# Euclidean self-dual gravity: Ashtekar variables without gauge fixing

J. Fernando Barbero G  Instituto de Estructura de la Materia, IEM-CSIC.
Serrano 123, 28006 Madrid, Spain  Grupo de Teorías de Campos y Física
Estadística. Instituto Gregorio Millán (UC3M). Unidad Asociada al Instituto de
Estructura de la Materia, CSIC  Marc Basquens  Departamento de Matemáticas,
Universidad Carlos III de Madrid, Avda. de la Universidad 30, 28911 Leganés,
Spain  Department of Energy Technology, Royal Institute of Technology (KTH),
10044 Stockholm, Sweden  Eduardo J.S. Villaseñor  Departamento de Matemáticas,
Universidad Carlos III de Madrid, Avda. de la Universidad 30, 28911 Leganés,
Spain  Grupo de Teorías de Campos y Física Estadística. Instituto Gregorio
Millán (UC3M). Unidad Asociada al Instituto de Estructura de la Materia, CSIC

###### Abstract

The GNH method is used in this paper to study the Hamiltonian formulation of
the Euclidean self-dual action. This action can be used to arrive at the
complex Ashtekar formulation of General Relativity or a real connection
formulation for Euclidean General Relativity. The main result of the paper is
a derivation of the Ashtekar formulation for Euclidean gravity without using
any gauge fixing. It is interesting to compare this derivation with the one
corresponding to the Holst action. In particular it is worth noting that no
“tertiary” constraints appear in the case considered in the present paper.

March 31, 2023

###### Contents

  1. 1 Introduction
  2. 2 Action and equations of motion
  3. 3 The Hamiltonian formulation in the GNH approach
  4. 4 Some details on the secondary constraints
  5. 5 The Hamiltonian vector fields
    1. 5.1 Solving equation 5.1
    2. 5.2 Solving equation 5.2
    3. 5.3 Solving equation 5.3
  6. 6 Tangency analysis
    1. 6.1 Checking condition 6.5
    2. 6.2 Checking condition 6.6
    3. 6.3 Checking condition 6.4
  7. 7 Hamiltonian formulation: Ahstekar variables and the time gauge
  8. 8 Conclusions and comments
  9. A Tangency analysis: some details
    1. A.1 Additional details on the tangency condition 6.6
    2. A.2 Additional details on the tangency condition 6.5
  10. B The Gauss law in the Ashtekar formulation: some computational details

Key Words: Self-dual action; Hamiltonian formulation; GNH method.

##  1 Introduction

The self-dual action [1, 2] played a very important role in the establishment
of the Ashtekar formulation for General Relativity (GR) as a new starting
point for the quantization of gravity. The self-dual formulation has also
drawn interest in the context of the quantization proposal for GR put forward
by Smolin in [3] and discussed recently in [4, 5, 6, 7, 8]. At variance with
the Hamiltonian treatment of the Hilbert-Palatini action—which leads to a
version of standard geometrodynamics endowed with an $SO(3)$ internal
symmetry—the self-dual version of the Hilbert-Palatini action leads to the
Ashtekar formulation for either complex or real-Euclidean GR. Although at the
present moment the standard way to arrive at the real Ashtekar formulation
makes use of the Holst action [9], the self-dual action is still widely
considered as a means to address a number of problems, in particular whenever
the complex formulation is useful (see, for instance, [10, 11]). For this
reason it is interesting to revisit the Hamiltonian formulation derived from
it and consider several issues that have not completely cleared out in the
past, in particular regarding the stability of all the constraints when the
Dirac method is used [12]. A very powerful way to address these issues is to
rely on the geometric perspective provided by the GNH approach to Hamiltonian
dynamics [13, 14]. In this setting the main consistency condition is
translated into a tangency requirement whose verification does not need the
computation of Poisson brackets and, thus, is significantly easier than its
counterpart in the Dirac “algorithm”.

It is interesting to compare the GNH analysis of the Holst action (discussed,
for instance, in [15] from the standard Dirac point of view and also in [17]
from the perspective of the GNH method). As we show here, on one hand, the
analysis of the self-dual action is shorter because no new secondary
constraints appear after the first batch of them show up in the analysis,
whereas, in the Holst case, secondary constraints appear at two different
stages. On the other hand, the explicit verification of the tangency of the
Hamiltonian vector fields to the phase space submanifold determined by the
secondary constraints in the case of the self-dual action is rather intricate
and requires careful consideration.

One interesting consequence of the analysis that we present in the paper is
the possibility of deriving the Ashtekar formulation for Euclidean GR without
any gauge fixing. The key insight to arrive at this result comes from the form
of the pullback of the canonical symplectic form to the primary constraint
submanifold. In fact it is possible to write this pullback in a way that leads
to the introduction of a function in the phase space that takes the sympletic
form to a simple “canonical” form (i.e. the analogous of
$\Sigma_{i}\mathrm{d}q_{i}\wedge\mathrm{d}p^{i}$). It is very interesting at
this point to rewrite the remaining elements of the Hamiltonian formulation
(constraints and Hamiltonian vector fields) with the help of this new object.
It is also worth to compare the result with the one derived by the usual time-
gauge fixing. As we will show the results are compatible in an appealing way
and the role of one of the $SO(3)$ factors of the $SO(4)$ symmetry is
noteworthy.

The structure of the paper is the following. After this introduction we will
quickly discuss some basic facts about the self-dual (actually the anti-delf-
dual) action for Euclidean GR in section 2. After that we will discuss in
Section 3 the Hamiltonian formulation for this action and perform the first
steps of the GNH analysis. Before completing the crucial tangency tests to
verify the consistency of the dynamics obtained by using the GNH method we
will discuss some features of the secondary constraints (Section 4) and solve
the equations that determine the components of the Hamiltonian vector fields
on the primary constraint submanifold (Section 5). With all this information
we will give a detailed account in Section 6 of the tangency analysis,
although we will leave some details for Appendix A. Although the main ideas
involved are simple, the computation itself is quite intricate. In Section 7
we show how the Ashtekar formulation appears quite naturally by following our
approach, in fact, this is probably the cleanest way to arrive at it from an
action principle. Furthermore, as we will explain there is no need in
principle to use any gauge fixing (at variance with the situation in the case
of the Holst action, see, for instance [16, 17]). We end the paper with some
conclusions in Section 8. Appendix B gives some computations related to the
Gauss law in the Ashtekar formulation.

Some comments about our notation. As a general rule we will employ boldface
characters to denote four dimensional geometric objects and non-boldface
letters for the 3-dimensional ones. The totally antisymmetric Levi-Civita
symbol in three dimensions will be denoted as $\epsilon_{ijk}$. The “internal”
indices $i,j,k,...$ will be raised and lowered with the Euclidean metric
$\mathrm{Diag}(+++)$ (so, in practice, their position upstairs or downstairs
is irrelevant). We will also employ boldface characters to denote canonical
momenta. The exterior differentials in $\mathcal{M}$, $\Sigma$ and the phase
space $T^{*}Q$ will be respectively denoted as $\bm{\mathrm{d}}$, $\mathrm{d}$
and $\mathrm{d\\!l}$. The interior product of a vector field $X$ and a
differential form $\beta$ will be denoted either as $\imath_{X}\beta$ or
$X\raisebox{8.0pt}{\scalebox{1.0}[-2.5]{$\neg\,$}}\beta$. Throughout the paper
$\pounds_{X}$ denotes the Lie derivative along a vector field $X$. Finally,
the scalar field $\phi$ satisfying $\sigma=\phi\,\mathsf{vol}$ for a top-form
$\sigma$ and a volume form $\mathsf{vol}$ will be often written as
$\left(\frac{\sigma}{\mathsf{vol}}\right)$.

##  2 Action and equations of motion

Let $\Sigma$ be a closed (i.e. compact without boundary), orientable,
3-dimensional manifold (this implies that $\Sigma$ is paralellizable) and
$\mathcal{M}=\mathbb{R}\times\Sigma$. The basic fields that we will use to
write the action are:

$\displaystyle{\bm{\mathrm{e}}}^{i}\in\Omega^{1}(\mathcal{M})\,,$
$\displaystyle i=1,2,3$
$\displaystyle{\bm{\omega}}^{i}\in\Omega^{1}(\mathcal{M})\,,$ $\displaystyle
i=1,2,3$ $\displaystyle{\bm{\alpha}}\in\Omega^{1}(\mathcal{M})\,.$

The fields ${\bm{\alpha}}$ and ${\bm{\mathrm{e}}}^{i}$ are chosen in such a
way that
${\bm{\alpha}}\otimes{\bm{\alpha}}+{\bm{\mathrm{e}}}_{i}\otimes{\bm{\mathrm{e}}}^{i}$
is a Euclidean metric of signature $(+\,+\,+\,\,+)$. As a consequence
$({\bm{\alpha}},{\bm{\mathrm{e}}}^{i})$ defines a non-degenerate tetrad.

Important geometric objects defined with the help of the fields introduced
above are the covariant exterior differential ${\bm{\mathrm{D}}}$, which acts
on the ${\bm{\mathrm{e}}}_{i}$ according to

${\bm{\mathrm{D}}}{\bm{\mathrm{e}}}_{i}:={\bm{\mathrm{d}}}{\bm{\mathrm{e}}}_{i}+\varepsilon_{ijk}{\bm{\omega}}^{j}\wedge{\bm{\mathrm{e}}}^{k}\,,$

and the curvature 2-form

${\bm{\mathrm{F}}}^{i}:={\bm{\mathrm{d}}}{\bm{\omega}}^{i}+\frac{1}{2}{\bm{\omega}}^{j}\wedge{\bm{\omega}}^{k}\,.$

The Euclidean self-dual action for GR can be written in the form [12]

$S({\bm{\mathrm{e}}},{\bm{\omega}},{\bm{\alpha}}):=\int_{\mathcal{M}}\left(\frac{1}{2}\varepsilon_{ijk}{\bm{\mathrm{e}}}^{i}\wedge{\bm{\mathrm{e}}}^{j}\wedge{\bm{\mathrm{F}}}^{k}-{\bm{\alpha}}\wedge{\bm{\mathrm{e}}}_{i}\wedge{\bm{\mathrm{F}}}^{i}\right)\,.$
(2.1)

Notice that the first term is the Husain-Kuchař action [18]. The indices
$i,j,k=1,2,3$ can be considered as “$SO(3)$ indices” because (2.1) is
invariant under the infinitesimal gauge transformations

$\displaystyle\begin{split}\delta_{1}{\bm{\omega}}^{i}&={\bm{\mathrm{D}}}{\bm{\Lambda}}^{i}\,,\\\
\delta_{1}{\bm{\alpha}}&=0\,,\\\
\delta_{1}{\bm{\mathrm{e}}}^{i}&=\varepsilon^{i}_{\phantom{i}jk}{\bm{\mathrm{e}}}^{j}{\bm{\Lambda}}^{k}\,,\end{split}$
(2.2)

with ${\bm{\Lambda}}^{k}\in C^{\infty}(\mathcal{M})$. The action is also
invariant under the infinitesimal transformations

$\displaystyle\begin{split}\delta_{2}{\bm{\omega}}^{i}&=0\,,\\\
\delta_{2}{\bm{\alpha}}&={\bm{\Upsilon}}_{i}{\bm{\mathrm{e}}}^{i}\,,\\\
\delta_{2}{\bm{\mathrm{e}}}^{i}&=-{\bm{\Upsilon}}^{i}{\bm{\alpha}}+\varepsilon^{i}_{\phantom{i}jk}{\bm{\mathrm{e}}}^{j}{\bm{\Upsilon}}^{k}\,,\end{split}$
(2.3)

where ${\bm{\Upsilon}}^{k}\in C^{\infty}(\mathcal{M})$. It is important to
point out that $\delta_{1}$ and $\delta_{2}$ are independent. Also, it is
worth noting that these transformation do not commute, in fact they satisfy

$[\delta_{1}({\bm{\Lambda}}),\delta_{2}({\bm{\Upsilon}})]=\delta_{2}({\bm{\Lambda}}\times{\bm{\Upsilon}})\,.$

The internal symmetries of the action can be written in other forms, for
instance

$\left\\{\begin{array}[]{l}\delta^{-}({\bm{\Lambda}}){\bm{\omega}}_{i}={\bm{D}}{\bm{\Lambda}}_{i}\\\
\delta^{-}({\bm{\Lambda}}){\bm{\alpha}}=-\frac{1}{2}{\bm{\Lambda}}_{i}{\bm{\mathrm{e}}}^{i}\\\
\delta^{-}({\bm{\Lambda}}){\bm{\mathrm{e}}}_{i}=\frac{1}{2}{\bm{\Lambda}}_{i}{\bm{\alpha}}+\frac{1}{2}\varepsilon_{ijk}{\bm{\mathrm{e}}}^{j}{\bm{\Lambda}}^{k}\end{array}\right.\,,\quad\left\\{\begin{array}[]{l}\delta^{+}({\bm{\Upsilon}}){\bm{\omega}}_{i}=0\\\
\delta^{+}({\bm{\Upsilon}}){\bm{\alpha}}=\frac{1}{2}{\bm{\Upsilon}}_{i}{\bm{\mathrm{e}}}^{i}\\\
\delta^{+}({\bm{\Upsilon}}){\bm{\mathrm{e}}}_{i}=-\frac{1}{2}{\bm{\Upsilon}}_{i}{\bm{\alpha}}+\frac{1}{2}\varepsilon_{ijk}{\bm{\mathrm{e}}}^{j}{\bm{\Upsilon}}^{k}\end{array}\right.\,.$
(2.4)

At variance with $\delta_{1}$ and $\delta_{2}$ these transformations do
commute, i.e.

$[\delta^{-}({\bm{\Lambda}}),\delta^{+}({\bm{\Upsilon}})]=0\,.$

Taking also into account that

$[\delta^{-}({\bm{\Lambda}}),\delta^{-}({\bm{M}})]=\delta^{-}({\bm{\Lambda}}\times{\bm{M}})\,,\quad[\delta^{+}({\bm{\Lambda}}),\delta^{+}({\bm{M}})]=\delta^{+}({\bm{\Lambda}}\times{\bm{M}})\,,$

we see that (2.4) provides an explicit realization of $SO(4)=SO(3)\otimes
SO(3)$ as the symmetry group of the self-dual action (2.1) [12]. Notice that

$\delta_{1}({\bm{\Lambda}})=\delta^{-}({\bm{\Lambda}})+\delta^{+}({\bm{\Lambda}})\,,\quad\delta_{2}({\bm{\Upsilon}})=\delta^{+}(2{\bm{\Upsilon}})\,.$

The field equations coming from the action (2.1) are

$\displaystyle{\bm{\mathrm{D}}}({\bm{\alpha}\wedge\bm{\mathrm{e}}_{k}})+\epsilon_{ijk}\bm{\mathrm{e}}^{i}\wedge{\bm{\mathrm{D}}}\bm{\mathrm{e}}^{j}=0\,,$
(2.5)
$\displaystyle\epsilon_{ijk}\bm{\mathrm{e}}^{j}\wedge\bm{\mathrm{F}}^{k}+\bm{\alpha}\wedge\bm{\mathrm{F}}_{i}=0\,,$
(2.6) $\displaystyle\bm{\mathrm{e}}^{i}\wedge\bm{\mathrm{F}}_{i}=0\,.$ (2.7)

They are equivalent to the Euclidean Einstein equations in vacuum.

##  3 The Hamiltonian formulation in the GNH approach

In order to get the Lagrangian and Hamiltonian formulations given by the self-
dual action (2.1) we take advantage of the foliation naturally associated with
$\mathcal{M}=\mathbb{R}\times\Sigma$. The spatial sheets of this foliation are
$\Sigma_{t}:=\\{t\\}\times\Sigma$ ($t\in\mathbb{R}$). The foliation also
defines a canonical evolution vector field $\partial_{t}$ given by the tangent
vectors to the curves
$c_{p}:\mathbb{R}\rightarrow\mathbb{R}\times\Sigma:\tau\mapsto(\tau,p)$ with
$p\in\Sigma$. For each $t\in\mathbb{R}$ we define the embedding
$\jmath_{t}:\Sigma\rightarrow\mathcal{M}:p\mapsto(t,p)$ and denote its
pullback as $\jmath_{t}^{\ast}$. Notice that $\Sigma_{t}=\jmath_{t}(\Sigma)$.

By remembering that

$\int_{\mathbb{R}\times\Sigma}\mathcal{L}=\int_{\mathbb{R}\times\Sigma}\mathrm{d}t\wedge\imath_{\partial_{t}}\mathcal{L}=\int_{\mathbb{R}}\mathrm{d}t\int_{\Sigma_{t}}\imath_{\partial_{t}}\mathcal{L}=\int_{\mathbb{R}}\mathrm{d}t\int_{\Sigma}\jmath_{t}^{*}\imath_{\partial_{t}}\mathcal{L}\,.$

we can compute the Lagrangian
$L:=\int_{\Sigma}\jmath_{t}^{*}\imath_{\partial_{t}}\mathcal{L}$ defined on
$\Sigma$ from the 4-form $\mathcal{L}$ appearing in the action. The Lagrangian
thus obtained is defined on the configuration space

$Q=\mathcal{C}^{\infty}(\Sigma)^{3}\times\Omega^{1}(\Sigma)^{3}\times\mathcal{C}^{\infty}(\Sigma)^{3}\times\Omega^{1}(\Sigma)^{3}\times\mathcal{C}^{\infty}(\Sigma)\times\Omega^{1}(\Sigma)\,,$

with elements of the form
$(e_{\mathrm{t}}^{i},e^{i},\omega_{\mathrm{t}}^{i},\omega^{i},\alpha_{\mathrm{t}},\alpha)$,
by interpreting the objects

$\displaystyle
e_{\mathrm{t}}^{i}(t)\,:=\jmath_{t}^{\ast}\imath_{\partial_{t}}\bm{\mathrm{e}}^{i}\,,$
$\displaystyle e^{i}(t)\,:=\jmath_{t}^{\ast}\bm{\mathrm{e}}^{i}\,,$
$\displaystyle\omega_{\mathrm{t}}^{i}(t):=\jmath_{t}^{\ast}\imath_{\partial_{t}}\bm{\omega}^{i}\,,$
$\displaystyle\omega^{i}(t):=\jmath_{t}^{\ast}\bm{\omega}^{i}\,,$
$\displaystyle\alpha_{\mathrm{t}}(t):=\jmath_{t}^{\ast}\imath_{\partial_{t}}\bm{\alpha}\,,$
$\displaystyle\alpha(t)\,\,:=\jmath_{t}^{\ast}\bm{\alpha}\,.$

as defining curves in the configuration space $Q$ and considering also their
velocities

$\displaystyle
v_{e_{\mathrm{t}}}^{i}(t):=\jmath_{t}^{\ast}\pounds_{\partial_{t}}(\imath_{\partial_{t}}{\bm{\mathrm{e}}}^{i})=\frac{\mathrm{d}e_{\mathrm{t}}^{i}}{\mathrm{d}\tau}(t)\,,$
$\displaystyle
v_{e}^{i}(t):=\jmath_{t}^{\ast}\pounds_{\partial_{t}}\bm{\mathrm{e}}^{i}=\frac{\mathrm{d}}{\mathrm{d}\tau}(\jmath_{\tau}^{\ast}\bm{\mathrm{e}}^{i})\Big{|}_{\tau=t}=\frac{\mathrm{d}e^{i}}{\mathrm{d}\tau}(t)\,,$
$\displaystyle
v_{\omega_{\mathrm{t}}}^{i}(t):=\jmath_{t}^{\ast}\pounds_{\partial_{t}}(\imath_{\partial_{t}}{\bm{\mathrm{\omega}}}^{i})=\frac{\mathrm{d}\omega_{\mathrm{t}}^{i}}{\mathrm{d}\tau}(t)\,,$
$\displaystyle
v_{\omega}^{i}(t):=\jmath_{t}^{\ast}\pounds_{\partial_{t}}\bm{\mathrm{\omega}}^{i}=\frac{\mathrm{d}}{\mathrm{d}\tau}(\jmath_{\tau}^{\ast}\bm{\mathrm{\omega}}^{i})\Big{|}_{\tau=t}=\frac{\mathrm{d}\omega^{i}}{\mathrm{d}\tau}(t)\,,$
$\displaystyle
v_{\alpha_{\mathrm{t}}}(t):=\jmath_{t}^{\ast}\pounds_{\partial_{t}}(\imath_{\partial_{t}}{\bm{\alpha}})=\frac{\mathrm{d}\alpha_{\mathrm{t}}}{\mathrm{d}\tau}(t)\,,$
$\displaystyle
v_{\alpha}(t):=\jmath_{t}^{\ast}\pounds_{\partial_{t}}\bm{\alpha}=\frac{\mathrm{d}}{\mathrm{d}\tau}(\jmath_{\tau}^{\ast}{\bm{\alpha}})\Big{|}_{\tau=t}=\frac{\mathrm{d}\alpha}{\mathrm{d}\tau}(t)\,,$

defined in terms of the Lie derivative $\pounds_{\partial_{t}}$ along the
vector field $\partial_{t}$. Here we are leaving aside functional
issues—necessary for the complete definition of the configuration space—that
play an important role in the rigorous Hamiltonian formulation of this model.
In any case, the results that we will obtain here are in par (and equivalent
to) with those derived by using Dirac’s method. The Lagrangian is

$\displaystyle L(\mathrm{v})=$
$\displaystyle\int_{\Sigma}\Big{(}\left(\frac{1}{2}\epsilon_{ijk}e^{i}\wedge
e^{j}+e_{k}\wedge\alpha\right)\wedge
v_{\omega}^{k}+\omega_{\mathrm{t}}^{i}D\left(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge
e^{k}+e_{i}\wedge\alpha\right)$ $\displaystyle\hskip
190.63338pt-\alpha_{\mathrm{t}}e_{i}\wedge
F^{i}+e_{\mathrm{t}}^{i}\left(\epsilon_{ijk}e^{j}\wedge F^{k}+\alpha\wedge
F^{i}\right)\Big{)}\,,$

where we denote the velocity in the fiber of $TQ$ corresponding to
$(e_{\mathrm{t}}^{i},e^{i},\omega_{\mathrm{t}}^{i},\omega^{i},\alpha_{\mathrm{t}},\alpha)$
as
$\mathrm{v}:=(v_{e_{\mathrm{t}}}^{i},v_{e}^{i},v_{\omega_{\mathrm{t}}}^{i},v_{\omega}^{i},v_{\alpha_{\mathrm{t}}},v_{\alpha})$
(as a consequence $L:TQ\rightarrow\mathbb{R}$ depends both on the components
of the velocity and the fields defining the configuration manifold $Q$). We
also define
$F^{i}:=d\omega^{i}+\frac{1}{2}\epsilon^{ijk}\omega_{j}\wedge\omega_{k}$ and
$De^{i}:=de^{i}+\epsilon^{ijk}\omega_{j}\wedge e_{k}$ (with the usual
generalization to forms of other degrees). An important consequence of the
non-degeneracy of the tetrad $(\bm{\alpha},{\bm{\mathrm{e}}}^{i})$ is the non-
degeneracy of the triad $e^{i}$ on any $\Sigma_{t}$.

The fiber derivative that defines the canonical momenta is given by

$FL(\mathrm{v})(\mathrm{w})=\int_{\Sigma}\left(\frac{1}{2}\epsilon_{ijk}e^{i}\wedge
e^{j}+e_{k}\wedge\alpha\right)\wedge w_{\omega}^{k}\,,$ (3.1)

hence, the primary constraint submanifold $\mathsf{M}_{0}$ in the phase space
is defined by the conditions

$\displaystyle\mathbf{p}_{e_{\mathrm{t}}}$ $\displaystyle=0\,,$
$\displaystyle\mathbf{p}_{e}$ $\displaystyle=0\,,$
$\displaystyle\mathbf{p}_{\omega_{\mathrm{t}}}$ $\displaystyle=0\,,$
$\displaystyle\mathbf{p}_{\omega}(\mathrm{w})$
$\displaystyle=\int_{\Sigma}\left(\frac{1}{2}\epsilon_{ijk}e^{i}\wedge
e^{j}+e_{k}\wedge\alpha\right)\wedge w_{\omega}^{k}\,,$
$\displaystyle\mathbf{p}_{\alpha_{\mathrm{t}}}$ $\displaystyle=0\,,$
$\displaystyle\mathbf{p}_{\alpha}$ $\displaystyle=0\,.$

The Hamiltonian is defined on the primary constraint submanifold by $H=E\circ
FL^{-1}$ where $E:=\langle FL(v),v\rangle-L$ is the energy (a real function in
$TQ$). As in this case the energy $E$ only depends on the configuration
variables the functional form of the Hamiltonian coincides with that of $E$,
hence

$H(\mathbf{p})=\int_{\Sigma}\Big{(}\alpha_{\mathrm{t}}e_{i}\wedge
F^{i}-\omega_{\mathrm{t}}^{i}D\left(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge
e^{k}+e_{i}\wedge\alpha\right)-e_{\mathrm{t}}^{i}\left(\epsilon_{ijk}e^{j}\wedge
F^{k}+\alpha\wedge F_{i}\right)\Big{)}\,.$ (3.2)

Here we denote the momenta in the fiber of $T^{*}Q$ over the point
$(e_{\mathrm{t}}^{i},e^{i},\omega_{\mathrm{t}}^{i},\omega^{i},\alpha_{\mathrm{t}},\alpha)\in
Q$ as
$\mathbf{p}:=(\mathbf{p}_{e_{\mathrm{t}}},\mathbf{p}_{e},\mathbf{p}_{\omega_{\mathrm{t}}},\mathbf{p}_{\omega},\mathbf{p}_{\alpha_{\mathrm{t}}},\mathbf{p}_{\alpha})$.
The Hamiltonian depends on $\mathbf{p}$ only through its base point.

A vector field in phase space $\mathbb{Y}\in\mathfrak{X}(T^{*}Q)$ has the
following component structure

$\mathbb{Y}=(Y_{e_{\mathrm{t}}}^{i},Y_{e}^{i},Y_{\omega_{\mathrm{t}}}^{i},Y_{\omega}^{i},Y_{\alpha_{\mathrm{t}}},Y_{\alpha},{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{e_{\mathrm{t}}}},{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{e}},{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\omega_{\mathrm{t}}}},{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\omega}},{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\alpha_{\mathrm{t}}}},{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\alpha}})\,,$

where we use boldface characters for those components of $\mathbb{Y}$ that are
dual objects. Notice that
$Y_{e}^{i}\,,Y_{\omega}^{i}\,,Y_{\alpha}\in\Omega^{1}(\Sigma)$ and
$Y_{e_{\mathrm{t}}}^{i}\,,Y_{\omega_{\mathrm{t}}}^{i}\,,Y_{\alpha_{\mathrm{t}}}\in\mathcal{C}^{\infty}(\Sigma)$.

We can then write the action of the exterior differential in phase space
$\mathrm{d\\!l}$ on a vector field $\mathbb{Y}$ as

$\displaystyle\mathrm{d\\!l}H(\mathbb{Y})=\int_{\Sigma}\Big{(}$
$\displaystyle-Y_{e_{\mathrm{t}}}^{i}(\epsilon_{ijk}e^{j}\wedge
F^{k}+\alpha\wedge F_{i})$
$\displaystyle+Y_{e}^{i}\wedge\big{(}\alpha_{\mathrm{t}}F_{i}+\epsilon_{ijk}(D\omega_{\mathrm{t}}^{j})\wedge
e^{k}-D\omega_{\mathrm{t}}^{i}\wedge\alpha+\epsilon_{ijk}e_{\mathrm{t}}^{j}F^{k}\big{)}$
$\displaystyle+Y_{\omega_{\mathrm{t}}}^{i}D(\alpha\wedge
e_{i}-\frac{1}{2}\epsilon_{ijk}e^{j}\wedge e^{k})$ (3.3)
$\displaystyle+Y_{\omega}^{i}\wedge\big{(}D(\alpha_{\mathrm{t}}e_{i}+\epsilon_{ijk}e_{j}e_{\mathrm{t}}^{k}-\alpha
e_{\mathrm{t}}^{i})-\omega_{\mathrm{t}}^{j}e_{j}\wedge
e_{i}-\epsilon_{ijk}\omega_{\mathrm{t}}^{j}\alpha\wedge e^{k}\big{)}$
$\displaystyle+Y_{\alpha_{\mathrm{t}}}e_{i}\wedge F^{i}$
$\displaystyle+Y_{\alpha}\wedge\big{(}D\omega_{\mathrm{t}}^{i}\wedge
e_{i}-e_{\mathrm{t}}^{i}F_{i}\big{)}\Big{)}\,.$

The action of the canonical symplectic form $\Omega$ on
$\mathbb{Y},\mathbb{Z}\in\mathfrak{X}(T^{*}Q)$ can be written in terms of the
components of these vector fields as

$\displaystyle\Omega(\mathbb{Z},\mathbb{Y})$
$\displaystyle={\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{e_{\mathrm{t}}}}(Z_{e_{\mathrm{t}}})-{\bm{\mathrm{Z}}}_{\bm{\mathrm{p}}_{e_{\mathrm{t}}}}(Y_{e_{\mathrm{t}}})+{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{e}}(Z_{e})-{\bm{\mathrm{Z}}}_{\bm{\mathrm{p}}_{e}}(Y_{e})+{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\omega_{\mathrm{t}}}}(Z_{\omega_{\mathrm{t}}})-{\bm{\mathrm{Z}}}_{\bm{\mathrm{p}}_{\omega_{\mathrm{t}}}}(Y_{\omega_{\mathrm{t}}})$
$\displaystyle+{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\omega}}(Z_{\omega})-{\bm{\mathrm{Z}}}_{\bm{\mathrm{p}}_{\omega}}(Y_{\omega})+{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\alpha_{\mathrm{t}}}}(Z_{\alpha_{\mathrm{t}}})-{\bm{\mathrm{Z}}}_{\bm{\mathrm{p}}_{\alpha_{\mathrm{t}}}}(Y_{\alpha_{\mathrm{t}}})+{\bm{\mathrm{Y}}}_{\\!\\!\bm{\mathrm{p}}_{\alpha}}(Z_{\alpha})-{\bm{\mathrm{Z}}}_{\bm{\mathrm{p}}_{\alpha}}(Y_{\alpha})\,.$

The vector fields $\mathbb{Y}\in\mathfrak{X}(T^{*}Q)$ tangent to the primary
constraint submanifold $\mathsf{M}_{0}$ have the form

$\mathbb{Y}=(Y_{e_{\mathrm{t}}},Y_{e},Y_{\omega_{\mathrm{t}}},Y_{\omega},Y_{\alpha_{\mathrm{t}}},Y_{\alpha},{\bm{0}},{\bm{0}},{\bm{0}},{\bm{\mathrm{Y}}},{\bm{0}},{\bm{0}})$

with

${\bm{\mathrm{Y}}}(\cdot)=\int_{\Sigma}(\epsilon_{ijk}Y_{e}^{i}\wedge
e^{j}-\alpha\wedge Y_{e}^{k}-Y_{\alpha}\wedge e_{k})\wedge\cdot$

One of the central ideas in the GNH approach is to work directly on the
primary constraint submanifold $\mathsf{M}_{0}$. In the present case this is
especially easy as $\mathsf{M}_{0}$ is precisely the configuration space $Q$.
This means that vector fields on this manifold have the form

$\mathbb{Y}_{0}=(Y_{e_{\mathrm{t}}}^{i},Y_{e}^{i},Y_{\omega_{\mathrm{t}}}^{i},Y_{\omega}^{i},Y_{\alpha_{\mathrm{t}}},Y_{\alpha})\,.$

The pullback of $\Omega$ to $\mathsf{M}_{0}$—another of the basic elements of
the GNH approach—can be written as

$\displaystyle\omega(\mathbb{Z}_{0},\mathbb{Y}_{0})=\int_{\Sigma}\Big{(}Y_{e}^{i}\wedge(\epsilon_{ijk}e^{j}\wedge
Z_{\omega}^{k}+\alpha\wedge Z_{\omega i})$ (3.4) $\displaystyle\hskip
165.02606pt+Y_{\omega}^{i}\wedge(Z_{\alpha}\wedge
e_{i}-\epsilon_{ijk}Z_{e}^{j}\wedge
e^{k}-Z_{e}^{i}\wedge\alpha)+Y_{\alpha}\wedge Z_{\omega}^{i}\wedge
e_{i}\Big{)}\,.$

The basic equation of the GNH approach
$\imath_{\mathbb{X}_{0}}\omega=\mathrm{d\\!l}H$ can now be easily solved by
equating the terms proportional to the different components of the vector
field $\mathbb{Y}$ in (3.3) and (3.4). By doing this one finds the secondary
constraints

$\displaystyle\epsilon_{ijk}e^{j}\wedge F^{k}+\alpha\wedge F_{i}=0\,,$ (3.5)
$\displaystyle D(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge
e^{k}+e_{i}\wedge\alpha)=0\,,$ (3.6) $\displaystyle e_{i}\wedge F^{i}=0\,,$
(3.7)

and the following equations for the components of the Hamiltonian vector field
$\mathbb{Z}_{0}$

$\displaystyle(\epsilon_{ijk}e^{j}+\delta_{ik}\alpha)\wedge(Z_{\omega}^{k}-D\omega_{\mathrm{t}}^{k})=(\delta_{ik}\alpha_{\mathrm{t}}+\epsilon_{ijk}e_{\mathrm{t}}^{j})F^{k}\,,$
(3.8)
$\displaystyle(\epsilon_{ijk}e^{j}-\delta_{ik}\alpha)\wedge(Z_{e}^{k}-De_{\mathrm{t}}^{k}-\epsilon^{k}_{\phantom{k}\ell
m}e^{\ell}\omega_{\mathrm{t}}^{m})+e_{i}\wedge(Z_{\alpha}-\mathrm{d}\alpha_{\mathrm{t}})$
(3.9) $\displaystyle\hskip
227.62204pt=e_{\mathrm{t}}^{i}\mathrm{d}\alpha+(\epsilon_{ijk}e_{\mathrm{t}}^{j}-\alpha_{\mathrm{t}}\delta_{ik})De^{k}\,$
$\displaystyle
e_{i}\wedge(Z_{\omega}^{i}-D\omega_{\mathrm{t}}^{i})=e_{\mathrm{t}}^{i}F_{i}\,.$
(3.10)

There are no conditions on $Z^{i}_{e_{\mathrm{t}}}$,
$Z^{i}_{\omega_{\mathrm{t}}}$ and $Z_{\alpha_{\mathrm{t}}}$ so they are
arbitrary and, hence, the dynamics of $e_{\mathrm{t}}^{i}$,
$\omega_{\mathrm{t}}^{i}$ and $\alpha_{\mathrm{t}}$ is also arbitrary.

The next step in the GNH method is to check if the vector fields whose
components satisfy (3.8)-(3.10) are tangent to the submanifold of
$\mathsf{M}_{0}$ defined by the secondary constraints (3.5)-(3.7). Before
performing this analysis—which turns out to be rather intricate—it is
necessary to analyse in detail the secondary constraints and also to solve for
$Z_{e}^{i}$, $Z_{\omega}^{i}$ and $Z_{\alpha}$ in (3.8)-(3.10). We devote the
next two sections to this issue.

##  4 Some details on the secondary constraints

In order to study the tangency of the Hamiltonian vector fields given by
(3.8)-(3.10) it is necessary to disentangle part of the content of the
constraints (3.5) and (3.7). We do it here. First we point out that the
curvature $F_{i}$ can be written as

$F_{i}=\frac{1}{2}\mathbb{F}_{ij}\epsilon^{jk\ell}e_{k}\wedge e_{\ell}$

with

$\mathbb{F}_{ij}:=\left(\frac{F_{i}\wedge e_{j}}{\mathsf{vol}_{e}}\right)\,.$

Here $\mathsf{vol}_{e}:=\frac{1}{3!}\epsilon_{ijk}e^{i}\wedge e^{j}\wedge
e^{k}$ is a volume form on $\Sigma$ because the $e^{i}$ are linearly
independent everywhere.

As the $e^{i}$ are linearly independent we can write $\alpha=\alpha_{i}e^{i}$
and put the secondary constraint (3.5) in the form

$\epsilon_{ijk}e^{j}\wedge F^{k}+\alpha\wedge
F_{i}=(\mathbb{F}_{ij}\alpha^{j}-\epsilon_{ijk}\mathbb{F}^{jk})\mathsf{vol}_{e}=0\,,$

which is equivalent to

$\mathbb{F}_{ij}\alpha^{j}-\epsilon_{ijk}\mathbb{F}^{jk}=0\,.$ (4.1)

It is also straightforward to rewrite (3.7) in terms of $\mathbb{F}_{ij}$,
indeed

$e^{i}\wedge
F_{i}=\frac{1}{2}e_{i}\wedge\mathbb{F}^{i}_{\,\,j}\epsilon^{jk\ell}e_{k}\wedge
e_{\ell}=\frac{1}{2}\epsilon_{ik\ell}\epsilon^{jkl}\mathbb{F}^{i}_{\,\,j}\mathsf{vol}_{e}=\mathbb{F}^{i}_{\,\,i}\mathsf{vol}_{e}\,,$

and, hence, the secondary constraint (3.7) is equivalent to

$\mathbb{F}^{i}_{\,\,i}=0\,.$ (4.2)

If we expand now $\mathbb{F}_{ij}$ in terms of a symmetric-traceless part
$S_{ij}$, a trace $S$ and an antisymmetric part $\epsilon_{ijk}A_{k}$

$\mathbb{F}_{ij}=S_{ij}+\frac{1}{3}\delta_{ij}S+\epsilon_{ijk}A^{k}\,,$

the constraint (4.2) implies $S=0$ and (4.1) becomes

$(2\delta_{ij}+\epsilon_{ijk}\alpha^{k})A^{j}=S_{ij}\alpha^{j}\,,$

which can be solved for $A_{i}$ in terms of the symmetric-traceless object
$S_{ij}$

$A_{i}=\frac{1}{2(4+\alpha^{2})}(4\delta_{ij}+\alpha_{i}\alpha_{j}-2\epsilon_{ijk}\alpha^{k})S^{j}_{\,\,\ell}\alpha^{\ell}\,,$

where $\alpha^{2}:=\alpha_{i}\alpha^{i}$. We then conclude that, when the
constraints hold

$\mathbb{F}_{ij}=S_{ij}+\frac{1}{2(4+\alpha^{2})}\epsilon_{ijk}(4\delta^{k}_{\,\,\ell}+\alpha^{k}\alpha_{\ell}-2\epsilon^{k}_{\,\,\ell
m}\alpha^{m})S^{\ell}_{\,\,p}\alpha^{p}\,,$ (4.3)

with $S_{ij}$ symmetric and traceless. Notice that this is not the full
solution to the constraints (3.5)-(3.7) but, rather, a restriction on the form
of the curvature $F_{i}$. In order to fully solve the constraits one should
find the connection $A_{i}$ from (4.3).

##  5 The Hamiltonian vector fields

In this section we will solve equations (3.8)-(3.10) for $Z_{e}^{i}$,
$Z_{\omega}^{i}$ and $Z_{\alpha}$. For this purpose it helps to define

$\displaystyle X_{e}^{k}$
$\displaystyle:=Z_{e}^{k}-De_{\mathrm{t}}^{k}-\epsilon^{k}_{\phantom{k}\ell
m}e^{\ell}\omega_{\mathrm{t}}^{m}\,,$ $\displaystyle X_{\omega}^{k}$
$\displaystyle:=Z_{\omega}^{k}-D\omega_{\mathrm{t}}^{k}\,,$ $\displaystyle
X_{\alpha}$ $\displaystyle:=Z_{\alpha}-\mathrm{d}\alpha_{\mathrm{t}}\,.$

so that (3.8)-(3.10) become

$\displaystyle(\epsilon_{ijk}e^{j}+\delta_{ik}\alpha)\wedge
X_{\omega}^{k}=(\delta_{ik}\alpha_{\mathrm{t}}+\epsilon_{ijk}e_{\mathrm{t}}^{j})F^{k}\,,$
(5.1) $\displaystyle(\epsilon_{ijk}e^{j}-\delta_{ik}\alpha)\wedge
X_{e}^{k}+e_{i}\wedge
X_{\alpha}=e_{\mathrm{t}i}\mathrm{d}\alpha+(\epsilon_{ijk}e_{\mathrm{t}}^{j}-\alpha_{\mathrm{t}}\delta_{ik})De^{k}\,$
(5.2) $\displaystyle e_{i}\wedge X_{\omega}^{i}=e_{\mathrm{t}}^{i}F_{i}\,.$
(5.3)

We solve now these equations taking into account that the secondary
constraints (3.8)-(3.10) hold.

###  5.1 Solving equation 5.1

By expanding the 1-forms $X_{\omega}^{k}=W^{k}_{\phantom{k}q}\,e^{q}$ and
$\alpha=\alpha_{i}e^{i}$ we can write (5.1) as the following linear,
inhomogeneous equation for $W_{ij}$

$W_{ij}-\delta_{ij}W^{k}_{\phantom{k}k}-\epsilon_{i}^{\phantom{i}pq}\alpha_{p}W_{jq}+\alpha_{\mathrm{t}}\mathbb{F}_{ji}+\epsilon_{j}^{\phantom{i}pq}e_{\mathrm{t}p}\mathbb{F}_{qi}=0\,.$
(5.4)

We expand now $W_{ij}$ in irreducible components as

$W_{ij}=w_{ij}+\frac{1}{3}\delta_{ij}w+\epsilon_{ijk}w^{k}\,,$ (5.5)

with $w_{ij}$ symmetric and traceless. From (5.5) we see that

$w^{i}=\frac{1}{2}\epsilon^{ijk}W_{jk}\,,\quad w=W^{i}_{\phantom{i}i}\,.$
(5.6)

Multiplying (5.4) by $\epsilon^{kij}$ and $\alpha^{i}$ respectively we get

$\displaystyle\epsilon^{ijk}W_{jk}-\alpha_{j}W^{ji}+\alpha^{i}W^{j}_{\phantom{j}j}-\alpha_{\mathrm{t}}\epsilon^{ijk}\mathbb{F}_{jk}-\mathbb{F}^{i}_{\phantom{i}j}e_{\mathrm{t}}^{j}=0$
$\displaystyle\alpha_{j}W^{ji}-\alpha^{i}W^{j}_{\phantom{j}j}+\alpha_{\mathrm{t}}\mathbb{F}_{ij}\alpha^{j}+\epsilon_{ijk}e_{\mathrm{t}}^{j}\mathbb{F}^{k\ell}\alpha_{\ell}=0\,.$

Adding both expressions, taking into account (5.6) and using the constraints
in the form (4.1)-(4.2) we find

$w^{i}=\frac{1}{2}e_{\mathrm{t}j}\mathbb{F}^{ji}\,.$

In order to find $w$ we take the trace of (5.4), which gives

$-2W^{i}_{\phantom{i}i}+\epsilon^{ijk}\alpha_{i}W_{jk}+\epsilon^{ijk}e_{\mathrm{t}i}\mathbb{F}_{jk}=0\,,$

after making use of the constraint $\mathbb{F}^{i}_{\phantom{i}i}=0$. From the
previous expression, the value of $w^{i}$ computed above and the constraint
(4.1) we get

$w=\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{F}^{jk}\,.$

A crucial simplification of equation (5.4) can be achieved by using
$W^{i}_{\phantom{n}i}=w$, noting that $W_{jq}=W_{qj}-2\epsilon_{qjk}w^{k}$ and
introducing this expression in the third term of (5.4). By doing this we find
that (5.4) can be rewritten as

$(\delta_{iq}+\epsilon_{iqp}\alpha^{p})W^{q}_{\phantom{q}j}+\alpha_{\mathrm{t}}\mathbb{F}_{ji}-\alpha_{j}e_{\mathrm{t}k}\mathbb{F}^{k}_{\phantom{k}i}+\epsilon_{jpq}e_{\mathrm{t}}^{p}\mathbb{F}^{q}_{\phantom{q}i}=0\,,$
(5.7)

which can be easily solved by inverting the $2\times 2$ matrix
$\delta_{iq}+\epsilon_{iqp}\alpha^{p}$ as the index $j$ in
$W^{q}_{\phantom{q}j}$ is a mere spectator. The result is

$W_{ij}=\frac{1}{1+\alpha^{2}}\big{(}\delta_{i}^{\phantom{n}k}+\alpha_{i}\alpha^{k}-\epsilon_{i}^{\phantom{n}k\ell}\alpha_{\ell}\big{)}M_{kj}\,,$
(5.8)

with

$M_{kj}:=\alpha_{j}e_{\mathrm{t}q}\mathbb{F}^{q}_{\phantom{i}k}-\epsilon_{jpq}e_{\mathrm{t}}^{p}\mathbb{F}^{q}_{\phantom{i}k}-\alpha_{\mathrm{t}}\mathbb{F}_{jk}\,.$
(5.9)

###  5.2 Solving equation 5.2

By following exactly the same steps as in the previous subsection it is
possible to solve equation (5.2). In this case we expand
$X_{e}^{k}=E^{k}_{\phantom{k}q}\,e^{q}$ and $X_{\alpha}=X_{i}e^{i}$. We also
take into account that the constraint (3.6) is equivalent to

$\mathrm{d}\alpha\wedge e_{i}=\alpha\wedge De_{i}-\epsilon_{ijk}e^{j}\wedge
De^{k}\,.$

This means that on the constraint submanifold defined by the secondary
constraints we have

$\mathrm{d}\alpha=\frac{1}{2}\epsilon_{ijk}\left(\frac{\mathrm{d}\alpha\wedge
e_{i}}{\mathsf{vol}_{e}}\right)e^{j}\wedge
e^{k}=\left(\frac{1}{2}\mathbb{B}^{i}_{\phantom{i}\ell}\alpha^{\ell}\epsilon_{ijk}+\mathbb{B}_{jk}\right)e^{j}\wedge
e^{k}\,,$ (5.10)

where $\mathbb{B}_{ij}:=\left(\frac{De_{i}\wedge
e_{j}}{\mathsf{vol}_{e}}\right)$.

Taking all this into account, on the constraint hypersurface equation (5.2)
can be written as

$E_{ij}-\delta_{ij}E^{k}_{\phantom{k}k}+\epsilon_{i}^{\phantom{i}pq}\alpha_{p}E_{jq}-\epsilon_{ijk}X^{k}-\alpha_{\mathrm{t}}\mathbb{B}_{ji}+\epsilon_{jk\ell}e_{\mathrm{t}}^{k}\mathbb{B}^{\ell}_{\phantom{i}i}+\mathbb{B}_{ik}\alpha^{k}e_{\mathrm{t}j}+\epsilon_{i}^{\phantom{i}pq}\mathbb{B}_{pq}e_{\mathrm{t}j}=0\,,$
(5.11)

where the $X_{i}$ can be taken to be completely arbitrary.

The resolution of this equation follows the same steps as that of (5.4) so we
just quote the result here:

$E_{ij}=\frac{1}{1+\alpha^{2}}\big{(}\delta_{i}^{\phantom{n}k}+\alpha_{i}\alpha^{k}+\epsilon_{i}^{\phantom{n}k\ell}\alpha_{\ell}\big{)}N_{kj}\,,$
(5.12)

with

$\displaystyle N_{ij}$ $\displaystyle:=$ $\displaystyle\delta_{ij}\Big{(}$
$\displaystyle\epsilon_{k\ell m}e_{\mathrm{t}}^{k}\mathbb{B}^{\ell
m}+\epsilon_{k\ell
m}\alpha^{k}e_{\mathrm{t}}^{\ell}\mathbb{B}^{mn}\alpha_{n}+\frac{1}{2}(e_{\mathrm{t}}\cdot\alpha)(\alpha\mathbb{B}\alpha)+\frac{1}{2}(e_{\mathrm{t}}\cdot\alpha)\epsilon_{k\ell
m}\alpha^{k}\mathbb{B}^{\ell m}$
$\displaystyle-\frac{1}{2}(e_{\mathrm{t}}\cdot\alpha)\mathbb{B}+(\alpha\mathbb{B}e_{\mathrm{t}})+\alpha\cdot
X-\frac{1}{2}\alpha_{\mathrm{t}}(\alpha\mathbb{B}\alpha)-\frac{1}{2}\alpha_{\mathrm{t}}\epsilon_{k\ell
m}\alpha^{k}\mathbb{B}^{\ell
m}-\frac{1}{2}\alpha_{\mathrm{t}}\mathbb{B}\Big{)}$
$\displaystyle+\alpha_{j}\Big{(}\mathbb{B}e_{\mathrm{t}i}-2\epsilon_{ipq}e_{\mathrm{t}}^{p}\mathbb{B}^{q}_{\phantom{q}r}\alpha^{r}-e_{\mathrm{t}i}(\alpha\mathbb{B}\alpha)-e_{\mathrm{t}i}\epsilon_{k\ell
m}\alpha^{k}\mathbb{B}^{\ell
m}-2\mathbb{B}_{ik}e_{\mathrm{t}}^{k}+e_{\mathrm{t}k}\mathbb{B}^{k}_{\phantom{i}i}-2X_{i}$
$\displaystyle\hskip 162.18062pt-\epsilon_{i\ell
m}\alpha^{\ell}X^{m}+\alpha_{\mathrm{t}}\mathbb{B}_{i\ell}\alpha^{\ell}+\alpha_{\mathrm{t}}\epsilon_{i\ell
m}\mathbb{B}^{\ell m}\Big{)}$
$\displaystyle-\epsilon_{jpq}e_{\mathrm{t}}^{p}\mathbb{B}^{q}_{\phantom{i}i}-\mathbb{B}_{i\ell}\alpha^{\ell}e_{\mathrm{t}j}-\epsilon_{ipq}\mathbb{B}^{pq}e_{\mathrm{t}j}+\epsilon_{ijk}X^{k}+\alpha_{\mathrm{t}}\mathbb{B}_{ji}\,,$
(5.13)

where the following shorthand notation has been used:

$e_{\mathrm{t}}\cdot\alpha:=e_{\mathrm{t}i}\alpha^{i}\,,\,\,\,\alpha\mathbb{B}\alpha:=\alpha_{i}\mathbb{B}^{ij}\alpha_{j}\,,\,\,\,\mathbb{B}:=\mathbb{B}^{i}_{\phantom{n}i}\,,\,\,\,\alpha\mathbb{B}e_{\mathrm{t}}:=\alpha_{i}\mathbb{B}^{ij}e_{\mathrm{t}j}\,,\,\,\,\alpha\cdot
X:=\alpha_{i}X^{i}\,.$

###  5.3 Solving equation 5.3

Equation (3.10) can be easily written in the form
$\epsilon^{ijk}W_{jk}=e_{\mathrm{t}j}\mathbb{F}^{ji}$. It is obvious from the
discussion presented in subsection 5.1 (and straightforward to check) that
$W_{ij}$, as given in (5.8), satisfies (3.10).

##  6 Tangency analysis

The consistency of the Hamiltonian dynamics requires that the Hamiltonian
vector fields obtained by solving (5.1)-(5.3) must be tangent to the
submanifold of $\mathsf{M}_{0}$ defined by the secondary constraints
(3.5)-(3.7). These tangency conditions can be easily obtained by computing the
derivatives of the functions defining the constraints along the field (i.e.
$\imath_{\mathbb{X}}\mathrm{d\\!l}$). By doing this one gets

$\displaystyle(\epsilon_{ijk}Z_{e}^{j}+\delta_{ik}Z_{\alpha})\wedge
F^{k}+(\epsilon_{ijk}e^{j}+\delta_{ik}\alpha)\wedge DZ_{\omega}^{k}=0\,,$
(6.1) $\displaystyle D(\epsilon_{ijk}e^{j}\wedge Z_{e}^{k}-Z_{\alpha}\wedge
e_{i}-\alpha\wedge Z_{ei})+Z_{\omega}^{k}\wedge(e^{i}\wedge
e_{k}-\epsilon^{i}_{\phantom{i}km}\alpha\wedge e^{m})=0\,,$ (6.2)
$\displaystyle Z_{e}^{i}\wedge F_{i}+e_{i}\wedge DZ_{\omega}^{i}=0\,.$ (6.3)

Before checking if the Hamiltonian vector fields given by (5.8) and (5.12)
satisfy the conditions (6.1)-(6.3) it is convenient to simplify them, in
particular by removing the covariant differential $D$ of the components of the
vector field $\mathbb{Z}$. As we show next (6.1)-(6.3) can be written in the
more convenient form

$\displaystyle\epsilon_{ijk}X_{e}^{j}\wedge F^{k}+X_{\alpha}\wedge
F_{i}+D(\epsilon_{ijk}e^{j}+\delta_{ik}\alpha)\wedge X_{\omega}^{k}=0\,,$
(6.4) $\displaystyle(\delta_{ij}e_{k}-\epsilon_{ijk}\alpha)\wedge e^{j}\wedge
X_{\omega}^{k}+(\alpha_{\mathrm{t}}\delta_{ik}+\epsilon_{ijk}e_{\mathrm{t}j})\alpha\wedge
F^{k}+e_{\mathrm{t}}^{j}e_{i}\wedge F_{j}=0\,,$ (6.5) $\displaystyle
X_{e}^{i}\wedge F_{i}+X_{\omega}^{i}\wedge De_{i}=0\,.\phantom{\big{[}}$ (6.6)

In the case of (6.1) one first takes the covariant exterior differential $D$
of (3.8) to get

$\displaystyle(\epsilon_{ijk}e_{j}+\delta_{ik}\alpha)\wedge DZ_{\omega}^{k}=$
$\displaystyle(\epsilon_{ijk}De^{j}+\delta_{ik}\mathrm{d}\alpha)\wedge(Z_{\omega}^{k}-D\omega_{\mathrm{t}}^{k})$
$\displaystyle+D\omega_{\mathrm{t}m}\epsilon^{mk\ell}(\epsilon_{ijk}e^{j}+\delta_{ik}\alpha)\wedge
F_{\ell}-(\delta_{ik}\mathrm{d}\alpha_{\mathrm{t}}+\epsilon_{ijk}De_{\mathrm{t}}^{j})\wedge
F^{k}\,,$

then introduces this expression into (6.1) and simplifies the result by using
the constraints (3.5) and (3.7) to get (6.4).

In order to rewrite the tangency condition (6.2) in a simpler way one first
writes (3.9) in the form

$\epsilon_{ijk}e^{j}\wedge Z_{e}^{k}-Z_{\alpha}\wedge e_{i}-\alpha\wedge
Z_{ei}=D(\epsilon_{ijk}e_{\mathrm{t}}^{j}e^{k}+\alpha
e_{\mathrm{t}}^{i}-\alpha_{\mathrm{t}}e^{i})+\omega_{\mathrm{t}}^{k}e_{k}\wedge
e_{i}-\epsilon^{i}_{\phantom{i}jk}\alpha_{\ell}\omega_{\mathrm{t}}^{k}e^{\ell}\wedge
e^{j}\,,$

and then takes the covariant differential of this expression which, by using
the constraints (3.6,3.7), gives

$D(\epsilon_{ijk}e^{j}\wedge Z_{e}^{k}-Z_{\alpha}\wedge e_{i}-\alpha\wedge
Z_{ei})=\alpha_{\mathrm{t}}\epsilon_{ijk}e^{j}\wedge
F^{k}-e_{\mathrm{t}}^{j}(\epsilon_{ijk}\alpha+\delta_{jk}e_{i})\wedge
F^{k}+(e_{k}\delta_{ij}-\epsilon_{ijk}\alpha)\wedge e^{j}\wedge
D\omega_{\mathrm{t}}^{k}\,.$

Introducing this now into (6.2) gives (6.5).

Finally, in order to arrive at (6.6) one first takes the exterior covariant
differential of (3.10) to get

$e_{i}\wedge DZ_{\omega}^{i}=De_{i}\wedge Z_{\omega}^{i}-De_{i}\wedge
D\omega_{\mathrm{t}}^{i}+\epsilon_{ijk}e^{i}\wedge
F^{j}\omega_{\mathrm{t}}^{k}-De_{\mathrm{t}}^{i}\wedge F_{i}\,,$

and then introduces this expression into (6.3).

In the remaining of this section we will check that the tangency conditions
(6.4)-(6.6) hold on the submanifold of $\mathsf{M}_{0}$ defined by the
secondary constraints (3.5)-(3.7). We will start from the easiest to the
hardest ones. As we will see, the computations are quite involved. Although,
by necessity, we will have to skip many details we will provide enough
information to enable the motivated readers to complete them.

One source of difficulties is the possibility of having many different ways to
write a particular expression by making use of the constraints (3.5)-(3.7).
One possible way to avoid this problem is to use (4.3) and write everything in
terms of $S_{ij}$. Although this is possible in principle, in practice the
computations are very long. A better strategy—that ultimately works—is to use
the fact that the constraints imply
$\mathbb{F}_{ij}=S_{ij}+\epsilon_{ijk}A^{k}$, with $S_{ij}$ symmetric and
traceless, and use
$S_{ij}\alpha^{j}=(2\delta_{ij}+\epsilon_{ijk}\alpha^{k})A^{j}$ [equivalent to
(4.1)], whenever possible, to write everything in terms of $A^{k}$.

###  6.1 Checking condition 6.5

The only tangency condition which is easy to check is (6.5). In order to see
that it holds it suffices to left-wedge-multiply (5.1)—the equation that must
be solved to obtain $X_{\omega}^{i}$—by $\alpha$ to get

$\alpha\wedge(\epsilon_{ijk}e^{j}+\delta_{ik}\alpha)\wedge
X_{\omega}^{k}=(\delta_{ik}\alpha_{\mathrm{t}}+\epsilon_{ijk}e_{\mathrm{t}}^{j})\alpha\wedge
F^{k}\,.$

Plugging this into (6.5) leads to

$e_{k}\wedge e_{i}\wedge X_{\omega}^{k}+e_{\mathrm{t}}^{j}e_{i}\wedge
F_{j}=0\,.$

which can be immediately seen to hold as a consequence of (5.3).

###  6.2 Checking condition 6.6

In order to check the tangency requirement expressed by (6.6) we will first
rewrite it as an equivalent condition in terms of the $M_{ij}$ and $N_{ij}$
introduced above [see (5.9) and (5.13)]:

$\mathbb{B}^{ij}M_{ij}+(\alpha^{i}\mathbb{B}_{ij})(\alpha_{k}M^{kj})+\epsilon^{ijk}\alpha_{i}M_{j\ell}\mathbb{B}_{k}^{\phantom{k}\ell}+\mathbb{F}^{ij}N_{ij}+(\alpha^{i}\mathbb{F}_{ij})(\alpha_{k}N^{kj})-\epsilon^{ijk}\alpha_{i}N_{j\ell}\mathbb{F}_{k}^{\phantom{k}\ell}=0\,.$

In principle one just has to substitute (5.9) and (5.13) in the previous
expression and show that the result is zero. A possible way to do this is to
use tensor manipulating packages such as xAct [19]. However, in our opinion it
is instructive to do the computation by hand as some important simplifications
are quite non-trivial. To this end it is helpful to separately consider the
terms depending on $X_{i}$, $\alpha_{\mathrm{t}}$ and $e_{\mathrm{t}}^{i}$. In
the first two instances ($X_{i}$ and $\alpha_{\mathrm{t}}$) the computations
are quite direct and the cancelations of the different terms obvious. The only
hints worth mentioning here are

  * •

Use $\mathbb{F}_{ij}=S_{ij}+\epsilon_{ijk}A^{k}$, taking into account,
whenever necessary, that $S_{i}^{\phantom{i}i}=0$.

  * •

Use the constraints in the form (4.1) to write
$\mathbb{F}^{ij}\alpha_{j}=2A^{i}$. Also use
$S_{ij}\alpha^{j}=2A_{i}+\epsilon_{ijk}A^{j}\alpha^{k}$ whenever the
combination $S_{ij}\alpha^{j}$ appears.

  * •

Replace $\mathbb{B}_{ij}-\mathbb{B}_{ji}$ by $\epsilon_{ijk}\epsilon^{k\ell
m}\mathbb{B}_{\ell m}$ whenever possible.

By doing this all the terms involving $X_{i}$, and those proportional to
$\alpha_{\mathrm{t}}$ cancel.

The computation of the terms proportional to $e_{\mathrm{t}}^{i}$ is
significantly harder, so we will give more details about it. In this case the
terms coming from the direct substitution of (5.9) and (5.13) into the
tangency condition do not cancel automatically in an obvious way. In fact, the
result is

$\displaystyle-\epsilon_{ijk}e_{\mathrm{t}}^{i}S^{j}_{\phantom{i}\ell}\mathbb{B}^{\ell
k}-\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{j\ell}S_{\ell}^{\phantom{i}k}-\epsilon_{ijk}\mathbb{B}^{ij}S^{k}_{\phantom{k}\ell}e_{\mathrm{t}}^{\ell}\hskip
227.62204pt$

$\displaystyle-4\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}\alpha_{\ell})A^{k}-3(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+3\epsilon_{ijk}\alpha^{i}A^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})+5\epsilon_{ijk}\alpha^{i}(\mathbb{B}^{j\ell}e_{\mathrm{t}\ell})A^{k}$
$\displaystyle-2\mathbb{B}\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}A^{k}\\!\\!+\\!2\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}A_{\ell})\\!+\\!(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\\!\\!+\\!2(A\cdot\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{jk}\\!\\!-\\!2\epsilon_{ijk}(\alpha_{\ell}\mathbb{B}^{\ell
i})e_{\mathrm{t}}^{j}A^{k}$

$\displaystyle+2(\alpha\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}+2\alpha^{2}\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})-2(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}(\mathbb{B}^{i\ell}\alpha_{\ell})A^{j}\alpha^{k}\hskip
91.04872pt$
$\displaystyle-2(A\cdot\alpha)\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})\hskip
91.04872pt$

Several features of the previous expression stand out, in particular, all the
terms involve $\epsilon_{ijk}$, some of them depend on $S_{ij}$, and the
remaining ones are either linear or cubic in $\alpha_{i}$.

In order to show that the $S$-dependent terms cancel out it suffices to
realize that $0=\epsilon_{[ijk}S^{j}_{\phantom{i}\ell]}$ implies
$\epsilon_{ijk}S^{j}_{\phantom{i}\ell}-\epsilon_{jk\ell}S^{j}_{\phantom{i}i}-\epsilon_{\ell
ij}S^{j}_{\phantom{i}k}=0$, where we have used $S^{i}_{\phantom{i}i}=0$.

The strategy to show that the other terms cancel is similar to this but more
involved so we leave some of the details for Appendix A.

###  6.3 Checking condition 6.4

As in the previous subsection we will write the tangency condition (6.4) as an
equivalent expression in terms of the $M_{ij}$ and $N_{ij}$. To avoid having
“dangling indices” we will multiply it by an arbitrary object $C_{i}$ (which
can be removed at the end). By doing this we get

$\displaystyle\phantom{-\,\,}\epsilon^{ijk}C_{i}\mathbb{B}_{j}^{\phantom{j}\ell}M_{k\ell}+\epsilon^{ijk}\alpha_{i}C_{j}\mathbb{B}_{k}^{\phantom{k}\ell}(\alpha^{m}M_{m\ell})-(\alpha_{i}\mathbb{B}^{ij})(C^{k}M_{kj})+(C\cdot\alpha)\mathbb{\mathbb{B}}^{ij}M_{ij}$
$\displaystyle-\epsilon^{ijk}C_{i}\mathbb{F}_{j}^{\phantom{j}\ell}N_{k\ell}\,-\,\epsilon^{ijk}\alpha_{i}C_{j}\mathbb{F}_{k}^{\phantom{k}\ell}(\alpha^{m}N_{m\ell})\,-\,(\alpha_{i}\mathbb{F}^{ij})(C^{k}N_{kj})\,+(C\cdot\alpha)\mathbb{\mathbb{F}}^{ij}N_{ij}$
$\displaystyle+(C^{i}M_{ij})(\mathbb{B}^{jk}\alpha_{k})+(C\cdot\alpha)(\alpha^{i}M_{ij})(\mathbb{B}^{jk}\alpha_{k})-\epsilon^{ijk}\alpha_{i}C_{j}M_{k\ell}(\mathbb{B}^{\ell
m}\alpha_{m})$
$\displaystyle+(C^{i}M_{ij})\epsilon^{jk\ell}\mathbb{B}_{k\ell}+(C\cdot\alpha)(\alpha^{i}M_{ij})\epsilon^{jk\ell}\mathbb{B}_{k\ell}-\epsilon^{ijk}\alpha_{i}C_{j}M_{k\ell}\epsilon^{\ell
mn}\mathbb{B}_{mn}$ $\displaystyle+C_{i}\mathbb{F}^{ij}X_{j}=0\,,$ (6.7)

where we have used the shorthand $C\cdot\alpha=C_{i}\alpha^{i}$. As above, it
is helpful to separately consider the terms depending on $X_{i}$,
$\alpha_{\mathrm{t}}$ and $e_{\mathrm{t}}^{i}$.

The computation showing that the terms proportional to $X_{i}$ in (6.7) cancel
is straightforward. The only hint worth mentioning here is to make use of the
identity

$\alpha^{\ell}\alpha_{\ell}\epsilon_{ijk}C^{i}A^{j}X^{k}=(\alpha\cdot
C)\epsilon_{ijk}A^{i}X^{j}\alpha^{k}-(\alpha\cdot
A)\epsilon_{ijk}X^{i}\alpha^{j}C^{k}+(\alpha\cdot
X)\epsilon_{ijk}C^{i}A^{j}\alpha^{k}\,,$

which can be derived from $\alpha_{[\ell}\epsilon_{ijk]}=0$.

The terms proportional to $\alpha_{\mathrm{t}}$ in (6.7) can be written as

$\displaystyle-\epsilon^{ijk}\eta_{i}\mathbb{B}_{j\ell}S^{\ell}_{\phantom{\ell}k}+\epsilon^{ijk}\eta_{i}\mathbb{B}_{\ell
j}S^{\ell}_{\phantom{\ell}k}-\epsilon^{ijk}(S_{i}^{\phantom{i}\ell}\eta_{\ell})\mathbb{B}_{jk}\hskip
162.18062pt$

$\displaystyle-2\epsilon^{ijk}\alpha_{i}\eta_{j}(\mathbb{B}_{k\ell}A^{\ell})+2\epsilon^{ijk}\eta_{i}A_{j}(\alpha^{\ell}\mathbb{B}_{\ell
k})+2(\eta\cdot
A)\epsilon^{ijk}\alpha_{i}\mathbb{B}_{jk}+4\epsilon^{ijk}\eta_{i}(\mathbb{B}_{j}^{\phantom{j}\ell}\alpha_{\ell})A_{k}$
$\displaystyle+2\mathbb{B}\epsilon^{ijk}\eta_{i}A_{j}\alpha_{k}-2(A\cdot\alpha)\epsilon^{ijk}\eta_{i}\mathbb{B}_{jk}-4\epsilon^{ijk}(\mathbb{B}_{i\ell}\eta^{\ell})A_{j}\alpha_{k}+2\epsilon^{ijk}(\eta^{\ell}\mathbb{B}_{\ell
i})A_{j}\alpha_{k}\,,$

where we have introduced the shorthand $\eta_{i}:=\alpha_{\mathrm{t}}C_{i}$.

In order to see that the $S$-dependent terms cancel out we make use of
$\epsilon^{[ijk}S_{i}^{\phantom{i}\ell]}=0$. Checking that the terms linear in
$\alpha_{i}$ in the previous expression cancel out is a little bit more
involved, so we leave some of the details for Appendix A.

Finally, we discuss the computation of the terms proportional to
$e_{\mathrm{t}}^{i}$ in (6.7). This is, by far, the longest computation
necessary to complete the Hamiltonian analysis of the Euclidean self-dual
action. The first step is to introduce the $e_{\mathrm{t}}^{i}$-proportional
terms of $M_{ij}$ and $N_{ij}$ into (6.7) and simplify the result by using, in
particular, $S_{ij}\alpha^{j}=2A_{i}+\epsilon_{ijk}A^{j}\alpha^{k}$. After
doing this some terms involving $S_{ij}$ still remain. This is somewhat
disturbing because this could force us to write everything in terms of
$S_{ij}$ (rather than $A_{i}$) which would make the computations much longer.
Fortunately, by replacing $\mathbb{B}_{ij}-\mathbb{B}_{ji}$ by
$\epsilon_{ijk}\epsilon^{k\ell m}\mathbb{B}_{\ell m}$, whenever possible, it
is possible to show that all the $S$-dependent terms can be grouped in the
expression

$(CS\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{jk}+(e_{\mathrm{t}}S\alpha)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}-(e_{\mathrm{t}}\cdot
C)S^{i\ell}\alpha_{\ell}\epsilon_{ijk}\mathbb{B}^{jk}\,,$ (6.8)

where we have used the shorthand notation $CS\alpha:=C^{i}S_{ij}\alpha^{j}$,
$e_{\mathrm{t}}S\alpha:=e_{\mathrm{t}}^{i}S_{ij}\alpha^{j}$ and
$e_{\mathrm{t}}\cdot C:=e_{\mathrm{t}}^{i}C_{i}$. As we see, the combination
$S_{ij}\alpha^{j}$ appears in all the terms of (6.8) so we can write it in
terms of $A_{i}$ as explained above. The result is

$\displaystyle\,\,2(C\cdot
A)\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{jk}+(C\cdot
e_{\mathrm{t}})(A\mathbb{B}\alpha)+(\alpha\cdot
e_{\mathrm{t}})(C\mathbb{B}A)+(e_{\mathrm{t}}\cdot
A)(\alpha\mathbb{B}C)-(e_{\mathrm{t}}\cdot A)(C\mathbb{B}\alpha)$
$\displaystyle-(C\cdot e_{\mathrm{t}})(\alpha\mathbb{B}A)-(\alpha\cdot
e_{\mathrm{t}})(A\mathbb{B}C)+2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}+(\alpha\cdot
C)(e_{\mathrm{t}}\mathbb{B}A)+(A\cdot C)(\alpha\mathbb{B}e_{\mathrm{t}})$
$\displaystyle-(C\cdot A)(e_{\mathrm{t}}\mathbb{B}\alpha)-(\alpha\cdot
C)(A\mathbb{B}e_{\mathrm{t}})-2(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\,,$

where, as in previous instances, we have used some self-explanatory notation.
By adding this to the rest of the terms (the ones that can be directly written
in terms of $A_{i}$) the final result for the terms proportional to
$e_{\mathrm{t}}^{i}$ in (6.7) is

$\displaystyle\,\,3(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}-2\epsilon_{ijk}C^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}A_{\ell})-3(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\hskip 45.5244pt$
$\displaystyle+2\mathbb{B}\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}C^{k}+3\epsilon_{ijk}A^{i}C^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})+2\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}-5\epsilon_{ijk}(\mathbb{B}^{i\ell}e_{\mathrm{t}\ell})A^{j}C^{k}$

$\displaystyle+2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}C^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+2(\alpha\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}C^{j}A^{k}-3(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})$
$\displaystyle+2(C\cdot
e_{\mathrm{t}})\epsilon_{ijk}(\mathbb{B}^{i\ell}\alpha_{\ell})A^{j}\alpha^{k}-(A\cdot
C)(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}$
$\displaystyle+\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}(\alpha\mathbb{B}C)+2(A\cdot\alpha)\epsilon_{ijk}C^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+(A\cdot\alpha)(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}$
$\displaystyle+(C\cdot\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}(\alpha_{\ell}\mathbb{B}^{\ell
j})A^{k}+\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}-\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(C_{\ell}\mathbb{B}^{\ell
j})A^{k}$
$\displaystyle+2(e_{\mathrm{t}}\mathbb{B}\alpha)\epsilon_{ijk}\alpha^{i}A^{j}C^{k}\,,$

where it is worth pointing out that all the linear, cubic and quartic terms in
$\alpha_{i}$ are zero as a consequence of direct cancellations. This last
expression can be shown to vanish. As in previous instances we leave some
details about how this happens for Appendix A.

##  7 Hamiltonian formulation: Ahstekar variables and the time gauge

The Hamiltonian formulation obtained after completing the GNH procedure is
formulated in a manifold $\mathsf{M}_{0}$ spanned by the fields
$(e_{\mathrm{t}},e^{i},\omega_{\mathrm{t}},\omega^{i},\alpha_{\mathrm{t}},\alpha)$.
The vector fields in this manifold have components
$\mathbb{Y}_{0}=(Y_{e_{\mathrm{t}}}^{i},Y_{e}^{i},Y_{\omega_{\mathrm{t}}}^{i},Y_{\omega}^{i},Y_{\alpha_{\mathrm{t}}},Y_{\alpha})$.

By using the standard notation for 2-forms in field spaces the presymplectic
2-form on $\mathsf{M}_{0}$ can be written as

$\omega=\int_{\Sigma}\mathrm{d\\!l}\omega^{i}{\wedge\\!\\!\\!\wedge}\,\mathrm{d\\!l}\left(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge
e^{k}+e_{i}\wedge\alpha\right)\,,$ (7.1)

which acting on vector fields $\mathbb{Y}\,,\mathbb{Z}$ in $\mathsf{M}_{0}$
gives (3.4). The secondary constraints can be written as

$\displaystyle\epsilon_{ijk}e^{j}\wedge F^{k}+\alpha\wedge F_{i}=0\,,$
$\displaystyle D\left(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge
e^{k}+e_{i}\wedge\alpha\right)=0\,,$ $\displaystyle e_{i}\wedge F^{i}=0\,,$

and the Hamiltonian vector fields are

$\displaystyle
Z_{e}^{i}=De_{\mathrm{t}}^{i}-\epsilon^{ijk}\omega_{\mathrm{t}j}e_{k}+E^{i}_{\phantom{i}j}e^{j}\,,$
$\displaystyle
Z_{\omega}^{i}=D\omega_{\mathrm{t}}^{i}+W^{i}_{\phantom{i}j}e^{j}\,,$
$\displaystyle Z_{\alpha}\quad\,\mathsf{arbitrary}\,,$ $\displaystyle
Z_{e_{\mathrm{t}}}^{i}\quad\mathsf{arbitrary}\,,$ $\displaystyle
Z_{\omega_{\mathrm{t}}}^{i}\quad\mathsf{arbitrary}\,,$ $\displaystyle
Z_{\alpha_{\mathrm{t}}}\quad\mathsf{arbitrary}\,,$

with $W_{ij}$ given by (5.8),(5.9) and $E_{ij}$ by (5.12),(5.13).

The forms of the pullback of the symplectic form (7.1) and the fiber
derivative (3.1) strongly suggest the introduction of the object

$H_{i}:=\frac{1}{2}\epsilon_{ijk}e^{j}\wedge e^{k}+e_{i}\wedge\alpha\,,$

which would be canonically conjugate to $\omega_{i}$ in the sense that:

$\omega=\int_{\Sigma}\mathrm{d\\!l}\omega^{i}{\wedge\\!\\!\\!\wedge}\,\mathrm{d\\!l}H_{i}\,.$

An important observation at this point is the following. Notice that the
number of independent components in $H_{i}$ and $e_{i}$ are the same, hence it
makes sense to write $e_{i}$ in terms of $H^{i}$ (or a suitably dualized
object as we discuss below) to get a cleaner Hamiltonian description of
Euclidean gravity. In fact, by proceeding in this way one arrives at the
Ashtekar formulation for Euclidean gravity without having to use any gauge
fixing. This is in marked contrast with the situation in the case of the Holst
action [16] and is significantly simpler.

To begin with it is convenient to define the following object

$\vec{H}_{i}:=\frac{1}{\sqrt{1+\alpha^{2}}}\left(\frac{\cdot\wedge
H_{i}}{\mathsf{vol}_{e}}\right)\,,$ (7.2)

that should be understood as an element of the double dual of the tangent
space at each point of $\Sigma$. As the double dual of a finite dimensional
vector space $V$ is canonically isomorphic to $V$, (7.2) determines a unique
vector field on $\Sigma$ that we also call $\vec{H}_{i}$. Given a 1-form
$\beta\in\Omega^{1}(\Sigma)$ we have

$\imath_{\vec{H}_{i}}\beta=\frac{1}{\sqrt{1+\alpha^{2}}}\left(\frac{\beta\wedge
H_{i}}{\mathsf{vol}_{e}}\right)\,.$

Taking this into account it is immediate to see that

$\imath_{\vec{H}_{i}}e^{j}=\frac{1}{\sqrt{1+\alpha^{2}}}(\delta_{i}^{\phantom{i}j}-\epsilon_{i}^{\phantom{k}jk}\alpha_{k})\,.$

In the following it will be useful to introduce the 1-forms $h_{i}$ defined by

$h^{i}:=\frac{1}{\sqrt{1+\alpha^{2}}}\big{(}e^{i}+\alpha^{i}\alpha+\epsilon^{ijk}\alpha_{j}e_{k}\big{)}\,.$

These satisfy the following important properties

$\displaystyle H_{i}=\frac{1}{2}\epsilon_{ijk}h^{j}\wedge h^{k}\,,$
$\displaystyle\imath_{\vec{H}_{i}}h_{j}=\delta_{ij}\,,$
$\displaystyle\mathsf{vol}_{h}=\sqrt{1+\alpha^{2}}\,\,\mathsf{vol}_{e}\,,$

where $\mathsf{vol}_{h}:=\frac{1}{3!}\epsilon_{ijk}h^{i}\wedge h^{j}\wedge
h^{k}$ is a volume form on $\Sigma$. Notice that the last property implies
that

$\vec{H}_{i}:=\left(\frac{\cdot\wedge H_{i}}{\mathsf{vol}_{h}}\right)\,.$
(7.3)

We will now write the constraints in terms of $\omega_{i}$ and $\vec{H}_{i}$.
First we compute

$\imath_{\vec{H}_{i}}F^{i}=\frac{1}{\sqrt{1+\alpha^{2}}}(\mathbb{F}^{ij}\alpha_{j}-\epsilon^{ijk}\mathbb{F}_{jk}-\mathbb{F}^{j}_{\phantom{i}j}\alpha^{i})e_{i}\,,$
(7.4)

and

$\epsilon^{ijk}\imath_{\vec{H}_{i}}\imath_{\vec{H}_{j}}F_{k}=\frac{2}{\sqrt{1+\alpha^{2}}}\big{(}\alpha_{i}(\epsilon^{ijk}\mathbb{F}_{jk}-\mathbb{F}^{ij}\alpha_{j})-\mathbb{F}^{k}_{\phantom{k}k}\big{)}\,.$
(7.5)

Obviously, the constraints (3.5),(3.7) in the form (4.1),(4.2) imply

$\displaystyle\imath_{\vec{H}_{i}}F^{i}=0\,,$ (7.6)
$\displaystyle\epsilon^{ijk}\imath_{\vec{H}_{i}}\imath_{\vec{H}_{j}}F_{k}=0\,,$
(7.7)

which look very much like the vector and scalar constraints in the Ashtekar
formulation for Euclidean gravity.

Conversely, the conditions (7.6)-(7.7) are equivalent to

$\displaystyle\mathbb{F}^{ij}\alpha_{j}-\epsilon^{ijk}\mathbb{F}_{jk}-\mathbb{F}^{j}_{\phantom{i}j}\alpha^{i}=0\,,$
(7.8)
$\displaystyle\alpha_{i}(\epsilon^{ijk}\mathbb{F}_{jk}-\mathbb{F}^{ij}\alpha_{j})-\mathbb{F}^{j}_{\phantom{j}j}=0\,.$
(7.9)

Multiplying the first of these two equations by $\alpha_{i}$ and adding it to
the second leads to $(1+\alpha^{2})\mathbb{F}^{j}_{\phantom{j}j}=0$ i.e.
(4.2), and introducing this into (7.8) we find (4.1).

Finally, in order to get the remaining constraint—equivalent to the usual
Gauss law as will be shown later—we first compute (remember that for a vector
field $X\in\mathfrak{X}(\Sigma)$ its divergence with respect to a given volume
form $\mathsf{vol}$ is defined by
$(\mathrm{div}X)\mathsf{vol}=\pounds_{X}\mathsf{vol}$ where $\pounds_{X}$
denotes the Lie derivative along $X$)

$\big{(}\mathrm{div}_{h}\vec{H}^{i}\big{)}{\mathsf{vol}_{h}}:=\pounds_{\vec{H}^{i}}\mathsf{vol}_{h}=\mathrm{d}\imath_{\vec{H}^{i}}\mathsf{vol}_{h}=\mathrm{d}\left(\frac{1}{2}\epsilon_{ijk}h^{j}\wedge
h^{k}\right)=\mathrm{d}H_{i}\,,$

whereas $\epsilon_{ijk}\omega^{j}\wedge H^{k}$ can easily be seen to be equal
to
$\epsilon_{ijk}\big{(}\imath_{\vec{H}^{k}}\omega^{j}\big{)}\mathsf{vol}_{h}$.
Hence, we conclude that the constraint (3.6) written in the form $DH_{i}=0$ is
equivalent to

$\mathrm{div}_{h}\vec{H}_{i}+\epsilon_{ijk}\imath_{\vec{H}^{k}}\omega^{j}=0\,.$
(7.10)

In order to make contact with the standard Ashtekar variables we introduce a
fiducial volume form $\mathsf{vol}_{0}$ (which, if one wishes, may even be
defined locally in terms of coordinates $x^{i}$, $i=1,2,3$) and write

$\omega=\int_{\Sigma}\mathrm{d\\!l}\omega^{i}{\wedge\\!\\!\\!\wedge}\,\mathrm{d\\!l}H_{i}=\int_{\Sigma}\big{(}\mathrm{d\\!l}\omega^{i}{\wedge\\!\\!\\!\wedge}\,\mathrm{d\\!l}\widetilde{H}_{i}\big{)}\mathsf{vol}_{0}\,.$

This should be understood as

$\omega(\mathbb{X},\mathbb{Y})=\int_{\Sigma}\big{(}Y_{\widetilde{H}_{i}}\raisebox{8.0pt}{\scalebox{1.0}[-2.5]{$\neg\,$}}X_{\omega}^{i}-X_{\widetilde{H}_{i}}\raisebox{8.0pt}{\scalebox{1.0}[-2.5]{$\neg\,$}}Y_{\omega}^{i}\big{)}\mathsf{vol}_{0}\,,$

with

$Y_{\widetilde{H}_{i}}:=\left(\frac{\cdot\wedge\epsilon_{ijk}h^{j}\wedge
Y_{h}^{k}}{\mathsf{vol}_{0}}\right)$

and

$Y_{\widetilde{H}_{i}}\raisebox{8.0pt}{\scalebox{1.0}[-2.5]{$\neg\,$}}\alpha=\left(\frac{\alpha\wedge\epsilon_{ijk}h^{j}\wedge
Y_{h}^{k}}{\mathsf{vol}_{0}}\right)\,,$

for any 1-form $\alpha\in\Omega^{1}(\Sigma)$.

The relation between $\vec{H}_{i}$ and $\widetilde{H}_{i}$ is
$\widetilde{H}_{i}=(\mathrm{det}\,h)\vec{H}_{i}$ with

$\mathrm{det}\,h:=\left(\frac{\mathsf{vol}_{h}}{\mathsf{vol}_{0}}\right)\,.$

In terms of $\widetilde{H}_{i}$ the constraints (7.6) and (7.7) can be
immediately seen to be equivalent to

$\displaystyle\imath_{\widetilde{H}_{i}}F^{i}=0\,,$ (7.11)
$\displaystyle\epsilon^{ijk}\imath_{\widetilde{H}_{i}}\imath_{\widetilde{H}_{j}}F_{k}=0\,,$
(7.12)

which are the vector and scalar constraints written in terms of the Ashtekar
variables for Euclidean gravity.

As explained in Appendix B, the constraint (7.10) in terms of
$\widetilde{H}_{i}$ and $\omega_{i}$ becomes

$\mathrm{div}_{0}\widetilde{H}_{i}+\epsilon_{ijk}\imath_{\widetilde{H}^{k}}\omega^{j}=0\,,$
(7.13)

which, again, is exactly the usual Gauss law in the Ashtekar formulation.
Notice that the divergence in (7.13) is actually independent of the choice of
the volume form $\mathsf{vol}_{0}$ as the argument presented in Appendix B
shows.

Another useful way to understand the constraints can be gained by introducing
the objects

${}^{h}\mathbb{F}_{ij}:=\left(\frac{F_{i}\wedge
h_{j}}{\mathsf{vol}_{h}}\right)\,,\quad^{h}\mathbb{B}_{ij}:=\left(\frac{Dh_{i}\wedge
h_{j}}{\mathsf{vol}_{h}}\right)\,.$

In terms of them the constraints become

${}^{h}\mathbb{F}_{[ij]}=0\,,$ ${}^{h}\mathbb{B}_{[ij]}=0\,,$
${}^{h}\mathbb{F}_{i}^{\phantom{i}i}=0\,.$

Let us look now at the Hamiltonian vector fields in terms of the new
variables. A direct computation using the definition of $h_{i}$ and
${}^{h}\mathbb{F}_{ij}$ gives the following expression for $Z_{\omega}^{k}$ on
the final constraint submanifold (which means that we can make use of the
constraints to simplify it)

$Z_{\omega}^{k}=D\omega_{\mathrm{t}}^{k}-\widehat{\alpha}_{t}\,\,^{h}\mathbb{F}^{k}_{\phantom{k}\ell}e^{\ell}-\epsilon_{\ell
mn}\widehat{e}_{\mathrm{t}}^{\,\,m}{\,\,{}^{h}\mathbb{F}}^{nk}h^{\ell}\,,$
(7.14)

with

$\widehat{\alpha}_{\mathrm{t}}:=\frac{\alpha_{\mathrm{t}}-(e_{\mathrm{t}}\cdot\alpha)}{\sqrt{1+\alpha^{2}}}\,,\quad\widehat{e}_{\mathrm{t}}^{\,\,i}:=\frac{e_{\mathrm{t}}^{i}+\alpha_{\mathrm{t}}\alpha^{i}-\epsilon^{ijk}e_{\mathrm{t}j}\alpha_{k}}{\sqrt{1+\alpha^{2}}}\,.$
(7.15)

Although it is possible to get $Z_{h}^{k}$ by a direct, brute force approach,
there is a much better—albeit slightly indirect—way to do it. The starting
point is the identity

$\epsilon_{ijk}h^{j}\wedge Z_{h}^{k}=\epsilon_{ijk}e^{j}\wedge
Z_{e}^{k}+Z_{ei}\wedge\alpha+e_{i}\wedge Z_{\alpha}\,,$

which comes directly from the definition of $H_{i}$. On the other hand, from
(3.9) we find

$\displaystyle\epsilon_{ijk}e^{j}\wedge
Z_{e}^{k}+Z_{ei}\wedge\alpha+e_{i}\wedge Z_{\alpha}$
$\displaystyle=\epsilon_{ijk}e^{j}\wedge
De_{\mathrm{t}}^{k}+\omega_{\mathrm{t}}^{j}(e_{j}\wedge e_{i})-\alpha\wedge
De_{\mathrm{t}i}-\epsilon_{ijk}(\alpha\wedge e^{j})\omega_{\mathrm{t}}^{k}$
$\displaystyle+e_{i}\wedge\mathrm{d}\alpha_{\mathrm{t}}+e_{\mathrm{t}}^{i}\mathrm{d}\alpha+\epsilon_{ijk}e_{\mathrm{t}}^{j}De^{k}-\alpha_{\mathrm{t}}De_{i}\,.$

By writing the terms on the right hand side of the previous expression in
terms of $h_{i}$ and using (7.15) we obtain the following equation for
$Z_{h}^{k}$:

$\epsilon_{ijk}h^{j}\wedge
Z_{h}^{k}=D\big{(}\epsilon_{ijk}\widehat{e}_{\mathrm{t}}^{\,\,j}h^{k}-\widehat{\alpha}_{\mathrm{t}}h_{i}\big{)}+\omega_{\mathrm{t}}^{j}h_{j}\wedge
h_{i}\,.$

This equation involving differential forms can be solved by using the method
explained in Appendix C of [17]. The solution on the final constraint
submanifold is

$Z_{h}^{k}=D\widehat{e}_{\mathrm{t}}^{\,\,k}+\epsilon^{k}_{\phantom{k}\ell
m}h^{\ell}\omega_{\mathrm{t}}^{m}-\frac{1}{2}\widehat{\alpha}_{\mathrm{t}}\,^{h}\mathbb{B}\,h^{k}-\epsilon_{\ell
mn}\widehat{e}_{\mathrm{t}}^{\,\,m}\,{}^{h}\mathbb{B}^{nk}h^{\ell}+\epsilon^{k}_{\phantom{k}\ell
m}\widehat{X}^{m}h^{\ell}+\widehat{\alpha}_{\mathrm{t}}\,\,^{h}\mathbb{B}^{k\ell}h_{\ell}\,,$
(7.16)

with

$\widehat{X}_{i}:=-\frac{1}{2}\epsilon_{ijk}\left(\frac{\mathrm{d}\widehat{\alpha}_{\mathrm{t}}\wedge
h^{j}\wedge h^{k}}{\mathsf{vol}_{h}}\right)\,.$

At this point it is very interesting to compare the formulation that we have
obtained in terms of the $h_{i}$ with the original one in the time gauge
$\alpha=0$. The latter can be immediately obtained by substituting $\alpha=0$
in the pre-symplectic form (7.1), the constraints (3.5)-(3.7) and the
Hamiltonian vector fields $Z_{\omega}^{k}$, $Z_{e}^{k}$.

The role of a gauge fixing is to reduce (or eliminate) the arbitrariness due
to the presence of arbitrary components in the Hamiltonian vector fields by
fixing some or all of them. This can be done directly, or by demanding that
the dynamics must be confined to a submanifold of $\mathsf{M}_{0}$ obtained by
adding a gauge fixing condition to the secondary constraints. In the latter
case we have to see what happens with the Hamiltonian vector fields. For the
time gauge that we are considering here we must have $Z_{\alpha}=0$, or,
equivalently, $X_{\alpha}=-\mathrm{d}\alpha_{\mathrm{t}}$.

The formulation in the time gauge can be summarized as follows: The
presymplectic form is

$\omega=\int_{\Sigma}\mathrm{d\\!l}\omega^{i}{\wedge\\!\\!\\!\wedge}\,\mathrm{d\\!l}\left(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge
e^{k}\right)\,,$ (7.17)

The secondary constraints become

$\displaystyle\epsilon_{ijk}e^{j}\wedge F^{k}=0\,,$ $\displaystyle
D\left(\frac{1}{2}\epsilon_{ijk}e^{j}\wedge e^{k}\right)=0\,,$ $\displaystyle
e_{i}\wedge F^{i}=0\,,$

or, equivalently,

$\displaystyle\mathbb{F}_{[ij]}=0\,,$ $\displaystyle\mathbb{B}_{[ij]}=0\,,$
$\displaystyle\mathbb{F}_{i}^{\phantom{i}i}=0\,.$

Finally, the Hamiltonian vector fields are

$\displaystyle Z_{\omega}^{k}$
$\displaystyle=D\omega_{\mathrm{t}}^{k}-{\alpha}_{t}\,\,\mathbb{F}^{k}_{\phantom{k}\ell}e^{\ell}-\epsilon_{\ell
mn}{e}_{\mathrm{t}}^{\,\,m}{\,\,\mathbb{F}}^{nk}e^{\ell}\,,$ $\displaystyle
Z_{e}^{k}$
$\displaystyle=D{e}_{\mathrm{t}}^{\,\,k}+\epsilon^{k}_{\phantom{k}\ell
m}e^{\ell}\omega_{\mathrm{t}}^{m}-\frac{1}{2}{\alpha}_{\mathrm{t}}\,\mathbb{B}\,e^{k}-\epsilon_{\ell
mn}{e}_{\mathrm{t}}^{\,\,m}\,\mathbb{B}^{nk}e^{\ell}+\epsilon^{k}_{\phantom{k}\ell
m}{X}^{m}e^{\ell}+{\alpha}_{\mathrm{t}}\,\,\mathbb{B}^{k\ell}e_{\ell}\,,$

with

${X}_{i}:=-\frac{1}{2}\epsilon_{ijk}\left(\frac{\mathrm{d}{\alpha}_{\mathrm{t}}\wedge
e^{j}\wedge e^{k}}{\mathsf{vol}_{e}}\right)\,.$

As we can see a remarkable thing happens: the form of the presymplectic form,
the constraints and the Hamiltonian vector fields obtained either by working
with the $h_{i}$ variables or going to the time gauge in the original
formulation is exactly the same once we replace the arbitrary objects
$\alpha_{\mathrm{t}}$ and $e_{\mathrm{t}}^{i}$ by the, also arbitrary,
$\widehat{\alpha}_{\mathrm{t}}$ and $\widehat{e}_{\mathrm{t}}^{\,\,i}$. An
interesting observation regarding this replacement of parameters is the fact
that this comes from one of the $SO(3)$ factors of the $SO(4)$ symmetry of the
action. Indeed, the infinitesimal transformations (2.3) imply

$\displaystyle\begin{split}&\delta_{2}\alpha_{\mathrm{t}}={\bm{\Upsilon}}_{i}e_{\mathrm{t}}^{i}\,,\\\
&\delta_{2}e_{\mathrm{t}}^{i}=-{\bm{\Upsilon}}^{i}\alpha_{\mathrm{t}}+\epsilon^{i}_{\phantom{i}jk}e_{\mathrm{t}}^{j}{\bm{\Upsilon}}^{k}\,,\end{split}$

which is given by the matrix

$\tau({\bm{\Upsilon}}_{i})=\left[\begin{array}[]{rrrr}0&{\bm{\Upsilon}}_{1}&{\bm{\Upsilon}}_{2}&{\bm{\Upsilon}}_{3}\\\
-{\bm{\Upsilon}}_{1}&0&{\bm{\Upsilon}}_{3}&-{\bm{\Upsilon}}_{2}\\\
-{\bm{\Upsilon}}_{2}&-{\bm{\Upsilon}}_{3}&0&{\bm{\Upsilon}}_{1}\\\
-{\bm{\Upsilon}}_{3}&{\bm{\Upsilon}}_{2}&-{\bm{\Upsilon}}_{1}&0\\\
\end{array}\right]\,.$

The exponential of this matrix gives the matrix corresponding to a finite
transformation

$T({\bm{\Upsilon}})=\cos{\bm{\Upsilon}}\cdot\mathrm{Id}_{4}+\frac{\sin{\bm{\Upsilon}}}{{\bm{\Upsilon}}}\cdot\tau({\bm{\Upsilon}})\,,\quad{\bm{\Upsilon}}:=\sqrt{{\bm{\Upsilon}}_{1}^{2}+{\bm{\Upsilon}}_{2}^{2}+{\bm{\Upsilon}}_{3}^{2}}\,.$

If we write ${\bm{\Upsilon}}_{i}=-\frac{\alpha_{i}}{\alpha}\arctan\alpha$ with
$\alpha:=\sqrt{\alpha_{1}^{2}+\alpha_{2}^{2}+\alpha_{3}^{2}}$, which is
equivalent to
$\alpha_{i}=-\frac{{\bm{\Upsilon}}_{i}}{{\bm{\Upsilon}}}\tan{\bm{\Upsilon}}$,
with $\alpha_{i}\in(-1,1)$; the form of the previous finite transformation
becomes

$T(\alpha_{i})=\frac{1}{\sqrt{1+\alpha^{2}}}\left[\begin{array}[]{cccc}1&-\alpha_{1}&-\alpha_{2}&-\alpha_{3}\\\
\alpha_{1}&1&-\alpha_{3}&\alpha_{2}\\\ \alpha_{2}&\alpha_{3}&1&-\alpha_{1}\\\
\alpha_{3}&-\alpha_{2}&\alpha_{1}&1\\\ \end{array}\right]\,,$

which gives (7.15).

Several comments are in order now:

  1. 1.

Both the 1-form $\omega_{i}$ and the 2-form $H_{i}$ are invariant under the
transformations coming from (2.3). This immediately allows us to perform a
symmetry reduction and eliminate one of the two original $SO(3)$ symmetries
form the final Hamiltonian formulation.

  2. 2.

The primary constraint hypersurface $\mathsf{M}_{0}$ is spanned by
$(e_{\mathrm{t}},h^{i},\omega_{\mathrm{t}},\omega^{i},\alpha_{\mathrm{t}},\alpha)$
or equivalently by the set of fields obtained by replacing $h_{i}$ by $H_{i}$,
$\vec{H}_{i}$ or $\widetilde{H}_{i}$. In the latter case we arrive at the
usual Ashtekar formulation, but the other $h$-variables also provide
interesting and equivalent phase space approaches to the dynamics of Euclidean
GR.

  3. 3.

The field dynamics given by the vector fields obtained above, in particular
that of $\omega_{i}$ and $h_{i}$ can be disentangled by introducing the
uniquely defined vector field $\xi\in\mathfrak{X}(\Sigma)$ obtained by solving
$\imath_{\xi}h^{i}=\widehat{e}_{\mathrm{t}}^{\,\,i}$ (uniqueness is a
consequence of the non-degeneracy of the triads). A straightforward
computation then gives

$Z_{\omega}^{k}=D(\omega_{\mathrm{t}}^{k}-\imath_{\xi}\omega^{k})+\pounds_{\xi}\omega^{k}-\widehat{\alpha}_{\mathrm{t}}\,\,^{h}\mathbb{F}^{k\ell}h_{\ell}\,.$

As expected, a part of the dynamics corresponds to the infinitesimal
diffeomorphisms defined by $\xi$ and $SO(3)$ gauge transformations
parametrized by $\omega_{\mathrm{t}}^{k}-\imath_{\xi}\omega^{k}$. The non-
trivial dynamics of Euclidean GR comes from the
$-\widehat{\alpha}_{\mathrm{t}}\,\,^{h}\mathbb{F}^{k\ell}h_{\ell}$ term.

  4. 4.

Proceding in a similar way one can rewrite $Z_{h}^{k}$ in a similar fashion
and interpret part of the dynamics, again, as infinitesimal diffeomorphisms
and local $SO(3)$ transformation given by the parameters $\xi$ and
$\omega_{\mathrm{t}}^{k}-\imath_{\xi}\omega^{k}$. The cleanest way to see this
is by introducing $Z_{H}^{k}$ as giving the dynamics of the 2-form $H_{i}$. In
this case it is possible to show that

$Z_{H}^{i}=\epsilon^{i}_{\phantom{i}jk}H^{j}(\omega_{\mathrm{t}}^{k}-\imath_{\xi}\omega^{k})+\pounds_{\xi}H^{i}-D(\widehat{\alpha}_{\mathrm{t}}h^{i})\,,$

where $h^{i}$ should be written in terms of $H^{i}$. Notice that the GR
dynamics is given here by the very simple term
$-D(\widehat{\alpha}_{\mathrm{t}}h^{i})$.

  5. 5.

Finally, the dynamics of $\widetilde{H}^{i}$ can also be written in the same
way. The result is exactly the one corresponding to the usual Ashtekar
variables.

  6. 6.

The rest of the fields $(e_{\mathrm{t}}$, $\omega_{\mathrm{t}}$, $\omega^{i}$,
$\alpha_{\mathrm{t}}$, $\alpha)$ are arbitrary as there are no restrictions on
the components of the Hamiltonian vector field giving their dynamics. This
implies that we can choose them as any function of the dynamical fields and
treat them as arbitrary external objects subject to the sole restriction of
providing non-trivial dynamics for the system (for instance,
$\widehat{\alpha}_{\mathrm{t}}$ should be different from zero everywhere on
$\Sigma$).

##  8 Conclusions and comments

The main result of this paper is to show how the Ashtekar formulation for
Euclidean gravity can be obtained from the self-dual action without
introducing any gauge fixing (compare with reference [16]). This means that
one can explicitly reduce the symmetry generated by one of the $SO(3)$ factors
in $SO(4)=SO(3)\times SO(3)$ and remove the arbitrariness associated with the
arbitrary components of the Hamiltonian vector fields that show up when using
the GNH method. It should be clear at this point that, a completely analogous
argument leading to the Ashtekar formulation without gauge fixing should apply
in the case of using Dirac’s approach. A secondary purpose of the paper is to
complete the discussion of the consistency of the Hamiltonian formulation for
Euclidean GR, which one of the authors has to admit, was not finalized in
previous work on the subject starting from the action used here [12]. In this
sense it is instructive to compare the computations needed to complete the
Hamiltonian analysis of the Holst action in the GNH framework [17] with the
ones presented here. Quite surprisingly for us, the complexity of the self-
dual case is far greater than that of the case of using the Holst action. This
is so even considering that the Hamiltonian analysis of the Holst action
produces secondary constraints in two stages and not in one as it happens in
the case analyzed in this paper.

In our opinion, the main use of the insights gained here may be to look for an
action for Lorentzian GR which shares some of the nice features of the
Euclidean self-dual action analyzed here. We hope that the clarification of
the inner workings of the internal $SO(4)$ symmetry of the Euclidean model may
help to better understand the much more relevant Lorentzian case.

## Acknowledgments

We would like to thank Wolfgang Wieland and Javier Olmedo for some interesting
comments. This work has been supported by the Spanish Ministerio de Ciencia
Innovación y Universidades-Agencia Estatal de Investigación grant
AEI/PID2020-116567GB-C22. E.J.S. Villaseñor is supported by the Madrid
Government (Comunidad de Madrid-Spain) under the Multiannual Agreement with
UC3M in the line of Excellence of University Professors (EPUC3M23), and in the
context of the V PRICIT (Regional Programme of Research and Technological
Innovation).

##  Appendix A Tangency analysis: some details

###  A.1 Additional details on the tangency condition 6.6

The terms proportional to $e_{\mathrm{t}}^{i}$ and linear in $\alpha^{i}$ in
this tangency condition are

$\displaystyle-4\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}\alpha_{\ell})A^{k}-3(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+3\epsilon_{ijk}\alpha^{i}A^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})+5\epsilon_{ijk}\alpha^{i}(\mathbb{B}^{j\ell}e_{\mathrm{t}\ell})A^{k}$
$\displaystyle-2\mathbb{B}\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}A^{k}\\!\\!+\\!2\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}A_{\ell})\\!+\\!(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\\!\\!+\\!2(A\cdot\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{jk}\\!\\!-\\!2\epsilon_{ijk}(\alpha_{\ell}\mathbb{B}^{\ell
i})e_{\mathrm{t}}^{j}A^{k}\,.$

By using $\epsilon_{[ijk}\alpha_{\ell]}=0$ to transform the last term in the
previous expression we get

$\displaystyle-2\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}\alpha_{\ell})A^{k}+3(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}-3(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+3\epsilon_{ijk}\alpha^{i}A^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})$
$\displaystyle-2\mathbb{B}\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}A^{k}+2\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}A_{\ell})+5\epsilon_{ijk}\alpha^{i}(\mathbb{B}^{j\ell}e_{\mathrm{t}\ell}A^{k})\,.$

Using now $\epsilon_{[ijk}e_{\mathrm{t}\ell]}=0$ to transform the last term in
the previous expression we find

$\displaystyle-2(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}-2\epsilon_{ijk}\alpha^{i}A^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})+2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}-2\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}\alpha_{\ell})A^{k}$
$\displaystyle+2\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}A_{\ell})-2\mathbb{B}\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}A^{k}\,.$

By writing now the last term as
$-2\mathbb{B}^{\ell}_{\phantom{\ell}\ell}\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}A^{k}$
and using $\mathbb{B}^{\ell}_{\phantom{\ell}[\ell}\epsilon_{ijk]}=0$ we get

$-2(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}+2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+2\epsilon_{ijk}A^{i}(\mathbb{B}^{j\ell}e_{\mathrm{t}\ell})\alpha^{k}-2\epsilon_{ijk}\alpha^{i}A^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})\,,$

which is zero as a consequence of $\epsilon_{[ijk}e_{\mathrm{t}\ell]}=0$.

The terms proportional to $e_{\mathrm{t}}^{i}$ and cubic in $\alpha^{i}$ are

$\displaystyle+2(\alpha\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}+2\alpha^{2}\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})-2(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}(\mathbb{B}^{i\ell}\alpha_{\ell})A^{j}\alpha^{k}$
$\displaystyle-2(A\cdot\alpha)\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})\,.\hskip
91.04872pt$

By writing now the first term in the previous expression as $2\mathbb{B}^{\ell
m}\alpha_{m}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}\alpha_{\ell}\epsilon_{ijk}$ and
using $\alpha_{[\ell}\epsilon_{ijk]}=0$ we immediately see that it cancels.

###  A.2 Additional details on the tangency condition 6.5

The terms proportional to $\alpha_{\mathrm{t}}$ (equivalently to $\eta_{i}$)
in this tangency condition are

$\displaystyle-2\epsilon^{ijk}\alpha_{i}\eta_{j}(\mathbb{B}_{k\ell}A^{\ell})+2\epsilon^{ijk}\eta_{i}A_{j}(\alpha^{\ell}\mathbb{B}_{\ell
k})+2(\eta\cdot
A)\epsilon^{ijk}\alpha_{i}\mathbb{B}_{jk}+4\epsilon^{ijk}\eta_{i}(\mathbb{B}_{j}^{\phantom{j}\ell}\alpha_{\ell})A_{k}$
$\displaystyle+2\mathbb{B}\epsilon^{ijk}\eta_{i}A_{j}\alpha_{k}-2(A\cdot\alpha)\epsilon^{ijk}\eta_{i}\mathbb{B}_{jk}-4\epsilon^{ijk}(\mathbb{B}_{i\ell}\eta^{\ell})A_{j}\alpha_{k}+2\epsilon^{ijk}(\eta^{\ell}\mathbb{B}_{\ell
i})A_{j}\alpha_{k}\,,$

Using $\epsilon^{[ijk}\mathbb{B}_{k}^{\phantom{k}\ell]}=0$ we can transform
the first term in the previous expression to get

$\displaystyle 2\epsilon^{ijk}\eta_{i}A_{j}(\alpha^{\ell}\mathbb{B}_{\ell
k})+2(\eta\cdot
A)\epsilon^{ijk}\alpha_{i}\mathbb{B}_{jk}+2\epsilon^{ijk}\eta_{i}(\mathbb{B}_{j}^{\phantom{j}\ell}\alpha_{\ell})A_{k}-2(A\cdot\alpha)\epsilon^{ijk}\eta_{i}\mathbb{B}_{jk}$
$\displaystyle-2\epsilon^{ijk}(\mathbb{B}_{i\ell}\eta^{\ell})A_{j}\alpha_{k}+2\epsilon^{ijk}(\eta^{\ell}\mathbb{B}_{\ell
i})A_{j}\alpha_{k}\,.$

By using now $\epsilon^{[ijk}\eta^{\ell]}=0$ to transform the last term, the
previous expression becomes

$2\epsilon^{ijk}\eta_{i}A_{j}(\alpha^{\ell}\mathbb{B}_{\ell
k})+2\epsilon^{ijk}\eta_{i}(\mathbb{B}_{j}^{\phantom{j}\ell}\alpha_{\ell})A_{k}-2(A\cdot\alpha)\epsilon^{ijk}\eta_{i}\mathbb{B}_{jk}+2(\eta\cdot\alpha)\epsilon^{ijk}A_{i}\mathbb{B}_{jk}\,,$

which vanishes as a consequence of $\epsilon^{[ijk}\alpha^{\ell]}=0$.

The terms proportional to $e_{\mathrm{t}}^{i}$ and independent of $\alpha_{i}$
in the tangency condition are

$\displaystyle\,\,3(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}-2\epsilon_{ijk}C^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}A_{\ell})-3(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\hskip 45.5244pt$
$\displaystyle+2\mathbb{B}\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}C^{k}+3\epsilon_{ijk}A^{i}C^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})+2\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}-5\epsilon_{ijk}(\mathbb{B}^{i\ell}e_{\mathrm{t}\ell})A^{j}C^{k}\,.$

Using $\epsilon_{[ijk}e_{\mathrm{t}\ell]}=0$ we can transform the last term in
the previous expression to get

$\displaystyle-2\epsilon_{ijk}A^{i}C^{j}(e_{\mathrm{t}\ell}\mathbb{B}^{\ell
k})-2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}+2(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}+2\mathbb{B}\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}C^{k}$
$\displaystyle+2\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}-2\epsilon_{ijk}C^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k}_{\phantom{k}\ell}A^{\ell})\,.$

By using now $\epsilon_{[ijk}\mathbb{B}^{k}_{\phantom{k}\ell]}=0$ to transform
the last term we arrive at

$2\epsilon_{ijk}A^{i}C^{j}(\mathbb{B}^{k\ell}-\mathbb{B}^{\ell
k})e_{\mathrm{t}\ell}-2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}+2(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\,,$

which can be seen to cancel because

$2\epsilon_{ijk}A^{i}C^{j}(\mathbb{B}^{k\ell}-\mathbb{B}^{\ell
k})e_{\mathrm{t}\ell}=2\epsilon_{ijk}\epsilon^{k\ell
m}\epsilon_{mpq}\mathbb{B}^{pq}A^{i}C^{j}=2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}C^{i}\mathbb{B}^{jk}-2(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\,.$

Finally, the terms proportional to $e_{\mathrm{t}}^{i}$ and quadratic in
$\alpha_{i}$ in the tangency condition are

$\displaystyle+2(e_{\mathrm{t}}\cdot
A)\epsilon_{ijk}\alpha^{i}C^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+2(\alpha\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}C^{j}A^{k}-3(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})$
$\displaystyle+2(C\cdot
e_{\mathrm{t}})\epsilon_{ijk}(\mathbb{B}^{i\ell}\alpha_{\ell})A^{j}\alpha^{k}-(A\cdot
C)(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}$
$\displaystyle+\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}(\alpha\mathbb{B}C)+2(A\cdot\alpha)\epsilon_{ijk}C^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+(A\cdot\alpha)(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}$
$\displaystyle+(C\cdot\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}(\alpha_{\ell}\mathbb{B}^{\ell
j})A^{k}+\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}-\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(C_{\ell}\mathbb{B}^{\ell
j})A^{k}$
$\displaystyle+2(e_{\mathrm{t}}\mathbb{B}\alpha)\epsilon_{ijk}\alpha^{i}A^{j}C^{k}\,.$

By using $e_{\mathrm{t}[\ell}\epsilon_{ijk]}=0$ we can transform the last term
in the previous expression written in the form

$2(e_{\mathrm{t}}\mathbb{B}\alpha)\epsilon_{ijk}\alpha^{i}A^{j}C^{k}=2(\mathbb{B}^{\ell
m}\alpha_{m})e_{\mathrm{t}\ell}\epsilon_{ijk}\alpha^{i}A^{j}C^{k}$

to get

$\displaystyle\,\,2(\alpha\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}C^{j}A^{k}-3(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})-(A\cdot
C)(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}$
$\displaystyle+2(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}A^{i}C^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}(\alpha\mathbb{B}C)+2(A\cdot\alpha)\epsilon_{ijk}C^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})$
$\displaystyle+(A\cdot\alpha)(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+(C\cdot\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}(\alpha_{\ell}\mathbb{B}^{\ell
j})A^{k}+\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}$
$\displaystyle-\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(C_{\ell}\mathbb{B}^{\ell
j})A^{k}-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}\,.$

The first term in this expression can be transformed by writing it in the form
$2(\mathbb{B}^{\ell
n}\alpha_{n})\alpha_{\ell}\epsilon_{ijk}e_{\mathrm{t}}^{i}C^{j}A^{k}$ and
using $\alpha_{[\ell}\epsilon_{ijk]}=0$. By doing this we get

$\displaystyle\,\,(C\cdot\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}(\alpha_{\ell}\mathbb{B}^{\ell
j})A^{k}+\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}(\alpha\mathbb{B}C)-(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})$
$\displaystyle-(A\cdot
C)(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+(A\cdot\alpha)(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}$
$\displaystyle-\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(C_{\ell}\mathbb{B}^{\ell
j})A^{k}-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}\,.$

Now, we transform the first term by writing it as
$\alpha^{\ell}e_{\mathrm{t}}^{i}(\alpha_{n}\mathbb{B}^{nj})A^{k}C_{\ell}\epsilon_{ijk}$
and using $C_{[\ell}\epsilon_{ijk]}=0$ to get

$\displaystyle(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}(\alpha_{\ell}\mathbb{B}^{\ell
i})A^{j}\alpha^{k}-(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})-(A\cdot
C)(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}$
$\displaystyle+(A\cdot
C)\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\alpha_{n}\mathbb{B}^{nk})+(A\cdot\alpha)(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}$
$\displaystyle+\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}C_{\ell})A^{k}-\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(C_{\ell}\mathbb{B}^{\ell
j})A^{k}\,.$

In the next step we use $\epsilon_{[ijk}\alpha_{\ell]}=0$ to transform the
first term and also rewrite the last two terms as

$\alpha^{2}\epsilon_{ijk}e_{\mathrm{t}}^{i}(\mathbb{B}^{j\ell}-\mathbb{B}^{\ell
j})C_{\ell}A^{k}=\alpha^{2}(C\cdot
A)\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{jk}-\alpha^{2}(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\mathbb{B}^{jk}\,,$

this way we obtain

$\displaystyle-(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})-(A\cdot
C)(e_{\mathrm{t}}\cdot\alpha)\epsilon_{ijk}\alpha^{i}\mathbb{B}^{jk}+(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\alpha^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})$
$\displaystyle+(A\cdot
C)\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\alpha_{n}\mathbb{B}^{nk})-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}+\alpha^{2}(C\cdot
A)\epsilon_{ijk}e_{\mathrm{t}}^{i}\mathbb{B}^{jk}\,.$

We can now transform the last term by writing it as $(C\cdot
A)e_{\mathrm{t}}^{i}\mathbb{B}^{jk}\alpha^{\ell}\alpha_{\ell}\epsilon_{ijk}$
and using $\alpha_{[\ell}\epsilon_{ijk]}=0$. This way the previous expression
becomes

$\displaystyle-(C\cdot\alpha)\epsilon_{ijk}A^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+(e_{\mathrm{t}}\cdot
C)\epsilon_{ijk}A^{i}\alpha^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})+(A\cdot
C)\epsilon_{ijk}\alpha^{i}e_{\mathrm{t}}^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})$
$\displaystyle-(C\mathbb{B}\alpha)\epsilon_{ijk}e_{\mathrm{t}}^{i}A^{j}\alpha^{k}=4e_{\mathrm{t}}^{m}A^{i}\alpha^{j}(\mathbb{B}^{k\ell}\alpha_{\ell})\epsilon_{[ijk}C_{\ell]}=0\,.$

##  Appendix B The Gauss law in the Ashtekar formulation: some computational
details

By multiplying the constraint (7.10) by $\mathrm{det}\,h$ (which is non-zero
at every point of $\Sigma$) we get the following equivalent expression:

$(\mathrm{det}\,h)\cdot\mathrm{div}_{h}\left(\frac{\widetilde{H}^{i}}{\mathrm{det}\,h}\right)+\epsilon_{ijk}\imath_{\widetilde{H}^{k}}\omega^{j}=0\,.$
(B.1)

We now prove that

$(\mathrm{det}\,h)\cdot\mathrm{div}_{h}\left(\frac{\widetilde{H}^{i}}{\mathrm{det}\,h}\right)=\mathrm{div}_{0}\widetilde{H}^{i}\,,$
(B.2)

for any field independent volume form $\mathsf{vol}_{0}$ [remember that, for a
given volume form $\mathsf{vol}$, the divergence of a vector field $X$ is
defined as $\left(\frac{\pounds_{X}\mathsf{vol}}{\mathsf{vol}}\right)$].
Notice that this implies that the right hand side of (B.2) is actually
independent on the choice of $\mathsf{vol}_{0}$.

In order to prove (B.2) we need the identity

$(\mathsf{vol}_{0})\cdot\pounds_{\widetilde{H}^{i}}\mathrm{det}\,h=\mathrm{d}(\mathrm{det}\,h)\wedge\imath_{\widetilde{H}^{i}}\mathsf{vol}_{0}\,,$
(B.3)

which can be obtained by taking the interior product
$\imath_{\widetilde{H}^{i}}$ of both sides of the trivial identity
$0=\mathrm{d}(\mathrm{det}\,h)\wedge{\mathsf{vol}}_{0}$.

For $X\in\mathfrak{X}(\Sigma)$, $\varphi\in\Omega^{0}(\Sigma)$ and any volume
form $\mathsf{vol}\in\Omega^{3}(\Sigma)$ we have

$\mathrm{div}(\varphi X)=\pounds_{X}\varphi+\varphi\,\mathrm{div}X\,.$

Using this expression one gets

$(\mathrm{det}\,h)\mathrm{div}_{h}\left(\frac{\widetilde{H}^{i}}{\mathrm{det}\,h}\right)=(\mathrm{det}\,h)\pounds_{\widetilde{H}^{i}}\left(\frac{1}{\mathrm{det}\,h}\right)\\!+\mathrm{div}_{h}\widetilde{H}^{i}\\!=-\frac{1}{\mathrm{det}\,h}\pounds_{\widetilde{H}^{i}}\mathrm{det}\,h+\mathrm{div}_{h}\widetilde{H}^{i}\,.$
(B.4)

Finally, making use of (B.3) we find

$\displaystyle\mathrm{div}_{h}\widetilde{H}^{i}$
$\displaystyle=\left(\frac{\pounds_{\widetilde{H}^{i}}{\mathsf{vol}}_{h}}{\mathsf{vol}_{h}}\right)=\left(\frac{\mathrm{d}[(\mathrm{det}\,h)\cdot(\imath_{\widetilde{H}^{i}}\mathsf{vol}_{0})]}{\mathsf{vol}_{h}}\right)=\left(\frac{\mathrm{d}(\mathrm{det\,h})\wedge\imath_{\widetilde{H}^{i}}\mathsf{vol}_{0}}{\mathsf{vol}_{h}}\right)$
$\displaystyle+\left(\frac{(\mathrm{det}\,h)\pounds_{\widetilde{H}^{i}}\mathsf{vol}_{0}}{\mathsf{vol}_{h}}\right)=\left(\frac{(\mathsf{vol}_{0})\cdot\pounds_{\widetilde{H}^{i}}\mathrm{det}\,h}{\mathsf{vol}_{h}}\right)+\left(\frac{(\mathrm{det}\,h)\pounds_{\widetilde{H}^{i}}\mathsf{vol}_{0}}{\mathsf{vol}_{h}}\right)$
$\displaystyle=\frac{1}{\mathrm{det}\,h}\pounds_{\widetilde{H}^{i}}\mathrm{det}\,h+\left(\frac{\pounds_{\widetilde{H}^{i}}\mathsf{vol}_{0}}{\mathsf{vol}_{0}}\right)=\frac{1}{\mathrm{det}\,h}\pounds_{\widetilde{H}^{i}}\mathrm{det}\,h+\mathrm{div}_{0}\widetilde{H}^{i}\,,$

which introduced in (B.4) gives (B.2) .

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