pi^{i}T_{ij}^{n}\varphi^{j}$,
where $\pi^{i}$ is conjugate momentum to $\varphi_{i}$. Like that, the
equation (18) gives
$J^{0i}=2i(D_{\mu}\psi_{k})\varepsilon{{}^{ki}}_{j}\Sigma^{\mu 0j}$(for
$\mu=a=1,2,3$), so here $D_{a}\psi_{i}$ is conjugate momentum to
$\Sigma^{a0i}$ with noting that the indices raising in $\Sigma^{\mu\nu j}$ is
done by using a metric $g_{\mu\nu}$.

The equation of motion of the action (13) with respect to $B^{i}$ in
constrained BF (like deriving the equation (8)) is

$F^{i}(A)+\varphi^{i}{{}_{j}}\Sigma^{j}+\varepsilon^{i}{{}_{jk}}\psi^{j}F^{k}(A)+T^{i}{{}_{j}}\Sigma^{j}+\xi^{i}{{}_{j}}\bar{\Sigma}^{j}=0.$
(19)

Multiplying with $\Sigma_{i}^{\mu\nu}$, summing over the indices and using
$\Sigma_{i}^{\mu\nu}\bar{\Sigma}_{\mu\nu}^{j}=0$, we obtain

$\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{i}+\varphi^{i}{{}_{j}}\Sigma_{i}^{\mu\nu}\Sigma_{\mu\nu}^{j}+\varepsilon^{i}{{}_{jk}}\psi^{j}\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{k}+T^{i}{{}_{j}}\Sigma_{i}^{\mu\nu}\Sigma_{\mu\nu}^{j}=0,$
(20)

then using $\Sigma_{i}^{\mu\nu}\Sigma_{\mu\nu}^{j}=\delta_{i}^{j}$ to obtain

$\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{i}+tr(\varphi)+\varepsilon^{i}{{}_{jk}}\psi^{j}\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{k}+tr(T)=0.$

Since $tr(\varphi)=0$ and
$\varepsilon^{i}{{}_{jk}}\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{k}=0$(by the equations
(14) and (16)), we obtain

$\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{i}+tr(T)=tr(F)+tr(T)=0.$ (21)

The equation (19) leads us to write $F^{i}(A)$ in terms of $\Sigma^{i}$ and
$\bar{\Sigma}^{i}$, and since
$\varepsilon^{i}{{}_{jk}}\Sigma_{i}^{\mu\nu}F_{\mu\nu}^{k}=0$, we can write

$F^{i}(A)=\chi^{i}{{}_{j}}\Sigma^{j}+{\chi^{\prime}}^{i}{{}_{j}}\bar{\Sigma}^{j},$
(22)

for some symmetric matrix $(\chi^{ij})$ and skew-hermitian matrix
$({\chi^{\prime}}^{ij})$. Using this equation in the equation (21), we obtain

$tr(\chi^{i}{{}_{j}})+tr(T^{i}{{}_{j}})=0.$ (23)

In addition to this relation, there is another relation between the vector
field $\psi^{i}$ and the symmetric matrix $\chi^{i}{{}_{j}}$ when
$T^{i}{{}_{j}}\neq 0$ and $J^{\mu i}\neq 0$, from the conservation of the
current (18), $D_{\nu}J^{\nu i}=0$, we have (for $D_{\mu}\Sigma^{\mu\nu i}=0$)

$\begin{split}\frac{i}{2}D_{\nu}J^{\nu
i}&=\left({D_{\nu}D_{\mu}\psi_{k}}\right)\varepsilon{{}^{ki}}_{j}\Sigma^{\mu\nu
j}=\frac{1}{2}\left(\left[{D_{\nu},D_{\mu}}\right]\psi_{k}\right)\varepsilon{{}^{ki}}_{j}\Sigma^{\mu\nu
j}\\\ &=-\frac{1}{2}F_{\mu\nu}^{\ell}(A)\varepsilon_{\ell
k}{{}^{m}}\psi_{m}\varepsilon{{}^{ki}}_{j}\Sigma^{\mu\nu
j}=-\frac{1}{2}F_{\mu\nu}^{\ell}(A)\varepsilon_{\ell
km}\psi^{m}\varepsilon^{kij}\Sigma_{j}^{\mu\nu}\\\
&=\frac{1}{2}F_{\mu\nu}^{\ell}(A)\varepsilon_{k\ell
m}\psi^{m}\varepsilon^{kij}\Sigma_{j}^{\mu\nu}=\frac{1}{2}F_{\mu\nu}^{\ell}(A)\psi^{m}\left({\delta_{\ell}^{i}\delta_{m}^{j}-\delta_{m}^{i}\delta_{\ell}^{j}}\right)\Sigma_{j}^{\mu\nu}\\\
&=\frac{1}{2}F_{\mu\nu}^{i}(A)\psi^{j}\Sigma_{j}^{\mu\nu}-\frac{1}{2}F_{\mu\nu}^{j}(A)\psi^{i}\Sigma_{j}^{\mu\nu},\end{split}$

and using the equation (22), we get

$\begin{split}\frac{i}{2}D_{\nu}J^{\nu
i}&=\frac{1}{2}\chi^{i}{{}_{m}}\Sigma_{\mu\nu}^{m}\psi^{j}\Sigma_{j}^{\mu\nu}-\frac{1}{2}\chi^{j}{{}_{m}}\Sigma_{\mu\nu}^{m}\psi^{i}\Sigma_{j}^{\mu\nu}=\frac{1}{2}\chi^{i}{{}_{m}}\psi^{j}\delta_{j}^{m}-\frac{1}{2}\chi^{j}{{}_{m}}\psi^{i}\delta_{j}^{m}\\\
&=\frac{1}{2}\chi^{i}{{}_{j}}\psi^{j}-\frac{1}{2}\psi^{i}tr\left({\chi^{i}{{}_{j}}}\right)=0,\quad\text{
for }J^{\mu i}\neq 0.\end{split}$ (24)

This is another relation between the vector field $\psi^{i}$ and the symmetric
matrix $\chi^{i}{{}_{j}}$ in existence of matter $T^{i}{{}_{j}}\neq 0$ with
$J^{\mu i}\neq 0$. In this case, the matrix $\chi^{i}{{}_{j}}$ has to satisfy
$\det\left(\chi^{i}{{}_{j}}-tr({\chi^{i}{{}_{j}}})\right)=0$ in order to get
$\psi^{i}\neq 0$, of course we do not need this condition in the vacuum
$T^{i}{{}_{j}}=0$, $J^{\mu i}=0$.

Using the equation (22) in (19), we obtain

$\chi^{i}{{}_{j}}\Sigma^{j}+\varphi^{i}{{}_{j}}\Sigma^{j}+\varepsilon^{i}{{}_{jk}}\psi^{j}\chi^{k}{{}_{\ell}}\Sigma^{\ell}+T^{i}{{}_{j}}\Sigma^{j}+(...)^{i}{{}_{j}}\bar{\Sigma}^{j}=0.$

That yields

$\chi^{i\ell}+\varphi^{i\ell}+\varepsilon^{i}{{}_{jk}}\psi^{j}\chi^{k\ell}+T^{i\ell}=0.$
(25)

This equation relates to the equation of motion $\delta S/\delta B=0$, it
includes the Lagrangian multiplier $\varphi_{ij}$ 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 $\chi$ and $\psi$ using
the other equations of motion we got before. We see that we do not require the
traceless matrix $\varphi$ 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 $\psi$, the symmetric matrix $\chi$(with (23)) and
the traceless matrix $\varphi$. The formula (25) gives $9$ equations,
therefore we have $16-9=7$ unknown variables, but when $J^{\mu i}\neq 0$, they
reduce to $6$ unknown variables(regarding the equation (24)). But if we choose
a solution for which the symmetric matrix $\chi^{ij}$ becomes diagonal, like

$\chi=\left(K^{i}\delta^{i}_{j}\right)=diag(K^{1},K^{2},K^{3}),$ (26)

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 $J^{\mu i}\neq 0$.

######  Remark 3.

The field $\psi^{i}$ is solution of $D^{\mu}D_{\mu}\psi^{i}=0$ (equation
(34)), so if $D_{\mu}v^{i}=0$, then $\psi^{i}+v^{i}$ is another solution, and
that makes the components $\psi^{1},\psi^{2}$ and $\psi^{3}$ of the vector
field $\psi^{i}$ 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 $\psi^{i}\mapsto\psi^{i}+v^{i}$ ($Dv^{i}=0$) does not
change the current $J^{\nu
i}=2i\left({D_{\mu}\psi_{k}}\right)\varepsilon{{}^{ki}}_{j}\Sigma^{\mu\nu j}$.

The Bianchi identity $DF^{i}=0$ implies
$\left(D\chi^{i}{{}_{j}}\right)\wedge\Sigma^{j}=0$ (for $D\Sigma^{i}=0$),
hence $(dK^{i})\delta^{i}{{}_{j}}\wedge\Sigma^{j}=0$, where we used
$D\delta^{ij}=0$, with using the covariant derivative
$Dv^{i}=dv^{i}+\varepsilon^{i}{{}_{jk}}A^{j}v^{k}$. Therefore we obtain

$\varepsilon^{\mu\nu\rho\sigma}(\partial_{\nu}K^{i})\Sigma_{\rho\sigma}^{i}=0,\quad\text{for}\quad\varepsilon^{0123}=-\varepsilon_{0123}=1.$
(27)

In $3+1$ decomposition of the space-time manifold $M=\Sigma\times\mathbb{R}$,
let $\Sigma_{t}$ 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
$\varepsilon^{\mu\nu\rho\sigma}D_{\mu}\Sigma_{\nu\rho}^{i}=0$ ($DB^{i}=0$)
decomposes to two equations,

$D_{a}E^{ai}=0\quad\text{ and }\quad\varepsilon^{abc}D_{b}B_{c}^{i}=0,$ (28)

in which we introduce the vector field $E^{i}$ and the 1-form $B^{i}$,

$E^{ai}=\epsilon^{0abc}\Sigma^{i}_{bc}/2=\varepsilon^{abc}\Sigma^{i}_{bc}/2,\quad\text{
and }\quad B_{c}^{i}=\Sigma^{i}_{0c},$ (29)

on the space-like slice $\Sigma_{t}$(the field $E^{ai}$ is conjugate to the
connection $A^{i}_{a}$). The covariant derivative on
$TM\otimes\mathfrak{so}(3,\mathbb{C})_{P}$ is $D_{\mu}=\nabla_{\mu}+A_{\mu}$,
for $A_{\mu}^{ij}=\varepsilon^{ij}{{}_{k}}A_{\mu}^{k}$, therefore on the 3d
surface $\Sigma_{t}$, it becomes $D_{a}=\nabla_{a}+A_{a}$.

The equation (27) decomposes to (for $\partial_{0}K^{i}=0$)

$\begin{split}&\varepsilon^{0abc}(\partial_{a}K^{i})\Sigma_{bc}^{i}=(\partial_{a}K^{i})\varepsilon^{abc}\Sigma_{bc}^{i}=2(\partial_{a}K^{i})E^{ai}=0,\\\
and&\\\
&\varepsilon^{abc0}(\partial_{b}K^{i})\Sigma_{c0}^{i}=-\varepsilon^{abc}(\partial_{b}K^{i})B_{c}^{i}=0.\end{split}$
(30)

We can solve them by writing (for non-zero curvature $F^{i}(A)$)

$E^{ai}=\frac{1}{2}\varepsilon^{abc}(\partial_{b}K^{i})r_{c}^{i},\quad\quad
B_{c}^{i}=(\partial_{c}K^{i})u^{i},$

for some $r^{i}\in\Omega^{1}\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$
and $u^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$. The
functions $K^{i}$ are scalars, the indices are just for distinguishing each
from the others. Thus we get the solutions

$\Sigma^{i}_{ab}=-\Sigma^{i}_{ba}=(\partial_{[a}K^{i})r_{b]}^{i},\quad\text{
and }\quad\Sigma^{i}_{0a}=-\Sigma^{i}_{a0}=(\partial_{a}K^{i})u^{i}.$ (31)

The equation (28) implies $Dr^{i}=0$ and $Du^{i}=0$.

In the static case $J^{a}_{i}=0$(zero current) with $J^{0}_{i}\neq 0$(non-zero
charge), the spin current formula
$J^{\nu}_{k}=2i\varepsilon_{kij}\left(D_{\mu}\psi^{i}\right)\Sigma^{\mu\nu
j}$, equation (18), decomposes to two equations

$\begin{split}&J^{b}_{k}=2i\varepsilon_{kij}\left(D_{a}\psi^{i}\right)\Sigma^{abj}=0,\\\
and&\\\
&J^{0}_{k}=2i\varepsilon_{kij}\left(D_{a}\psi^{i}\right)\Sigma^{a0j}\neq
0.\end{split}$ (32)

We can solve the first equation in terms of $D_{a}\psi^{i}$ by writing

$\Sigma^{abi}=-\Sigma^{bai}=fv^{[a}D^{b]}\psi^{i},$

for some vector $v\in\Gamma\left(M;T\Sigma\right)$ that satisfies
$v^{a}D_{a}\psi^{i}=0$, and $f$ is scalar function on $M$. We can include $f$
in $v$, so we just write $\Sigma^{abi}=v^{[a}D^{b]}\psi^{i}$. We note that we
have $D^{a}\psi^{i}=g^{ab}D_{b}\psi^{i}$ without needing specifying the used
metric $g_{ab}$(the solution is satisfied by using an arbitrary metric). If
$J^{a}_{k}\neq 0$, we let $\Sigma^{abi}=v^{[a}D^{b]}G^{i}$ with some field
$G^{i}\neq\psi^{i}$.

Regarding the second equation of (32), when $J^{0i}=0$, we get the solution
$\Sigma^{0ai}=-\Sigma^{a0i}=fg^{ab}D_{b}\psi^{i}$. And when $J^{0i}\neq 0$, we
let $\Sigma^{0ai}=fg^{ab}D_{b}\xi^{i}$ for some vector field
$\xi^{i}\neq\psi^{i}$, and without needing specifying the used metric $g_{ab}$
because the solution is satisfied by using an arbitrary metric. Using the
equation (17), $D_{\mu}\Sigma^{\mu\nu i}=0$, and for $\nu=0$, we have
$D_{a}\Sigma^{a0i}=0$. Therefore the solution
$\Sigma^{0ai}=-\Sigma^{a0i}=fg^{ab}D_{b}\xi^{i}$ implies

$g^{ab}D_{a}D_{b}\xi^{i}=0,$ (33)

for constant $f$ and with using $\nabla_{a}g_{bc}=0$ which defines affine
connection on $\Sigma_{t}$ in terms of the arbitrary metric $g_{ab}$.
Therefore in the vacuum, where $\xi^{i}=\psi^{i}$, we obtain
$g^{ab}D_{a}D_{b}\psi^{i}=0$. And to satisfy
$D_{0}\Sigma^{0ai}+D_{b}\Sigma^{bai}=0$, we use the previous solutions and let
$D_{0}$ be determined in terms of them. And in order to satisfy the invariance
under coordinates transformations on $M$, we let the equation
$g^{ab}D_{a}D_{b}\psi^{i}=0$ on $\Sigma_{t}$ come from pulling-back of the
equation $g^{\mu\nu}D_{\mu}D_{\nu}\psi^{i}=0$ on $M$, because the 3d space-
like surface $\Sigma_{t}$ is immersed in the 4d space-time manifold $M$, thus
pulling-back of

$g^{\mu\nu}D_{\mu}D_{\nu}\psi^{i}=0,\quad\nabla_{\mu}g_{\nu\rho}=0,$ (34)

under the immersion map $i:\Sigma_{t}\to M$ gives
$g^{ab}D_{a}D_{b}\psi^{i}=0$, therefore $g_{\mu\nu}$ is also arbitrary metric
on $M$. Regarding the discussion in the introduction, we let the metric in
$g^{\mu\nu}D_{\mu}D_{\nu}\psi^{i}=0$ be flat, at least locally, in order to
avoid the influences of geometry of $M$ on the field $\psi^{i}$ which relates
with the spin current by the formula (18).

######  Lemma 3.

By comparing the solutions $\Sigma^{abi}=v^{[a}D^{b]}\psi^{i}$ and
$\Sigma^{0ai}=fD^{a}\xi^{i}$ of the equations (32) with the solutions (31),
and in order to get a correspondence between that solutions, e.g, by using a
metric, we find that

$\begin{split}&\psi^{i}=K^{i}b^{i},\quad
r_{c}^{i}=v_{c}b^{i},\quad\xi^{i}=K^{i}u^{i},\\\ and&\\\ &Db^{i}=0,\quad
Du^{i}=0,\quad dv=0,\quad f=1,\end{split}$ (35)

for some vector fields
$b^{i},u^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$.

By that we obtain the solutions

$\begin{split}&E^{i}=\frac{1}{2}\epsilon^{abc}\Sigma_{ab}^{i}\partial_{c}=\frac{1}{2}\epsilon^{abc}v_{a}\left(\partial_{b}K^{i}\right)b^{i}\partial_{c}\in\Gamma\left(M;T\Sigma\otimes\mathfrak{so}(3,\mathbb{C})_{P}\right),\\\
and&\\\
&B^{i}=\Sigma_{0a}^{i}dx^{a}=\left(dK^{i}\right)u^{i}\in\Gamma\left(M;T^{*}\Sigma\otimes\mathfrak{so}(3,\mathbb{C})_{P}\right),\end{split}$
(36)

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 $v^{a}\partial_{a}K^{i}$=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 $\Sigma_{ab}^{i}$ and $\Sigma^{abi}$, we get a metric
$g_{ab}$ satisfying
$\Sigma_{ab}^{i}=g_{aa^{\prime}}g_{bb^{\prime}}\Sigma^{a^{\prime}b^{\prime}i}$.
Also we note that the metric used in
$\Sigma_{ab}^{i}=g_{aa^{\prime}}g_{bb^{\prime}}\Sigma^{a^{\prime}b^{\prime}i}$
is not necessary the same metric used in $D^{a}\psi^{i}=g^{ab}D_{b}\psi^{i}$
for getting the solutions of (32). Therefore the metric in BF theory is a
derived quantity from the solutions of the 2-form
$B\in\Omega^{2}(M;\mathfrak{so}(3,\mathbb{C}))$.

######  Remark 5.

In the solution (36), we see that $\Sigma^{IJ}$ can be written as $e^{I}\wedge
e^{J}$, 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 $K^{i}b^{i}=P^{i}_{IJ}b^{I}(K^{J}b^{J})$, therefore
$\Sigma^{IJ}_{ab}=\left(b^{I}v_{[a}\right)D_{b]}\left(K^{J}b^{J}\right)$, then
we can write $e^{I}_{a}=v_{a}b^{I}$ and
$e^{J}_{b}=\left(\partial_{b}K^{J}\right)b^{J}$. And from
$\Sigma_{0a}^{IJ}=u^{I}\left(\partial_{a}K^{J}\right)u^{J}$, we get
$e^{J}_{a}=\left(\partial_{a}K^{J}\right)u^{J}$ and $e^{I}_{0}=v_{0}u^{I}$,
for $v_{0}=1$. A more general case is to find three vector fields $b_{1}^{I}$,
$b_{2}^{I}$ and $K^{I}$ satisfying
$K^{i}b^{i}=P^{i}_{IJ}b_{1}^{I}(K^{J}b_{2}^{J})$, therefore
$\Sigma^{IJ}_{ab}=\left(b_{1}^{I}v_{[a}\right)D_{b]}\left(K^{J}b_{2}^{J}\right)$,
then we can write $e^{I}_{a}=v_{a}b_{1}^{I}$ and
$e^{J}_{b}=\left(\partial_{b}K^{J}\right)b_{2}^{J}$. And from
$\Sigma_{0a}^{IJ}=u_{1}^{I}\left(\partial_{a}K^{J}\right)u_{2}^{J}$, we get
$e^{J}_{a}=\left(\partial_{a}K^{J}\right)u_{2}^{J}$ and
$e^{I}_{0}=v_{0}u_{1}^{I}$, for $v_{0}=1$. By that the (36) can be written as
$\Sigma^{i}=P^{i}_{IJ}\Sigma^{IJ}$ for $\Sigma^{IJ}=e^{I}\wedge e^{J}$. 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
$\Sigma^{0ai}=-\Sigma^{a0i}=g^{ab}D_{b}\xi^{i}=g^{ab}(\partial_{b}K^{i})u^{i}$
in the second equation of (32), implies

$\begin{split}J^{0}_{k}=Q_{k}&=-2i\epsilon_{kij}g^{ab}\left(b^{i}\partial_{a}K^{i}\right)\left(u^{j}\partial_{b}K^{j}\right)\\\
&=-i\epsilon_{kij}g^{ab}\left({b^{i}\partial_{a}K^{i}u^{j}\partial_{b}K^{j}-b^{j}\partial_{a}K^{j}u^{i}\partial_{b}K^{i}}\right)\\\
&=-i\epsilon_{kij}g^{ab}\left(\partial_{a}K^{i}\right)\left(\partial_{b}K^{j}\right)(b^{i}u^{j}-b^{j}u^{i}).\end{split}$
(37)

We see that $J^{0}_{k}\neq 0$ takes place only when $b\neq u$. Therefore in
the vacuum it must be $b=u$. Regarding the equation (34), the field
$\psi^{i}=K^{i}b^{i}$ satisfies $D^{2}\psi^{i}=0$, so $Db^{i}=0$ implies
$\nabla^{2}K^{i}=0$. If $J^{a}_{k}\neq 0$, we let
$\Sigma^{abi}=v^{[a}(\partial^{b]}g^{i})u^{i}$ with some field $g^{i}\neq
K^{i}$.

If the charges $J^{0i}\neq 0$ are given as a functions on $M$, and in order to
get a solution using them, we let
$b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$ be constant
field on $M$, so we can determine the scalar functions $K^{i}$ using the
equation (37), and so obtaining the vector $v\in\Gamma\left(M;T\Sigma\right)$
using $v^{a}\partial_{a}K^{i}=0$. But to satisfy
$Db^{i}=db^{i}+\varepsilon^{i}_{jk}A^{j}b^{k}=0$ for a constant vector field
$b^{i}$, the connection $A^{i}_{\mu}$ must be written as
$A^{i}_{\mu}=A_{\mu}b^{i}$. And we choose $u^{i}=b^{i}+f(x)b^{i}+a^{i}$, for a
constant $a^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$
satisfying $a_{i}b^{i}=0$, the function $f$ is needed for satisfying
$Du^{i}=0$. We will see examples of determining $b^{i}$ and $u^{i}$ in
spherical and cylindrical symmetries.

Another way of getting the solutions using the charges $J^{0i}\neq 0$ is by
using the solutions of $K^{i}$(obtained from $\nabla^{2}K^{i}=0$) 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 $\chi$ using the equation (26), so obtaining the curvature
$F=\chi\Sigma$. We note that $v^{a}E^{i}_{a}=0$, $v_{a}B^{ai}=0$ and
$v^{a}\partial_{a}K^{i}=0$ 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 $(\Sigma^{0ai},\Sigma^{abi})$ and $(\Sigma^{i}_{0a},\Sigma^{i}_{ab})$
according the remark (4).

##  3 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
$\psi^{i}=K^{i}b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$
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 $v\in\Gamma\left(M;T\Sigma\right)$
that satisfies $v^{a}D_{a}\psi^{i}=0$, $v^{a}E^{i}_{a}=0$ and
$v^{a}B^{i}_{a}=0$. For example, in spherical symmetry, we use the spherically
coordinates $(r,\theta,\varphi)$ on the space-like slice
$\Sigma_{t}=\Sigma=\mathbb{R}^{3}$. And according to the equation (33), we
have $D^{2}\psi^{i}=0$ on $\Sigma_{t}$ 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 $\psi^{i}$ depend
only on the radius $r$, we get (for $Db^{i}=0$)

$D^{2}\psi^{i}=D^{2}(K^{i}b^{i})=b^{i}\nabla^{2}K^{i}=b^{i}\frac{1}{{r^{2}}}\frac{\partial}{{\partial
r}}\left({r^{2}\frac{\partial}{{\partial r}}K^{i}}\right)=0\Rightarrow
K^{i}=\frac{{c^{i}}}{r},$ (38)

so $\psi^{i}=c^{i}b^{i}/r$, for some constants $c^{i}\in\mathbb{R}$. Actually
we can include $c^{i}$ in $b^{i}$ and just write $\psi^{i}=b^{i}/r$. Therefore

$D\psi^{i}=b^{i}(dK^{i})=b^{i}\left({dr\partial_{r}+d\theta\partial_{\theta}+d\varphi\partial_{\varphi}}\right)\frac{{1}}{r}=-\frac{{b^{i}}}{{r^{2}}}dr,$

thus the 1-form $v$ ($dv=0$, $g^{ab}v_{a}D_{b}\psi^{i}=0$) is

$v=a_{1}d\theta+a_{2}d\varphi,\text{ }a_{1},a_{2}\in\mathbb{R},$

where in the spherical symmetry we let $a_{1}$ and $a_{2}$ do not depend on
the coordinates $\theta$ and $\varphi$. 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 $E^{i}$,

$B^{i}=(dK^{i})u^{i}=-\frac{{b^{i}}}{{r^{2}}}dr,\quad for\quad u=b,$
$\begin{split}E^{i}=\frac{1}{2}\varepsilon^{abc}\Sigma_{bc}^{i}\partial_{a}=\frac{1}{2}\varepsilon^{abc}v_{b}\left(\partial_{c}K^{i}\right)b^{i}\partial_{a}&=-\varepsilon^{\varphi\theta
r}\frac{{b^{i}}}{{2r^{2}}}\partial_{\varphi}-\varepsilon^{\theta\varphi
r}\frac{{b^{i}}}{{2r^{2}}}\partial_{\theta}\\\ &=\varepsilon^{\varphi
r\theta}\frac{{b^{i}}}{{2r^{2}}}\partial_{\varphi}+\varepsilon^{\theta
r\varphi}\frac{{b^{i}}}{{2r^{2}}}\partial_{\theta}.\end{split}$

By that we get

$\Sigma_{0r}^{i}=-\frac{{b^{i}}}{{r^{2}}},\quad\Sigma_{r\theta}^{i}=\frac{{b^{i}}}{{2r^{2}}},\quad\Sigma_{r\varphi}^{i}=\frac{{b^{i}}}{{2r^{2}}},$
(39)

while the other components like
$\Sigma_{0\theta}^{i},\Sigma_{0\varphi}^{i},...$, can be obtained by using
gravitational fields $e^{I}_{\mu}$ derived from the solutions (39). That is
according to self-dual map, there are at least two constant fields $b_{1}^{I}$
and $b_{2}^{I}$ satisfying $b^{i}=P^{i}_{IJ}b_{1}^{I}b_{2}^{J}$. So from
$\Sigma_{\mu\nu}^{i}=P^{i}_{IJ}e_{\mu}^{I}e_{\nu}^{J}$(regarding remark (5)),
we get the gravitational fields

$e_{0}^{I}=-\frac{b_{1}^{I}}{r},\quad e_{r}^{I}=\frac{b_{2}^{I}}{r},\quad
e_{\theta}^{I}=-\frac{b_{1}^{I}}{2r},\quad
e_{\varphi}^{I}=-\frac{b_{1}^{I}}{2r}.$

We obtain the matrix $\chi$ using the equation (26) with the solution (38),

$\chi=\left(K_{j}\delta^{i}_{j}\right)=\frac{1}{{r}}\text{diag}(c^{1},c^{2},c^{3}),$

where the constants $c^{i}$ have to be determined in order to satisfy the
condition $tr\chi=0$(in the vacuum), so $\sum\limits_{i=1}^{3}{c_{i}=0}$. Thus
we get the curvature $F=\chi\Sigma+\chi^{\prime}\bar{\Sigma}$ (with setting
$\chi^{\prime}=0$ in the vacuum [15]),

$\begin{split}&F_{0r}^{i}=\chi^{i}{{}_{j}}\Sigma_{0r}^{j}=-\frac{{c^{i}b^{i}}}{{r^{3}}},\quad\quad
F_{r\theta}^{i}=\chi^{i}{{}_{j}}\Sigma_{r\theta}^{j}=\frac{{c^{i}b^{i}}}{{2r^{3}}},\\\
and&\\\
&F_{r\varphi}^{i}=\chi^{i}{{}_{j}}\Sigma_{r\varphi}^{j}=\frac{{c^{i}b^{i}}}{{2r^{3}}}.\end{split}$
(40)

Now we calculate the connection $A^{i}$ and the field $b^{i}$ which satisfies
$Db^{i}=0$. Using $F^{i}=dA^{i}+\varepsilon^{i}{{}_{jk}}A^{j}\wedge A^{k}$, we
obtain

$\begin{split}&F_{0r}^{i}=\frac{1}{2}\left({\partial_{0}A_{r}^{i}-\partial_{r}A_{0}^{i}}\right)+\varepsilon^{i}{{}_{jk}}A_{0}^{j}A_{r}^{k},\\\
&F_{\theta
r}^{i}=\frac{1}{2}\left({\partial_{\theta}A_{r}^{i}-\partial_{r}A_{\theta}^{i}}\right)+\varepsilon^{i}{{}_{jk}}A_{\theta}^{j}A_{r}^{k},\\\
&F_{\varphi
r}^{i}=\frac{1}{2}\left({\partial_{\varphi}A_{r}^{i}-\partial_{r}A_{\varphi}^{i}}\right)+\varepsilon^{i}{{}_{jk}}A_{\varphi}^{j}A_{r}^{k}.\end{split}$

Since we study a spherical symmetric system, we let the connection $A^{i}$
depend only on $r$. If we choose the gauge $A_{r}^{i}=0,i=1,2,3$, we get

$F_{0r}^{i}=-\frac{1}{2}\partial_{r}A_{0}^{i},\quad\quad F_{\theta
r}^{i}=-\frac{1}{2}\partial_{r}A_{\theta}^{i},\quad\quad F_{\varphi
r}^{i}=-\frac{1}{2}\partial_{r}A_{\varphi}^{i},$

therefore by using the solution (40), we obtain

$-\frac{1}{2}\partial_{r}A_{0}^{i}=-\frac{{c^{i}b^{i}}}{{r^{3}}},\quad\quad\frac{1}{2}\partial_{r}A_{\theta}^{i}=\frac{{c^{i}b^{i}}}{{2r^{3}}},\quad\quad\frac{1}{2}\partial_{r}A_{\varphi}^{i}=\frac{{c^{i}b^{i}}}{{2r^{3}}}.$

But $Db^{i}=db^{i}+\varepsilon^{i}_{jk}A^{j}b^{k}=0$ and
$\partial_{\mu}b^{i}=0$ for $\mu\neq r$, therefore

$\begin{split}&\partial_{r}b^{i}+\varepsilon^{i}{{}_{jk}}A_{r}^{j}b^{k}=\partial_{r}b^{i}=0,\quad\quad\partial_{0}b^{i}+\varepsilon^{i}{{}_{jk}}A_{0}^{j}b^{k}=\varepsilon^{i}{{}_{jk}}A_{0}^{j}b^{k}=0,\\\
&\partial_{\theta}b^{i}+\varepsilon^{i}{{}_{jk}}A_{\theta}^{j}b^{k}=\varepsilon^{i}{{}_{jk}}A_{\theta}^{j}b^{k}=0,\quad\quad\partial_{\varphi}b^{i}+\varepsilon^{i}{{}_{jk}}A_{\varphi}^{j}b^{k}=\varepsilon^{i}{{}_{jk}}A_{\varphi}^{j}b^{k}=0.\end{split}$

Therefore the field
$b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$ is constant and
we get the solution

$A_{0}^{i}=-\frac{{c^{i}b^{i}}}{{r^{2}}},\quad\quad
A_{\theta}^{i}=-\frac{{c^{i}b^{i}}}{{2r^{2}}},\quad\quad
A_{\varphi}^{i}=-\frac{{c^{i}b^{i}}}{{2r^{2}}},$

where we used $\varepsilon^{i}{{}_{jk}}b^{j}b^{k}=0$. Thus in this solution
the field $b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$ is
constant on $M=\Sigma\times\mathbb{R}$. Next we try to find $b^{i}$ in the
case of matter located at a point.

###  3.1 Solutions for matter located at a point

If we have matter located at a point in $\Sigma_{t}=\Sigma=\mathbb{R}^{3}$,
thus we have a spherical symmetry system in a static case $J^{0}_{i}\neq 0$,
$J^{a}_{i}=0$. We let that point be the origin $(0,0,0)\in\mathbb{R}^{3}$,
therefore the charge (37) is given by $Q^{i}(x)=Q^{i}_{0}\delta^{3}(x)$, so
$\int\limits_{\mathbb{R}^{3}}{Q^{i}_{0}\delta^{3}(x)}=Q^{i}_{0}=constant$(conservation
of the charges). In order to get a same solution, as in the equations (39) and
(40), we keep the field $b^{i}$ be constant, and in the formula (37)

$\begin{split}J^{0}_{k}(x)=Q_{k}(x)=-i\epsilon_{kij}\left(\partial_{a}K^{i}\right)\left(\partial^{a}K^{j}\right)(b^{i}u^{j}-b^{j}u^{i}),\end{split}$
(41)

we use $Q^{i}(x)=Q^{i}_{0}\delta^{3}(x)$.

In spherical symmetry, the functions $K^{i}$ are given by $K^{i}=c^{i}/r$, the
equation (38), therefore

$Q_{k}(x)=-2ig^{rr}\left(\partial_{r}K^{i}\partial_{r}K^{j}\right)\left(\epsilon_{kij}b^{i}u^{j}\right)=-i2\frac{c^{i}c^{j}}{r^{4}}\left(\epsilon_{kij}b^{i}u^{j}\right).$

Therefore in order to get $Q^{i}=Q^{i}_{0}\delta^{3}(x)$, we replace
$1/{r^{4}}$ with $1/(r^{4}+\epsilon^{4})$, for some infinitesimal parameter
$\epsilon\to 0^{+}$, and we choose a solution for the field $u^{j}$ like

$u^{j}=b^{j}+\epsilon fb^{j}+\epsilon a^{j}/({-2i\pi^{2}\sqrt{2}}),$

for some function $f$ on $M$ that is needed for satisfying $Du^{i}=0$ and a
constant vector field
$a^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$. With that we
obtain (for $i,j\neq k$)

$Q_{k}(r)=-2i\frac{c^{i}c^{j}}{r^{4}+\epsilon^{4}}\epsilon_{kij}b^{i}\left(b^{j}+\epsilon
fb^{j}+\frac{\epsilon}{-2i\pi^{2}\sqrt{2}}a^{j}\right)=\frac{1}{\pi^{2}\sqrt{2}}\frac{\epsilon}{r^{4}+\epsilon^{4}}\epsilon_{kij}(c^{i}b^{i})(c^{j}a^{j}).$

Comparing with $Q^{i}(x)=Q^{i}_{0}\delta^{3}(x)$, we find
$\epsilon^{k}{{}_{ij}}(c^{i}b^{i})(c^{j}a^{j})=Q_{0}^{k}=constant$, and by
imposing $(c_{i}a_{i})(c^{i}a^{i})=1$ with $(c_{i}b_{i})(c^{i}a^{i})=0$, we
find

$(cb)^{2}=(c_{i}b_{i})(c^{i}b^{i})=Q_{0k}Q_{0}^{k},$

thus we choose $c^{i}b^{i}=e^{i}\sqrt{Q_{0k}Q_{0}^{k}}$ for
$\left\|e^{i}\right\|=1$, so the constant field $b^{i}$ is determined by
$\sqrt{Q_{0k}Q_{0}^{k}}$ with free ${SO}(3,\mathbb{C})$ rotation.

By that, for $r>0$, $\epsilon\to 0^{+}$, we obtain the same solutions as in
the equations (39) and (40), but with $c^{i}b^{i}=e^{i}\sqrt{Q_{0k}Q_{0}^{k}}$
for $\left\|e^{i}\right\|=1$ and $\sum\limits_{i=1}^{3}{c_{i}=0}$. Since
$c^{i}b^{i}$ is finite value, it is not sufficient to let the constants
$c_{i}$ take arbitrary values, so we choose them to be $(c_{i})=(1,1,-2)$.

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
$\psi^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$ which is
defined in the spin current
$J^{\mu}_{k}=2i\varepsilon_{kij}\left(D_{\nu}\psi^{i}\right)\Sigma^{\mu\nu j}$
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 $v^{a}\partial_{a}K^{i}=0$, $v^{a}E^{i}_{a}=0$,
$v_{a}B^{ai}=0$.

##  4 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
$\psi^{i}=K^{i}b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$,
then we calculate $v,E^{i},B^{i}$ and $J^{0i}$ according to the equations
(35), (36) and (37). The vector $v\in\Gamma\left(M;T\Sigma\right)$ satisfies
$v^{a}D_{a}\psi^{i}=0$, $v^{a}E^{i}_{a}=0$ 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 $(\rho,\varphi,z)$ on the space-like slice
$\Sigma_{t}=\Sigma=\mathbb{R}^{3}$ of constant time $t$. We let the vector
field $\psi^{i}$ depend only on the radius $\rho$, we get (for
$D^{2}\psi^{i}=0$ and $Db^{i}=0$)

$D^{2}\psi^{i}=D^{2}(K^{i}b^{i})=b^{i}\nabla^{2}K^{i}=b^{i}\frac{1}{{\rho}}\frac{\partial}{{\partial\rho}}\left({\rho\frac{\partial}{{\partial\rho}}K^{i}}\right)=0\Rightarrow
K^{i}={c^{i}}\log(\rho),$ (42)

so $\psi^{i}=c^{i}b^{i}\log(\rho)$, for some constants $c^{i}\in\mathbb{R}$.
Therefore

$D\psi^{i}=c^{i}b^{i}(dK^{i})=c^{i}b^{i}\left({d\rho\partial_{\rho}+d\varphi\partial_{\varphi}+dz\partial_{z}}\right)\log(\rho)=\frac{{c^{i}b^{i}}}{{\rho}}d\rho,$

thus the 1-form $v$ ($dv=0$, $g^{ab}v_{a}D_{b}\psi^{i}=0$) is

$v=a_{1}d\varphi+a_{2}dz,\text{ }a_{1},a_{2}\in\mathbb{R},$

where in the cylindrical symmetry we let $a_{1}$ and $a_{2}$ do not depend on
the coordinates $z$ and $\varphi$. 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 $E^{i}$,

$B^{i}=\frac{{c^{i}u^{i}}}{{\rho}}d\rho=\frac{{c^{i}b^{i}}}{{\rho}}d\rho,\quad
for\quad u=b,$
$E^{i}=\frac{1}{2}\varepsilon^{abc}\Sigma_{bc}^{i}\partial_{a}=\frac{1}{2}\varepsilon^{abc}v_{b}\left(\partial_{c}K^{i}\right)b^{i}\partial_{a}=\varepsilon^{\varphi
z\rho}\frac{{c^{i}b^{i}}}{{2\rho}}\partial_{\varphi}+\varepsilon^{z\varphi\rho}\frac{{c^{i}b^{i}}}{{2\rho}}\partial_{z}.$

By that we get

$\Sigma_{0\rho}^{i}=-\Sigma_{\rho
0}^{i}=\frac{{c^{i}b^{i}}}{{\rho}},\quad\Sigma_{\rho
z}^{i}=-\Sigma_{z\rho}^{i}=-\frac{{c^{i}b^{i}}}{{2\rho}},\quad\Sigma_{\rho\varphi}^{i}=-\Sigma_{\varphi\rho}^{i}=-\frac{{c^{i}b^{i}}}{{2\rho}},$
(43)

while the other components like $\Sigma_{0z}^{i},\Sigma_{0\varphi}^{i},...$
are zeros.

We obtain the matrix $\chi$ using the equation (26) with the solution (42),

$\chi=\left(K_{j}\delta^{i}_{j}\right)=\log(\rho)\text{diag}(c^{1},c^{2},c^{3}),$

where the constants $c^{i}$ have to be determined in order to satisfy the
condition $tr\chi=0$(in the vacuum), so $\sum\limits_{i=1}^{3}{c_{i}=0}$. Thus
we get the curvature $F=\chi\Sigma+\chi^{\prime}\bar{\Sigma}$ (with setting
$\chi^{\prime}=0$ in the vacuum [15]),

$\begin{split}&F_{0\rho}^{i}=\chi^{i}{{}_{j}}\Sigma_{0\rho}^{j}=\frac{{\log(\rho)}}{{\rho}}c^{i}c^{i}b^{i},\quad\quad
F_{\rho z}^{i}=\chi^{i}{{}_{j}}\Sigma_{\rho
z}^{j}=-\frac{\log(\rho)}{{2\rho}}c^{i}c^{i}b^{i},\\\ and&\\\
&F_{\rho\varphi}^{i}=\chi^{i}{{}_{j}}\Sigma_{\rho\varphi}^{j}=-\frac{{\log(\rho)}}{{2\rho}}c^{i}c^{i}b^{i}.\end{split}$
(44)

Using the gauge $A_{\rho}^{i}=0,i=1,2,3$, with letting $A^{i}$ depend only on
$\rho$, we obtain

$A_{0}^{i}=-(\log(\rho))^{2}(c^{i})^{2}b^{i},\quad\quad
A_{z}^{i}=-\frac{1}{2}(\log(\rho))^{2}(c^{i})^{2}b^{i},\quad\quad
A_{\varphi}^{i}=-\frac{1}{2}(\log(\rho))^{2}(c^{i})^{2}b^{i},$

where the field $b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$
is constant on $M=\Sigma\times\mathbb{R}$.

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
$Q^{i}(\rho,\varphi,z)$ is given by
$Q^{i}(\rho,\varphi,z)=Q^{i}_{0}\delta(\rho)/2\pi\rho$, which yields
$\int\limits_{0}^{\infty}\int\limits_{0}^{2\pi}\rho d\varphi d\rho
Q^{i}_{0}\delta(\rho)/2\pi\rho=Q^{i}_{0}$ for each point of $Z$. 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)

$\begin{split}J^{0}_{k}(x)=Q_{k}(x)=-i\epsilon_{kij}\left(\partial_{a}K^{i}\right)\left(\partial^{a}K^{j}\right)(b^{i}u^{j}-b^{j}u^{i}),\end{split}$
(45)

we use $Q^{i}(x)=Q^{i}_{0}\delta(\rho)/2\pi\rho$.

In cylindrical symmetry, the functions $K^{i}$ are given by
$K^{i}=c^{i}\log(\rho)$, the equation (42), therefore

$Q_{k}(x)=-2ig^{\rho\rho}\left(\partial_{\rho}K^{i}\partial_{\rho}K^{j}\right)\left(\epsilon_{kij}b^{i}u^{j}\right)=-2i\frac{c^{i}c^{j}}{\rho^{2}}\left(\epsilon_{kij}b^{i}u^{j}\right).$

Therefore in order to get $Q^{i}=Q^{i}_{0}\delta(\rho)/2\pi\rho$, we replace
$1/{\rho^{2}}$ with $1/{\rho\rho^{1-\epsilon}}$, for some infinitesimal
parameter $\epsilon\to 0^{+}$, and we choose a solution for the field $u^{j}$
like

$u^{j}=b^{j}+\epsilon fb^{j}+\epsilon a^{j}/({-4i\pi}),$

for some scalar function $f$ on $M$ that is needed for satisfying $Du^{i}=0$,
with a constant vector field
$a^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})_{P}\right)$. By that we
obtain (for $i,j\neq k$)

$Q_{k}(r)=-2i\frac{c^{i}c^{j}}{{\rho\rho^{1-\epsilon}}}\epsilon_{kij}b^{i}\left(b^{j}+\epsilon
fb^{j}+\frac{\epsilon}{{-4i\pi}}a^{j}\right)=\frac{1}{2\pi\rho}\frac{\epsilon}{\rho^{1-\epsilon}}\epsilon_{kij}(c^{i}b^{i})(c^{j}a^{j}).$

Comparing with $Q^{i}=Q^{i}_{0}\delta(\rho)/2\pi\rho$, we find
$\epsilon^{k}{{}_{ij}}(c^{i}b^{i})(c^{j}a^{j})=Q_{0}^{k}=constant$, and by
imposing $(c_{i}a_{i})(c^{i}a^{i})=1$ with $(c_{i}b_{i})(c^{i}a^{i})=0$, we
find

$(cb)^{2}=(c_{i}b_{i})(c^{i}b^{i})=Q_{0k}Q_{0}^{k},$

thus we choose $c^{i}b^{i}=e^{i}\sqrt{Q_{0k}Q_{0}^{k}}$ for
$\left\|e^{i}\right\|=1$, so the constant field $c^{i}b^{i}$ is determined by
$\sqrt{Q_{0k}Q_{0}^{k}}$ with free ${SO}(3,\mathbb{C})$ rotation.

By that, for $\rho>0$, $\epsilon\to 0^{+}$, we obtain the same solutions as in
the equations (43) and (44), but with $c^{i}b^{i}=e^{i}\sqrt{Q_{0k}Q_{0}^{k}}$
for $\left\|e^{i}\right\|=1$ and $\sum\limits_{i=1}^{3}{c_{i}=0}$. Since
$c^{i}b^{i}$ is finite value, it is not sufficient to let the constants
$c_{i}$ take arbitrary values, so we choose them to be $(c_{i})=(1,1,-2)$.

##  5 Conclusions

We have studied the BF theory including matter by redefinition the 2-form
$B^{i}$, as $B^{i}+\varepsilon^{i}{{}_{jk}}\psi^{j}B^{k}$, or redefinition the
Lagrangian multipliers $\varphi_{ij}$ as $\varphi_{ij}+\varepsilon_{\ell
ik}\psi^{\ell}\chi^{k}{{}_{j}}$, so that we can get $DB^{i}=0$, in the case of
non-zero spin current of matter fields. The new field $\psi^{i}$ is defined
using the spin current vector $J^{\nu
i}=(D_{\mu}\psi_{k})\varepsilon{{}^{ki}}_{j}\Sigma^{\mu\nu j}$. 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 $\delta S/\delta A^{i}=0$, $DB^{i}=0$ and
$J^{\nu i}=(D_{\mu}\psi_{k})\varepsilon{{}^{ki}}_{j}\Sigma^{\mu\nu j}$ without
using a gravitational metric on $M$ and without needing solving the equation
$\delta S/\delta B^{i}=0$ which includes the Lagrangian multiplier
$\varphi_{ij}$(a non-physical variable), so we get $\varphi_{ij}$ by using the
solutions in $\delta S/\delta B^{i}=0$. We found that to obtain the solutions
of BF theory, it is enough to use(find) the field $\psi^{i}$ and the Killing
vector $v$(satisfies $v^{a}D_{a}\psi^{i}=0$) 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
$J^{0i}\neq 0$ 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 $\psi^{i}$, that relates to the idea that the spin current $J^{\mu
i}$ is the source for $\psi^{i}$, therefore $\psi^{i}$ 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
$(\Sigma^{i}_{ab},\Sigma^{i}_{0a})$ and $(\Sigma^{abi},\Sigma^{0ai})$
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 $e^{I}\wedge e^{J}$ (remark 5). Finally we applied the solutions of
BF theory in a spherical and cylindrical symmetric systems in static case of
matter.

##  6 Acknowledgements

I am grateful to Professor Yoshihiro Fukumoto at Ritsumeikan University for
many useful notes on the manuscript.

##  7 Appendix A

$\rm{I}$\- We test satisfying of the equation of motion of classical general
relativity in the vacuum using the solutions in BF theory, the equations have
not to include the non-physical variable $\varphi$. The canonical formalism of
GR in the vacuum gives the constraints [9]

$D_{a}E_{i}^{a}=0,\quad E_{i}^{a}F_{ab}^{i}=0,\quad
C=\varepsilon_{ijk}E^{ai}E^{bj}F_{ab}^{k}=0.$ (46)

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
$\varepsilon^{\mu\nu\rho\sigma}D_{\nu}B_{\rho\sigma}^{i}=0$ by setting $\mu=0$
to get $\varepsilon^{0abc}D_{a}B_{bc}^{i}=0$, and using
$\varepsilon^{0abc}\equiv\varepsilon^{abc}$ to get
$\varepsilon^{abc}D_{a}B_{bc}^{i}=2D_{a}E^{ai}=0$, where
$E^{ai}=\varepsilon^{abc}B_{bc}^{i}/2$ is conjugate to the connection
$A_{a}^{i}$ 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 ${\chi^{\prime}}^{i}{{}_{j}}=0$, it
reduces to $F^{i}(A)=\chi^{i}{{}_{j}}\Sigma^{j}$, on the spacelike surface of
constant time $\Sigma_{t}$, it becomes
$F^{i}(A)_{ab}=\chi^{i}{{}_{j}}\Sigma^{j}_{ab}$. Using
$E^{ai}=\varepsilon^{abc}\Sigma^{i}_{bc}$, we obtain

$F^{i}(A)_{ab}=\chi^{i}{{}_{j}}\Sigma^{j}_{ab}=\chi^{i}{{}_{j}}\varepsilon_{abc}E^{jc}.$

Multiplying it by $E_{i}^{a}$ and summing over the contracted indices, we get

$E_{i}^{a}F^{i}(A)_{ab}=E_{i}^{a}\chi^{i}{{}_{j}}\varepsilon_{abc}E^{jc}=\chi_{ij}\varepsilon_{abc}E^{ai}E^{jc}=0,$

where we used the fact that the matrix $\chi_{ij}$ is symmetric. Therefore the
second constraint of (46) is satisfied. Using
$F^{i}(A)_{ab}=\chi^{i}{{}_{j}}\varepsilon_{abc}E^{jc}$ in $C$, yields

$C=\varepsilon_{ijk}E^{ai}E^{bj}\chi^{k}{{}_{i^{\prime}}}\varepsilon_{abc}E^{i^{\prime}c}=\chi{{}_{k}}^{i^{\prime}}\varepsilon^{ijk}\varepsilon_{abc}E^{a}_{i}E^{b}_{j}E^{c}_{i^{\prime}}.$

Then using $E^{a}_{i}=ee^{a}_{i}$, where $e^{a}_{i}$ is the inverse of the
gravitational field $e_{a}^{i}$ and $e=\det(e_{a}^{i})$, we obtain

$C=e^{3}\chi{{}_{k}}^{i^{\prime}}\varepsilon^{ijk}\varepsilon_{abc}e^{a}_{i}e^{b}_{j}e^{c}_{i^{\prime}}=e^{3}\chi{{}_{k}}^{i^{\prime}}\varepsilon^{ijk}e^{-1}\varepsilon_{iji^{\prime}}=2e^{2}tr\chi.$

But in the vacuum $tr\chi=0$(the equation (23)), thus
$C=\varepsilon_{ijk}E^{ai}E^{bj}F_{ab}^{k}=0$ 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 $\chi$ and $\varphi$ and do not include
the new field $\psi^{i}$.

$\rm{II}$\- 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:

$P\times G\to P,\quad(p,g)\mapsto pg.$

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 $\pi:P\to M$ and for every $p\in M$, the submanifold
$\pi^{-1}(p)\subset P$ is fibre over $M$. Let $\left\\{{U_{\alpha}}\right\\}$
be open cover of $M$, the local trivialization is G-equivariant
diffeomorphisms

$\psi_{\alpha}:\pi^{-1}(U_{\alpha})\to U_{\alpha}\times G,$

given by $\psi_{\alpha}(p)=\left({\pi(p),g_{\alpha}(p)}\right)$ for some
G-equivariant map $g_{\alpha}:\pi^{-1}(U_{\alpha})\to G$. Equivariance means
that $g_{\alpha}(pg)=g_{\alpha}(p)g$. We say that the bundle is trivial if
there exists a diffeomorphism $\psi:P\to M\times G$ such that
$\psi(p)=\left({\pi(p),\psi(p)}\right)$ and such that $\psi(pg)=\psi(p)g$.
This last condition is simply the G-equivariance of $\psi$.

We separate $T_{p}P$ to vertical and horizontal vector spaces at each point
$p\in P$, we get the vertical vector fields by acting of group $G$ on $P$ by

$\sigma_{p}(X)=\frac{d}{{dt}}\left.{\left({pe^{tX}}\right)}\right|_{t=0},$

for every vector $X\in\mathfrak{g}$ (where $\mathfrak{g}$ is Lie algebra of
$G$), this satisfies

$\pi_{*}\sigma_{p}(X)=\frac{d}{{dt}}\left.{\left({\pi\left({pe^{tX}}\right)}\right)}\right|_{t=0}=\frac{d}{{dt}}\left.{\left({\pi\left(p\right)}\right)}\right|_{t=0}=0,$

thus $\sigma_{p}(X)$ is vertical vector field at $p\in P$. In this bundle the
connection is defined as a map

$D:\Gamma(TP)\to\Gamma(T_{e}G),\quad T_{e}G=lie(G),$

as usual, the connection is a map from the tangent space to itself. And since
$P$ is locally product $U\times G$, then $\left.TP\right|_{U}=TU\times
T_{e}G$, so locally

$D:\Gamma(TU)\times\Gamma(T_{e}G)\to\Gamma(T_{e}G).$

We can get this map by letting $D=\nabla+A$, for $A\in\Gamma(T^{*}U\otimes
T_{e}G)$, and $\nabla$ is connection on $TU$.

##  8 Appendix B: Calculating $\frac{{\delta S_{m}}}{{\delta\Sigma^{i}}}$

Starting from

$\left.{\frac{{\delta S_{m}}}{{\delta
B_{\sigma_{1}\rho_{1}}^{\ell}}}}\right|_{constriant}=\frac{{\delta
S_{m}}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}=\frac{{-2}}{{\sqrt{-g}}}\frac{{\delta
S_{m}}}{{\delta g_{\mu_{1}\nu_{1}}}}\frac{{\sqrt{-g}}}{{-2}}\frac{{\delta
g_{\mu_{1}\nu_{1}}}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}=T^{\mu_{1}\nu_{1}}\frac{{\sqrt{-g}}}{{-2}}\frac{{\delta
g_{\mu_{1}\nu_{1}}}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}},$ (47)

where $T^{\mu\nu}$ is energy-momentum tensor, and $S_{m}$ is matter action
without specifying. Using

$\delta
g_{\mu_{1}\nu_{1}}=\frac{1}{{\sqrt{-g}}}\delta\left({\sqrt{-g}g_{\mu_{1}\nu_{1}}}\right)-\frac{1}{{\sqrt{-g}}}g_{\mu_{1}\nu_{1}}\delta\left({\sqrt{-g}}\right),$

we obtain

$\sqrt{-g}\frac{{\delta
g_{\mu_{1}\nu_{1}}}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}=\frac{{\delta\left({\sqrt{-g}g_{\mu_{1}\nu_{1}}}\right)}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}-g_{\mu_{1}\nu_{1}}\frac{{\delta\left({\sqrt{-g}}\right)}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}=I_{1}-I_{2}.$
(48)

We use the Urbantke formula [12]

$\sqrt{-g}g_{\mu_{1}\nu_{1}}=\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\Sigma_{\sigma\nu_{1}}^{k},$
(49)

from which we get

$\begin{split}&\delta\left({\sqrt{-g}g_{\mu_{1}\nu_{1}}}\right)=\delta\left({\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\Sigma_{\sigma\nu_{1}}^{k}}\right)\\\
&=\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\left({\delta\Sigma_{\mu_{1}\mu}^{i}}\right)\Sigma_{\nu\rho}^{j}\Sigma_{\sigma\nu_{1}}^{k}+\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\mu_{1}\mu}^{i}\left({\delta\Sigma_{\nu\rho}^{j}}\right)\Sigma_{\sigma\nu_{1}}^{k}+\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta\Sigma_{\sigma\nu_{1}}^{k}}\right).\end{split}$

We use it for calculating $I_{1}$ in (48),

$\begin{split}I_{1}&=\frac{\delta}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}\left({\sqrt{-g}g_{\mu_{1}\nu_{1}}}\right)\\\
&=\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\nu\rho}^{j}\Sigma_{\sigma\nu_{1}}^{k}\left({\delta_{\ell}^{i}\delta_{\mu_{1}}^{\sigma_{1}}\delta_{\mu}^{\rho_{1}}}\right)+\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\sigma\nu_{1}}^{k}\left({\delta_{\ell}^{j}\delta_{\nu}^{\sigma_{1}}\delta_{\rho}^{\rho_{1}}}\right)\\\
&\quad\quad\quad+\varepsilon^{\mu\nu\rho\sigma}\varepsilon_{ijk}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta_{\ell}^{k}\delta_{\sigma}^{\sigma_{1}}\delta_{\nu_{1}}^{\rho_{1}}}\right)\\\
&=\varepsilon^{\rho_{1}\nu\rho\sigma}\varepsilon_{\ell
jk}\Sigma_{\nu\rho}^{j}\Sigma_{\sigma\nu_{1}}^{k}\left({\delta_{\mu_{1}}^{\sigma_{1}}}\right)+\varepsilon^{\mu\sigma_{1}\rho_{1}\sigma}\varepsilon_{i\ell
k}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\sigma\nu_{1}}^{k}+\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{ij\ell}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta_{\nu_{1}}^{\rho_{1}}}\right)\\\
&=\varepsilon^{\rho_{1}\nu\rho\sigma}\varepsilon_{\ell
ij}\Sigma_{\nu\rho}^{i}\Sigma_{\sigma\nu_{1}}^{j}\left({\delta_{\mu_{1}}^{\sigma_{1}}}\right)+\varepsilon^{\mu\sigma_{1}\rho_{1}\sigma}\varepsilon_{i\ell
j}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\sigma\nu_{1}}^{j}+\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{ij\ell}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta_{\nu_{1}}^{\rho_{1}}}\right)\\\
&=\varepsilon^{\rho_{1}\nu\rho\sigma}\varepsilon_{\ell
ij}\Sigma_{\nu\rho}^{i}\Sigma_{\sigma\nu_{1}}^{j}\left({\delta_{\mu_{1}}^{\sigma_{1}}}\right)-\varepsilon^{\mu\sigma_{1}\rho_{1}\sigma}\varepsilon_{\ell
ij}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\sigma\nu_{1}}^{j}+\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{\ell
ij}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta_{\nu_{1}}^{\rho_{1}}}\right).\\\
\end{split}$

To calculate $I_{2}$, we use

$\sqrt{-g}=\frac{i}{6}\varepsilon^{\mu\nu\rho\sigma}\delta_{ij}\Sigma_{\mu\nu}^{i}\Sigma_{\rho\sigma}^{j},$

hence

$\begin{split}I_{2}&=g_{\mu_{1}\nu_{1}}\frac{\delta}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}\left({\sqrt{-g}}\right)=\frac{i}{6}g_{\mu_{1}\nu_{1}}\varepsilon^{\mu\nu\rho\sigma}\delta_{ij}\left({\delta_{\ell}^{i}\delta_{\mu}^{\sigma_{1}}\delta_{\nu}^{\rho_{1}}}\right)\Sigma_{\rho\sigma}^{j}+\frac{i}{6}g_{\mu_{1}\nu_{1}}\varepsilon^{\mu\nu\rho\sigma}\delta_{ij}\Sigma_{\mu\nu}^{i}\left({\delta_{\ell}^{j}\delta_{\rho}^{\sigma_{1}}\delta_{\sigma}^{\rho_{1}}}\right)\\\
&=\frac{i}{6}g_{\mu_{1}\nu_{1}}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\delta_{i\ell}\Sigma_{\rho\sigma}^{i}+\frac{i}{6}g_{\mu_{1}\nu_{1}}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\delta_{i\ell}\Sigma_{\mu\nu}^{i}=\frac{i}{3}g_{\mu_{1}\nu_{1}}{\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\delta_{i\ell}\Sigma_{\mu\nu}^{i}}.\end{split}$

Therefore

$\begin{split}&\sqrt{-g}\frac{{\delta
g_{\mu_{1}\nu_{1}}}}{{\delta\Sigma_{\sigma_{1}\rho_{1}}^{\ell}}}=I_{1}-I_{2}\\\
&=\varepsilon^{\rho_{1}\nu\rho\sigma}\varepsilon_{\ell
ij}\Sigma_{\nu\rho}^{i}\Sigma_{\sigma\nu_{1}}^{j}\left({\delta_{\mu_{1}}^{\sigma_{1}}}\right)-\varepsilon^{\mu\sigma_{1}\rho_{1}\sigma}\varepsilon_{\ell
ij}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\sigma\nu_{1}}^{j}+\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{\ell
ij}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta_{\nu_{1}}^{\rho_{1}}}\right)\\\
&-\frac{i}{3}g_{\mu_{1}\nu_{1}}\left({\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\delta_{i\ell}\Sigma_{\mu\nu}^{i}}\right).\end{split}$
(50)

By this the equation (47) becomes

$\begin{split}&\left.{\frac{{\delta S_{m}}}{{\delta
B_{\sigma_{1}\rho_{1}}^{\ell}}}}\right|_{constriant}=\frac{{-1}}{2}T^{\mu_{1}\nu_{1}}\left({I_{1}-I_{2}}\right)\\\
&=\frac{1}{2}T^{\mu_{1}\nu_{1}}\varepsilon^{\mu\sigma_{1}\rho_{1}\sigma}\varepsilon_{\ell
ij}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\sigma\nu_{1}}^{j}-\frac{1}{2}T^{\mu_{1}\nu_{1}}\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{\ell
ij}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}\left({\delta_{\nu_{1}}^{\rho_{1}}}\right)\\\
&-\frac{1}{2}T^{\mu_{1}\nu_{1}}\varepsilon^{\rho_{1}\nu\rho\sigma}\varepsilon_{\ell
ij}\Sigma_{\nu\rho}^{i}\Sigma_{\sigma\nu_{1}}^{j}\left({\delta_{\mu_{1}}^{\sigma_{1}}}\right)+\frac{i}{6}T^{\mu_{1}\nu_{1}}g_{\mu_{1}\nu_{1}}{\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\delta_{i\ell}\Sigma_{\mu\nu}^{i}}\\\
&=I_{1}+I_{2}+I_{3}+I_{4}.\end{split}$ (51)

Using $T^{\mu_{1}\nu_{1}}=T^{IJ}e_{I}^{\mu_{1}}e_{J}^{\nu_{1}}$, the first
term of (51) becomes

$\begin{split}2I_{1}&=T^{\mu_{1}\nu_{1}}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\varepsilon_{ij\ell}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\nu_{1}}^{j}=T^{IJ}e_{I}^{\mu_{1}}e_{J}^{\nu_{1}}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\varepsilon_{ij\ell}P_{KL}^{i}P_{K_{1}L_{1}}^{j}e_{\mu_{1}}^{K}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\nu_{1}}^{L_{1}}\\\
&=T^{IJ}\delta_{I}^{K}\delta_{J}^{L_{1}}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\varepsilon_{ij\ell}P_{KL}^{i}P_{K_{1}L_{1}}^{j}e_{\mu}^{L}e_{\nu}^{K_{1}}=T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\varepsilon_{ij\ell}P_{IL}^{i}P_{K_{1}J}^{j}e_{\mu}^{L}e_{\nu}^{K_{1}}\\\
&=T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\varepsilon_{ij\ell}P_{IL}^{i}P_{K_{1}J}^{j}\left(P_{n}^{LK_{1}}\Sigma_{\mu\nu}^{n}+\bar{P}_{n}^{LK_{1}}\bar{\Sigma}_{\mu\nu}^{n}\right),\end{split}$

where we used

$e_{[\mu}^{L}e_{\nu]}^{K_{1}}=P_{n}^{LK_{1}}\Sigma_{\mu\nu}^{n}+\bar{P}_{n}^{LK_{1}}\bar{\Sigma}_{\mu\nu}^{n}.$

We use the property of the self-dual projection

$-\frac{1}{2}\varepsilon_{ij\ell}P_{IJ}^{i}P_{KL}^{j}=\frac{1}{4}\left(\eta_{JK}P_{\ell
IL}-\eta_{JL}P_{\ell IK}\right)-\frac{1}{4}\left({I\leftrightarrow J}\right),$
(52)

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 $SO(3,1)\times
SO(3,\mathbb{C})$ invariance asserts that this property is also satisfied when
$I,J,K,L$ are all spatial indices. By using this property, we obtain

$\begin{split}2I_{1}&=T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\frac{-1}{2}\left({\eta_{LK_{1}}P_{\ell
IJ}-\eta_{LJ}P_{\ell IK_{1}}-\eta_{IK_{1}}P_{\ell LJ}+\eta_{IJ}P_{\ell
LK_{1}}}\right)e_{\mu}^{L}e_{\nu}^{K_{1}}\\\
&=T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\frac{-1}{2}\left({P_{\ell
IJ}g_{\mu\nu}-\eta_{LJ}P_{\ell
IK_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}-\eta_{IK_{1}}P_{\ell
LJ}e_{\mu}^{L}e_{\nu}^{K_{1}}+\eta_{IJ}\Sigma_{\ell\mu\nu}}\right)\\\
&=T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\frac{-1}{2}\left(-\eta_{LJ}P_{\ell
IK_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}-\eta_{IK_{1}}P_{\ell
LJ}e_{\mu}^{L}e_{\nu}^{K_{1}}+\eta_{IJ}\Sigma_{\ell\mu\nu}\right)\\\
&=\frac{1}{2}T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\left(\eta_{LJ}P_{\ell
IK_{1}}+\eta_{IK_{1}}P_{\ell
LJ}\right)e_{[\mu}^{L}e_{\nu]}^{K_{1}}-\frac{1}{2}\eta_{IJ}T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\ell\mu\nu},\end{split}$

hence

$\begin{split}2I_{1}&=\frac{1}{2}T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\left({\eta_{LJ}P_{\ell
IK_{1}}+\eta_{IK_{1}}P_{\ell
LJ}}\right)\left({P_{i}^{LK_{1}}\Sigma_{\mu\nu}^{i}+\bar{P}_{i}^{LK_{1}}\bar{\Sigma}_{\mu\nu}^{i}}\right)\\\
&-\frac{1}{2}\eta_{IJ}T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\ell\mu\nu}.\end{split}$

Finally

$\begin{split}I_{1}&=\frac{1}{4}T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\left({\eta_{LJ}P_{\ell
IK_{1}}+\eta_{IK_{1}}P_{\ell
LJ}}\right)\left({P_{i}^{LK_{1}}\Sigma_{\mu\nu}^{i}+\bar{P}_{i}^{LK_{1}}\bar{\Sigma}_{\mu\nu}^{i}}\right)\\\
&-\frac{1}{4}\eta_{IJ}T^{IJ}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\ell\mu\nu}.\end{split}$

And the second term of (51) becomes

$\begin{split}2I_{2}&=-T^{\mu_{1}\nu_{1}}\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{ij\ell}\Sigma_{\mu_{1}\mu}^{i}\Sigma_{\nu\rho}^{j}{\delta_{\nu_{1}}^{\rho_{1}}}=-T^{IJ}e_{I}^{\mu_{1}}e_{J}^{\rho_{1}}\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{ij\ell}P_{KL}^{i}P_{K_{1}L_{1}}^{j}e_{\mu_{1}}^{K}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}\\\
&=-T^{IJ}\delta_{I}^{K}e_{J}^{\rho_{1}}\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{ij\ell}P_{KL}^{i}P_{K_{1}L_{1}}^{j}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}=-T^{IJ}e_{J}^{\rho_{1}}\varepsilon^{\mu\nu\rho\sigma_{1}}\varepsilon_{ij\ell}P_{IL}^{i}P_{K_{1}L_{1}}^{j}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}\\\
&=-T^{IJ}\varepsilon^{\mu\nu\rho\sigma_{1}}\frac{-1}{2}\left({\eta_{LK_{1}}P_{\ell
IL_{1}}-\eta_{LL_{1}}P_{\ell IK_{1}}-\eta_{IK_{1}}P_{\ell
LL_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}}\right)e_{J}^{\rho_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}=\\\
\frac{1}{2}T^{IJ}&\varepsilon^{\mu\nu\rho\sigma_{1}}\left({P_{\ell
IL_{1}}g_{\mu\nu}e_{\rho}^{L_{1}}-P_{\ell
IK_{1}}g_{\mu\rho}e_{\nu}^{K_{1}}-\eta_{IK_{1}}P_{\ell
LL_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}}\right)e_{J}^{\rho_{1}}\\\
&=\frac{1}{2}T^{IJ}\varepsilon^{\mu\nu\rho\sigma_{1}}\left({-\eta_{IK_{1}}P_{\ell
LL_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}}\right)e_{J}^{\rho_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}.\end{split}$

Using
$\varepsilon^{\mu\nu\rho\sigma_{1}}e_{\mu}^{L}e_{\nu}^{K_{1}}e_{\rho}^{L_{1}}=e\varepsilon^{LK_{1}L_{1}M}e_{M}^{\sigma_{1}}$,
where $e$ is the determinant of $(e^{I}_{\mu})$, we obtain

$\begin{split}2I_{2}&=\frac{1}{2}eT^{IJ}\left({-\eta_{IK_{1}}P_{\ell
LL_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}}\right)e_{J}^{\rho_{1}}\varepsilon^{LK_{1}L_{1}M}e_{M}^{\sigma_{1}}\\\
&=\frac{1}{2}eT^{IJ}\left({-\eta_{IK_{1}}P_{\ell LL_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}}\right)\varepsilon^{LK_{1}L_{1}M}e_{J}^{[\rho_{1}}e_{M}^{\sigma_{1}]}\\\
&=\frac{1}{2}eT^{IJ}\left({-\eta_{IK_{1}}P_{\ell LL_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}}\right)\varepsilon^{LK_{1}L_{1}M}e_{[J}^{\rho_{1}}e_{M]}^{\sigma_{1}}\\\
&=\frac{1}{2}eT^{IJ}\left({-\eta_{IK_{1}}P_{\ell LL_{1}}+\eta_{IL_{1}}P_{\ell
LK_{1}}}\right)\varepsilon^{LK_{1}L_{1}M}\left({P_{iJM}\Sigma^{i\rho_{1}\sigma_{1}}+\bar{P}_{iJM}\bar{\Sigma}^{i\rho_{1}\sigma_{1}}}\right).\end{split}$

We use the selfdual property

$P_{IJ}^{i}\varepsilon^{IJKL}=-2iP^{iKL},$

to get

$2I_{2}=-ieT^{IJ}\left({\eta_{IK_{1}}P_{\ell}^{K_{1}M}+\eta_{IL_{1}}P_{\ell}^{L_{1}M}}\right)\left({P_{iJM}\Sigma^{i\rho_{1}\sigma_{1}}+\bar{P}_{iJM}\bar{\Sigma}^{i\rho_{1}\sigma_{1}}}\right).$

And using

$\frac{1}{{2!}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\rho\sigma}^{i}=\left({*\Sigma^{i}}\right)^{\mu\nu}=\left({-i\Sigma^{i}}\right)^{\mu\nu}=-i\Sigma^{\mu\nu
i},\quad e=\sqrt{-g},$

we obtain

$I_{2}=-\frac{1}{4}T^{IJ}\left({\eta_{IK_{1}}P_{\ell}^{K_{1}M}+\eta_{IL_{1}}P_{\ell}^{L_{1}M}}\right)\left({-P_{iJM}\varepsilon^{\rho_{1}\sigma_{1}\rho\sigma}\Sigma_{\rho\sigma}^{i}+\bar{P}_{iJM}\varepsilon^{\rho_{1}\sigma_{1}\rho\sigma}\bar{\Sigma}_{\rho\sigma}^{i}}\right).$

We get the third term of (51) by the replacing
$\sigma_{1}\leftrightarrow\rho_{1}$ in $I_{2}$ with reversing its sign, we
obtain

$I_{3}=+\frac{1}{4}T^{IJ}\left({\eta_{IK_{1}}P_{\ell}^{K_{1}M}+\eta_{IL_{1}}P_{\ell}^{L_{1}M}}\right)\left({-P_{iJM}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\Sigma_{\rho\sigma}^{i}+\bar{P}_{iJM}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\bar{\Sigma}_{\rho\sigma}^{i}}\right)=I_{2},$

therefore

$I_{2}+I_{3}=+\frac{1}{2}T^{IJ}\left({\eta_{IK_{1}}P_{\ell}^{K_{1}M}+\eta_{IL_{1}}P_{\ell}^{L_{1}M}}\right)\left({-P_{iJM}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\Sigma_{\rho\sigma}^{i}+\bar{P}_{iJM}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\bar{\Sigma}_{\rho\sigma}^{i}}\right).$

The fourth term of (51) is

$I_{4}=\frac{i}{6}T^{\mu_{1}\nu_{1}}g_{\mu_{1}\nu_{1}}{\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\mu\nu\ell}}.$

Using $T^{\mu\nu}g_{\mu\nu}=T^{IJ}\eta_{IJ}=T$, the equation (51) becomes

$\begin{split}&\left.{\frac{{\delta S_{m}}}{{\delta
B_{\sigma_{1}\rho_{1}}^{\ell}}}}\right|_{constriant}=I_{1}+I_{2}+I_{3}+I_{4}\\\
&=\frac{1}{4}T^{IJ}\left({\eta_{LJ}P_{\ell IK}+\eta_{IK}P_{\ell
LJ}}\right)\left({P_{i}^{LK}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\mu\nu}^{i}+\bar{P}_{i}^{LK}\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\bar{\Sigma}_{\mu\nu}^{i},}\right)-\frac{1}{4}T\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\ell\mu\nu}\\\
&+\frac{1}{2}T^{IJ}\left({\eta_{IK}P_{\ell}^{KM}+\eta_{IL}P_{\ell}^{LM}}\right)\left({-P_{iJM}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\Sigma_{\rho\sigma}^{i}+\bar{P}_{iJM}\varepsilon^{\sigma_{1}\rho_{1}\rho\sigma}\bar{\Sigma}_{\rho\sigma}^{i}}\right)\\\
&+\frac{i}{6}T\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\Sigma_{\mu\nu\ell}.\end{split}$

We write this for short as

$\left.{\frac{{\delta S_{m}}}{{\delta
B_{\sigma_{1}\rho_{1}}^{\ell}}}}\right|_{constriant}=\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}T_{\ell
i}\Sigma_{\mu\nu}^{i}+\varepsilon^{\mu\nu\sigma_{1}\rho_{1}}\xi_{\ell
i}\bar{\Sigma}_{\mu\nu}^{i},$

or

$\left.{\frac{{\delta S_{m}}}{{\delta
B^{i}}}}\right|_{constriant}={\frac{{\delta
S_{m}}}{{\delta\Sigma^{i}}}}=T_{ij}\Sigma^{j}+\xi_{ij}\bar{\Sigma}^{j},$ (53)

the complex matrices $T$ and $\xi$ are given by

$\begin{split}T_{ij}&=\frac{1}{4}T^{IJ}\left({\eta_{LJ}P_{iIK}+\eta_{IK}P_{iLJ}}\right)P_{j}^{LK}-\frac{1}{2}T^{IJ}\left({\eta_{IK}P_{i}^{KM}+\eta_{IL}P_{i}^{LM}}\right)P_{jJM}\\\
&\quad\quad+T\left({\frac{i}{6}-\frac{1}{4}}\right)\delta_{ij}\\\
&=-\frac{1}{2}\left(P_{iIK}P_{jJ}{{}^{K}}\right)T^{IJ}+\left({\frac{i}{6}-\frac{1}{4}}\right)T\delta_{ij},\end{split}$
(54)

and

$\begin{split}\xi_{ij}&=\frac{1}{4}T^{IJ}\left({\eta_{LJ}P_{iIK}+\eta_{IK}P_{iLJ}}\right)\bar{P}_{j}^{LK}+\frac{1}{2}T^{IJ}\left({\eta_{IK}P_{i}^{KM}+\eta_{IL}P_{i}^{LM}}\right)\bar{P}_{jJM}\\\
&=\frac{3}{2}\left(P_{iIK}\bar{P}_{jJ}{{}^{K}}\right)T^{IJ}.\end{split}$

The self-dual projection matrices $P_{IJ}^{i}$ are given in the equation (1),

$P^{i}_{IJ}=\frac{1}{2}\varepsilon^{i}{{}_{jk}},\text{ for }I=i,J=j,\text{ and
}P^{i}_{0j}=-P^{i}_{j0}=-\frac{i}{2}\delta^{i}_{j},\text{ for }I=0,J=j\neq 0.$
(55)

and $\bar{P}_{IJ}^{i}$ are their complex conjugate. We get

$\begin{split}&\left({P_{iIK}P_{jJ}^{K}}\right)T^{IJ}=\left({P_{iI0}P_{jJ}^{0}+P_{iI\ell}P_{jJ}^{\ell}}\right)T^{IJ}\\\
&=P_{i\ell 0}P_{jm}^{0}T^{\ell
m}+P_{i0\ell}P_{j0}^{\ell}T^{00}+P_{im\ell}P_{j0}^{\ell}T^{m0}+P_{i0\ell}P_{jm}^{\ell}T^{0m}+P_{in\ell}P_{jm}^{\ell}T^{nm}\\\
&=-P_{i\ell 0}P_{jm0}T^{\ell
m}+P_{i0\ell}P_{j0}^{\ell}T^{00}+P_{im\ell}P_{j0}^{\ell}T^{m0}+P_{i0\ell}P_{jm}^{\ell}T^{0m}+P_{in\ell}P_{jm}^{\ell}T^{nm}\\\
&=-\frac{i}{2}\delta_{i\ell}\frac{i}{2}\delta_{jm}T^{\ell
m}+\frac{{-i}}{2}\delta_{i\ell}\frac{{-i}}{2}\delta_{j}^{\ell}T^{00}+\frac{1}{2}\varepsilon_{im\ell}\frac{{-i}}{2}\delta_{j}^{\ell}T^{m0}+\frac{{-i}}{2}\delta_{i\ell}\frac{1}{2}\varepsilon_{jm}^{\ell}T^{0m}+\frac{1}{2}\varepsilon_{in\ell}\frac{1}{2}\varepsilon_{jm}^{\ell}T^{nm}\\\
&=\frac{1}{4}T_{ij}-\frac{1}{4}\delta_{ij}T^{00}+\frac{{-i}}{4}\varepsilon_{imj}T^{m0}+\frac{{-i}}{4}\varepsilon_{jmi}T^{0m}+\frac{1}{4}\left({\delta_{ij}\delta_{nm}-\delta_{im}\delta_{nj}}\right)T^{nm}\\\
&=\frac{1}{4}T_{ij}-\frac{1}{4}\delta_{ij}T^{00}+\frac{{-i}}{4}\varepsilon_{imj}T^{0m}+\frac{{-i}}{4}\varepsilon_{jmi}T^{0m}+\frac{1}{4}\delta_{ij}\delta_{nm}T^{nm}-\frac{1}{4}T_{ji}\\\
&=-\frac{1}{4}\delta_{ij}T^{00}+\frac{1}{4}\delta_{ij}\delta_{nm}T^{nm}=\frac{1}{4}\delta_{ij}\eta_{00}T^{00}+\frac{1}{4}\delta_{ij}\eta_{nm}T^{nm}=\frac{1}{4}\delta_{ij}\eta_{IJ}T^{IJ}=\frac{1}{4}\delta_{ij}T.\end{split}$

Using this in the equation (54), we get

$T_{ij}=-\frac{1}{8}\delta_{ij}T+\left({\frac{i}{6}-\frac{1}{4}}\right)T\delta_{ij}=\left({\frac{i}{6}-\frac{3}{8}}\right)T\delta_{ij}.$
(56)

##  9 Appendix C

$\rm{I}$\- We verify that the (0,2) tensor field $\Sigma^{\mu\nu i}$ defined
in

$-i\Sigma^{\mu\nu
i}=\frac{1}{{2!}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\rho\sigma}^{i},\quad\text{for}\quad\varepsilon^{0123}=-\varepsilon_{0123}=1,\quad
e=\det(e_{\mu}^{I}),$

is inversion of the 2-form $\Sigma_{\mu\nu}^{i}$, that is
$\Sigma_{\mu\nu}^{i}\Sigma^{\mu\nu}_{j}=\delta^{i}_{j}$. Multiplying with
$\Sigma_{\mu\nu j}$ and sum over contracted indices, we get

$-i\Sigma_{\mu\nu j}\Sigma^{\mu\nu
i}=\frac{1}{{2}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\mu\nu
j}\Sigma_{\rho\sigma}^{i}.$

Then using $\Sigma^{i}_{\mu\nu}=P_{IJ}^{i}e^{I}_{\mu}e^{J}_{\nu}$, implies

$-i\Sigma_{\mu\nu j}\Sigma^{\mu\nu
i}=\frac{1}{{2}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}P_{jIJ}P^{i}_{KL}e^{I}_{\mu}e^{J}_{\nu}e^{K}_{\rho}e^{L}_{\sigma}.$

Using
$\varepsilon^{\mu\nu\rho\sigma}e^{I}_{\mu}e^{J}_{\nu}e^{K}_{\rho}e^{L}_{\sigma}=e\varepsilon^{IJKL}$,
to get

$-i\Sigma_{\mu\nu j}\Sigma^{\mu\nu
i}=\frac{1}{{2}}e^{-1}P_{jIJ}P^{i}_{KL}e\varepsilon^{IJKL}=\frac{1}{{2}}P_{jIJ}P^{i}_{KL}\varepsilon^{IJKL}.$

Then we use the self-dual projection property

$P_{IJ}^{i}\varepsilon^{IJKL}=-2iP^{iKL}$

to obtain

$-i\Sigma_{\mu\nu j}\Sigma^{\mu\nu
i}=\frac{1}{{2}}P_{jIJ}(-2iP^{iIJ})=P_{jIJ}(-iP^{iIJ}).$

Therefore

$\Sigma_{\mu\nu j}\Sigma^{\mu\nu i}=P_{jIJ}P^{iIJ}=\delta^{i}_{j},$

where we used the self-dual projection property
$P_{jIJ}P^{iIJ}=\delta^{i}_{j}$. The sum is over the contracted indices.

$\rm{II}$\- 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 $\varphi$ is ([16], section 22)

$M^{IJK}=x^{J}T^{IK}-x^{K}T^{IJ},$ (57)

the conservation law is $\partial_{I}M^{IJK}=0$, it follows from the
conservation of energy-momentum tensor $\partial_{I}T^{IJ}=0$ of the field
$\varphi$. When the field carries spinor indices, like $\varphi^{\alpha}$,
this adds a term like

$M^{IJK}=x^{J}T^{IK}-x^{K}T^{IJ}+\pi_{\alpha}^{I}(S^{JK})^{\alpha}{{}_{\beta}}\varphi^{\beta}.$

In arbitrary coordinates $(x^{\mu})$, we write $x^{I}=e^{I}_{\mu}x^{\mu}$,
where $x^{I}$ becomes tangent vector and $e^{I}_{\mu}$ are gravitational
fields, also $T^{\mu\nu}=e_{I}^{\mu}e_{J}^{\nu}T^{IJ}$. One can write

$M^{\mu IJ}=e^{I}_{\nu}x^{\nu}T^{\mu J}-e^{J}_{\nu}x^{\nu}T^{\mu I}.$ (58)

Therefore $D_{\mu}M^{\mu IJ}=0$ is satisfied when $D_{\mu}e^{I}_{\nu}=0$ and
$D_{\mu}T^{\mu I}=0$. Here $e^{I}_{\mu}$ and $T^{\mu I}$ are functions of the
coordinates $(x^{\mu})$, so also $M^{\mu IJ}$ are functions of $(x^{\mu})$.
The current $M^{\mu IJ}$ couples to the spin connection $\omega^{IJ}_{\mu}$
for local symmetry of Lorentz group $SO(3,1)$. The equation
$D_{\mu}e^{I}_{\nu}=0$ defines affine connection $\nabla$ on $TM\times TM$ and
spin connection $\omega$ on $TM\times\mathfrak{so}(3,1)$. When the coordinates
$(x^{\mu})$ are flat, so $e^{I}_{\mu}=\delta^{I}_{\mu}$, by that the equation
(58) becomes (57).

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