††institutetext: Rudolf Peierls Centre for Theoretical Physics, Clarendon
Laboratory, Parks Road, University of Oxford, Oxford OX1 3PU, UK

# Towards a dual formulation of quantum gravity via metric-curvature
bijections

Praveen D. Xavier<EMAIL_ADDRESS>
###### Abstract

We prove that Riemannian metrics in General Relativity in the _‘normal-
coordinates’_ gauge are in one-to-one correspondence with curvature 2-forms.
We discuss how this can be used as a change of variables in the operator
formalism to construct a dual formulation of quantum gravity pertinent in the
context of asymptotic safety-like approaches to quantum gravity.

###### Keywords:

dual formulations, quantum gravity, curvature, non-perturbative methods,
general relativity, asymptotic safety-like approaches, bijections, operator
formalism

##  1 Introduction

###  1.1 Revisiting a classical problem…

In Yang-Mills (YM) theory, Wu and Yang have shown that there exists
connections, unrelated by gauge transformation, that have the same curvature
wu-yang . This is known as the _field copy problem_ mostow . Halpern pointed
out that the curvature map restricted to connections in the _axial gauge_ is,
however, injective halpern . Later, Durand and Mendel showed this also in the
_Fock-Shwinger gauge_ durand . (The Fock-Schwinger gauge is preferred to the
axial gauge because of its relative simplicity.) To be precise, Durand and
Mendel showed that connections in the Fock-Shwinger gauge are mapped, by the
curvature map, _bijectively_ to curvature 2-forms, $F$, satisfying the ‘YM
Bianchi identity for curvature’:

$\begin{split}&dF+ig[A\wedge F]=0,\\\
\text{where}\quad&A_{\mu}(x)=\int_{0}^{1}tx^{\nu}F_{\nu\mu}(tx)dt.\end{split}$
(1)

Just as in YM, in General Relativity, Riemannian metrics unrelated by
coordinate transformation may have the same curvature. Müller, Schubert and
van de Ven have shown, however, that the curvature map restricted to metrics
in the _‘normal-coordinates’_ gauge is _injective_ muller . So that we can set
up a bijection, we pose the following question: what is the image of this map?
In this paper we prove that the image (of the curvature map restricted to
metrics in the ‘normal coordinates’ gauge) is the restricted set of curvature
2-forms, $R^{a}_{b}$, satisfying what we call the _‘1 st and 2nd Bianchi
identities for curvature’_:

$\begin{split}&R^{a}_{b}\wedge e^{b}=0,\\\
\text{where}\quad&e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}t(1-t)x^{b}x^{\nu}R^{a}_{b\nu\mu}(tx)dt\end{split}$
(2)

and

$\begin{split}&dR^{a}_{b}+\omega^{a}_{c}\wedge
R^{c}_{b}-R^{a}_{c}\wedge\omega^{c}_{b}=0,\\\
\text{where}\quad&\omega^{a}_{b\mu}(x)=\int_{0}^{1}tx^{\nu}R^{a}_{b\nu\mu}(tx)dt\end{split}$
(3)

respectively. This result proves that there is a bijection between metrics in
normal coordinates and curvature 2-forms satisfying the 1st and 2nd Bianchi
identities for curvature. We discuss, below, how this result can be used to
derive a dual formulation of quantum gravity.

###  1.2 …and its quantum application

A quantum theory is usually formulated in terms of operators $\vec{q}$ (which
denotes the collection $\\{q_{\alpha}^{i}\\}$ – where $\alpha$ may be a
continuous index and $i$ may be a discrete index) and their canonical
conjugates $\vec{p}$, with dynamics governed by a Hamiltonian
$H(\vec{q},\vec{p})$. If one makes a change of variables
$\vec{q}\to\vec{Q}(\vec{q})$ (these are known as point canonical
transformations in Hamiltonian mechanics), one can reformulate the quantum
theory in terms of $\vec{Q}$ and its canonical conjugate $\vec{P}$. The theory
is governed by the _dual_ Hamiltonian: $H(\vec{q},\vec{p})\to
H_{\text{dual}}(\vec{Q},\vec{P})$. The corresponding phase-space path integral
formulation will involve integrations over paths $\vec{Q}(t)$ and
$\vec{P}(t)$. (If either $\vec{Q}(t)$ or $\vec{P}(t)$ appears at most
quadratically in the phase-space action, then it can be integrated out of the
path integral – as it happens in certain cases of interest.) In this way one
can arrive at a _dual_ path integral formulation of the theory.

In soliton physics, a change of variables has been used to calculate
transition amplitudes involving solitons Gervals1976 ; gervais1976point ;
mandelstam1975soliton ; tomboulis1975canonical ; christ1975quantum . In the
statistical mechanics problem of the Coulomb gas, a quantum change of
variables has been used to prove the occurrence of a mass gap Brydges  (see
also the discussion in clay  §6.6).

As an application of this method to the non-perturbative dynamics of Yang-
Mills (YM), consider the Durand-Mendel result of §1.1. Using this result, the
connection on a spatial slice in the Fock-Shwinger gauge can be mapped
bijectively to the magnetic field subject to the YM Bianchi identity for
curvature. The field that is canonically conjugate to the magnetic field is
the _dual connection_ hooft  on the spatial slice halpern  (see Fig. 1). So
the change of variables allows one to reformulate the theory in terms of the
magnetic field and spatial dual connection. The magnetic field can be
integrated out of the phase-space path integral completely (because it appears
only quadratically in the phase-space action), allowing the path integral to
be expressed solely in terms of the dual connection. Now, a formulation of YM
in terms of the dual connection has long been a candidate for solving the non-
perturbative dynamics of YM kondo : it is established that if the dual
connection acquires a v.e.v. Seiberg ; Giacomo  or a mass Tong  (§8.3.2), clay
, then confinement follows.

Figure 1: The chain of relations.

In quantum gravity, the effective field theory (EFT) formalism arises
naturally because counter-terms to the Einstein action generate all
interactions allowed by symmetry. In the EFT of gravity there are an infinite
number of allowed interactions. The space of couplings associated with these
interactions is, obviously, infinite-dimensional. The question of the high
energy behaviour of the theory comes down to the renormalization group flow of
the couplings with the energy scale – the initial conditions for this flow are
determined by the requirement that at low-energies the theory reproduces
general relativity, i.e., at low energies, all but Newton’s constant and the
cosmological constant are zero (these themselves being determined by low-
energy, classical experiments). Since Newton’s constant, $G$, has negative
mass dimension (which is the source of gravity’s nonrenormalizability), the
free-fixed-point (with all couplings equal to zero) is ultraviolet repulsive
in the direction of $G$. Therefore the couplings grow as the energy increases,
and perturbation theory can no longer describe the flow accurately.

Nonrenormalizability is in itself not disastrous (in fact, from this point of
view, nonrenormalizability is inconsequential for the UV finiteness of the
theory), but what would be disastrous for the theory (in the sense that the
theory cannot be physical) is if reaction rates develop singularities at
finite, but very high, energies (otherwise reaction rates are finite at all
energies and everything is dandy). It was Wienberg weinbergcritical  who first
proposed this more general criteria. Notice that it is vastly more
accommodating than the criteria of renormalizability. We will say that a
theory satisfying Weinberg’s criteria is _UV finite_. In addition, Weinberg
weinberg ; ambjorn ; xavier  suggested that one way of, fairly surely,
guaranteeing that this condition is met is to require that the couplings
asymptotically approach fixed values at high energies – in this case, reaction
rates can, reasonably, be expected to be finite at all energies. This goes
under the name of _asymptotic safety_. Asymptotic safety appears to be a
_sufficient_ condition for UV finiteness but it is certainly not a _necessary_
condition. In fact, even if the couplings diverged at finite, but high, energy
(let alone asymptotically), it is not clear whether reactions rates would too.

So we learn that ‘nonrenormalizability’ cannot be conflated with the theory
being unphysical; it is simply an indication that we must look beyond
perturbation theory. What is more important than the pronouncement that
gravity is nonrenormalizable is to determine the flow of the coupling
constants and the reaction rates as functions of the couplings beyond
perturbation theory. What is required, then, is a formulation of quantum
gravity in which non-perturbative calculations are made possible.

The result of this paper can be used to derive a dual formulation of quantum
gravity as follows: one can bijectively map the _spatial metric_ in the
normal-coordinates gauge to the spatial curvature 2-form subject to the 1st
and 2nd Bianchi identities for curvature. In line with the discussions above,
this change of variables allows one to reformulate quantum gravity in terms of
the spatial curvature and its canonical conjugate. The change of variables
from the metric to the curvature is in analogy with the change of variables
from the connection to the curvature in YM. (We should also point out that, as
in YM, it may happen that the spatial curvature can be integrated out of the
phase-space path integral in the dual formulation of quantum gravity, allowing
a complete formulation in terms of its canonical conjugate.) The expectation
that this dual formulation of quantum gravity will be non-perturbative appeals
to the fact that, as we have discussed, the analogous variables change in YM
provides a formulation which has strong reasons to be non-perturbative.

###  1.3 Plan

The plan of the paper is as follows. In §3 we review the Durand-Mendel result
in YM. In §4 we review the Müller-Schubert-van de Ven result in gravity. In
§5-§7 – which are our original contributions – we prove the following
statements respectively:

  * •

Curvatures satisfying the 1st and 2nd Bianchi identities for curvature are
bijective with spin-connections satisfying the Fock-Shwinger gauge and
torsionless conditions.

  * •

Spin-connections satisfying the Fock-Shwinger gauge and torsionless conditions
are bijective with vielbeins satisfying (45)-(47).

  * •

Vielbeins satisfying (45)-(47) are bijective with metrics in normal
coordinates.

These statements, taken together, allow us to prove that: _curvatures
satisfying the 1 st and 2nd Bianchi identities for curvature are bijective
with metrics in normal coordinates_. Fig. 2 below summarizes the proof.

Figure 2: A summary of the proof, with references to equation numbers.

##  2 Conventions and Basics

Let $x^{\mu}$ denote the coordinates. We adopt Greek indices for the
components of tensors in the coordinate basis:

$T^{\alpha_{1}...\alpha_{p}}_{\beta_{1}...\beta_{q}}\equiv
T(dx^{\alpha_{1}},...,dx^{\alpha_{p}},\partial_{\beta_{1}},...,\partial_{\beta_{q}})$
(4)

We adopt Wald’s wald  (appendix B) conventions for the standard operations on
differential forms:

$\displaystyle\text{exterior derivative:}\quad(d\omega)_{\mu_{1}...\mu_{p+1}}$
$\displaystyle=(p+1)\partial_{[\mu_{1}}\omega_{\mu_{2}...\mu_{p+1}]}$ (5)
$\displaystyle\text{wedge
product:}\quad(\omega\wedge\sigma)_{\mu_{1}...\mu_{p+q}}$
$\displaystyle=\frac{(p+q)!}{p!q!}\omega_{[\mu_{1}...\mu_{p}}\sigma_{\mu_{p+1}...\mu_{p+q}]}$
(6) $\displaystyle\text{interior
product:}\quad(i_{X}\omega)_{\mu_{2}...\mu_{p}}$
$\displaystyle=X^{\mu_{1}}\omega_{\mu_{1}\mu_{2}...\mu_{p}}$ (7)

where antisymmetrization is defined in such a way that
$\omega_{\mu_{1}...\mu_{n}}=\omega_{[\mu_{1}...\mu_{n}]}$.

The Lie derivative is defined as

$(\mathcal{L}_{X}T)_{\mu_{1}...\mu_{p}}=X(T_{\mu_{1}...\mu_{p}})-\sum_{i=1}^{p}T(\partial_{\mu_{1}},...,\partial_{\mu_{i-1}},[X,\partial_{\mu_{i}}],\partial_{\mu_{i+1}},...,\partial_{\mu_{p}})$
(8)

and satisfies

$\displaystyle\mathcal{L}_{X}$ $\displaystyle=\\{d,i_{X}\\}.$ (9)

As a result,

$[\mathcal{L}_{X},d]=0.$ (10)

We define the _radial vector field_

$\mathfrak{r}\equiv x^{\mu}\partial_{\mu}~{}.$ (11)

Since $[\mathfrak{r},\partial_{\mu}]=-\partial_{\mu}$

$(\mathcal{L}_{\mathfrak{r}}\sigma)_{\mu_{1}...\mu_{p}}=(\mathfrak{r}+p)\sigma_{\mu_{1}...\mu_{p}}$
(12)

for any $p$-form $\sigma$.

If $\eta$ is a $p$-form, less singular than $|x|^{-p}$ near $x=0$, then

$\mathcal{L}_{\mathfrak{r}}\eta=0\Longleftrightarrow\eta=0.$ (13)

This follows from the fact that the l.h.s. of (13) is equivalent to (using
(12))

$\frac{d}{dt}(t^{p}\eta_{\mu_{1}...\mu_{p}}(tx))=0$ (14)

which integrates to $\eta=0$ assuming $\eta$ is less singular than $|x|^{-p}$
near $x=0$.

##  3 The Durand-Mendel result in Yang-Mills

In this section we review the results of durand  which introduces several of
the notions relevant for the latter discussions on gravity.

Connections in Yang-Mills (YM) theory are Lie algebra valued 1-forms $A$.
Without loss of generality we may assume that the Lie algebra is a matrix
algebra. The wedge product of matrix valued forms, of degree $p$ and $q$, is
defined as

$[\omega\wedge\eta]^{a}_{b}:=\omega^{a}_{c}\wedge\eta^{c}_{b}+(-)^{pq+1}\eta^{a}_{c}\wedge\omega^{c}_{b}$
(15)

(where $a,b$ and $c$ are the ‘matrix’ indices). Consider the _Fock-Schwinger
gauge_ cronstrom ; kummer

$i_{\mathfrak{r}}A=0$ (16)

where $\mathfrak{r}$ is the radial vector field (11). This is a completely
fixed gauge111Which means that no infinitesimal gauge transformation preserves
the gauge condition. To see this, note that an infinitesimal gauge
transformation $A\to A+i[A,\omega]-\frac{1}{g}d\omega$ (where the matrix
$\omega(x)$ is the infinitesimal gauge transformation parameter) preserves the
gauge condition iff $\frac{d}{dt}\omega(tx)=0$. But since $\omega(x)=0$ when
$|x|=\infty$ (this is because bona fide gauge transformations must approach
the identity at infinity), we must have $\omega(x)=0$.. When we map
connections in this gauge to field strengths via

$F=dA+\frac{ig}{2}[A\wedge A]$ (17)

there are two questions which arise:

  1. 1.

is the map injective?

  2. 2.

what is the image of the map?

The answer to Q1 is yes. To see this, apply $i_{\mathfrak{r}}$ to (17) and
make use of (9) and (16):

$i_{\mathfrak{r}}F=\mathcal{L}_{\mathfrak{r}}A.$ (18)

Using (12) this becomes

$x^{\mu}F_{\mu\nu}=(\mathfrak{r}+1)A_{v}$ (19)

which is easily integrated, assuming the connection is not singular at $x=0$,
to give

$A_{\mu}(x)=\int_{0}^{1}tx^{\nu}F_{\nu\mu}(tx)dt.$ (20)

(Note that $i_{\mathfrak{r}}A=0$ automatically from above.) To answer Q2 we
first note that the Bianchi identity for curvature

$\begin{split}&dF+ig[A\wedge F]=0,\\\
\text{where}\quad&A_{\mu}(x)=\int_{0}^{1}tx^{\nu}F_{\nu\mu}(tx)dt\end{split}$
(21)

is a necessary condition for a field strength to lie in the image. But it is
also a sufficient condition, which is seen by applying $i_{\mathfrak{r}}$ to
$\eqref{BI}$ and using (10) and (18):

$\begin{split}&\mathcal{L}_{\mathfrak{r}}(F-dA-\frac{ig}{2}[A\wedge A])=0,\\\
\text{where}\quad&A_{\mu}(x)=\int_{0}^{1}tx^{\nu}F_{\nu\mu}(tx)dt.\end{split}$
(22)

From (13), we find that this is equivalent to

$\begin{split}&F=dA+\frac{ig}{2}[A\wedge A]\\\
\text{where}\quad&A_{\mu}(x)=\int_{0}^{1}tx^{\nu}F_{\nu\mu}(tx)dt,\end{split}$
(23)

This shows that if $F$ satisfies the Bianchi identity for curvature then it is
indeed in the image of the curvature map restricted to connections in the
Fock-Shwinger gauge. This completes the proof that the space of connections in
YM satisfying the Fock-Shwinger gauge condition is bijective with the space of
field strengths satisfying the Bianchi identity for curvature.

##  4 The Müller-Schubert-van de Ven result in gravity

In this section we review the results of muller , crucial for the latter
developments.

###  4.1 Vielbein formalism

See jost  for the basics of the vielbein formalism.

Consider the vielbein 1-forms $e^{a}$, and their vector duals $e_{a}$,
satisfying

$\displaystyle g(e_{a},e_{b})$ $\displaystyle=\delta_{ab}$ (24) $\displaystyle
e^{a}(e_{b})$ $\displaystyle=\delta^{a}_{b}$ (25)

(Note that we have taken the metric signature to be Euclidean because,
ultimately, what we have in mind is to perform the change of variables on the
spatial metric.) In the coordinate basis, $e^{a}_{\mu}$ and $e_{a}^{\mu}$ are
matrix inverses of each other so they satisfy

$e^{a}_{\mu}e_{a}^{\nu}=\delta^{\nu}_{\mu}$ (26)

We use Roman indices for the vielbein basis and Greek indices for the
coordinate basis. The vielbeins obviously possess an $SO(d)$ gauge freedom (in
$d$ dimensions). Assuming the torsion is zero, the spin-connection and
curvature 2-form are defined as

$\displaystyle\omega^{a}_{b}$ $\displaystyle=\frac{1}{2}(-e^{c}\wedge
i_{b}i_{a}de^{c}+i_{b}de^{a}-i_{a}de^{b}),$ (27) $\displaystyle R^{a}_{b}$
$\displaystyle=d\omega^{a}_{b}+\omega^{a}_{c}\wedge\omega^{c}_{b}$ (28)

respectively, where $i_{a}$ is the interior product w.r.t. $e_{a}$.

We should point out that (27) is equivalent to the vanishing of the torsion:

$de^{a}+\omega^{a}_{b}\wedge e^{b}=0.$ (29)

To prove this use the fact that $e^{b}\wedge i_{b}(\eta)=p\eta$ for any
$p$-form $\eta$. To see that (29) implies (27) use
$de^{a}=-\omega^{a}_{b}\wedge e^{b}$ to simplify the r.h.s. of (27).

###  4.2 Recovering the spin-connection from the curvature

Impose the Fock-Shwinger gauge condition on the connection:

$i_{\mathfrak{r}}\omega^{a}_{b}=0.$ (30)

This fixes the $SO(d)$ gauge freedom completely. Then, in just the same way as
in §3, we can prove the following: the space of spin-connections satisfying
the Fock-Shwinger gauge condition is in bijection with the space of curvatures
satisfying the _2 nd Bianchi identity for curvature_:

$\begin{split}&dR^{a}_{b}+\omega^{a}_{c}\wedge
R^{c}_{b}-R^{a}_{c}\wedge\omega^{c}_{b}=0,\\\
\text{where}\quad&\omega^{a}_{b\mu}(x)=\int_{0}^{1}tx^{\nu}R^{a}_{b\nu\mu}(tx)dt.\end{split}$
(31)

The forward map for this bijection is given by (28); and the reverse map is
given by

$\omega^{a}_{b\mu}(x)=\int_{0}^{1}tx^{\nu}R^{a}_{b\nu\mu}(tx)dt.$ (32)

###  4.3 Recovering the vielbein from the spin-connection

Applying $i_{\mathfrak{r}}$ to (29) gives

$-di_{\mathfrak{r}}e^{a}+(\mathfrak{r}+1)e^{a}-\omega^{a}_{b}\wedge
i_{\mathfrak{r}}e^{b}=0.$ (33)

Assume we are working in _‘normal-coordinates’_ , i.e. a coordinate system in
which

$x^{\mu}g_{\mu\nu}(x)=x^{\nu}$ (34)

(note that this implies $g_{\mu\nu}(0)=\delta_{\mu\nu}$). As a result,

$\displaystyle x^{\mu}e^{a}_{\mu}(x)$ $\displaystyle=x^{a}$ (35)
$\displaystyle x^{a}e^{a}_{\mu}(x)$ $\displaystyle=x^{\mu}$ (36)

(note that this implies $e^{a}_{\mu}(0)=\delta^{a}_{\mu}$). So (33) becomes

$(\mathfrak{r}+1)e^{a}_{\mu}=\delta^{a}_{\mu}+\omega^{a}_{b\mu}x^{b}$ (37)

which is easily integrated to give

$e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}\omega^{a}_{b\mu}(tx)tx^{b}dt.$
(38)

Using (32) then gives

$e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}\int_{0}^{1}t_{1}t_{2}^{2}x^{b}x^{\nu}R^{a}_{b\nu\mu}(t_{1}t_{2}x)dt_{1}dt_{2}$
(39)

Fixing $t_{1}t_{2}$ and doing the $t_{2}$ integral first gives, finally,

$e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}t(1-t)x^{b}x^{\nu}R^{a}_{b\nu\mu}(tx)dt.$
(40)

Note that $x^{\mu}e^{a}_{\mu}(x)=x^{a}$ and $x^{a}e^{a}_{\mu}(x)=x^{\mu}$
follow automatically from this.

##  5 Bijection between the curvature and spin-connection

######  Theorem 1.

Curvatures satisfying the 1st and 2nd Bianchi identities for curvature are
bijective with spin-connections satisfying the Fock-Shwinger gauge and
torsionless conditions.

###### Proof.

We pointed out in §4.2 that the space of curvatures satisfying the 2nd Bianchi
identity for curvature is in bijection with the space of connections
satisfying $i_{\mathfrak{r}}\omega^{a}_{b}=0$. We now place the _1 st Bianchi
identity for curvature_

$\begin{split}&R^{a}_{b}\wedge e^{b}=0,\\\
\text{where}\quad&e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}t(1-t)x^{b}x^{\nu}R^{a}_{b\nu\mu}(tx)dt\end{split}$
(41)

as an additional restriction on the space of curvatures. Via the bijection,
this descends into an extra restriction on the space of connections:

$\begin{split}&R^{a}_{b}\wedge e^{b}=0,\\\
\text{where}\quad&R^{a}_{b}=d\omega^{a}_{b}+\omega^{a}_{c}\wedge\omega^{c}_{b}\\\
\text{and}\quad&e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}\omega^{a}_{b\mu}(tx)tx^{b}dt.\end{split}$
(42)

Applying $i_{\mathfrak{r}}$ to this and manipulating gives

$\begin{split}&\mathcal{L}_{\mathfrak{r}}(de^{a}+\omega^{a}_{b}\wedge
e^{b})=0,\\\
\text{where}\quad&e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}\omega^{a}_{b\mu}(tx)tx^{b}dt\end{split}$
(43)

which implies (using (13)) that

$\begin{split}&de^{a}+\omega^{a}_{b}\wedge e^{b}=0,\\\
\text{where}\quad&e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}\omega^{a}_{b\mu}(tx)tx^{b}dt.\end{split}$
(44)

We will refer to (44) as the _torsionless conditon_ on connections (c.f.
(29)).

∎

##  6 Bijection between the spin-connection and vielbein

######  Theorem 2.

Spin-connections satisfying the Fock-Shwinger gauge and torsionless conditions
are bijective with vielbeins satisfying (45)-(47).

###### Proof.

Consider the space of vielbeins satisfying

$\displaystyle x^{\mu}e^{a}_{\mu}(x)$ $\displaystyle=x^{a}$ (45)
$\displaystyle x^{a}e^{a}_{\mu}(x)$ $\displaystyle=x^{\mu}$ (46)
$\displaystyle i_{a}\mathcal{L}_{\mathfrak{r}}e^{b}$
$\displaystyle=i_{b}\mathcal{L}_{\mathfrak{r}}e^{a}$ (47)

We leave it as an exercise to the reader to show that connections arising from
such vielbeins satisfy the Fock-Shwinger gauge and torsionless conditions
automatically.

Conversely, consider the image of connections satisfying the Fock-Shwinger
gauge and torsionless conditions under the map (38). It is immediately clear
that such vielbeins satisfy (45)-(46). But, in addition, such vielbeins
satisfy (47), which can be proved as follows: we pointed out, in §4.1, that
(29) is equivalent to (27). This means that the torsionless condition is
equivalent to

$\begin{split}&\omega^{a}_{b}=\frac{1}{2}(-e^{c}\wedge
i_{b}i_{a}de^{c}+i_{b}de^{a}-i_{a}de^{b})\\\
\text{where}\quad&e^{a}_{\mu}(x)=\delta^{a}_{\mu}+\int_{0}^{1}\omega^{a}_{b\mu}(tx)tx^{b}dt.\end{split}$
(48)

Applying $i_{\mathfrak{r}}$ to this and using $x^{a}\wedge
e^{a}=x^{\mu}dx^{\mu}$ shows that (47) is indeed satisfied.

From (48), it is clear that mapping a connection – satisfying the torsionless
condition – to a vielbein (via (38)) and back is equivalent to the identity
operation. ∎

##  7 Bijection between the vielbein and metric

######  Theorem 3.

Vielbeins satisfying (45)-(47) are bijective with metrics in normal
coordinates.

###### Proof.

The metric is obtained from the vielbein using

$g_{\mu\nu}=e^{a}_{\mu}e^{a}_{\nu}$ (49)

Conversely, to obtain the vielbein from the metric, we differentiate (49) and
use (47) (which, using (12), can be shown to be equivalent to
$e^{a}_{[\mu}(tx)\frac{d}{dt}e^{a}_{\nu]}(tx)=0$, and which completely fixes
the $SO(d)$ gauge freedom the vielbeins possess):

$\frac{d}{dt}e^{a}_{\mu}(tx)=\frac{1}{2}e^{\nu}_{a}(tx)\frac{d}{dt}g_{\nu\mu}(tx).$
(50)

Given the metric, this can be integrated uniquely for the vielbeins (a fact
essentially guaranteed by the Picard-Lindelöf theorem Teschl  (§2.2)) using
the initial condition $e^{a}_{\mu}(0)=\delta^{a}_{\mu}$ (note that this
initial condition is consistent with (45)-(46)). This is the forward map,
taking metrics to vielbeins.

Metrics in normal coordinates satisfy (see (34))

$x^{\mu}g_{\mu\nu}(x)=x^{\nu}$ (51)

From (49) it is clear that vielbeins satisfying (45)-(46) map to metrics in
normal coordinates. The question is whether metrics in normal coordinates map
to vielbeins satisfying (45)-(47). (47) is easily shown by multiplying (50) by
$e_{b}^{\mu}(tx)$ and using the symmetry of the metric.

From (50) we have

$\frac{d}{dt}(tx^{\mu}e^{a}_{\mu}(tx))=x^{\mu}e^{a}_{\mu}(tx)$ (52)

which implies

$x^{\mu}e^{a}_{\mu}(tx)=\text{const.}=x^{a}$ (53)

proving (45).

(53) implies $x^{a}e^{\mu}_{a}(x)=x^{\mu}$, which together with (50) gives

$\frac{d}{dt}(tx^{a}e^{a}_{\mu}(tx))=x^{a}e^{a}_{\mu}(tx)$ (54)

which implies

$x^{a}e^{a}_{\mu}(tx)=\text{const.}=x^{\mu}$ (55)

proving (46).

We leave it as an simple exercise to the reader to convince themselves that
(49) and (50) are indeed inverses of each other when restricted to the
subspaces in question. ∎

##  8 Conclusions

Combining Theorems 1, 2 and 3 finally proves that

######  Theorem 4.

The space of curvatures satisfying the 1st and 2nd Bianchi identities for
curvature is bijective with the space of metrics in normal coordinates.

The proof is schematically represented in Fig 2.

###### Acknowledgements.

The author thanks the Rudolf Peierls Centre for their hospitality. For the
purpose of Open Access, the author has applied a CC BY public copyright
licence to any Author Accepted Manuscript version arising from this
submission.

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# Spectral topologies on skew braces

Themba Dube  (1) Department of Mathematical Sciences, University of South
Africa, P.O. Box 392, 0003 Pretoria, South Africa. (2) National Institute for
Theoretical and Computational Sciences (NITheCS), South Africa.
<EMAIL_ADDRESS>Amartya Goswami  (1) Department of Mathematics and Applied
Mathematics, University of Johannesburg, P.O. Box 524, Auckland Park 2006,
South Africa. (2) National Institute for Theoretical and Computational
Sciences (NITheCS), South Africa<EMAIL_ADDRESS>
###### Abstract.

Using a new definition of a prime ideal of a skew brace $A,$ on set
$\mathrm{Spec}\,A$ of prime ideals of $A$ we endow a spectral topology (in the
sense of Grothendieck). We characterize irreducible closed subsets of
$\mathrm{Spec}\,A$ and prove every irreducible closed subset of the space has
a unique generic point. We give a sufficient condition for the space to be
Noetherian. We study continuous maps between such spaces, and finally, we
prove that $\mathrm{Spec}(\mathrm{Idl}\,A)$ is a spectral space, where
$\mathrm{Idl}\,A$ is the set of all ideals of $A$.

###### Key words and phrases:

skew brace, prime ideal, irreducibility, spectral space

###### 1991 Mathematics Subject Classification:

16T25; 08A99

##  1\. Introduction

To study non-degenerate involutive set-theoretic solutions of the Yang–Baxter
equation, Rump introduced a new algebraic structure in [15] called a (left)
brace, which was generalized into noncommutative setting by Guarnieri &
Vendramin in [6] and called a (left) skew brace. Since the introduction of
skew braces, several algebraic properties of them have been studied. Jespers,
Kubat,Van Antwerpen, & Vendramin studied factorization of skew left braces
through strong left ideals in [10] and proved analogs of Itô’s theorem in the
context of skew left braces, whereas in [11], they obtained a Wedderburn type
decomposition for Artinian skew left braces and also proved analogues of a
theorem of Wiegold, a theorem of Schur and its converse. In [2], Cedó,
Smoktunowicz & Vendramin studied series of left ideals of skew left braces
that are analogs of upper central series of groups and using that they defined
left and right nilpotent skew left braces. The concepts like ideals, series of
ideals, prime, and semiprime ideals, Baer and Wedderburn radicals and
solvability for skew braces have been explored in [13] by Konovalov,
Smoktunowicz & Vendramin.

Skew braces are fairly new algebraic structures and much more study of their
algebraic aspects yet to come. Compared to algebraic properties studied so
far, little has been explored from the topological side. The aim of this paper
is to initiate that aspect and we do so by studying some of the topological
properties of the spectrum of prime ideals of a skew brace. It is worth it to
mention that in [13], it has been pointed out that “in the conference ‘Groups,
rings, and the Yang–Baxter equation,’ Spa, 2017, Louis Rowen suggested that it
could be interesting to study prime ideals of skew braces.” To meet our
requirements and to make analogy with the prime spectrum of a (commutative)
ring, we propose a new definition of prime ideals of a skew brace. We discuss
separation axioms and irreducubility. Using a skew brace homomorphism, we
obtain continuous maps between spaces of spectra and study properties of these
maps. Finally, we prove a result on spectrality. We end the paper with some
open questions.

##  2\. Preliminaries

A _skew left brace_ is a triple $\left(A,+,\circ\right)$ such that
$\left(A,+\right)$ and $\left(A,\circ\right)$ are groups, and the operations
satisfy the identity:

$a\circ\left(b+c\right)=a\circ b-a+a\circ c,$

for all $a,b,c$ in $A$, where $-a$ denotes the inverse of $a$ with respect to
the $+$ operation. By a skew brace, we will always mean a skew left brace. We
say that $\left(A,+\right)$ is the additive group of $A$, whereas
$\left(A,\circ\right)$ is the multiplicative group of $A$. The identity of the
additive group of $A$ coincides with that of the multiplicative group of $A,$
and we denote the common identity element by $e$. The multiplicative inverse
of an element $a$ in $A$ is denoted by
$a^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}$. We write $A$ as a skew
brace to mean the triple $\left(A,+,\circ\right)$.

Given a skew brace $A$, a new operation $*$ is defined as follows:

$a*b=-a+a\circ b-b,$

for all $a,b$ in $A$. In particular, $a*b=\lambda_{a}\left(b\right)-b$, for
all $a,b$ in $A$, where $\lambda_{a}$ is an automorphism of $(A,+)$ and is
defined by:

$\lambda_{a}(b)=-a+a\circ b.$

An _ideal_ $I$ of a skew brace $A$ is a normal subgroup of both the group
structures $(A,+)$ and $(A,\circ)$ such that $\lambda_{a}(I)\subseteq I$ for
all $a\in A$. We define the _trivial_ or _zero_ ideal of $A$ as the singleton
set $\\{e\\}$ of $A$ and denote it by $0$, whereas by a _proper_ ideal $I$, we
mean $I$ is an ideal of $A$ and $I\neq A.$ If $S$ is a subset of $A$, then
$\langle S\rangle$ denotes the ideal generated by $S$. A _maximal ideal_ $M$
is a proper ideal of $A$ and not properly containing in another proper ideal
of $A$. We denote the set of all ideals of $A$ by $\mathrm{Idl}\,A$, whereas
the set of all maximal of a skew brace $A$ is denoted by $\mathrm{Max}\,A$.
Involving the $*$ operation, a characterisation (see Cedó, Smoktunowicz, &
Vendramin [2, Lemma 1.8, Lemma 1.9]) of normal subgroups of any skew brace to
be an ideal is given in the following

######  Proposition 2.1.

Suppose $A$ is a skew brace and $I$ is a normal subgroup of
$\left(A,+\right)$. Then $I$ is an ideal of $A$ if and only if $A*I\subseteq
I$ and $I*A\subseteq I$.

######  Lemma 2.2.

If $\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ is a family of ideals of $A$, then
$\bigcap_{\lambda\in\Lambda}I_{\lambda}$ is also an ideal of $A$. If $I$ and
$J$ are ideals of $A$, so is $I+J,$ where $I+J$ is defined as the additive
subgroup of $A$ generated by $\\{i+j\mid i\in I,\,j\in J\\}.$

###### Proof.

Since $\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ is a family of ideals,
$\lambda_{a}(x)$ is in $\bigcap_{\lambda\in\Lambda}I_{\lambda}$ for all $a\in
A$ and for all $x\in\bigcap_{\lambda\in\Lambda}I_{\lambda}.$ Since arbitrary
intersection of normal subgroups is normal, it follows that
$\bigcap_{\lambda\in\Lambda}I_{\lambda}$ is an ideal of $A$. To prove that
$I+J$ is an ideal, we first observe that $I+J$ is a normal subgroup of
$(A,+)$. By Proposition 2.1, it is sufficient to show that $A*(I+J)\subseteq
I+J$ and $(I+J)*A\subseteq I+J$. To prove the first inclusion, let $a\in A$,
$i\in I$, $j\in J.$ Then we have

$\begin{split}a*(i+j)&=-a+a\circ(i+j)-i-j\\\ &=-a+a\circ i-a+a\circ j-i-j\\\
&=\lambda_{a}(i)+\lambda_{a}(j)-(i+j)\in I+J,\end{split}$

proving $A*(I+J)\subseteq I+J$. The proof of the containment $(I+J)*A\subseteq
I+J$ follows from Konovalov, Smoktunowicz, & Vendramin [13, Lemma 3.7]. ∎

The binary sum of ideals of a skew brace can be extended to an “infinite sum”,
which we shall see in the next corollary. But before that, let us first define
this infinite sum of ideals of a skew brace $A$. If
$\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ is a family of ideals of $A$, then by
the _sum_ $\sum_{\lambda\in\Lambda}I_{\lambda}$ of ideals of $A$, we mean each
term of the sum is over finitely many indices.

######  Corollary 2.3.

If $\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ is a family of ideals of $A$, then
$\sum_{\lambda\in\Lambda}I_{\lambda}$ is also an ideal of $A$.

######  Remark 2.4.

_$(1).$ Recall that an associative ring $(R,+,\circ)$ is called a _radical
ring_ if $(R,*)$ is a group, where the operation $*$ is defined by
$a*b=a+b-a\circ b.$ Rump [15] showed that $(R,+,*)$ is an example of a (left)
_brace_. It is now natural to ask what should be “ring-type” example of a skew
brace, and here the problem starts. Since the addition operation of a ring is
commutative, we can not expect an example even from a noncommutative ring. We
rather need to consider a near-ring and apply the above method to have the
expected example._

_$(2).$ Note that if a ring $R$ is a radical ring, then every element of $R$
is quasi-regular (see Jacobson [9]) and such a ring can not have a
multiplicative identity $1$. Indeed, for $x\in R$,_

$x*1=x+1-x\circ 1=x+1-x=1\neq 0,$

_a contradiction. Therefore, it follows that we can not have an “unital” brace
or a “unital” skew brace (unless the braces or the skew braces are trivial).
Furthermore, we observe that the identity element $e$ of a skew brace $A$ can
not play the role of an unital element. Otherwise, $A$ will not have any
proper ideals (as by definition $e\in I$ for all ideals $I$ of $A$)._

##  3\. Spectra of skew braces

There are various ways to define prime ideals of a skew brace. Here we
describe three of them.

######  Definition 3.1.

Let $I$ and $J$ be two ideals of a skew brace $A.$ A proper ideal $P$ of $A$
is called _prime_ if

$I*J\equiv 0\,(\mathrm{mod}\;\,P)\implies I\equiv
0\,(\mathrm{mod}\;\,P)\;\;\text{or}\;\;J\equiv 0\,(\mathrm{mod}\;\,P),$

where $I*J$ is the additive subgroup of $A$ generated by $\\{i*j\mid i\in
I,\,j\in J\\}.$

The Definition 3.1 above follows from Konovalov, Smoktunowicz, & Vendramin
[13, Definition 5.8], whereas Definition 3.2 below is based on Bourn,
Facchini, & Pompili [1, Proposition 3.18], and has been proposed to the second
author by Alberto Facchini in a private communication.

######  Definition 3.2.

_Let $I$ and $J$ be any two ideals of $A.$ A proper ideal $P$ of $A$ is called
_prime_ if_

$[I,J]\subseteq P\implies I\subseteq P\;\;\text{or}\;\;J\subseteq P,$

_where $[I,J]$ is the (Huq) commutator of the ideals $I$ and $J$ generated by
the union of the following three sets:_

  1. (i)

$\\{i+j-i-j\mid i\in I,j\in J\\};$

  2. (ii)

$\\{i\circ j\circ i^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}\circ
j^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}\mid i\in I,j\in J\\};$

  3. (iii)

$\\{i\circ j-j-i\mid\in I,j\in J\\},$

_where the sets in ( i) and (ii) respectively generate the commutators
$[I,J]_{(A,+)}$ and $[I,J]_{(A,\circ)}$ of the normal subgroups $I$ and $J$ of
the groups $(A,+)$ and $(A,\circ).$ _

Finally, we propose the following definition of a prime ideal of a skew brace.

######  Definition 3.3.

_A proper ideal $P$ of $A$ is called _prime_ if, for any subsets I and J of
A,_

$I*J\subseteq P\implies I\subseteq P\;\;\text{or}\;\;J\subseteq P,$

_where $I*J=\\{i*j\mid i\in I,\,j\in J\\}$. _

We denote the set of all prime ideals of $A$ by $\mathrm{Spec}\,A$. For the
rest of the work, by a prime ideal of a skew brace, we shall mean it in the
sense of Definition 3.3 and this definition helps us in proving the
preservation of prime ideals under inverse image of a skew brace homomorphism
(see Proposition 4.16(iii)). Although in Definition 3.3 we have considered $I$
and $J$ just as subsets of $A$, nevertheless, we have the following
implication, where with an abuse of notation, we shall also write $I*J$ even
when $I$ and $J$ are ideals of a skew brace $A$.

######  Proposition 3.4.

Suppose a skew brace $A$ is a skew brace and $P$ is a prime ideal of $A$. If
$I$ and $J$ are ideals of $A$, then $I*J\subseteq P$ implies $I\subseteq P$ or
$J\subseteq P,$ where $*$ is the operation as in Definition 3.3 and $I*J$ is
the ideal generated by the set $\\{i*j\mid i\in I,j\in J\\}.$

###### Proof.

Suppose $\hat{I}$ and $\hat{J}$ respectively denote the underlying sets of the
ideals $I$ and $J$. Then

$\hat{I}*\hat{J}\subseteq I*J\subseteq P.$

Since $P$ is prime, either $\hat{I}\subseteq P$ or $\hat{J}\subseteq P$, and
that implies either $I=\langle\hat{I}\rangle\subseteq P$ or
$J=\langle\hat{J}\rangle\subseteq P$. ∎

######  Lemma 3.5.

If $I$ and $J$ are any two ideals of a skew brace $A$, then $I*J\subseteq
I\cap J.$

###### Proof.

If $i*j\in I*J,$ then $i*j=\lambda_{i}(j)-j\in J$ as $J$ is an ideal. On the
other hand,

$i*j=-i+i\circ
j-j=-i+j\circ\left(j^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}\circ
i\circ
j\right)-j=-i+j+\lambda_{j}\left(j^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}\circ
i\circ j\right)-j\in I.\qed$

######  Proposition 3.6.

Suppose $A$ is a skew brace. Then $(\mathrm{Idl}\,A,\subseteq,*)$ is a
multiplicative lattice.

###### Proof.

Notice that under subset inclusion $\subseteq$ relations, the emptyset and $A$
are respectively the bottom and top elements of $\mathrm{Idl}\,A$. That the
pair $(\mathrm{Idl}\,A,\subseteq)$ is a complete lattice now follows from
Lemma 2.2 and Corollary 2.3, whereas by Lemma 3.5 it follows that the complete
lattice $(\mathrm{Idl}\,A,*)$ is also multiplicative. ∎

In a skew brace, maximal ideals need not be prime. For example, in a finite
two sided brace $A$ with more than one element, each maximal ideal of $A$
cannot be prime. What we have instead is the following characterisation (see
Jespers, Kubat, Van Antwerpen, & L. Vendramin [11, Proposition 3.6]) of
maximal ideals that are prime.

######  Proposition 3.7.

A maximal ideal $M$ of $A$ is prime if and only if $A^{\scriptscriptstyle
2}\nsubseteq M,$ where $A^{\scriptscriptstyle 2}=A*A.$

Since the operation $*$ is neither associative nor commutative, an expression
like $a^{\scriptscriptstyle n}=a*a^{\scriptscriptstyle n-1}$ does not make
sense for us. Therefore, we cannot adapt the “elementwise” definition of
radical of an ideal as we have for commutative rings. Following the definition
of the radical of an ideal of a noncommutative ring (see Lam [14, Theorem
10.7]), we propose

######  Definition 3.8.

_The _radical_ of an ideal $I$ in a skew brace $A$, denoted by
$\mathrm{Rad}\,I$, is defined as_

$\mathrm{Rad}\,I=\bigcap_{I\subseteq P\,\in\,\mathrm{Spec}\,A}P.$

_The _nil radical_ $\mathrm{Nil}\,A$ of $A$ is the radical of the zero ideal
(see Section 2) of $A$. So, $\mathrm{Nil}\,A$ is the intersection of all prime
ideals of $A$._

##  4\. Spectral topology

Let $S$ be a subset of a skew brace $A$. If $H(S)=\\{P\in\mathrm{Spec}\,A\mid
S\subseteq P\\},$ then it is easy to see that $H(S)=H(\langle S\rangle).$
Therefore, it is sufficient to consider the set of all ideals of a skew brace
brace $A$ to define closed sets of spectral topology on $\mathrm{Spec}\,A$.
The terminology “spectral topology” was first used by Grothendieck in [5].

######  Definition 4.1.

_A _spectral_ topology on $\mathrm{Spec}\,A$ is imposed by considering the
collection of sets $\\{H(I)\\}_{I\in\mathrm{Idl}\,A}$ as closed sets, where_

$H(I)=\\{P\in\mathrm{Spec}\,A\mid I\subseteq P\\}.$

The following proposition shows that Definition 4.1 indeed makes sense for
skew braces.

######  Proposition 4.2.

The collection $\\{H(I)\\}_{I\in\mathrm{Idl}\,A}$ of sets satisfies the
following properties:

  1. (i)

$H(A)=\emptyset$ and $H(0)=\mathrm{Spec}\,A.$

  2. (ii)

$H(I)\cup H(J)=H(I\cap J)=H(I*J)$ for all $I,J\in\mathrm{Idl}\,A.$

  3. (iii)

$\bigcap_{\lambda\in\Lambda}H(I_{\lambda})=H\left(\sum_{\lambda\in\Lambda}I_{\lambda}\right)$
for all $I_{\lambda}\in\mathrm{Idl}\,A$ and $\lambda\in\Lambda.$

###### Proof.

(i). Since by Definition 3.3, $A\notin\mathrm{Spec}\,A,$ we have
$H(A)=\emptyset.$ Since $\\{e\\}\subseteq P$ for all $P$ in
$\mathrm{Spec}\,A$, we have $H(0)=\mathrm{Spec}\,A.$

(ii). By Definition 4.1, it follows that $H(I)\cup H(J)\subseteq H(I\cap J)$
for any two ideals $I$ and $J$ of $A.$ On the other hand, $H(I)\cup
H(J)\supseteq H(I\cap J)$ follows by Lemma 3.5. If $P\in H(I*J),$ then
$I*J\subseteq P$, and by Proposition 3.4, $I\subseteq P$ or $J\subseteq P,$
and that gives $P\in H(I)\cup H(J)$. The inclusion $H(I*J)\supseteq H(I)\cup
H(J)$ follows from Lemma 3.5.

(iii). Note that if $\\{I_{\lambda}\\}_{\lambda\in\Lambda}$ is a family of
ideals of $A$, then by Corollary 2.3, $\sum_{\lambda\in\Lambda}I_{\lambda}$ is
an ideal of $A$. By Definition 4.1 we also have
$\bigcap_{\lambda\in\Lambda}H(I_{\lambda})\supseteq
H\left(\sum_{\lambda\in\Lambda}I_{\lambda}\right).$ Conversely, if $P$ is in
$\bigcap_{\lambda\in\Lambda}H(I_{\lambda}),$ then $I_{\lambda}\subseteq P$ for
all $\lambda\in\Lambda$, and hence
$\sum_{\lambda\in\Lambda}I_{\lambda}\subseteq P.$ This proves that
$\bigcap_{\lambda\in\Lambda}H(I_{\lambda})\subseteq
H\left(\sum_{\lambda\in\Lambda}I_{\lambda}\right).$ ∎

######  Lemma 4.3.

If $I\in\mathrm{Idl}\,A$, then $H(I)=H(\mathrm{Rad}\,I).$

###### Proof.

The proof follows by repeated applications of Definition 3.8. ∎

######  Theorem 4.4.

For a subset $S$ of $\mathrm{Spec}\,A$, let $K(S)=\bigcap_{I\in S}I.$ The
operator $K$ has the following properties.

  1. (i)

$K(\emptyset)=R$ and
$K\left(\bigcup_{\lambda\in\Lambda}T_{\lambda}\right)=\bigcap_{\lambda\in\Lambda}K\left(T_{\lambda}\right).$

  2. (ii)

If $T$ is a subset of $X$ and $I$ is an ideal of $R$ then
$KH(I)=\mathrm{Rad}\,I,$ and $HK(S)$ is the closure of $S$ in
$\mathrm{Spec}\,A$.

###### Proof.

(i). The first assertion follows from the empty intersection property. For the
second, the fact $S_{\lambda}\subseteq\bigcup_{\lambda\in\Lambda}S_{\lambda}$
implies $K(S_{\lambda})\supseteq K(\bigcup_{\lambda\in\Lambda}S_{\lambda}),$
and hence $\bigcap_{\lambda\in\Lambda}K(S_{\lambda})\supseteq
K(\bigcup_{\lambda\in\Lambda}S_{\lambda}).$ For the other half of the
inclusion, let $S_{\lambda}=\\{I_{\alpha,\lambda}\\}_{\alpha\in L}$ and let
$x\in\bigcap_{\lambda\in\Lambda}K(S_{\lambda})$. Then
$x\in\bigcap_{\lambda\in\Lambda}(\bigcap_{\alpha\in L}I_{\alpha,\lambda});$
whence $x\in K(\bigcup_{\lambda\in\Lambda}S_{\lambda}).$

(ii). For the first assertion we observe that

$a\in KH(I)\Leftrightarrow a\in P\;\text{for every}\;P\;\text{with}\;P\in
H(I)\Leftrightarrow a\in P\;\text{for every}\;P\supseteq I.$

Therefore, $KH(I)=\bigcap_{P\supseteq I}=\mathrm{Rad}\,I.$ For the second
claim, if a closed set $H(X)$ (for some subset $X$ of $A$) contains $S,$ then
$X\subseteq I$ for all $I\in S;$ which subsequently implies $X\subseteq K(S)$
and hence $H(X)\supset HK(S)$. Since $T\subseteq HK(S),$ and $HK(S)$ is the
smallest closed set of $X$ containing $S$, we have the desired claim. ∎

Following Kock [12], we can represent the relation between $H$ and $K$
categorically as follows: We observe that the poset map $K$ is a right adjoint
of the map $H$. The unit of the adjunction $H\dashv K$ is
$\eta\colon\left\langle S\right\rangle\mapsto KH(S)=\left\langle
S\right\rangle.$ Hence the full subcategory
$\mathrm{Fix}\,\eta=\\{S\in\mathfrak{P}(A)\mid\eta_{S}\;\text{is an
isomorphism}\\}$ is the set of ideals of $R$. The counit of the adjunction
$H\dashv K$ is $\epsilon\colon T\mapsto HK(T)=\overline{T}.$ Therefore
$\mathrm{Fix}\,\epsilon=\\{T\in\mathfrak{P}(\mathrm{Spec}\,A)^{\text{op}}\mid\epsilon_{T}\;\text{is
an isomorphism}\\}$ is the set of closed subsets. Therefore the adjunction
$H\dashv K$ restricts to an adjoint equivalence between categories
$\mathrm{Idl}\,A$ and $\mathrm{Closed}(\mathrm{Spec}\,A).$ By considering open
sets of the topology on $\mathrm{Spec}\,A$, the above isomorphism of
categories become $\mathrm{Idl}\,A\approx\mathrm{Open}(\mathrm{Spec}\,A).$ The
following diagram summarises the above inter-relations.

$\textstyle{\mathfrak{P}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{H}$$\textstyle{\mathfrak{P}(\mathrm{Spec}\,A)^{\text{op}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{K}$$\scriptstyle{\bot}$$\textstyle{\mathrm{Fix}\,\eta=}$$\textstyle{\mathrm{Idl}\,A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Closed}(\mathrm{Spec}\,A)^{\text{op}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\approx}$$\textstyle{=\mathrm{Fix}\,\epsilon.}$

######  Lemma 4.5.

Suppose $A$ is a skew brace and $P,Q\in\mathrm{Spec}\,A$. Then

  1. (i)

$\mathrm{cl}{\\{P\\}}=H(P)$.

  2. (ii)

$P\in\mathrm{cl}{\\{Q\\}}$ if and only if $P\in H(Q)$.

###### Proof.

(i). Since $H(P)$ is a closed set containing $P$ and $\mathrm{cl}{\\{P\\}}$ is
the smallest closed set containing $P$, we immediately have
$\mathrm{cl}{\\{P\\}}\subseteq H(P).$ For the converse, suppose
$\mathrm{cl}{\\{P\\}}=H(I)$ for some ideal $I$ of $A$. Clearly,
$P\in\mathrm{cl}{\\{P\\}}$, hence $I\subseteq P$. Thus any prime ideal that
contains $P$ will also contain $I.$ Hence, $H(P)\subseteq
H(I)=\mathrm{cl}{\\{P\\}}$.

(ii). Observe that by (i), $P\in\mathrm{cl}{\\{Q\\}}$ if and only if $P\in
H(Q)$, and this occurs if and only if $Q\subseteq P$. ∎

######  Proposition 4.6.

Every $\mathrm{Spec}\,A$ is a $T_{0}$-space.

###### Proof.

Let $P$ and $Q$ be two distinct points of $\mathrm{Spec}\,A$ such that
$P\nsubseteq Q.$ By Lemma 4.5 we have $Q\notin\mathrm{cl}{\\{P\\}}=H(P)\ni P,$
which implies there exists a closed set $H(P)$ containing $P$ that does not
contain $Q$. ∎

With a further restriction on a skew brace, we obtain the following

######  Proposition 4.7.

If $A$ is a skew brace with $A^{\scriptscriptstyle 2}\nsubseteq M$ for all
$M\in\mathrm{Max}\,A,$ then $\mathrm{Spec}\,A$ is a $T_{1}$-space if and only
if $\mathrm{Spec}\,A=\mathrm{Max}\,A$.

###### Proof.

By Theorem 3.7 it follows that $\mathrm{Spec}\,A\supseteq\mathrm{Max}\,A$. Let
$\mathrm{Spec}\,A\neq\mathrm{Max}\,A$. Then there exists
$P\in\mathrm{Spec}\,A$ such that $P\notin\mathrm{Max}\,A$, and that implies
there exists $M\in\mathrm{Max}\,A$ such that $P\subsetneq M$. Then $M\in
H(P)=\mathrm{cl}{\\{P\\}}$, where the equality follows from Lemma 4.5(i).
Hence $\\{P\\}$ is not closed and hence $\mathrm{Spec}\,A$ is not a
$T_{1}$-space. Conversely, suppose that $\mathrm{Spec}\,A=\mathrm{Max}\,A$.
Then $H(P)=\\{P\\}$ for all $P\in\mathrm{Spec}\,A$ and hence every singleton
set is closed, and that implies $\mathrm{Spec}\,A$ is a $T_{1}$-space. ∎

Let us recall the notion of irreducible topological spaces and some of their
properties. Readers may consult Görtz & Wedhorn [7] for further details on
irreducibility.

######  Definition 4.8.

_If $X$ is a topological space, a closed subset $S$ is _irreducible_ if $S$ is
not the union of two properly smaller closed subsets $S_{1},S_{2}\subsetneq
S.$ A maximal irreducible subset of a topological space $X$ is called an
_irreducible component_ of $X.$ A point $x$ in a closed subset $S$ is called a
_generic point_ of $S$ if $S=\mathrm{cl}(\\{x\\}),$ the closure of $\\{x\\}$._

Recall the following basic properties on irreducibility.

######  Proposition 4.9.

Let $X$ be a topological space.

  1. (i)

A subspace $Y$ of $X$ is irreducible if and only if $\mathrm{cl}(Y)$ is
irreducible.

  2. (ii)

The irreducible components of $X$ are closed and cover $X$.

######  Lemma 4.10.

$\\{H(P)\\}_{P\in\mathrm{Spec}\,A}$ are precisely the irreducible closed
subsets of $\mathrm{Spec}\,A$.

###### Proof.

Let $P\in\mathrm{Spec}\,A$. Since $\\{P\\}$ is irreducible, by Proposition
4.9(i) and Lemma 4.5(i), so is $H(P).$ Let $H(I)$ be an irreducible closed
subset of of $\mathrm{Spec}\,A$ and if possible, $I\notin\mathrm{Spec}\,A.$
This implies there exist $I_{1}$ and $I_{2}$ such that $I_{\scriptscriptstyle
1}\nsubseteq I$ and $I_{\scriptscriptstyle 2}\nsubseteq I$, but
$I_{\scriptscriptstyle 1}*I_{\scriptscriptstyle 2}\subseteq I$. Then we have:

$H(\langle I,I_{\scriptscriptstyle 1}\rangle)\cup H(\langle
I,I_{\scriptscriptstyle 2}\rangle)=H(\langle I,I_{\scriptscriptstyle
1}\rangle\cap\langle I,I_{\scriptscriptstyle 2}\rangle)=H(\langle
I,I_{\scriptscriptstyle 1}*I_{\scriptscriptstyle 2}\rangle)=H(I),$

where the first two equalities follow from Proposition 4.2(ii). Since
$H(\langle I,I_{\scriptscriptstyle 1}\rangle)\neq H(I)$ and $H(\langle
I,I_{\scriptscriptstyle 2}\rangle)\neq H(I),$ the closed set $H(I)$ is not
irreducible, a contradiction. ∎

######  Theorem 4.11.

Every irreducible closed subset of $\mathrm{Spec}\,A$ has a unique generic
point.

###### Proof.

The existence of generic point follows from Lemma 4.5 and Lemma 4.10, whereas
uniqueness part follows from Proposition 4.6. ∎

Note that if we have a decreasing chain of prime ideals, then the intersection
of the ideals in the chain gives the minimal element, which is either a prime
ideal or the zero ideal, and in the second case, with a vacuous argument, it
is also a prime ideal. This confirms the existence of the minimal prime ideals
of $A$ and hence allows us to obtain the following

######  Lemma 4.12.

The irreducible components of $\mathrm{Spec}\,A$ are the closed sets $H(P)$,
where $P$ is a minimal prime ideal of $A$.

###### Proof.

By Proposition 4.9(ii), irreducible components of $\mathrm{Spec}\,A$ are
closed. If $P$ is a minimal prime ideal, then by Lemma 4.10, $H(P)$ is
irreducible. If $H(P)$ is not a maximal irreducible subset of
$\mathrm{Spec}\,A$, then there exists a maximal irreducible subset $H(Q)$ with
$Q\in\mathrm{Spec}\,A$ such that $H(P)\subsetneq H(Q)$. This implies that
$P\in H(Q)$ and hence $Q\subsetneq P$, contradicting the minimality property
of $P$. ∎

######  Proposition 4.13.

$\mathrm{Spec}\,A$ is irreducible if and only if the $\mathrm{Nil}\,A$ is
prime.

###### Proof.

If $\mathrm{Nil}\,A$ is prime, then it is the minimum prime ideal. Therefore,
$\mathrm{Spec}\,A$ has only one maximal irreducible component, namely
$H(\mathrm{Nil}\,A),$ and since by Proposition 4.9(ii) the maximal irreducible
components cover $\mathrm{Spec}\,A,$ the irreducible component
$H(\mathrm{Nil}\,A)$ must be equal to $\mathrm{Spec}\,A$. Hence
$\mathrm{Spec}\,A$ is irreducible by Lemma 4.12. Conversely, suppose
$\mathrm{Spec}\,A$ is irreducible. Then by Proposition 4.9(ii), there exists
$P\in\mathrm{Spec}\,A$ such that $H(P)=\mathrm{Spec}\,A.$ Hence $P\subseteq Q$
for all $Q\in\mathrm{Spec}\,A,$ and so is $\mathrm{Nil}\,A,$ i.e.,
$P\subseteq\mathrm{Nil}\,A$. By Definition 3.8, $\mathrm{Nil}\,A\subseteq P$,
and hence $\mathrm{Nil}\,A$ is prime. ∎

Before we discuss continuous maps, let us first recall some of the basic facts
about skew brace homomorphisms and quotient skew braces.

######  Definition 4.14.

_A _skew subbrace_ $S$ of a skew brace $A$ is a subgroup of $(A,+)$ and
$(A,\circ).$ If $A$ and $B$ are skew braces, a _skew brace homomorphism_ is a
map $f\colon A\to B$ of $A$ into $B$ such that
$f(a+a^{\prime})=f(a)+f(a^{\prime}),$ $f(a\circ a^{\prime})=f(a)\circ
f(a^{\prime})$ for all $a,$ $a^{\prime}\in A$. The _kernel_ of $f$ is the
subset $\mathrm{ker}\,f=\\{a\in A\mid f(a)=e\\}$ of $A$. The _image_ of a skew
brace homomorphism $f$ is the subset $\mathrm{im}\,f=\\{b\in B\mid
b=f(a)\;\text{for some}\;a\in A\\}$ of $B$. Let $f\colon A\to B$ be a skew
brace homomorphism and let $I,$ $J$ be ideals of $B$ and $A$ respectively. The
_contraction_ of $I$ and the _extension_ of $J$ are respectively denoted by
$I^{c}$ and $J^{e}$._

######  Lemma 4.15.

Let $I$ and $J$ respectively be ideals of skew left braces $A$ and $B.$ Let
$f\colon A\to B$ be a skew left brace homomorphism. Then

  1. (i)

$\mathrm{ker}\,f,$ $f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(J)$ are
ideals of $A$ and $\mathrm{im}\,f$ is a skew subbrace of $B;$

  2. (ii)

$a\circ I=a+I$ for every $a\in A$ and $A/I=\\{a+I\mid a\in A\\}$ is a skew
left brace under the operation:
$(a+I)+(b+I)=(a+b)+I\;\;\text{and}\;\;(a+I)\circ(b+I)=(a\circ b)+I;$

  3. (iii)

there is a bijection between ideals of $A/I$ and ideals of $A$ containing $I;$

  4. (iv)

$f(I*J)=f(I)*f(J)$ for all $I,J\in\mathrm{Idl}\,A.$

######  Proposition 4.16.

Suppose $f\colon A\to B$ is a skew brace homomorphism and define the map
$f^{!}\colon\mathrm{Spec}\,B\to\mathrm{Spec}\,A$ by
$f^{!}(P)=f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(P)$, where
$P\in\mathrm{Spec}\,B.$ Then

  1. (i)

every prime ideal of $A$ is a contracted ideal if and only if $f^{!}$ is
surjective;

  2. (ii)

if every prime ideal of $B$ is an extended ideal then $f^{!}$ is injective;

  3. (iii)

$f^{!}$ is continuous;

  4. (iv)

if $f$ is surjective, then $\mathrm{im}(f^{!})$ is homeomorphic to
$H(\mathrm{ker}\,f)$;

  5. (v)

the image of $f^{!}$ is dense in $\mathrm{Spec}\,A$ if and only if
$\mathrm{ker}\,f\subseteq\mathrm{Nil}\,A.$

###### Proof.

(i). Let $P\in\mathrm{Spec}\,A$ and $P=Q^{c}$ for some $Q\in\mathrm{Spec}\,B.$
Then, $f^{!}(Q)=f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(Q)=P.$ Hence
$f^{!}$ is surjective. Conversely, if $f^{!}$ is surjective, then for any
$P\in\mathrm{Spec}\,A,$ we have $P=f^{!}(Q)=Q^{c}$, as required.

(ii). Let $Q\in\mathrm{Spec}\,B$ and $Q=P^{e}.$ Then,
$f^{!}(Q)=P^{ec}\supseteq P.$ Suppose $f^{!}(Q)=f^{!}(Q^{\prime}).$ Then,
$P^{ec}=(P^{\prime})^{ec}$, where $Q^{\prime}=(P^{\prime})^{e},$ this implies
$P^{ece}=(P^{\prime})^{ece},$ which implies $P^{e}=(P^{\prime})^{e}$ and that
$Q=Q^{\prime}.$

(iii). To show $f^{!}$ is continuous, we first show that
$f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(P)\in\mathrm{Spec}\,A,$
whenever $P\in\mathrm{Spec}\,B$. Let $I*J\subseteq
f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(P),$ where
$I,J\in\mathrm{Idl}\,A.$ Then $f(I*J)\subseteq P,$ and by Lemma 4.15(iv) we
have $f(I)*f(J)\subseteq P$. Since $P$ is prime, by Definition 3.3 either
$f(I)\subseteq P$ or $f(J)\subseteq P,$ and that implies either $I\subseteq
f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(P)$ or $J\subseteq
f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(P)$. If $H(I)$ is a closed
subset of $\mathrm{Spec}\,A$, then for any $Q\in\mathrm{Spec}\,B,$ we have:

$\displaystyle
Q\\!\in\\!(f^{!})^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(H(I))\\!\Leftrightarrow\\!f^{!}(Q)\in
H(I)\\!\Leftrightarrow\\!f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(Q)\\!\in\\!H(I)\\!\Leftrightarrow\\!I\subseteq
f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(Q)\\!\Leftrightarrow\\!Q\\!\in\\!H(\langle
f(I)\rangle).$

(iv). By Lemma 4.15(i), $\mathrm{im}(f^{!})$ and $\mathrm{ker}\,f$ are ideals
of $A$. We first show that
$\mathrm{cl}(f^{!}(H(I)))=H(f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(I))$
for any $I\in\mathrm{Idl}\,B$ and for that it is sufficient to show:
$f^{!}(H(I))=H(f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(I))$. We
observe:

$P\in H(f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(I))\Leftrightarrow
f(f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(I))\subseteq
f(P)\Leftrightarrow
P=f^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}(f(P))=f^{!}(f(P))\in
f^{!}(H(I)),$

and by taking $I=0$, we obtain $\mathrm{im}(f^{!})=H(\mathrm{ker}\,f).$ This
implies $f^{!}$ induces a continuous bijection between $\mathrm{im}(f^{!})$
and $H(\mathrm{ker}\,f).$ The continuity of
$(f^{!})^{\raisebox{0.60275pt}{$\scriptscriptstyle-1$}}$ also now follows from
the above. Hence we have the desired homeomorphism.

(v). Note that $\mathrm{im}(f^{!})$ is dense in $\mathrm{Spec}\,A$ if and only
if $\mathrm{cl}(\mathrm{im}\,f^{!})=H(\mathrm{ker}\,f)=\mathrm{Spec}\,A$, and
this occurs if and only if $\mathrm{ker}\,f\subseteq P$ for all
$P\in\mathrm{Spec}\,A,$ and that holds if and only if
$\mathrm{ker}\,f\subseteq\mathrm{Nil}\,A$. ∎

######  Corollary 4.17.

For a skew brace $A$, the spaces $\mathrm{Spec}\,A$ and
$\mathrm{Spec}(A/\mathrm{Nil}\,A)$ are homeomorphic.

###### Proof.

Note that by Lemma 4.15(ii), $A/\mathrm{Nil}\,A$ is a skew brace. On applying
Proposition 4.16(iv) on the homomorphism $A\to A/\mathrm{Nil}\,A$ we obtain
the desired result. ∎

######  Proposition 4.18.

Let $I$ be an ideal of a skew brace $A$ and $J=I^{c}$ be an ideal of a skew
brace $A^{\prime}.$ Let $\bar{\phi}\colon A/I\to A^{\prime}/J$ be the skew
brace homomorphism induced by the skew brace homomorphism $f\colon A\to
A^{\prime}.$ Then the restriction of the map
$\phi_{*}\colon\mathrm{Spec}\,A^{\prime}\to\mathrm{Spec}\,A$ to $H(J)$ is the
map $\bar{\phi}_{*}\colon\mathrm{Spec}(A^{\prime}/J)\to\mathrm{Spec}(A/I).$

###### Proof.

Note that below the left commutative diagram induces the right commutative
diagram.

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\scriptstyle{\pi^{I}}$$\textstyle{A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{J}}$$\textstyle{A/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\bar{\phi}}$$\textstyle{A^{\prime}/J}$
$\textstyle{\mathrm{Spec}\,A}$$\textstyle{\mathrm{Spec}\,A^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{*}}$$\textstyle{\mathrm{Spec}(A/I)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{I}_{*}}$$\scriptstyle{\bar{\phi_{*}}}$$\textstyle{\mathrm{Spec}(A^{\prime}/J).\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{J}_{*}}$

Moreover, $K\in H(J)$ implies $\phi_{*}(K)=K^{c}\supseteq
J^{c}=I^{ec}\supseteq I,$ that is, $\phi_{*}(H(J))\subseteq H(I).$ Now the
desired result follows from Proposition 4.16(iv). ∎

######  Definition 4.19.

_According to Jespers, Kubat, Van Antwerpen, & Vendramin [11, Definition 4.1]
the weight $\omega(A)$ of non-zero skew brace $A$ is defined as the minimal
number of elements of A needed to generate $A$ (as an ideal). By convention,
we put $\omega(A)=1$ if $A=0.$ _

Recall that a skew left brace A is said to be _Noetherian_ if every ascending
chain of ideals of A is eventually stationary. It follows immediately that a
skew left brace is Noetherian if and only if all its ideals have finite
weight. Also recall that a topological space $X$ is said to be _Noetherian_ if
the closed subsets of $X$ satisfy the descending chain condition.

######  Proposition 4.20.

If all ideals of $A$ have finite weight, then $\mathrm{Spec}\,A$ is a
Noetherian space.

###### Proof.

It suffices to show that a collection of closed sets in $\mathrm{Spec}\,A$
satisfy descending chain condition. Let $H(I_{1})\supseteq
H(I_{2})\supseteq\cdots$ be a descending chain of closed sets in
$\mathrm{Spec}\,A$. Then by Lemma 4.3, $I_{1}\subseteq I_{2}\subseteq\cdots$
is an ascending chain of ideals in $A.$ Since $A$ is Noetherian, it stabilizes
at some $n\in\mathds{N}.$ Hence, $H(I_{n})=H(I_{n+k})$ for any $k.$ Thus
$\mathrm{Spec}\,A$ is Noetherian. ∎

######  Corollary 4.21.

The set of minimal prime ideals in a Noetherian skew brace is finite.

###### Proof.

By Proposition 4.20, $\mathrm{Spec}\,A$ is Noetherian, thus $\mathrm{Spec}\,A$
has finitely many irreducible components. By Lemma 4.12, every irreducible
closed subset of $\mathrm{Spec}\,A$ is of the form $H(P),$ where $P$ is a
minimal prime ideal. Thus $H(P)$ is irreducible components if and only if $P$
is minimal prime. Hence, $A$ has only finitely many minimal prime ideals. ∎

The following well-known result holds for any topological space and hence for
$\mathrm{Spec}\,A$.

######  Proposition 4.22.

If $A$ is a skew brace, then the following are equivalent:

  1. (i)

$\mathrm{Spec}\,A$ is Noetherian.

  2. (ii)

Every open subspace of $\mathrm{Spec}\,A$ is compact.

  3. (iii)

Every subspace of $\mathrm{Spec}\,A$ is compact.

Recall that in the sense of Hochster [8], a topological space is called
_spectral_ if it is quasi-compact, sober, admits a basis of quasi-compact open
subspaces that is closed under finite intersections.

######  Theorem 4.23.

$\mathrm{Spec}(\mathrm{Idl}\,A)$ is a spectral space.

###### Proof.

Note that by Proposition 3.6, $\mathrm{Idl}\,A$ is a multiplicative lattice.
Since every skew brace $A$ is finitely generated (by $e$) as an ideal of
itself, by Facchini, Finocchiaro, & Janelidze [4, Theorem 11.5] we have the
desired claim. ∎

######  Concluding remarks 1.

_The following are some natural topological questions that arise when we study
spectral topology on a prime spectrum of an algebraic structure and we do not
have answers to these questions for skew braces._

_$\bullet$ Although the partition of unity property is a sufficient condition
for compactness of prime spectrum of a ring with identity (see Dube & Goswami
[3]), we can not apply the same argument for skew braces. In the context of a
skew brace $A$, when we try to apply the finite intersection property,
although we obtain a finite sum representation of the identity $e$ of $A$, but
from that we cannot conclude the same for an arbitrary element of $A$._

_$\bullet$ If $a$ is an element of a skew brace $A$ such that $a^{2}=a$, then
we immediately see that $a=e$. This means a skew brace does not have
nontrivial idempotent elements. For a commutative ring $R$ (with identity),
without having any nontrivial idempotetnt elements implies $\mathrm{Spec}\,R$
is connected. Therefore, one may expect the same for a skew brace. Once again
not having zero divisors and “multiplicative identity” in a skew brace lead to
the problem of checking the connectivity._

_$\bullet$ Hochster [8] proved that a prime spectra of a commutative ring
(with 1) endowed with spectral topology is spectral. We proved
$\mathrm{Spec}(\mathrm{Idl}\,A)$ is spectral (see Theorem 4.23), but it could
be interesting to know whether $\mathrm{Spec}\,A$ itself is spectral._

_$\bullet$ Proposition 4.20 gives a sufficiant condition for
$\mathrm{Spec}\,A$ to be Noetherian. But it would be nice to characterize
Notherian spaces as well as obtain a similar result for Artinian spaces._

## Acknowledgement

The second author wishes to express his gratitude to Leandro Vendramin and
Paola Stefanelli for some fruitful discussions.

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