#  Mass-coupled relativistic spontaneous collapse models

C. Jones  Department of Physics, University of Trieste, Strada Costiera 11,
34151 Trieste, Italy  Istituto Nazionale di Fisica Nucleare, Trieste Section,
Via Valerio 2, 34127 Trieste, Italy<EMAIL_ADDRESS>G.
Gasbarri  Física Teòrica: Informació i Fenòmens Quàntics, Department de
Física, Universitat Autònoma de Barcelona, 08193 Bellaterra (Barcelona), Spain
<EMAIL_ADDRESS>A. Bassi  Department of Physics, University of Trieste,
Strada Costiera 11, 34151 Trieste, Italy  Istituto Nazionale di Fisica
Nucleare, Trieste Section, Via Valerio 2, 34127 Trieste, Italy

###### Abstract

Currently there is not a satisfactory relativistic spontaneous collapse model.
Here we show the impossibility of a simple generalization of the continuous
spontaneous collapse (CSL) model to the relativistic framework. We consider a
mass coupled model as in the non-relativistic limit this gives the CSL model.
We show that a Lorentz covariant collapse equation cannot simultaneously
satisfy the following conditions: i) To avoid a divergent rate of energy, ii)
To prevent superluminal signaling.

##  1 Introduction

The transition between the quantum and classical realm has long been a problem
faced by quantum mechanics [1]. Quantum theory describes microscopic objects
which can be found in superpositions and classical mechanics describes
macroscopic objects which cannot. Explaining the transition between these two
regimes has been a problem quantum mechanics has faced since its inception,
sometimes called the macroscopic objectification problem [2].

Spontaneous collapse models [3, 4, 5, 6, 7, 8, 9] are an attempt to solve this
problem by modifying the Schrödinger equation to give a non-unitary and
stochastic time evolution. This modification has two important properties. The
first is that it causes superpositions of the state to spontaneously collapse
in a particular basis. Since macroscopic objects are never found in spatial
superpositions, most collapse models localise the state in the position basis.
The second property is that the rate of collapse scales with the mass, and
this is known as the amplification mechanism. It is exactly this mechanism
that guarantees that collapse models recover the quantum mechanical behaviour
for microscopic systems, and the classical behaviour for macroscopic objects.

Since their introduction there have been a plethora of different collapse
models proposed, see [6] and references within. As mentioned above most of
these models cause collapse in the position basis and hence are coupled to the
mass or number density operator. It is this class of spontaneous collapse
models will we study in this paper. Continuous Spontaneous Localisation (CSL)
[5] and its non-white noise generalisation coloured CSL [10] are the most
studied of such mass-coupled models. In this class, another spontaneous
collapse model of note is the Diósi-Penrose (DP) model [7, 11] which proposes
gravity induced wavefunction collapse. As they offer predictions which differ
from conventional quantum mechanics, there are experimental projects aimed at
testing the validity of these theories [12, 13, 14]. Much of the possible
parameter space for these models has been explored, and research is still
ongoing.

As originally formulated neither CSL, coloured CSL nor the DP model are
compatible with special relativity. In order to be fundamental, a spontaneous
collapse model is expected to be in agreement with special relativity,
assuming special relativity is correct. There have been many attempts to
construct a relativistic collapse model [15, 16, 17, 18, 19, 20, 21, 22, 23,
24], but none of them proved to be satisfactory. Every model so far proposed
has encountered some issues, for example having a divergence in rate of change
of energy [17] or not being able to describe indistinguishable particles [15,
19, 24] etc.

The purpose of this work is to understand why these attempts at relativistic
collapse models fail. This is done by starting from the most general form of a
continuous time spontaneous collapse model and then applying a set of minimal
requirements for a successful relativistic collapse model. We will consider
the case of scalar bosons but the results of this work are not expected to
differ for fermions. We assume that state vector reduction occurs in the mass
density basis. These requirements are; that the map is Lorentz covariant, that
the rate of change of energy is finite, and that microcausality is respected.
It is found that it is not possible for all the requirements to be
simultaneously satisfied. We also find that the requirement for finite energy
ensures that the theory must have a well defined non-relativistic limit.

This work is laid out as follows: in section 2 the general structure of a
spontaneous collapse model is given and it is shown how the dynamics is
characterised by a collapse operator and a two point correlation function. In
section 3 it will be shown that the requirements for this map to be Lorentz
covariant are that given the collapse operator is a Lorentz scalar, the two
point correlation function is a function of the invariant 4 distance. The
collapse operator is specified in section 4. In section 5 it is shown that the
rate of change of the energy under the map is only finite if the two point
correlation function is zero in the limit of the two points coinciding.
Finally in section 6 it is shown how under the above requirements such a
dynamics permits superluminal signalling.

##  2 General Form of a Spontaneous Collapse Model

Spontaneous collapse models are described by non-linear stochastic Schrödinger
equations with the usual deterministic evolution associated with the unitary
dynamics and a non-linear term coupled to a gaussian stochastic noise field
[10, 6, 3]. For the purposes of this article however it is only necessary to
consider the evolution of the statistical operator $\hat{\rho}$ which is
averaged over the stochastic noise. Throughout this work we set $\hbar=c=1$.
For a collapse models the evolution of $\hat{\rho}$ is given by[25, 26, 10]:

$\displaystyle\hat{\rho}_{t}=\mathcal{U}_{t}^{0}[\mathcal{M}_{t}[\hat{\rho}_{0}]]$
(1)

where $\mathcal{U}_{t}^{0}[\cdot]=e^{-i\hat{H}t}\cdot e^{i\hat{H}t}$ is the
standard quantum evolution, with $\hat{H}$ as the standard free Hamiltonian
and $\mathcal{M}_{t}$ is the contribution due to the presence of the
stochastic noise and has the following expression

$\mathcal{M}_{t}=\overleftarrow{\text{T}}\exp\left(\gamma{\mathcal{L}_{t}}\right)$
(2)

with

$\mathcal{L}_{t}\\!=\\!\\!\int_{\Omega_{t}}\\!\\!\\!d^{4}x\\!\\!\int_{\Omega_{t}}\\!\\!\\!d^{4}yD(x,y)\big{[}\hat{Q}^{L}(y)\hat{Q}^{R}(x)-\theta(x^{0}-y^{0})\hat{Q}^{L}(x)\hat{Q}^{L}(y)-\theta(y^{0}-x^{0})\hat{Q}^{R}(y)\hat{Q}^{R}(x)\big{]}$
(3)

$\hat{Q}(\mathbf{x})$ is an arbitrary self-adjoint operator,
$\hat{Q}(x)=\mathcal{U}_{x^{0}}^{0\dagger}[\hat{Q}(\mathbf{x})]$,
$\Omega_{t}=\\{x=(\mathbf{x},x^{0})\,|\,x^{0}\in[0,t],\,\mathbf{x}\in\mathbb{R}^{3}\\}$,
the superscript L (R) denotes operator acting on the statistical operator
$\hat{\rho}$ from left (right), i.e.
$\hat{Q}^{R}(x)\hat{Q}^{L}(x)\hat{\rho}=\hat{Q}(y)\hat{\rho}\hat{Q}(y)$ and
$\overleftarrow{\text{T}}$ is the chronological time ordering acting on the
L/R operators and is defined as

$\displaystyle
Q^{L/R}(y)Q^{L/R}(x)=\left\\{\begin{array}[]{c}Q^{L/R}(x)Q^{L/R}(y)\,\,\text{if}\,\,x^{0}>y^{0}\\\
Q^{L/R}(y)Q^{L/R}(x)\,\,\text{if}\,\,y^{0}>x^{0}\end{array}\right.$ (6)

Equation 2 suppresses the off diagonal elements of $\hat{\rho}$ in the basis
of the eigenstates of $\hat{Q}(\mathbf{x})$. For collapse models this is a
consequence of the collapse of the wavefunction, which again occurs in the
eigenbasis of $\hat{Q}(\mathbf{x})$ when the free Hamiltonian dynamics is
neglected, see [26] for a proof. It is for this reason that many collapse
models have $\hat{Q}(\mathbf{x})$ as the mass density or number density
operator, since this means macroscopic objects do not remain in spatial
superposition.

As one can see from Eq. (3), all collapse models are defined by specifying
$D(x,y)$, $\hat{Q}(x)$, and the coupling $\gamma$. For example the average map
of the coloured CSL model is characterized by the above equation with:

$\displaystyle D(x,y)$ $\displaystyle=\frac{1}{m_{0}^{2}(4\pi
r_{C}^{2})^{3/2}}\exp[-(\mathbf{x}-\mathbf{y})^{2}/2r_{C}^{2}]F(t-s)$
$\displaystyle\hat{Q}(x)$ $\displaystyle=\hat{M}(x)$ (7)

where $F(t-s)$ is an arbitrary positive function of time, $\hat{M}(x)$ is the
non-relativistic mass density operator, $m_{0}$ a reference mass (usually the
nucleon mass), and $r_{C}$ is a parameter of the model which determines the
length scale of the collapse. Original CSL is the recovered when
$F(t-s)=\delta(t-s)$. This choice of $\hat{Q}(x)$ ensures that the
wavefunction is driven to collapse in the mass density basis, which correctly
predicts that macroscopic objects are not found in spatial superpositions. The
form of $D(x,y)$ means that the wave-function is localised by a Gaussian of
width $r_{C}$ where the amplitude of the wavefunction is maximised, which
means that the theory effectively replicates the Born probability rule. The
standard CSL model is Galilei invariant [6], guaranteeing that, in a non-
relativistic setting, the dynamical law are the same in all inertial frames.

##  3 Lorentz Covariance of Non-unitary Dynamics

In a probabilistic theory consistent with special relativity the transition
probabilities between point-like events in spacetime is Poincaré covariant.
For this work it is sufficient to check whether the conditional probability
distribution is invariant under the Lorentz group, instead of the larger
Poincaré group. So from on we will consider only Lorentz group transforms. We
can reasonably do this as this work shows that a Lorentz covariant CSL model
is not possible, and as Poincaré covariance is a strictly stronger condition
it follows that a Poincaré covariant CSL model is also not possible.
Furthermore we will consider only the orthochronous Lorentz group as the
dynamics is for positive time translations only.

So for a theory to be consistent with special relativity the transition
probability between point-like events in spacetime

$P(A(x_{1}),B(x_{2})\,...|C(y_{1}),D(y_{2})\,...)$ (8)

must be invariant under Lorentz transformations, where $A(x_{1})$ is an event
at position $x_{1}$ in space-time and so on. For the case of collapse models
the covariance of the transition probabilities is guaranteed only if the
dynamics of the statistical operator is covariant, as already noticed in [6,
27]. Since the map $\mathcal{U}_{t}^{0}$ describing the standard free quantum
evolution is Lorence convariant we are left to verify the covariance of the
collaspe contribution $\mathcal{M}_{t}$. Therefore, in the rest of the paper
we will work with the map $\mathcal{M}_{t}$.

We call the map $\mathcal{M}_{t}$ Lorentz covariant if:

$\mathcal{U}_{\Lambda}\mathcal{M}_{t}\mathcal{U}_{\Lambda}^{-1}=\mathcal{M}_{\Lambda(t)},$
(9)

where
$\mathcal{U}_{\Lambda}[\cdot]=\hat{U}(\Lambda)\,\cdot\,\hat{U}(\Lambda)^{\dagger}$
and $\hat{U}(\Lambda)$ is the standard unitary representation of Lorentz group
on the Hilbert space of the system and $\Lambda$ is an element of the
orthochronous Lorentz group [28]. Then with the help of Eq. (2) and Eq. (3)
and by exploiting the unitarity of $\mathcal{U}_{\Lambda}$, it is
straightforward to show that Eq. (9) is satisfied if and only if the following
condition is satisfied:

$\displaystyle\overleftarrow{\text{T}}\exp\Bigg{\\{}\int_{\Omega_{t}}d^{4}x\int_{\Omega_{t}}d^{4}yD(x,y)\Big{[}\mathcal{U}_{\Lambda}[\hat{Q}^{L}(y)]\mathcal{U}_{\Lambda}[\hat{Q}^{R}(x)]$
$\displaystyle\hskip
39.83368pt-\theta(x^{0}-y^{0})\mathcal{U}_{\Lambda}[\hat{Q}^{L}(x)]\mathcal{U}_{\Lambda}[\hat{Q}^{L}(y)]-\theta(y^{0}-x^{0})\mathcal{U}_{\Lambda}[\hat{Q}^{R}(y)]\mathcal{U}_{\Lambda}[\hat{Q}^{R}(x)]\Big{]}\Bigg{\\}}$
$\displaystyle=\overleftarrow{\text{T}}\exp\Bigg{\\{}\int_{\Lambda(\Omega_{t})}\\!\\!\\!\\!\\!d^{4}x\int_{\Lambda(\Omega_{t})}\\!\\!\\!\\!\\!d^{4}yD(x,y)$
$\displaystyle\hskip
39.83368pt\Big{[}\hat{Q}^{L}(y)\hat{Q}^{R}(x)-\theta(x^{0}-y^{0})\hat{Q}^{L}(x)\hat{Q}^{L}(y)-\theta(y^{0}-x^{0})\hat{Q}^{R}(y)\hat{Q}^{R}(x)\Big{]}\Bigg{\\}}$
(10)

Let us consider the case when the operator $\hat{Q}$ is a Lorentz scalar:

$\mathcal{U}_{\Lambda}[\hat{Q}(y)]=\hat{Q}(\Lambda(y)).$ (11)

In this case condition in Eq. (3) is met only if $D(x,y)=D(|x-y|)$ [29], in
which case $D(x^{\prime},y^{\prime})=D(x,y)$, where $x^{\prime}=\Lambda(x)$.

Finally we note that as $D(x,y)$ is a function of the 4-distance, then it is
symmetric under exchange of $x$ and $y$. This allows Eq. (3) to be simplified
to:

$\mathcal{L}_{t}=\int_{\Omega_{t}}\\!\\!\\!d^{4}x\int_{\Omega_{x^{0}}}\\!\\!\\!d^{4}y\,D(x,y)\big{[}\hat{Q}^{L}(y)\hat{Q}^{R}(x)+\hat{Q}^{L}(x)\hat{Q}^{R}(y)-\hat{Q}^{L}(x)\hat{Q}^{L}(y)-\hat{Q}^{R}(x)\hat{Q}^{R}(y)\big{]}$
(12)

which is a form which makes calculations easier to perform.

##  4 Choice of the collapse operator

As discussed in Section II, the choice of the operator $\hat{Q}(x)$ determines
the basis that spontaneous collapse occurs in. We seek a model that is
relativistically invariant and at the same time is able to reproduce the CSL
dynamics in the non relativistic limit. A natural choice is to select
$\hat{Q}(x)$ to be the relativistic mass density operator, i.e.

$\hat{Q}(x)=m\hat{a}^{\dagger}(x)\hat{a}(x)$ (13)

where $\hat{a}^{\dagger}(x)$ is the the positive energy part of the Klein-
Gorden field operator. From now on we take Eq. (13) to be the collapse
operator.

Previous attempts to derive relativistic extensions of CSL model have
encountered the problem of an infinite energy rate in the dynamics. In the
next section we will find a necessary condition to avoid this unpleasant
feature from a relativistic collapse model.

##  5 Finiteness of rate of change of energy

An important physical requirement for the dynamics to satisfy is that the rate
of change of energy is finite. Some previous attempts to find a relativistic
collapse model have suffered from this divergence [23, 6]. Here we show that
the dynamics given by Eq. (3) with the choice of Eq. (13) as the collapse
operator does not lead to a divergent rate of change of energy provided that
the correlator prevents large transfers of momentum between momentum
eigenstates. We show this for a single particle system in the weak coupling
regime.

The rate of change of the energy is given by:

$\frac{d}{dt}\text{Tr}(\hat{H}\mathcal{M}_{t}[\hat{\rho}])=\frac{d}{dt}\text{Tr}(\hat{H}\>\overleftarrow{\text{T}}\exp{\Lambda\mathcal{L}_{t}[\hat{\rho}]}),$
(14)

where $\hat{H}$ is the Hamiltonian for the free dynamics [30]:

$\hat{H}=\int\frac{d\mathbf{p}}{(2\pi)^{3}}\>E_{\mathbf{p}}\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}}$
(15)

where $E_{\mathbf{p}}=\sqrt{m^{2}+\mathbf{p}^{2}}$ and $\mathbf{p}$ is the
three momentum and $\hat{a}_{\textbf{p}}$ and $\hat{a}^{\dagger}_{\textbf{p}}$
are the creation and annihilation operators in momentum eigenbasis and satisfy
the following commutation relation:

$[\hat{a}_{\textbf{p}},\hat{a}^{\dagger}_{\textbf{q}}]=(2\pi)^{3}\delta^{3}(\textbf{p}-\textbf{q})$
(16)

We expand the map up to the first order in $\gamma$, to obtain:

$\overleftarrow{\text{T}}\exp\gamma\mathcal{L}_{t}[\hat{\rho}]=\mathbb{I}+\gamma\mathcal{L}_{t}[\hat{\rho}]+\mathcal{O}(\gamma^{2}).$
(17)

Exploiting this and dropping terms of orders higher than $\gamma$ gives:

$\frac{d}{dt}\text{Tr}(\hat{H}\mathcal{M}_{t}[\hat{\rho}])\approx\gamma\text{Tr}(\hat{H}\mathcal{L}_{t}[\hat{\rho}])$
(18)

We evaluate Eq. (18) in the single particle sector, where the state of the
system can be expressed as:

$\hat{\rho}=\int d\mathbf{p}\int
d\mathbf{q}\>A(\mathbf{p},\mathbf{q})\>\hat{a}^{\dagger}_{\textbf{p}}|0\rangle\langle
0|\hat{a}_{\textbf{q}}$ (19)

where $A(\mathbf{p},\mathbf{q})$ is an arbitrary function such that:

$\text{Tr}(\hat{\rho})=\int d\mathbf{p}\>A(\mathbf{p},\mathbf{p})=1$ (20)

In order to calculate Eq. (18) we expand the correlation function and the
collapse operator in Fourier components, i.e.

$\displaystyle D(\mathbf{x},t)=\frac{1}{(2\pi)^{3}}\int
d\mathbf{q}\,e^{i\mathbf{x}\cdot\mathbf{q}}\tilde{D}(\mathbf{q},t)$
$\displaystyle\hat{Q}(\mathbf{x},t)=\frac{m}{(2\pi)^{6}}\int d\mathbf{q}\int
d\mathbf{p}\;\frac{1}{2\sqrt{E_{\mathbf{p}}E_{\mathbf{q}}}}\,e^{i\mathbf{x}\cdot(\mathbf{q}-\mathbf{p})}e^{-i(E_{\mathbf{q}}-E_{\mathbf{p}})t}\hat{a}^{\dagger}_{\mathbf{q}}\hat{a}_{\mathbf{p}}$
(21)

substituting these into Eq. (12) with $D(x,y)=D(|x-y|)$ and then integrating
over the spatial variables $d\mathbf{x}$ and $d\mathbf{y}$ lets
$\mathcal{L}_{t}$ be written as:

$\displaystyle\mathcal{L}_{t}=\frac{1}{(2\pi)^{12}}\int\\!\\!d\mathbf{q}\int_{0}^{t}ds\int_{0}^{s}d\tau\,\tilde{D}(\mathbf{q},s-\tau)$
$\displaystyle\big{\\{}\hat{K}^{L}(\mathbf{q},s)\hat{K}^{\dagger,L}(\mathbf{q},\tau)+\hat{K}^{R}(\mathbf{q},s)\hat{K}^{\dagger,R}(\mathbf{q},\tau)$
$\displaystyle-\hat{K}^{\dagger,L}(\mathbf{q},\tau)\hat{K}^{R}(\mathbf{q},s)-\hat{K}^{L}(\mathbf{q},s)\hat{K}^{\dagger,R}(\mathbf{q},\tau)\big{\\}}$
(22)

where:

$\displaystyle\hat{K}(\mathbf{q},t)=m\int\frac{d\mathbf{p}}{2\sqrt{E_{\mathbf{p}}E_{\mathbf{p}-\mathbf{q}}}}e^{i\Delta
E(\mathbf{p},\mathbf{q})t}\hat{a}^{\dagger}_{\mathbf{p}}\hat{a}_{\mathbf{p}-\mathbf{q}}$
(23)

with $\Delta E(\mathbf{p},\mathbf{q})=E_{\mathbf{p-q}}-E_{\mathbf{p}}$. Note
that $\hat{K}(\mathbf{q},t)=\hat{K}^{\dagger}(-\mathbf{q},-t)$.

Then using Eq. (5) and Eq. (5) we have that Eq. (18) evaluates to:

$\displaystyle\text{Tr}(\hat{H}\mathcal{L}_{t}[\hat{\rho}])=$
$\displaystyle-\frac{m^{2}}{(2\pi)^{3}}\int\\!d\mathbf{q}\\!\int\\!d\mathbf{p}\\!\int_{0}^{t}\\!\\!ds\\!\int_{0}^{s}\\!\\!d\tau\tilde{D}(\mathbf{q},s-\tau)\cos(\Delta
E(\mathbf{p},\mathbf{q})(s-\tau))A(\mathbf{p},\mathbf{p})\left(\frac{1}{2E_{\mathbf{p}-\mathbf{q}}}-\frac{1}{2E_{\mathbf{p}}}\right)\qquad$
(24)

We are interested to see if this integral is finite. The integrals over both q
and p are over an infinite range, but if the system is assumed to initially
have a finite energy then the integral over $\mathbf{p}$ will automatically
converge to a finite value. Then in order for the integral over $\mathbf{q}$
to converge a necessary condition is that:

$\lim_{|\mathbf{q}|\to\infty}\tilde{D}(|\mathbf{q}|,s-\tau)=0.$ (25)

Note here that if the noise is white, i.e. if $D(x,y)=\delta^{4}(x-y)$ then:

$\tilde{D}(|\mathbf{q}|,s-\tau)=(2\pi)^{3}\delta(s-\tau)$ (26)

and the energy rate in Eq. (5) diverges. Therefore, relativistic white-noise
models are physically inconsistent, as already noticed in chapter 13 of [6]
and [23]. Through a similar calculation it can be seen that if the correlation
function $D(x-y)$ further satisfies the following condition

$\tilde{D}(|\mathbf{q}|,t)\approx 0\quad\text{for}\>\>\mathbf{q}>\kappa m$
(27)

where $\kappa\gg 1$ is a fixed constant, then the map described by Eq. (3) is
well behaved in the non relativistic sector, i.e. it leaves non-relativistic
particles non relativistic. This is shown in A.1. Notice that condition Eq.
(25) is a weaker condition than Eq.(27), hence if the model has a NR limit
then it also has a finite energy rate.

##  6 Superluminal Signalling and Micro-causality

We have given the minimal requirements for the map in Eq. (12) to guarantee
that the wave-function collapses and has a finite energy rate. Here we check
conditions under which the model does not allow superluminal signalling by
checking if the model satisfies the microcausality condition. We will show
that given the requirements from sections 3, 4 and 5 the dynamics described by
Eq. (12) violates microcausality. For this section we will work in the
Heisenberg picture, therefore operators evolve with the dual map
$\mathcal{M}_{t}^{*}[\mathcal{U}^{0*}_{t}[\,\cdot\,]]$.

In standard quantum mechanics where the dynamics $\mathcal{U}_{t}$ is unitary
the microcausality condition reads

$[\hat{A}(z_{1}),\hat{B}(z_{2})]=0\hskip 56.9055pt\forall\hskip
8.5359pt|z_{1}-z_{2}|<0$ (28)

where $\hat{A}(z_{1})=\mathcal{U}_{t_{1}}^{0*}\hat{A}[(\mathbf{z}_{1},0)]$ and
$\hat{B}(z_{2})=\mathcal{U}_{t_{2}}^{0*}[\hat{B}(\mathbf{z}_{2},0)]$ are local
operators, and $z_{1}$ and $z_{2}$ are two points in spacetime such that
$z_{1}=(\mathbf{z}_{1},t_{1})$ and $z_{2}=(\mathbf{z}_{2},t_{2})$ with
$\mathbf{z}_{1}$ and $\mathbf{z}_{2}$ are two points in 3D space. The above
condition is synonymous with no superluminal signalling, because it guarantees
that local measurements at space-like separated points do not influence each
other. For the case of a non unitary dynamics it is not possible to evaluate
Eq. (28) because:

$\displaystyle\mathcal{M}^{*}_{t_{1}}[\mathcal{U}_{t_{1}}^{0*}\hat{A}(\mathbf{z}_{1},0)\mathcal{U}_{t_{1}}^{0*}\hat{B}(\mathbf{z}_{2},0)]\neq\mathcal{M}^{*}_{t_{1}}[\mathcal{U}_{t_{1}}^{0*}\hat{A}(\mathbf{z}_{1},0)]\mathcal{M}^{*}_{t_{1}}[\mathcal{U}_{t_{1}}^{0*}\hat{B}(\mathbf{z}_{2},0)].$
(29)

In words, for non-unitary dynamics the evolution of the product of two
operators $\hat{A}$ and $\hat{B}$ cannot be described by the product of the
independently evolved operators.

One way around this problem is if the non unitary map admits a unitary
stochastic unravelling $\tilde{\mathcal{U}}_{t}$, i.e.
$\mathbb{E}(\tilde{\mathcal{U}}_{t})=\mathcal{U}_{t}^{0}\mathcal{M}_{t}$ as in
the case of the map in Eq. (12) (for an explicit expression of the unitary
stochastic unravelling see [10, 6]).
In this case one could define a microcausality condition as:

$\displaystyle\mathbb{E}([\hat{A}(z_{1}),\hat{B}(z_{2})])=0\hskip
56.9055pt\forall\hskip 8.5359pt|z_{1}-z_{2}|<0$ (30)

where
$\hat{A}(z_{1})\equiv\mathcal{\tilde{U}}^{*}_{t_{1}}[\hat{A}(\mathbf{z}_{1},0)]$
and
$\hat{B}(z_{2})\equiv\mathcal{\tilde{U}}^{*}_{t_{2}}[\hat{B}(\mathbf{z}_{2},0)]$
are local operators.

However the presence of the stochastic average makes the equation hard to
evaluate in full generality, but for our purpose it is sufficient to restrict
the study to the less general case $\hat{A}=\hat{B}=\hat{\phi}(x)$, $t_{1}=t$
and $t_{2}=0$. Under these assumptions Eq. (30) reduces to:

$\displaystyle\mathbb{E}([\tilde{\mathcal{U}}_{t}^{*}[\hat{\phi}(\mathbf{z}_{1},0)],\hat{\phi}(\mathbf{z}_{2},0)])=[\mathcal{M}^{*}_{t}[\hat{\phi}(\mathbf{z}_{1},t)],\hat{\phi}(\mathbf{z}_{2},0)]=0$
(31)

where
$\hat{\phi}(\mathbf{z}_{1},t)=\mathcal{U}_{t}^{0*}[\hat{\phi}(\mathbf{z}_{1},0)]$
is the field operator evolved under the standard free evolution dynamics. We
expand the map $\mathcal{M}^{*}_{t}$ to first order in $\gamma$. By
substituting the Fourier expansion for $\hat{\phi}(x)$ (see A.2) we arrive at:

$\displaystyle[\mathcal{M}_{t}^{*}[\hat{\phi}(\mathbf{z}_{1},t)],\hat{\phi}(\mathbf{z}_{2},0)]\approx[\hat{\phi}(\mathbf{z}_{1},t)+\gamma\mathcal{L}_{t}^{*}[\hat{\phi}(\mathbf{z}_{1},t)],\hat{\phi}(\mathbf{z}_{2},0)]$
$\displaystyle\hskip
14.22636pt=[\hat{\phi}(\mathbf{z}_{1},t),\hat{\phi}(\mathbf{z}_{2},0)]+\gamma\int_{\Omega_{t}}\\!\\!\\!d^{4}x\int_{\Omega_{x^{0}}}\\!\\!\\!\\!d^{4}y\,D(x-y)[[\hat{Q}(y),[\hat{\phi}(\mathbf{z}_{1},t),\hat{Q}(x)]],\hat{\phi}(\mathbf{z}_{2},0)]$
$\displaystyle\hskip
14.22636pt=[\hat{\phi}(\mathbf{z}_{1},t),\hat{\phi}(\mathbf{z}_{2},0)]+\gamma
m^{2}\\!\\!\int_{0}^{t}\\!\\!d\tau\\!\\!\int
d\mathbf{x}D(\mathbf{x},\tau)\left\\{F(\mathbf{z}_{1}-\mathbf{z}_{2},\mathbf{x},t,\tau)-F(\mathbf{z}_{2}-\mathbf{z}_{1},-\mathbf{x},t,\tau)\right\\}$
(32)

where

$\displaystyle
F(\mathbf{z}_{1}-\mathbf{z}_{2},\mathbf{x},t,\tau)=\int\frac{d\mathbf{k}}{2E_{\mathbf{k}}}\\!\int\\!\frac{d\mathbf{k^{\prime}}}{4E_{\mathbf{k^{\prime}}}^{2}}e^{i[\mathbf{k}\cdot(\mathbf{z}_{1}-\mathbf{z}_{2})-E_{\mathbf{k}}t]}e^{-i[(\mathbf{k}-\mathbf{k}^{\prime})\cdot\mathbf{x}-(E_{\mathbf{k}}-E_{\mathbf{k^{\prime}}})\tau]}$
(33)

The first term is the microcausality condition for standard quantum fields
while the second term is due to the non-unitary evolution. From this equation
one immediately notices that the microcausality condition could be satisfied
only if: 1\. $D(x)=0$, i.e. the dynamics is unitary and no collapse mechanism
is present, but this is not the working hypothesis of this paper; 2\.
$F(\mathbf{z}_{1}-\mathbf{z}_{2},\mathbf{x},t,\tau)=F(\textbf{z}_{2}-\textbf{z}_{1},-\mathbf{x},t,\tau)$,
but a simple inspection at Eq. (33) shows that in general this is not true 3\.
$D(x-y)=\delta^{(4)}(x-y)$,111exploiting the invariance of integral measure
$\int\frac{d\mathbb{k}}{2E_{\mathbb{k}}}$ it is not difficult to show that
$F(\mathbb{z},0,t,\tau)=F(0,0,t,\tau)$ for any value of $z$ however in this
case the model produces an infinite energy rate and cannot be reduced to the
CSL model in the non relativistic limit, as shown in Section 5.

##  7 Discussion and Conclusion

In this article we have analysed the possibility of a relativistic extension
of the CSL model. The study was done by constructing a candidate relativistic
generalization.

Starting from the prototypical structure of a generic continuous collapse
model it was required that the model was Lorentz covariant, did not have a
divergent energy rate, and prohibited superluminal signalling. As discussed in
section 3 Lorentz covariance is defined as the evolution equation for the
density operator transforming covariantly under the Lorentz group. The
prohibition of superluminal signalling was checked by ensuring that the
microcausality condition was satisfied, see section 6. It was found that it is
not possible to construct a model that fulfilled all of these requirements.
This is because the need for a finite rate of change of energy implies that
the stochastic noise associated with the model must be coloured, however this
requirement implies that the microcausality condition is violated. As CSL is a
mass coupled spontaneous collapse model, this result implies that a
relativistic CSL or coloured CSL is not possible.

This approach has the advantage that the restrictions are applied in a way
that makes it clear how the various physical requirements on the proposed
model constrain the form.

The main limitation of the present work is that it only considers relativistic
collapse models with the number density operator as the collapse operator.
Further work would be to consider if there exists another choice of Lorentz
covariant collapse operator aside from this, which in the NR limit sector
gives the number density operator. This would establish if any mass coupled
model (like CSL) is the NR limit of any relativistic collapse model.

CJ thanks L. Asprea and J.L. Gaona Reyes for their comments and insights. CJ
and AB acknowledge financial support from the H2020 FET Project TEQ (grant n.
766900). AB acknowledges the COST Action QTSpace (CA15220). AB and CJ
acknowledge financial support from INFN and the University of Trieste. GG
acknowledge support from the Spanish Agencia Estatal de Investigación, project
PID2019-107609GB-I00, Spanish MINECO FIS2016-80681-P (AEI/FEDER, UE),
Generalitat de Catalunya CIRIT 2017-SGR-1127, from QuantERA grant “C’MON-
QSENS!”, by Spanish MICINN PCI2019-111869-2, and the Leverhulme Trust
(RPG-2016- 046).

##  Appendix A Appendices

###  A.1 Existence of the Non-relativistic Sector

In this appendix we will provide necessary conditions under which the map in
Eq. (12) does not produce relativistic phenomena such as creation or
annihilation of NR particles, nor accelerate NR particles to relativistic
velocities. We restrict our analysis to the one particle sector.

For a single particle system $\hat{\rho}$ has the form given in Eq. (19). We
consider a one particle system to be in a non-relativistic state
$\hat{\rho}_{NR}$ if the state satisfies the following condition:

$\langle\mathbf{p}_{L}|\hat{\rho}_{NL}|\mathbf{p}_{R}\rangle\simeq
0\quad\text{if}\quad|\mathbf{p}_{L}|>\kappa
m\quad\text{or}\quad|\mathbf{p}_{R}|>\kappa m$ (34)

where $|\mathbf{p}\rangle$ is a one particle state with 3-momentum
$\mathbf{p}$ and $\kappa\in\mathcal{R}^{+}$ that acts as a momentum cut off
$\kappa\ll 1$. In other words the system is characterised by momentum much
less than the rest energy $m$.

Substituting Eq. (19) into Eq. (34) shows that condition Eq. (34) is met if:

$A(\mathbf{p}_{L},\mathbf{p}_{R})=0\quad\text{if}\quad|\mathbf{p}_{L}|>\kappa
m\quad\text{or}\quad|\mathbf{p}_{R}|>\kappa m$ (35)

With this in hand we can say that the map $\mathcal{M}_{t}$ is “well behaved”
in the NR scenario if the following conditions are met:

$\langle\mathbf{p}_{L}|\mathcal{M}_{t}[\hat{\rho}_{NL}]|\mathbf{p}_{R}\rangle\simeq
0\quad\text{if}\quad|\mathbf{p}_{R}|>\kappa\,m\hskip 5.69046pt\text{or}\hskip
5.69046pt|\mathbf{p}_{L}|>\kappa\,m$ (36)

and

$\langle\mathbf{p}_{L},\mathbf{q}_{L}|\mathcal{M}_{t}[\hat{\rho}_{NL}]|\mathbf{p}_{R},\mathbf{q}_{R}\rangle\approx
0$ (37)

Where $|\mathbf{p}_{R},\mathbf{q}_{R}\rangle$ is a two particle state.
Equation (36) guarantees that a NR one particle state is not driven to a
relativistic states by the map $\mathcal{M}_{t}$. Equation (37) forbids the
creation of particles in the NR regime.

We check these conditions by expanding in $\gamma$ using Eq. (17), which
simplifies (36) and (37) to:

$\langle\mathbf{p}_{L}|\mathcal{L}_{t}[\hat{\rho}_{NL}]|\mathbf{p}_{R}\rangle\simeq
0\hskip 5.69046pt\text{if}\quad|\mathbf{p}_{L}|>\kappa\,m\hskip
5.69046pt\text{or}\hskip 5.69046pt|\mathbf{p}_{R}|>\kappa\,m$ (38)

and

$\langle\mathbf{p}_{L},\mathbf{q}_{L}|\mathcal{L}_{t}[\hat{\rho}_{NL}]|\mathbf{q}_{R},\mathbf{p}_{R}\rangle\approx
0$ (39)

With this equation in hand it is straightforward to verify that condition Eq.
(39) is always satisfied 222 Indeed writing
$|\mathbf{p}_{1},\mathbf{p}_{2}\dots,\mathbf{p}_{n}\rangle=\hat{a}_{\mathbf{p}_{1}},\hat{a}_{\mathbf{p}_{2}}\dots\hat{a}_{\mathbf{p}_{n}}|0\rangle$
one immediately notices that the expression will always contain an odd number
of creation and annihilation operators.. To evaluate condition Eq. (38) we use
Eq. (5) and Eq. (5) and Eq. (19) and find that:

$\displaystyle\langle\mathbf{p}_{L}|\mathcal{L}_{t}[\hat{\rho}_{NL}]|\mathbf{p}_{R}\rangle=-\frac{m^{2}}{4}\int
d\mathbf{q}\int_{0}^{t}ds\int_{0}^{s}d\tau\tilde{D}(\mathbf{q},s-\tau)$
$\displaystyle\hskip
56.9055pt\Bigg{\\{}{A}(\mathbf{p}_{L},\mathbf{p}_{R})\Bigg{(}\frac{e^{i\Delta
E(\mathbf{p}_{L},\mathbf{q})(x_{0}-y_{0})}}{E_{\mathbf{p}_{L}}E_{\mathbf{p}_{L}-\mathbf{q}}}+\frac{e^{-i\Delta
E(\mathbf{p}_{R},\mathbf{q})(s-\tau)}}{E_{\mathbf{p}_{R}}E_{\mathbf{p}_{R}-\mathbf{q}}}\Bigg{)}-$
$\displaystyle\frac{A(\mathbf{p}_{R}-\mathbf{q},\mathbf{p}_{L}-\mathbf{q})}{\sqrt{E_{\mathbf{p}_{L}}E_{\mathbf{p}_{R}}E_{\mathbf{p}_{R}-\mathbf{q}}E_{\mathbf{p}_{L}-\mathbf{q}}}}\Bigg{(}e^{-i\Delta
E(\mathbf{p}_{R},\mathbf{q})s}e^{i\Delta
E(\mathbf{p}_{L},\mathbf{q})\tau}+e^{i\Delta
E(\mathbf{p}_{L},\mathbf{q})s}e^{-i\Delta
E(\mathbf{p}_{R},\mathbf{q})\tau}\Bigg{)}\Bigg{\\}}$ (40)

In order for the map to have a well behaved NR sector this quantity must be
negligible when $\mathbf{p}_{L}$ or $\mathbf{p}_{R}$ are relativistic, i.e.
when $|\mathbf{p}_{R}|>\kappa m$ or $|\mathbf{p}_{L}>\kappa m$. For the term
in the first curly brackets this condition is automatically met due to Eq.
(35) which ensures that ${A}(\mathbf{p}_{L},\mathbf{p}_{R})\approx 0$ for a
non-relativistic state $\rho_{NL}$. However, condition in Eq. (35) is not
enough to guarantee that the term in the second pair of curly brackets
vanishes for $\mathbf{p}_{R}>\kappa m$ or $\mathbf{p}_{L}>\kappa m$, due to
the explicit dependency of
$A(\mathbf{p}_{L}-\mathbf{q},\mathbf{p}_{R}-\mathbf{q})$ on the momentum
transfer $\mathbf{q}$ which can be only bounded by the form of the correlation
function $\tilde{D}(\mathbf{q},s-\tau)$. This means that assuming that
$\tilde{D}(\mathbf{q},s-\tau)$ vanishes for relativistic momentum transfer

$\tilde{D}(\mathbf{q},s-\tau)\approx 0\quad\text{for}\>\>\mathbf{q}>\kappa m.$
(41)

will be sufficient to guarantee that Eq. (A.1) vanishes when when
$|\mathbf{p}_{L}|>\kappa m$ or $|\mathbf{p}_{R}>\kappa m$. To summarise, a
sufficient condition for the map specified by Eq. (12) to be well behaved in
the NR regime is if $\tilde{D}(\mathbf{q},t)$ satisfies Eq. (41), or in other
words if the Fourier transform of the noise correlation function is only
characterized by non-relativistic momentum transfer.

###  A.2 Calculation details for checking necessary condition for
microcausality

Here we evaluate Eq. (31). We expand the map $\mathcal{M}^{*}_{t}$ to first
order in $\gamma$ to obtain:

$\displaystyle[\mathcal{M}_{t}^{*}[\hat{\phi}(\mathbf{z}_{1},t)],\hat{\phi}(\mathbf{z}_{2},0)]\approx[\hat{\phi}(\mathbf{z}_{1},t)+\mathcal{L}_{t}[\hat{\phi}(\mathbf{z}_{1},t)]\big{)},\hat{\phi}(\mathbf{z}_{2},0)]$
$\displaystyle=[\hat{\phi}(\mathbf{z}_{1},t),\hat{\phi}(\mathbf{z}_{2},0)]+\gamma\int_{\Omega_{t}}\\!\\!\\!\\!dsd\mathbf{x}\int_{\Omega_{s}}\\!\\!\\!\\!d\tau
d\mathbf{y}D(s,\tau,\mathbf{x},\mathbf{y})[[\hat{Q}(\mathbf{y},\tau),[\hat{\phi}(\mathbf{z}_{1},t),\hat{Q}(\mathbf{x},s)]],\hat{\phi}(\mathbf{z}_{2},0)]$

The first term is the normal microcausality condition that we will leave as it
is and only consider the second term.

We evaluate this expression from the inner commutator outwards. Recalling that

$\displaystyle\hat{\phi}(\mathbf{x},t)$
$\displaystyle=a(\mathbf{x},t)+a^{\dagger}(\mathbf{x},t)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int\frac{d\mathbf{k}}{\sqrt{2E_{k}}}\left(e^{i(\mathbf{k}\cdot\mathbf{x}-E_{k}t)}\hat{a}_{\mathbf{k}}+e^{-i(\mathbf{k}\cdot\mathbf{x}-E_{k}t)}\hat{a}_{\mathbf{k}}^{\dagger}\right)$
(43)

and exploiting commutation relations in Eq. (16) its easy to obtain

$\displaystyle[\hat{\phi}(\mathbf{z}_{1},t),\hat{Q}(\mathbf{x},s)]=m\int\frac{d\mathbf{k}}{2E_{\mathbf{k}}}\left(e^{i[\mathbf{k}\cdot(\mathbf{z}_{1}-\mathbf{x})-E_{k}(t-s)]}\hat{a}^{\dagger}(\mathbf{x},s)-h.c.\right)$
(44)

This expression can be used to find that

$\displaystyle[[\hat{Q}(\mathbf{y},\tau),[\hat{\phi}(\mathbf{z}_{1},t),\hat{Q}(\mathbf{x},s)]],\hat{\phi}(\mathbf{z}_{2},0)]=$
$\displaystyle=m^{2}\\!\\!\\!\int\frac{d\mathbf{k}}{2E_{\mathbf{k}}}\\!\int\\!\frac{d\mathbf{k^{\prime}}}{2E_{\mathbf{k^{\prime}}}}\\!\int\\!\frac{d\mathbf{k^{\prime\prime}}}{2E_{\mathbf{k^{\prime\prime}}}}\\!\\!\left(\\!e^{i[\mathbf{k}\cdot(\mathbf{z}_{1}-\mathbf{x})-E_{\mathbf{k}}(t-s)]}e^{-i[\mathbf{k}^{\prime}\cdot(\mathbf{x}-\mathbf{y})-E_{\mathbf{k}^{\prime}}(\tau-s)]}e^{i[\mathbf{k}^{\prime\prime}\cdot(\mathbf{y}-\mathbf{z}_{2})-E_{\mathbf{k}^{\prime\prime}}(\tau)]}-h.c.\\!\right)$
(45)

Finally, substituting this expression back into Eq. (A.2), sending
$\mathbf{x}\to\mathbf{x}+\mathbf{y}$, $\tau\to s+\tau$ and integrating over
$d\mathbf{y}$ gives:

$\displaystyle[\mathcal{L}_{t}[\hat{\phi}(\mathbf{z}_{1},t)],\hat{\phi}(\mathbf{z}_{2},0)]$
(46) $\displaystyle=\gamma
m^{2}\int_{\Omega_{t}}dsd\mathbf{x}\int_{\Omega_{s}}d\tau
d\mathbf{y}\,D(\mathbf{x}-\mathbf{y},s-\tau)[[\hat{Q}(\mathbf{y},\tau),[\hat{\phi}(\mathbf{z}_{1},t),\hat{Q}(\mathbf{x},s)]],\hat{\phi}(\mathbf{z}_{2},0)]$
$\displaystyle=\gamma m^{2}\int_{0}^{t}ds\int_{0}^{s}d\tau\int
d\mathbf{x}\,D(\mathbf{x},\tau)\left\\{F(\mathbf{z}_{1}-\mathbf{z}_{2},t,\tau,\mathbf{x})-F(\mathbf{z}_{1}-\mathbf{z}_{2},-t,-\tau,\mathbf{x})\right\\}$
(47)

with

$\displaystyle
F(\mathbf{z}_{1}-\mathbf{z}_{2},t,\tau,\mathbf{x})=\int\frac{d\mathbf{k}}{2E_{\mathbf{k}}}\\!\int\\!\frac{d\mathbf{k^{\prime}}}{4E_{\mathbf{k^{\prime}}}^{2}}e^{i[\mathbf{k}\cdot(\mathbf{z}_{1}-\mathbf{z}_{2})-E_{\mathbf{k}}t]}e^{-i[(\mathbf{k}-\mathbf{k}^{\prime})\cdot\mathbf{x}-(E_{\mathbf{k}}-E_{\mathbf{k^{\prime}}})\tau]}$
(48)

Further making the transformation $(\mathbf{x},\tau)\to(\mathbf{x},-\tau)$ in
the second term of Eq. (A.2) and making use of the symmetry
$D(\mathbf{x},\tau)=D(-\mathbf{x},\tau)$ and
$F(\mathbf{z}_{2}-\mathbf{z}_{1},t,\tau,-\mathbf{x})=F(\mathbf{z}_{1}-\mathbf{z}_{2},-t,-\tau,\mathbf{x})$
we get

$\displaystyle=\gamma
m^{2}\int_{\Omega_{t}}dsd\mathbf{x}\int_{\Omega_{s}}d\tau
d\mathbf{y}\,D(\mathbf{x}-\mathbf{y},s-\tau)[[\hat{Q}(\mathbf{y},\tau),[\hat{\phi}(\mathbf{z}_{1},t),\hat{Q}(\mathbf{x},s)]],\hat{\phi}(\mathbf{z}_{2},0)]$
$\displaystyle=\gamma m^{2}\int_{0}^{t}ds\int_{0}^{s}d\tau\int
d\mathbf{x}\,D(\mathbf{x},\tau)\left\\{F(\mathbf{z}_{1}-\mathbf{z}_{2},t,\tau,\mathbf{x})-F(\mathbf{z}_{2}-\mathbf{z}_{1},t,\tau,-\mathbf{x})\right\\}.$
(49)

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# Solving sums of squares in global fields

Przemysław Koprowski<EMAIL_ADDRESS>Institite of Mathematics,
University of Silesia, ul. Bankowa 14, Katowice, Poland

###### Abstract.

The problem of writing a totally positive element as a sum of squares has a
long history in mathematics, going back to Bachet and Lagrange. While for some
specific rings (like integers or polynomials over the rationals), there are
known methods for decomposing an element into a sum of squares, in general,
for many other important rings and fields, the problem is still widely open.
In this paper, we present an explicit algorithm for decomposing an element of
an arbitrary global field (either a number field or a global function field)
into a sum of squares of minimal length.

###### Key words and phrases:

Sums of squares, algorithms, global fields, number fields, quadratic forms.

###### 2020 Mathematics Subject Classification:

11E25, 11Y40, 11E12

##  1\. Introduction

One of the fundamental results in classical number theory is the famous
Lagrange Four Squares Theorem, stating that every positive integer is a sum of
four squares. Over centuries, Lagrange theorem has been generalized in
numerous directions. A remarkable result by Siegel says that a totally
positive element of any global field is a sum of four squares. Hence, from the
existential standpoint, the theory over an arbitrary number field is as
complete as over the rationals. It is no longer so if we consider the
computational aspect of the problem. There is a number of algorithms for
decomposing a positive rational number into a sum of four (or less) squares,
see for example [2, 18, 19, 20, 21]. On the other hand, for number fields, or
more generally global fields, there is only the recent paper by M.-K. Darkey-
Mensah and B. Rothkegel [7], where the authors give an algorithm for computing
the _length_ (i.e. the minimal number of summands needed to represent given
$a$ as a sum of squares) of an element of a global field. The author is not
aware of any algorithm that would actually solve the problem of representing
an algebraic number as a sum of squares of minimal length. Thus, the goal of
this paper is to fill this gap and propose such algorithms, working over
arbitrary global fields (so either number fields or global function fields).

Recall that over a field of characteristic $2$, every sum of squares is a
square. Hence the problem of decomposing an element into a sum of squares is
meaningful only over fields of characteristic distinct from $2$. Therefore, in
what follows the term _global field_ will always mean “global field of
characteristic $\neq 2$”. It can be either a number field (then the
characteristic is just zero) or an algebraic function field over a finite
field $\mathbb{F}_{p^{n}}$, for some odd prime $p$.

Throughout this paper we use the following notation and terminology. By $K$ we
always denote a global field and $K^{\times}$ stands for the group of nonzero
elements. We write $\langle a_{1},\dotsc,a_{n}\rangle$ for the quadratic form
$a_{1}x_{1}^{2}+\dotsb+a_{n}x_{n}^{2}$. The _length_ of an element $a\in
K^{\times}$, denoted $\ell(a)$, is the minimal number of summands needed to
express $a$ as a sum of squares. If $a$ is not a sum of squares, we put
$\ell(a):=\infty$. Since a representation of $-1$ as a sum of squares plays a
special role, the length of $-1$ is called the _level_ of $K$ and denoted
$s(K)$. In particular, $s(K)=\infty$ if and only if $K$ is formally real. It
is known (see e.g., [15, Example XI.2.4 (8)]) that the level of a global field
can take only one of the following values: $1$, $2$, $4$ or $\infty$. Another
invariant related to sums of squares is the _Pythagoras number_ of $K$,
denoted $P(K)$ hereafter. By definition, the Pythagoras number is the supremum
of the lengths of all the elements of K that are sums of squares.

An equivalency class of valuations is called a _place_ of $K$. We always use
fraktur letters $\mathfrak{d}$, $\mathfrak{p}$, $\mathfrak{q}$, $\mathfrak{r}$
to denote places. To avoid a repeated use of the word “place”, non-archimedean
places will be called also _primes_ (or _finite primes_ , when we wish to
emphasize the fact that they are not archimedean). The completion of $K$ with
respect to a place $\mathfrak{p}$ is denoted $K_{\mathfrak{p}}$. If
$\mathfrak{p}$ is non-archimedean, then
$\operatorname{ord}_{\mathfrak{p}}:K\to\mathbb{Z}\cup\\{\infty\\}$ stands for
the corresponding discrete valuation on $K$. The valuation
$\operatorname{ord}_{\mathfrak{p}}$ induces a map
$\nicefrac{{K^{\times}}}{{K^{\times
2}}}\to\nicefrac{{\mathbb{Z}}}{{2\mathbb{Z}}}$ on the square class group of
$K$. Abusing the notation harmlessly, we denote the latter map by
$\operatorname{ord}_{\mathfrak{p}}$, again.

Let $S$ be a finite set of places of $K$, containing all archimedean places. A
square-class $\alpha\in\nicefrac{{K^{\times}}}{{K^{\times 2}}}$ is said to be
_$S$ -singular_ if $\operatorname{ord}_{\mathfrak{p}}\alpha$ is even for every
prime $\mathfrak{p}\notin S$. The set of $S$-singular square classes form a
subgroup of $\nicefrac{{K^{\times}}}{{K^{\times 2}}}$, that we denote
$\mathbb{E}_{S}$. Observe that $\mathbb{E}_{S}$ is an elementary $2$-group,
hence it can be treated as a vector space over $\mathbb{F}_{2}$. We will use
this fact extensively.

For a real place $\mathfrak{r}$, by $\operatorname{sgn}_{\mathfrak{r}}$ we
denote the sign function on $K$, associated to the ordering of $K$, induced by
the unique ordering of the completion $K_{\mathfrak{r}}$. Given a finite prime
$\mathfrak{p}$ and two elements $a$ and $b$ of $K$, we write
$(a,b)_{\mathfrak{p}}$ for their Hilbert symbol at $\mathfrak{p}$. Further, if
$q=\langle a_{1},\dotsc,a_{n}\rangle$ is a quadratic form (with coefficients
in $K$), then $q\otimes K_{\mathfrak{p}}$ stands for the associated quadratic
form over $K_{\mathfrak{p}}$. We write $s_{\mathfrak{p}}q$ for the Hasse
invariant of $q$ at $\mathfrak{p}$. By definition (see e.g., [15, Definition
V.3.17]):

$s_{\mathfrak{p}}q:=\prod_{i<j}(a_{i},a_{j})_{\mathfrak{p}}.$

Our algorithms proposed in this paper rely on a number of already existing
tools. The three main ingredients are:

  1. (1)

Factorization of (fractional) ideals in global fields. There are a couple of
methods for achieving this (see e.g., [4, 6]). At the moment of writing the
fastest ideal factorization method known to the author is the one described in
[11].

  2. (2)

Solution of a norm equation of a form $N_{L/K}\left(x\right)=b$, where $b\in
K^{\times}$ and $L$ is an extension of $K$. Algorithms for solving the norm
equations are described in [5, 8, 9, 10, 22]. In fact, in this paper, $L/K$
will be always a quadratic extension.

  3. (3)

Construction of a basis (over $\mathbb{F}_{2}$) of the group $\mathbb{E}_{S}$
of $S$-singular square classes, for a given set $S$. The author is aware of
three algorithms for this task. One approach, outlined in Magma manual [3], is
to find a set $S^{\prime}$ containing $S$ and such that the $S^{\prime}$-class
number is odd. Then $\mathbb{E}_{S^{\prime}}$ coincides with the group of
$S^{\prime}$-units modulo squares. The sought group $\mathbb{E}_{S}$ can be
subsequently constructed as a subspace of $\mathbb{E}_{S^{\prime}}$. Another
method due to A. Czogała has been recently described in [14]. Yet another
algorithm is published in [13]. The last one is probably best suited for our
purpose, since it is incremental in nature, and our solution is based on a
procedure of gradual enlarging a certain set $S$.

##  2\. Algorithms

We may assume that the length of $a$ is known a priori, since it can be
calculated using [7]. Of course, if the length is $\ell(a)=1$, then $a$ is a
square and it suffices to compute its square root in $K$. Below we present
separate algorithms for all other possible lengths of $a$. We begin with sums
of just two squares.

######  Algorithm 1.

Given an element $a\in K^{\times}$ of length $\ell(a)=2$, this algorithm
outputs $c_{1},c_{2}\in K^{\times}$ such that $a=c_{1}^{2}+c_{2}^{2}$.

  1. (1)

If $-1$ is a square in $K$, then output

$c_{1}:=\frac{a+1}{2},\qquad c_{2}:=\frac{a-1}{2}\cdot\sqrt{-1}$

and quit.

  2. (2)

Otherwise set $L:=K\bigl{(}\sqrt{-1}\bigr{)}$ and solve the norm equation
$N_{L/K}\left(x\right)=a$. Denote a solution by $c_{1}+c_{2}\cdot\sqrt{-1}$.

  3. (3)

Output $c_{1},c_{2}$.

###### Proof of correctness.

If $-1$ is a square in $K$, then the correctness of the output of step (1)
follows by a direct calculation. Otherwise, when $-1$ is not a square, then
$L:=K\bigl{(}\sqrt{-1}\bigr{)}$ is a proper extension of $K$ and we have

$a=c_{1}^{2}+c_{2}^{2}=N_{L/K}\left(c_{1}+c_{2}\cdot\sqrt{-1}\right)$

as desired. ∎

We may now focus on sums of three squares. First of all, observe that if
$\ell(a)=3$ for some $a\in K$, then the Pythagoras number of $K$ must be at
least three. It follows from [15, Theorem XI.5.6] that the level of $K$ is
strictly grater than $1$. Hence it is either $2$, or $4$, or infinity if $K$
is a formally real number field. In particular, $-1\notin K^{\times 2}$ and
the field $L:=K\bigl{(}\sqrt{-1}\bigr{)}$ is always a proper extension of $K$.
We need to distinguish between two cases. For the sake of clarity of the
exposition, we present them as two separate algorithms. The first one is
rather trivial.

######  Algorithm 2.

Let $K$ be a global field of level $s(K)=2$. Given an element $a\in
K^{\times}$ of length $\ell(a)=3$, this algorithm outputs
$c_{1},c_{2},c_{3}\in K^{\times}$ such that $a=c_{1}^{2}+c_{2}^{2}+c_{3}^{2}$.

  1. (1)

Let $L:=K\bigl{(}\sqrt{-1}\bigr{)}$.

  2. (2)

Solve the norm equation $N_{L/K}\left(x\right)=-1$ and let
$d_{1}+d_{2}\cdot\sqrt{-1}$ be a solution.

  3. (3)

Output

$c_{1}:=\frac{a+1}{2},\qquad c_{2}:=\frac{a-1}{2}\cdot d_{1},\qquad
c_{3}:=\frac{a-1}{2}\cdot d_{2}.$

###### Proof of correctness.

By assumption, $-1$ is a sum of two squares in $K$ and so the norm equation is
step (2) is solvable. By a direct computation we obtain that $a$ is the sum of
squares of the elements $c_{1},c_{2},c_{3}$ constructed in step (3). This
proves the correctness of the algorithm. ∎

We can now concentrate on fields of level strictly greater than $2$ (so either
$4$ or infinity). Observe that the level of a global function field equals the
level of its full field of constants and so it cannot exceed $2$. Thus, if the
level is greater than $2$, the field in question must be a number field.

######  Algorithm 3.

Let $K$ be a number field of level $s(K)\neq 1,2$ and let $a\in K^{\times}$ be
an element of length $\ell(a)=3$. This algorithm outputs $c_{1},c_{2},c_{3}\in
K^{\times}$ such that $a=c_{1}^{2}+c_{2}^{2}+c_{3}^{2}$.

  1. (1)

Construct a set $S:=\\{\mathfrak{p}_{1},\dotsc,\mathfrak{p}_{2}\\}$ of finite
primes of $K$ such that:

     * •

$S$ contains all the dyadic primes of $K$;

     * •

for every prime $\mathfrak{p}$, if $\operatorname{ord}_{\mathfrak{p}}a\equiv
1\pmod{2}$, then $\mathfrak{p}\in S$.

  2. (2)

Construct a basis $\mathscr{B}=\\{\kappa_{1},\dotsc,\kappa_{k}\\}$ of the
group $\mathbb{E}_{S}$ of $S$-singular square classes.

  3. (3)

Construct a set $S_{\infty}=\\{\mathfrak{r}_{1},\dotsc,\mathfrak{r}_{r}\\}$ of
all the real places of $K$.

  4. (4)

Repeat the following steps:

    1. (a)

Build a vector $u=(1,\dotsc,1)$ of length $r=|S_{\infty}|$.

    2. (b)

Build a vector $v=(v_{1},\dotsc,v_{s})$, indexed by the primes in $S$,
setting:

$v_{i}:=\begin{cases}1&\text{if }(-1,-1)_{\mathfrak{p}_{i}}=-1,\\\ 0&\text{if
}(-1,-1)_{\mathfrak{p}_{i}}=1.\end{cases}$

    3. (c)

Build a vector $w=(0,\dotsc,0)$ of length $s=|S|$.

    4. (d)

Construct a matrix $A=(a_{ij})$ with $r$ rows (indexed by the real places of
$K$) and $k$ columns (indexed by elements in $\mathscr{B}$), setting

$a_{ij}:=\begin{cases}1&\text{if
}\operatorname{sgn}_{\mathfrak{r}_{i}}\kappa_{j}=-1,\\\ 0&\text{if
}\operatorname{sgn}_{\mathfrak{r}_{i}}\kappa_{j}=1.\end{cases}$

    5. (e)

Construct a matrix $B=(b_{ij})$ with $s$ rows and $k$ columns, setting

$b_{ij}:=\begin{cases}1&\text{if }(-1,\kappa_{j})_{\mathfrak{p}_{i}}=-1,\\\
0&\text{if }(-1,\kappa_{j})_{\mathfrak{p}_{i}}=1.\end{cases}$

    6. (f)

Construct a matrix $C=(c_{ij})$ with $s$ rows and $k$ columns, setting

$c_{ij}:=\begin{cases}1&\text{if }(a,\kappa_{j})_{\mathfrak{p}_{i}}=-1,\\\
0&\text{if }(a,\kappa_{j})_{\mathfrak{p}_{i}}=1.\end{cases}$

    7. (g)

Check if the following system of $\mathbb{F}_{2}$-linear equations is
solvable:

($\clubsuit$) $\left(\begin{array}[]{c}A\\\ \hline\cr B\\\ \hline\cr
C\end{array}\right)\cdot\begin{pmatrix}x_{1}\\\ \vdots\\\
x_{k}\end{pmatrix}=\left(\begin{array}[]{c}u\\\ \hline\cr v\\\ \hline\cr
w\end{array}\right).$

If it is, denote a solution by $\xi_{1},\dotsc,\xi_{k}$ and exit the loop.

    8. (h)

If system ($\clubsuit$ ‣ 4g) is not solvable, find a new prime
$\mathfrak{q}\notin S$, replace $S$ by $S\cup\\{\mathfrak{q}\\}$ and update
the basis $\mathscr{B}$ accordingly.

    9. (i)

Reiterate the loop.

  5. (5)

Set

$b:=\kappa_{1}^{\xi_{1}}\dotsm\kappa_{k}^{\xi_{k}}.$

  6. (6)

Set $L:=K(\sqrt{-1})$ and find a solution $d_{1}+d_{2}\cdot\sqrt{-1}\in L$ to
the norm equation:

$N_{L/K}\left(x\right)=-b.$

  7. (7)

Set $M:=K(\sqrt{a})$ and find a solution $d_{3}+d_{4}\cdot\sqrt{a}\in M$ to
the norm equation:

$N_{M/K}\left(y\right)=b.$

  8. (8)

Output

$c_{1}:=\frac{d_{1}}{d_{4}},\qquad c_{2}:=\frac{d_{2}}{d_{4}},\qquad
c_{3}:=\frac{d_{3}}{d_{4}}.$

###### Proof of correctness.

First we need to show that the loop in step (4) terminates. That is, that
after appending enough primes to the set $S$, systen ($\clubsuit$ ‣ 4g)
eventually becomes solvable. We know that $a$ is a sum of three squares, hence
the quadratic form $\langle 1,1,1,-a\rangle$ is isotropic over $K$. It
follows, that there is some $c\in K^{\times}$ such that $(-c)$ is represented
by the form $\langle 1,1\rangle$ and $c$ is represented by $\langle
1,-a\rangle$. In other words, the two forms $\langle 1,1,c\rangle$ and
$\langle 1,-a,-c\rangle$ are both isotropic. In particular, $c$ is totally
negative.

Let $T$ be a (finite) set of places of $K$, consisting of

  * •

all archimedean places,

  * •

all dyadic primes,

  * •

all non-dyadic primes, at which $a$ has odd valuation.

In other words, $T$ is the union of the sets $S$, $S_{\infty}$ and all the
complex places of $K$. Then [16, Lemma 2.1] says that there is a prime
$\mathfrak{q}\notin T$ and a
$\bigl{(}T\cup\\{\mathfrak{q}\\}\bigr{)}$-singular element $b\in K^{\times}$
such that:

  1. ($\textup{C}_{1}$)

$\operatorname{sgn}_{\mathfrak{r}}b=\operatorname{sgn}_{\mathfrak{r}}c$ for
every real place $\mathfrak{r}$ of $K$,

  2. ($\textup{C}_{2}$)

$b\equiv c\pmod{\mathfrak{p}^{1+\operatorname{ord}_{\mathfrak{p}}4}}$ for
every finite prime $\mathfrak{p}\in T$,

  3. ($\textup{C}_{3}$)

$\operatorname{ord}_{\mathfrak{q}}b=1$.

Consider the following two quadratic forms:

$q_{1}:=\langle 1,1,b\rangle\qquad\text{and}\qquad q_{2}:=\langle
1,-a,-b\rangle$

and let $\mathscr{B}=\\{\kappa_{1},\dotsc,\kappa_{k}\\}$ be a basis, over
$\mathbb{F}_{2}$, of the group
$\mathbb{E}_{T\cup\\{\mathfrak{q}\\}}=\mathbb{E}_{S\cup\\{\mathfrak{q}\\}}$ of
$\bigl{(}S\cup\\{\mathfrak{q}\\}\bigr{)}$-singular square classes. Write the
square class of $b$ in a form
$\kappa_{1}^{\xi_{1}}\dotsm\kappa_{k}^{\xi_{k}}$. We will show that the
coordinates $\xi_{1},\dotsc,\xi_{k}$ form a solution to ($\clubsuit$ ‣ 4g),
and that every solution makes the corresponding forms $q_{1}$ and $q_{2}$
isotropic.

Of course, $q_{1}$, $q_{2}$ are locally isotropic at complex places of $K$.
Likewise, they are locally isotropic at every finite prime $\mathfrak{p}\notin
S\cup\\{\mathfrak{q}\\}$ by [15, Corollary VI.2.5], since $\mathfrak{p}$ is
non-dyadic and all three coefficients of each form have even valuations at
$\mathfrak{p}$. Therefore, there are only finitely many places of $K$ that
must be considered. These are precisely the places in $S_{\infty}\cup
S\cup\\{\mathfrak{q}\\}$. We should deal first with the real places. Observe
that $b$, having the same signs as $c$, is totally negative. Therefore we have

$A\cdot\begin{pmatrix}\xi_{1}\\\ \vdots\\\ \xi_{k}\end{pmatrix}=u.$

Conversely, every solution to the above equation gives a totally negative
element $b$. Therefore, the forms $q_{1}\otimes K_{\mathfrak{r}}$ and
$q_{2}\otimes K_{\mathfrak{r}}$ are isotropic for every real place
$\mathfrak{r}$. To prove that the latter one is isotropic, we use the fact
that $a$ is totally positive as a sum of squares.

We may now focus on non-archimedean places. Take a prime $\mathfrak{p}\in S$.
By ($\textup{C}_{2}$) and Local Square Theorem (see e.g. [15, Theorem
VI.2.19]) the local squares classes $b\cdot K_{\mathfrak{p}}^{\times 2}$ and
$c\cdot K_{\mathfrak{p}}^{\times 2}$ coincide. This yields the isometries:

$q_{1}\otimes K_{\mathfrak{p}}\cong\langle 1,1,c\rangle\otimes
K_{\mathfrak{p}}\qquad\text{and}\qquad q_{2}\otimes
K_{\mathfrak{p}}\cong\langle 1,-a,-c\rangle\otimes K_{\mathfrak{p}}.$

The right hand sides are isotropic and so are the left hand sides. Finally,
take the prime $\mathfrak{q}$ that we appended to $T$. By the previous part,
the two Hilbert symbols:

$(-1,-b)_{\mathfrak{p}}\qquad\text{and}\qquad(a,b)_{\mathfrak{p}}$

vanish for every $\mathfrak{p}\neq\mathfrak{q}$. It follows from Hilbert
Reciprocity Law (see e.g., [17, Chapter VII]) that

$(-1,-b)_{\mathfrak{q}}=(a,b)_{\mathfrak{q}}=1.$

Thus, the forms $q_{1}\otimes K_{\mathfrak{q}}$ and $q_{2}\otimes
K_{\mathfrak{q}}$ are also isotropic. This way, we proved that $q_{1}$ and
$q_{2}$ are locally isotropic everywhere, hence they are isotropic over $K$ by
the local-global principle (see e.g., [15, Theorem VI.3.1]).

Toke again a prime $\mathfrak{p}_{i}\in S\cup\\{\mathfrak{q}\\}$. We have
already proved that $\mathfrak{q}_{1}\otimes K_{\mathfrak{p}_{i}}$ is
isotropic. Hence, by [15, Proposition V.3.22], its Hasse invariant
$s_{\mathfrak{p}_{i}}(q_{1})$ equals the Hilbert symbol $(-1-\det
q_{1})_{\mathfrak{p}_{i}}$. This leads to:

$\displaystyle(-1)^{v_{i}}$ $\displaystyle=(-1,-1)_{\mathfrak{p}_{i}}\cdot 1$
$\displaystyle=(-1,-1)_{\mathfrak{p}_{i}}\cdot(1,1)_{\mathfrak{p}_{i}}\cdot(1,b)_{\mathfrak{p}_{i}}^{2}$
$\displaystyle=(-1,-1)_{\mathfrak{p}_{i}}\cdot s_{\mathfrak{p}_{i}}(q_{1})$
$\displaystyle=(-1,-1)_{\mathfrak{p}_{i}}\cdot(-1,-\det
q_{1})_{\mathfrak{p}_{i}}$
$\displaystyle=(-1,-1)_{\mathfrak{p}_{i}}\cdot(-1,-b)_{\mathfrak{p}_{i}}$
$\displaystyle=(-1,b)_{\mathfrak{p}_{i}}$ $\displaystyle=\prod_{j\leq
k}(-1,\kappa_{j})^{\xi_{j}}_{\mathfrak{p}_{i}}$ $\displaystyle=\prod_{j\leq
k}(-1)^{\xi_{j}b_{ij}}.$

Consequently we have

$\sum_{j\leq k}\xi_{j}b_{ij}=v_{i}.$

Conversely, if the above equality holds then
$s_{\mathfrak{p}_{i}}(q_{1})=(-1,-\det q_{1})_{\mathfrak{p}_{i}}$ and so
$q_{1}\otimes K_{\mathfrak{p}_{i}}$ is isotropic.

Analogously, $q_{2}\otimes K_{\mathfrak{p}_{i}}$ is isotopic, hence
$s_{\mathfrak{p}_{i}}(q_{2})=(-1,-\det q_{2})_{\mathfrak{p}_{i}}$. Write

(1) $\begin{split}(-1)^{0}&=(-1,-ab)_{\mathfrak{p}_{i}}^{2}\\\ &=(-1,-\det
q_{2})_{\mathfrak{p}_{i}}\cdot(-1,-ab)_{\mathfrak{p}_{i}}\\\
&=s_{\mathfrak{p}_{i}}q_{2}\cdot(-1,-ab)_{\mathfrak{p}_{i}}\\\
&=(-a,-b)_{\mathfrak{p}_{i}}\cdot(-1,-a)_{\mathfrak{p}_{i}}\cdot(-1,b)_{\mathfrak{p}_{i}}\\\
&=(a,b)_{\mathfrak{p}_{i}}\\\ &=\prod_{j\leq
k}(a,\kappa_{j})_{\mathfrak{p}_{i}}^{\xi_{j}}\\\ &=\prod_{j\leq
k}(-1)^{\xi_{j}c_{ij}}.\end{split}$

This way we obtained the “$C$-part” of system ($\clubsuit$ ‣ 4g):

$\sum_{j\leq k}\xi_{j}c_{ij}=w_{i}.$

Conversely, if the above equality holds for some
$\xi_{1},\dotsc,\xi_{k}\in\\{0,1\\}$ and
$b=\kappa_{1}^{\xi_{1}}\dotsm\kappa_{k}^{\xi_{k}}$, then
$s_{\mathfrak{p}_{i}}q_{2}=(-1,-\det q_{2})_{\mathfrak{p}_{i}}$ and so
$q_{2}\otimes K_{\mathfrak{p}_{i}}$ is isotopic.

This way we proved that once we appended $\mathfrak{q}$ to the initial set
$S$, the system ($\clubsuit$ ‣ 4g) becomes solvable and every solution to this
system results in an element $b\in\mathbb{E}_{S\cup\\{\mathfrak{q}\\}}$ for
which the forms $q_{1}$ and $q_{2}$ are isotropic.

The existence of the prime $\mathfrak{q}$ in [16] is proved using Chebotarev
density theorem. Although it is not explicit in the statement of [16, Lemma
2.1], the set of such primes has positive density. This means that the loop in
step (4) terminates (see Remark 1 at the end of the paper).

Now, let $b=\kappa_{1}^{\xi_{1}}\dotsm\kappa_{k}^{\xi_{k}}$ for some
$\xi_{1},\dotsc,\xi_{k}\in\\{0,1\\}$ forming a solution to ($\clubsuit$ ‣ 4g).
Then, $q_{1}=\langle 1,1,b\rangle$ is isotropic, hence $(-b)$ is a sum of two
squares in $K$. Therefore, the norm equation in step (6) has a solution. Say,

(2) $-b=N_{L/K}\left(d_{1}+d_{2}\cdot\sqrt{-1}\right)=d_{1}^{2}+d_{2}^{2}.$

Likewise, the norm equation $b=N_{M/K}\left(y\right)$ also has a solution,
since $b$ is represented by the form $\langle 1,-a\rangle$. Write

(3) $b=N_{M/K}\left(d_{3}+d_{4}\cdot\sqrt{a}\right)=d_{3}^{2}-a\cdot
d_{4}^{2}.$

Combining (2) with (3) we arrive at the sought representation of $a$ as a sum
of three squares:

$a=\left(\frac{d_{1}}{d_{4}}\right)^{2}+\left(\frac{d_{2}}{d_{4}}\right)^{2}+\left(\frac{d_{3}}{d_{4}}\right)^{2}.\qed$

We may now turn our attention to sums of four squares. Recall that the level
of a global function field is either $1$ or $2$, hence the Pythagoras number
of such field is at most $3$ by [15, Theorem XI.5.6]. Consequently, the
$4$-squares problem is meaningful only for number fields.

######  Algorithm 4.

Let $a\in K^{\times}$ be an algebraic number of length $\ell(a)=4$. This
algorithm finds $c_{1},\dotsc,c_{4}\in K^{\times}$ such that
$a=c_{1}^{2}+\dotsb+c_{4}^{2}$.

  1. (1)

Construct a set $D:=\\{\mathfrak{d}_{1},\dotsc,\mathfrak{d}_{d}\\}$ consisting
of the dyadic primes of $K$, whose ramification indices and inertia degrees
are simultaneously odd, i.e.

$e(\mathfrak{d}_{i}/2)\equiv f(\mathfrak{d}_{i}/2)\equiv 1\pmod{2}$

for $i\leq d$.

  2. (2)

Construct a set $S\supseteq D$ consisting of all the dyadic primes of $K$ and
all the primes where $a$ has odd valuation.

  3. (3)

Construct a set $S_{\infty}=\\{\mathfrak{r}_{1},\dotsc,\mathfrak{r}_{r}\\}$ of
all the real places of $K$.

  4. (4)

For every prime ${\mathfrak{d}_{i}}\in D$ find elements $g_{i},h_{i}\in
K^{\times}$ such that

$(a,g_{i})_{\mathfrak{d}_{i}}=1\qquad\text{and}\qquad(g_{i},h_{i})_{\mathfrak{d}_{i}}=-1.$

  5. (5)

Construct a basis $\mathscr{B}=\\{\kappa_{1},\dotsc,\kappa_{k}\\}$ of the
group $\mathbb{E}_{S}$ of $S$-singular square classes.

  6. (6)

Repeat the following steps:

    1. (a)

Build a vector $u=(1,\dotsc,1)$ of length $r=|S_{\infty}|$.

    2. (b)

Build a vector $v=(1,\dotsc,1)$ of length $d=|D|$.

    3. (c)

Build a vector $w=(0,\dotsc,0)$ of length $s=|S|$.

    4. (d)

Construct a matrix $A=(a_{ij})$ with $r$ rows (indexed by the real places of
$K$) and $k$ columns (indexed by elements of $\mathscr{B}$), setting:

$a_{ij}:=\begin{cases}1&\text{if
}\operatorname{sgn}_{\mathfrak{r}_{i}}\kappa_{j}=-1,\\\ 0&\text{if
}\operatorname{sgn}_{\mathfrak{r}_{i}}\kappa_{j}=1.\end{cases}$

    5. (e)

Construct a matrix $B=(b_{ij})$ with $d$ rows (indexed by the primes in $D$)
and $k$ columns, setting:

$b_{ij}:=\begin{cases}1&\text{if }(h_{i},\kappa_{j})_{\mathfrak{d}_{i}}=-1\\\
0&\text{if }(h_{i},\kappa_{j})_{\mathfrak{d}_{i}}=1.\end{cases}$

    6. (f)

Construct a matrix $C=(c_{ij})$ with $s$ rows (indexed by the primes in $S$)
and $k$ columns, setting

$c_{ij}:=\begin{cases}1&\text{if }(a,\kappa_{j})_{\mathfrak{p}_{i}}=-1\\\
0&\text{if }(a,\kappa_{j})_{\mathfrak{p}_{i}}=1.\end{cases}$

    7. (g)

Check if the following system of $\mathbb{F}_{2}$-linear equations is
solvable:

($\spadesuit$) $\left(\begin{array}[]{c}A\\\ \hline\cr B\\\ \hline\cr
C\end{array}\right)\cdot\begin{pmatrix}x_{1}\\\ \vdots\\\
x_{k}\end{pmatrix}=\left(\begin{array}[]{c}u\\\ \hline\cr v\\\ \hline\cr
w\end{array}\right).$

If it is, denote a solution by $\xi_{1},\dotsc,\xi_{k}$ and exit the loop.

    8. (h)

Otherwise, find a new prime $\mathfrak{q}\notin S$, append it to $S$ and
update the basis $\mathscr{B}$.

    9. (i)

Reiterate the loop.

  7. (7)

Set $b:=\kappa_{1}^{\xi_{1}}\dotsm\kappa_{m}^{\xi_{m}}$.

  8. (8)

Execute Algorithm 3 to obtain $d_{1},d_{2},d_{3}\in K^{\times}$ such that

$-b=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}.$

  9. (9)

Set $L:=K\bigl{(}\sqrt{a}\bigr{)}$ and find a solution
$d_{4}+d_{5}\cdot\sqrt{a}\in L$ to the norm equation

$b=N_{L/K}\left(x\right).$

  10. (10)

Output
$c_{1}:=\nicefrac{{d_{1}}}{{d_{5}}},\dotsc,c_{4}:=\nicefrac{{d_{4}}}{{d_{5}}}$.

###### Proof of correctness.

The proof is very similar to that of Algorithm 3. Since $a$ is a sum of four
squares, there is some $c\in K^{\times}$ such that the two forms $\langle
1,1,1,c\rangle$ and $\langle 1,-a,-c\rangle$ are isotropic. In particular, $c$
must be totally negative. Again, there is a positive-density set of primes of
$K$ such that appending to $S$ any prime $\mathfrak{q}$ from this set, we may
find an $\bigl{(}S\cup\\{\mathfrak{q}\\}\bigr{)}$-singular element $b$ such
that:

  1. ($\textup{C}_{1}$)

$\operatorname{sgn}_{\mathfrak{r}}b=\operatorname{sgn}_{\mathfrak{r}}c$ for
every real place $\mathfrak{r}$ of $K$,

  2. ($\textup{C}_{2}$)

$b\equiv c\pmod{\mathfrak{p}^{1+\operatorname{ord}_{\mathfrak{p}}4}}$ for
every prime $\mathfrak{p}\in S\setminus D$,

  3. ($\textup{C}_{3}$)

$b\equiv
g_{i}\pmod{\mathfrak{d}_{i}^{1+\operatorname{ord}_{\mathfrak{d}_{i}}4}}$ for
the primes $\mathfrak{d}_{1},\dotsc,\mathfrak{d}_{d}\in D$,

  4. ($\textup{C}_{4}$)

$\operatorname{ord}_{\mathfrak{q}}b=1$.

By the Local Square Theorem we have

  * •

$b\cdot K_{\mathfrak{p}}^{\times 2}=c\cdot K_{\mathfrak{p}}^{\times 2}$ for
every $\mathfrak{p}\in S\setminus D$,

  * •

$b\cdot K_{\mathfrak{d}_{i}}^{\times 2}=g_{i}\cdot
K_{\mathfrak{d}_{i}}^{\times 2}$ for
$\mathfrak{d}_{1},\dotsc,\mathfrak{d}_{d}\in D$.

Denote

$q_{1}:=\langle 1,1,1,b\rangle\qquad\text{and}\qquad q_{2}:=\langle
1,-a,-b\rangle.$

As in the proof of correctness of Algorithm 3, we will show that the
coordinates $\xi_{1},\dotsc,\xi_{k}$ of $b$ in
$\mathbb{E}_{S\cup\\{\mathfrak{q}\\}}$ form a solution to ($\spadesuit$ ‣ 6g).
And every solution to ($\spadesuit$ ‣ 6g) results in an element $b$ for which
$q_{1}$ and $q_{2}$ are isotropic.

For archimedean places of $K$, as well as the non-archimedean ones, but
distinct from $\mathfrak{q}$ and not sitting in $S$, the arguments are the
same as used in the previous proof. Take a prime ${\mathfrak{p}_{i}}\in
S\setminus D$. Then $q_{1}\otimes K_{\mathfrak{p}_{i}}$ is isotropic since it
contains three units. For the form $q_{2}$ we use the same arguments as in the
previous proof to show that $q_{2}\otimes K_{\mathfrak{p}_{i}}$ is isotropic,
since $b\cdot K_{\mathfrak{p}_{i}}^{\times 2}=c\cdot
K_{\mathfrak{p}_{i}}^{\times 2}$, and this isotropy is equivalent to the
condition

$\sum_{j\leq k}\xi_{j}c_{ij}=w_{i}.$

What really distinguishes sums of four squares from sums of three squares is
the local behavior of $q_{1}$ and $q_{2}$ at the dyadic primes
$\mathfrak{d}_{1},\dotsc,\mathfrak{d}_{d}\in D$. Fix a prime
${\mathfrak{d}_{i}}\in D$. We know that $b\cdot K_{\mathfrak{d}_{i}}^{\times
2}=g_{i}\cdot K_{\mathfrak{d}_{i}}^{\times 2}$ and
$(g_{i},h_{i})_{\mathfrak{d}_{i}}=-1$. Therefore we have

$(-1)^{v_{i}}=-1=(h_{i},b)_{\mathfrak{d}_{i}}=\prod_{j\leq
k}(h_{i},\kappa_{j})^{\xi_{j}}_{\mathfrak{d}_{i}}=\prod_{j\leq
k}(-1)^{\xi_{j}b_{ij}}.$

This means that the following $\mathbb{F}_{2}$-linear condition holds

$\sum_{j\leq k}\xi_{j}b_{ij}=v_{i}.$

On the other hand, if $\xi_{1},\dotsc,\xi_{k}\in\\{0,1\\}$ satisfy the
following equality, and $b=\kappa_{1}^{\xi_{1}}\dotsm\kappa_{k}^{\xi_{k}}$,
then $(h_{i},b)=-1$ and so $b$ is not a local square at ${\mathfrak{d}_{i}}$.
Consequently the form $q_{1}\otimes K_{\mathfrak{d}_{i}}$ is isotropic by [15,
Corollary VI.2.15].

Similar arguments apply to the form $q_{2}\otimes K_{\mathfrak{d}_{i}}$. By
assumption, $e({\mathfrak{d}_{i}}/2)$ and $f({\mathfrak{d}_{i}}/2)$ are odd.
Hence [15, Example XI.2.4] asserts that $s(K_{\mathfrak{d}_{i}})=4$ and
consequently $(-1,-1)_{\mathfrak{d}_{i}}=-1$. Using the facts that $b\cdot
K_{\mathfrak{d}_{i}}=g_{i}\cdot K_{\mathfrak{d}_{i}}$ and
$(a,g_{i})_{\mathfrak{d}_{i}}=1$ we may write

$\displaystyle s_{\mathfrak{d}_{i}}q_{2}$
$\displaystyle=(-a,-b)_{\mathfrak{d}_{i}}$
$\displaystyle=(-1,-ab)_{\mathfrak{d}_{i}}\cdot(a,b)_{\mathfrak{d}_{i}}$
$\displaystyle=(-1,-\det
q_{2})_{\mathfrak{d}_{i}}\cdot(a,g_{i})_{\mathfrak{d}_{i}}$
$\displaystyle=(-1,-\det q_{2}).$

Therefore, $q_{2}\otimes K_{\mathfrak{d}_{i}}$ is isotropic by [15,
Proposition V.3.22]. Repeating the same calculations that we did in Eq. (1) we
see that

$\sum_{j\leq k}\xi_{j}c_{ij}=w_{i}.$

Conversely, if the above equality holds for some
$\xi_{1},\dotsc,\xi_{k}\in\\{0,1\\}$ and
$b=\kappa_{1}^{\xi_{1}}\dotsm\kappa_{k}^{\xi_{k}}$, then
$s_{\mathfrak{d}_{i}}q_{2}=(-1,-\det q_{2})_{\mathfrak{d}_{i}}$ and so
$q_{2}\otimes K_{\mathfrak{d}_{i}}$ is isotropic, as desired.

Finally, take the prime $\mathfrak{q}$. The form $q_{1}\otimes
K_{\mathfrak{q}}$ is isotropic because it contains three units and
$\mathfrak{q}$ is non-dyadic. On the other hand, using Hilbert Reciprocity Law
in the same way as we did in the proof of correctness of Algorithm 3, we show
that $q_{2}\otimes K_{\mathfrak{q}}$ is isotropic, as well.

All in all, we proved that the coordinates of the square class of $b$ in the
group $\mathbb{E}_{S\cup\\{\mathfrak{q}\\}}$, treated as a linear space over
$\mathbb{F}_{2}$, form a solution to ($\spadesuit$ ‣ 6g). And that for every
such solution the corresponding forms $q_{1}$ and $q_{2}$ are locally
isotropic at every completion of $K$, hence the local-global principle says
that they are isotropic over $K$. This implies that the algorithm terminates.
The rest of the proof is fully analogous to the previous one, where we dealt
with sums of three squares. ∎

######  Remark 1.

In Algorithms 3 and 4 we keep enlarging the set $S$ until we happen to come
across a prime $\mathfrak{q}$ from a certain positive-density set. There are
two basic strategies to do it, with different worst-case scenarios and average
behaviors. One method is to select primes of $K$ at random. The probability
that we will find a “good” prime $\mathfrak{q}$ is given by the density of
this set, and so is positive. However, in the worst case scenario, we may
hypothetically keep picking the primes that are constantly not in this set.
Hence, while this method is better-behaved in practice, it does not warrant
that the algorithms terminate in finite time. In theory it may take
indefinitely long to find a correct prime $\mathfrak{q}$. The other approach
uses a purely deterministic exhaustive search. Start with a (rational) prime
$p=3$ and try all the primes of $K$ extending $p$. Then go to the next prime
number, take the primes of $K$ extending it, and so on. Although slower in
practice, this procedure guaranties that the two algorithms terminate in
finite time.

######  Remark 2.

In step (4) of Algorithm 4, for every ${\mathfrak{d}_{i}}\in D$ with
$i\leq|D|$, we need to find elements $g_{i},h_{i}\in K$ such that

$(a,g_{i})_{\mathfrak{d}_{i}}=1\qquad\text{and}\qquad(g_{i},h_{i})_{\mathfrak{d}_{i}}=-1.$

The element $g_{i}$ is not used again in the algorithm. It is needed only to
find $h_{i}$ and can be discarded afterwards. (On the other hand it is very
important in the proof of correctness of the algorithm.)

Here again, a fast and efficient method is a probabilistic one. Let
$n_{i}:=(K_{\mathfrak{d}_{i}}:\mathbb{Q}_{2})$. If $a\notin
K_{\mathfrak{d}_{i}}^{\times 2}$ then the probability that a random element
$g_{i}\in K$ satisfies the first of the above two conditions is
$\nicefrac{{1}}{{2}}$. However, the probability that $g_{i}$ itself is a local
square (and so the second condition can never be satisfied) is $2^{-n_{i}-2}$.
Thus, the probability that we pick a “good” $g_{i}$ at random equals
$(\nicefrac{{1}}{{2}}-\nicefrac{{1}}{{2^{n_{i}+2}}})\geq\nicefrac{{1}}{{4}}$.
On the other hand, if $a\in K_{\mathfrak{d}_{i}}^{\times 2}$, then _every_
$g_{i}\in K$ satisfies the first condition. Nevertheless, $g_{i}$ still cannot
be a local square at ${\mathfrak{d}_{i}}$, for the second condition to be
satisfiable. Therefore, the probability of finding a “good” $g_{i}$ at random
is $(1-\nicefrac{{1}}{{2^{n_{i}+2}}})\geq\nicefrac{{3}}{{4}}$. In both cases,
once we selected $g_{i}$, an element $h_{i}$ satisfying the second condition
can be found by a random search with probability $\nicefrac{{1}}{{2}}$.

##  3\. Conclusion

In this paper we present an algorithmic method for finding a representation of
an element of a global field as a sum of squares with the minimal number of
summands. The described algorithms have been implemented by the author using
computer algebra system Magma [1] as a part of package CQF [12].

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