# Power efficiency of Hall-like devices: comparison between reciprocal and
anti-reciprocal Onsager relations

Jean-Eric Wegrowe<EMAIL_ADDRESS>LSI, École
Polytechnique, CEA/DRF/IRAMIS, CNRS, Institut Polytechnique de Paris, 91120
Palaiseau, France  Luqian Zhou  LSI, École Polytechnique, CEA/DRF/IRAMIS,
CNRS, Institut Polytechnique de Paris, 91120 Palaiseau, France  Sariah Al
Saati  Centre de Physique Théorique, 91120 Palaiseau, France

###### Abstract

Two well-known Hall-like effects are occurring in ferromagnets: the Anomalous
Hall effect and the Planar Hall effect. The former is analogous to the
classical Hall effect and is defined by the Onsager reciprocity relation of
the second kind (antisymmetric conductivity matrix), while the latter is
defined by the Onsager reciprocity relation of the first kind (symmetric
conductivity matrix). The difference is fundamental, as it is based on time-
invariance symmetry breaking at the microscopic scale. We study the Hall
current generated in both cases, together with the power that can be extracted
from the edges of Hall device. The expressions of the distribution of the
electric currents, the distribution of electric carriers, and the power
efficiencies (i.e. the power that can be injected into a load circuit) are
derived at stationary regime from a variational method based on the second law
of thermodynamics. It is shown that the distribution of the transverse Hall-
current is identical in both cases but the longitudinal current and the power
dissipated differ at the second order in the Hall angle.

##  I Introduction

Two hundred years ago exactly, in 1824, Sadi Carnot published a seminal work
that contained the first formulation of the second law of thermodynamics
Carnot . Independently, fifty five years letter, in 1879, Edwin Hall published
a report about the effect that bears his name Hall . And fifty two years
later, in 1931, Lars Onsager published two papers Onsager1 ; Onsager2  on the
reciprocity relations that describe the cross-effects occurring in transport
phenomena. This breakthrough was crowned by a Nobel prize in 1968
Nobel_Lecture . More recently, and related to the three discoveries recalled
above, a series of fascinating studies have been devoted to the anomalous Hall
effect (AHE), and its link with topological materials. This series started
with a paper of Karplus and Luttinger in 1954 Luttinger , ending fifty years
latter Kondo ; Nozieres ; Berger ; Bruno  with a paper of Haldane Haldane  and
a Noble prize in 2016. This subject is still very productive todayAHE ,
especially within the context of new topological materials exhibiting
unconventional AHE Altermag ; Mazin .

In a first paper, Onsager derived the reciprocity relation of the first kind
(called reciprocal) from the microscopic time-reversal symmetry Onsager1 . The
second paper deals with the reciprocity relation of the second kind (called
antireciprocal) derived from the partial breaking of this time-reversal
symmetry, typically generated by an axial vectorOnsager2 . The first
reciprocity relation leads to a symmetric matrix of the transport
coefficients, while the second reciprocity relation leads to an antisymmetric
transport matrix. Because of his simplicity, seniority, and ubiquity Dyakonov
; Onoda , Hall-like effects can be considered as an archetype of
antireciprocal Onsager relations, applied to electric transport. The Hall-like
effects are characterized by an antisymmetric conductivity matrix, for which
the off-diagonal coefficients represent the coupling between the two
directions $x$ and $y$ of the plane of the Hall device. In a ferromagnet (or
unconventional magnets), two well-known Hall-like effects are observed: the
Anomalous Hall effect (AHE) and the Planar Hall effect (PHE) PHE_Review ;
Schuhl . In contrast to AHE, the PHE is due to the anisotropic
magnetoresistance, and is defined by the reciprocal Onsager relation
(symmetric conductivity matrix). Like for AHE, the lateral Hall-current
produced by PHE can be used in order to switch an adjacent ferromagnetic
layerKrivorotov  (an effect called spin-orbit torque, or SOT SOT ). In the
present report, we study the electric power that can be injected into a load
circuit through the Hall-current. We used the PHE as an archetype of the
reciprocal Onsager symmetry relation, while the AHE is used as an archetype of
the reciprocal relation.

In the context of the theoretical studies about dissipation in the AHE, the
discussion focused on the microscopic mechanisms responsible for the effect
AHE . Yet, at the macroscopic scale, dissipative transport phenomena do not
follow the microscopic rules, since there are first governed by the second law
of thermodynamics, for which the global constraints (the position of the power
generators, the configuration of the contacts, the topology of the device,
etc) play a major role Benda .

The role of the second law of thermodynamics in the Hall effect has been
overlooked Popovic ; Putley  because, as pointed-out in reference Ref. Moelter
, the “voltage differences between suitable pairs of points in the current
flow suffice to characterize the transport processes”. However, such a
characterization is incomplete because the Hall bar is not analogous to a
capacitor. Indeed, a specificity of the Hall effect is that the accumulation
of the electric carriers at the edges are produced by the system itself, in
order to counter the action of the magnetic field on the electric carriers
(Lenz law). The values of both the density and the current at the edges cannot
be taken as independent local boundary conditions. An immediate consequence is
that both the density of electric carriers and the current distribution inside
a perfect Hall bar are not homogeneous. More puzzling: Hall-like systems seem
not to be reducible to a classical lumped-element circuit obeying the
Kirchhoff’s laws Gyrator . Indeed, if it were the case, we would have the same
lumped-element circuit for both PHE and AHE (both effects are described by the
same set of parameters: resistances $R$ and $R_{\ell}$, Hall angle $\Theta$,
and the current $J_{x}^{0}$ injected by the generator, as sketched in Fig.1),
which is in contradiction with the fact that the stationary states are
different in both systems (as shown below).

In previous works, a variational method based on the second law of
thermodynamics Benda ; JAP1 ; JAP2 ; JAP3 ; SHE  has been developed in order
to calculate the power dissipated by the Hall current in a perfect Hall bar in
contact with a lead circuit JAP3 . Due to the small value of the Hall angle
(below 1%), the properties of the dissipated power are found to be
surprisingly close to that expected for of a usual lumped-element circuit. The
validity of the results have however been confirmed by an experimental study
performed on the AHE in GdCo ferrimagnets ArXiv_2024 .

These results raised the question about the difference of stationary states
generated between reciprocal vs. antireciprocal Onsager relations. The goal of
the present work is to answer this question for AHE vs. PHE devices, and to
present a synoptic calculation based on the stationary variational method, all
other parameters being equivalent. It is shown that the transverse Hall-
current is identical in both cases, while the difference for the power is of
the second order with respect to the Hall angle. These results confirm the
observation that the Joule dissipation is surprisingly similar - but not equal
- for both AHE and PHE. In particular, both effects find the maximum of the
efficiency for the resistance matching condition. The synoptic presentation
proposed below allows the differences and similarities to be better
understood.

##  II Model

Figure 1: Left: sketch of the Hall bar contacted to a load circuit (resistance
$R_{\ell}$) that preserves the translation invariance along $x$. The sign of
the electric-charge accumulation is indicated by red (-) and blue (+) colors
(see the calculated profile in Fig.3). Right: Integration loop ABCDA defined
on one of the $yz$ vertical planes. A magnetic vector $\vec{m}$ is present but
it is not represented in the picture (see definition of the Hall angles in the
text).

The system is a thin homogeneous magnetized conducting layer of length $L$ and
width $\ell$ contacted to an electric generator. The material used can be a
conducting ferromagnet, or antiferromagnet, or altermagnet or any kind of
materials that can be described with an effective magnetic field. We assume
that the conducting layer is planar, invariant by translation along the
$x$-axis $\ell\ll L$ (this excludes the region in contact with the power
generator), and the two lateral edges are symmetric.

We define the distribution of electric charge carriers by $n(y)=n_{0}+\delta
n(y)$, where $\delta n(y)$ is the charge accumulation and $n_{0}$ the
homogeneous density in the electrically neutral system. Due to the symmetry of
the device and the global charge conservation we have
$\int_{-\ell}^{+\ell}\delta n\differential{y}=0$, and the total charge carrier
density is constant $n_{\text{tot}}=\frac{1}{2\ell}\int n\differential{y}$.
For the sake of simplicity, we assume a global charge neutrality so that
$n_{\text{tot}}=n_{0}$. On the other hand, the external generator injects a
current along the $x$ direction, so that the global current flowing in the $x$
direction throughout the device is also constant along $x$, by definition of
the galvanostatic condition. This is expressed as
$\int_{-\ell}^{\ell}J_{x}(y)\differential{y}=2\ell J_{x}^{0}$.

Furthermore, in Section V, we will consider the case where a Hall-current is
injected into the load circuit contacted at the lateral edges. We then have to
introduce the Joule dissipation due to the load resistance. Unlike for the $x$
direction, there is no external generator in the $y$ direction, so that the
loop integral of current in the $\\{yz\\}$ plan has to be $0$ :
$\oint_{ABCDA}J\differential{l}=0$, where the integration path $ABCD$ is
sketched in Fig.1b. This can be written equivalently as
$\int_{AB}J\differential{l}+\int_{BCDA}J\differential{l}=\int_{-\ell}^{\ell}J_{y}(y)\differential{y}+\int_{BCDA}J\differential{l}=0$.
The definite integral $\int_{BCDA}J\differential{l}$ is a constant that we do
not know for the moment, and we will represent the constant as
$\int_{BCDA}J\differential{l}=-2lJ_{y}^{0}$, where $J_{y}^{0}$ is determined
in Section V. In summary, we can write the global constraint on the $y$
direction as $\int_{-\ell}^{\ell}J_{y}(y)\differential{y}=2lJ_{y}^{0}$.

##  III Joule dissipation including screening

The charge accumulation is governed by the Poisson’s equation
$\nabla^{2}V=-\frac{q}{\varepsilon}\delta n$, where $V$ is the electrostatic
potential, $q$ is the electric charge, and $\varepsilon$ is the electric
permittivity. The local electrochemical potential $\mu(x,y)$ \- that takes
into account not only the electrostatic potential $V$ but also the energy (or
the entropy) responsible for the diffusion - is given by the expression
DeGroot ; Rubi  (local equilibrium is assumed everywhere):

$\mu=\frac{kT}{q}\ln\left(\frac{n}{n_{0}}\right)+V,$ (1)

where $k$ is the Boltzmann constant and the temperature $T$ is the temperature
of the heat bath in the case of a non-degenerate semiconductor, or the Fermi
temperature $T_{F}$ in the case of a fully degenerate conductor. Poisson’s
equation now reads

$\nabla^{2}\mu-\lambda_{D}^{2}\frac{q}{\varepsilon}n_{0}\nabla^{2}\ln\left(\frac{n}{n_{0}}\right)+\frac{q}{\varepsilon}\delta
n=0,$ (2)

where $\lambda_{D}=\sqrt{\frac{kT\varepsilon}{q^{2}n_{0}}}$ is the Debye-Fermi
length ($T$ is the temperature and $k$ the Boltzmann constant). On the other
hand, the transport equation under a magnetic field is given by Ohm’s law:

$\vec{J}=-\hat{\sigma}\vec{\nabla}\mu=-qn\hat{\eta}\vec{\nabla}\mu,$ (3)

where the transport coefficients are the conductivity tensor $\hat{\sigma}$ or
the mobility tensor $\hat{\eta}$.

We define the resistivity of the isotropic material $\rho$, the anisotropic
magnetoresistance $\Delta\rho/\rho$, and the Hall resistivity $\rho_{H}$.
Based on the symmetry property of the magnetic system and taking into account
the Onsager reciprocity relations, the resistivity matrix takes the following
form Seitz ; Matrix :

$\displaystyle\hat{\rho}=\left(\begin{array}[]{ccc}\rho+\Delta\rho
m_{x}^{2}&\Delta\rho m_{x}m_{y}-\rho_{H}m_{z}&\Delta\rho
m_{x}m_{z}+\rho_{H}m_{y}\\\ \Delta\rho
m_{y}m_{x}+\rho_{H}m_{z}&\rho+\Delta\rho m_{y}^{2}&\Delta\rho
m_{y}m_{z}-\rho_{H}m_{x}\\\ \Delta\rho m_{z}m_{x}-\rho_{H}m_{y}&\Delta\rho
m_{z}m_{y}+\rho_{H}m_{x}&\rho+\Delta\rho m_{z}^{2}\end{array}\right)$ (7)

where the unit vector $\vec{m}$ gives the direction of the magnetization with
respect to the orthogonal basis attached to the Hall-bar. The unit vector
$\vec{m}$ can be described as a function of the radial angle $\theta$ and the
orthoradial angle $\varphi$, such that
$m_{x}=sin(\theta)cos(\varphi),\,m_{y}=sin(\theta)sin(\varphi),\,m_{z}=cos\theta$.
In order to compare the Planar Hall effect with the Anomalous Hall effect, we
choose the material of the Hall bar so that $\rho\gg\Delta\rho\gg\rho_{H}$ for
the planar Hall effect (this is typically the case for $NiFe$ alloys Madon )
and $\rho\gg\rho_{H}\gg\Delta\rho$ for the anomalous Hall effect (this is
typically the case for $GdCo$ alloys Madon ; ArXiv_2024 ). In the following,
the index $pl$ stands for PHE (left) and the index $an$ stands for AHE
(right). In addition, we only consider the behavior of the system on the
$xy$-plane (translation invariance along $z$ axis) thus the matrix Eq.(7) can
be reduced to a $2\times 2$ matrix:

$\displaystyle\begin{aligned}
\hat{\rho}_{pl}=\left(\begin{array}[]{cc}\rho&\Delta\rho\,m_{x}m_{y}\\\
\Delta\rho\,m_{y}m_{x}&\rho\\\ \end{array}\right)\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
\hat{\rho}_{an}=\left(\begin{array}[]{cc}\rho&-\rho_{H}\,m_{z}\\\
\rho_{H}\,m_{z}&\rho\\\ \end{array}\right)\end{aligned}$ (8)

We then define the ratio of the off-diagonal coefficient over the diagonal
coefficient:

$\displaystyle\begin{aligned}
\Theta_{pl}=\frac{\Delta\rho}{\rho}\,sin^{2}(\theta)\,sin(2\varphi)\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
\Theta_{an}=-\frac{\rho_{H}}{\rho}\,cos(\theta)\end{aligned}$ (9)

The angular dependence of Eqs.(9) appears directly in the Hall voltage
(through the charge accumulation at the edges), and is an unambiguous
signature that allows the AHE to be distinguished from PHE Madon ; ArXiv_2024
. However, in our context (theory and experiment), we will choose the
magnetization direction such that $\theta=\pi/2$ and $\varphi=-\pi/4$ for the
PHE and $\theta=0$ for AHE. If possible, we will also choose two materials
such that $\Delta\rho/\rho\approx\rho_{H}/\rho$ in order to have
$\Theta=\Theta_{an}=-\Theta_{pl}$. Furthermore, since $\Delta\rho/\rho$ and
$\rho_{H}/\rho$ are of the order of 1% maximum, we have
$sin(\Theta)\approx\Theta$ and we call $\Theta$ the “Hall angle”.

Then the conductivity matrix follows as the inverse of resistivity matrix
$\hat{\sigma}=\hat{\rho}^{-1}$ and the mobility tensor can be derived from
$\hat{\sigma}=qn\hat{\eta}$.

$\displaystyle\begin{aligned} \hat{\eta}_{pl}=\begin{pmatrix}\eta&\eta_{pl}\\\
\eta_{pl}&\eta\end{pmatrix}=\eta\begin{pmatrix}1&-\Theta\\\
-\Theta&1\end{pmatrix}\end{aligned}\qquad\text{ and }\qquad\begin{aligned}
\hat{\eta}_{an}=\begin{pmatrix}\eta&\eta_{an}\\\
-\eta_{an}&\eta\end{pmatrix}=\eta\begin{pmatrix}1&\Theta\\\
-\Theta&1\end{pmatrix}\end{aligned}$ (10)

In the following, we want to express the joule dissipation density in terms of
the current vector components $J_{x}$, $J_{y}$. The Joule dissipation power
density is formally given by

$\displaystyle p_{J}=\vec{J}\cdot\vec{\nabla}\mu$ (11)

Using Ohm’s law expressed in Eq.(3), the gradients of the chemical potentials
$\vec{\nabla}\mu$ read

$\displaystyle\begin{aligned}
\vec{\nabla}\mu^{pl}&=-\dfrac{1}{qn}\hat{\eta}^{-1}_{pl}\vec{J}\\\
&=-\dfrac{1}{qn\eta(1-\Theta^{2})}\begin{pmatrix}1&\Theta\\\
\Theta&1\end{pmatrix}\begin{pmatrix}J_{x}\\\ J_{y}\end{pmatrix}\\\
\vec{\nabla}\mu^{pl}&=-\dfrac{n_{0}}{n\sigma_{pl}}\begin{pmatrix}J_{x}+\Theta
J_{y}\\\ \Theta J_{x}+J_{y}\end{pmatrix}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
\vec{\nabla}\mu^{an}&=-\dfrac{1}{qn}\hat{\eta}^{-1}_{an}\vec{J}\\\
&=-\dfrac{1}{qn\eta(1+\Theta^{2})}\begin{pmatrix}1&-\Theta\\\
\Theta&1\end{pmatrix}\begin{pmatrix}J_{x}\\\ J_{y}\end{pmatrix}\\\
\vec{\nabla}\mu^{an}&=-\dfrac{n_{0}}{n\sigma_{an}}\begin{pmatrix}J_{x}-\Theta
J_{y}\\\ \Theta J_{x}+J_{y}\end{pmatrix}\end{aligned}$ (12)

where, for convenience, we have introduced the two constant conductivities
(i.e; the bulk conductivities for $y\gg\lambda_{D}$),:

$\displaystyle\begin{aligned}
\sigma_{pl}&=qn_{0}\eta(1-\Theta^{2})\end{aligned}\qquad\text{ and
}\qquad\begin{aligned} \sigma_{an}&=qn_{0}\eta(1+\Theta^{2})\end{aligned}$
(13)

Note that since $\rho$ is the intrinsic value of the resistivity of the
material Madon ; ArXiv_2024  (that does not depend on $\Theta$) and assuming
identical material for both AHE and PHE, we have
$\sigma_{pl}=\frac{1}{\rho(1-\Theta^{2})}$ and
$\sigma_{ah}=\frac{1}{\rho(1+\Theta^{2})}$. This gives us

$\displaystyle\begin{aligned}
p_{J}^{pl}&=-\dfrac{n_{0}}{n\sigma_{pl}}\left(J_{x}^{2}+J_{y}^{2}+2\Theta
J_{x}J_{y}\right)\end{aligned}\qquad\text{ and }\qquad\begin{aligned}
p_{J}^{an}&=-\dfrac{n_{0}}{n\sigma_{an}}\left(J_{x}^{2}+J_{y}^{2}\right)\end{aligned}$
(14)

The expression of the Joule power dissipated by the system reads:

$\displaystyle P_{J}$
$\displaystyle=S_{\text{lat}}\int_{-\ell}^{\ell}p_{J}(y)\differential{y}$ (15)

which gives respectively

$\displaystyle\begin{aligned}
P_{J}^{pl}&=\frac{S_{\text{lat}}}{\sigma_{pl}}\int_{-\ell}^{\ell}\frac{n_{0}}{n}\left(J_{x}^{2}+J_{y}^{2}+2\Theta
J_{x}J_{y}\right)\differential{y}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
P_{J}^{an}&=\frac{S_{\text{lat}}}{\sigma_{pl}}\int_{-\ell}^{\ell}\frac{n_{0}}{n}\left(J_{x}^{2}+J_{y}^{2}\right)\differential{y}\end{aligned}$
(16)

where $S_{\text{lat}}$ is the lateral surface of the Hall bar (product of the
length $L$ by the thickness), and $2\ell$ is the width. As expected the two
expressions of the Joule heating for PHE and AHE are significantly different.

##  IV Currents and charge densities without load circuit

The stationary state is defined by the least dissipation principle that states
that the current distributes itself so as to minimize Joule heating $P_{J}$
compatible with the constraints Onsager_Diss ; Bruers ; MinDiss . Without load
circuit, in the notation of Section III, the integration path ABCD is reduced
to AB and the global constraint along the $y$ direction is zero:
$\int_{-\ell}^{\ell}J_{y}(y)\differential{y}=0$.

The global constraints read:

$\int_{-\ell}^{\ell}n(y)\differential{y}=2\ell
n_{0}\quad\textrm{and}\quad\int_{-\ell}^{\ell}J_{x}(y)\differential{y}=2\ell
J_{x}^{0}\quad\textrm{and}\quad\int_{-\ell}^{\ell}J_{y}(y)\differential{y}=0$
(17)

We define for convenience the reduced power
$\tilde{P_{J}}=\frac{q\eta(1\pm\Theta^{2})}{S_{\text{lat}}}\,P_{J}$. Let us
introduce the two Lagrange multiplayers $\lambda_{x}$ and $\lambda_{n}$
corresponding to the two constraints Eqs.(17). The functional to be minimized
then reads:

$\displaystyle\begin{aligned}
\tilde{\mathcal{P}}_{J}^{pl}=\int_{-\ell}^{\ell}\left(\frac{J_{x}^{2}+J_{y}^{2}+2\Theta
J_{x}J_{y}}{n}-\lambda_{x}\,J_{x}-\lambda_{y}\,J_{y}-\lambda_{n}\,n\right)\differential{y}\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
\tilde{\mathcal{P}}_{J}^{an}=\int_{-\ell}^{\ell}\left(\frac{J_{x}^{2}+J_{y}^{2}}{n}-\lambda_{y}\,J_{y}-\lambda_{x}\,J_{x}-\lambda_{n}\,n\right)\differential{y}\end{aligned}$
(18)

The minimum corresponds to

$\displaystyle\frac{\boldsymbol{\delta}\tilde{\mathcal{P}}_{J}}{\boldsymbol{\delta}J_{x}}=0\quad\textrm{and}\quad\frac{\boldsymbol{\delta}\tilde{\mathcal{P}}_{J}}{\boldsymbol{\delta}J_{y}}=0\quad\textrm{and}\quad\frac{\boldsymbol{\delta}\tilde{\mathcal{P}}_{J}}{\boldsymbol{\delta}(n)}=0$
(19)

which gives respectively

$\displaystyle\begin{aligned} \begin{cases}2J_{x}+2\Theta
J_{y}=n\lambda_{x}\\\ 2J_{y}+2\Theta J_{x}=n\lambda_{y}\\\
J_{x}^{2}+J_{y}^{2}+2\Theta
J_{x}J_{y}=-\lambda_{n}n^{2}\end{cases}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned} \begin{cases}2J_{x}=n\lambda_{x}\\\
2J_{y}=n\lambda_{y}\\\
J_{x}^{2}+J_{y}^{2}=-\lambda_{n}n^{2}\end{cases}\end{aligned}$ (20)

Integrating the two first equations on both sides from $-l$ to $l$, and then
applying the constraints, we can solve $\lambda_{x}$ and $\lambda_{y}$
explicitly. Thus, we can obtain $J_{x}$ and $J_{y}$ by re-inserting this
result into the equations above. The solution then reads respectively:

$\displaystyle\begin{aligned}
\begin{cases}J_{x}^{pl}(y)=J_{x}^{0}\dfrac{n(y)}{n_{0}}\\\
J_{y}^{pl}(y)=0\end{cases}\end{aligned}\qquad\text{ and }\qquad\begin{aligned}
\begin{cases}J_{x}^{an}(y)=J_{x}^{0}\dfrac{n(y)}{n_{0}}\\\
J_{y}^{an}(y)=0\end{cases}\end{aligned}$ (21)

As can be seen, the density of current is not homogeneous throughout the
sample (since $n(y)\neq n_{0}$), and it has the same form for both AHE and
PHE.

Inserting the solution (21) into the transport equations Eq.(12), we deduce

$\displaystyle\begin{aligned}
\begin{cases}\partial_{x}\mu^{pl}(y)=-\frac{J_{x}^{0}}{\sigma_{pl}}\\\
\partial_{y}\mu^{pl}(y)=-\frac{\Theta
J_{x}^{0}}{\sigma_{pl}}\end{cases}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
\begin{cases}\partial_{x}\mu^{an}(y)=-\frac{J_{x}^{0}}{\sigma_{an}}\\\
\partial_{y}\mu^{an}(y)=-\frac{\Theta
J_{x}^{0}}{\sigma_{an}}\end{cases}\end{aligned}$ (22)

As a consequence, the electrochemical potential of the stationary state is
harmonic in both cases: $\nabla^{2}\mu=0$. Since the profile of the lateral
current $J_{y}(y)$ is defined by the charge density $n(y)$, the Poisson’s
equation Eq.(2) for $\nabla^{2}\mu=0$ gives the equation:

$\lambda_{D}^{2}\partial_{y}^{2}\ln\left(1+\frac{\delta
n}{n_{0}}\right)=\frac{\delta n}{n_{0}}.$ (23)

Assuming $\delta n\ll n_{0}$, we have, at the first order:

$\lambda_{D}^{2}\partial_{y}^{2}\left(\frac{\delta
n}{n_{0}}\right)=\frac{\delta n}{n_{0}}.$ (24)

which gives us

$\frac{\delta
n(y)}{n_{0}}=A\operatorname{ch}(y/\lambda_{D})+B\operatorname{sh}(y/\lambda_{D})$
(25)

Once again, the boundary conditions for the density $n$ are not defined
locally but globally by Eq.(17). We can specify the boundary conditions for
$\partial_{y}\frac{\delta n(y)}{n_{0}}$ noting that for the stationary
solutions of Eq.(21) gives us for both systems

$\displaystyle E_{y}$ $\displaystyle=-\partial_{y}V$ (26)
$\displaystyle=-\partial_{y}\mu+\dfrac{kT}{q}\partial_{y}\left(\frac{\delta
n}{n_{0}}\right)$ (27) $\displaystyle E_{y}$ $\displaystyle=-\dfrac{\Theta
J_{x}^{0}}{\sigma_{h}}+\dfrac{kT}{q}\partial_{y}\left(\frac{\delta
n}{n_{0}}\right)$ (28)

where $\sigma_{h}$ is given by $\sigma_{pl}$ or $\sigma_{an}$.

Specifying these conditions at the edges of the sample for
$E_{y}(\pm\ell)=E_{y}(\pm\infty)$, which correspond to the electric field at
the exterior of the bar, assuming that the bar is surrounded by vacuum. If we
define $2\Delta^{\pm}E_{\infty}=E_{y}(\infty)\pm E_{y}(-\infty)$, then the
approximated solution of $\delta n$ is given for both cases by:

$\displaystyle\dfrac{q\lambda_{D}}{\varepsilon}\delta n(y)$
$\displaystyle=\Delta^{-}E_{\infty}\dfrac{\operatorname{ch}(y/\lambda_{D})}{\operatorname{sh}(l/\lambda_{D})}+\left(\Delta^{+}E_{\infty}+\dfrac{\Theta
J_{x}^{0}}{\sigma_{h}}\right)\dfrac{\operatorname{sh}(y/\lambda_{D})}{\operatorname{ch}(l/\lambda_{D})}$
(29)

where $\sigma_{h}$ is here again given by $\sigma_{pl}$ or $\sigma_{an}$.

As can be seen, the the profiles of the screening are similar for both AHE and
PHE in the perfect Hall-like device. A difference can be seen (at the second
order in $\Theta$) in terms of the conductivities $\sigma_{pl}$ and
$\sigma_{an}$ given in Eq.(13).

##  V Current Injection in the load circuit

The solution found in the preceding section is valid as long as the
dissipation due to charge leakage at the edges is negligible with respect to
the dissipation inside the device. However, if it is no longer the case, the
stationary regime should be reconsidered by introducing the dissipation due to
the resistance of a lateral load circuit that connects the edges of the Hall
bar. In order to take into account this supplementary dissipation, we
introduce the load resistivity $R_{\ell}$ ($\Omega\cdot\mathrm{m}^{2}$) of the
lateral circuit. The power dissipated in the lateral circuit is:

$P_{\text{lat}}\ =\ \frac{S_{\text{lat}}\,\Delta\mu^{2}}{R_{\ell}}$

where $\Delta\mu=\mu(+\ell)-\mu(-\ell)$ is the difference of the chemical
potential between both edges. Note that due to our hypothesis of the
invariance along $x$, we do not treat the case of a resistance contacted with
a usual contact to the two edges of the Hall bar. Indeed, such a contact would
break the translation invariance symmetry along $x$, and would distort the
current lines in a specific manner that depends on the details of the contact
geometry and resistivity. Such a contact-specific effect is not related to the
generic problem studied here. The relevant contact, as sketched in Fig. 1
(left), is an idealization for the “perfect Hall bar”, but it can be
approximated experimentallyArXiv_2024 ; Madon . Incidentally, it is well known
that the main advantage of the Corbino disk with respect to Hall-bar devices
is precisely that it is much easier to design two quasi-perfect concentric
equipotentials (circular symmetry) instead of two quasi-perfect longitudinal
equipotentials (translational symmetry).

###  V.1 Stationary Currents

The difference of chemical potential can be expressed as a function of the
current:

$\displaystyle\begin{aligned} \Delta\mu^{pl}\ &=\
\int_{-\ell}^{+\ell}\differential{y}\,\partial_{y}\mu\ \\\ &=\
-\int_{-\ell}^{+\ell}\differential{y}\,\frac{J_{y}+\Theta
J_{x}}{qn\eta(1-\Theta^{2})}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned} \Delta\mu^{an}\ &=\
\int_{-\ell}^{+\ell}\differential{y}\,\partial_{y}\mu\ \\\ &=\
-\int_{-\ell}^{+\ell}\differential{y}\,\frac{J_{y}+\Theta
J_{x}}{qn\eta(1+\Theta^{2})}\end{aligned}$ (30)

so that

$\displaystyle\begin{aligned} P_{\text{lat}}^{pl}\ =\
\frac{S_{\text{lat}}}{R_{\ell}(q\eta)^{2}(1-\Theta^{2})^{2}}\left(\int_{-\ell}^{+\ell}\differential
y\frac{J_{y}+\Theta J_{x}}{n}\right)^{2}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned} P_{\text{lat}}^{an}\ =\
\frac{S_{\text{lat}}}{R_{\ell}(q\eta)^{2}(1+\Theta^{2})^{2}}\left(\int_{-\ell}^{+\ell}\differential
y\frac{J_{y}+\Theta J_{x}}{n}\right)^{2}\end{aligned}$ (31)

As in the preceding section, we define the reduced power
$\tilde{P}=\frac{q\eta(1\pm\Theta^{2})}{S_{\text{lat}}}P$. The total power
dissipated is then:

$\displaystyle\begin{aligned} \tilde{P}^{pl}\ &=\
\tilde{P}_{J}+\tilde{P}_{\text{lat}}\\\ &=\ \int_{-\ell}^{+\ell}\differential
y\frac{J_{x}^{2}+J_{y}^{2}+2\Theta J_{x}J_{y}}{n}\ +\\\
&\phantom{=}\alpha^{pl}\frac{2l}{n_{0}}\left(\frac{n_{0}}{2\ell}\int_{-\ell}^{+\ell}\differential
y\frac{J_{y}+\Theta J_{x}}{n}\right)^{2}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned} \tilde{P}^{an}\ &=\
\tilde{P}_{J}+\tilde{P}_{\text{lat}}\\\ &=\ \int_{-\ell}^{+\ell}\differential
y\frac{J_{x}^{2}+J_{y}^{2}}{n}\ +\\\
&\phantom{=}\alpha^{an}\frac{2l}{n_{0}}\left(\frac{n_{0}}{2\ell}\int_{-\ell}^{+\ell}\differential
y\frac{J_{y}+\Theta J_{x}}{n}\right)^{2}\end{aligned}$ (32)

where we have introduced the dimensionless control parameter
$\alpha^{h}=\frac{2\ell}{R_{\ell}\sigma_{h}}$. Note that this control
parameter $\alpha$ is defined by the ratio $\alpha=\frac{R}{R_{\ell}}$ of the
resistance of the material
$R\equiv\frac{V}{J_{x}^{0}}=\frac{2\ell}{\sigma_{h}}$ over the load resistance
$R_{\ell}$ ArXiv_2024 . The case $\alpha\rightarrow 0$ corresponds to perfect
Hall bar while the case $\alpha\rightarrow\infty$ correspond to the perfect
Corbino disk Benda ; Madon . The exponent of index $h$ denotes respectively
$h=pl$ for the planar Hall system and $h=an$ for the anomalous Hall system. We
define for convenience the constant $A^{h}$:

$\displaystyle A^{h}\equiv\frac{n_{0}}{2\ell}\int_{\ell}^{\ell}\frac{\Theta
J_{x}+J_{y}}{n}\differential{y}.$ (33)

At this stage, we can adopt the constraint on $J_{y}(y)$ as discussed in the
section above:

$\displaystyle\int_{-\ell}^{\ell}J_{y}(y)\differential{y}=2\ell J_{y}^{0,h},$
(34)

where $J_{y}^{0,h}$ is some constant for each system which will be determined
later.

The minimization of the corresponding functional $\tilde{\mathcal{P}}$ now
reads:

$\displaystyle\begin{aligned} \alpha^{pl}\,A^{pl}\,\Theta+J_{x}+\Theta
J_{y}=\frac{\lambda_{x}n}{2},\\\ \alpha^{pl}\,A^{pl}\,+J_{y}+\Theta
J_{x}=\frac{\lambda_{y}n}{2},\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
\alpha^{an}\,A^{an}\,\Theta+J_{x}=\frac{\lambda_{x}n}{2},\\\
\alpha^{an}\,A^{an}\,+J_{y}=\frac{\lambda_{y}n}{2}.\end{aligned}$ (35)

Once again, applying the global constraints given by Eqs.(17) along with the
third replaced by Eq. (34), we can immediately solve from these two equations

$\displaystyle\begin{aligned}
\lambda_{x}&=\frac{2}{n_{0}}\left(\alpha^{pl}A^{pl}\Theta+J_{x}^{0}+\Theta
J_{y}^{0,pl}\right)\\\
\lambda_{y}&=\frac{2}{n_{0}}\left(\alpha^{pl}A^{pl}+\Theta
J_{x}^{0}+J_{y}^{0,pl}\right)\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
\lambda_{x}&=\frac{2}{n_{0}}\left(\alpha^{an}A^{an}\Theta+J_{x}^{0}\right)\\\
\lambda_{y}&=\frac{2}{n_{0}}\left(\alpha^{an}A^{an}+J_{y}^{0,an}\right)\end{aligned}$
(36)

Then it follows that:

$\displaystyle\begin{aligned} J_{x}&=\frac{n}{n_{0}}J_{x}^{0}\\\
J_{y}&=\frac{n}{n_{0}}J_{y}^{0,pl}+\alpha^{pl}A^{pl}\left(\frac{n}{n_{0}}-1\right)\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
J_{x}&=\frac{n}{n_{0}}J_{x}^{0}+\alpha^{an}A^{an}\Theta\left(\frac{n}{n_{0}}-1\right)\\\
J_{y}&=\frac{n}{n_{0}}J_{y}^{0,an}+\alpha^{an}A^{an}\left(\frac{n}{n_{0}}-1\right)\end{aligned}$
(37)

For the sake of simplicity, let us focus on a system which obeys the local
conservation of electric charges: $\mathbf{\nabla}\cdot\mathbf{J}=0$, which
leads to $\frac{\partial J_{x}}{\partial x}+\frac{\partial J_{y}}{\partial
y}=0$. Since the system is invariant along the $x$ direction, so
$\frac{\partial J_{x}}{\partial x}=0$. Then this will lead to the result that
$\frac{\partial J_{y}}{\partial y}=0$. Thus, this requires that coefficient in
factor of $n(y)$ vanishes, which leads to $J_{y}^{0}=-\alpha A$. We get
eventually:

$\displaystyle\begin{aligned} J_{x}&=\frac{n}{n_{0}}J_{x}^{0}\\\
J_{y}&=J_{y}^{0,pl}\end{aligned}\qquad\text{ and }\qquad\begin{aligned}
J_{x}&=\frac{n}{n_{0}}J_{x}^{0}-J_{y}^{0,an}\Theta\left(\frac{n}{n_{0}}-1\right)\\\
J_{y}&=J_{y}^{0,an}\end{aligned}$ (38)

Re-inserting Eq.(38) into the definition of $A^{pl/an}$ Eq.(33), we obtain the
following equation

$\displaystyle\begin{aligned}
J_{y}^{0,pl}\,\left(1+\alpha\int\frac{n_{0}}{n}\frac{\differential{y}}{2\ell}\right)&=-\alpha\Theta
J_{x}^{0}\end{aligned}\qquad\text{ and }\qquad\begin{aligned}
J_{y}^{0,an}\,\left(1+\alpha\Theta^{2}+\alpha(1-\Theta^{2})\int\frac{n_{0}}{n}\frac{\differential{y}}{2\ell}\right)&=-\alpha\Theta
J_{x}^{0}\end{aligned}$ (39)

Assuming that $\delta n\ll n_{0}$, we have at the first order
$\frac{n_{0}}{2\ell}\int_{-\ell}^{+\ell}\frac{\differential{y}}{n(y)}\simeq
1$, thus we can solve

$\displaystyle\begin{aligned} J_{x}&=\frac{n}{n_{0}}J_{x}^{0}\\\
J_{y}&=-\frac{\alpha}{1+\alpha}\Theta J_{x}^{0}\end{aligned}\qquad\text{ and
}\qquad\begin{aligned}
J_{x}&=\frac{n}{n_{0}}J_{x}^{0}+\Theta^{2}\frac{\alpha}{1+\alpha}\left(\frac{n}{n_{0}}-1\right)J_{x}^{0}\\\
J_{y}&=-\frac{\alpha}{1+\alpha}\Theta J_{x}^{0}\end{aligned}$ (40)

The results Eqs.(40) show that, at the first order in the charge accumulation,
the distributions of the transverse Hall-current $J_{y}$ are identical for the
both the AHE and the PHE, and a difference can be seen at the second order in
the Hall angle $\Theta$ for the longitudinal current $J_{x}$. The well-known
expressions are recovered for the extreme cases, that are the perfect Hall bar
and the Corbino disk. For an ideal Hall bar, the load resistance is infinite,
and this implies $\alpha=0$. We thus see that Eq.(40) converges to Eq.(21) as
$\alpha\to 0$.

Conversely, in a Corbino disk, we have $\alpha\rightarrow+\infty$ and the
solutions converge to the well-known result

$\displaystyle J_{x}^{Cor}=\frac{n}{n_{0}}J_{x}^{0}\quad\text{and}\quad
J_{y}^{Cor}=-\Theta J_{x}^{0}$ (41)

for both PHE and AHE. Once again, we see that the second law of thermodynamics
levels out the fundamental differences that define the two processes - PHE vs.
AHE - at the microscopic scales.

###  V.2 Power Injected in the load circuit

The total power $\tilde{P}=\tilde{P}_{J}+\tilde{P}_{\text{lat}}$ \- given in
Eq.(32) - is the sum of the Joule heating $\tilde{P}_{J}$ dissipated inside
the Hall device, and the power $\tilde{P}_{\text{lat}}$ dissipated into the
lateral passive circuit. Inserting the stationary state Eqs.(40) and using the
first global condition in Eqs.(17) (assuming that $\delta n\ll n_{0}$, we have
$\int_{-\ell}^{+\ell}\frac{\differential{y}}{n(y)}\simeq 2\ell/n_{0}$), the
power dissipated in the lateral circuit reads:

$\displaystyle\begin{aligned} P^{pl}_{\text{lat}}(\alpha)&\simeq\frac{2\ell
S_{\text{lat}}(J_{x}^{0})^{2}}{\sigma_{pl}}\dfrac{\alpha}{(1+\alpha)^{2}}\,\Theta^{2}\end{aligned}\,\,\,\text{
and }\,\,\,\,\,\,\begin{aligned} P^{an}_{\text{lat}}(\alpha)&\simeq\frac{2\ell
S_{\text{lat}}(J_{x}^{0})^{2}}{\sigma_{an}}\dfrac{\alpha}{(1+\alpha)^{2}}\,\Theta^{2}\left(1+2\Theta^{2}\alpha\left(\frac{n}{n_{0}}-1\right)\right)\end{aligned}$
(42)

Figure 2: Power efficiency $P_{lat}/P_{0}$ for both AHE (red lines) and PHE
(blue dashed line). Left: Efficiency as a function of the ratio
$\alpha=R/R_{\ell}$ for values of $\Theta$ ranging from 0 to 0.5 with
increments of 0.05. Right: Efficiency as a function of the Hall angle
$\Theta$.

When we consider the power dissipated in the material in the absence of the
Hall-like effect (this can be performed experimentally by adjusting the
magnetization state according to Eq.(9))

$\displaystyle P_{0}\equiv P_{J}(\Theta=0)=\dfrac{2\ell
S_{\text{lat}}(J_{x}^{0})^{2}}{q\eta n_{0}}$ (43)

This leads to a very simple expression of the power efficiency due to the
injection of the Hall-current:

$\displaystyle\begin{aligned}
\frac{P^{pl}_{\text{lat}}}{P_{0}}&\simeq\dfrac{\alpha}{(1+\alpha)^{2}}\frac{\Theta^{2}}{1-\Theta^{2}}\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
\frac{P^{an}_{\text{lat}}}{P_{0}}&\simeq\dfrac{\alpha}{(1+\alpha)^{2}}\frac{\Theta^{2}}{1+\Theta^{2}}\left(1+2\Theta^{2}\alpha\left(\frac{n}{n_{0}}-1\right)\right)\\\
&\simeq\dfrac{\alpha}{(1+\alpha)^{2}}\frac{\Theta^{2}}{1+\Theta^{2}}\end{aligned}$
(44)

The function $\alpha\mapsto\frac{\alpha}{(1+\alpha)^{2}}$ reaches a maximum at
$\alpha=1$ decreases to $0$ at $\alpha=+\infty$. The two expressions
$P^{an}_{\text{lat}}$ and $P^{pl}_{\text{lat}}$ coincide at the lowest order
in $\Theta$.

Equations (44) shows that power efficiency for both effects is the same as a
function of $\alpha$, but the dependence as a function of $\Theta$ differs.
This behavior can be observed in Fig.2 left panel (as a function of the
variable $\alpha$) and Fig.2 right panel (as a function of the variable
$\Theta$) for AHE (red continuous lines) and PHE (blue dashed lines).
Particularly, both curves have a unique maximum for the resistance matching
$\alpha=1$ (i.e., $R=R_{\ell}$) between the two sub-circuits. In other terms,
the maximum transfer theorem is verified in both Hall-like systems. But this
result does not implies that the Kirchhoff’s circuit law is valid. Indeed,
this is not the case, since both stationary states are different while the
lumped-element circuits for AHE and PHE would be identical (see Fig.1).

Besides, in standard materials, Anomalous-Hall angles and planar-Hall angles
are small (of the order of 1% in usual materials), so that the difference of
power efficiency between AHE and PHE as a function of the Hall angle is small,
but easy to access experimentally (incidentally, the difference of dissipated
power between the Hall-bar and the Corbino disk is also at the second order in
$\Theta_{an}$). Note that if - instead of Eq.(43) - we define the reference
power as $P_{0}^{pl,an}(\alpha=0)=2\ell
S_{latt}\,(J_{x}^{0})^{2}/\sigma_{pl,an}$ (i.e. the power of the perfect Hall
bar for AHE and PHE separately), there is no difference between the two power
efficiencies.

###  V.3 Charge accumulation

It is now necessary to calculate the expression of the charge accumulation
$\delta n(y)$ at the edges of a Hall bar contacted to the load circuit. The
simple result obtained in section III is modified.

Assuming the solutions given by Eqs.(40), we rewrite $\vec{\nabla}\mu$ given
by Eq.(12) as

$\displaystyle\begin{aligned}
&\begin{cases}\partial_{x}\mu^{pl}=-\dfrac{n_{0}}{n\sigma_{pl}}(J_{x}+\Theta
J_{y})\\\ \partial_{y}\mu^{pl}=-\dfrac{n_{0}}{n\sigma_{pl}}(\Theta
J_{x}+J_{y})\end{cases}\\\
&\begin{cases}\partial_{x}\mu^{pl}=-\dfrac{J_{x}^{0}}{\sigma_{pl}}(1-\frac{n_{0}}{n}\frac{\alpha}{1+\alpha}\Theta^{2})\\\
\partial_{y}\mu^{pl}=-\dfrac{J_{x}^{0}\Theta}{\sigma_{pl}}(1-\frac{n_{0}}{n}\frac{\alpha}{1+\alpha})\end{cases}\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
&\begin{cases}\partial_{x}\mu^{an}=-\dfrac{n_{0}}{n\sigma_{an}}(J_{x}-\Theta
J_{y})\\\ \partial_{y}\mu^{pl}=-\dfrac{n_{0}}{n\sigma_{an}}(\Theta
J_{x}+J_{y})\end{cases}\\\
&\begin{cases}\partial_{x}\mu^{an}=-\dfrac{J_{x}^{0}}{\sigma_{an}}(1+\Theta^{2}\frac{\alpha}{1+\alpha})\\\
\partial_{y}\mu^{pl}=-\dfrac{J_{x}^{0}\Theta}{\sigma_{an}}(1+\Theta^{2}\frac{\alpha}{1+\alpha}-\frac{n_{0}}{n}\frac{\alpha}{1+\alpha}\left(1+\Theta^{2}\right))\end{cases}\end{aligned}$
(45)

We get

$\displaystyle\begin{aligned}
\nabla^{2}\mu^{pl}&=-\dfrac{J_{x}^{0}\Theta}{\sigma_{pl}}\dfrac{\alpha}{1+\alpha}\dfrac{\partial_{y}(n/n_{0})}{(n/n_{0})^{2}}\\\
&=-\dfrac{J_{x}^{0}}{qn_{0}\eta}\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}\dfrac{\partial_{y}(n/n_{0})}{(n/n_{0})^{2}}\\\
&=-\dfrac{q\lambda_{D}}{\varepsilon}n_{pl}\dfrac{\partial_{y}(n/n_{0})}{(n/n_{0})^{2}}\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
\nabla^{2}\mu^{an}&=-\dfrac{J_{x}^{0}\Theta(1+\Theta^{2})}{\sigma_{an}}\dfrac{\alpha}{1+\alpha}\dfrac{\partial_{y}(n/n_{0})}{(n/n_{0})^{2}}\\\
&=-\dfrac{J_{x}^{0}}{qn_{0}\eta}\Theta\dfrac{\alpha}{1+\alpha}\dfrac{\partial_{y}(n/n_{0})}{(n/n_{0})^{2}}\\\
&=-\dfrac{q\lambda_{D}}{\varepsilon}n_{an}\dfrac{\partial_{y}(n/n_{0})}{(n/n_{0})^{2}}\end{aligned}$
(46)

The density of carriers reads:

$\displaystyle\begin{aligned}
n_{pl}&=\dfrac{\varepsilon}{q\lambda_{D}}\dfrac{J_{x}^{0}}{qn_{0}\eta}\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}\\\
&=\nu\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}\end{aligned}\qquad\text{
and }\qquad\begin{aligned}
n_{an}&=\dfrac{\varepsilon}{q\lambda_{D}}\dfrac{J_{x}^{0}}{qn_{0}\eta}\Theta\dfrac{\alpha}{1+\alpha}\\\
&=\nu\,\Theta\dfrac{\alpha}{1+\alpha}\end{aligned}$ (47)

where we have set $\nu=\varepsilon\,J_{x}^{0}/(q^{2}\lambda_{D}n_{0}\eta)$. In
this case, we find that the chemical potential is not harmonic. The resulting
formulation of Poisson’s equation Eq.(2) is thus the same for both system and
is given by

$\displaystyle-\lambda_{D}^{2}n_{0}\nabla^{2}\ln\left(1+\frac{\delta
n}{n_{0}}\right)-\lambda_{D}n_{h}\dfrac{\partial_{y}(\delta
n/n_{0})}{(n/n_{0})^{2}}+\delta n=0$ (48)

where we recall that the index $h$ in $n_{h}$ is given by $h=pl$ for the
planar Hall case and by $h=an$ for the anomalous Hall case. We can linearize
the equations in the regime $\delta n/n_{0}\ll 1$ to get

$\displaystyle-\lambda_{D}^{2}\partial_{y}^{2}\left(\frac{\delta
n}{n_{0}}\right)-\frac{n_{h}}{n_{0}}\lambda_{D}\partial_{y}\left(\frac{\delta
n}{n_{0}}\right)+\frac{\delta n}{n_{0}}=0$ (49)

which can be solved exactly to give in both cases

$\displaystyle\begin{aligned} \dfrac{q\lambda_{D}}{\varepsilon}\delta
n(y)&=e^{\frac{r_{1}y}{\lambda_{D}}}\dfrac{E_{+\infty}e^{-\frac{r_{2}\ell}{\lambda_{D}}}-E_{-\infty}e^{\frac{r_{2}\ell}{\lambda_{D}}}+2\operatorname{sh}(\frac{r_{2}\ell}{\lambda_{D}})\frac{J_{x}^{0}\Theta}{\sigma_{pl}(1+\alpha)}}{2r_{1}\operatorname{sh}((r_{1}-r_{2})\frac{\ell}{\lambda_{D}})}+e^{\frac{r_{2}y}{\lambda_{D}}}\dfrac{E_{+\infty}e^{-\frac{r_{1}\ell}{\lambda_{D}}}-E_{-\infty}e^{\frac{r_{1}\ell}{\lambda_{D}}}+2\operatorname{sh}(\frac{r_{1}\ell}{\lambda_{D}})\frac{J_{x}^{0}\Theta}{\sigma_{pl}(1+\alpha)}}{2r_{2}\operatorname{sh}((r_{2}-r_{1})\frac{\ell}{\lambda_{D}})}\end{aligned}$
(50)

with

$\displaystyle\begin{aligned}
r_{1}^{pl}&=\dfrac{1}{2}\left(-\dfrac{\nu}{n_{0}}\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}-\sqrt{\left(\dfrac{\nu}{n_{0}}\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}\right)^{2}+4}\right)\\\
r_{2}^{pl}&=\dfrac{1}{2}\left(-\dfrac{\nu}{n_{0}}\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}+\sqrt{\left(\dfrac{\nu}{n_{0}}\dfrac{\Theta}{1-\Theta^{2}}\dfrac{\alpha}{1+\alpha}\right)^{2}+4}\right)\end{aligned}\text{
and }\begin{aligned}
r_{1}^{an}&=\dfrac{1}{2}\left(-\dfrac{\nu}{n_{0}}\Theta\dfrac{\alpha}{1+\alpha}-\sqrt{\left(\dfrac{\nu}{n_{0}}\Theta\dfrac{\alpha}{1+\alpha}\right)^{2}+4}\right)\\\
r_{2}^{an}&=\dfrac{1}{2}\left(-\dfrac{\nu}{n_{0}}\Theta\dfrac{\alpha}{1+\alpha}+\sqrt{\left(\dfrac{\nu}{n_{0}}\Theta\dfrac{\alpha}{1+\alpha}\right)^{2}+4}\right)\end{aligned}$
(51)

Without other external source of electric fields, we have

$\displaystyle\dfrac{\delta n}{n_{0}}(y)$
$\displaystyle=\dfrac{\nu}{n_{0}}\dfrac{\Theta}{1\pm\Theta^{2}}\dfrac{1}{1+\alpha}\dfrac{1}{\operatorname{sh}((r_{2}-r_{1})\frac{\ell}{\lambda_{D}})}\left(\operatorname{sh}(\frac{r_{1}\ell}{\lambda_{D}})\dfrac{e^{\frac{r_{2}y}{\lambda_{D}}}}{r_{2}}-\operatorname{sh}(\frac{r_{2}\ell}{\lambda_{D}})\dfrac{e^{\frac{r_{1}y}{\lambda_{D}}}}{r_{1}}\right)$
(52)

which, in the limit $\alpha\rightarrow 0$, adequately gives us back the
distribution of Eq.(29). The profiles of the charge accumulations $n(y)$ are
functions of the parameters $\lambda_{D}/\ell$, $\nu$, $\alpha$ and $\Theta$.
Despite the complexity of the expressions Eqs(51) and Eqs.(52), the only
difference between PHE and AHE is due to the difference as a function of
$\Theta$ (see Fig.3). We can see that changing the value of the Hall angle
$\Theta$ will only change the amplitude of the distribution.

Figure 3: The profiles of the charge accumulations $\delta n(y)/\nu$ for both
AHE (blue) and PHA (red) for different values of the Hall angle $\Theta$ at
the maximum $\alpha=1$ (i.e., $R=R_{\ell}$) and for $\nu=0.1n_{0}$. Left: over
the whole sample. Right: close to the right boundary. In both panels, the
angle $\Theta$ range from $0$ to $0.5$ by increments of $0.05$.

##  VI Conclusion

In the context of the study of Hall currents injected into a lateral load
circuit (Fig.1), we have compared the stationary states of the planar Hall
effect (PHE) with that of the anomalous Hall effect (AHE) - all other
parameters being equivalent. The calculation is based on the principle of
minimum power dissipation under the global constraints imposed to the system,
and taking into account the screening effect. We derived analytical
expressions of the distribution of charge carriers and currents for arbitrary
values of the load resistance and Hall angle, at the first order in the charge
accumulation $\delta n/n_{0}$. The well-known limiting cases corresponding to
the perfect Hall bar (infinite load resistance) and the Corbino disk (zero
load resistance) are recovered.

Despite the fundamental difference between the two effects at the microscopic
scales - namely time-reversal-symmetry breaking - the accumulation of charge
carriers, the distribution of currents, and the power efficiency of the
injected Hall currents are surprisingly similar for both AHE and PHE (after
having fixed the direction of the magnetization $\vec{m}$, according to
Eqs.(9)). The difference is then due to the Hall-angle dependence of the
conductivities (Eq.(13)), i.e. observable at the second order in $\Theta$
($\Theta$ is of the order of 1% in ordinary materials). In the case of the
power efficiency, both curves have a maximum that corresponds to the
resistance matching between the load circuit and the Hall bar. In other terms,
the maximum transfer theorem is satisfied for both AHE and PHE. This property
is not trivial because the system (shown in Fig.1) cannot be reduced to a
simple lumped-element circuit. These results are an illustration of the
importance of the second law of thermodynamics for dissipative transport
processes, that levels out the peculiarities found at the microscopic scales,
and that invalidates the prediction of dissipationless transport for
macroscopic Hall-like systems.

It is worth noting that the results found are valid whatever the mechanisms
responsible for the AHE: either due to spin-orbit scattering, topological
properties of the band structure (Berry connection), as long as it leads to an
effective magnetic field. This is also true for the PHE: it is valid whatever
the mechanisms leading to axial anisotropy. Yet, the main limitations of the
model are the quasi-classical approach and the local equilibrium hypothesis.
The case of quantum anomalous Hall effect or unconventional anomalous Hall
effect could be beyond the limit of validity of this model. However, the
specificity of nanoscopic or quantum characteristics would now be easily
determined and quantified with a comparison of the measurements performed on
these systems with that presented here.

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#  Exclusions of smooth actions on spheres

of the non-split extension of $C_{2}$ by $\operatorname{SL}(2,5)$

Piotr Mizerka

###### Abstract

There are four groups $G$ fitting into a short exact sequence
$1\rightarrow\operatorname{SL}(2,5)\rightarrow G\rightarrow C_{2}\rightarrow
1,$ where $\operatorname{SL}(2,5)$ is the special linear group of $(2\times
2)$-matrices with entries in the field of five elements. Except for the direct
product of $\operatorname{SL}(2,5)$ and $C_{2}$, there are two other
semidirect products of these two groups and just one non-semidirect product
$\operatorname{SL}(2,5).C_{2}$, considered in this paper. It is known that
each finite nonsolvable group can act on spheres with arbitrary positive
number of fixed points. Clearly, $\operatorname{SL}(2,5).C_{2}$ is a
nonsolvable group. Moreover, it turns out that $\operatorname{SL}(2,5).C_{2}$
possesses a free representation and as such, can potentially act pseudofreely
with nonempty fixed point set on manifolds of arbitrarily large dimension. We
prove that $\operatorname{SL}(2,5).C_{2}$ cannot act effectively with odd
number of fixed points on low-dimensional spheres. In the special case of
effective one fixed point actions, we are able to exclude a wider class of
spheres. Moreover, we prove that specific pseudofree one fixed point actions
of $\operatorname{SL}(2,5).C_{2}$ on spheres do not exist.††$2020$ Mathematics
Subject Classification. Primary $57S25$; Secondary $55M35$.

††journal: Journal Name

##  0 Introduction

A smooth action of finite group $G$ on a smooth manifold $M$ is a homomorphism
$\varphi:G\rightarrow\operatorname{Diff}(M)$, where $\operatorname{Diff}(M)$
denotes the group of diffeomorphisms of $M$. All groups appearing in this
paper are assumed to be finite and actions to be smooth. This article concerns
exclusions of actions on spheres with exactly one fixed point or with odd
number of fixed points. We refer to such actions as _one fixed point actions_
or _odd fixed point actions_ respectively.

In [15, p. 55] Montgomery and Samelson wrote the following: _”If a compact Lie
group $G$ acts smoothly on the $n$-sphere $S^{n}$ in such a way as to have one
stationary point, it is likely that there must be a second stationary point.
This appears difficult to prove.”_. It is already known which groups admit one
fixed point actions on spheres. It turns out that such groups are precisely
the groups admitting fixed point free actions on disks. The characterization
of groups admitting fixed point free actions on disks was provided by Oliver
[20] in $1975$. Such groups $G$ are characterized in the following way: $G$
does not contain subgroups $P,H\leq G$ fitting into a sequence
$P\trianglelefteq H\trianglelefteq G$ such that $P$ and $G/H$ are groups of
prime power order and the quotient $H/P$ is cyclic. We refer to such groups
$G$ as _Oliver groups_. The first examples of groups admitting one fixed point
$G$-actions go back to Stein [22] for $G=\operatorname{SL}(2,5)$ and Petrie
[21] for abelian groups with three non-cyclic Sylow subgroups. In $1995$,
Laitinen, Morimoto and Pawałowski [12] proved that each nonsolvable group
admits a one fixed point action on a sphere. Later, in $1998$, by modifying
the techniques descibed in the paper concerning nonsolvable groups, Laitinen
and Morimoto [11] generalized this result and showed that Oliver groups
coincide with the groups admitting one fixed point actions on spheres.

Further research has shown that the minimal dimension of a sphere admitting
one fixed point actions is $6$. First, the combined work of De Michelis [6],
Buchdahl et. al. [5] and Furuta [8] has shown that $S^{n}$ does not admit one
fixed point actions for $n\leq 5$. Later, on the other hand, Bak and Morimoto
[1, 2, 16, 17, 18] proved that $A_{5}$, the alternating group on five letters,
admits one fixed point actions on $S^{n}$ for every $n\geq 6$.

This leads us to the following question: given an Oliver group $G$ and an
integer $n\geq 6$, is it true that there exists a one fixed point action of
$G$ on $S^{n}$? Apart from the results described in the last paragraph
concerning this question, other have been obtained as well. Let us discuss
them. In $1977$, Stein [22] proved that $\operatorname{SL}(2,5)$ can act on
$S^{7}$ with exactly on fixed point. On the other hand, in $2016$, Borowiecka
[3] excluded effective one fixed point actions of $\operatorname{SL}(2,5)$ on
$S^{8}$. Two years later, in the joint work of Borowiecka and the author [4],
new exclusions of effective one fixed point actions have been obtained for
Oliver groups of order up to $216$ and spheres of dimensions varying from $6$
to $10$. In $2020$, Morimoto and Tamura [19] excluded odd fixed point actions
of $S_{5}$, the symmetric group on five letters, and $\operatorname{SL}(2,5)$
on spheres of dimensions varying from $0$ to $13$. Further exclusion results
concerning odd fixed point actions have been recently obtained by Tamura [23].
These results concern six Oliver groups (including symmetric groups, special
linear and projective special linear groups, automorphism groups and Mathieu
groups) and cover many of the dimensions from the set $\\{0,...,50\\}$. The
methods used in the aticles of Borowiecka, Morimoto, Tamura and the author
have been used to extend exclusion results in the PhD thesis of the author
[14, Theorem 1.4, Theorem 5.29] by the application of GAP [9] software.

A general scheme in the study of transformation groups is to consider
prescibed isotropy subgroups. The simpliest case constitute semifree actions
(i. e. actions with trivial isotropy subgroups outside of the fixed point
set). One can relax this criterion however and allow a weaker version of
semifree actions. Such actions have restrictive dimensions of fixed point
sets. After Illman [10], we call an action of a group $G$ on a smooth manifold
$M$ a $k$-_pseudofree_ action if $\dim M^{H}\leq k$ for any nontrivial
subgroup $H\leq G$. Lately, Morimoto announced a result that $3$-pseudofree
one fixed point actions of $G$ on even-dimensional homotopy spheres
$\Sigma^{n}$ are possible only for $G=A_{5},A_{5}\times C_{2}$ or $G=S_{5}$
and $n=6$. On the other hand, if a $k$-pseudofree action is required to exist
for arbitrarily large spheres, the group in question must possess a free
representation. The classification of such groups was provided in [24, 6.3.1
Theorem]. Following this classification, there exists a group of order $240$
possessing a free representation which contains $\operatorname{SL}(2,5)$ as a
subgroup of index $2$ (this group is denoted by $\operatorname{TL}_{2}(5)$ in
[13]). In the GAP [9] SmallGroup library, the group
$\operatorname{SL}(2,5).C_{2}$ we consider appears with
$\operatorname{IdGroup}=[240,89]$. It turns out that
$\operatorname{SL}(2,5).C_{2}$ possesses a free representation (see
Proposition 1.1) and therefore
$\operatorname{SL}(2,5).C_{2}\cong\operatorname{TL}_{2}(5)$.

In a private correspondence with the author, Morimoto suggested that there may
exist $6$-pseudofree effective one fixed point actions of
$\operatorname{SL}(2,5).C_{2}$ on $S^{n}$ whenever $n=6+8k$ for some $k\geq
1$. Our main results concern the converse statement. Before, we formulate
these results, let us recall a notion of a homology sphere. Given a
commutative ring $R$ with unity, we call a closed smooth manifold $\Sigma$ of
dimension $n$ an $R$-_homology sphere_ if the homology groups of $\Sigma$ with
coefficients in $R$ coincide with the groups $H_{k}(S^{n};R)$ for
$k=0,\ldots,n$. The two main theorems of the paper are the following.

######  Theorem 0.1.

_[cf. Theorem 3.1]_ $\operatorname{SL}(2,5).C_{2}$ cannot act effectively with
odd number of fixed points on $n$-dimensional $\mathbb{Z}$-homology sphere
provided $n\in\\{0,1,\ldots,13\\}$. Moreover, $\operatorname{SL}(2,5).C_{2}$
cannot act effectively with exactly one fixed point on an $S^{n}$ provided
$n\in\\{0,1,\ldots,13\\}\cup\\{15,16,17\\}$.

######  Theorem 0.2.

_[cf. Theorem 3.2]_ There are no $5$-pseudofree one fixed point actions of
$\operatorname{SL}(2,5).C_{2}$ on $\mathbb{Z}$-homology spheres. In case
$\operatorname{SL}(2,5).C_{2}$ acts $6$-pseudofreely with exactly one fixed
point on an $n$-dimensional $\mathbb{Z}$-homology sphere $\Sigma$, then
$n=6+8k$ or $n=18+8k$ for $k\geq 0$. Moreover, if the action in question is
effective and $n=6+8k$, then $k\geq 1$.

The article is organized as follows. In the first section we provide the
necessary algebraic data for $\operatorname{SL}(2,5).C_{2}$. We compute the
fixed point dimensions for irreducible
$\mathbb{R}\operatorname{SL}(2,5).C_{2}$-modules for specific subgroups of
$\operatorname{SL}(2,5).C_{2}$ and prove useful facts about these subgroups by
means of the subgroup lattice of $\operatorname{SL}(2,5).C_{2}$. The next
section contains the main technical exclusion results which the proofs of
theorems 0.1 and 0.2 rely on. In the last section we prove the two main
theorems of this paper.

##  1 The necessary algebraic data for $\operatorname{SL}(2,5).C_{2}$

In this section, we show the existence of subgroups of
$\operatorname{SL}(2,5).C_{2}$ with suitable fixed point dimensions and
forming generating sets for $\operatorname{SL}(2,5).C_{2}$. For that purpose,
we present the real irreducible characters of $\operatorname{SL}(2,5).C_{2}$
and the fixed point dimensions for the corresponding representations of the
actions of subgroups of our interest. The poset of conjugacy classes of
$\operatorname{SL}(2,5).C_{2}$ shall be important as it provides the necessary
information concerning the subgroup lattice of $\operatorname{SL}(2,5).C_{2}$.

###  1.1 Nontrivial real irreducible characters of
$\operatorname{SL}(2,5).C_{2}$

The centre, $Z$, of $\operatorname{SL}(2,5).C_{2}$ is isomorphic to $C_{2}$
and $\operatorname{SL}(2,5).C_{2}/Z\cong S_{5}$. Among nontrivial real
irreducible characters of $\operatorname{SL}(2,5).C_{2}$, $6$ characters are
not faithful and $5$ are faithful. Let $U_{1}$, $U_{4,1}$, $U_{4,2}$,
$U_{5,2}$, $U_{5,1}$ and $U_{6}$ be nontrivial irreducible
$\mathbb{R}\operatorname{SL}(2,5).C_{2}$-modules with the non-faithful
characters (these modules are faithful as
$\mathbb{R}(\operatorname{SL}(2,5).C_{2}/Z=S_{5})$-modules), and $W_{8,1}$,
$W_{8,2}$, $W_{8,3}$, $W_{12,1}$, and $W_{12,2}$ irreducible faithful
$\mathbb{R}\operatorname{SL}(2,5).C_{2}$-modules. The characters are presented
in the table below (they can be derived from complex irreducible characters of
$\operatorname{SL}(2,5).C_{2}$ which can be found at [7] for example). The
notation of conjugacy classes in the table is as follows. The class is denoted
by a number indicating the order of its representative. If there is more than
one class whose representative has a given order, then subsequent capital
letters are used to distinguish between such classes.

$\begin{array}[]{c|rrrrrrrrrrrr}\rm class&\rm 1&\rm
2&\rm(3)&\rm(4A)&\rm(4B)&\rm(5)&\rm(6)&\rm(8A)&\rm(8B)&\rm(10)&\rm(12A)&\rm(12B)\cr\rm
size&1&1&20&20&30&24&20&30&30&24&20&20\cr\hline\cr
U_{1}&1&1&1&-1&1&1&1&-1&-1&1&-1&-1\cr U_{4,1}&4&4&1&-2&0&-1&1&0&0&-1&1&1\cr
U_{4,2}&4&4&1&2&0&-1&1&0&0&-1&-1&-1\cr U_{5,1}&5&5&-1&-1&1&0&-1&1&1&0&-1&-1\cr
U_{5,2}&5&5&-1&1&1&0&-1&-1&-1&0&1&1\cr U_{6}&6&6&0&0&-2&1&0&0&0&1&0&0\cr
W_{8,1}&8&-8&-4&0&0&-2&4&0&0&2&0&0\cr
W_{8,2}&8&-8&2&0&0&-2&-2&0&0&2&2\sqrt{3}&-2\sqrt{3}\cr
W_{8,3}&8&-8&2&0&0&-2&-2&0&0&2&-2\sqrt{3}&2\sqrt{3}\cr
W_{12,1}&12&-12&0&0&0&2&0&-2\sqrt{2}&2\sqrt{2}&-2&0&0\cr
W_{12,2}&12&-12&0&0&0&2&0&2\sqrt{2}&-2\sqrt{2}&-2&0&0\cr\end{array}$ Table
1.1: Character table of $\operatorname{SL}(2,5).C_{2}$.

###  1.2 Fixed point dimensions for nontrivial irreducible
$\mathbb{R}\operatorname{SL}(2,5).C_{2}$-modules

The conjugacy classes of subgroups of $\operatorname{SL}(2,5).C_{2}$ are
described in the poset below (see [7]). The classes are denoted by their
representatives (either by a well-known alias or by the id from the GAP [9]
$\operatorname{SmallGroup}$ library). If there is more than one class with
isomorphic representatives, then we add an appropriate letter to distinguish
between these classes. The normal subgroups are marked by blue color.

$C_{1}$$C_{2}$$(C_{3})$$(C_{5})$$(C_{4,A})$$(C_{4,B})$$(C_{6})$$(C_{10})$$(Q_{8,B})$$(Q_{8,A})$$(C_{8})$$([12,1])$$([12,1])$$(C_{12})$$([20,1])$$(Q_{16})$$(\operatorname{SL}(2,3))$$([24,4])$$(C_{5}{\rtimes}C_{8}$)$([48,28])$$\operatorname{SL}(2,5)$$\operatorname{SL}(2,5).C_{2}$
Figure 1: Subgroups of $\operatorname{SL}(2,5).C_{2}$.

Let us show first that $\operatorname{SL}(2,5).C_{2}$ is indeed isomorphic to
the group $\operatorname{TL}_{2}(5)$ mentioned in [13, Remark on p. 154]. To
this aim, we have to show that $\operatorname{SL}(2,5).C_{2}$ possesses a free
representation. It turns out that the real irreducible
$\operatorname{SL}(2,5).C_{2}$-representation $W_{8,1}$ is free. This follows
from the proposition below, since the only primes dividing $240$, which is the
order of $\operatorname{SL}(2,5).C_{2}$, are $2$, $3$, and $5$.

######  Proposition 1.1.

Suppose $\mathcal{U}_{4}=\\{U_{4,k}|k=1,2\\}$,
$\mathcal{U}_{5}=\\{U_{5,k}|k=1,2\\}$,
$\mathcal{W}_{8}=\\{W_{8,k}|k=1,2,3\\}$,
$\mathcal{W}_{12}=\\{W_{12,k}|k=1,2\\}$. Then, for any
$U_{4}\in\mathcal{U}_{4}$, $U_{5}\in\mathcal{U}_{5}$,
$W_{8}\in\mathcal{W}_{8}$, $W_{12}\in\mathcal{W}_{12}$,
$W\in\mathcal{W}_{8}\cup\mathcal{W}_{12}$, and $i\in\\{2,3\\}$, the following
equalities hold.

$\dim W^{C_{2}}=\dim W_{8,1}^{C_{3}}=\dim W_{8}^{C_{5}}=\dim
W_{12}^{C_{2}}=0,$ $\dim U_{1}^{C_{2}}=\dim U_{1}^{C_{3}}=\dim
U_{1}^{C_{5}}=\dim U_{5}^{C_{3}}=\dim U_{5}^{C_{5}}=1,$ $\dim
U_{4}^{C_{3}}=\dim U_{6}^{C_{3}}=\dim U_{6}^{C_{5}}=2,$ $\dim
U_{4}^{C_{2}}=\dim W_{8,i}^{C_{3}}=\dim W_{12}^{C_{3}}=\dim W_{12}^{C_{5}}=4,$
$\dim U_{5}^{C_{2}}=5,\text{ and }\dim U_{6}^{C_{2}}=6.$

###### Proof.

The equalities concerning the subgroup $C_{2}\leq\operatorname{SL}(2,5).C_{2}$
are straightforward as $\operatorname{Ker}U=C_{2}$ for $U=U_{k}$ for
$k=1,4,5,6$ and $W$ is a faithful representation.

Since there are unique conjugacy classes of elements of order $3$, $5$, we get
the desired equalities from Table 1.1 by using the formula for the fixed point
dimension,

$\dim V^{H}=\frac{1}{|H|}\sum_{h\in
H}\raisebox{0.0pt}[4.30554pt][4.30554pt]{$\chi$}_{V}(h),$

where $H\leq G$, $V$ is an $\mathbb{R}G$-module and
$\raisebox{0.0pt}[4.30554pt][4.30554pt]{$\chi$}_{V}$ is its character. ∎

The next proposition provides fixed point dimensions for $4$ conjugacy classes
of subgroups which are of our particular interest.

######  Proposition 1.2.

Assuming the notations from Proposition 1.1, and
$H\in(C_{4,A})\cup(Q_{8,A})\cup(Q_{16})\cup([24,4])$, the following equalities
hold.

$\dim U_{1}^{H}=\dim W^{H}=\dim U_{4,1}^{Q_{8,A}}=\dim U_{4,1}^{Q_{16}}=\dim
U_{4,1}^{[24,4]}=\dim U_{5,1}^{[24,4]}=\dim U_{6}^{Q_{16}}=\dim
U_{6}^{[24,4]}=0,$ $\dim U_{4,1}^{C_{4,A}}=\dim U_{4,2}^{Q_{16}}=\dim
U_{4,2}^{[24,4]}=\dim U_{5,1}^{Q_{8,A}}=\dim U_{5,1}^{Q_{16}}=\dim
U_{5,2}^{Q_{16}}=\dim U_{5,2}^{[24,4]}=\dim U_{6}^{Q_{8,A}}=1,$ $\dim
U_{4,2}^{Q_{8,A}}=\dim U_{5,1}^{C_{4,A}}=\dim U_{5,2}^{Q_{8,A}}=2,\text{ and
}\dim U_{4,2}^{C_{4,A}}=\dim U_{5,2}^{C_{4,A}}=\dim U_{6}^{C_{4,A}}=3.$

###### Proof.

The equality $\dim W^{H}=0$ follows immediately from Proposition 1.1.

Let $C_{4,A}^{\prime}$ be the cyclic subgroup of order $4$, generated by an
element from the conjugacy class $(4A)$. It follows from Table 1.1 that

$1>\frac{1}{4}\cdot(1-1+1+1)\geq\dim U_{1}^{C_{4,A}^{\prime}}\geq 0.$

Thus $\dim U_{1}^{C_{4,A}^{\prime}}=0$. On the other hand,

$1\geq\dim U_{1}^{\operatorname{SL}(2,3)}\geq\frac{1}{24}\cdot(1+1+8\cdot
1+6\cdot(-1)+8\cdot 1)>0,$

as $\operatorname{SL}(2,3)$ contains precisely: one element of order $1$ and
$2$, eight elements of order $3$, six elements of order $4$, and eight
elements of order $6$. Hence, $\dim U_{1}^{\operatorname{SL}(2,3)}=1$, which
in connection with $\dim U_{1}^{C_{4,A}^{\prime}}=0$ and Figure 1 yields
$C_{4,A}^{\prime}\in(C_{4,A})$. Hence $\dim U_{1}^{H}=0$ as $H$ contains as a
subgroup a representative of the class $(C_{4,A})$. Moreover, as we know now
that the elements of $C_{4,A}$ of order $2$ belong to $(4A)$, all equalities
involving $C_{4,A}$ follow directly from Table 1.1 and we skip it.

The subgroup $Q_{8,A}$ is isomorphic to the quaternion group, and therefore
contains one element of order $1$ and $2$, and six elements of order $4$
(moreover, at least two of them are contained in the conjugacy class of
elements $(4A)$ since a representative of $(C_{4,A})$ is contained in
$(Q_{8,A})$). Thus,

$\dim U_{4,1}^{Q_{8,A}}\leq\frac{1}{8}\cdot(4+4+2\cdot(-2)+4\cdot 0)<1,$

which yields $\dim U_{4,1}^{Q_{8,A}}=\dim U_{4,1}^{Q_{16}}=\dim
U_{4,1}^{[24,4]}=0$. The subgroup $Q_{16}$ contains one element of order $1$
and $2$, ten elements of order $4$, and four elements of order $8$, while
$[24,4]$ contains one element of order $1$ and $2$, two elements of order $3$,
fourteen elements of order $4$, two elements of order $6$, and four elements
of order $12$. Using this knowledge, The equalities concerning $U_{5,2}$ can
be derived directly from Table 1.1, just as in the proof of Proposition 1.1.
We skip these computations. Moreover,

$\dim U_{5,1}^{[24,4]}<\frac{1}{24}\cdot(5+5+2\cdot(-1)+14\cdot
1+2\cdot(-1)+4\cdot(-1))<1,$ $\dim
U_{6}^{Q_{16}}<\frac{1}{16}\cdot(6+6+10\cdot 0+4\cdot 0)<1,$

and

$\dim U_{6}^{[24,4]}<\frac{1}{24}\cdot(6+6+2\cdot 0+14\cdot 0+2\cdot 0+4\cdot
0)<1,$

which means that $\dim U_{5,1}^{[24,4]}=\dim U_{6}^{Q_{16}}=\dim
U_{6}^{[24,4]}=0$.

Further proof is the repetition of the already used arguments. We are left
with the proofs of equalities concerning dimensions which are to be equal $1$
or $2$. The following inequalities conclude then the proof.

$2>\frac{1}{16}\cdot(4+4+10\cdot 2+4\cdot 0)\geq\dim
U_{4,2}^{Q_{16}}\geq\frac{1}{16}\cdot(4+4+10\cdot 0+4\cdot 0)>0.$
$2>\frac{1}{24}\cdot(4+4+2\cdot 1+14\cdot 2+2\cdot 1+4\cdot(-1))\geq\dim
U_{4,2}^{[24,4]}\geq\frac{1}{24}\cdot(4+4+2\cdot 1+14\cdot 0+2\cdot
1+4\cdot(-1))>0,$ $2>\frac{1}{8}\cdot(5+5+2\cdot(-1)+4\cdot 1)\geq\dim
U_{5,1}^{Q_{8,A}}\geq\frac{1}{8}\cdot(5+5+6\cdot(-1))>0,$ $1=\dim
U_{5,1}^{Q_{8,A}}\geq\dim
U_{5,1}^{Q_{16}}\geq\frac{1}{16}\cdot(5+5+10\cdot(-1)+4\cdot 1)>0,$
$2>\frac{1}{8}\cdot(6+6+6\cdot 0)\geq\dim
U_{6}^{Q_{8,A}}\geq\frac{1}{8}\cdot(6+6+2\cdot 0+4\cdot(-2))>0,$
$3>\frac{1}{8}\cdot(4+4+6\cdot 2)\geq\dim
U_{4,2}^{Q_{8,A}}\geq\frac{1}{8}\cdot(4+4+2\cdot 2+4\cdot 0)>1.$

∎

Another important fact is that the fixed point set,
$U_{1}^{\operatorname{SL}(2,5)}$, has positive dimension. The group
$\operatorname{SL}(2,5)$ contains one element of order $1$ and $2$, twenty
elements of order $3$, thirty elements of order $4$, twenty four elements of
order $5$, twenty elements of order $6$, and twenty four elements of order
$10$. Hence,

$\dim U_{1}^{\operatorname{SL}(2,5)}\geq\frac{1}{120}\cdot(1+1+20\cdot
1+30\cdot(-1)+24\cdot 1+20\cdot 1+24\cdot 1)>0,$

which means that $\dim U_{1}^{\operatorname{SL}(2,5)}=1$.

###  1.3 Suitable generating subgroups of $\operatorname{SL}(2,5).C_{2}$

Recall that we denoted the centre of $\operatorname{SL}(2,5).C_{2}$ by $Z$. In
the following two lemmas, let
$\pi:\operatorname{SL}(2,5).C_{2}\rightarrow\operatorname{SL}(2,5).C_{2}/Z\cong
S_{5}$ be the quotient epimorphism. As there shall be no confusion, from now
on, we identify $\operatorname{SL}(2,5).C_{2}/Z$ with $S_{5}$, the group of
all permutations on $5$ letters. The following figure is the poset of
conjugacy classes of subgroups of $S_{5}$. As in the case of Figure 1, we mark
the normal subgroups by blue color. In the figure below, $D_{2n}$ denotes the
dihedral group of order $2n$, and $F_{5}$ is the Frobenius group of order
$20$. This poset can be also found at [7]. The convention for notation is the
same as in Figure 1.

$C_{1}$$(C_{2})$$(C_{2})$$(C_{3})$$(C_{5})$$(C_{2}^{2})$$(C_{2}^{2})$$(C_{4})$$(C_{6})$$(S_{3})$$(S_{3})$$(D_{10})$$(D_{8})$$(A_{4})$$(D_{12})$$(F_{5})$$(S_{4})$$A_{5}$$S_{5}$
Figure 2: Subgroups of $S_{5}$.

Define the subgroups $L,K_{1},K_{2}\leq S_{5}$ as follows.

$L=\langle(\begin{matrix}1&3\end{matrix})\rangle\cong C_{2}\text{,
}K_{1}=\langle(\begin{matrix}1&3\end{matrix}),(\begin{matrix}4&5\end{matrix})\rangle\cong
C_{2}^{2}\text{ and
}K_{2}=\langle(\begin{matrix}1&2&3&4\end{matrix}),(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix})\rangle.$

######  Lemma 1.3.

We have $K_{2}\cong D_{8}$ and $L=K_{1}\cap K_{2}$.

###### Proof.

Let $D_{8}=\langle a,b|a^{4}=b^{2}=1,bab=a^{-1}\rangle$. Define
$\varphi:D_{8}\rightarrow K_{2}$ by
$\varphi(a)=(\begin{matrix}1&2&3&4\end{matrix})$ and
$\varphi(b)=(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix})$.
Then

$\varphi(a^{4})=\varphi(b^{2})=1$

and

$\varphi(bab)=\varphi(b)\varphi(a)\varphi(b)=(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix})(\begin{matrix}1&2&3&4\end{matrix})(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix})=(\begin{matrix}1&4&3&2\end{matrix})=\varphi(a^{-1}).$

Therefore $\varphi$ is a well-defined homomorphism. As it maps generators to
generators, it is an epimorphism. Now, we have

$\operatorname{Ker}\varphi=\\{b^{\varepsilon}a^{k}|\varphi(b^{\varepsilon}a^{k})=\operatorname{id}\\}=\\{b^{\varepsilon}a^{k}|((\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix}))^{\varepsilon}(\begin{matrix}1&2&3&4\end{matrix})^{k}=\operatorname{id}\\}.$

Notice that $(\begin{matrix}1&2&3&4\end{matrix})^{k}=\operatorname{id}$ iff
$4|k$. Thus, if $\varepsilon=0$, then
$\varphi(b^{\varepsilon}a^{k})=\operatorname{id}$ if and only if
$b^{\varepsilon}a^{k}=1$. In case $\varepsilon=1$, we have

$\varphi(b^{\varepsilon})=\varphi(b)=(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix}),$
$\varphi(b^{\varepsilon}a)=\varphi(ba)=(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix})(\begin{matrix}1&2&3&4\end{matrix})=(\begin{matrix}2&4\end{matrix}),$
$\varphi(b^{\varepsilon}a^{2})=\varphi(ba^{2})=\varphi(ba)\varphi(a)=(\begin{matrix}2&4\end{matrix})(\begin{matrix}1&2&3&4\end{matrix})=(\begin{matrix}1&4\end{matrix})(\begin{matrix}2&3\end{matrix}),$
$\varphi(b^{\varepsilon}a^{3})=\varphi(ba^{3})=\varphi(ba^{2})\varphi(a)=(\begin{matrix}1&4\end{matrix})(\begin{matrix}2&3\end{matrix})(\begin{matrix}1&2&3&4\end{matrix})=(\begin{matrix}1&3\end{matrix}).$

Thus $\varphi(b^{\varepsilon}a^{k})=\operatorname{id}$ if and only if
$b^{\varepsilon}a^{k}=1$, whence $\varphi$ has the trivial kernel. Therefore
$\varphi$ is an isomorphism and $K_{2}\cong D_{8}$.

We show now that $L=K_{1}\cap K_{2}$. Notice that

$(\begin{matrix}1&2&3&4\end{matrix})(\begin{matrix}1&2\end{matrix})(\begin{matrix}3&4\end{matrix})=(\begin{matrix}1&3\end{matrix}).$

Thus $L\leq K_{1}\cap K_{2}$. This shows (as $|K_{1}|=4$) that $|K_{1}\cap
K_{2}|\in\\{2,4\\}$, whence $K_{1}\cap
K_{2}=\langle(\begin{matrix}1&3\end{matrix})\rangle$ or $K_{1}\cap
K_{2}=K_{1}$. On the other hand, $(\begin{matrix}4&5\end{matrix})\in
K_{1}\setminus K_{2}$, whence $K_{1}\cap
K_{2}=\langle(\begin{matrix}1&3\end{matrix})\rangle=L$. ∎

######  Lemma 1.4.

Using the notations from Figure 1,

$\pi^{-1}(L)\in(C_{4,A})\text{, }\pi^{-1}(K_{1})\in(Q_{8,A})\text{, and
}\pi^{-1}(K_{2})\in(Q_{16}).$

Moreover,

$\pi^{-1}(L)=\pi^{-1}(K_{1})\cap\pi^{-1}(K_{2})\text{ and
}\langle\pi^{-1}(K_{1}),\pi^{-1}(K_{2})\rangle=\operatorname{SL}(2,5).C_{2}.$

###### Proof.

Let us show first that $\pi^{-1}(L)\in(C_{4,A})$. We just have to exclude the
case $\pi^{-1}(L)\in(C_{4,B})$. To this aim, we study the behaviour of
subgroups from the class $(C_{4,B})$. Take $M=\langle g\rangle\in(C_{4,B})$
and let $V=U_{5,1}$. By Table 1.1, $\dim V^{M}=3$. Therefore we can arrange
the basis of $V$ to be $\mathcal{B}=\\{v_{1},\ldots,v_{5}\\}$ such that
$v_{i}$ remains fixed under the action of $g$ for $i=1,2,3$. Thus, the action
of $g$ on $V$ is the order $4$ orthogonal transformation a $2$-dimensional
real vector space. Hence, $g$ preserves the orientation of $V$ given by the
basis $\mathcal{B}$. Note that $\pi(g)$ is the generator of $\pi(M)\cong
C_{2}$. Therefore, $\pi(g)$ preserves the orientation of $V$ as well since $V$
can be viewed as an $\mathbb{R}S_{5}$-module as
$\operatorname{SL}(2,5).C_{2}/\operatorname{Ker}(V)\cong S_{5}$. We show that
$\pi(g)\in A_{5}$ which would mean, by Figure 1 and Figure 2, that
$\pi^{-1}(L)\notin(C_{4,B})$. We see by Figure 2 that there are two conjugacy
classes of elements of order $2$ in $S_{5}$. One of these conjugacy classes
consists of a single element contained in $A_{5}$, while the other class has
empty intersection with $A_{5}$. Take $\sigma$ which is the representative of
the second class. Thus, $\sigma=\pi(g_{\sigma})$ for some $g_{\sigma}\in(4A)$,
and $\raisebox{0.0pt}[4.30554pt][4.30554pt]{$\chi$}_{V}(\sigma)=-1$ by Table
1.1. This means that the eigenvalues of the matrix determined by $\sigma$ are
$1,1,-1,-1,-1$, and therefore $\sigma$ reverses the orientation. As $\pi(g)$
preserves the orientation, it follows then that $\pi(g)\in A_{5}$.

Since the conjugacy class $(Q_{16})$ from Figure 1 is the unique conjugacy
class of subgroups of $\operatorname{SL}(2,5).C_{2}$ containing subgroups of
order $16$, and $|\pi^{-1}(K_{2})|=16$ (as $|K_{2}|=8$ by Lemma 1.3), it
follows that $\pi^{-1}(K_{2})\in(Q_{16})$. Moreover, since $(Q_{8,A})$ is the
only class with a representative of order $8$ and containing $C_{4,A}$, it
follows that $\pi^{-1}(K_{1})\in(Q_{8,A})$.

In order to prove the second part, take any
$g_{1}\in\pi^{-1}(\\{(\begin{matrix}4&5\end{matrix})\\})$ and
$g_{2}\in\pi^{-1}(\\{(\begin{matrix}1&2&3&4\end{matrix})\\})$. Then

$\pi(g_{2}g_{1})=\pi(g_{2})\pi(g_{1})=(\begin{matrix}1&2&3&4\end{matrix})(\begin{matrix}4&5\end{matrix})=(\begin{matrix}1&2&3&4&5\end{matrix}).$

Thus, the order of the element $g_{2}g_{1}$ is $10$, and we conclude by Figure
1 that
$\langle\pi^{-1}(K_{1}),\pi^{-1}(K_{2})\rangle=\operatorname{SL}(2,5).C_{2}$.
Finally, by Lemma 1.3, we have $\pi^{-1}(L)=\pi^{-1}(K_{1}\cap
K_{2})=\pi^{-1}(K_{1})\cap\pi^{-1}(K_{2})$. ∎

##  2 The main exclusion results

The following table contains useful information concerning fixed point
dimensions for the four subgroups of $\operatorname{SL}(2,5).C_{2}$ considered
in the previous section. This is a summary of Proposition 1.2. In the table
below, $i=2,3$ and $j=1,2$.

class | $(C_{4,A})$ | $(Q_{8,A})$ | $(Q_{16})$ | $([24,4])$
---|---|---|---|---
$U_{1}$ | $0$ | $0$ | $0$ | $0$
$U_{4,1}$ | $1$ | $0$ | $0$ | $0$
$U_{4,2}$ | $3$ | $2$ | $1$ | $1$
$U_{5,1}$ | $2$ | $1$ | $1$ | $0$
$U_{5,2}$ | $3$ | $2$ | $1$ | $1$
$U_{6}$ | $3$ | $1$ | $0$ | $0$
$W_{8,1}$ | $0$ | $0$ | $0$ | $0$
$W_{8,i}$ | $0$ | $0$ | $0$ | $0$
$W_{12,j}$ | $0$ | $0$ | $0$ | $0$
Table 2.1: Fixed point dimension table for generating subgroups of
$\operatorname{SL}(2,5).C_{2}$.

In the following four lemmas, $G$ can be taken to be any finite group acting
on a $\mathbb{Z}$-homology sphere $\Sigma$.

######  Lemma 2.1.

_[19, cf. Proposition 2.4]_ If $\Sigma^{G}$ is nonempty, then for any non-
Oliver subgroup $H\leq G$ we have
$\raisebox{0.0pt}[4.30554pt][4.30554pt]{$\chi$}(\Sigma^{H})\neq 1$.

######  Lemma 2.2.

Suppose that $H_{1}$ and $H_{2}$ are non-Oliver subgroups of $G$ which
generate $G$ and suppose $P$ is a prime power order subgroup of $H_{1}\cap
H_{2}$. If there exists $x\in\Sigma^{G}$ with $\dim T_{x}(\Sigma^{P})=0$, then
$\Sigma^{G}$ is a two point set.

###### Proof.

By the Smith theory, it follows that $\Sigma^{P}$ is a
$\mathbb{Z}_{p}$-homology sphere. As it is a finite set, we have
$|\Sigma^{P}|=2$. Since $\Sigma^{H_{1}},\Sigma^{H_{2}}\subseteq\Sigma^{P}$,
the Euler characteristics of $\Sigma^{H_{1}}$ and $\Sigma^{H_{2}}$ are equal
to their cardinalities. On the other hand, by Lemma 2.1, we conclude that
$|\Sigma^{H_{i}}|\neq 1$ for $i=1,2$. Thus $|\Sigma^{H_{i}}|=2$ for $i=1,2$,
and, in conclusion, $\Sigma^{H_{1}}=\Sigma^{H_{2}}=\Sigma^{P}$. As $\langle
H_{1},H_{2}\rangle=G$, we have
$\Sigma^{G}=\Sigma^{H_{1}}\cap\Sigma^{H_{2}}=\Sigma^{P}$, and $\Sigma^{G}$ is
therefore a two point set. ∎

The following lemma is a modified version of the previous one.

######  Lemma 2.3.

Suppose $\Sigma$ is a standard sphere and $H_{1}$, $H_{2}$ are non-Oliver
subgroups generating $G$. Let $x_{0}\in\Sigma^{G}$ and put
$V=T_{x_{0}}\Sigma$. Suppose $H\leq H_{1}\cap H_{2}$ is such that

$\dim V^{K}+\dim V^{L}=\dim V^{H}$

and $\Sigma^{H}\cong S^{k}$ for some $k\in\\{0,1,2\\}$. Then
$\Sigma^{G}\neq\\{x_{0}\\}$.

###### Proof.

Suppose the assumptions of the Lemma hold. Consider the case $\dim
V^{H_{1}}=\dim V^{H_{2}}=1$ as it is a bit different. Let $C(H_{1})$ and
$C(H_{2})$ be the connected components containing $x_{0}$ of $\Sigma^{H_{1}}$
and $\Sigma^{H_{2}}$ respectively. Suppose for the converse that
$\Sigma^{G}=\\{x_{0}\\}$. Then, as $\langle H_{1},H_{2}\rangle=G$, we have
$\Sigma^{G}=C(H_{1})\cap C(H_{2})$. On the other hand the intersection number
of $C(H_{1})$ and $C(H_{2})$ in $\Sigma^{H}\cong S^{k}$ is zero. A
contradiction.

Suppose now that at least one of the dimensions, $\dim V^{H_{1}}$ or $\dim
V^{H_{2}}$, is not equal to $1$. As $\dim V^{H}=k$, where $k\in\\{0,1,2\\}$,
this means that precisely one dimension from $\dim V^{H_{1}}$ and $\dim
V^{H_{2}}$ is zero and the second one is $\dim V^{H}$. The case $k=0$ can be
excluded by the previous lemma. Without loss of generality, suppose that $\dim
V^{H_{1}}=\dim V^{H}$. This means that $\Sigma^{H_{1}}$ contains a connected
component which is a closed submanifold of $\Sigma^{H}\cong S^{k}$ of
dimension $k\in\\{1,2\\}$. Therefore this component must be entire
$\Sigma^{H}$, which yields $\Sigma^{H_{1}}=\Sigma^{H}$. Thus, as
$\Sigma^{H_{2}}\subseteq\Sigma^{H}$,

$\Sigma^{G}=\Sigma^{H_{1}}\cap\Sigma^{H_{2}}=\Sigma^{H}\cap\Sigma^{H_{2}}=\Sigma^{H_{2}}.$

Therefore, if $\Sigma^{G}=\\{x_{0}\\}$, then $\Sigma^{H_{2}}=\\{x_{0}\\}$. The
latter, however, cannot hold since $H_{2}$ is a non-Oliver group and as such
it cannot act on spheres with exactly one fixed point. ∎

######  Lemma 2.4.

_[19, Corollary 2.8]_ Let $G_{2}$ be the intersection of all subgroups $H\leq
G$ with $[G:H]\leq 2$ and suppose $x\in\Sigma^{G}$. If $\dim
T_{x}(\Sigma)^{G_{2}}>0$, then $\Sigma^{G}\neq\\{x\\}$.

The theorem below is similar to Theorem $5.1$ of [23].

######  Theorem 2.5.

Let $\Sigma$ be a $\mathbb{Z}$-homology sphere with
$\operatorname{SL}(2,5).C_{2}$-action and $|\Sigma^{SL(2,5).C_{2}}|\equiv 1\
(\mathrm{mod}\ 2)$. Then the following statements hold.

  1. 1.

If for each $x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$, $T_{x}(\Sigma)$
contains at least one of the modules $U_{4,2}$ or $U_{5,2}$, then the set of
all points $x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$ such that
$T_{x}(\Sigma)$ contains $U_{6}$ consists of odd number of points.

  2. 2.

If for at least one point $x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$,
$T_{x}(\Sigma)$ contains neither of $U_{4,2}$ and $U_{5,2}$, then
$T_{x}(\Sigma)$ contains $U_{5,1}$ or $U_{6}$.

###### Proof.

Let us prove first the first statement. Assume therefore that for each
$x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$, $T_{x}(\Sigma)$ contains at least
one of the modules $U_{4,2}$ or $U_{5,2}$. Denote by $F$ the set consisting of
all points from $x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$ such that the
tangential representation $T_{x}(\Sigma)$ contains $U_{6}$ and put
$F^{\prime}=\Sigma^{\operatorname{SL}(2,5).C_{2}}\setminus F$. We show that
$|F^{\prime}|\equiv 0\ (\mathrm{mod}\ 2)$.

If $F^{\prime}$ is empty, then the assertion follows. Hence, we may assume
$F^{\prime}$ is non-empty and take $y\in F^{\prime}$. Suppose now $F$ is non-
empty and take $x\in F$. Then,

$T_{x}(\Sigma)=U_{6}^{\oplus k}\oplus U_{1}^{\oplus a}\oplus U_{4,1}^{\oplus
b_{1}}\oplus U_{4,2}^{\oplus b_{2}}\oplus U_{5,1}^{\oplus c_{1}}\oplus
U_{5,2}^{\oplus c_{2}}\oplus W,$ $T_{y}(\Sigma)=U_{1}^{\oplus r}\oplus
U_{4,1}^{\oplus s_{1}}\oplus U_{4,2}^{\oplus s_{2}}\oplus U_{5,1}^{\oplus
t_{1}}\oplus U_{5,2}^{\oplus t_{2}}\oplus W^{\prime},$

where $W$ and $W^{\prime}$ contain only the irreducibles denoted with the
capital ”W” and some index (these are precisely faithful real irreducibles of
$\operatorname{SL}(2,5).C_{2}$) and
$a,b_{1},b_{2},c_{1},c_{2},r,s_{1},s_{2},t_{1},t_{2}$ are non-negative
integers and $k$ is a positive integer. Moreover, by our assumption, we know
that $s_{2}$ or $t_{2}$ is positive (and, similarly, $b_{2}$ or $c_{2}$ is
positive).

Since $\langle Q_{16},[24,4]\rangle={\operatorname{SL}(2,5).C_{2}}$, we have
$\Sigma^{Q_{16}}\cap\Sigma^{[24,4]}=\Sigma^{\operatorname{SL}(2,5).C_{2}}.$
Note that $0<s_{2}+t_{2}=\dim T_{y}(\Sigma^{[24,4]})\leq\dim
T_{y}(\Sigma^{Q_{8,A}})$, and $\dim
T_{y}(\Sigma^{Q_{16}})=s_{2}+t_{1}+t_{2}>0$. Therefore both $\Sigma^{Q_{8,A}}$
and $\Sigma^{Q_{16}}$ are connected as positive dimensional
$\mathbb{Z}_{2}$-homology spheres (by the Smith Theory) because $Q_{8,A}$ and
$Q_{16}$ are $2$-groups. Summing up, by Table 2.1, we get

$\dim\Sigma^{Q_{16}}=b_{2}+c_{1}+c_{2},\dim\Sigma^{Q_{16}}=s_{2}+t_{1}+t_{2},$
$\dim T_{x}(\Sigma^{[24,4]})=b_{2}+c_{2},\dim
T_{y}(\Sigma^{[24,4]})=s_{2}+t_{2},$

and

$\dim\Sigma^{Q_{8,A}}=2b_{2}+c_{1}+2c_{2}+k,\dim\Sigma^{Q_{8,A}}=2s_{2}+t_{1}+2t_{2}.$

Thus,

$\dim\Sigma^{Q_{16}}+\dim T_{x}(\Sigma^{[24,4]})=\dim\Sigma^{Q_{8,A}}-k,$
$\dim\Sigma^{Q_{16}}+\dim T_{y}(\Sigma^{[24,4]})=\dim\Sigma^{Q_{8,A}}.$

Denote by $\Sigma^{[24,4]}_{0}$ the union of all connected components $C$ of
$\Sigma^{[24,4]}$ such that
$C\cap\Sigma^{\operatorname{SL}(2,5).C_{2}}\neq\emptyset$ and
$\dim\Sigma^{Q_{16}}+\dim C=\dim\Sigma^{Q_{8,A}}$. Then
$F\cap\Sigma^{[24,4]}_{0}=\emptyset$ and
$F^{\prime}=\Sigma^{[24,4]}_{0}\cap\Sigma^{\operatorname{SL}(2,5).C_{2}}$.
Thus, the $\ (\mathrm{mod}\ 2)$-intersection number of $\Sigma^{Q_{16}}$ and
$\Sigma^{[24,4]}_{0}$ in $\Sigma^{Q_{8,A}}$ is equal to $|F^{\prime}|$ in
$\mathbb{Z}_{2}$. On the other hand, the canonical $\ (\mathrm{mod}\
2)$-intersection form on $\Sigma^{Q_{8,A}}$,

$\varphi:H_{s_{2}+t_{1}+t_{2}}(\Sigma^{Q_{8,A}};\mathbb{Z}_{2})\times
H_{s_{2}+t_{1}}(\Sigma^{Q_{8,A}};\mathbb{Z}_{2})\rightarrow\mathbb{Z}_{2}$

is trivial as $\Sigma^{Q_{8,A}}$ is a $\mathbb{Z}_{2}$-homology sphere and
$s_{2}+t_{2}>0$. Thus, the intersection number of $\Sigma^{Q_{16}}$ and
$\Sigma^{[24,4]}_{0}$ in $\Sigma^{Q_{8,A}}$ is zero, whence
$|F^{\prime}|\equiv 0\ (\mathrm{mod}\ 2)$ in the case $F$ is non-empty. If $F$
is empty, then $\Sigma^{[24,4]}_{0}=\Sigma^{[24,4]}$ and the same argument
with intersection numbers holds (in fact $F$ cannot be empty as by this
argument, $|F^{\prime}|$ is even which means that
$|\Sigma^{\operatorname{SL}(2,5).C_{2}}|$ is even as well, as
$\Sigma^{\operatorname{SL}(2,5).C_{2}}=F^{\prime}$ in this case). This
concludes the proof of the first statement.

Let us prove the second statement. Suppose therefore that for some
$z\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$, $T_{z}(\Sigma)$ contains neither
of $U_{4,2}$ and $U_{5,2}$. If $T_{z}(\Sigma)$ contains $U_{5,1}$ or $U_{6}$,
there is nothing to prove. Thus, we can assume that $T_{z}(\Sigma)$ does not
contain neither of $U_{4,2}$, $U_{5,2}$, $U_{5,1}$, and $U_{6}$. Hence,
$T_{z}(\Sigma)\cong U_{1}^{\oplus r}\oplus U_{4,1}^{\oplus s_{1}}\oplus W$ for
$W$ containing only direct summands from the set
$\\{W_{8,1},W_{8,2},W_{8,3},W_{12,1},W_{12,2}\\}$. This means, however, that
$\dim T_{z}(\Sigma^{Q_{8,A}})=0$. As $Q_{16}$ and $[24,4]$ are non-Oliver and
generate $\operatorname{SL}(2,5).C_{2}$, we conclude by Lemma 2.2 that
$|\Sigma^{\operatorname{SL}(2,5).C_{2}}|=2$ which is a contradiction. ∎

##  3 Proofs of Theorem 0.1 and Theorem 0.2

In the light of Lemma 1.4, we can assume that $C_{4,A}\leq Q_{8,A}\cap Q_{16}$
and $\langle Q_{8,A},Q_{16}\rangle={\operatorname{SL}(2,5).C_{2}}$.

######  Theorem 3.1.

_[cf. Theorem 0.1]_ $\operatorname{SL}(2,5).C_{2}$ cannot act effectively with
odd number of fixed points on $n$-dimensional $\mathbb{Z}$-homology sphere
provided $n\in\\{0,1,\ldots,13\\}$. Moreover, $\operatorname{SL}(2,5).C_{2}$
cannot act effectively with exactly one fixed point on an $S^{n}$ provided
$n\in\\{0,1,\ldots,13\\}\cup\\{15,16,17\\}$.

###### Proof.

Suppose $\operatorname{SL}(2,5).C_{2}$ acts on a $\mathbb{Z}$-homology
$n$-sphere $\Sigma$ with odd number of fixed points. Suppose first that for
each $x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$, $T_{x}(\Sigma)$ contains at
least one of the modules $U_{4,2}$ or $U_{5,2}$. By Theorem 2.5 we conclude
that there exists $x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$ such that
$T_{x}(\Sigma)$ contains $U_{6}$. As the action is faithful, $T_{x}(\Sigma)$
must contain one of the modules from the set
$\\{W_{8,1},W_{8,2},W_{8,3},W_{12,1},W_{12,2}\\}$. This would mean that $n\geq
6+8=14$.

Assume that for at least one point
$x\in\Sigma^{\operatorname{SL}(2,5).C_{2}}$, $T_{x}(\Sigma)$ contains neither
of $U_{4,2}$, $U_{5,2}$, and $U_{6}$. By Theorem 2.5 we conclude that
$T_{x}(\Sigma)$ contains $U_{5,1}$. Thus (by Table 2.1), in order to prove the
assertion concerning actions with odd number of fixed points, we have to
exclude three possibilities: $T_{x}(\Sigma)\cong U_{5,1}\oplus U_{8,i}$ for
$i=1,2,3$. Fortunately, we will be able to exclude these three possibilities
at once - just by using the fact that $U_{5,1}$ appears as a unique non-
faithful direct summand of $T_{x}(\Sigma)$. The proof of the first part can be
finished by considering the $\ (\mathrm{mod}\ 2)$-intersection numbers of
$\Sigma^{Q_{8,A}}$ and $\Sigma^{Q_{16}}$ in $\Sigma^{C_{4,A}}$ as
$\Sigma^{Q_{8,A}}$, $\Sigma^{Q_{16}}$ and $\Sigma^{C_{4,A}}$ are connected by
the Smith theory (because they have positive dimension by Table 2.1) and
$\dim\Sigma^{Q_{8,A}}+\dim\Sigma^{Q_{16}}=1+1=2=\dim\Sigma^{C_{4,A}}$. Indeed,
the intersection number argument yields
$|\Sigma^{\operatorname{SL}(2,5).C_{2}}|$ even as
$\Sigma^{Q_{8,A}}\cap\Sigma^{Q_{16}}=\Sigma^{\operatorname{SL}(2,5).C_{2}}$
for $Q_{8,A}$ and $Q_{16}$ generate $\operatorname{SL}(2,5).C_{2}$.

Assume now that $\operatorname{SL}(2,5).C_{2}$ acts effectively with exactly
one fixed point on an $S^{n}$ for $n\in\\{15,16,17\\}$. Denote the fixed point
by $x$ and put $V=T_{x}\Sigma$. By Lemma 2.4 and the last paragraph of
subsection 1.2, we conclude that $V$ cannot contain $U_{1}$ as a direct
summand. As $V$ contains an effective module as a direct summand, and $\dim
V>13$ (which has been shown in the first part of the proof), it follows that
$\dim V\geq 8+5+4=17$. On the other hand, by Theorem 2.5, $V$ must contain
$U_{5,1}$ or $U_{6}$ as a direct summand. Thus, by the effectiveness of $V$
and Table 2.1, one of the following possibilities must hold: $V\cong
U_{4,1}\oplus U_{5,1}\oplus W_{8,i}$, $V\cong U_{4,2}\oplus U_{5,1}\oplus
W_{8,i}$ or $V\cong U_{5,1}\oplus W_{12,j}$ for $i\in\\{1,2,3\\}$ and
$j\in\\{1,2\\}$. If $V\cong U_{4,1}\oplus U_{5,1}\oplus W_{8,i}$, then, by
Table 2.1, we have $\dim V^{Q_{8,A}}=1=1+0=\dim V^{Q_{16}}+\dim V^{[24,4]}$
and we can exclude this case by Lemma 2.3. If $V\cong U_{4,2}\oplus
U_{5,1}\oplus W_{8,i}$, then $\dim V^{C_{4,A}}=5=3+2=\dim V^{Q_{8,A}}+\dim
V^{Q_{16}}$, whereas if $V\cong U_{5,1}\oplus W_{12,j}$, then $\dim
V^{C_{4,A}}=2=1+1=\dim V^{Q_{8,A}}+\dim V^{Q_{16}}$. Both situations can be
excluded by the intersection number argument. ∎

######  Theorem 3.2.

_[cf. Theorem 0.2]_ There are no $5$-pseudofree one fixed point actions of
$\operatorname{SL}(2,5).C_{2}$ on $\mathbb{Z}$-homology spheres. In the case
$\operatorname{SL}(2,5).C_{2}$ acts $6$-pseudofreely with exactly one fixed
point on an $n$-dimensional $\mathbb{Z}$-homology sphere $\Sigma$, then
$n=6+8k$ or $n=18+8k$ for $k\geq 0$. Moreover, if the action in question is
effective and $n=6+8k$, then $k\geq 1$.

###### Proof.

Suppose first that $\operatorname{SL}(2,5).C_{2}$ acts $6$-pseudofreely with
exactly one fixed point on some $\mathbb{Z}$-homology sphere $\Sigma$. Let
$\Sigma^{\operatorname{SL}(2,5).C_{2}}=\\{x\\}$ and put $V=T_{x}(\Sigma)$. By
Theorem 3.1, it follows that $V$ contains either $U_{5,1}$ or $U_{6}$ as a
direct summand. On the other hand, by Lemma 2.4, $V$ cannot contain $U_{1}$ as
a direct summand since, by the last paragraph of subsection 1.2, $\dim
U_{1}^{\operatorname{SL}(2,5)}=1$. As $\operatorname{SL}(2,5).C_{2}$ acts
$6$-pseudofreely, it follows by Proposition 1.1 that $V\cong U_{5,1}\oplus W$
or $V\cong U_{6}\oplus W$, where $W$ contains only $W_{8,i}$ and $W_{12,j}$ as
direct summands. The former case can be however excluded by the intersection
number argument as $\dim V^{C_{4,A}}=2=1+1=\dim V^{Q_{8,A}}+\dim V^{Q_{16}}$.
This shows immediately that $\operatorname{SL}(2,5).C_{2}$ cannot act
$5$-pseudofreely. Moreover, if $\operatorname{SL}(2,5).C_{2}$ acts
$6$-pseudofreely, then, by Proposiotion 1.2, $V\cong U_{6}\oplus W$, where $W$
consists only of modules $W_{8,i}$ and at most one of the modules $W_{12,j}$.
This shows $n=6+8k$ or $n=18+8k$ for $k\geq 0$. In case the action in question
is effective and $n=6+8k$, $W$ must by nontrivial and thus $k\geq 1$. This
concludes the proof. ∎

## Acknowledgements

The author would like to thank Prof. Masaharu Morimoto for the suggestion to
consider the problem presented in this article and many helpful comments
during the work on it. I would like to thank also Prof. Krzysztof Pawałowski
for important suggestions which essentially improved the presentation of this
paper. My sincere thanks go as well to Mr. Shunsuke Tamura for enlightening
discussions on the subject.

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fsf
_Faculty of Mathematics and Computer Science_
_Adam Mickiewicz University in Poznań_
_ul Uniwersytetu Poznańskiego 4_
_61-614 Poznań, Poland_
_Email address:_<EMAIL_ADDRESS>