# Yang–Mills Theory of Gravity

Malik AL Matwi

( Department of Mathematical Science, Ritsumeikan University, 4-2-28, Kusatsu-
Shi, Shiga, 525-0034, Japan
<EMAIL_ADDRESS>
This version includes some corrections to the
published paper: physics 2019, 1(3), 339-359)

###### Contents

  1. 1 Solutions using spin currents
  2. 2 Modifying by adding Yang-Mills Lagrangian
  3. 3 Decomposition Space–Time Manifold $M$ into $R\times\Sigma$
  4. 4 Conclusions
  5. 5 Appendix A
  6. 6 Appendix B

## Abstract

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

Keywords: Lorentz invariance; spin currents; complex connection $A^{i}$.

## Introduction

Gravity can be formulated based on gauge theory by gauging the Lorentz group
$SO(3,1)$ [1]. For this purpose, we need to fix some base space and consider
that the Lorentz group $SO(3,1)$ acts locally on Lorentz frames which are
regarded as a frame bundle over a fixed base space $M$. We can consider this
base space as an arbitrary space–time manifold $M$ with coordinates
$(x^{\mu})$, and consider the local Lorentz frame as an element in the tangent
frame bundle over $M$. By that we have two symmetries; invariance under
continuous transformations of local Lorentz frame, $SO(3,1)$ group, and
invariance under diffeomorphism of the space–time $M$, which is originally
considered as a base space [2].

The local Lorentz invariance gives a conserved current, call it spin current,
or Lorentz current. In flat coordinates $(x^{I})$, the spin current for
arbitrary field $\varphi$ is ([15], section 22)

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

the conservation law is $\partial_{I}M^{IJK}=0$, with conservation of energy-
momentum tensor $\partial_{I}T^{IJ}=0$. 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, also
$T^{\mu\nu}=e_{I}^{\mu}e_{J}^{\nu}T^{IJ}$. One can write

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

Therefore $D_{\mu}M^{\mu IJ}=0$ when $D_{\mu}e^{I}_{\nu}=0$ and $D_{\mu}T^{\mu
I}=0$. 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 $TM\times\mathfrak{so}(3,1)$. In
this paper, we try to solve the equation of motion of the Plebanski Lagrangian

$L=B^{\mu\nu}_{i}F^{i}_{\mu\nu}+\phi_{ij}B^{\mu\nu
i}B^{j}_{\mu\nu}+L_{matter},$ (1)

using only the spin connection, without needing using the affine connection
$\nabla$(avoiding using it).

##  1 Solutions using spin currents

By regarding the local Lorentz symmetry as a gauge symmetry with spin
connection $\omega\in\Omega^{1}(M,\mathfrak{so}(3,1))$ (or
$A\in\Omega^{1}(M,\mathfrak{so}(3,\mathbb{C}))$) as gauge fields, we recognize
Yang–Mills theory in gravity. But not full gravity, since in the Yang–Mills
theory, the variables are connections and conserved currents, while in the
gravity the metric is also variable. The local Lorentz symmetry produces
locally conserved currents, and those currents are coupled to spin connection
$\omega^{IJ}$. This makes the local Lorentz symmetry a gauge symmetry with the
Lorentz group as a gauge group. Also, these currents must be conserved and
vanish in the vacuum. We try in this section to solve the equations of motion
of general relativity in self-dual formalism using only the spin
current(Lorentz current), in static case, and without needing solving the
Einstein’s equation. We give an example, matter located at a point, so we have
spherical symmetric system.

In self-dual formalism of general relativity, the independent variables are
$B^{\mu\nu}_{i}$ and $A^{i}_{\mu}$, they take values in Lie algebra
$\mathfrak{so}(3,\mathbb{C})$. The non-zero spin current $\delta
L_{matter}/\delta A^{i}_{\mu}$ implies $D_{\mu}B^{\mu\nu i}\neq 0$(here
$D_{\mu}=\nabla_{\mu}+A_{\mu}$). Therefore we write $B^{\mu\nu
i}=K{{}^{i}}_{j}\Sigma^{\mu\nu i}$, for some $3\times 3$ complex matrix
$K{{}^{i}}_{j}$ of scalar functions on $M$ in order to get
$D_{\mu}\Sigma^{\mu\nu i}=0$ and $\Sigma^{\mu\nu
i}=P^{iIJ}e_{I}^{\mu}e_{J}^{\nu}$, with gravitational field $e^{I}_{\mu}$
satisfying $D_{\nu}e^{I}_{\mu}=0$, here $D_{\mu}=\nabla_{\mu}+\omega_{\mu}$,
where $\nabla$ is affine connection on $M$, we avoid using it. The
gravitational field $e^{I}_{\mu}$ defines a metric
$g_{\mu\nu}=e^{J}_{\mu}e_{J\nu}$, it satisfies
$\nabla_{\rho}g_{\mu\nu}=(D_{\rho}e^{J}_{\mu})e_{J\nu}+e^{J}_{\mu}(D_{\rho}e_{J\nu})=0$.
The field $e_{I}^{\mu}$ is inversion of $e^{I}_{\mu}$, that is
$e^{I}_{\mu}e_{J}^{\mu}=\delta^{I}_{J}$ and
$e^{I}_{\mu}e_{I}^{\nu}=\delta^{\nu}_{\mu}$.

We try to find the solutions of $K{{}^{i}}_{j}$ and $\Sigma^{\mu\nu i}$, so
obtaining the spin connection $A^{i}_{\mu}$ from $D_{\mu}\Sigma^{\mu\nu i}=0$.
By that we obtain the curvature of $A^{i}_{\mu}$ and the field $B^{\mu\nu
i}=K{{}^{i}}_{j}\Sigma^{\mu\nu i}$.

The general relativity Lagrangian in self-dual formalism including matter is

$L_{dual\text{ }GR}+L_{matter}=B^{\mu\nu}_{i}F^{i}_{\mu\nu}+\phi_{ij}B^{\mu\nu
i}B^{j}_{\mu\nu}+L_{matter},$ (2)

where $\phi$ is traceless matrix. The equation of motion of $B^{\mu\nu}_{i}$
is

$F^{i}_{\mu\nu}+\phi{{}^{i}}_{j}B^{j}_{\mu\nu}+\frac{\delta L_{matter}}{\delta
B^{\mu\nu}_{i}}=0.$

The two forms $F^{i}_{\mu\nu}$ and $B^{i}_{\mu\nu}$ takes values in Lie
algebra $\mathfrak{so}(3,\mathbb{C})$, so there are always matrices $\psi$ and
$\psi^{\prime}$ of scalar functions, so writing

$F^{i}=\psi{{}^{i}}_{j}B^{j}+\xi{{}^{i}}_{j}\bar{B}^{j}.$ (3)

The equations of motion asserts that the matrices $\psi$ and $\xi$ are
symmetric.

The equation of motion of the spin connection $A^{i}_{\mu}$ is

$-D_{\mu}B^{\mu\nu i}+J^{\nu i}=0.$ (4)

Now we try to solve the equation (4) using $B^{\mu\nu
i}=K{{}^{i}}_{j}\Sigma^{\mu\nu i}$ with $D_{\mu}\Sigma^{\mu\nu i}=0$.
Therefore

$-(D_{\mu}K{{}^{i}}_{j})\Sigma^{\mu\nu i}+J^{\nu i}=0.$ (5)

Using $B^{i}=K{{}^{i}}_{j}\Sigma^{i}$ in (3), one finds that the matrix $K$
must also be symmetric in order to satisfy the general relativity constraints
using $\Sigma^{i}$ and $F^{i}=(\psi K){{}^{i}}_{j}\Sigma^{j}$ in the
vacuum($\psi^{\prime}{{}^{i}}_{j}=0$), where we let $\Sigma^{i}$ be a
solution. By $SO(3,\mathbb{C})$ transformation using an invertable matrix
$U{{}^{i}}_{j}$, we can get $K{{}^{i}}_{j}=K^{i}\delta^{i}_{j}$, for some
scalar functions $K^{i}$. Latter we find that $K^{i}$ satisfies
$\partial^{2}K^{i}=0$.

The equation (5) means that the spin current becomes a source for
$D_{\mu}K{{}^{i}}_{j}$ instead of being source for $D_{\mu}B^{\mu\nu i}$. The
new basis $\Sigma^{\mu\nu i}$ has to be given using gravitational field
$e^{I}_{\mu}$ with the inversion $e_{I}^{\mu}$, so writing $\Sigma^{\mu\nu
i}=P^{iIJ}e_{I}^{\mu}e_{J}^{\nu}$, the field $e^{I}_{\mu}$ satisfies
$D_{\nu}e^{I}_{\mu}=0$. The matrix $K{{}^{i}}_{j}$ has to be symmetric as
mentioned before.

Using $\Sigma^{\mu\nu i}=P^{iIJ}e_{I}^{\mu}e_{J}^{\nu}$ and $J^{\nu
i}=P_{IJ}^{i}J^{\nu IJ}$ in (5), we obtain

$\left({D_{\mu}K{{}^{i}}_{j}}\right)P_{IJ}^{j}e^{I\mu}e^{J\nu}=P_{IJ}^{i}J^{\nu
IJ},$

and using $J^{\nu IJ}=e^{I\mu}e^{\rho J}J_{\mu\rho}^{\nu}$ which is possible
since $e_{I\mu}e_{J\rho}J^{\nu IJ}=J_{\mu\rho}^{\nu}$ exists, we find

$\left({D_{\mu}K{{}^{i}}_{j}}\right)P_{IJ}^{j}e^{I\mu}e^{J\nu}=P_{IJ}^{i}e^{I\mu}e^{\rho
J}J_{\mu\rho}^{\nu},$

so

$\left({D_{\mu}K{{}^{i}}_{j}}+v_{\mu}^{i}{{}_{j}}\right)P_{IJ}^{j}e^{J\nu}=P_{IJ}^{i}e^{\rho
J}J_{\mu\rho}^{\nu},\quad\text{ for
}v_{\mu}^{i}{{}_{j}}P_{IJ}^{j}e^{I\mu}e^{J\nu}=0.$

Since $K{{}^{i}}_{j}$ is real symmetric matrix, we let $v_{\mu}^{i}{{}_{j}}$
be also real and symmetric in $i,j$. The equation
$v_{\mu}^{i}{{}_{j}}P_{IJ}^{j}e^{I\mu}e^{J\nu}=0$ implies
$v_{\mu}^{i}{{}_{j}}P_{IJ}^{j}e^{I\mu}=0$. We can make $SO(3)\times SO(3,1)$
transformation on $v_{\mu}^{i}{{}_{j}}P_{IJ}^{j}e^{I\mu}=0$ to get
$v_{\mu}^{i}{{}_{j}}=v_{\mu}^{i}\delta^{i}_{j}$. And according to the
calculations below, we find that
$v_{\mu}^{i}\delta^{i}_{j}P_{IJ}^{j}e^{I\mu}=0$ implies $v_{\mu}^{i}=0$. So we
drop $v_{\mu}^{i}{{}_{j}}$ from now.

From the conservation $D_{\nu}J^{\nu IJ}=D_{\nu}(e^{I\mu}e^{\rho
J}J_{\mu\rho}^{\nu})=e^{I\mu}e^{\rho J}D_{\nu}J_{\mu\rho}^{\nu}=0$(using
$D_{\nu}e^{I}_{\mu}=0$), one obtains

$P_{IJ}^{j}e^{J\nu}D_{\nu}\left({D_{\mu}K{{}^{i}}_{j}}\right)=P_{IJ}^{i}e^{\rho
J}D_{\nu}J_{\mu\rho}^{\nu}=0,$

so

$P_{IJ}^{j}e^{J\nu}\left({D_{\nu}D_{\mu}K{{}^{i}}_{j}}\right)=0,\quad\text{
for each }I=0,1,2,3\text{ and }i=1,2,3.$

The matrix $K^{i}{{}_{j}}$ is symmetric, so we can make $SO(3)\times SO(3,1)$
transformation using a matrix $U{{}^{i}}_{j}$ to get
$K^{i}{{}_{j}}=K^{i}\delta{{}^{i}}_{j}$, so

$D_{\mu}K{{}^{i}}_{j}=\left({\partial_{\mu}K^{i}}\right)\delta{{}^{i}}_{j}+K^{i}D_{\mu}\delta{{}^{i}}_{j}.$

But $D_{\mu}\delta{{}^{i}}_{j}=0$, thus

$D_{\mu}K{{}^{i}}_{j}=\left({\partial_{\mu}K^{i}}\right)\delta{{}^{i}}_{j},$

Therefore

$P_{IJ}^{i}e^{J\nu}\left({\partial_{\nu}\partial_{\mu}K^{i}}\right)=0\quad\text{
for each }I=0,1,2,3\text{ and }i=1,2,3.$

For $I=0$, we have $P_{0j}^{i}=-i\delta{{}^{i}}_{j}/2$, it implies
$e^{i\nu}{\partial_{\nu}\partial_{\mu}K^{i}}=0$ for each $i=1,2,3$. And for
$I=j$, we have $P_{jk}^{i}=\epsilon{{}^{i}}_{jk}/2$, it implies
$e^{i\nu}{\partial_{\nu}\partial_{\mu}K^{j}}=e^{j\nu}{\partial_{\nu}\partial_{\mu}K^{i}}$
for each $i,j=1,2,3$. The first equation
$e^{i\nu}{\partial_{\nu}\partial_{\mu}K^{i}}=0$ implies the proportionality
${\partial_{\nu}\partial_{\mu}K^{i}}\sim\epsilon{{}^{i}}_{jk}e^{j}_{\nu}$(or
$\sim\epsilon_{\nu\rho\sigma\gamma}e^{\rho i}$), using it in the second
equation leads to

$e^{i\nu}{\partial_{\nu}\partial_{\mu}K^{j}}\sim\epsilon{{}^{j}}_{\ell
k}e^{i\nu}e^{\ell}_{\nu}=\epsilon{{}^{j}}_{\ell k}\delta^{\ell
i}=-\epsilon^{ij}{{}_{k}}$

which is anti-symmetric in $i,j$, this is contradict with the symmetric in the
second equation
$e^{i\nu}{\partial_{\nu}\partial_{\mu}K^{j}}=e^{j\nu}{\partial_{\nu}\partial_{\mu}K^{i}}$.
Therefore we choose
$e^{I\nu}\left({\partial_{\nu}\partial_{\mu}K^{i}}\right)=0$, so

$e^{\mu}_{I}e^{I\nu}\left({\partial_{\nu}\partial_{\mu}K^{i}}\right)=g^{\mu\nu}\partial_{\nu}\partial_{\mu}K^{i}=\partial^{2}K^{i}=0,$
(6)

where we used the metric $g_{\mu\nu}=e^{J}_{\mu}e_{J\nu}$, it satisfies
$\nabla_{\rho}g_{\mu\nu}=(D_{\rho}e^{J}_{\mu})e_{J\nu}+e^{J}_{\mu}(D_{\rho}e_{J\nu})=0$.

We note that we used a metric $g_{\mu\nu}$ defined by $e_{J\mu}e^{J}_{\nu}$,
but the equation $\partial^{2}K^{i}=0$ is invariant under any coordinates
transformation $x\to x^{\prime}$,
${\psi^{\prime}}^{i}(x^{\prime})=\psi^{i}(x)$, so the equation
$\partial^{2}K^{i}=0$ is valid for any other metric ${g^{\prime}}_{\mu\nu}$(at
least locally). Also we let the affine connection $\nabla_{\mu}$ be flat
connection, therefore we can make coordinates transformation to obtain
euclidean metric on $M$, and so $\nabla_{\mu}=\partial_{\mu}$. Another
treatment of $e^{I\nu}{\partial_{\nu}\partial_{\mu}K^{i}}=0$ is multiplying it
with $e_{I\rho}$ and sum over $I$ to get
${\partial_{\rho}\partial_{\mu}K^{i}}=0$ which can be solved by letting
$K^{i}$ depend only on one coordinate(like $r$ in spherical coordinates), we
obtain $\eta^{\mu\nu}{\partial_{\mu}\partial_{\nu}K^{i}}=0$, for Lorentz
metric $\eta$. By that the functions $K^{i}$ do not depend on the metric
$e_{J\mu}e^{J}_{\nu}$ on $M$, but they have to depend only on one coordinate
on $M$.

To get a solution, we first solve $\partial^{2}K^{i}=0$(by letting $K^{i}$
depend only on one coordinate), then using it in
$-(\partial_{\mu}K^{i})\Sigma^{\mu\nu i}+J^{\nu i}=0$, the equation (5), to
get the solution of $\Sigma^{i}_{\mu\nu}$ in terms of given spin current
$J^{\nu i}$, and so getting $B^{i}_{\mu\nu}$, where the spin current
$J^{i\mu}$ is assumed to be given as a function on $M$. Using the solution of
$\Sigma^{i}_{\mu\nu}$, we obtain the connection
$A\in\Omega^{1}\left(M;\mathfrak{so}(3,\mathbb{C})\right)$ using
$D\Sigma^{i}=0$. So obtaining the curvature
$F^{i}(A)=dA^{i}+\varepsilon{{}^{i}}_{jk}A^{j}\wedge A^{k}$.

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
$D_{\mu}\Sigma^{\mu\nu i}=0$ decomposes to two equations,

$D_{a}E^{ai}=0\quad\text{ and }\quad\epsilon^{abc}D_{b}B_{c}^{i}=0,\quad\text{
for }\quad\epsilon^{abc}=\epsilon^{0abc},$ (7)

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

$E^{ai}=\Sigma^{0ai},\quad\text{ and
}\quad\Sigma^{abi}=\epsilon^{abc}B_{c}^{i},$ (8)

on the space-like slice $\Sigma_{t}$(the field $E^{ai}$ is conjugate to the
connection $A^{i}_{a}$).

By that the equation $-(\partial_{\mu}K^{i})\Sigma^{\mu\nu i}+J^{\nu i}=0$
yields (with $\partial_{0}K^{i}=0$)

$\begin{split}&J^{ai}=(\partial_{b}K^{i})\Sigma^{bai}=(\partial_{b}K^{i})\epsilon^{bac}B_{c}^{i}=\epsilon^{bac}(\partial_{b}K^{i})B_{c}^{i},\\\
and&\\\
&J^{0i}=(\partial_{a}K^{i})\Sigma^{a0i}=-(\partial_{a}K^{i})\Sigma^{0ai}=-(\partial_{a}K^{i})E^{ai},\end{split}$
(9)

where $J^{ai}$ are vector currents and $J^{0i}=Q^{i}$ are charges.

In the static case $J^{ai}=0$, we can solve the first equation in terms of
$\partial_{a}K^{i}$ by writing

$B_{a}^{i}=f(\partial_{a}K^{i})b^{i},\quad\text{ for
}b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})\right),$ (10)

and $f$ is scalar function on $M$. But when $J^{ai}\neq 0$, we let
$B_{a}^{i}=f(\partial_{a}G^{i})b^{i}$ for some scalar functions $G^{i}\neq
K^{i}$. The equation $\epsilon^{abc}D_{b}B_{c}^{i}=0$ implies $Db^{i}=0$ and
$df=0$.

Regarding the second equation of (9), when $J^{0i}=0$(in the vacuum), we can
solve it in terms of $\partial_{a}K^{i}$ by writing

$E^{ai}=g\epsilon^{abc}(\partial_{b}K^{i})v^{i}_{c},$ (11)

for some 1-form
$v\in\Gamma\left(M;T^{*}\Sigma\times\mathfrak{so}(3,\mathbb{C}\right)$, and
$g$ is scalar function on $M$. The equation $D_{a}E^{ai}=0$ implies $Dv=0$ and
$dg=0$.

And when $J^{0i}\neq 0$(existence of matter), we let
$E^{ai}=g^{ab}(\partial_{b}K^{i})u^{i}$, for some vector field
$u\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})\right)$. By that

$\begin{split}Q^{i}=J^{0i}=-g^{ab}(\partial_{a}K^{i})(\partial_{b}K^{i})u^{i}.\end{split}$
(12)

Thus there are two solutions of $E^{ai}$(for vacuum and for matter), the total
solution is
$E^{ai}=g^{ab}(\partial_{b}K^{i})u^{i}+g\epsilon^{abc}(\partial_{b}K^{i})v^{i}_{c}$.
The equation $D_{a}E^{ai}=0$ implies $Du=0$ since $\partial^{2}K^{i}=0$ as
showed before.

Using the self-duality [22, 11]

$\frac{1}{{2!}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\rho\sigma}^{i}=\left({*\Sigma^{i}}\right)^{\mu\nu}=\left({-i\Sigma^{i}}\right)^{\mu\nu}=-i\Sigma^{\mu\nu
i},\quad e=\det(e_{\mu}^{I}),$ (13)

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. Regarding the equations (7), (10), (11) and (13), one can take
the solutions

$\begin{split}\Sigma_{0a}^{i}=-\Sigma_{a0}^{i}=(\partial_{a}K^{i})b^{i},\quad\text{
and
}\quad\Sigma_{ab}^{i}=-\Sigma_{ba}^{i}=(\partial_{[a}K^{i})v^{i}_{b]}.\end{split}$
(14)

In order to write $\Sigma^{IJ}$ as $e^{I}\wedge e^{J}$ in the solutions (14),
as required to get gravity theory, we have to let $v^{i}_{a}=v_{a}b^{i}$, for
some 1-form $v\in\Gamma\left(M;T^{*}\Sigma\right)$. Therefore the solutions
(14) become

$\begin{split}&\Sigma_{0a}^{i}=-\Sigma_{a0}^{i}=(\partial_{a}K^{i})b^{i},\quad\text{
and }\quad\Sigma_{ab}^{i}=-\Sigma_{ba}^{i}=(\partial_{[a}K^{i})v_{b]}b^{i},\\\
\text{ for }&\\\
&b^{i}\in\Gamma\left(M;\mathfrak{so}(3,\mathbb{C})\right),\quad
Db^{i}=0,\text{ and }dv=0.\end{split}$ (15)

And because of $v_{a}E^{ai}=0$, we let $v^{a}=g^{ab}v_{b}$ be Killing vector
satisfying $v^{a}\partial_{a}K^{i}=0$, so $v^{a}B^{i}_{a}=0$. By that and
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 solutions (15) 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 (15) is a general solution and
we have to find a special solution, like to choose $b^{i}=u^{i}$ and let it be
constant field, as we will do in the following study.

If the charges $J^{0i}\neq 0$ are given as a functions on $M$, and in order to
get a solution using them, we use the solutions of $K^{i}$(obtained from
$\partial^{2}K^{i}=0$) in the equation (12), so obtaining the field
$u^{i}=b^{i}$. And obtaining the vector $v$ from $v^{a}\partial_{a}K^{i}=0$ so
obtaining the 1-form by $v_{a}=g_{ab}v^{b}$. We obtain $\Sigma^{i}_{0a}$ and
$\Sigma^{i}_{ab}$ using the solutions (15), so getting the connection $A^{i}$
from $D\Sigma^{i}=0$. 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.

Using the solutions (15), we get the connection
$A\in\Omega^{1}\left(M;\mathfrak{so}(3,\mathbb{C})\right)$ using
$D\Sigma^{i}=0$, where $D=d+A$ is exterior covariant derivative, the good
thing with it is that it does not include the affine connection $\nabla$ on
$TM$. The equation $D\Sigma^{i}=0$ yields

$\varepsilon^{\mu\nu\rho\sigma}\partial_{\nu}\Sigma_{\rho\sigma}^{i}+\varepsilon^{\mu\nu\rho\sigma}\varepsilon{{}^{i}}_{jk}A_{\nu}^{j}\Sigma_{\rho\sigma}^{k}=0.$

The solution of the spin connection $A^{i}$ is

$A_{\mu}^{i}=\varepsilon_{\mu\nu\rho\sigma}V^{\nu}\Sigma^{\rho\sigma i},$ (16)

with

$V_{\gamma}=\frac{{-1}}{4}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{i\gamma\mu}\partial_{\nu}\Sigma_{\rho\sigma}^{i},\quad
for\quad\varepsilon^{0123}=-1,\quad\varepsilon_{0123}=1.$

The used metric is $g_{\mu\nu}=e^{I}_{\mu}e^{J}_{\nu}\eta_{IJ}$ and
$\Sigma^{\mu\nu}_{i}$ is inversion of $\Sigma_{\mu\nu}^{i}$(see appendix A for
more details). By using

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

we get the inversion tensor $\Sigma^{\mu\nu i}$, and $e_{\mu}^{I}$ is obtained
from $\Sigma^{i}_{\mu\nu}=P^{i}_{IJ}e^{I}_{\mu}e^{J}_{\nu}$. We obtain the
curvature $F^{i}(A)=dA^{i}+\varepsilon{{}^{i}}_{jk}A^{j}\wedge A^{k}$, then we
satisfy the equation $\Sigma_{i}^{\mu\nu}F^{i}_{\mu\nu}(A)=0$ in the vacuum.
Since $A_{\mu}^{i}\sim b^{i}$, the equation $Db^{i}=0$ implies that $b^{i}$ is
constant field.

We try to give an example for obtaining that solutions. In spherical symmetry,
where the matter located at a point. We use the spherically coordinates
$(r,\theta,\varphi)$ on the space-like slice
$\Sigma_{t}=\Sigma=\mathbb{R}^{3}$. We let the solutions depend only on the
radius $r$, and start with solution of $\partial^{2}K^{i}=0$(the equation
(6)),

$\nabla^{2}K^{i}=\frac{1}{{r^{2}}}\frac{\partial}{{\partial
r}}\left({r^{2}\frac{\partial}{{\partial r}}K^{i}}\right)=0\Rightarrow
K^{i}=\frac{{c^{i}}}{r},$ (17)

for some constants $c^{i}\in\mathbb{R}$. By that the functions $K^{i}$ depend
only on one coordinate, $r$, as discussed below of equation ((6)). Therefore

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

thus the 1-form $v$ ($dv=0$, $g^{ab}v_{a}\partial_{b}K^{i}=0$) 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 $a^{i}$ is Killing vector of
$K^{i}$.

We get the solutions of the 1-form $B^{i}$ and the vector field $E^{i}$ using
(15),

$B^{i}=(dK^{i})b^{i}=-\frac{{b^{i}}}{{r^{2}}}dr,$

and

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

By that we get

$\Sigma_{0r}^{i}=-\frac{{b^{i}}}{{r^{2}}},\quad\Sigma_{r\theta}^{i}=\frac{{b^{i}}}{{2r^{2}}},\quad\Sigma_{r\varphi}^{i}=\frac{{b^{i}}}{{2r^{2}}}.$
(18)

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
$\Sigma_{\mu\nu}^{i}=P^{i}_{IJ}e_{\mu}^{I}e_{\nu}^{J}$, we get the
gravitational fields

$e_{0}^{I}=-\frac{b_{1}^{I}}{r},\quad e_{r}^{I}=\frac{b_{2}^{I}}{r},\quad
e_{\theta}^{I}=-\frac{b_{1}^{I}}{2r},\quad
e_{\varphi}^{I}=-\frac{b_{1}^{I}}{2r}.$ (19)

We let that matter be located at the origin $(0,0,0)\in\mathbb{R}^{3}$,
therefore the charge (12) 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). Using the equation (12)

$\begin{split}Q^{i}=J^{0i}=-g^{ab}(\partial_{a}K^{i})(\partial_{b}K^{i})u^{i}.\end{split}$

Using $K^{i}=c^{i}/r$, one finds

$Q^{i}(r)=-\frac{c^{i}c^{i}}{r^{4}}u^{i}.$

And in order to get $Q^{i}(x)=Q^{i}_{0}\delta^{3}(x)$, we let

$Q^{i}(x)=\frac{1}{\pi^{2}\sqrt{2}}\frac{\epsilon}{r^{4}+\epsilon^{4}}Q^{i}_{0},\quad
for\quad\epsilon\to 0^{+}.$

Therefore

$-c^{i}c^{i}u^{i}=\frac{1}{\pi^{2}\sqrt{2}}\epsilon Q^{i}_{0}.$

So we choose $c^{i}=1$ and $u^{i}=-\epsilon Q^{i}_{0}/(\pi^{2}\sqrt{2})$, and
as we saw before the total solution of $E^{ai}$ (for vacuum and for matter) is
$E^{ai}=g^{ab}(\partial_{b}K^{i})u^{i}+g\epsilon^{abc}(\partial_{b}K^{i})v^{i}_{c}$,
discussion below of (12). Therefore in the limit $\epsilon\to 0^{+}$, the
remaining solution is $E^{ai}=\epsilon^{abc}(\partial_{b}K^{i})v_{c}b^{i}$.

There is multiplicity in the gravitational fields (19), therefore we let the
non-zero fields be

$e_{0}^{0}=-\frac{b_{1}^{0}}{r},\quad\quad
e_{r}^{1}=\frac{b_{2}^{1}}{r},\quad\quad
e_{\theta}^{2}=-\frac{b_{1}^{2}}{2r},\quad\quad
e_{\varphi}^{3}=-\frac{b_{1}^{3}}{2r}.$

We obtain the connection $A^{i}_{\mu}$ using (16), so getting the curvature
$F^{i}(A)=dA^{i}+\varepsilon{{}^{i}}_{jk}A^{j}\wedge A^{k}$. The constants
$b_{1}^{I}$ and $b_{2}^{I}$ have to be fixed in order to satisfy
$\Sigma_{i}^{\mu\nu}F^{i}_{\mu\nu}(A)=0$ in the vacuum.

##  2 Modifying by adding Yang-Mills Lagrangian

In this section we get $D\Sigma^{i}=0$($De^{I}=0$) by adding a new term to the
Lagrangian in self-dual formalism. We add Yang-Mills Lagrangian of the
connection $A^{i}$ to the Lagrangian to get

$\begin{split}L&=L_{gravity}(\Sigma,A)+L_{Y-M}(A)+L_{matter}\\\
&=\Sigma^{\mu\nu}_{i}F^{i}_{\mu\nu}+\phi_{ij}\Sigma^{\mu\nu
i}\Sigma^{j}_{\mu\nu}+\frac{k}{4}F^{\mu\nu}_{i}F^{i}_{\mu\nu}+L_{matter},\end{split}$
(20)

the equation of motion of $A^{i}_{\nu}$ is ($k$ is constant)

$-P_{iIJ}D_{\mu}\Sigma^{\mu\nu i}-kD_{\mu}F^{\mu\nu i}+J^{\nu i}=0.$

Thus we can choose $D_{\mu}\Sigma^{\mu\nu i}=0$, by that we get

$-kD_{\mu}F^{\mu\nu i}+J^{\nu i}=0,\quad\text{for}\quad D=\nabla+A.$ (21)

This is same equation of motion in Yang-Mills theory of gauge fields. This
relates to the fact that we can regard the local Lorentz symmetry (internal
symmetry) as a gauge symmetry with spin connection $\omega^{IJ}$ (or $A^{i}$)
as a gauge field. The Lorentz current in 21 is conserved since

$D_{\nu}D_{\mu}F^{\mu\nu i}=\frac{1}{2}[D_{\nu},D_{\mu}]F^{\mu\nu
i}=-\frac{1}{2}\varepsilon{{}^{i}}_{jk}F_{\mu\nu}^{j}F^{\mu\nu k}=0\to
D_{\nu}J^{\nu i}=0.$

We can get same current by using Riemann curvature tensor
$R^{\mu}{{}_{\nu\rho\sigma}}$ given in terms of the Levi-Civita connection
$\nabla$. The Bianchi identity is

$\nabla_{\gamma}R^{\mu}{{}_{\nu\rho\sigma}}+\nabla_{\rho}R^{\mu}{{}_{\nu\sigma\gamma}}+\nabla_{\sigma}R^{\mu}{{}_{\nu\gamma\rho}}=0.$

Multiplying it by $\delta^{\rho}_{\mu}$, yields to

$\nabla_{\gamma}R^{\mu}{{}_{\nu\mu\sigma}}+\nabla_{\mu}R^{\mu}{{}_{\nu\sigma\gamma}}+\nabla_{\sigma}R^{\mu}{{}_{\nu\gamma\mu}}=0,$

or

$\nabla_{\gamma}R^{\mu}{{}_{\nu\mu\sigma}}+\nabla_{\mu}R^{\mu}{{}_{\nu\sigma\gamma}}-\nabla_{\sigma}R^{\mu}{{}_{\nu\mu\gamma}}=0,$

the Ricci tensor $R_{\nu\sigma}=R^{\mu}{{}_{\nu\mu\sigma}}$ vanishes in the
vacuum, so $\nabla_{\mu}R^{\mu}{{}_{\nu\sigma\gamma}}$ also vanishes in the
vacuum, so
$\nabla^{\mu}R_{\mu\nu\sigma\gamma}=\nabla^{\mu}R_{\sigma\gamma\mu\nu}\neq 0$
associates only with matter. The curvature of the spin connection $\omega$
relates with the Riemann curvature by

$R_{\mu\nu}^{IJ}(\omega(e))=e^{I\sigma}e^{J\gamma}R_{\sigma\gamma\mu\nu}(g(e)),\quad\text{with}\quad
g_{\mu\nu}(e)=\eta_{IJ}e^{I}_{\mu}e^{J}_{\nu},$

therefore

$\nabla^{\mu}R_{\sigma\gamma\mu\nu}\neq 0(=0)\to D^{\mu}R_{\mu\nu}^{IJ}\neq
0(=0),\text{ }\text{ }D_{\mu}e^{I}_{\nu}=0.$

Thus $D^{\mu}R_{\mu\nu}^{IJ}$ associates only with matter and vanishes in the
vacuum. Using the self-dual projection,
$F^{i}_{\mu\nu}(A)=P^{i}_{IJ}R_{\mu\nu}^{IJ}(\omega)$, we find that

$J^{i}_{\nu}=D^{\mu}F^{i}_{\mu\nu}(A)=P^{i}_{IJ}D^{\mu}R_{\mu\nu}^{IJ}(\omega)$

also vanishes in the vacuum. Therefore this current associates only with
matter and the equation (21) is well defined.

In Plebaniski formalism, we have the formula

$F^{i}(A)=\psi{{}^{i}}_{j}\Sigma^{j}+\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j},\quad\text{with}\quad\Sigma^{i}=P^{i}_{IJ}e^{I}\wedge
e^{J},$

in the space $(\Sigma^{i},F^{i})$, where $Tr(\psi)=0$,
$\bar{\psi}{{}^{i}}_{j}=0$ in the vacuum without cosmological constant [10].
By using this formula in Equation (21), we get (Appendix B)

$(D_{\mu}\psi{{}^{i}}_{j})\Sigma^{j\mu\nu}=\frac{3}{2}J^{i\nu},\quad\text{for}\quad
D_{\mu}\Sigma^{j\mu\nu}=0.$ (22)

This is same equation (5), so its solutions are same solutions we obtained in
the previous section. By that we obtain the solutions of $\Sigma^{i}$ and
$F^{i}(A)$, and by inserting them in the equation of motion $\delta
S/\delta\Sigma^{i}=0$, we obtain the Lagrangian multiplier functions
$\phi_{ij}$.

##  3 Decomposition Space–Time Manifold $M$ into $R\times\Sigma$

The formulation of general relativity based on decomposition space–time
manifold $M$ into $R\times\Sigma$ is needed for expressing the metric of
space–time as a solution of an equation for time evolution, such as in the
Hamiltonian formulation. Thus the time evolution is the changing of the
geometry of this surface. This decomposition preserves the continuous
symmetries (gauge invariance and diffeomorphism invariance) of general
relativity and its canonical quantization, so we can use it for the gauge
theory of general relativity [4, 12, 5, 6].

We get the energy density $H$ and we set $\frac{d}{dt}H=0$ when there is no
external source. We see that $\frac{d}{dt}H=0$ is same $d\theta=0$ for 3-form
$\theta$ on $\Sigma$. We define gravitational field as a one-form
$e^{I}=e^{I}_{\mu}(x)dx^{\mu}$ that is related with metric $g_{\mu\nu}(x)$ on
an arbitrary space–time manifold $M$ by
$g_{\mu\nu}=\eta_{IJ}e^{I}_{\mu}e^{J}_{\nu}$, with spin connection
$\omega^{IJ}(x)\in\Omega^{1}(M,\mathfrak{so}(3,1))$, where
$\mathfrak{so}(3,1)$ is Lie algebra of Lorentz group $SO(3,1)$. The spin
connection defines covariant derivative $D_{\mu}$ that acts on all fields
which have Lorentz indices $(I,J,...)$:

$D_{\mu}v^{I}=\partial_{\mu}v^{I}+\omega^{I}_{\mu J}v^{J}.$

We start with the general relativity Lagrangian of the form

$L(e,\omega)=(16\pi G)^{-1}e_{I}^{\mu}e_{J}^{\nu}(R_{\mu\nu})^{IJ}e,$ (23)

where

$R^{IJ}=d\omega^{IJ}+\omega{{}^{I}}_{K}\wedge\omega^{KJ}$

is the Riemannian curvature tensor and $e_{I}^{\mu}$ satisfies
$e^{I}_{\mu}e_{J}^{\mu}=\delta^{I}_{J}$.

By the decomposition $M\to R\times\Sigma$, we decompose this Lagrangian into

$L(e,\omega)=(16\pi G)^{-1}e_{I}^{a}e_{J}^{b}(R_{ab})^{IJ}e+(16\pi
G)^{-1}e_{I}^{0}e_{J}^{a}(R_{0a})^{IJ}e.$ (24)

The part

$H=(16\pi G)^{-1}e_{I}^{a}e_{J}^{b}(R_{ab})^{IJ}e$ (25)

takes the role of the energy density on the surface $\Sigma_{t}$ of constant
time, it does not include the time derivative, but it includes the spatial
derivatives with respect to coordinates on $\Sigma_{t}$. It is invariant under
the $SO(3,1)$ transformation, thus we let $\frac{d}{dt}H=0$ when there are no
external sources. The second term

$e_{I}^{0}e_{J}^{a}(R_{0a})^{IJ}e=ee_{I}^{0}e_{J}^{a}\left(\partial_{0}\omega_{a}^{IJ}-\partial_{a}\omega_{0}^{IJ}+\omega{{}^{I}}_{0K}\omega^{KJ}_{a}\right),$

has the term $ee_{I}^{0}e_{J}^{a}\partial_{0}\omega_{a}^{IJ}$ with represents
the variables of the phase space
$\left(ee_{I}^{0}e_{J}^{a},\omega_{a}^{IJ}\right)$ on the surface
$\Sigma_{t}$, where we let $ee_{I}^{0}e_{J}^{a}=E^{a}_{IJ}$ be conjugate
momentum to $\omega_{a}^{IJ}$. The final term is

$ee_{I}^{0}e_{J}^{a}\left(-\partial_{a}\omega_{0}^{IJ}+\omega{{}^{I}}_{0K}\omega^{KJ}_{a}-\omega{{}^{I}}_{aK}\omega^{KJ}_{0}\right).$

Integrating by parts, this term becomes

$\begin{split}&\omega_{0}^{IJ}\partial_{a}\left(ee_{I}^{0}e_{J}^{a}\right)+\left(\omega^{I}_{0K}\omega^{KJ}_{a}-\omega^{I}_{aK}\omega^{KJ}_{0}\right)ee_{I}^{0}e_{J}^{a}\\\
&=\omega_{0}^{IJ}\partial_{a}\left(ee_{I}^{0}e_{J}^{a}\right)+\left(\omega^{I}_{0K}\omega^{KJ}_{a}-\omega^{JK}_{0}\omega_{Ka}{{}^{I}}\right)ee_{I}^{0}e_{J}^{a}\\\
&=\omega_{0}^{IJ}\partial_{a}\left(ee_{I}^{0}e_{J}^{a}\right)+\omega^{IK}_{0}\omega_{Ka}{{}^{J}}\left(ee_{I}^{0}e_{J}^{a}\right)+\omega^{KJ}_{0}\omega_{Ka}{{}^{I}}\left(ee_{I}^{0}e_{J}^{a}\right)\\\
&=\omega_{0}^{IJ}\partial_{a}\left(ee_{I}^{0}e_{J}^{a}\right)+\omega^{IJ}_{0}\omega_{Ja}{{}^{K}}\left(ee_{I}^{0}e_{K}^{a}\right)+\omega^{IJ}_{0}\omega_{Ia}{{}^{K}}\left(ee_{K}^{0}e_{J}^{a}\right)\\\
&=\omega_{0}^{IJ}D_{a}\left(ee_{I}^{0}e_{J}^{a}\right).\end{split}$

The variable $\omega_{0}^{IJ}$ has no time derivative, so it is Lagrangian
multiplier, it gives the constraint
$D_{a}\left(ee_{I}^{0}e_{J}^{a}\right)=D_{a}E^{a}_{IJ}=0$.

In self-dual formalism of GR, we use the self-duality [22, 11]

$\frac{1}{{2!}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\rho\sigma}^{i}=\left({*\Sigma^{i}}\right)^{\mu\nu}=\left({-i\Sigma^{i}}\right)^{\mu\nu}=-i\Sigma^{\mu\nu
i},\quad e=\det(e_{\mu}^{I}).$ (26)

The Lie algebra $\mathfrak{so}(3,1)$(with indices $I,J,...$) is replaced by
the Lie algebra $\mathfrak{so}(3,\mathbb{C})$(with indices $i,j,...$). The
field $E^{a}_{IJ}$ becomes complex given by [9, 10]

$E^{ia}=\frac{1}{2}\varepsilon^{abc}P^{i}_{IJ}e^{I}_{b}e^{J}_{c},\text{ with
}E^{i}_{ab}=\Sigma^{i}_{ab}=P^{i}_{IJ}e^{I}_{a}e^{J}_{b}$ (27)

where $P^{i}_{IJ}$ is a self-dual map given by

$P^{i}_{IJ}=\frac{1}{2}\varepsilon^{i}_{jk},\text{ for }I=i,J=j,\text{ and
}P^{i}_{0j}=-P^{i}_{j0}=\frac{i}{2}\delta^{i}_{j},\text{ for }I=0,J=j\neq 0.$
(28)

For example
$E^{1a}=\frac{1}{2}\varepsilon^{abc}(e^{2}_{b}e^{3}_{c}+ie^{0}_{b}e^{1}_{c})$.
This map relates to the decomposition of Lie algebra of the Lorentz group
$SO(1,3)$ into two copies of Lie algebra of $SL(2,R)$ [11].

In the equation (27), we used the formula (26) to get

$\varepsilon^{0abc}P^{i}_{IJ}e^{I}_{b}e^{J}_{c}/2=-ieP^{i}_{IJ}e^{0I}e^{aJ}.$

Regarding the previous definition of the conjugate momentum
$E^{a}_{IJ}=ee_{I}^{0}e_{J}^{a}$ of $\omega_{a}^{IJ}$, we ignore $-i$ and just
write $E^{ia}=eP^{i}_{IJ}e^{0I}e^{aJ}$ which becomes conjugate momentum to
complex connection $A^{i}$.

The obtained connection $A^{i}$ is a three-complex one-form given by

$\omega^{IJ}\mapsto A^{i}=P{{}^{i}}_{IJ}\omega^{IJ},$ (29)

The curvature which associates with this connection is

$F^{i}=dA^{i}+\varepsilon{{}^{i}}_{jk}A^{j}\wedge A^{k}.$

On the surface $\Sigma_{t}(\sigma^{a})$, it is

$F^{i}_{ab}=\frac{1}{2}\left(\partial_{b}A_{b}^{i}-\partial_{b}A_{a}^{i}+\varepsilon{{}^{i}}_{jk}A_{a}^{j}A_{b}^{k}\right).$

Thus we have self-dual plus anti-self-dual map:

$e^{a}_{I}e^{b}_{J}R(\omega)^{IJ}_{ab}\to\varepsilon^{abc}E_{ia}F^{i}_{bc}+\varepsilon^{abc}{\bar{E}}_{ia}{\bar{F}}^{i}_{bc},$
(30)

where ${\bar{E}}_{ia}$ and ${\bar{F}}^{i}_{bc}$ are the Hermitian conjugate of
${E}_{ia}$ and ${F}^{i}_{bc}$.

For the Lagrangian part $H$ on $\Sigma_{t}$, the equation (25), it becomes

$H=\varepsilon^{abc}E_{ai}F^{i}_{bc}(A)e,$ (31)

or

$Hd^{3}\sigma=\varepsilon^{abc}E_{ai}F^{i}_{bc}(A)ed^{3}\sigma,$

where $\sigma^{a}$ is coordinates on $\Sigma_{t}$. We can write $Hd^{3}\sigma$
as 3-form on $\Sigma_{t}$, like

$\theta=Hd^{3}\sigma=E_{i}\wedge F^{i}(A).$ (32)

By that the condition $\frac{d}{dt}H=0$ becomes

$({\Sigma(\sigma^{a})},d\theta)=dt\frac{d}{dt}H=0.$

Here $(\Sigma,V)$ is a pairing of 4-form $V\in\wedge^{4}T^{*}_{p}M$ with a
surface $\Sigma$, the pairing of $V$ with tangent basis in $T_{p}\Sigma$,
defined below in equation (40).

$\left({\Sigma(\sigma^{a})},dt\wedge\frac{d}{dt}\theta\right)=dt\frac{d}{dt}H=0.$
(33)

Since $\theta$ is three-form on $\Sigma(\sigma^{a})$, so
$d\sigma^{a}\wedge\frac{\partial}{\partial\sigma^{a}}\theta=0$, if we add it
to the last formula, we get

$\left({\Sigma(\sigma^{a})},dt\wedge\frac{\partial}{\partial
t}\theta+d\sigma^{a}\wedge\frac{\partial}{\partial\sigma^{a}}\theta\right)=\left({\Sigma(\sigma^{a})},d\theta\right)=0.$
(34)

Under arbitrary transformation $(t,\sigma^{a})\to x^{\mu}$,
$d\sigma^{a}=\frac{\partial\sigma^{a}}{\partial x^{\mu}}dx^{\mu}$, the basis
on $\Sigma$ transforms as

$\partial_{a}\wedge\partial_{b}\wedge\partial_{c}\to\frac{\partial
x^{\mu}}{\partial\sigma^{a}}\frac{\partial
x^{\nu}}{\partial\sigma^{b}}\frac{\partial
x^{\rho}}{\partial\sigma^{c}}\partial_{\mu}\wedge\partial_{\nu}\wedge\partial_{\rho}.$
(35)

Therefore, the components of three-form $\theta$ transforms as

$\theta_{abc}\to\theta_{\mu\nu\rho}=\theta_{abc}\frac{\partial\sigma^{a}}{\partial
x^{\mu}}\frac{\partial\sigma^{b}}{\partial
x^{\nu}}\frac{\partial\sigma^{c}}{\partial x^{\rho}}.$

To keep the invariance under this transformation, we let the equation (34),
$\left(\Sigma,d\theta\right)=0$ still hold. We write the components of the
curvature as

$F_{ab}^{i}=\frac{1}{2}\left(D_{a}A_{b}^{i}-D_{b}A_{a}^{i}\right)=\frac{1}{2}\left(\partial_{a}A_{b}^{i}-\partial_{b}A_{a}^{i}+\varepsilon{{}^{i}}_{jk}A_{a}^{j}A_{b}^{k}\right),$
(36)

where

$D_{a}A_{b}^{i}=\partial_{a}A_{b}^{i}+\frac{1}{2}\varepsilon{{}^{i}}_{jk}A^{j}_{a}A^{k}_{b},$

which motivates introducing a notation of the covariant derivative like [12]

$DV^{i}=dV^{i}+\frac{1}{2}\varepsilon{{}^{i}}_{jk}A^{j}\wedge V^{k},$

or

$D_{\mu}V^{i}_{\nu}=\partial_{\mu}V^{i}_{\nu}+\frac{1}{2}\varepsilon{{}^{i}}_{jk}A^{j}_{\mu}V^{k}_{\nu}=\partial_{\mu}V^{i}_{\nu}-\frac{i}{2}A^{j}_{\mu}(T^{j}_{A})^{ik}V^{k}_{\nu},$

so

$D_{\mu}=\partial_{\mu}-\frac{i}{2}A^{j}_{\mu}(T^{j}_{A}),$

where the matrix elements $(T^{j}_{A})^{ik}=-i\varepsilon^{jik}$ are the
elements of the generators $T^{j}_{A}$ in the adjoint representation of the
group $SU(2)$ [13]. The coupling constant here is $g=1$. In general we write
this covariant derivative as

$D_{\mu}=\partial_{\mu}-\frac{i}{2}gA^{j}_{\mu}(T^{j}_{A}).$ (37)

By that, we write the 3-form (32) as

$\theta(E,A)=(16\pi G)^{-1/2}E_{i}\wedge DA^{i}.$ (38)

Its pairing with the surface $\Sigma_{t}(\sigma^{a})$ is the energy

$H=(\Sigma_{t}(\sigma^{a}),\theta)=(16\pi
G)^{-1/2}\varepsilon^{abc}{E_{ia}}F^{i}_{bc}(A).$

Regarding the equations (33) and (34), we let
$(\Sigma_{t}(\sigma^{a}),d\theta)=0$, this gives a continuity equation, as we
see next.

Our condition $\left({\Sigma(\sigma^{a})},d\theta\right)=0$ makes sense here
because of the decomposition $R\times\Sigma$ and fixing a coordinate system
$\sigma^{a}$ on the hypersurface $\Sigma$, this yields to an equation of
continuity on this surface. For this purpose, we take the pairing of the four-
form $d\theta$ with a tangent basis on the surface $\Sigma_{t}(\sigma^{a})$ at
an arbitrary point, we get a one-form co-vector $(\Sigma(\sigma^{a}),d\theta)$
in the direction of the normal to this surface at that point. Then we set
$(\Sigma(\sigma^{a}),d\theta)=0$, we obtain an equation of continuity on
$\Sigma_{t}(\sigma^{a})$.

Now taking the exterior derivative of Equation (38), we obtain

$(16\pi G)^{1/2}d\theta=d\left({E_{i}\wedge
DA^{i}}\right)=\left({DE_{i}}\right)\wedge DA^{i}-E_{i}\wedge DDA^{i},$ (39)

where $E_{0i}=0$. The tri-tangent basic on $\Sigma_{t}(\sigma^{a})$ is
$\partial_{a}\wedge\partial_{b}\wedge\partial_{c}$, we rewrite it as
$(1/3!)\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}$. The pairing of
$d\theta$ with this basis is

$(16\pi
G)^{1/2}\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=\left(({DE_{i}})\wedge
DA^{i},\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)-\left(E_{i}\wedge
DDA^{i},\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)$

where $\left(\cdot,\cdot\right)$ is contraction pairing defined by

$\left(V_{\mu\nu\rho\sigma}dx^{\mu}\wedge dx^{\nu}\wedge dx^{\rho}\wedge
dx^{\sigma},\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=\varepsilon^{abc}V_{\mu\nu\rho\sigma}dx^{[\mu}\delta^{\nu}_{c}\delta^{\rho}_{b}\delta^{\sigma]}_{a},$
(40)

where the bracket $[....]$ is anti-symmetrization of the indices. Although
$DDA^{i}$ is zero when the sum is over all indices like
$\varepsilon^{abc}D_{a}D_{b}A_{c}^{i}=0$, but here we do not sum over all
indices of $DDA^{i}$ alone, since we take common functions out of the sum.

For cotangent basis $\\{dx^{\mu}\\}$ and tangent basis $\\{\partial_{a}\\}$,
this pairing can be defined simply by using inner product like [14]

$\left(dx^{\mu},\partial_{a}\right)=\delta^{\mu}_{a},$

in which we consider $dx^{a}=d\sigma^{a}$ for $a=1,2,3$, so $E_{0i}=0$
regarding to our gauge.

Starting with the first term

$\left({({DE_{i}})\wedge
DA^{i}},{\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}}\right)=\varepsilon^{abc}D_{\mu}E_{\nu
i}D_{\rho}A_{\sigma}^{i}\left({dx^{\mu}\wedge dx^{\nu}\wedge dx^{\rho}\wedge
dx^{\sigma}},\partial_{a}\partial_{b}\partial_{c}\right)$

we get

$\displaystyle 4\left({\left({DE_{i}}\right)\wedge
DA^{i}},\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=$
$\displaystyle-\varepsilon^{abc}D_{a}E_{bi}D_{c}A_{\mu}^{i}dx^{\mu}+\varepsilon^{abc}D_{a}E_{bi}D_{\mu}A_{c}^{i}dx^{\mu}-\varepsilon^{abc}D_{a}E_{\mu
i}D_{b}A_{c}^{i}dx^{\mu}+\varepsilon^{abc}D_{\mu}E_{ai}D_{b}A_{c}^{i}dx^{\mu}.$

The second term is

$\left({E_{i}\wedge
DDA^{i}},\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=\varepsilon^{abc}E_{\mu
i}D_{\nu}D_{\rho}A_{\sigma}^{i}\left(dx^{\mu}\wedge dx^{\nu}\wedge
dx^{\rho}\wedge dx^{\sigma},\partial_{a}\partial_{b}\partial_{c}\right).$

Doing the same thing, we get

$\displaystyle 4\left(E_{i}\wedge
DDA^{i},\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=$
$\displaystyle-\varepsilon^{abc}E_{ai}D_{b}D_{c}A_{\mu}^{i}dx^{\mu}+\varepsilon^{abc}E_{ai}D_{b}D_{\mu}A_{c}^{i}dx^{\mu}-\varepsilon^{abc}E_{ai}D_{\mu}D_{b}A_{c}^{i}dx^{\mu}+\varepsilon^{abc}E_{\mu
i}D_{a}D_{b}A_{c}^{i}dx^{\mu}.$

Adding the two terms, we obtain

$\displaystyle 4(16\pi
G)^{1/2}\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=$
$\displaystyle-$
$\displaystyle\varepsilon^{abc}D_{a}E_{bi}D_{c}A_{\mu}^{i}dx^{\mu}+\varepsilon^{abc}D_{a}E_{bi}D_{\mu}A_{c}^{i}dx^{\mu}-\varepsilon^{abc}D_{a}E_{\mu
i}D_{b}A_{c}^{i}dx^{\mu}+\varepsilon^{abc}D_{\mu}E_{ai}D_{b}A_{c}^{i}dx^{\mu}$
$\displaystyle+$
$\displaystyle\varepsilon^{abc}E_{ai}D_{b}D_{c}A_{\mu}^{i}dx^{\mu}-\varepsilon^{abc}E_{ai}D_{b}D_{\mu}A_{c}^{i}dx^{\mu}+\varepsilon^{abc}E_{ai}D_{\mu}D_{b}A_{c}^{i}dx^{\mu}-\varepsilon^{abc}E_{\mu
i}D_{a}D_{b}A_{c}^{i}dx^{\mu}.$

We define the curvature by using the covariant derivative from Equation (36)
as

$F_{\mu
c}^{i}=\frac{1}{2}\left(D_{\mu}A_{c}^{i}-D_{c}A_{\mu}^{i}\right)=\frac{1}{2}\left(\partial_{\mu}A_{c}^{i}-\partial_{c}A_{\mu}^{i}+g\varepsilon{{}^{i}}_{jk}A_{\mu}^{j}A_{c}^{k}\right),$
(41)

therefore

$F^{\rho\nu i}=g^{\rho\mu}g^{\nu c}F_{\mu
c}^{i}=\frac{1}{2}\left(D^{\rho}A^{\nu i}-D^{\nu}A^{\rho i}\right);\text{
}D^{\rho}g^{\mu\nu}=0.$

Its Hodge dual on the surface $\Sigma$ with respect to the coordinates
$(\sigma^{a})$ is

$F^{ai}=\varepsilon^{abc}F_{bc}^{i}.$

Also we define the complex two-form field from Equation (27) as

$\Sigma^{bc}_{i}=E^{bc}_{i}=\varepsilon^{bca}E_{ai}=\varepsilon^{abc}E_{ai}.$

Using them in the last formula, we get

$\displaystyle 4(16\pi G)^{1/2}$
$\displaystyle\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=2\varepsilon^{abc}(D_{a}E_{bi})F_{\mu
c}^{i}dx^{\mu}-2(D_{a}E_{\mu i})F^{ai}dx^{\mu}$
$\displaystyle+2(D_{\mu}E_{ai})F^{ai}dx^{\mu}+2E^{bc}_{i}D_{b}F_{c\mu}^{i}dx^{\mu}+2E_{ai}D_{\mu}F^{ai}dx^{\mu}-2E_{\mu
i}D_{a}F^{ai}dx^{\mu}.$

And using

$\varepsilon^{abc}(D_{a}E_{bi})F_{\mu
c}^{i}dx^{\mu}=-\varepsilon^{acb}(D_{a}E_{bi})F_{\mu
c}^{i}dx^{\mu}=-(D_{a}E^{ac}_{i})F_{\mu c}^{i}dx^{\mu},$

we obtain

$\displaystyle 2(16\pi
G)^{1/2}\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=$
$\displaystyle-(D_{a}E^{ac}_{i})F_{\mu c}^{i}dx^{\mu}-(D_{a}E_{\mu
i})F^{ai}dx^{\mu}+$
$\displaystyle(D_{\mu}E_{ai})F^{ai}dx^{\mu}-E^{bc}_{i}D_{b}F_{\mu
c}^{i}dx^{\mu}$ $\displaystyle+$ $\displaystyle
E_{ai}D_{\mu}F^{ai}dx^{\mu}-E_{\mu i}D_{a}F^{ai}dx^{\mu}.$

With

$-(D_{a}E^{ac}_{i})F_{\mu c}^{i}dx^{\mu}-E^{bc}_{i}D_{b}F_{\mu
c}^{i}dx^{\mu}=-D_{a}\left(E^{ac}_{i}F_{\mu c}^{i}dx^{\mu}\right),$

it becomes

$2(16\pi
G)^{1/2}\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=-D_{a}(E^{ac}_{i}F_{\mu
c}^{i})dx^{\mu}-D_{a}(E_{\mu i}F^{ai})dx^{\mu}+D_{\mu}(E_{ai}F^{ai})dx^{\mu}.$

As we suggested before, we let the normal of the surface
$\Sigma_{t}(\sigma^{a})$ be in direction of the time $dx^{0}$, so
$\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)$
is in direction of the time. Therefore we set $\mu=0$, thus we get

$2(16\pi
G)^{1/2}\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=-D_{a}(E^{ac}_{i}F_{0c}^{i})dx^{0}-2D_{a}(E_{0i}F^{ai})dx^{0}+D_{0}(E_{ai}F^{ai})dx^{0}.$

The vector
$\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)$
is one-form in the direction of the normal to the surface
$\Sigma_{t}(\sigma^{a})$. It is zero as we mentioned before, thus we get

$-D_{a}(E^{ac}_{i}F_{0c}^{i})dx^{0}-D_{a}(E_{0i}F^{ai})dx^{0}+D_{0}(E_{ai}F^{ai})dx^{0}=0,$

or

$-D_{a}(E^{ac}_{i}F_{0c}^{i})-D_{a}(E_{0i}F^{ai})+D_{0}(E_{ai}F^{ai})=0.$

For $De=0$ on the surface $\Sigma$, where $e=det(e^{i}_{a})$, We write it as

$D_{a}(e^{2}E^{ac}_{i}F_{c0}^{i})-D_{a}(E_{0i}F^{ai})+D_{0}(e^{2}E_{ai}F^{ai})=0.$

The term $(E_{ai}F^{ai})$ is scalar, so
$D_{0}(E_{ai}F^{ai})=\partial_{0}(E_{ai}F^{ai})$, the vector
$E^{ab}_{i}F_{b0}^{i}$ is a usual vector field on $\Sigma$, it does not carry
a Lorentz index, so
$D_{a}(E^{ab}_{i}F_{b0}^{i})=\partial_{a}(E^{ab}_{i}F_{b0}^{i})$ and
$E_{0i}=0$, thus we get

$\partial_{a}(e^{2}E^{ab}_{i}F_{b0}^{i})+\partial_{0}(e^{2}E_{ai}F^{ai})=0,$

using $F^{ai}=\varepsilon^{abc}F_{bc}^{i}$,
$\Sigma^{bc}_{i}=E^{bc}_{i}=\varepsilon^{bca}E_{ai}$ and $E^{ai}=ee^{ai}$, we
have

$e^{2}E_{ai}F^{ai}=\frac{1}{2}e^{2}\Sigma^{ab}_{i}F_{ab}^{i}=\frac{1}{4}e^{2}e^{a}_{i}e^{b}_{j}F_{ab}^{ij}=\frac{1}{4}E^{a}_{i}E^{b}_{j}F_{ab}^{ij}=H,$

so we obtain

$\partial_{a}\left(e^{2}\Sigma^{ab}_{i}F_{b0}^{i}\right)+\partial_{0}H=2(16\pi
G)^{1/2}\left(d\theta,\varepsilon^{abc}\partial_{a}\partial_{b}\partial_{c}\right)=0.$
(42)

We have regarded $\frac{1}{4}E^{a}_{i}E^{b}_{j}F_{ab}^{ij}$ as energy density.

The equation

$\partial_{a}\left(e^{2}E^{ab}_{i}F_{b0}^{i}\right)+\partial_{0}\left(\frac{1}{4}E^{a}_{i}E^{b}_{j}F_{ab}^{ij}\right)=0.$

We regard this equation as an equation of continuity with respect to a
Lagrangian like $L(F^{0ai},DA^{i})$ that satisfies the action principle
$\delta S(DA^{i})=0$ and the invariance under continuous symmetries of general
relativity. Therefore we regard $\frac{1}{2}c\Sigma^{ab}_{i}F_{ab}^{i}$ as
energy density $T^{00}$, and $c\Sigma^{ab}_{i}F_{b0}^{i}$ as momentum density
$T^{0a}$, where $c$ is constant for satisfying the units.

##  4 Conclusions

We have considered the 4-manifold $M$ as a base space and consider that the
Lorentz group $SO(3,1)$ act locally on Lorentz frames which are regarded as a
frame bundle over a fixed base space, and consider the local Lorentz frame as
an element in the tangent frame bundle over $M$. By that we have two
symmetries; invariance under continuous transformations of local Lorentz
frame, $SO(3,1)$ group, and invariance under diffeomorphism of the space–time
$M$. The local invariance of matter Lagrangian under $SO(3,1)$ transformation
produces a current, called spin current $J=\delta S_{matter}/\delta A$, this
current couples to the spin connection, so we have gauge theory, that makes
the general relativity similar to Yang-Mills theory of gauge fields. We have
solved the equations of motion of general relativity in self-dual formalism
using only the spin currents(Lorentz currents), in static case, and without
needing using the Einstein’s equation. We give an example, matter located at a
point, so we have spherical symmetric system. We solved the equation $\delta
S/\delta A_{\nu}^{i}=-D_{\mu}B^{\mu\nu}_{i}+J^{\nu}_{i}=0$ by using $B^{\mu\nu
i}=K{{}^{i}}_{j}\Sigma^{\mu\nu i}$ for $D_{\mu}\Sigma^{\mu\nu i}=0$.

##  5 Appendix A

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

$-i\Sigma^{\mu\nu
i}=\frac{1}{{2!}}e^{-1}\varepsilon^{\mu\nu\rho\sigma}\Sigma_{\rho\sigma}^{i},\quad
e=\det(e_{\mu}^{I}),\quad\varepsilon^{0123}=-1,\quad\varepsilon_{0123}=1,$
(43)

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.

##  6 Appendix B

We calculate $D^{\nu}F^{i}_{\nu\mu}$ by using

$(*D*F^{i})_{\mu}=D^{\nu}F^{i}_{\nu\mu},$

where the Hodge star operator

$*F^{i}=*\left(\psi{{}^{i}}_{j}\Sigma^{j}+\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j}\right)=\psi{{}^{i}}_{j}(*\Sigma^{j})+\bar{\psi}{{}^{i}}_{j}(*\bar{\Sigma}^{j}),$

is with respect to the metric $g_{\mu\nu}$. By using the properties of self-
duality:

$*\Sigma^{j}=-i\Sigma^{j},\text{ }*\bar{\Sigma}^{j}=i\bar{\Sigma}^{j},$

we obtain

$D^{\nu}F^{i}_{\nu\mu}=i\left(*(-D\psi{{}^{i}}_{j}\Sigma^{j})+*(D\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j})\right)_{\mu}.$

The first term becomes

$D(\psi{{}^{i}}_{j}\Sigma^{j})=D(\psi{{}^{i}}_{j}\Sigma^{j}+\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j})-D(\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j}),\text{
so
}D(\psi{{}^{i}}_{j}\Sigma^{j})=DF^{i}-D(\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j}),$

and by Bianchi identity $DF^{i}=0$, we obtain
$D(\psi{{}^{i}}_{j}\Sigma^{j})=-D(\bar{\psi}{{}^{i}}_{j}\bar{\Sigma}^{j})$.
Therefore

$D^{\nu}F^{i}_{\nu\mu}=i\left(*(-D(\psi{{}^{i}}_{j}\Sigma^{j})-*(D\psi{{}^{i}}_{j}\Sigma^{j})\right)_{\mu}=-2i(*D(\psi{{}^{i}}_{j})\wedge\Sigma^{j}))_{\mu},$

where we used $D\Sigma^{j}=0$. Thus we obtain

$D^{\nu}F^{i}_{\nu\mu}=-2i\frac{1}{3!}\epsilon_{\mu}{{}^{\nu\rho\sigma}}(D_{\nu}\psi{{}^{i}}_{j})\Sigma^{j}_{\rho\sigma},$

or

$D^{\nu}F^{i}_{\nu\mu}=-\frac{2i}{3!}\epsilon_{\mu\nu\rho\sigma}(D^{\nu}\psi{{}^{i}}_{j})\Sigma^{j\rho\sigma},$

the raising and lowering is done by using the metric $g_{\mu\nu}$. But

$\frac{1}{2!}\epsilon_{\mu\nu\rho\sigma}\Sigma^{j\rho\sigma}=(*\Sigma^{j})_{\mu\nu}=(-i\Sigma^{j})_{\mu\nu},$

so we obtain

$D^{\nu}F^{i}_{\nu\mu}=\frac{-2i}{3}(D^{\nu}\psi{{}^{i}}_{j})(-i\Sigma^{j})_{\mu\nu}=\frac{-2}{3}(D^{\nu}\psi{{}^{i}}_{j})\Sigma^{j}_{\mu\nu}=\frac{2}{3}(D^{\nu}\psi{{}^{i}}_{j})\Sigma^{j}_{\nu\mu}.$

From $D^{\nu}F^{i}_{\nu\mu}=J^{i}_{\mu}$, we get

$(D^{\nu}\psi{{}^{i}}_{j})\Sigma^{j}_{\nu\mu}=\frac{3}{2}J^{i}_{\mu},\text{ so
}\quad(D_{\nu}\psi{{}^{i}}_{j})\Sigma^{j\nu\mu}=\frac{3}{2}J^{i\mu}.$

This is a linear equation in the basis $\\{\Sigma^{i}\\}$.

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Relation Extraction (RE) remains a challenging task, especially when considering realistic out-of-domain evaluations.
One of the main reasons for this is the limited training size of current RE datasets: obtaining high-quality (manually annotated) data is extremely expensive and cannot realistically be repeated for each new domain.
An intermediate training step on data from related tasks has shown to be beneficial across many NLP tasks.
However, this setup still requires supplementary annotated data, which is often not available.
In this paper, we investigate intermediate pre-training specifically for RE. We exploit the affinity between syntactic structure and semantic RE, and identify the syntactic relations which are closely related to RE by being on the shortest dependency path between two entities.
We then take advantage of the high accuracy of current syntactic parsers in order to automatically obtain large amounts of low-cost pre-training data.
By pre-training our RE model on the relevant syntactic relations, we are able to outperform the baseline in five out of six cross-domain setups, without any additional annotated data.

§ INTRODUCTION

Relation Extraction (RE) is the task of extracting structured knowledge, often in the form of triplets, from unstructured text.
Despite the increasing attention this task received in recent years, the performance obtained so far are very low <cit.>. This happens in particular when considering realistic scenarios which include out-of-domain setups, and deal with the whole task—in contrast to the simplified Relation Classification which assumes that the correct entity pairs are given <cit.>.
One main challenge of RE and other related Information Extraction tasks is the "domain-specificity": Depending on the text domain, the type of information to extract changes.
For example, while in the news domain we can find entities like person and city, and relations like city of birth <cit.>, in scientific texts, we can find information about metrics, tasks and comparisons between computational models <cit.>.
While high-quality, domain-specific data for fine-tuning the RE models would be ideal, as for many other NLP tasks, annotating data is expensive and time-consuming.[For example,  report a cost of 19K USD ( $\approx$ 1$ per annotated relation) and seven months of annotation work for an RE dataset including 5.3K sentences.]
A recent approach that leads to improved performance on a variety of NLP tasks is intermediate task training. It consists of a step of training on one or more NLP tasks between the general language model pre-training and the specific end task fine-tuning (STILT, Supplementary Training on Intermediate Labeled-data Tasks; ).
However, STILT assumes the availability of additional high quality training data, annotated for a related task.

[...] in a number of & European & countries & including the & United & Kingdom & [...]

[group style=fill=orange!40, draw=brown]123entity2
[group style=fill=orange!40, draw=brown]156entity1
[edge height=2ex]36nmod
[edge below, edge style = ultra thick, brown, arc edge, arc angle=25, label style = below, text=brown, text only label]entity1entity2TYPE-OF1ex

[...] & John & Madsen & , the & Professor & of Electrical Engineering at the & University & of Sydney & [...]

[group style=fill=green!85!blue!30!white, draw=green!60!black]123entity1
[group style=fill=green!85!blue!30!white, draw=green!60!black]178entity2
[edge height=2ex]25appos
[edge height=2ex]57nmod
[edge below, edge style = ultra thick, green!60!black, arc edge, arc angle=20, label style = below, text=green!60!black, text only label]entity1entity2ROLE1ex

Syntactic and Semantic Structures Affinity. Shortest dependency path (above), and semantic relation (below) between two semantic entities.

[column sep=0.3cm]
Linear-fractional & programming & ( & LFP & ) & is & a & generalization & of & linear & programming & ( & LP & ) & .

[edge height=2ex]21amod
[edge height=6ex]82nsubj
[edge height=2ex]43puntc
[edge height=4.5ex]24appos
[edge height=2ex]45punct
[edge height=4.5ex]86cop
[edge height=2ex]87det
[edge height=4.5ex]119case
[edge height=2ex]1110amod
[edge height=6ex]811nmod
[edge height=2ex]1312punct
[edge height=4.5ex]1113appos
[edge height=2ex]1314punct
[edge height=7.5ex]815punct

[column sep=0.3cm]
Linear-fractional & programming & ( & LFP & ) & is & a & generalization & of & linear & programming & ( & LP & ) & .

[edge height=4ex, edge style=ultra thick, label style=font=,thick]82nsubj
[edge height=2ex, edge style=ultra thick, label style=font=,thick]24appos
[edge height=2ex, edge style=ultra thick, label style=font=,thick]811nmod
[edge height=2ex, edge style=ultra thick, label style=font=,thick]1113appos
[group style=fill=yellow!40, draw=brown!50!yellow]112entity1
[group style=fill=yellow!40, draw=brown!50!yellow]144entity2
[group style=fill=yellow!40, draw=brown!50!yellow]11011entity3
[group style=fill=yellow!40, draw=brown!50!yellow]11313entity4
[edge below, edge style = ultra thick, brown!50!yellow, arc edge, arc angle=20, label style = above, text=brown!50!yellow, text only label]entity2entity1NAMED1ex
[edge below, edge style = ultra thick, brown!50!yellow, arc edge, arc angle=20, label style = below, text=brown!50!yellow, text only label]entity3entity1TYPE-OF1ex
[edge below, edge style = ultra thick, brown!50!yellow, arc edge, arc angle=20, label style = below, text=brown!50!yellow, text only label]entity4entity3NAMED1ex

Pre-training Example. Given the dependency tree (above), we filter in for pre-training only the UD labels which are on the shortest dependency path between two semantic entities (below).

In this paper, we explore intermediate pre-training specifically for cross-domain RE and look for alternatives which avoid the need of external manually annotated datasets to pre-train the model on.
In particular, we analyze the affinity between syntactic structure and semantic relations, by considering the shortest dependency path between two entities <cit.>.
We replace the traditional intermediate pre-training step on additional annotated data, with a syntax pre-training step on silver data.
We exploit the high accuracy of current syntax parsers, for obtaining large amount of low-cost pre-training data.
The use of syntax has a long tradition in RE <cit.>. Recently, work has started to infuse syntax during language model pre-training <cit.> showing benefits for RE as well. We instead investigate dependency information as silver data in intermediate training, which is more efficient. To the best of our knowledge, the use of syntax in intermediate pre-training for RE is novel.
We aim to answer the following research questions: 1 Does syntax help RE via intermediate pre-training (fast and cheap approach)? and 2 How does it compare with pre-training on additional labeled RE data (expensive)? We release our model and experiments.[<https://github.com/mainlp/syntax-pre-training-for-RE>]

§ SYNTAX PRE-TRAINING FOR RE

Syntactic parsing is a structured prediction task aiming to extract the syntactic structure of text, most commonly in the form of a tree.
RE is also a structured prediction task, but with the aim of extracting the semantics expressed in a text in the form of triplets—entity A, entity B, and the semantic relation between them.[In this project, we follow previous work, and assume gold entities, leaving end-to-end RE for future work.]
We exploit the affinity of these two structures by considering the shortest dependency path between two (semantic) entities (see Figure <ref>).

The idea we follow in this work is to pre-train an RE baseline model over the syntactic relations—Universal Dependency (UD) labels—which most frequently appear on the shortest dependency paths between two entities (black bold arrows in Figure <ref>).
We assume these labels to be the most relevant with respect to the final target task of RE.
In order to feed the individual UD relations into the RE baseline (model details in Section <ref>) we treat them similarly as the semantic connections.
In respect to Figure <ref>, we can formalize the semantic relations as the following triplets:

* .

Accordingly, we define the syntax pre-training instances as:

* .

In the next section we describe the detailed training process.

§ EXPERIMENTS

§.§ Setup

UD Label Distribution Over the Shortest Dependency Paths. Statistics of the UD labels which are on the shortest dependency path between two entities over the six train sets of CrossRE <cit.>.

Shortest Dependency Path Length. Statistics of the shortest dependency path length between two entities over the train sets of CrossRE <cit.>.

In order to evaluate the robustness of our method over out-of-domain distributions, we experiment with CrossRE <cit.>,[Released with a GNU General Public License v3.0.] a recently published multi-domain dataset.
CrossRE includes 17 relation types spanning over six diverse text domains: news, politics, natural science, music, literature and artificial intelligence (AI). The dataset was annotated on top of a Named Entity Recognition dataset—CrossNER <cit.>—which comes with an unlabeled domain-related corpora.[Released with an MIT License.]
We used the latter for the syntax pre-training phase.

UD Label Selection
In order to select the UD labels which most frequently appear on the shortest dependency path between two semantic entities, we parsed the training portions of CrossRE.
Our analysis combines RE annotations and syntactically parsed data.
We observe that the syntactic distance between two entities is often higher than one (see Figure <ref>), meaning that the shortest dependency path between two entities includes multiple dependencies—in the examples in Figure <ref>, the one above has distance one, the one below has distance two.
However, the shortest dependency paths contain an high frequency of just a few UD labels (see Figure <ref>) which we use for syntax pre-training: , , , , .
See Appendix <ref> for additional data analysis.

Our RE model follows the current state-of-the-art architecture by  which augments the sentence with four entity markers $e_1^{start}$, $e_1^{end}$, $e_2^{start}$, $e_2^{end}$ before feeding it into a pre-trained encoder (BERT; ).
The classification is then made by a 1-layer feed-forward neural network over the concatenation of the start markers $[\hat{s}_{e_1^{start}}, \hat{s}_{e_2^{start}}]$.
We run our experiments over five random seeds and report the average performance. See Appendix <ref> for reproducibility and hyperparameters settings of our model.

The training of our RE model is divided into two phases.
In the first one—which we are going to call syntax pre-training—we use the unlabeled corpora from CrossNER for pre-training the baseline model over the RE-relevant UD labels.
To do so, 1 we sample an equal amount of sentences from each domain[Regarding the news domain, which does not have a corresponding unlabeled corpus available, we sampled from the train set of CrossNER which is not included in CrossRE.] (details in Section <ref>), and 2 use the MaChAmp toolkit <cit.> for inferring the syntactic tree of each of them.
We apply an additional sub-step for disentangling the  dependency, as illustrated in Appendix <ref>.
Then, 3 we filer in only the , , , , and  UD labels and 4 feed those connections to the RE model (as explained in the previous section).
Within the RE model architecture described above, each triplet corresponds to one instance. In this phase, in order to assure more variety, we randomly select a maximum of five triplets from each pre-train sentence.

In the second training phase—the fine-tuning one—we replace the classification head (i.e. the feed-forward layer) with a new one, and individually train six copies of the model over the six train sets of CrossRE. Note that the encoder is fine-tuned in both training phases.
Finally, we test each model on in- and out-of-domain setups.

§.§ Results

Table <ref> reports the results of our cross-domain experiments in terms of Macro-F1.
We compare our proposed approach which adopts syntax pre-training with the zero-shot baseline model.[While utilizing the model implementation by , our score range is lower because we include the no-relation case, while they assume gold entity pairs.] Five out of six models outperform the average of the baseline evaluation, including in- and out-of-domain assessments. The average improvement—obtained without any additional annotated RE data—is 0.71, which considering the low score range given by the challenging dataset (with limited train sets, see dataset size in Appendix <ref>), and the cross-domain setup, is considerable.
The model fine-tuned on the news domain is the only one not outperforming the baseline. However, the performance scores on this domain are already extremely low for the baseline, because news comes from a different data source with respect to the other domains, has a considerable smaller train set, and present a sparse relation types distribution, making it a bad candidate for transferring to other domains <cit.>.

train     1cnews     1cpolitics     1cscience     1cmusic     1cliterature     1cAI     1cavg.

6*90baseline     news     10.98     1.32     1.24     1.01     1.49     1.42     2.91

politics     16.07     11.30     6.74     7.24     7.29     5.54     9.03

science     6.54     5.95     8.57     7.13     6.65     7.29     7.02

music     3.99     9.91     9.22     19.01     10.43     8.53     10.18

literature     11.30     9.60     9.79     12.49     17.17     9.79     11.69

AI     6.58     7.42     11.03     7.11     6.15     15.57     8.98

6*90syntax     news     6.67     1.15     0.72     0.61     1.13     0.75     1.84

politics     13.72     12.09     7.47     7.15     7.78     6.24     9.08

science     8.46     7.08     8.69     8.19     7.52     8.91     8.14

music     3.35     10.65     9.35     18.63     11.62     10.34     10.66

literature     11.85     9.84     10.35     13.58     18.64     9.94     12.37

AI     8.87     8.59     11.87     8.29     7.68     15.93     10.21

6*90scierc     news     11.88     2.30     2.09     1.13     1.82     2.16     3.56

politics     14.25     13.55     6.52     7.12     7.42     7.07     9.32

science     8.27     10.31     13.59     9.09     7.78     11.11     10.03

music     5.41     11.84     10.85     21.39     12.26     11.22     12.16

literature     12.36     8.05     8.87     13.13     16.44     9.40     11.37

AI     11.00     10.12     14.03     8.93     8.50     18.89     11.91

Performance Scores. Macro-F1 scores of the baseline model, compared with the proposed syntax pre-training approach, and—as comparison—with the traditional pre-training over the manually annotated SciERC dataset <cit.>.

As comparison, we report the scores obtained with the traditional intermediate pre-training which includes additional annotated data. We pre-train the language encoder on SciERC <cit.>, a manually annotated dataset for RE.
SciERC contains seven relation types, of which three overlap with the CrossRE relation set. In this setup, the improvement over the baseline includes the news, but not the literature domain. Nevertheless, while the gain is on average slightly higher with respect to the proposed syntax pre-training approach, it comes at a much higher annotation cost.

Pre-train Data Quantity Analysis. Average (dev) performance of the six models when pre-trained over increasing amounts of syntactic instances.

Per-Domain Pre-train Data Quantity Analysis. Individual (dev) performance of the model fine-tuned on AI when pre-trained over increasing amounts of syntactic instances.

§ PRE-TRAINING DATA QUANTITY ANALYSIS

We inspect the optimal quantity of syntactic data to pre-train our RE model on by fine-tuning this hyperparameter over the dev sets of CrossRE. The plot in Figure <ref> reports the average performance of the six models when pre-trained on increasing amounts of syntactic dependencies.[Pre-training performance in Appendix <ref>.] Starting from 8.4K instances onward, the performance stabilizes above the baseline. We select the peak (20.4K, albeit results are similar between 18-20.4K) for reporting our test set results in Table <ref>.
While we are interested in the robustness of our method across multiple domains, and therefore consider the average (Figure <ref>), domain-optima could be achieved by examining individual domain performance.
As example, we report in Figure <ref> the plot relative to the model fine-tuned on AI, which is the one  obtaining the highest gain.
The model fine-tuned on AI generally gains a lot from the syntax pre-training step, with its peak on 15.6K pre-training instances.

§ CONCLUSION

We introduce syntax pre-training for RE as an alternative to the traditional intermediate training which uses additional manually annotated data.
We pre-train our RE model over silver UD labels which most frequently connect the semantic entities via the shortest dependency path.
We test the proposed method over CrossRE and outperform the baseline in five out of six cross-domain setups.
Pre-training over a manually annotated dataset, in comparison, only slightly increases our scores in five out of six evaluations, but at a much higher cost.

§ LIMITATIONS

While we already manage to outperform the baseline, the pre-training data quantity is relatively small ($\sim$20K instances).
Given the computational cost of training 30 models—six train sets, over five random seeds each—and testing them within in- and cross- domain setups, we break the inspection of the optimal pre-training data amount at 24K instances.
However we do not exclude that more pre-training instances would be even more beneficial for improving even more over the baseline.

Related to computation cost constrains, we test our syntax pre-training approach over one set of UD labels only (, , , , ). Different sets could be investigated, e.g. including  and , which present a lower, but still considerable amount of instances (see Figure <ref>).

Finally, while approaching RE by assuming that the gold entities are given is a common area of research, we leave for future work the inspection of the proposed method over end-to-end RE.

§ ACKNOWLEDGMENTS

We thank the NLPnorth and the MaiNLP groups for feedback on an earlier version of this paper, and TurkuNLP for hosting EB for a research stay.

EB and BP are supported by the Independent Research Fund Denmark (Danmarks Frie Forskningsfond; DFF) Sapere Aude grant 9063-00077B. BP is in parts supported by the European Research Council (ERC)  (grant agreement No. 101043235). FG and SP were supported by the Academy of Finland.

§ UD ANALYSIS FOR RE

We inspect the same statistics as Figure <ref> and Figure <ref>—UD labels on the shortest dependency paths, and shortest dependency path lengths respectively—but instead of at domain level, at semantic relation type level.
Table <ref> and Table <ref> report this analysis, revealing similar trends over the 17 types.

2c|     rel-to     artifact     cause-eff     compare     gen-aff     named     opposite     origin     part-of     physical     role     social     temporal     topic     type-of     usage     win-def

21*9055Universal Dependencies
nsubj     9089     80106     502     9012     90120     6554     10061     9053     7075     50115     90248     10033     5054     3010     5018     9030     7068

obj       7078     3051     001     506     5076     4036     8048     6041     8083     1055     50129     209     4048     5017     3014     10037     10086

iobj      000     000     000     000     000     001     000     000     000     001     003     000     000     000     000     000     000

ccomp     005     007     000     304     007     0010     008     002     002     002     009     000     000     000     000     000     000

xcomp     006     009     000     103     0015     005     005     109     005     0011     0017     001     1016     003     001     002     1010

obl       8088     5062     1005     10014     7092     6553     4025     7044     7077     80202     80224     6019     100121     5017     7026     76     1311

advcl     0010     009     904     708     3047     1521     3019     1010     1018     0014     341     003     1015     002     106     107     002

advmod     000     003     000     000     001     000     000     000     001     000     000     000     001     000     000     000     000

nmod      100100     100140     502     9012     100181     7057     7047     10058     100148     100276     100386     8529     7072     10043     10035     5019     5048

appos     2026     6589     000     002     6085     100108     1011     3023     3041     3072     50112     209     512     006     106     001     3020

nummod     001     000     000     000     000     000     000     000     000     000     000     000     000     000     000     000     000

acl       5040     1024     000     000     1039     3030     1010     4025     4048     533     1074     000     511     7024     002     3013     2015

amod      005     001     000     002     831     005     003     003     005     002     003     000     003     002     000     003     004

det       000     000     000     000     000     000     000     000     000     000     001     000     000     000     000     000     000

conj      001     004     000     000     003     001     000     001     002     003     0011     000     001     001     000     000     000

flat      002     003     000     000     001     0012     008     000     002     0011     0037     108     007     001     000     000     003

compound     3029     1024     000     405     4570     2827     005     007     5054     1053     557     002     009     002     005     1010     3022

list      000     001     000     000     002     002     000     000     000     000     002     000     002     000     000     000     000

parataxis     005     007     000     000     830     814     000     000     0014     005     0017     001     008     001     005     000     001

orphan     001     000     000     000     000     000     000     000     000     000     000     000     000     000     000     000     000

punct     000     001     000     000     000     000     000     000     000     000     000     000     000     000     000     000     000

UD Label Distribution Over the Shortest Dependency Paths per Relation Type. Statistics of the UD labels which are on the shortest dependency path between two entities divided by the 17 relation types of CrossRE <cit.>.

2c|     related-to     artifact     cause-eff     compare     gen-aff     named     opposite     origin     part-of     physical     role     social     temporal     topic     type-of     usage     win-def

11*9055Shortest Dependency Path Length

1     3     5043     0     0     9073     9092     1     7019     3026     6072     3064     0     105     507     9017     0     3012

2     5029     5043     0     503     7059     1023     9025     8028     7051     9085     70127     7010     9041     306     5011     506     7031

3     7036     9071     902     101     3033     14     7021     9029     9057     7075     90136     9017     7030     9014     3     9014     9034

4     9042     3024     902     704     5052     1020     5013     3012     5037     5041     50104     507     6028     306     9017     7012     5017

5     3017     1012     0     905     3036     12     5014     108     1018     3028     1033     305     4010     708     2     304     3

6     109     7     0     704     1018     6     104     5     8     1012     10     0     2     1     2     1     2

7     4     0     0     0     4     3     0     0     2     6     4     0     105     0     0     0     105

8     0     0     0     0     0     3     1     0     0     1     1     0     3     0     0     0     0

9     0     1     0     0     0     1     0     0     0     0     2     0     0     0     0     0     0

10     0     0     0     0     0     2     0     0     0     0     0     0     0     0     0     0     0

11     0     0     0     0     0     2     0     0     0     0     0     0     0     0     0     0     0

Shortest Dependency Path Length per Relation Type. Statistics of the shortest dependency path length between two semantic entities divided by the 17 relation types of CrossRE <cit.>.

§ REPRODUCIBILITY

We report in Table <ref> the hyperparameter setting of our RE model (see Section <ref>).
All experiments were ran on an NVIDIA® A100 SXM4 40 GB GPU and an AMD EPYC™ 7662 64-Core CPU.
Within this computation infrastructure the baseline converges in $\sim7$ minutes. The the syntax pre-training step takes $\sim10$ minutes, to which we have to add $\sim7$ minutes in order to obtain the complete training time.

Parameter     Value

Classifier     1-layer FFNN

Loss     Cross Entropy

Optimizer     Adam optimizer

Batch size     12, 24

Learning rate     $1e^{-5}$ (pre-train)

Learning rate     $2e^{-5}$ (fine-tuning)

Seeds     4012, 5096, 8878, 8857, 9908

Hyperparameters Setting. Model details for reproducibility of the baseline.

We train MaChAmp v0.4 on the English Web Treebank v2.10 with XLM-R large <cit.> as language model with all default hyperparameters of MaChAmp.

§ HANDLING OF

In UD, the first element in a conjuncted list governs all other elements of the list via a  dependency and represents the list syntactically w.r.t. the remainder of the sentence.  CrossRE <cit.> relations, on the other hand, directly link the two entities involved in the semantic structure. To account for this difference, we propagate the conjunction dependencies in order to reflect the semantic relations, as shown in Figure <ref>.

[...] opinion-based recommender system utilize various & techniques & including & text & mining & , & information & retrieval & , & sentiment & analysis [...]

[edge height=2ex]25nmod
[edge height=2ex]58conj
[edge height=4ex]511conj

[...] opinion-based recommender system utilize various & techniques & including & text & mining & , & information & retrieval & , & sentiment & analysis [...]

[edge height=2ex]25nmod
[edge height=4ex]28nmod
[edge height=6ex]211nmod

Example of  Alteration. Original UD dependencies (above), and disentangled conjunction dependencies reflecting the semantic relation annotations (below).

§ CROSSRE SIZE

We report in Table <ref> the dataset statistics of CrossRE <cit.> including the number of sentences and of relations.

4c|sentences     4crelations

train     dev     test     tot.     train     dev     test     tot.

news     164     350     400     914     175     300     396     871

politics     101     350     400     851     502     1,616     1,831     3,949

science     103     351     400     854     355     1,340     1,393     3,088

music     100     350     399     849     496     1,861     2,333     4,690

literature     100     400     416     916     397     1,539     1,591     3,527

AI     100     350     431     881     350     1,006     1,127     2,483

tot.     668     2,151     2,446     5,265     2,275     7,662     8,671     18,608

CrossRE Statistics. Number of sentences and number of relations for each domain of CrossRE <cit.>.

§ SYNTAX PRE-TRAINING PERFORMANCE

Figure <ref> reports the performance of the RE model during the syntax pre-training phase, over increasing amounts of pre-training dependency instances. The scores are computed on a set including 600 sentences (100 per domain) not overlapping with the train set used in the syntax pre-training phase.

Pre-train Performance. Pre-train performance of the RE model over increasing amounts of dependency instances