# Dynamic and Thermodynamic Stability of Charged Perfect Fluid Stars

Kai Shi1, Yu Tian2, Xiaoning Wu3, Hongbao Zhang1, Jingchao Zhang 1
1Department of Physics, Beijing Normal University, Beijing 100875, China
2 School of Physics, University of Chinese Academy Sciences, Beijing 100049,
China
3 Institute of Mathematics, Chinese Academy of Sciences, Beijing 100190, China
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###### Abstract

We perform a thorough analysis of the dynamic and thermodynamic stability for
the charged perfect fluid star by applying the Wald formalism to the
Lagrangian formulation of Einstein-Maxwell-charged fluid system. As a result,
we find that neither the presence of the additional electromagnetic field nor
the Lorentz force experienced by the charged fluid makes any obstruction to
the key steps towards the previous results obtained for the neutral perfect
fluid star. Therefore, the criterion for the dynamic stability of our charged
star in dynamic equilibrium within the symplectic complement of the trivial
perturbaions with the ADM $3$-momentum unchanged is given by the non-
negativity of the canonical energy associated with the timelike Killing field,
where it is further shown for both non-axisymmetric and axisymmetric
perturbations that the dynamic stability against these restricted
perturbations also implies the dynamic stability against more generic
perturbations. On the other hand, the necessary condition for the
thermodynamic stability of our charged star in thermodynamic equilibrium is
given by the positivity of the canonical energy of all the linear on-shell
perturbations with the ADM angular momentum unchanged in the comoving frame,
which is equivalent to the positivity of the canonical energy associated with
the timelike Killing field when restricted onto the axisymmetric
perturbations. As a by-product, we further establish the equivalence of the
dynamic and thermodynamic stability with respect to the spherically symmetric
perturbations of the static, spherically symmetric isentropic charged star.

##  1 Introduction

Partly motivated by the Gubser-Mitra conjecture on the relationship between
the dynamic instability and thermodynamic instability of the black branes in
AdS spacetimes[1, 2], Hollands and Wald established a criterion for the
dynamic stability of black holes as well as its close connection with the
thermodynamic stability in [3] by working with the Wald formalism[4, 5, 6].
Such a strategy was further extended successfully to the relativistic neutral
perfect fluid stars[7], where not only were the dynamic and thermodynamic
stability analysed in a transparent and unified manner, demonstrating its very
advantage over the case by case explicit calculations[8, 9, 10, 11], but also
some new observations were made for the axisymmetric perturbations, compared
to the previous explorations[12, 13, 14, 15, 16].

Although most of the real life stars are believed to be nearly electrically
neutral111Among other possible exceptions, quark stars are suspected to be
generically charged[17]., it is at least of academic interest to inquire about
whether the above results for the neutral perfect fluid stars also hold for
the charged perfect fluid stars. Moreover, it is noteworthy that such a
charged perfect fluid star in an AdS spacetime also plays an important role in
addressing the holographic metallic criticality[18]. But nevertheless, there
is no a priori answer to the above inquiry, since the electromagnetic field as
well as the Lorentz force acting on the charged perfect fluid kicks in the
game. The purpose of this paper is intended to provide a definite answer by
performing a comprehensive and detailed analysis of the dynamic and
thermodynamic stability of the charged perfect fluid star in an asymptotically
flat spacetime. As a result, we find that neither the presence of the
electromagnetic field nor the electromagnetic force experienced by the charged
fluid makes any obstruction to the key steps towards the results obtained in
[7] for the neutral star. To this end, we are especially required to derive an
explicit expression of the canonical energy at null infinity for the
electromagnetic part and argue for its non-negativity as well as its zero
value for the physically stationary perturbations. Among others, we are also
required to show that the angular momentum per particle, vorticity, and
circulation share the same properties as those for the neutral perfect fluid
stars, although their expressions get modified in the presence of the
electromagnetic field. Thus we are able to obtain exactly the same results for
the dynamic and thermodynamic stability of our charged stars as those
presented in [7] for the neutral stars. In passing, not only do we provide
more detailed derivations and arguments in some places where they are only
sketched or completely skipped in [7], but also we present a thorough
alternative derivation for Eq. (80) to that suggested in [7].

The rest of this paper is structured as follows. In the next section, we shall
provide a brief review of the dynamics of the charged perfect fluid, the
definition of the resulting star in dynamic equilibrium, and the criterion for
its thermodynamic equilibrium as well as the corresponding first law of
thermodynamics. In Section 3, we devote ourselves to developing the Lagrangian
formulation of the Einstein-Maxwell-charged fluid system. To this end, we
first set up the dynamical fields for our system, pin down the associated
redundancies, define the Eulerian and Lagrangian variations and point out
their relations in the subsection 3.1. Then we propose the Lagrangian for our
system and apply the Wald formalism to it in the subsection 3.2, where we
particularly introduce the symplectic form and define the three Noether
current related charges for both the diffeomorphisms and $U(1)$ gauge
transformations. In the subsection 3.3, we figure out the phase space of our
system, resort to the canonical variables to introduce an inner product on the
subspace of perturbations, on which the symplectic complement is defined and
the double symplectic complement of a subspace is shown to be itself. Finally,
we introduce the canonical energy for the linear on-shell perturbations in the
subsection 3.4, where not only do we present the explicit expression of the
canonical energy at null infinity for both the electromagnetic and
gravitational parts, but also show the relation of the canonical energy with
the second order perturbations. With such a preparation, we take advantage of
the physically stationary perturbations to establish the criterion for the
dynamic stability of our charged star within some subspace of linear on-shell
perturbations in Section 4, where the implication of such a dynamic stability
for both non-axisymmetric and axisymmetric perturbations are further explored,
respectively. In Section 5, we further establish the criterion for the
thermodynamic stability of our charged star against generic perturbations as
well as axisymmetric perturbations, and show the equivalence of the dynamic
and thermodynamic stability with respect to the spherically symmetric
perturbations of the static, spherically symmetric isentropic charged star. We
conclude our paper with some discussions.

The notations and conventions of [19] will be followed by us, except that the
indices are not required to be balanced in our equations if no confusion
arises. In particular, early Latin letters $(a,b,c,...)$ denote abstract
spacetime indices while middle Latin letters $(i,j,k,...)$ denote concrete
spatial indices on a spacelike Cauchy surface unless specified otherwise. In
addition, the differential forms with the indices omitted are indicated in
bold typeface and a vector contracting with the first index of a differential
form is denoted by the dot with it.

##  2 Charged perfect fluid star, dynamic equilibrium, and thermodynamic
equilibrium

By a charged perfect fluid in gravitational and electromagnetic fields, we
mean that its energy momentum and charge current satisfy the following
equations of motion

$\nabla_{a}T_{m}^{ab}=F^{bc}J_{c},\quad\nabla_{a}J^{a}=0$ (1)

with $\bm{F}=d\bm{A}$ the electromagnetic field strength and

$T_{m}^{ab}=(\rho+p)u^{a}u^{b}+pg^{ab},\quad J^{a}=nu^{a}.$ (2)

Here the charge of per particle has been set to be unity such that the charge
density and number density are conflated, and $u^{a}$ is the $4$-velocity
satisfying $u_{a}u^{a}=-1$. In addition, the energy density is determined by
the equation of state $\rho(n,s)$ and the pressure can be obtained as

$p=n\frac{\partial\rho}{\partial n}-\rho,$ (3)

where we have used Euler’s equation

$\rho+p=(Ts+\mu)n$ (4)

with $s$ the entropy per particle and $\mu$ the chemical potential, as well as
the local version of the first law of thermodynamics

$d\rho=(Ts+\mu)dn+Tnds.$ (5)

Furthermore, one can show that the entropy current is also conserved, i.e.,
$\nabla_{a}(sJ^{a})=0$, which is equivalent to the statement that the entropy
per particle along the $4$-velocity is constant.

The resulting charged star is said to be in dynamic equilibrium if

$\mathscr{L}_{t}A_{a}=\mathscr{L}_{\varphi}A_{a}=\mathscr{L}_{t}u^{a}=\mathscr{L}_{\varphi}u^{a}=\mathscr{L}_{t}n=\mathscr{L}_{\varphi}n=\mathscr{L}_{t}s=\mathscr{L}_{\varphi}s=0$
(6)

for the timelike and axial Killing vector fields $t^{a}$ and $\varphi^{a}$, as
well as the $4$-velocity $u^{a}$ taking the following circular form

$u^{a}=(t^{a}+\Omega\varphi^{a})/|v|$ (7)

with $\Omega$ the angular velocity and

$|v|^{2}=-g_{ab}(t^{a}+\Omega\varphi^{a})(t^{b}+\Omega\varphi^{b}).$ (8)

Note that $v^{a}\nabla_{a}(|v|^{2})=2v^{a}v^{b}\nabla_{a}v_{b}=0$ with
$v^{a}=t^{a}+\Omega\varphi^{a}$, we have $\nabla_{a}u^{a}=0$. Namely, for the
charged star in dynamic equilibrium, the corresponding $4$-velocity is
divergence free, which further implies that the particle number density along
the $4$-velocity is also constant.

As proven in [20], the above charged star in dynamic equilibrium is in
thermodynamic equilibrium if and only if

$\tilde{T}\equiv|v|T=\text{const.},\quad\tilde{\mu}\equiv|v|\mu=\text{const.},\quad\Omega=\text{const.}$
(9)

throughout the whole star222Note that we have $\nabla_{a}u^{a}=0$,
$\sigma^{ab}\equiv
q^{ac}q^{bd}(\nabla_{(c}u_{d)}-\frac{1}{4}g_{cd}\nabla_{e}u^{e})=0$, and
$\nabla_{a}(\frac{\mu}{T})=0$ for a star in thermodynamic equilibrium, so its
energy momentum tensor, charge current, and entropy current receive no
correction individually from the bulk viscosity, shear viscosity, and electric
conductivity in the first order gradient expansion of dissipative
hydrodynamics as it should be the case[21].. Moreover, the charged star in
thermodynamic equilibrium satisfies the first law of thermodynamics as

$\delta\mathcal{M}=\tilde{T}\delta S+\tilde{\mu}\delta
N+\Omega\delta\mathcal{J}$ (10)

with $\mathcal{M}$ the ADM mass, $\mathcal{J}$ the ADM angular momentum, $S$
the total entropy, and $N$ the total particle number.

##  3 Lagrangian formulation of the Einstein-Maxwell-charged fluid system

###  3.1 Dynamical fields, redundancies, and variations

To develop a Lagrangian description of the Einstein-Maxwel-charged fluid
system, not only are we required to have the spacetime manifold $M$, on which
the metric $g_{ab}$ and the electromagnetic potential $A_{a}$ are defined, but
also we like to introduce a fiducial manifold $M^{\prime}$, called fluid
spacetime, which is diffeomorphic to $M$. Then with a fixed scalar field
$s^{\prime}$ and a fixed $3$-form $\bm{\mathcal{N}}^{\prime}$ on $M^{\prime}$,
satisfying

$d\bm{\mathcal{N}}^{\prime}=d(s^{\prime}\bm{\mathcal{N}}^{\prime})=0,$ (11)

one can define the physical fluid fields on $M$ by pushing forward with a
diffeomorphism $\chi$ as

$nu\cdot\bm{\epsilon}\equiv\bm{\mathcal{N}}=\chi_{*}\bm{\mathcal{N}}^{\prime},\quad
s=\chi_{*}s^{\prime},$ (12)

where $\bm{\epsilon}$ is the associated spacetime volume element. Accordingly,
both the charge current conservation and the entropy current conservation are
automatically satisfied.

With the above prescription, we can take $\phi=(g_{ab},A_{a},\chi)$ as the
dynamical fields for our Einstein-Maxwell-charged fluid system, where $\chi$
can be understood equivalently as a collection of 4 scalar fields
$x^{\prime\mu}\circ\chi^{-1}$ on $M$ with $x^{\prime}$ the local coordinates
on $M^{\prime}$. But nevertheless, it is noteworthy that such a description
has an additional redundancy besides the usual diffeomorphism and $U(1)$ gauge
redundancies. Namely, two field configurations $\phi=(g_{ab},A_{a},\chi)$ and
$\tilde{\phi}=(g_{ab},A_{a},\tilde{\chi})$ are physically equivalent to each
other, or trivially related if
$\chi_{*}\bm{\mathcal{N}}^{\prime}=\tilde{\chi}_{*}\bm{\mathcal{N}}^{\prime}$
and $\chi_{*}s^{\prime}=\tilde{\chi}_{*}s^{\prime}$, which amounts to saying
that $\bm{\mathcal{N^{\prime}}}$ and $s^{\prime}$ are invariant under
$\tilde{\chi}^{-1}\circ\chi$, or equivalently $\bm{\mathcal{N}}$ and $s$ are
invariant under $\tilde{\chi}\circ\chi^{-1}$.

The variation around an arbitrary field configuration $\phi$ can be formulated
by introducing a one-parameter family of dynamical fields
$\phi(\lambda)=(g_{ab}(\lambda),A_{a}(\lambda),\chi_{\lambda})$ with
$\phi(0)=\phi$. Note that
$\varphi_{\lambda}\equiv\chi_{\lambda}\circ\chi_{0}^{-1}$ gives rise to a one-
parameter family of diffeomorphisms on $M$, so the first order perturbation is
completely specified by a triple

$\delta\phi\equiv(\delta
g_{ab}\equiv\frac{dg_{ab}(\lambda)}{d\lambda}|_{\lambda=0},\delta
A_{a}\equiv\frac{dA_{a}(\lambda)}{d\lambda}|_{\lambda=0},\xi^{a}\equiv\frac{dx^{\mu}\circ\varphi_{\lambda}}{d\lambda}(\frac{\partial}{\partial
x^{\mu}})^{a}|_{\lambda=0}),$ (13)

where the vector field $\xi^{a}$ is known as the Lagrangian displacement with
$x$ the local coordinate on $M$. A first order perturbation
$\delta\mathcal{Q}$ of an arbitrary tensor field $\mathcal{Q}$ induced by
$\delta\phi$ is usually called the Eulerian perturbation. More generally, the
$k^{\text{th}}$-order Eulerian perturbation of $\mathcal{Q}$ is defined as

$\delta^{k}\mathcal{Q}\equiv\frac{d^{k}\mathcal{Q}(\lambda)}{d\lambda^{k}}|_{\lambda=0}.$
(14)

However, it turns out to be convenient to consider the so called Lagrangian
perturbation, in which the variation is performed onto the diffeomorphism
equivalent field configuration
$\hat{\phi}(\lambda)=(\varphi_{\lambda}^{*}g_{ab}(\lambda),\varphi_{\lambda}^{*}A_{a}(\lambda),\chi_{0})$.
Put it another way, the $k^{\text{th}}$-order Lagrangian perturbation of
$\mathcal{Q}$ is defined as

$\Delta^{k}\mathcal{Q}\equiv\frac{d^{k}\hat{\mathcal{Q}}(\lambda)}{d\lambda^{k}}|_{\lambda=0}$
(15)

with $\hat{\mathcal{Q}}(\lambda)=\varphi_{\lambda}^{*}\mathcal{Q}(\lambda)$.
Whence we obviously have

$\Delta^{k}\phi=(\Delta^{k}g_{ab},\Delta^{k}A_{a},0),\quad\Delta^{k}\bm{\mathcal{N}}=0,\quad\Delta^{k}s=0$
(16)

for any $k\geq 1$. In addition, it is easy to show

$\Delta\mathcal{Q}=\delta\mathcal{Q}+\mathscr{L}_{\xi}\mathcal{Q}$ (17)

for the first order perturbation, whereby we further have

$\delta\bm{\mathcal{N}}=-\mathscr{L}_{\xi}\bm{\mathcal{N}},\quad\delta
s=-\mathscr{L}_{\xi}s.$ (18)

$\xi^{a}$ is called a trivial displacement if $\delta\bm{\mathcal{N}}=0$ and
$\delta s=0$. In what follows, we shall denote such a trivial displacement as
$\eta^{a}$. It is not hard to show that $\eta^{a}$ can always be written in
the form of the superposition of two trivial displacements as follows[7]

$\eta^{a}=fu^{a}+\tilde{\eta}^{a},$ (19)

where $fu^{a}$ is called the flowline trivial with $f$ an arbitrary function
and

$\tilde{\eta}=\frac{1}{n^{2}}\mathcal{N}^{abc}\nabla_{b}Z_{c},$ (20)

with $Z_{c}$ satisfying

$u^{c}Z_{c}=0,\quad\mathscr{L}_{u}Z_{c}=0,\quad\nabla_{[a}s\nabla_{b}Z_{c]}=0.$
(21)

Furthermore, one can show $\bm{Z}$ can be expressed as

$\bm{Z}=Fds$ (22)

with $\mathscr{L}_{u}F=0$ for $\nabla_{a}s\neq 0$ and as a sum of terms of the
form $F_{1}dF_{2}$ with $\mathscr{L}_{u}F_{1}=\mathscr{L}_{u}F_{2}=0$ for
$\nabla_{a}s=0$.

###  3.2 Lagrangian, symplectic form, and Noether current related charges

Next we like to apply the Wald formalism to the Einstein-Maxwell-charged fluid
system with the Lagrangian $4$-form given by

$\bm{L}=R\bm{\epsilon}-\frac{1}{4}F_{ab}F^{ab}\bm{\epsilon}+A_{a}J^{a}\bm{\epsilon}-\rho\bm{\epsilon}$
(23)

with $J^{a}=nu^{a}$. Note that

$\displaystyle\delta(R\bm{\epsilon})$ $\displaystyle=-G^{ab}\delta
g_{ab}\bm{\epsilon}+d\bm{\theta}_{g},$ (24)
$\displaystyle\delta(-\frac{1}{4}F_{ab}F^{ab}\bm{\epsilon})$
$\displaystyle=\nabla_{b}F^{ba}\delta
A_{a}\bm{\epsilon}+\frac{1}{2}T^{ab}_{EM}\delta
g_{ab}\bm{\epsilon}+d\bm{\theta}_{EM}$

with the energy momentum tensor for the electromagnetic field given by

$T_{EM}^{ab}=F^{a}{}_{c}F^{bc}-\frac{1}{4}F^{cd}F_{cd}g^{ab}$ (25)

and

$\displaystyle\bm{\theta}_{g}$ $\displaystyle=g^{ab}g^{cd}(\nabla_{d}\delta
g_{bc}-\nabla_{b}\delta g_{cd})\epsilon_{aefg},$ (26)
$\displaystyle\bm{\theta}_{EM}$ $\displaystyle=-F^{ab}\delta
A_{b}\epsilon_{aefg}.$

In addition, we can derive

$\displaystyle\delta(A_{a}J^{a}\bm{\epsilon})$
$\displaystyle=(\Delta-\mathscr{L}_{\xi})(J^{a}A_{a}\bm{\epsilon})$ (27)
$\displaystyle=\Delta
A_{a}J^{a}\bm{\epsilon}-\mathscr{L}_{\xi}(A_{a}J^{a}\bm{\epsilon})$
$\displaystyle=J^{a}\delta
A_{a}\bm{\epsilon}+J^{a}\mathscr{L}_{\xi}A_{a}\bm{\epsilon}-d(J^{a}A_{a}\xi\cdot\bm{\epsilon})$
$\displaystyle=J^{a}\delta
A_{a}\bm{\epsilon}+J^{a}\xi^{b}F_{ba}\bm{\epsilon}+\nabla_{a}(\xi^{b}A_{b}J^{a})\bm{\epsilon}-d(A_{a}J^{a}\xi\cdot\bm{\epsilon})$
$\displaystyle=J^{a}\delta
A_{a}\bm{\epsilon}+J^{a}\xi^{b}F_{ba}\bm{\epsilon}+d\bm{\theta}_{m1}$

with

$\bm{\theta}_{m1}=\xi^{a}A_{a}J\cdot\bm{\epsilon}-A_{a}J^{a}\xi\cdot\bm{\epsilon},$
(28)

where we have used the Cartan identity $\mathscr{L}_{w}\bm{F}=w\cdot
d\bm{F}+d(w\cdot\bm{F})$ as well as the fact that

$\displaystyle\Delta\bm{\epsilon}$ $\displaystyle=\frac{1}{2}g^{ab}\Delta
g_{ab}\bm{\epsilon}$ (29) $\displaystyle\Delta u^{a}$
$\displaystyle=\frac{1}{2}u^{a}u^{b}u^{c}\Delta g_{bc}$ $\displaystyle\Delta
n$ $\displaystyle=-\frac{1}{2}nq^{bc}\Delta g_{bc}.$

in the second step with $q^{bc}=g^{bc}+u^{b}u^{c}$ and $\nabla_{a}J^{a}=0$ in
the fourth step. With the help of $\Delta\rho=\frac{\rho+p}{n}\Delta n$ and by
the same token, one can show

$\displaystyle\delta(-\rho\bm{\epsilon})$
$\displaystyle=\frac{1}{2}T^{ab}_{m}\delta
g_{ab}\bm{\epsilon}-\xi_{b}\nabla_{a}T_{m}^{ab}\bm{\epsilon}+d\bm{\theta}_{m2}$
(30)

with the energy momentum tensor for our fluid taking the familiar form

$T_{m}^{ab}=(\rho+p)u^{a}u^{b}+pg^{ab}$ (31)

and

$\bm{\theta}_{m2}=(\xi_{b}T_{m}^{ba}+\rho\xi^{a})\epsilon_{aefg}.$ (32)

Therefore the variation of the full Lagrangian $4$-form can be encapsulated as

$\delta\bm{L}=\bm{E}(\phi)\delta\phi+d\bm{\theta}(\phi;\delta\phi).$ (33)

Here $\bm{E}=0$ gives rise to the following equations of motion

$\displaystyle E^{ab}$ $\displaystyle\equiv-G^{ab}+\frac{1}{2}T^{ab}=0,$ (34)
$\displaystyle E_{EM}^{a}$ $\displaystyle\equiv\nabla_{b}F^{ba}+J^{a}=0,$
$\displaystyle E_{mb}$ $\displaystyle\equiv-\nabla^{a}T_{mab}+F_{ba}J^{a}=0$

with the total energy momentum tensor $T^{ab}=T_{EM}^{ab}+T_{m}^{ab}$, and the
total pre-symplectic potential $3$-form is given by

$\bm{\theta}=\bm{\theta}_{g}+\bm{\theta}_{EM}+\bm{\theta}_{m}$ (35)

with

$\bm{\theta}_{m}=\bm{\theta}_{m1}+\bm{\theta}_{m2}=\xi^{a}P_{abcd},$ (36)

where we have defined

$P_{abcd}=[(\rho+p)q_{a}{}^{e}-A_{f}J^{f}\delta_{a}{}^{e}+A_{f}J^{e}\delta_{a}{}^{f}]\epsilon_{ebcd}=[(\rho+p)q_{a}{}^{e}-A_{f}J^{f}q_{a}{}^{e}+A_{f}J^{e}q_{a}{}^{f}]\epsilon_{ebcd}.$
(37)

Now with $\delta\phi$ formally viewed as a vector at the point $\phi$ in the
configuration space $\mathcal{F}$, denoted as $\delta\phi^{A}$, one can define
a $1$-form field $\Theta_{A}$ and the pre-symplectic form on $\mathcal{F}$ as
follows

$\Theta_{A}\delta\phi^{A}=\int_{\Sigma}\bm{\theta}(\phi;\delta\phi),\quad\Omega_{AB}=(D\Theta)_{AB},$
(38)

where $\Sigma$ denotes a Cauchy surface in $M$ and $D$ represents the exterior
derivative on $\mathcal{F}$. A convenient way to evaluate
$\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}$ at the point
$\phi\in\mathcal{F}$ is to extend $\delta_{1}\phi^{A}$ and
$\delta_{2}\phi^{A}$ off of $\phi$ in an arbitrary manner and use the
following formula

$\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}=\mathscr{L}_{\delta_{1}\phi}(\Theta_{A}\delta_{2}\phi^{A})-\mathcal{L}_{\delta_{2}\phi}(\Theta_{A}\delta_{1}\phi^{A})-\Theta_{A}[\delta_{1}\phi,\delta_{2}\phi]^{A}.$
(39)

This amounts to saying

$\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}=\int_{\Sigma}\bm{\omega}(\phi;\delta_{1}\phi,\delta_{2}\phi),$
(40)

with the pre-symplectic current $3$-form on $M$ defined as

$\bm{\omega}(\phi;\delta_{1}\phi,\delta_{2}\phi)=\delta_{1}\bm{\theta}(\phi;\delta_{2}\phi)-\delta_{2}\bm{\theta}(\phi;\delta_{1}\phi)-\bm{\theta}(\phi;\delta_{1}\delta_{2}\phi-\delta_{2}\delta_{1}\phi).$
(41)

For our purpose, we choose $(\delta_{1}g_{ab},\delta_{1}A_{a})$ and
$(\delta_{2}g_{ab},\delta_{2}A_{a})$ as variations along a two-parameter
family of metrics and gauge fields
$(g_{ab}(\lambda_{1},\lambda_{2}),A_{a}(\lambda_{1},\lambda_{2}))$, which
implies

$\delta_{1}\delta_{2}g_{ab}=\delta_{2}\delta_{1}g_{ab},\quad\delta_{1}\delta_{2}A_{a}=\delta_{2}\delta_{1}A_{a}.$
(42)

On the other hand, we like to keep $\xi_{1}^{a}$ and $\xi_{2}^{a}$ fixed,
i.e.,

$\delta_{1}\xi_{2}^{a}(x)=\delta_{2}\xi_{1}^{a}(x)=0,$ (43)

which implies

$\delta_{1}\delta_{2}\chi=\delta_{1}\xi_{2}^{a}(\chi)=\xi_{1}^{b}\partial_{b}\xi_{2}^{a}(\chi),\quad\delta_{2}\delta_{1}\chi=\delta_{2}\xi_{1}^{a}(\chi)=\xi_{2}^{b}\partial_{b}\xi_{1}^{a}(\chi).$
(44)

Whence we have

$\delta_{1}\delta_{2}\chi-\delta_{2}\delta_{1}\chi=[\xi_{1},\xi_{2}]^{a}.$
(45)

As a result, the pre-symplectic form can be written explicitly as

$\displaystyle\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}$
$\displaystyle=$
$\displaystyle\int_{\Sigma}(\delta_{2}h_{ij}\delta_{1}\bm{\pi}^{ij}-\delta_{1}h_{ij}\delta_{2}\bm{\pi}^{ij})+\int_{\Sigma}(\delta_{2}A_{i}\delta_{1}\bm{\pi}^{i}-\delta_{1}A_{i}\delta_{2}\bm{\pi}^{i})$
(46)
$\displaystyle+\int_{\Sigma}(\xi_{2}^{a}\delta_{1}P_{aefg}-\xi_{1}^{a}\delta_{2}P_{aefg}-[\xi_{1},\xi_{2}]^{a}P_{aefg}),$

where

$\bm{\pi}^{ij}=(K^{ij}-Kh^{ij})\hat{\bm{\epsilon}},\quad\bm{\pi}^{i}=\nu_{a}F^{ai}\hat{\bm{\epsilon}}$
(47)

with $K_{ij}$ the extrinsic curvature and
$\hat{\bm{\epsilon}}=\nu\cdot\bm{\epsilon}$ the induced volume on $\Sigma$ by
the future directed normalized normal vector $\nu^{a}$.

But nevertheless, no matter whether
$\delta_{1}\delta_{2}\phi-\delta_{2}\delta_{1}\phi$ vanishes or not, one can
show that we always have

$d\bm{\omega}(\phi;\delta_{1}\phi,\delta_{2}\phi)=\delta_{2}\bm{E}\delta_{1}\phi-\delta_{1}\bm{E}\delta_{2}\phi,$
(48)

which means that $\bm{\omega}$ is closed if both $\delta_{1}\phi$ and
$\delta_{2}\phi$ satisfy the linearized equations of motion
$\delta_{1}\bm{E}=\delta_{2}\bm{E}=0$. Accordingly,
$\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}$ does not depends on the
choice of the Cauchy surface $\Sigma$ if not only do $\delta_{1}\phi$ and
$\delta_{2}\phi$ satisfy the linearized equations of motion but also have an
appropriate fall-off behavior at the spatial infinity. In addition, for the
local symmetry associated with the diffeomorphisms generated by an arbitrary
vector field $X^{a}$, we can define the corresponding Noether current $3$-form
as

$\bm{J}_{X}=\bm{\theta}(\phi;\mathscr{L}_{X}\phi)-X\cdot\bm{L},$ (49)

with
$\mathscr{L}_{X}\phi=(\mathscr{L}_{X}g_{ab},\mathscr{L}_{X}A_{a},-X^{a})$,
whereby we have

$d\bm{J}_{X}=-\bm{E}\mathscr{L}_{X}\phi.$ (50)

Therefore the corresponding Noether charge

$Q_{X}=\int_{\Sigma}\bm{J}_{X}=0$ (51)

for the on-shell field configurations provided that $X^{a}$ vanishes
sufficiently rapidly at the spatial infinity. Furthermore, one can show

$\bm{\omega}(\phi;\delta\phi,\mathscr{L}_{X}\phi)=\delta\bm{J}_{X}+X\cdot\bm{E}(\phi)\delta\phi-d(X\cdot\bm{\theta}(\phi,\delta\phi)).$
(52)

Note that the Noether current can be written as follows

$\bm{J}_{X}=X^{a}\bm{C}_{a}+d\bm{Q}_{X}$ (53)

for our Einstein-Maxwell-charged fluid system with

$\bm{C}_{a}=(-2E^{b}{}_{a}-E_{EM}^{b}A_{a})\epsilon_{befg},\quad\bm{Q}_{X}=-*d\bm{X}-*\bm{F}A_{c}X^{c},$
(54)

where the star denotes the Hodge dual. So we end up with the following
fundamental identity

$\bm{\omega}(\phi;\delta\phi,\mathscr{L}_{X}\phi)=X^{b}\delta\bm{C}_{b}+X\cdot\bm{E}(\phi)\delta\phi+d(\delta\bm{Q}_{X}-X\cdot\bm{\theta}(\phi,\delta\phi)).$
(55)

One immediate implication of this identity is the diffeomorphism invariance of
the symplectic form. Speaking specifically, if $\phi$ is on-shell,
$\delta\phi$ satisfies the linearized constraint equations
$\delta\bm{C}_{a}=0$, and $X^{a}$ vanishes sufficiently rapidly as before, we
have

$\Omega_{AB}\delta\phi^{A}\mathscr{L}_{X}\phi^{B}=0,$ (56)

which amounts to saying that $\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}$
keeps invariant under the gauge shift
$\delta\phi\rightarrow\delta\phi+\mathscr{L}_{X}\phi$. On the other hand, for
a large gauge transformation, where $X^{a}$ approaches a nontrivial asymptotic
symmetry, we instead have

$\Omega_{AB}\delta\phi^{A}\mathscr{L}_{X}\phi^{B}=\int_{\Sigma}(X\cdot\bm{E}\delta\phi+X^{a}\delta\bm{C}_{a})+\int_{S_{\infty}}(\delta\bm{Q}_{X}-X\cdot\delta\bm{B}),$
(57)

where we have assumed that

$\int_{S_{\infty}}X\cdot\bm{\theta}(\phi;\delta\phi)=\int_{S_{\infty}}X\cdot\delta\bm{B}(\phi)$
(58)

for some $3$-form $\bm{B}$. By evaluating Eq. (57) at an on-shell $\phi$, we
have

$\Omega_{AB}\delta\phi^{A}\mathscr{L}_{X}\phi^{B}=\delta
H_{X}=\delta\phi^{A}D_{A}H_{X}$ (59)

with the Hamiltonian conjugate to $X^{a}$ defined as

$H_{X}=\int_{\Sigma}X^{a}\bm{C}_{a}+\int_{S_{\infty}}(\bm{Q}_{X}-X\cdot\bm{B}).$
(60)

By further restricting ourselves onto the phase space $\mathcal{P}$, which is
obtained by factoring out the degeneracy orbits of $\Omega_{AB}$ in the
configuration space $\mathcal{F}$, we wind up with a non-degenerate
$\Omega_{AB}$. Thus we have the familiar form of Hamilton’s equation on the
phase space as

$\mathscr{L}_{X}\phi^{A}=\Omega^{AB}D_{B}H_{X}$ (61)

with $\Omega^{AB}$ the inverse of $\Omega_{AB}$, satisfying
$\Omega^{AB}\Omega_{BC}=\delta^{A}{}_{C}$.

The ADM charge $\mathcal{Q}_{X}$ is defined as the surface term in $H_{X}$,
which is equal to $H_{X}$ when evaluated at an on-shell $\phi$. In particular,
for the asymptotic time translation $t^{a}$ and rotation $\varphi^{a}$, the
corresponding ADM mass and angular momentum are defined as

$\mathcal{M}=\int_{S_{\infty}}(\bm{Q}_{t}-t\cdot\bm{B})=\int_{S_{\infty}}(-*d\bm{t}-t\cdot\bm{B}),\quad\mathcal{J}=-\int_{S_{\infty}}\bm{Q}_{\varphi}=\int_{S_{\infty}}*d\bm{\varphi},$
(62)

where not only have we chosen $S_{\infty}$ such that $\varphi^{a}$ is tangent
to it but also have taken the gauge such that $A_{a}|_{S_{\infty}}=0$. If the
rotation $\varphi^{a}$ is a background symmetry generator, the ADM angular
momentum can be expressed as follows

$\displaystyle\mathcal{J}$
$\displaystyle=-2\int_{\Sigma}\nabla_{a}\nabla^{[a}\varphi^{b]}\epsilon_{befg}=-2\int_{\Sigma}\nabla_{a}\nabla^{a}\varphi^{b}\epsilon_{befg}=2\int_{\Sigma}R^{ab}\varphi_{a}\epsilon_{befg}=\int_{\Sigma}T^{ab}\varphi_{a}\epsilon_{befg}$
(63)
$\displaystyle=\int_{\Sigma}[\frac{\rho+p}{n}\varphi\cdot\bm{u}\bm{\mathcal{N}}+F^{cb}\nabla_{c}(\varphi\cdot\bm{A})\epsilon_{befg}]=\int_{\Sigma}[\varphi\cdot(\frac{\rho+p}{n}\bm{u}+\bm{A})\bm{\mathcal{N}}-d*\bm{F}(\varphi\cdot\bm{A})]$
$\displaystyle=\int_{\Sigma}\varphi\cdot(\frac{\rho+p}{n}\bm{u}+\bm{A})\bm{\mathcal{N}}=\int_{\Sigma}j\bm{\mathcal{N}}$

with $\Sigma$ so chosen that $\varphi^{a}$ is tangent to it and
$j\equiv\varphi\cdot(\frac{\rho+p}{n}\bm{u}+\bm{A})$ interpreted as the
angular momentum per particle. Here we have used
$d*\bm{\Lambda}=-\nabla_{a}\Lambda^{ab}\epsilon_{befg}$ for any $2$-form
$\bm{\Lambda}$ in the first step,
$\nabla_{a}\nabla_{b}\kappa_{c}=R_{cba}{}^{d}\kappa_{d}$ for any Killing
vector field $\kappa^{a}$ in the third step, and
$\mathscr{L}_{\varphi}\bm{A}=0$ as well as the Cartan identity in the fifth
step.

Similarly, associated with the local symmetry under the field configuration
independent $U(1)$ gauge transformation
$\delta_{\vartheta}\phi=(0,\nabla_{a}\vartheta,0)$, we can define the
corresponding Noether current as

$\bm{J}_{\vartheta}=\bm{\theta}(\phi;\delta_{\vartheta}\phi)-\vartheta\bm{\mathcal{N}},$
(64)

whereby we have

$\displaystyle\delta\bm{J}_{\vartheta}$
$\displaystyle=\bm{\omega}(\phi;\delta\phi,\delta_{\vartheta}\phi)+\xi^{a}\delta_{\vartheta}\bm{P}_{a}-\vartheta\delta\bm{\mathcal{N}}$
(65)
$\displaystyle=\bm{\omega}(\phi;\delta\phi,\delta_{\vartheta}\phi)+\xi^{a}[-\nabla_{f}(\vartheta
J^{f})\epsilon_{abcd}+\nabla_{a}\vartheta\bm{\mathcal{N}}]+\vartheta\mathscr{L}_{\xi}\bm{\mathcal{N}}$
$\displaystyle=\bm{\omega}(\phi;\delta\phi,\delta_{\vartheta}\phi)-\xi\cdot
d(\vartheta\bm{\mathcal{N}})+\mathscr{L}_{\xi}(\vartheta\bm{\mathcal{N}})$
$\displaystyle=\bm{\omega}(\phi;\delta\phi,\delta_{\vartheta}\phi)+d(\vartheta\xi\cdot\bm{\mathcal{N}})$

In addition, a straightforward calculation yields

$\bm{J}_{\vartheta}=\vartheta\bm{C}+d\bm{Q}_{\vartheta}$ (66)

with

$\bm{C}=-E^{EM}\cdot\bm{\epsilon},\quad\bm{Q}_{\vartheta}=-*\bm{F}\vartheta.$
(67)

Thus we obtain

$\bm{\omega}(\phi;\delta\phi,\delta_{\vartheta}\phi)=\vartheta\delta\bm{C}+d(\delta\bm{Q}_{\vartheta}-\vartheta\xi\cdot\bm{\mathcal{N}}),$
(68)

whereby one can obviously introduce the analogous charges and make the
analogous statements to those following Eq. (55). In particular, note that
$d*\bm{F}=\bm{\mathcal{N}}$, so we have the on-shell variation $\delta
H_{\vartheta}=0$ automatically for those gauge transformations which preserve
the gauge condition $A_{a}=0$ at $S_{\infty}$.

###  3.3 Phase space, inner product, and symplectic complement

As alluded to in the previous subsection, the phase space is obtained by
factoring out the degeneracy orbits in the configuration space. Put it another
way, $\delta\phi_{\text{p}}=0$ is the only degeneracy of the symplectic form
if and only if one can parameterize the phase space by $\phi_{\text{p}}$. To
proceed, we would like first to introduce the space of fiducial flowlines
$\Sigma^{\prime}$, namely the space of the integral curves of a non-vanishing
$u^{\prime a}$ with $u^{\prime}\cdot\bm{\mathcal{N}}^{\prime}$. Then we can
further introduce the diffeomorphism $\psi$ from $\Sigma^{\prime}$ to $\Sigma$
obtained by intersecting with $\Sigma$ the images of the fiducial flowlines
under $\chi$. Next we shall show that one can parameterize the phase space for
our Einstein-Maxwell-charged fluid system by
$\phi_{\text{p}}=(h_{ij},\bm{\pi}^{ij},A_{i},\bm{\pi}^{i},\psi,u^{i})$ on
$\Sigma$ with $u^{i}=h^{ia}u_{a}$ the fluid $3$-velocity, namely we are
required to show that $\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}=0$ for
any $\delta_{2}\phi$ necessitates $\delta_{1}\phi_{\text{p}}=0$. To this end,
we like to express

$\displaystyle\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}$
$\displaystyle=\Omega_{AB}\delta_{1}\phi^{A}\Delta_{2}\phi^{B}-\Omega_{AB}\delta_{1}\phi^{A}\mathscr{L}_{\xi_{2}}\phi^{B}$
(69)
$\displaystyle=\int_{\Sigma}[(\Delta_{2}h_{ij}\delta_{1}\bm{\pi}^{ij}-\delta_{1}h_{ij}\Delta_{2}\bm{\pi}^{ij}+\Delta_{2}A_{i}\delta_{1}\bm{\pi}^{i}-\delta_{1}A_{i}\Delta_{2}\bm{\pi}^{i}-\xi_{1}^{a}\Delta_{2}\bm{P}_{a})-(\xi_{2}\cdot\bm{E}\delta_{1}\phi+\xi_{2}^{a}\delta_{1}\bm{C}_{a})],$

where we have used Eq. (46) to evaluate the first term and Eq. (57) together
with the Lagrangian displacement of spatial compact support for the second
term. By $\Delta q_{a}{}^{e}=u^{e}q_{a}{}^{(b}u^{c)}\Delta g_{bc}$, one can
show

$\Delta\bm{P}_{a}=\bm{A}_{a}{}^{bc}\Delta g_{bc}+\bm{B}_{a}{}^{f}{}\Delta
A_{f}$ (70)

with

$\bm{A}_{a}{}^{bc}=-\frac{1}{2}(\rho+p)u^{b}u^{c}q_{a}{}^{e}\epsilon_{ebcd}-\frac{1}{2}c_{s}^{2}(\rho+p)q^{bc}q_{a}{}^{e}\epsilon_{ebcd}+\frac{\rho+p}{n}q_{a}{}^{(b}u^{c)}\bm{\mathcal{N}},\quad\bm{B}_{a}{}^{f}=2q_{a}{}^{[f}J^{e]}\epsilon_{ebcd},$
(71)

where $c_{s}$ is the sound speed, defined as $c_{s}^{2}\equiv(\frac{\partial
p}{\partial\rho})_{s}=\frac{\Delta p}{\Delta\rho}$ and assumed to satisfy
$0\leq c^{2}_{s}\leq 1$. Furthermore, with the $3+1$ decomposition in the
coordinates $(\tau,x^{i})$ with $\Sigma$ given by $\tau=0$, we have

$ds^{2}=-\alpha^{2}d\tau^{2}+h_{ij}(dx^{i}+\beta^{i}d\tau)(dx^{j}+\beta^{j}d\tau),\quad\bm{A}=Ad\tau+A_{i}(dx^{i}+\beta^{i}d\tau),$
(72)

whereby we can express

$\Delta
g_{ab}=-\frac{2}{\alpha}\nu_{a}\nu_{b}\Delta\alpha-\frac{1}{\alpha}\nu_{a}\Delta
N_{b}-\frac{1}{\alpha}\nu_{b}\Delta N_{a}+\Delta\gamma_{ab},\quad\Delta
A_{a}=-\frac{1}{\alpha}\nu_{a}\Delta\mathcal{A}+\Delta\mathcal{A}_{a}$ (73)

with $\Delta N_{a}=\Delta N_{i}(dx^{i}+\beta^{i}d\tau)_{a}\equiv
h_{ij}\Delta\beta^{j}(dx^{i}+\beta^{i}d\tau)_{a}$,
$\Delta\gamma_{ab}\equiv\Delta
h_{ij}(dx^{i}+\beta^{i}d\tau)_{a}(dx^{j}+\beta^{j}d\tau)_{b}$,
$\Delta\mathcal{A}\equiv\Delta A+A_{i}\Delta\beta^{i}$, and
$\Delta\mathcal{A}_{a}\equiv\Delta A_{i}(dx^{i}+\beta^{i}d\tau)_{a}$. The
degeneracy of our symplectic form is obtained by requiring the coefficients of
$\Delta\bm{\pi}^{ij}$, $\Delta_{2}\alpha$, $\Delta_{2}N_{i}$,
$\Delta_{2}h_{ij}$, $\Delta_{2}\bm{\pi}^{i}$, $\Delta_{2}\mathcal{A}$,
$\Delta_{2}A_{i}$, and $\xi_{2}^{a}$ in Eq. (69) vanish when pulled back onto
$\Sigma$, which gives rise to

$\displaystyle\delta_{1}h_{ij}$ $\displaystyle=0,$ (74)
$\displaystyle\xi_{1}^{a}\bm{A}_{a}{}^{bc}\nu_{b}\nu_{c}$ $\displaystyle=0,$
$\displaystyle\xi_{1}^{a}\bm{A}_{a}{}^{bi}\nu_{b}$ $\displaystyle=0,$
$\displaystyle\delta_{1}\bm{\pi}^{ij}-\xi_{1}^{a}\bm{A}_{a}{}^{ij}$
$\displaystyle=0,$ $\displaystyle\delta_{1}A_{i}$ $\displaystyle=0,$
$\displaystyle\xi_{1}^{a}\bm{B}_{a}{}^{b}\nu_{b}$ $\displaystyle=0,$
$\displaystyle\delta_{1}\bm{\pi}^{i}-\xi_{1}^{a}\bm{B}_{a}{}^{i}$
$\displaystyle=0,$
$\displaystyle\delta_{1}\bm{C}_{a}-\nu\cdot\bm{E}\delta_{1}\phi\nu_{a}$
$\displaystyle=0.$

Here the second and third equations can be encapsulated as

$0=\xi_{1}^{a}\bm{A}_{a}{}^{bc}\nu_{b}=-\frac{1}{2}(\rho+p)[(u^{b}\nu_{b})^{2}\delta^{c}{}_{d}-c_{s}^{2}\nu_{b}q^{bc}\nu_{d}]\xi_{1}^{a}q_{a}{}^{d}\hat{\bm{\epsilon}}.$
(75)

The contraction with $\nu_{c}$ implies $\nu_{d}\xi_{1}^{a}q_{a}{}^{d}=0$,
whereby one can further obtain $\xi_{1}^{a}q_{a}{}^{d}=0$ by the above
equation. That is to say $\xi_{1}^{a}\propto u^{a}$, namely
$\delta_{1}\psi=0$. In addition, by
$u^{a}\bm{A}_{a}{}^{bc}=u^{a}\bm{B}_{a}{}^{b}=0$, the fourth and seventh
equations in Eq. (74) yield
$\delta_{1}\bm{\pi}^{ij}=\delta_{1}\bm{\pi}^{i}=0$. So right now we are only
left with $\delta u^{i}=0$ to show. To achieve this, we first decompose
$\delta_{1}\phi$ as follows

$(\delta_{1}g_{ab},\delta_{1}A_{a},\xi_{1}^{a})=(0,0,fu^{a})+(\delta_{1}g_{ab},\delta_{1}A_{a},\tau\zeta^{a})$
(76)

with $u^{a}\zeta_{a}=0$. Note that the last equation in Eq. (74) is
automatically satisfied by $(0,0,fu^{a})$ since $u^{a}E_{ma}=0$ due to the
built-in conservation of the charge current and entropy current in our
Lagrangian description. Furthermore,
$\delta_{1}h_{ij}=\delta_{1}\bm{\pi}^{ij}=\delta_{1}A_{i}=\delta_{1}\bm{\pi}^{i}=0$
implies that the pure gravitational and electromagnetic parts do not
contribute to the left side of the last equation in Eq. (74). Accordingly, for
$(\delta_{1}g_{ab},\delta_{1}A_{a},\tau\zeta^{a})$, we have

$\displaystyle 0$
$\displaystyle=-\delta_{1}(T^{b}_{ma}\epsilon_{befg})-\frac{1}{2}T_{m}^{bc}\delta_{1}g_{bc}\nu_{a}\hat{\bm{\epsilon}}-\delta_{1}(A_{a}\bm{\mathcal{N}})-J^{b}\delta_{1}A_{b}\nu_{a}\hat{\bm{\epsilon}}$
(77)
$\displaystyle=-\delta_{1}(\frac{\rho+p}{n}u_{a}\bm{\mathcal{N}}+p\epsilon_{aefg})-\frac{1}{2}[(\rho+p)u^{b}u^{c}\delta_{1}g_{bc}+pg^{bc}\delta_{1}g_{bc}]\nu_{a}\hat{\bm{\epsilon}}-\delta_{1}(A_{a}\bm{\mathcal{N}})-J^{b}\delta_{1}A_{b}\nu_{a}\hat{\bm{\epsilon}}$
$\displaystyle=-\delta_{1}(\frac{\rho+p}{n}u_{a})\bm{\mathcal{N}}+\delta_{1}p\nu_{a}\hat{\bm{\epsilon}}+\frac{1}{2}(\rho+p)u_{b}u_{c}\delta_{1}g^{bc}\nu_{a}\hat{\bm{\epsilon}}-\delta_{1}A_{a}\bm{\mathcal{N}}-J^{b}\delta_{1}A_{b}\nu_{a}\hat{\bm{\epsilon}}$
$\displaystyle=[nu_{b}\nu^{b}\Delta_{1}(\frac{\rho+p}{n}u_{a})+\Delta_{1}p\nu_{a}-(\rho+p)u_{b}\nu^{b}u_{c}\delta_{1}\nu^{c}\nu_{a}]\hat{\bm{\epsilon}}$
$\displaystyle=-(\rho+p)(\frac{1}{2}c_{s}^{2}u_{b}\nu^{b}q^{cd}\Delta_{1}g_{cd}u_{a}-u_{b}\nu^{b}\Delta_{1}u_{a}+\frac{1}{2}c_{s}^{2}q^{cd}\Delta_{1}g_{cd}\nu_{a}+u_{b}\nu^{b}u_{c}\delta_{1}\nu^{c}\nu_{a})\hat{\bm{\epsilon}}$
$\displaystyle=-(\rho+p)(\frac{1}{2}c_{s}^{2}q^{cd}\Delta_{1}g_{cd}q_{ab}\nu^{b}-u_{b}\nu^{b}\Delta_{1}u_{a}+u_{b}\nu^{b}u_{c}\delta_{1}\nu^{c}\nu_{a})\hat{\bm{\epsilon}}$
$\displaystyle=-(\rho+p)[c_{s}^{2}(\nu_{c}\delta_{1}\nu^{c}+u_{c}\nu^{c}u_{d}\delta_{1}\nu^{d}-\frac{1}{\alpha}\zeta_{c}\nu^{c})q_{ab}\nu^{b}-u_{b}\nu^{b}\Delta_{1}u_{a}+u_{b}\nu^{b}u_{c}\delta_{1}\nu^{c}\nu_{a}]\hat{\bm{\epsilon}},$

where we have used
$\delta_{1}\bm{\mathcal{N}}=-\mathscr{L}_{\tau\zeta}\bm{\mathcal{N}}=-d\tau\wedge\zeta\cdot\bm{\mathcal{N}}=0$
when restricted onto $\Sigma$ in the third step, $\delta_{1}h^{ab}=0$ as well
as $\delta_{1}A_{a}\propto\nu_{a}$ in the fourth step, and
$\Delta_{1}g_{ab}=\delta_{1}g_{ab}-\frac{1}{\alpha}(\nu_{a}\zeta_{b}+\nu_{b}\zeta_{a})$
in the seventh step. To proceed, we compute

$\displaystyle\Delta_{1}u_{a}$
$\displaystyle=\nu_{a}u_{d}\delta_{1}\nu^{d}+u_{d}\nu^{d}g_{ac}\delta_{1}\nu^{c}-\frac{1}{\alpha}\zeta_{a}u_{d}\nu^{d}+u_{a}u_{c}\nu^{c}u_{d}\delta_{1}\nu^{d}$
(78)
$\displaystyle=h_{ab}\delta_{1}(h^{bc}u_{c})+\nu_{a}[u_{d}\delta_{1}\nu^{d}-u_{d}\nu^{d}\nu_{c}\delta_{1}\nu^{c}+\frac{1}{\alpha}\zeta_{c}\nu^{c}u_{d}\nu^{d}-(u_{c}\nu^{c})^{2}u_{d}\delta_{1}\nu^{d}],$

whereby we further have

$u_{b}\delta_{1}(h^{bc}u_{c})=u^{a}h_{ab}\delta_{1}(h^{bc}u_{c})=(u_{c}\nu^{c})^{2}(\nu_{d}\delta_{1}\nu^{d}-\frac{1}{\alpha}\zeta_{d}\nu^{d}+u_{b}\nu^{b}u_{d}\delta_{1}\nu^{d}).$
(79)

Then plugging Eq. (78) into Eq. (77) and using Eq. (79), we end up with

$0=-(\rho+p)B_{ab}\delta_{1}(h^{bc}u_{c})\hat{\bm{\epsilon}},$ (80)

where

$B_{ab}=\frac{c_{s}^{2}}{(u_{d}\nu^{d})^{2}}q_{ac}\nu^{c}u_{b}-2u^{c}h_{b[a}\nu_{c]}.$
(81)

By contracting Eq. (80) with $\nu^{a}$, we obtain
$u_{b}\delta_{1}(h^{bc}u_{c})=0$, whereby Eq. (80) further implies
$\delta_{1}(h^{ab}u_{b})=0$. Thus we have accomplished the proof that the
phase space for our Einstein-Maxwell-charged fluid system is described exactly
by $\phi_{\text{p}}$ on $\Sigma$.

However, the variables $(\psi,u^{i})$ are not canonically conjugate. To
construct the canonically conjugate variables for our charged fluid, we like
to choose the coordinates in $M^{\prime}$ such that the fiducial flowlines are
given by the integral curves of $(\frac{\partial}{\partial x^{\prime
0}})^{a}$. Then by $\delta x^{\prime\mu}=-\partial_{a}x^{\prime\mu}\xi^{a}$,
we can write

$\bm{\theta}_{m}=-\delta x^{\prime\mu}\bm{P}^{\prime}_{\mu}=-\delta x^{\prime
i}\bm{P}^{\prime}_{i}$ (82)

with $\bm{P}^{\prime}_{\mu}=(\frac{\partial}{\partial
x^{\prime\mu}})^{a}\bm{P}_{a}$. Whence we further have

$\bm{\omega}_{m}=-(\delta_{2}x^{\prime
i}\delta_{1}\bm{P}^{\prime}_{i}-\delta_{1}x^{\prime
i}\delta_{2}\bm{P}^{\prime}_{i}),$ (83)

which tells us that $(x^{\prime i},-\bm{P}^{\prime}_{i})$ can be regarded as
the canonically conjugate variables for our charged fluid. Accordingly, the
symplectic form for our Einstein-Maxwell-charged fluid system can be cast into
the following canonical manner

$\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}=\int_{\Sigma}(\delta_{2}q^{\alpha}\delta_{1}p_{\alpha}-\delta_{1}q^{\alpha}\delta_{2}p_{\alpha})$
(84)

with $q^{\alpha}=(h_{ij},A_{i},x^{\prime i})$ and
$p_{\alpha}=(\bm{\pi}^{ij},\bm{\pi}^{i},-\bm{P}^{\prime}_{i})$. Then by
working with the coordinates in which $h=1$, we can introduce an inner product

$\langle\delta_{1}\phi,\delta_{2}\phi\rangle=\int_{\Sigma}(h_{\alpha\beta}\delta_{1}q^{\alpha}\delta_{2}q^{\beta}+h^{\alpha\beta}\delta_{1}p_{\alpha}\delta_{2}p_{\beta}),$
(85)

where $h_{\alpha\beta}$ and $h^{\alpha\beta}$ should be understood as either
$h_{ij}$ or $h^{ij}$ depending on the tensor indices of
$(q^{\alpha},p_{\alpha})$. Associated with this inner product, we have the
Hilbert space $\mathcal{H}$ as a subspace of perturbations. As pointed out in
[7], the square integrability indicates that $\mathcal{H}$ does not include
those perturbations with $\delta\mathcal{M}\neq 0$, but includes all the
perturbations of interest with $\delta\mathcal{M}=0$. Then it is not hard to
see that our symplectic form can be expressed as a bounded linear operator $W$
on $\mathcal{H}$ as follows

$\Omega_{AB}\delta_{1}\phi^{A}\delta_{2}\phi^{B}=\langle\delta_{1}\phi,\hat{\Omega}\delta_{2}\phi)$
(86)

with

$\hat{\Omega}(\delta q^{\alpha},\delta p_{\alpha})=(-h^{\alpha\beta}\delta
p_{\beta},h_{\alpha\beta}\delta q^{\beta}).$ (87)

Whence we know that $\hat{\Omega}$ is an orthogonal map, since
$\hat{\Omega}^{2}=-1$ and $\hat{\Omega}^{\dagger}=-\hat{\Omega}$.

The symplectic complement of a subspace $\mathcal{S}$ in $\mathcal{H}$ is
defined as follows

$\mathcal{S}^{\perp_{\text{s}}}=\\{\delta^{\prime}\phi\in\mathcal{H}|\langle\delta^{\prime}\phi,\hat{\Omega}\delta\phi\rangle=0,\forall\delta\phi\in\mathcal{S}\\}.$
(88)

Then it is not hard to show
$\mathcal{S}^{\perp_{\text{s}}}=(\hat{\Omega}[\mathcal{S}])^{\perp}=\hat{\Omega}[\mathcal{S}^{\perp}]$,
which further gives

$(\mathcal{S}^{\perp_{\text{s}}})^{\perp_{\text{s}}}=(\hat{\Omega}[\mathcal{S}^{\perp}])^{\perp_{\text{s}}}=\hat{\Omega}[(\hat{\Omega}[\mathcal{S}^{\perp}])^{\perp}]=\hat{\Omega}[\hat{\Omega}[(\mathcal{S}^{\perp})^{\perp}]=(\mathcal{S}^{\perp})^{\perp}=\bar{\mathcal{S}},$
(89)

where the bar represents the closure in $\mathcal{H}$. Since any subspace is
dense in its closure, below we shall not bother ourselves by saying that the
double symplectic complement of any subspace is itself.

###  3.4 Canonical energy, explicit expression at null infinity, and relation
to second order perturbations

Finally, associated with an arbitrary background symmetry generated by
$\kappa^{a}$, we would like to introduce the corresponding canonical energy,
which is a bilinear form on the space of linear on-shell perturbations defined
as

$\mathcal{E}_{\kappa}(\delta_{1}\phi,\delta_{2}\phi)=\Omega_{AB}\delta_{1}\phi^{A}\mathscr{L}_{\kappa}\delta_{2}\phi^{B}.$
(90)

It is easy to show that not only is the canonical energy conserved and
symmetric, but also gauge invariant.

To obtain the expression of the canonical energy
$\mathcal{E}_{t}(\delta_{1}\phi,\delta_{2}\phi)$ at the future null infinity
$\mathscr{I}$ for later use, we like to introduce the unphysical metric
$\tilde{g}_{ab}=\Omega^{2}g_{ab}$ with the conformal factor $\Omega=0$
corresponding to the location of $\mathscr{I}$ and work in the gauge such that
$\tilde{\nabla}_{a}n_{b}|_{\mathscr{I}}=n^{b}A_{b}|_{\mathscr{I}}=0$ with
$n_{b}=\tilde{\nabla}_{b}\Omega$. Here the indices are raised or lowered by
our unphysical metric, which is smooth across $\mathscr{I}$. We further assume
that $\Omega$ near $\mathscr{I}$ and $\tilde{g}_{ab}$ at $\mathscr{I}$ are
universal quantities in the sense that $\delta\Omega=0$ near $\mathscr{I}$ and
$\delta\tilde{g}_{ab}|_{\mathscr{I}}=0$. Whence the symplectic potential at
$\mathscr{I}$ for the electromagnetic part can be written as

$\bm{\theta}_{EM}=-F^{ab}\delta A_{b}\tilde{\epsilon}_{aefg}=\Pi^{b}\delta
A_{b}\hat{\tilde{\bm{\epsilon}}}$ (91)

with $\Pi^{b}=n_{a}F^{ab}$ and the induced volume $3$-form given by
$\tilde{\bm{\epsilon}}=-\bm{n}\wedge\hat{\tilde{\bm{\epsilon}}}$. Whence the
symplectic current reads

$\bm{\omega}_{EM}=(\delta_{2}A_{b}\delta_{1}\Pi^{b}-\delta_{1}A_{b}\delta_{2}\Pi^{b})\hat{\tilde{\bm{\epsilon}}},$
(92)

where one should keep in mind that
$\delta\Pi^{b}=n_{a}\tilde{g}^{ac}\tilde{g}^{bd}\delta F_{cd}$. Note that
$t^{a}=cn^{a}$ at $\mathscr{I}$ with $c$ a positive constant for the future
directed timelike Killing vector field $t^{a}$, so we have
$\mathscr{L}_{t}\delta\bm{A}=d(t\cdot\delta\bm{A})+t\cdot
d\delta\bm{A}=\sigma\bm{n}+cn\cdot\delta\bm{F}$ at $\mathscr{I}$ for some
function $\sigma$. In addition,
$0=\tilde{\nabla}_{a}n^{a}\tilde{\bm{\epsilon}}=\mathscr{L}_{n}\tilde{\bm{\epsilon}}=-\bm{n}\wedge\mathscr{L}_{n}\hat{\tilde{\bm{\epsilon}}}$
at $\mathscr{I}$ implies that $\mathscr{L}_{n}\hat{\tilde{\bm{\epsilon}}}=0$
when restricted onto $\mathscr{I}$. As a result, we have

$\displaystyle\mathcal{E}^{EM}_{t}(\delta_{1}\phi,\delta_{2}\phi)$
$\displaystyle=\int_{\mathscr{I}_{12}}c[2\delta_{1}\Pi_{b}\delta_{2}\Pi^{b}-\mathscr{L}_{n}(\delta_{1}A_{b}\delta_{2}\Pi^{b})]\hat{\tilde{\bm{\epsilon}}}$
(93)
$\displaystyle=2c\int_{\mathscr{I}_{12}}\delta_{1}\Pi_{b}\delta_{2}\Pi^{b}\hat{\tilde{\bm{\epsilon}}}-c\int_{\mathscr{I}_{12}}\mathscr{L}_{n}(\delta_{1}A_{b}\delta_{2}\Pi^{b}\hat{\tilde{\bm{\epsilon}}})$
$\displaystyle=2c\int_{\mathscr{I}_{12}}\delta_{1}\Pi_{b}\delta_{2}\Pi^{b}\hat{\tilde{\bm{\epsilon}}}+c(\int_{\mathcal{B}_{2}}\delta_{1}A_{b}\delta_{2}\Pi^{b}\bm{\varepsilon}-\int_{\mathcal{B}_{1}}\delta_{1}A_{b}\delta_{2}\Pi^{b}\bm{\varepsilon})$
$\displaystyle=2c\int_{\mathscr{I}_{12}}\delta_{1}\Pi_{b}\delta_{2}\Pi^{b}\hat{\tilde{\bm{\epsilon}}}+c(\int_{\mathcal{B}_{2}}\delta_{1}A^{b}\mathscr{L}_{n}\delta_{2}A_{b}\bm{\varepsilon}-\int_{\mathcal{B}_{1}}\delta_{1}A^{b}\mathscr{L}_{n}\delta_{2}A_{b}\bm{\varepsilon})$

where the portion of the null infinity $\mathscr{I}_{12}$, as depicted in Fig.
1, is bounded by its two cross sections $\mathcal{B}_{1}$ and
$\mathcal{B}_{2}$ with
$\bm{\varepsilon}\equiv-n\cdot\hat{\tilde{\bm{\epsilon}}}$ the induced volume
$2$-form on them. Similarly, the canonical energy for the gravitational part
can be obtained as[3]

$\displaystyle\mathcal{E}^{g}_{t}(\delta_{1}\phi,\delta_{2}\phi)$
$\displaystyle=c\int_{\mathscr{I}_{12}}\delta_{1}N_{ab}\delta_{2}N^{ab}\hat{\tilde{\bm{\epsilon}}}-\frac{c}{2}(\int_{\mathcal{B}_{2}}\tau_{1}^{ab}\delta_{2}N_{ab}\bm{\varepsilon}-\int_{\mathcal{B}_{1}}\tau_{1}^{ab}\delta_{2}N_{ab}\bm{\varepsilon})$
(94)
$\displaystyle=c\int_{\mathscr{I}_{12}}\delta_{1}N_{ab}\delta_{2}N^{ab}\hat{\tilde{\bm{\epsilon}}}+\frac{c}{2}(\int_{\mathcal{B}_{2}}\tau_{1}^{ab}\mathscr{L}_{n}\tau_{2ab}\bm{\varepsilon}-\int_{\mathcal{B}_{1}}\tau_{1}^{ab}\mathscr{L}_{n}\tau_{2ab}\bm{\varepsilon}),$

where $N_{ab}$ is the well known Bondi news tensor and its variation at
$\mathscr{I}$ satisfies

$\delta
N_{ab}=-\mathscr{L}_{n}\tau_{ab}+4\Omega^{-1}n_{(a}\tau_{b)c}n^{c}-\Omega^{-1}n^{c}n_{c}\tau_{ab}$
(95)

with $\tau_{ab}\equiv\Omega\delta g_{ab}$. The canonical energy for the
charged fluid part is automatically zero at null infinity since our charged
star has a spatial compact support.

Figure 1: Penrose diagram for a perturbed relativistic star in an
asymptotically flat spacetime. $\Sigma$ and $\Sigma^{\prime}$ are spacelike
Cauchy surface terminating at the spatial infinity $i^{0}$ and the boundary of
the red region is given by spacelike hypersurfaces $\mathscr{S}_{1}$ and
$\mathscr{S}_{2}$ as well as the portion of the null infinity
$\mathscr{I}_{12}$, which is bounded by its two cross-sections
$\mathcal{B}_{1}$ and $\mathcal{B}_{2}$.

On the other hand, to develop the relation of the canonical energy with the
second order perturbations, we would like to work in the gauge with the
Lagrangian displacement vanishing to all orders. Accordingly, we have

$\displaystyle\mathcal{E}_{\kappa}(\delta\phi,\delta\phi)$
$\displaystyle=\int_{\Sigma}[\bm{\omega}^{g}(g_{ab};\Delta
g_{ab},\mathscr{L}_{\kappa}\Delta g_{ab})+\bm{\omega}^{EM}(A_{a};\Delta
A_{a},\mathscr{L}_{\kappa}\Delta A_{a})]$ (96)
$\displaystyle=\int_{\Sigma}[\Delta\bm{\omega}^{g}(g_{ab};\Delta
g_{ab},\mathscr{L}_{\kappa}g_{ab})+\Delta\bm{\omega}^{EM}(A_{a};\Delta
A_{a},\mathscr{L}_{\kappa}A_{a})]$
$\displaystyle=\Delta^{2}\mathcal{Q}_{\kappa}+\int_{\Sigma}\Delta\\{\kappa^{a}\Delta[(2G^{b}{}_{a}-T_{EM}^{b}{}_{a}-\nabla_{c}F^{cb}A_{a})\epsilon_{befg}]+\kappa^{a}[(-G^{bc}+\frac{1}{2}T_{EM}^{bc})\Delta
g_{bc}+\nabla_{b}F^{bc}\Delta A_{c}]\epsilon_{aefg}\\}$
$\displaystyle=\delta^{2}\mathcal{Q}_{\kappa}+\int_{\Sigma}\kappa^{a}\Delta\\{[\Delta(T_{m}^{b}{}_{a}\epsilon_{befg})-\frac{1}{2}T_{m}^{bc}\Delta
g_{bc}\epsilon_{aefg}]+[\Delta(J^{b}A_{a}\epsilon_{befg})-J^{c}\Delta
A_{c}\epsilon_{aefg}]\\}.$

Here we have used Eq. (55) in the third step by replacing $\bm{C}_{a}$ and
$\bm{E}$ with the corresponding terms purely from gravitational and
electromagnetic fields. In addition, we have also used the fact that the ADM
charge is gauge invariant in the last step.

##  4 Physically stationary perturbations and dynamical stability

Dynamical stability we are concerned with is mode stability. That is to say,
our charged star in dynamic equilibrium is mode stable if there does not exist
any non-pure-gauge linearized solution with the time dependence of the form
$e^{\Lambda t}$ with $\Lambda>0$. Otherwise, it is mode unstable. Compared to
a complete analysis of linearized perturbation equations, one favorable way of
proving mode stability is to construct a positive-definite conserved norm on
the space of linear on-shell perturbations, because this precludes those
perturbations with exponential growth. A nice candidate for such a norm is the
canonical energy $\mathcal{E}_{t}$. To be more specific, let us first
introduce the physically stationary perturbations. A linear on-shell
perturbation $\delta\phi$ off of a stationary background is called physically
stationary if

$\displaystyle\mathscr{L}_{t}(\delta g_{ab}+\mathscr{L}_{X}g_{ab})$
$\displaystyle=0,$ (97) $\displaystyle\mathscr{L}_{t}(\delta
A_{a}+\mathscr{L}_{X}A_{a}+\nabla_{a}\vartheta)$ $\displaystyle=0,$
$\displaystyle\mathscr{L}_{t}(\delta\bm{\mathcal{N}}+\mathscr{L}_{X}\bm{\mathcal{N}})=-\mathscr{L}_{[t,\xi-X]}\bm{\mathcal{N}}$
$\displaystyle=0$ $\displaystyle\mathscr{L}_{t}(\delta
s+\mathscr{L}_{X}s)=-\mathscr{L}_{[t,\xi-X]}s$ $\displaystyle=0$

with $X^{a}$ the asymptotic symmetry generator and $\vartheta$ satisfying
$n^{a}\tilde{\nabla}_{a}\vartheta|_{\mathscr{I}}=0$, or equivalently

$\mathscr{L}_{t}\delta\phi=-\mathscr{L}_{[t,X]}\phi-\delta_{\mathscr{L}_{t}\vartheta}\phi+\text{trivial}.$
(98)

For the gravitational part, we first note that the news tensor vanishes for
all stationary spacetimes[22, 23], so the perturbed news tensor induced by
$\delta g_{ab}+\mathscr{L}_{X}g_{ab}$ vanishes. In addition, the news tensor
keeps vanishing under the asymptotic symmetry transformation on a stationary
spacetime, so the perturbed news tensor induced by $\mathscr{L}_{X}g_{ab}$
also vanishes. As a result, the perturbed news tensor vanishes for a
physically stationary perturbation. While for the electromagnetic part, we
first note that $\mathscr{L}_{t}\bm{A}=0$ implies $n\cdot\bm{F}\propto\bm{n}$
at $\mathscr{I}$. Moreover, the commutator $[t,X]^{a}$, at most, leads to an
asymptotic spatial translation, which is proportional to $n^{a}$ at the null
infinity. So we have $\mathscr{L}_{[t,X]}\bm{A}\propto\bm{n}$ at
$\mathscr{I}$. On the other hand,
$\mathscr{L}_{t}\vartheta|_{\mathscr{I}}=cn^{a}\tilde{\nabla}_{a}\vartheta=0$,
so we have $\tilde{\nabla}_{a}\mathscr{L}_{t}\vartheta\propto n_{a}$ at
$\mathscr{I}$. Then by Eq. (98), for a physically stationary perturbation, we
have $\mathscr{L}_{n}\delta A_{a}|_{\mathscr{I}}\propto n_{a}$, which vanishes
when restricted onto $\mathscr{I}$. Accordingly, according to Eq. (93) and Eq.
(94), we have that the canonical energy
$\mathcal{E}_{t}(\delta\phi,\delta\phi)$ across the portion of the null
infinity is apparently non-negative for both the gravitational part and the
electromagnetic one if $\mathcal{B}_{2}$ is where a physically stationary
perturbation solution settles down and $\mathcal{B}_{1}$ is chosen at the
spatial infinity where the boundary term is supposed to vanish. In addition,
the canonical energy $\mathcal{E}_{t}(\delta\phi,\delta_{\text{ps}}\phi)$
across the null infinity vanishes for any physically stationary perturbation
$\delta_{\text{ps}}\phi$, which implies

$\mathcal{E}_{t}(\delta\phi,\delta_{\text{ps}}\phi)=\mathcal{E}_{t}(\mathscr{S}_{1};\delta\phi,\delta_{\text{ps}}\phi)=\mathcal{E}_{t}(\mathscr{S}_{2};\delta\phi,\delta_{\text{ps}}\phi)$
(99)

with the canonical energy evaluated on $\mathscr{S}_{i}$ denoted by
$\mathcal{E}_{t}(\mathscr{S}_{i};\delta\phi,\delta_{\text{ps}}\phi)$.

Next let us define the subspace $\mathcal{V}\subset\mathcal{H}$ of the linear
on-shell perturbations such that $\delta\phi\in\mathcal{V}$ satisfies

$\mathcal{E}_{t}(\delta\phi,\delta_{\text{ps}}\phi)=0$ (100)

for any physically stationary perturbation $\delta_{\text{ps}}\phi$. This
amounts to saying that $\mathcal{V}$ is the symplectic complement of the
subspace $\mathcal{W}$ spanned by the perturbations of the form
$\mathscr{L}_{t}\delta_{\text{ps}}\phi$, i.e.,

$\delta
H_{\mathscr{L}_{t}\vartheta}=\langle\delta\phi,\hat{\Omega}\delta_{\mathscr{L}_{t}\vartheta}\phi\rangle=0,\quad\delta
H_{[t,X]}=\langle\delta\phi,\hat{\Omega}\mathscr{L}_{[t,X]}\phi\rangle=0,\quad\langle\delta\phi,\hat{\Omega}[\text{trivial}]\rangle=0,$
(101)

where the first requirement is trivial since it is automatically satisfied,
and the second one requires the perturbation should not change the ADM
$3$-momentum. Note that the double symplectic complement of $\mathcal{W}$ is
itself, thus the symplectic completement of $\mathcal{V}$ in itself is
$\mathcal{V}\cap\mathcal{W}$, which implies that the degeneracy of the
canonical energy is given precisely by the physically stationary perturbations
when restricted onto $\mathcal{V}$. With this observation, we have the
following criterion for the dynamical stability of our charged star.

Criterion for dynamical stability: If
$\mathcal{E}_{t}(\delta\phi,\delta\phi)\geq 0$ for each
$\delta\phi\in\mathcal{V}$, then our charged star has mode stability with
respect to the perturbations within $\mathcal{V}$. On the other hand, if there
exists some $\delta\phi\in\mathcal{V}$ such that
$\mathcal{E}_{t}(\delta\phi,\delta\phi)<0$, then our charged star has
instability in the sense that such a perturbation cannot approach a physically
stationary perturbation at late times.

It is noteworthy that the physically stationary perturbations play a vital
role in the above criterion. For the stability criterion,
$\mathcal{E}_{t}(\delta\phi,\delta\phi)=0$ only for the physically stationary
perturbations in $\mathcal{V}$. This can be argued by contradiction. Suppose
that we have $\mathcal{E}_{t}(\delta\phi,\delta\phi)=0$ for a perturbation
$\delta\phi\in\mathcal{V}$, which is not a physically stationary one. Then
there exists another perturbation $\delta^{\prime}\phi\in\mathcal{V}$ such
that $\mathcal{E}_{t}(\delta^{\prime}\phi,\delta\phi)\neq 0$. Whence we have

$\mathcal{E}_{t}(\delta^{\prime}\phi+z\delta\phi,\delta^{\prime}\phi+z\delta\phi)=\mathcal{E}_{t}(\delta^{\prime}\phi,\delta^{\prime}\phi)+2z\mathcal{E}_{t}(\delta^{\prime}\phi,\delta\phi),$
(102)

which can become negative by an appropriate choice of $z$. But this
contradicts with the assumption that the canonical energy $\mathcal{E}_{t}$ is
non-negative in $\mathcal{V}$. Thus for those perturbations which are not
physically stationary ones, the canonical energy $\mathcal{E}_{t}$ provides a
positive definite conserved norm during the evolution from one Cauchy surface
$\Sigma$ to a later one $\Sigma^{\prime}$, as depicted in Fig. 1, guaranteeing
the mode stability. On the other hand, a physically stationary perturbation is
stable by definition, albeit its zero canonical energy. While for the
instability criterion, we can also argue for it by contradiction. Suppose that
the negative canonical energy mode $\delta\phi$ approaches a physically
stationary solution $\delta_{\text{ps}}\phi$ at late times, say
$\mathscr{S}_{2}$ in Fig. 1, then we have

$\mathcal{E}_{t}(\mathscr{S}_{2};\delta_{\text{ps}}\phi,\delta_{\text{ps}}\phi)\leq\mathcal{E}_{t}(\delta\phi,\delta\phi)<0$
(103)

due to the non-negative net flux across the null infinity. On the other hand,
Eq. (99) gives rise to

$\mathcal{E}_{t}(\mathscr{S}_{2};\delta_{\text{ps}}\phi,\delta_{\text{ps}}\phi)=\mathcal{E}_{t}(\delta\phi,\delta_{\text{ps}}\phi)=0$
(104)

according to the definition of $\mathcal{V}$. This leads to an obvious
contradiction. So we are done.

However, such a criterion has its apparent shortcoming because one can only
talk about the stability of the perturbations within the subspace
$\mathcal{V}$. Gratefully, the second condition in Eq. (101) imposed on
$\mathcal{V}$ can always be achieved by adding a perturbation generated by the
Lorentz boosts on the background. Since such an added perturbation is stable,
the above criterion for the stability of the perturbations in $\mathcal{V}$
also implies the criterion for the stability of the perturbations in the
larger subspace with the second condition in Eq. (101) disregarded. On the
other hand, regarding the third condition in Eq. (101), we also have good news
for both non-axisymmetric and axisymmetric perturbations, on which we shall
elaborate separately below.

###  4.1 Nonaxisymmetric perturbations

As to the neutral star, Friedman shows that the aforementioned third condition
does not lead to a real physical restriction for nonaxisymmetric
perturbations, because it can be achieved in the suitable background solution
by adding a trivial perturbation[15]. Here we focus only onto the background
in which $\nabla_{a}s\neq 0$ to show the key strategy developed in [15] for
the neutral star as well as its utterly equal applicability to our charged
star.

First, by working in the gauge $\delta\phi=(\Delta g_{ab},\Delta A_{a},0)$, we
have

$\displaystyle\langle\delta\phi,\hat{\Omega}(0,0,\tilde{\eta})$
$\displaystyle=\int_{\Sigma}\tilde{\eta}^{a}\Delta P_{aefg}$ (105)
$\displaystyle=\int_{\Sigma}\frac{1}{n^{2}}\mathcal{N}^{abc}\nabla_{b}Z_{c}\Delta
P_{aefg}$
$\displaystyle=\int_{\Sigma}\nabla_{b}Z_{c}\Delta(\frac{1}{n^{2}}\mathcal{N}^{abc}P_{aefg})$
$\displaystyle=6\int_{\Sigma}\nabla_{[e}Z_{f}\Delta(\frac{\rho+p}{n}u_{g]}+A_{g]})$
$\displaystyle=\int_{\Sigma}d\bm{Z}\wedge\Delta(\frac{\rho+p}{n}\bm{u}+\bm{A})$
$\displaystyle=\int_{\Sigma}\bm{Z}\wedge\Delta
d(\frac{\rho+p}{n}\bm{u}+\bm{A})$ $\displaystyle=\int_{\Sigma}F\Delta[ds\wedge
d(\frac{\rho+p}{n}\bm{u}+\bm{A})],$

where we have used
$\Delta(\frac{1}{n^{2}}\mathcal{N}^{abc})=u^{[a}\Lambda^{bc]}$ for some
antisymmetric tensor $\Lambda^{bc}$ and $u\cdot d\bm{Z}=u^{a}\bm{P}_{a}=0$ in
the third step. We define the vorticity and circulation in the presence of the
electromagnetic field as $\bm{\omega}=d(\frac{\rho+p}{n}\bm{u}+\bm{A})$ and
$\bm{\Gamma}=ds\wedge\bm{\omega}$, respectively. Then we have

$\displaystyle u\cdot\bm{\omega}$
$\displaystyle=u\cdot[d(\frac{\rho+p}{n})\wedge\bm{u}+\frac{\rho+p}{n}d\bm{u}+\bm{F}]$
(106)
$\displaystyle=d(\frac{\rho+p}{n})+\frac{\rho+p}{n}u^{a}\nabla_{a}u_{b}+u^{a}F_{ab}$
$\displaystyle=d(\frac{\rho+p}{n})-\frac{dp}{n}=Tds,$

where we have used the first equation of Eq. (1) as well as the fact that both
the particle number density and the entropy per particle along the divergence
free $4$-velocity are constant for our charged star in dynamic equilibrium.
Whence we obtain Ertel’s theorem

$\mathscr{L}_{u}\bm{\omega}=dT\wedge ds$ (107)

as well as $u\cdot\bm{\Gamma}=\mathscr{L}_{u}\bm{\Gamma}=0$ in the presence of
the electromagnetic field, which implies $\bm{\Gamma}=\gamma\bm{\mathcal{N}}$
and $\mathscr{L}_{u}\gamma=0$. On the other hand, note that

$\langle(0,0,\tilde{\eta}^{\prime}),\hat{\Omega}(0,0,\tilde{\eta})\rangle=\int_{\Sigma}F\mathscr{L}_{\tilde{\eta}^{\prime}}\bm{\Gamma}$
(108)

for any trivial displacement $\tilde{\eta}^{\prime
a}=\frac{1}{2n^{2}}\mathcal{N}^{abc}(dF^{\prime}\wedge ds)_{bc}$. So we are
required to ask whether there exists a trivial displacement such that
$\mathscr{L}_{\tilde{\eta}^{\prime}}\bm{\Gamma}=-\Delta\bm{\Gamma}$ on
$\Sigma$, i.e.,

$\displaystyle-\Delta\gamma$
$\displaystyle=\mathscr{L}_{\tilde{\eta}^{\prime}}\gamma=\frac{1}{n^{2}}\mathcal{N}^{abc}(dF^{\prime})_{b}(ds)_{c}(d\gamma)_{a}=\frac{1}{n}\epsilon^{abcd}(d\gamma)_{a}(ds)_{b}(dF^{\prime})_{c}u_{d}$
(109) $\displaystyle=\frac{\frac{\partial
F^{\prime}}{\partial\varphi}}{n}\epsilon^{abcd}(d\gamma)_{a}(ds)_{b}[(d\varphi)_{c}-\Omega(dt)_{c}]u_{d}=\frac{\frac{\partial
F^{\prime}}{\partial\varphi}}{n}\epsilon^{abcd}(d\gamma)_{a}(ds)_{b}(d\varphi)_{c}(dt)_{d}(u_{t}+\Omega
u_{\varphi})$ $\displaystyle=\frac{\frac{\partial
F^{\prime}}{\partial\varphi}}{n}\epsilon^{abcd}(d\gamma)_{a}(ds)_{b}(d\varphi)_{c}(dt)_{d}(g_{tt}+\Omega
g_{t\varphi}+\Omega g_{\varphi
t}+\Omega^{2}g_{\varphi\varphi})/|v|=-\frac{|v|\frac{\partial
F^{\prime}}{\partial\varphi}}{n}\epsilon^{abcd}(d\gamma)_{a}(ds)_{b}(d\varphi)_{c}(dt)_{d},$

where we have assumed that $(\gamma,s,t,\varphi)$ can offer a local atlas for
our charged star with $t^{a}(dt)_{a}=\varphi^{a}(d\varphi)_{a}=1$ and used
$\mathscr{L}_{u}F^{\prime}=0$. With this, the answer is definitely yes,
because one can always find a $F^{\prime}$ on $\Sigma$ such that

$\frac{\partial
F^{\prime}}{\partial\varphi}=\frac{n\Delta\gamma}{|v|\epsilon^{abcd}(d\gamma)_{a}(ds)_{b}(d\varphi)_{c}(dt)_{d}}.$
(110)

The upshot is that the third condition in Eq. (101) does not correspond to a
real physical restriction on the nonaxisymmetric perturbations in most of the
cases we are interested in, since it can be fulfilled by supplementing a
trivial perturbation.

###  4.2 Axisymmetric perturbations

For an arbitrary axisymmetric perturbation $\delta\mathcal{Q}=(\delta
g_{ab},\delta A_{a},\delta\bm{\mathcal{N}},\delta s)$, which may not be
described by our Lagrangian formulation, one can follow exactly the same
construction developed in [7] for the neutral star to show333The derivation
presented in [7] has used $u^{a}\nabla_{a}s=0$ and $u^{a}\nabla_{a}j=0$ for
axisymmetric perturbations. The validity of the former is obvious since it
holds actually for arbitrary perturbations. The validity of the latter also
for our charged star can be seen through
$0=\varphi_{b}(\nabla_{a}T_{m}^{ab}-F^{bc}J_{c})=\nabla_{a}(jJ^{a}+p\varphi^{a})=\nabla_{a}(jJ^{a})=J^{a}\nabla_{a}j$.

$\mathscr{L}_{t}\delta\bm{\mathcal{N}}=-\mathscr{L}_{\xi}\bm{\mathcal{N}},\quad\mathscr{L}_{t}\delta
s=-\mathscr{L}_{\xi}s,\quad\mathscr{L}_{t}\delta j=-\mathscr{L}_{\xi}j,$ (111)

where $\xi^{a}=|v|\delta u^{a}+\beta\varphi^{a}$ with an arbitrary
axisymmetric scalar $\beta$ satisfying $u^{a}\nabla_{a}\beta=\delta
u^{a}\nabla_{a}\Omega$. This means that $\mathscr{L}_{t}\delta\mathcal{Q}$ can
be realized in our Lagrangian description as
$\hat{\delta}\phi=(\mathscr{L}_{t}\delta g_{ab},\mathscr{L}_{t}\delta
A_{a},\xi^{a})$ with $\Delta j=0$. Now for an arbitrary axisymmetric trivial
displacement $\tilde{\eta}^{a}$, we have

$\displaystyle\langle\mathscr{L}_{t}\hat{\delta}\phi,\hat{\Omega}(0,0,\tilde{\eta})\rangle$
$\displaystyle=-\langle\hat{\delta}\phi,\hat{\Omega}(0,0,\mathscr{L}_{t}\tilde{\eta})\rangle=-\langle\hat{\delta}\phi,\hat{\Omega}(0,0,\gamma\varphi)\rangle=\int_{\Sigma}\gamma\varphi^{a}\Delta
P_{abcd}$ (112)
$\displaystyle=\int_{\Sigma}\gamma\varphi^{a}\Delta[\frac{\rho+p}{n}(n\epsilon_{abcd}+u_{a}\mathcal{N}_{bcd})-A_{f}J^{f}\epsilon_{abcd}+A_{a}\mathcal{N}_{bcd}]$
$\displaystyle=\int_{\Sigma}\gamma\mathcal{N}_{bcd}\Delta[(\frac{\rho+p}{n}u_{a}+A_{a})\varphi^{a}]=\int_{\Sigma}\gamma\bm{\mathcal{N}}\Delta
j=0,$

where we have used $\mathscr{L}_{t}\tilde{\eta}=\gamma\varphi^{a}$ for some
function $\gamma$[7]. Similarly, one can also have

$\langle\mathscr{L}_{t}\hat{\delta}\phi,\hat{\Omega}\mathscr{L}_{[t,X]}\phi\rangle=-\langle\hat{\delta}\phi,\hat{\Omega}\mathscr{L}_{Y}\phi\rangle=0,$
(113)

where we have used the fact that $Y^{a}=[t,[t,X]]^{a}$ vanishes at infinity
for any asymptotic symmetry generator $X^{a}$.

Eq. (112) and Eq. (113) imply that
$\mathscr{L}_{t}^{2}\delta\mathcal{Q}\in\mathcal{V}$ for an arbitrary
axisymmetric perturbation $\delta\mathcal{Q}$. On the other hand, the mode
stability for $\delta\mathcal{Q}$ is equivalent to the mode stability for
$\mathcal{L}_{t}^{2}\delta\mathcal{Q}$. Thus the mode stability for
perturbations within $\mathcal{V}$ implies the mode stability for all
perturbations in axisymmetric case.

##  5 Thermodynamic stability and axisymmetric perturbations

Our charged star in thermodynamic equilibrium is said to be linearly
thermodynamically stable if $\delta^{2}S$ is negative for all the linear on-
shell perturbations with $\delta\mathcal{M}=\delta
N=\delta\mathcal{J}=\delta^{2}\mathcal{M}=\delta^{2}N=\delta^{2}\mathcal{J}=0$.
Note that the variation of the first law of thermodynamics gives

$\delta^{2}\mathcal{M}-\tilde{T}\delta^{2}S-\tilde{\mu}\delta^{2}N-\Omega\delta^{2}\mathcal{J}=\int_{\Sigma}(\delta\tilde{T}\delta(s\bm{\mathcal{N}})+\delta\tilde{\mu}\delta\bm{\mathcal{N}}-\delta\Omega
d\delta\bm{Q}_{\varphi}),$ (114)

which implies that the left side, denoted by $\mathcal{C}$ later on, depends
solely on the first order variation of our system. Thus the criterion for the
thermodynamic stability of our charged star is the positivity of $\mathcal{C}$
for all the linear on-shell perturbations with $\delta\mathcal{M}=\delta
N=\delta\mathcal{J}=0$. In our Lagrangian description, we have $\delta
N=\delta S=\delta^{2}N=\delta^{2}S=0$ automatically444We also have
$\delta\mathcal{M}=0$ since we are always working within the subspace
$\mathcal{H}$.. Then by Eq. (96), we obtain

$\mathcal{C}=\mathcal{E}_{v}-\int_{\Sigma}v^{a}\Delta\\{[\Delta(T_{m}^{b}{}_{a}\epsilon_{befg})-\frac{1}{2}T_{m}^{bc}\Delta
g_{bc}\epsilon_{aefg}]+[\Delta(J^{b}A_{a}\epsilon_{befg})-J^{c}\Delta
A_{c}\epsilon_{aefg}]\\}$ (115)

for the Killing field $v^{a}$, where $\mathcal{E}_{v}$ can be understood as
the canonical energy in the comoving frame. We borrow directly from [20] the
following two identities

$\displaystyle u^{a}[J^{c}\Delta
A_{c}\epsilon_{aefg}-\Delta(J^{b}A_{a}\epsilon_{befg})]$
$\displaystyle=-u\cdot\bm{A}\Delta\bm{\mathcal{N}},$ (116) $\displaystyle
u^{a}[\frac{1}{2}T_{m}^{bc}\Delta
g_{bc}\epsilon_{aefg}-\Delta(T_{m}^{b}{}_{a}\epsilon_{befg})]$
$\displaystyle=\mu\Delta\bm{\mathcal{N}}+T\Delta(s\bm{\mathcal{N}}).$

By the Lagrangian variation of these two identities, we further have

$\displaystyle u^{a}\Delta[J^{c}\Delta
A_{c}\epsilon_{aefg}-\Delta(J^{b}A_{a}\epsilon_{befg})]$ $\displaystyle=0,$
(117) $\displaystyle u^{a}\Delta[\frac{1}{2}T^{bc}\Delta
g_{bc}\epsilon_{aefg}-\Delta(T^{b}{}_{a}\epsilon_{befg})]$ $\displaystyle=0,$

where we have used $\Delta u^{a}\propto u^{a}$ as well as the fact that any
order of the Lagrangian variation of $\bm{\mathcal{N}}$ and
$s\bm{\mathcal{N}}$ vanishes. Accordingly, we obtain

$\mathcal{C}=\mathcal{E}_{v},$ (118)

whereby we further have

$\mathcal{C}=\mathcal{E}_{t}$ (119)

for axisymmetric perturbations. With this observation, we obtain the following
criterion for the thermodynamic stability of our charged star.

Criterion for thermodynamic stability: For our charged star in thermodynamic
equilibrium, the necessary condition for its thermodynamic stability is the
positivity of $\mathcal{E}_{v}$ for all the linear on-shell perturbations with
$\delta\mathcal{J}=0$, and the necessary condition for its thermodynamic
stability with respect to all the linear on-shell axisymmetric perturbations
with $\delta\mathcal{J}=0$ is the positivity of the corresponding
$\mathcal{E}_{t}$.

Here the necessary condition can be replaced by the necessary and sufficient
condition if and only if $\frac{\delta s}{|D_{a}s|}$ is bounded for all
allowable perturbations of $s$, since only in this case can the perturbations
be implemented within our Lagrangian description[15]. One such an application
scenario is provided by the so-called isentropic star, which is regarded
generally as a good approximation at least for the convective part of a real
life star[24]. For such an isentropic star, we have a uniform entropy per
particle throughout the whole star. As a result, $\frac{\delta s}{|D_{a}s|}$
is bounded in the sense that not only do we have $D_{a}s=0$ but also the
allowable $\delta s=0$. In particular, let us consider the linear on-shell
spherically symmetric perturbations of a static, spherically symmetric
isentropic charged star, for which we have $\delta\mathcal{J}=0$
automatically. On the other hand, we also have no spherically symmetric
trivial displacement because of

$0=\mathscr{L}_{\xi\frac{\partial}{\partial
r}}\bm{\mathcal{N}}=d(\xi\frac{\partial}{\partial
r}\cdot\bm{\mathcal{N}})=d(n\sqrt{h}\xi d\theta\wedge
d\varphi)=\frac{\partial(n\sqrt{h}\xi)}{\partial r}dr\wedge d\theta\wedge
d\varphi$ (120)

subject to the boundary condition $\xi=0$ at $r=0$ in the spherical
coordinates. Together with the ADM $3$-momentum unchanged under the
spherically symmetric perturbations, we know that there is no restriction on
$\mathcal{V}$ for the spherically symmetric perturbations. Thus we end up with
the corollary that the dynamic stability is equivalent to the thermodynamic
stability for the spherically symmetric perturbations of the static,
spherically symmetric isentropic charged star, which has also been obtained
most recently in [25] instead by performing explicit calculations in the
spherical coordinates.

##  6 Discussions

As promised, we have accomplished a thorough analysis of the dynamic and
thermodynamic stability for the charged star. As we show explicitly in our
paper, neither the presence of the electromagnetic field nor the Lorentz force
experienced by the charged fluid makes any obstruction to the key steps
towards the results obtained previously in [7] for the neutral star. Thus our
main results for the charged star are in close parallel with those presented
in [7] for the neutral star, but with various improvements scattered in our
paper.

We conclude our paper with two interesting issues worthy of further
investigation. The first one is related to the Gubser-Mitra conjecture. As
alluded to in the introduction section, Hollands and Wald proved the Gubser-
Mitra conjecture for black branes in AdS spacetimes by resorting to the
criterion they established for the dynamic and thermodynamic stability of the
black holes[3]. Thus it is tempting to ask whether one can obtain the similar
equivalence between the dynamic instability and the thermodynamic instability
for the AdS planar star by the same token. On the other hand, new progress has
recently been made towards the Lagrangian formulation of dissipative fluids
and its various implications[26, 27, 28, 29, 30], so it is also interesting to
explore what new insights one can gain by applying the Wald formalism to it.
We hope to report some progress along these two lanes in the future.

## Acknowledgements

We are grateful to Bob Wald for his helpful communications regarding his work.
We also thank Hu Zhu for his inspiring discussions on some issues related to
this work. This work is partly supported by the National Key Research and
Development Program of China Grant No. 2021YFC2203001 as well as the NSFC
under Grant Nos. 11975239, 12005088, 12035016, 12075026, and 12275350. JZ is
also supported by the Beijing Research Fund for Talented Undergraduates.

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# The Buchweitz set of a numerical semigroup

S. Eliahou, J.I. García-García,
D. Marín-Aragón, A. Vigneron-Tenorio

###### Abstract

Let $A\subset\mathbb{Z}$ be a finite subset. We denote by
$\operatorname{\mathcal{B}}(A)$ the set of all integers $n\geq 2$ such that
$|nA|>(2n-1)(|A|-1)$, where $nA=A+\cdots+A$ denotes the $n$-fold sumset of
$A$. The motivation to consider $\operatorname{\mathcal{B}}(A)$ stems from
Buchweitz’s discovery in 1980 that if a numerical semigroup
$S\subseteq\mathbb{N}$ is a Weierstrass semigroup, then
$\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)=\emptyset$. By constructing
instances where this condition fails, Buchweitz disproved a longstanding
conjecture by Hurwitz (1893). In this paper, we prove that for any numerical
semigroup $S\subset\mathbb{N}$ of genus $g\geq 2$, the set
$\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)$ is finite, of unbounded
cardinality as $S$ varies.

> Keywords and phrases. Weierstrass numerical semigroup; gapset; additive
> combinatorics; sumset growth; Freiman’s $3k-3$ theorem.
>
> _MSC-class:_ 14H55, 11P70, 20M14

_To the memory of our friend and colleague
Fernando Torres (1961-2020)_

##  1 Introduction

Denote $\mathbb{N}=\\{0,1,2,3,\dots\\}$ and
$\mathbb{N}_{+}=\mathbb{N}\setminus\\{0\\}=\\{1,2,3,\dots\\}$. For
$a,b\in\mathbb{Z}$, let $[a,b[=\\{z\in\mathbb{Z}\mid a\leq z<b\\}$ and
$[a,\infty[=\\{z\in\mathbb{Z}\mid a\leq z\\}$ denote the integer intervals
they span. A _numerical semigroup_ is a subset $S\subseteq\mathbb{N}$
containing $0$, stable under addition and with finite complement in
$\mathbb{N}$. Equivalently, it is a subset of $\mathbb{N}$ of the form
$S=\langle a_{1},\dots,a_{n}\rangle=\mathbb{N}a_{1}+\dots+\mathbb{N}a_{n}$
where $\gcd(a_{1},\dots,a_{n})=1$. The set $\\{a_{1},\dots,a_{n}\\}$ is then
called a _system of generators_ of $S$, and the smallest such $n$ is called
the _embedding dimension_ of $S$.

For a numerical semigroup $S$, its corresponding _gapset_ is the complement
$G=\mathbb{N}\setminus S$, its _genus_ is $g=|G|$, its _multiplicity_ is
$m=\min S^{*}$ where $S^{*}=S\setminus\\{0\\}$, its _Frobenius number_ is
$f=\max(\mathbb{Z}\setminus S)$ and its _conductor_ is $c=f+1$. Thus
$[c,\infty[\,\subseteq S$ and $c$ is minimal for this property. Finally, the
_depth_ of $S$ is $q=\lceil c/m\rceil$.

Given a finite subset $A\subset\mathbb{N}$, we denote by $nA=A+\dots+A$ the
_$n$ -fold sumset of $A$._ See Section 2 for more details.

######  Definition 1.1.

Let $A\subset\mathbb{Z}$ be a finite subset. We associate to $A$ the function
$\beta=\beta_{A}\colon\mathbb{N}_{+}\to\mathbb{Z}$ defined for all $n\geq 1$
by

$\beta_{A}(n)=|nA|-(2n-1)(|A|-1).$

######  Notation 1.2.

We denote by $\operatorname{\mathcal{B}}(A)$ the _positive support_ in
$2+\mathbb{N}$ of the function $\beta_{A}$, i.e.

$\operatorname{\mathcal{B}}(A)=\\{n\geq 2\mid\beta_{A}(n)\geq 1\\}.$

For instance, $2\in\operatorname{\mathcal{B}}(A)$ if and only if $|2A|\geq
3|A|-2$. Interestingly, the failure of this condition, namely the inequality
$|2A|\leq 3|A|-3$, is the key hypothesis of the famous Freiman’s $3k-3$
Theorem in additive combinatorics [5].

######  Example 1.3.

If $|A|=0$ or $1$, then $\operatorname{\mathcal{B}}(A)$ is infinite. Indeed,
if $A=\emptyset$, then $|nA|=0$ and so $\beta_{\emptyset}(n)=2n-1$ for all
$n\geq 1$. Thus $\operatorname{\mathcal{B}}(\emptyset)=2+\mathbb{N}$ in that
case. Similarly, if $|A|=1$, then $\beta_{A}(n)=1$ for all $n\geq 1$. So here
again $\operatorname{\mathcal{B}}(A)=2+\mathbb{N}$.

In sharp contrast, Theorem 3.3 below states that _if $S\subset\mathbb{N}$ is a
numerical semigroup of genus $g\geq 2$, then
$\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)$ is finite_.

######  Example 1.4.

Let $S=\langle 3,7\rangle$. Then $\mathbb{N}\setminus S=\\{1,2,4,5,8,11\\}$
and $\beta_{\mathbb{N}\setminus S}(n)=0$ for all $n\geq 2$ as easily seen. In
particular, $\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)=\emptyset$.

More generally, it was shown in [8, 10] that $\beta_{\mathbb{N}\setminus
S}(n)=0$ for all _symmetric_ numerical semigroups $S$ of multiplicity $m\geq
3$ and all $n\geq 2$. We shall not use this result below, but instead give a
short self-contained proof of an immediate consequence, namely that
$\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)$ is empty in that case.

In fact, $\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)$ is empty in most
cases. Indeed, Buchweitz discovered in 1980 that the condition
$\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)=\emptyset$ is necessary for
$S$ to be a Weierstrass semigroup. By constructing instances where this
condition fails, Buchweitz [1] was able to negate the longstanding conjecture
by Hurwitz [6] according to which all numerical semigroups of genus $g\geq 2$
are Weierstrass semigroups. His first counterexample was $S=\langle
13,14,15,16,17,18,20,22,23\rangle$, with corresponding gapset

$G=\mathbb{N}\setminus S=[1,12]\cup\\{19,21,24,25\\}$

of cardinality $16$. Then $2G=[2,50]\setminus\\{39,41,47\\}$, so that
$|2G|=46$ and $\beta_{G}(2)=46-3\cdot 15=1$, implying
$2\in\operatorname{\mathcal{B}}(G)$ and thus impeding $S$ to be a Weierstrass
semigroup. For more information on Buchweitz’s condition and Weierstrass
semigroups, see e.g. [3, 7].

Here are the contents of this paper. In Section 2, we recall a result of
Nathanson in additive combinatorics and we use it to study the asymptotic
behavior of the function $\beta_{A}(n)$. In Section 3, we introduce the
Buchweitz set of a numerical semigroup and we prove our main results. Section
4 concludes the paper with open questions on the possible shapes of the sets
$\operatorname{\mathcal{B}}(A)$.

##  2 Sumset growth

Given finite subsets $A,B$ of a commutative monoid $(M,+)$, we denote as usual

$A+B=\\{a+b\mid a\in A,b\in B\\},$

the _sumset_ of $A,B$, and $2A=A+A$. More generally, if $n\geq 2$, we denote
$nA=A+(n-1)A$, where $1A=A$. The set $nA$ is called the $n$-fold sumset of
$A$.

A classical question in additive combinatorics is, how does $|nA|$ grow with
$n$? Here we only consider the case $M=\mathbb{Z}$. We shall need the
following result of Nathanson [9, Theorem 1.1].

######  Theorem 2.1.

Let $A_{0}\subset\mathbb{N}$ be a finite subset of cardinality $k\geq 2$,
containing $0$ and such that $\gcd(A_{0})=1$. Let $a_{0}=\max(A_{0})$. Then
there exist integers $c,d$ and subsets $C\subseteq[0,c-2],\,D\subseteq[0,d-2]$
such that

$nA_{0}=C\sqcup[c,a_{0}n-d]\sqcup(a_{0}n-D)$

for all $n\geq\max\\{(|A_{0}|-2)(a_{0}-1)a_{0},1\\}$.

As pointed out in [9], the hypotheses $0\in A_{0}$ and $\gcd(A_{0})=1$ are not
really restrictive. Indeed, for any finite set $A\subset\mathbb{Z}$ with
$|A|\geq 2$, the simple transformation $A\mapsto A_{0}=(A-\alpha)/d$, where
$\alpha=\min(A)$ and $d=\gcd(A-\alpha)$, yields a set $A_{0}$ satisfying these
hypotheses and such that $|nA_{0}|=|nA|$ for all $n$. In view of our
applications to gapsets, we shall need the following version.

######  Corollary 2.2.

Let $A\subset\mathbb{N}_{+}$ be a finite subset containing $\\{1,2\\}$. Let
$a=\max(A)$. Then there is an integer $b\leq 1$ such that

$|nA|=(a-1)n+b$

for all $n\geq(|A|-2)(a-2)(a-1)$.

###### Proof.

Set $A_{0}=A-1$ and $a_{0}=a-1$. Then $A_{0}$ contains $\\{0,1\\}$, hence it
satisfies the hypotheses of Theorem 2.1. Using the same notation, its
conclusion implies

(1) $|nA_{0}|=a_{0}n+b$

for all $n\geq\max\\{(|A_{0}|-2)(a_{0}-1)a_{0},1\\}$, where

$b=|C|+|D|-c-d+1.$

Note that $b\leq 1$ since $|C|\leq\max(0,c-1)$, $|D|\leq\max(0,d-1)$. The
desired statement follows from (1) since $|nA|=|nA_{0}|$ for all $n\geq 0$. ∎

###  2.1 Asymptotic behavior of $\beta_{A}(n)$

We now study the evolution of $\beta_{A}(n)$ as $n$ grows.

######  Theorem 2.3.

Let $A\subset\mathbb{N}_{+}$ be a finite set containing $\\{1,2\\}$. Let
$f=\max(A)$ and $g=|A|$. Then

$\lim_{n\to\infty}\beta_{A}(n)=\left\\{\begin{array}[]{rcl}-\infty&\textrm{if}&f\leq
2g-2,\\\ +\infty&\textrm{if}&f\geq 2g.\end{array}\right.$

Finally if $f=2g-1$, then $\beta_{A}(n)$ is constant and nonpositive for $n$
large enough.

###### Proof.

By Corollary 2.2, we have $|nA|=(f-1)n+b$ for some integer $b\leq 1$ and for
$n$ large enough. Hence

$\displaystyle\beta_{A}(n)$ $\displaystyle=$
$\displaystyle(f-1)n+b-(2n+1)(g-1)$ $\displaystyle=$
$\displaystyle(f-2g+1)n+b+1-g$

for $n$ large enough. The claims for $f\leq 2g-2$ and $f\geq 2g$ follow. If
$f=2g-1$, then $\beta_{A}(n)=b+1-g\leq 0$ for $n$ large enough, since $b\leq
1$ and $g\geq 2$. ∎

######  Corollary 2.4.

Let $A\subseteq\mathbb{N}_{+}$ be a finite set containing $\\{1,2\\}$. Let
$f=\max(A)$ and $g=|A|$. Then $\operatorname{\mathcal{B}}(A)$ is finite if and
only if $f\leq 2g-1$.

###### Proof.

If $f\geq 2g$, then $\lim_{n\to\infty}\beta_{A}(n)=\infty$ by the theorem,
whence $\beta_{A}(n)\geq 1$ for all large enough $n$. Thus
$\operatorname{\mathcal{B}}(A)$ is infinite in this case. If $f\leq 2g-1$, the
theorem implies $\beta_{A}(n)\leq 0$ for $n$ large enough, whence
$\operatorname{\mathcal{B}}(A)$ is finite in that case. ∎

##  3 Application to numerical semigroups

######  Definition 3.1.

Let $S\subseteq\mathbb{N}$ be a numerical semigroup. We define the _Buchweitz
set of $S$_ as
$\operatorname{Buch}(S)=\operatorname{\mathcal{B}}(\mathbb{N}\setminus S)$.
Explicitly, setting $G=\mathbb{N}\setminus S$, we have

$\displaystyle\operatorname{Buch}(S)$ $\displaystyle=$ $\displaystyle\\{n\geq
2\mid|nG|>(2n-1)(|G|-1)\\}$ $\displaystyle=$ $\displaystyle\\{n\geq
2\mid\beta_{G}(n)\geq 1\\}.$

In this section, we first prove that $\operatorname{Buch}(S)$ is _finite_ for
all numerical semigroups $S$ of genus $g\geq 2$. We then show, by explicit
construction, that the cardinality of $\operatorname{Buch}(S)$ may be
arbitrarily large.

###  3.1 Finiteness of $\operatorname{Buch}(S)$

We start with a well known inequality linking the Frobenius number and the
genus of a numerical semigroup.

######  Proposition 3.2.

Let $S\subset\mathbb{N}$ be a numerical semigroup with Frobenius number $f$
and genus $g\geq 1$. Then $f\leq 2g-1$.

###### Proof.

Let $x\in S\cap[0,f]$. Then $f-x\notin S$ since $S$ is stable under addition
and $x+(f-x)=f\notin S$. Hence, the map $x\mapsto f-x$ induces an injection

$S\cap[0,f]\hookrightarrow\mathbb{N}\setminus S.$

Since $|S\cap[0,f]|=(f+1)-g$, it follows that $f\leq 2g-1$, as claimed. ∎

Recall that $S$ is said to be _symmetric_ if
$|S\cap[0,f]|=|\mathbb{N}\setminus S|$, i.e. if $f=2g-1$. A classical result
of Sylvester states that any numerical semigroup of the form $S=\langle
a,b\rangle$ with $\gcd(a,b)=1$ is symmetric.

######  Theorem 3.3.

Let $S\subseteq\mathbb{N}$ be a numerical semigroup of genus $g\geq 2$. Then
$\operatorname{Buch}(S)$ is finite.

###### Proof.

Let $G=\mathbb{N}\setminus S$. Then
$\operatorname{Buch}(S)=\operatorname{\mathcal{B}}(G)$ by definition. We have
$g=|G|\geq 2$. Let $f=\max(G)$ be the Frobenius number of $S$. Let
$m=\min(S\setminus\\{0\\})$ be the multiplicity of $S$. Then $m\geq 2$ since
$g\geq 2$, and $[1,m-1]\subseteq G$.

Assume first $m\geq 3$. Then $\\{1,2\\}\subseteq G$. Hence Corollary 2.4
applies, and since $f\leq 2g-1$ by Proposition 3.2, it yields that
$\operatorname{\mathcal{B}}(G)$ is finite, as desired.

Assume now $m=2$. Then $S=\langle 2,b\rangle$ with $b$ odd and $b\geq 5$ since
$|G|\geq 2$. At this point, we might conclude the proof right away using what
is known in the symmetric case [8, 10]. However, for the convenience of the
reader, let us give a short self-contained argument. We have
$G=\\{1,3,\dots,b-2\\}$, i.e. all odd numbers from $1$ to $b-2$. Hence
$G-1=\\{0,2,\dots,b-3\\}$ and $\gcd(G-1)=2$. Set $A=(G-1)/2=[0,k]$, where
$k=(b-3)/2$. For all $n\geq 1$, we have

$|nG|=|nA|=|nk+1|=n(|G|-1)+1.$

Therefore $\beta_{G}(n)=-(n-1)|G|+n$, whence $\beta_{G}(n)\leq 0$ for all
$n\geq 2$. It follows that $\operatorname{\mathcal{B}}(G)=\emptyset$ and we
are done. ∎

###  3.2 Unboundedness of $|\operatorname{Buch}(S)|$

We show here, by explicit construction, that $|\operatorname{Buch}(S)|$ may be
arbitrarily large.

######  Proposition 3.4.

For any integer $b\geq 3$, there exists a numerical semigroup $S$ such that
$\operatorname{Buch}(S)=[2,b]$.

###### Proof.

Let $k=b-2$, and let $S$ be the numerical semigroup of multiplicity $m=6k+15$
and depth $q=2$ whose corresponding gapset $G=\mathbb{N}\setminus S$ is given
by

(2) $G=[1,m-1]\sqcup\\{2m-7,2m-5,2m-2,2m-1\\}.$

We claim that $\operatorname{Buch}(S)=[2,k+2]$. Indeed, we will show a more
precise statement, namely

$\beta_{G}(n)=\left\\{\begin{array}[]{ccl}1&\textrm{if}&n=2,\\\
2&\textrm{if}&3\leq n\leq k+2,\\\ -6(n-k-3)&\textrm{if}&n\geq
k+3.\end{array}\right.$

Let $A=(2m-1)-G$. Then $\beta_{G}(n)=\beta_{A}(n)$ since $|nG|=|nA|$ for all
$n\geq 1$. We have

$A=[0,1]\sqcup\\{4,6\\}\sqcup[m,2(m-1)].$

Let us compute $2A$ and $3A$. We obtain

$\displaystyle 2A$ $\displaystyle=$
$\displaystyle[0,2]\sqcup[4,8]\sqcup\\{10,12\\}\sqcup[m,4(m-1)],$
$\displaystyle 3A$ $\displaystyle=$
$\displaystyle[0,14]\cup\\{16,18\\}\cup[m,6(m-1)].$

In general, we have

(3) $nA=\left([0,6n-4]\sqcup\\{6n-2,6n\\}\right)\cup[m,2n(m-1)]$

for all $n\geq 3$, as easily verified by induction on $n$.

Let us determine $|nA|$ for all $n\geq 1$. Note first that _the union in ( 3)
is disjoint if and only if $6n+1\leq m$ _. Moreover, as $m=6k+15$, we have

$6n+1\leq m\Longleftrightarrow n\leq k+2.$

In contrast, if $n\geq k+3$, i.e. if $6n-3\geq m$, then the union in (3)
collapses to a single interval and we get

$nA=[0,2n(m-1)].$

Summarizing, we have

$|nA|=\left\\{\begin{array}[]{lcl}m+3&\textrm{if}&n=1,\\\
3m+7&\textrm{if}&n=2,\\\ (2n-1)(m-1)+6n-1&\textrm{if}&3\leq n\leq k+2,\\\
2n(m-1)+1&\textrm{if}&n\geq k+4.\end{array}\right.$

The stated formula for $\beta_{G}(n)=\beta_{A}(n)=|nA|-(2n-1)(|A|-1)$ follows.
Hence $\operatorname{Buch}(S)=[2,k+2]$, as claimed. ∎

This family of numerical semigroups was inspired by the _PF-semigroups_
introduced in [4].

###  3.3 More intervals

What are the possible shapes of $\operatorname{Buch}(S)$ when $S$ varies? We
do not know in general. By Proposition 3.4, any finite integer interval $I$
with $|I|\geq 2$ and $\min(I)=2$ may be realized as $I=\operatorname{Buch}(S)$
for some numerical semigroup $S$. Here we present families of numerical
semigroups $S$ realizing as $\operatorname{Buch}(S)$ all finite integer
intervals $I$ with $|I|\geq 2$ and $\min(I)\in\\{3,4,5,6\\}$.

######  Proposition 3.5.

Let $k\geq 1$. Let $S$ be the numerical semigroup of multiplicity $m=6k+19$
and depth $q=2$ whose corresponding gapset $G=\mathbb{N}\setminus S$ is given
by

(4) $G=[1,m-1]\sqcup\\{2m-7,2m-6,2m-2,2m-1\\}.$

Then $\operatorname{Buch}(S)=[3,k+3]$.

###### Proof.

Let again $A=(2m-1)-G=[0,1]\sqcup\\{5,6\\}\sqcup[m,2(m-1)]$. We then have

$\displaystyle 2A$ $\displaystyle=$
$\displaystyle[0,2]\sqcup[5,7]\sqcup[10,12]\sqcup[m,4(m-1)],$ $\displaystyle
3A$ $\displaystyle=$
$\displaystyle[0,3]\sqcup[5,8]\sqcup[10,13]\sqcup[15,18]\sqcup[m,6(m-1)],$
$\displaystyle 4A$ $\displaystyle=$ $\displaystyle[0,24]\cup[m,8(m-1)].$

It follows that $nA=[0,6n]\cup[m,2n(m-1)]$ for all $n\geq 4$. In particular,
if $6n\geq m$ then $nA=[0,2n(m-1)]$. Therefore,

$\beta_{G}(n)=\beta_{A}(n)=\left\\{\begin{array}[]{ccl}0&\textrm{if}&n=2,\\\
1&\textrm{if}&n=3,\\\ 4&\textrm{if}&4\leq n\leq k+3,\\\
6k-6n+22&\textrm{if}&n\geq k+4.\end{array}\right.$

Hence $\operatorname{Buch}(S)=[3,k+3]$, as claimed. ∎

######  Proposition 3.6.

For $k\geq 1$ and $i\in\\{1,2,3\\}$, let $S_{i}$ be the numerical semigroup
with $G_{i}=\mathbb{N}\setminus S_{i}$ given by

$\begin{array}[]{rcl}G_{1}&=&[1,m_{1}-1]\sqcup\\{2m_{1}-6,2m_{1}-2,2m_{1}-1\\},\\\
G_{2}&=&[1,m_{2}-1]\sqcup\\{2m_{2}-10,2m_{2}-4,2m_{2}-3,2m_{2}-2\\},\\\
G_{3}&=&[1,m_{3}-1]\sqcup\\{2m_{3}-10,2m_{3}-9,2m_{3}-2\\},\end{array}$

where $m_{1}=4k+22$, $m_{2}=7k+44$ and $m_{3}=5k+55$, respectively. Then

$\operatorname{Buch}(S_{1})=[4,k+4],\;\;\operatorname{Buch}(S_{2})=[5,k+5],\;\;\operatorname{Buch}(S_{3})=[6,k+6].$

###### Proof.

Similar to the proofs of Propositions 3.4 and 3.6. We omit it here. ∎

Having realized all finite integer intervals $I$ with $|I|\geq 2$ and
$\min(I)\in[2,6]$ as $I=\operatorname{Buch}(S)$ for a suitable numerical
semigroups $S$, is it possible to do the same for all finite integer intervals
$I$ with $\min(I)\geq 7$? We do not know in general. But here is a particular
case where $\min(I)$ can be arbitrarily large. It is based on a family of
numerical semigroups found in [8].

######  Proposition 3.7.

For any integer $k\geq 1$, there is a numerical semigroup $S$ such that
$\operatorname{Buch}(S)=[7+2k,7+4k]$.

###### Proof.

For $k\geq 1$, let $S$ be the numerical semigroup minimally generated by the
set $T_{1}\cup T_{2}\cup T_{3}$, where

$\displaystyle T_{1}$ $\displaystyle=$
$\displaystyle[44+27k+4k^{2},79+51k+8k^{2}],$ $\displaystyle T_{2}$
$\displaystyle=$ $\displaystyle[81+51k+8k^{2},84+53k+8k^{2}],$ $\displaystyle
T_{3}$ $\displaystyle=$ $\displaystyle[87+53k+8k^{2},87+54k+8k^{2}].$

The corresponding gapset $G=\mathbb{N}\setminus S$ is then given by

$G=[1,43+27k+4k^{2}]\cup\\{80+51k+8k^{2},85+53k+8k^{2},86+53k+8k^{2}\\}.$

Let $A=(86+53k+8k^{2})-G$. Then

$A=[0,1]\sqcup\\{6+2k\\}\sqcup[43+26k+4k^{2},85+53k+8k^{2}],$

of cardinality $|A|=46+27k+4k^{2}$. The $n$-fold sumsets of $A$ are then given
by

(5)
$nA=[0,n]\cup\left(\bigcup_{i=1}^{n}\left[i(6+2k),i(6+2k)+n-i\right]\right)\\\
\cup\left[43+26k+4k^{2},(85+53k+8k^{2})n\right].$

$\bullet$ Assume first $2\leq n<6+2k$. In this case, we have

$\begin{split}&0<n<6+2k<6+2k+n-1<\dots<(n-1)(6+2k)\\\
&<(n-1)(6+2k)+n-1<n(6+2k)<(7+2k)(6+2k)+1\\\
&=43+26k+4k^{2}<(85+53k+8k^{2})n.\end{split}$

Thus, all the sets appearing in (5) are disjoint and the cardinality of $nA$
is equal to

$(n+1)+\sum_{i=1}^{n}i+\left(\left(8k^{2}+53k+85\right)n-(4k^{2}+26k+43)+1\right)=\\\
-41-26k-4k^{2}+(173n)/2+53kn+8k^{2}n+n^{2}/2.$

Thus,

(6) $\beta_{G}(n)=(-41-26k-4k^{2}+(173n)/2+53kn+8k^{2}n+n^{2}/2)\\\
-\left(4k^{2}+27k+46-1\right)(2n-1)=\\\
(4+k)-(\frac{7}{2}+k)n+\frac{1}{2}n^{2}$

for every $n\in[2,5+2k]$.

The only difference between the case $n=6+2k$ and the previous one is that the
sets $[0,n]$ and $[6+2k,6+2k+n-1]$ have a nonempty intersection, equal to
$\\{6+2k\\}$. Replacing $n$ by $6+2k$ and subtracting one, we obtain
$\beta_{G}(6+2k)=0$.

$\bullet$ Assume now $6+2k<n\leq 11+4k$. The sequence of sets $[0,n]$ and

$[6+2k,6+2k+n-1],\dots,[(n-5-2k)(6+2k),(n-5-2k)(6+2k)+(5+2k)]$

verifies that the intersection of any two consecutive terms is nonempty.
Moreover, their union is the interval $[0,(n-5-2k)(6+2k)+(5+2k)]$ whose
cardinality is equal to $(n-5-2k)(6+2k)+(5+2k)+1$. For $i=n-4-2k,\dots,6+2k$
the intervals are disjoint with all the others sets appearing in the
expression (5); the cardinality of the union of these sets is equal to
$\displaystyle\sum_{i=n-2k-5}^{2k+5}i$. For every $i=7+2k,\dots,n$ the
intersection

$[i(6+2k),i(6+2k)+n-i]\cap[43+26k+4k^{2},(8k^{2}+53k+85)n]$

is nonempty, except for $n=7+2k$. Since $(7+2k)(6+2k)=42+26k+4k^{2}$, the set

$\left(\bigcup_{i=7+2k}^{n}\left[i(6+2k),i(6+2k)+n-i\right]\right)\\\
\cup\left[43+26k+4k^{2},(85+53k+8k^{2})n\right]$

is equal to $[42+26k+4k^{2},(85+53k+8k^{2})n]$, and the cardinality of this
set is $\left(\left(8k^{2}+53k+85\right)n-4k^{2}-26k-42\right)+1$. Putting all
the above together, we have that if $6+2k<n\leq 11+4k$, the set $nA$ has
cardinality equal to

$((n-5-2k)(6+2k)+(5+2k)+1)+\\\
\sum_{i=n-2k-5}^{2k+5}i+\left(\left(8k^{2}+53k+85\right)n-4k^{2}-26k-42\right)+1=\\\
-65-46k-8k^{2}+(193n)/2+57kn+8k^{2}n-n^{2}/2,$

and therefore

(7) $\beta_{G}(n)=-65-46k-8k^{2}+(193n)/2+57kn+8k^{2}n-n^{2}/2\\\
-(2n-1)(46+27k+4k^{2}-1)=\\\
(-20-19k-4k^{2})+(3k+\frac{13}{2})n-\frac{n^{2}}{2}$

for every $n\in[6+2k,11+4k]$.

$\bullet$ Finally, assume $11+4k<n$. The set

$[0,n]\cup(\bigcup_{i=1}^{6+2k}[i(6+2k),i(6+2k)+n-i])\cup[43+26k+4k^{2},(85+53k+8k^{2})n]$

is equal to $[0,(85+53k+8k^{2})n]$ and the remaining intervals are contained
in this union. So we have $nA=[0,(85+53k+8k^{2})n]$ and therefore

(8)
$\beta_{G}(n)=-\left(4k^{2}+27k+46-1\right)(2n-1)+\left(8k^{2}+53k+85\right)n+1=\\\
(4k^{2}+27k+46)-(k+5)n$

for every $n>11+4k$.

Combining $\beta_{G}(6+2k)=0$ with the formulation of (6), (7) and (8) for
$\beta_{G}(n)$, we get the following formulas:

$\beta_{G}(n)=\left\\{\begin{array}[]{lcl}(4+k)-(\frac{7}{2}+k)n+\frac{1}{2}n^{2}&\textrm{if}&2\leq
n<6+2k,\\\ 0&\textrm{if}&n=6+2k,\\\
(-20-19k-4k^{2})+(3k+\frac{13}{2})n-\frac{n^{2}}{2}&\textrm{if}&6+2k<n\leq
11+4k,\\\ (4k^{2}+27k+46)-(k+5)n&\textrm{if}&11+4k<n.\end{array}\right.$

Let $k\geq 1$ be fixed. For $2\leq n<6+2k$, the formula of $\beta_{G}(n)$ is a
degree two polynomial in $n$ with positive leading coefficient such that
$\beta_{G}(2)=-1-k<0$ and $\beta_{G}(5+2k)=-1-k<0$. We have therefore
$\beta_{G}(n)<0$ for every $n=2,\dots,5+2k$.

If $n>11+4k$, we now have that $\beta_{G}(n)$ is a degree one polynomial with
negative leading coefficient and such that $\beta_{G}(12+4k)=-14-5k<0$. So
$\beta_{G}(n)<0$ for every $n>11+4k$.

Finally, if $6+2k<n\leq 11+4k$ the function $\beta_{G}(n)$ is a degree two
polynomial in $n$ with negative leading coefficient. As in addition
$\beta_{G}(7+2k)=1+k$, $\beta_{G}(7+4k)=1$ and $\beta_{G}(8+4k)=-k$, the only
positive values that we have in this part are for $n\in[7+2k,7+4k]$.

Since $\beta_{G}(n)\leq 0$ except for $n\in[7+2k,7+4k]$, the set
$\operatorname{Buch}(S)$ is equal to $[7+2k,7+4k]$. ∎

##  4 Concluding remarks

The current knowledge on the structure of $\operatorname{\mathcal{B}}(A)$ for
finite subsets $A\subset\mathbb{Z}$ is very scarce, even for gapsets. Do they
have some special shape or property? We end this paper with three questions
based on the few currently available observations.

######  Question 4.1.

Let $A\subset\mathbb{Z}$ be a finite subset, or more specifically a gapset. Is
the set $\operatorname{\mathcal{B}}(A)$ always an interval of integers?

######  Question 4.2.

Even more so, is the function $\beta_{A}(n)$ unimodal?

######  Question 4.3.

In sharp contrast with the above questions, let $T\subset 2+\mathbb{N}$ be any
finite subset. Does there exist a finite subset $A\subset\mathbb{N}$, or more
specifically a gapset, such that $\operatorname{\mathcal{B}}(A)=T$?

Acknowledgement. Part of this paper was written during a visit of the first-
named author to the Universidad de Cádiz (Spain) which was partially supported
by Ayudas para Estancias Cortas de Investigadores (EST2019-039, Programa de
Fomento e Impulso de la Investigación y la Transferencia en la Universidad de
Cádiz). The second, third and fourth-named authors were partially supported by
Junta de Andalucía research group FQM-366 and by the project MTM2017-84890-P
(MINECO/FEDER, UE).

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Authors’ addresses:

Shalom Eliahoua,b

aUniv. Littoral Côte d’Opale, UR 2597 - LMPA - Laboratoire de Mathématiques
Pures et Appliquées Joseph Liouville, F-62100 Calais, France
bCNRS, FR2037, France
e-mail<EMAIL_ADDRESS>
Juan Ignacio García-Garcíac,d
e-mail<EMAIL_ADDRESS>
Daniel Marín-Aragónc
cDepartamento de Matemáticas, Universidad de Cádiz, E-11510 Puerto Real
(Cádiz, Spain).
e-mail<EMAIL_ADDRESS>
Alberto Vigneron-Tenoriod,e
eDepartamento de Matemáticas, Universidad de Cádiz, E-11406 Jerez de la
Frontera (Cádiz, Spain).
dINDESS (Instituto Universitario para el Desarrollo Social Sostenible),
Universidad de Cádiz, E-11406 Jerez de la Frontera (Cádiz, Spain).
e-mail<EMAIL_ADDRESS>