# DexTouch: Learning to Seek and Manipulate Objects
with Tactile Dexterity

Kang-Won Lee$1$, Yuzhe Qin$2$, Xiaolong Wang$2$ and Soo-Chul Lim$1$ $1$
Dongguk University, Seoul, South Korea$2$ University of California San Diego,
CA, USA

###### Abstract

The sense of touch is an essential ability for skillfully performing a variety
of tasks, providing the capacity to search and manipulate objects without
relying on visual information. Extensive research has been conducted over time
to apply these human tactile abilities to robots. In this paper, we introduce
a multi-finger robot system designed to search for and manipulate objects
using the sense of touch without relying on visual information. Randomly
located target objects are searched using tactile sensors, and the objects are
manipulated for tasks that mimic daily-life. The objective of the study is to
endow robots with human-like tactile capabilities. To achieve this, binary
tactile sensors are implemented on one side of the robot hand to minimize the
Sim2Real gap. Training the policy through reinforcement learning in simulation
and transferring the trained policy to the real environment, we demonstrate
that object search and manipulation using tactile sensors is possible even in
an environment without vision information. In addition, an ablation study was
conducted to analyze the effect of tactile information on manipulative tasks.
Our project page is available at https://lee-kangwon.github.io/dextouch/

figure We propose DexTouch, a novel dexterous manipulation robotic system to
perform three types of daily-life tasks with only tactile information. On the
left, we show our manipulation tasks in simulation. A robotic system,
consisting of a UR5e robotic arm and an Allegrohand with 16 attached touch
sensors, was studied in both simulation and real environments. Our trained
policy can be directly deployed in a real robotic arm-hand system and can
successfully perform the tasks.

††
$1$ Dongguk University, Seoul, Republic of Korea
$2$ University of California San Diego, CA, USA
This research was supported by the Technology Innovation Program (Grant No.
20016252) funded by the Ministry of Trade, Industry and Energy (MOTIE, Korea)
and the MOTIE, under the Fostering Global Talents for Innovative Growth
Program (P0017307) supervised by the Korea Institute for Advancement of
Technology (KIAT)

##  I Introduction

Consider a situation of retrieving something from inside of a cabinet or from
a tall cupboard where shelves are not visually observable. Despite the fact
that we cannot see the object in the high cupboard and don’t know its exact
position, we can easily seek and retrieve them. Likewise, we can seek a switch
in a dark room and turn it on, or turn off the light and open the door to
leave. The reason humans can perform these tasks in daily life is that we can
perceive our surroundings or interacting objects through the sense of touch,
even when visual information is unavailable. Human skin has precise tactile
sensors. These allow us to seek, grab, and manipulate objects without relying
on vision by feeling the environment and objects we are interacting with [1].
In general, tactile senses play an important role in manipulation, as there
are many occlusion even in environments with visual information. While these
tasks may seem easy for us, they are not equally feasible for robots. As in
the previous example, when taking an object out of a tall cupboard, we can
estimate its approximate location, but it is difficult to determine the exact
location of the object using only visual information.

To address this issue, many studies have been conducted to apply tactile
sensing to robots. The tactile approach enabled the robot to stably manipulate
objects without damage or slipping, and showed that a method that only detects
touch can also be effectively used to identify and manipulate objects [2].
Although recent literature has shown that the sense of touch is useful for
robot manipulation, it has two limitations. (i) Most current literature
focuses on precision manipulation of objects in the hand. For application to
extended tasks, it is necessary to learn the complex movements of robotic arms
and hands for object manipulation. (ii) Focus on detecting objects with
minimal touches using precise sensors to prevent objects from moving.
Generally objects may move during interaction, so we need to learn behaviors
that can take this into account. Additionally, the ability to seek objects
using the sense of touch shuld be learned during the interaction process.

We present DexTouch in which a tactile manipulation system can explore and
manipulate objects using only the sense of touch, even in environments where
objects move depending on interaction with a robot. As an extension of our
previous research [3], we utilize binary tactile state information, which is
effective in overcoming the Sim2Real gap. We utilize a robotic arm-hand
system. The tactile sensors are attached to one side of the hand, covering the
fingertips, links, and palm, and equipped with 16 the Force Sensing Resistor
(FSR) sensors, as illustrated in Fig. DexTouch: Learning to Seek and
Manipulate Objects  with Tactile Dexterity. Our insight is that touch can
compensate for gaps in information when approximate information about the
interaction is available. As in the previous example of grasping an object
located out of sight, when we know approximate location of the object but do
not know the exact location, the sense of touch can compensate for the loss of
this information and make the task possible.

With this system setup, we focus on solving the task of exploring and
manipulating objects in an environment without vision. To do this, we set
three types of tasks that are relevant to our daily lives. These tasks
typically rely on object pose data obtained from visual sensors. Nonetheless,
in instances of abnormal lighting conditions or the absence of visual sensors,
not having access to the object’s location information can significantly
increase the difficulty of these tasks. We believe that a tactile approach
will be useful in accomplishing these tasks. We learn policies on the IsaacGym
simulator [4] using Reinforcement Learning (RL) to learn complex movements of
the robot hand and arm and the trained policy can be deployed on the real
environments. Our RL policy uses both touch state information and robot
proprioception as input and predicts actions for closed-loop control. Through
the use of touch sensors, our hypothesis is that the policy implicitly learns
how to explore the target objects and perform targeted actions depending on
the task.

In our experiments, we test a real world-system for three types of the tasks.
Through the successful performance of the robot, the proposed method shows
that the touch-based approach can be effectively applied in real environments.
Additionally, we conduct ablations on our system to validate our design,
including tasks such as disabling the touch sensor and adjusting the
sensitivity of the binary force sensor.

##  II Related Work

Dexterous Manipulation Dexterous manipulation allows robots to perform a
variety of tasks, allowing them to be applied to a wide range of fields [5, 6,
7]. Particularly, the implementation of the dexterous manipulative ability of
robots that manipulate objects is receiving a lot of attention [6, 8, 9]. To
address the challenge of dexterous manipulation, model-based approaches to
modeling robots and interaction systems were initially proposed [10, 11].
However, approaches utilizing classical controls are based on designed
dynamical models, which limits their usefulness for scaling to more complex
tasks. To overcome these limitations, methods using deep reinforcement
learning have recently been receiving attention [12, 13, 14, 15]. Effective
learning and high task success rates can be achieved by utilizing large-scale
parallel learning using simulation. Additionally, imitation learning using
expert demonstrations improves sample efficiency and enables natural
manipulation motion generation [16, 17, 18, 19]. However, most manipulation
methods still rely heavily on visual input and thus focus on overcoming
limitations such as occlusion. Additionally, most of these studies have
focused on dexterous in-hand manipulation, often limited to small movements
within the hand. In order to expand to a variety of tasks, complex movements
of not only the fingers but also the arms should be considered in learning.
Compared to these works, our system focuses on learning complex movements of
robotic arms and hands using touch sensors. Through this, we aim to expand
into tasks that mimic daily lives.

Figure 1: Pipeline of the system. Left: visualization of the Allegro hand with
tactile sensors (red indicators) in simulation and real-world. Right:
visualization for all three tasks in simulation and training pipeline. The
blue areas and arrows represent the random range of the target object. The
state contains robot proprioception, tactile information, and task information
like goal position and random range from origin state. Then the policy uses
this state to get actions.

Tactile-based Manipulation The sense of touch contributes much to a human’s
ability in object manipulation. In particular, the sense of touch not only
enables stable object manipulation but can also supplement or replace visual
information depending on the situation [20, 21, 22, 23]. In general, more
accurate tactile sensors offer richer information, enabling finer
manipulation. Consequently, extensive research is concentrated on employing
advanced tactile sensors to achieve fine-grained and precise manipulation [24,
25, 26, 27, 28]. However, utilizing simulators for precise sensors is
challenging due to substantial Sim2Real gaps. To address this challenge,
approaches employing multiple binary force sensors have been proposed,
demonstrating the effective utilization of binary contact signals for object
manipulation, exploration, and recognition [29, 30, 31]. As an extension of
our previous research, binary contact sensors distributed on the hand are
utilized. This approach minimizes the Sim2Real gap that may arise from the
complexity of sensor modeling during transmission. Additionally, research has
recently been conducted on using the sense of touch to determine the location
of objects and to identify and manipulate specific objects [2]. This shows
that it is possible to recognize objects using only tactile feedback. However,
in order to explore objects, the robot moves only along a set trajectory and
operates to minimize the movement of the object when touched. In contrast, our
work does not establish a specific trajectory and allows the robot to learn
movements that can explore objects using touch.

##  III Tactile Robotic System

###  III-A System Setup

Our robot system is a multi degree-of-freedom (DoF) robot system with a total
of 22-DoF, consisting of a 6-DoF UR-5e robot arm and a 16-DoF AllegroHand. The
robot hand has a total of 16 FSR sensors, with 3 sensors attached to each
finger and 4 sensors on the palm, as illustrated by the white dots in Fig 1
and as suggested by [3]. Using an STM32F103 microcontroller, the sensor’s
voltage signal is measured at a sampling rate of 125 Hz, passed through a low-
pass filter to remove noise, and then transmitted to the main controller.
Binary signals are used to reduce the gap between simulation and real robots
and simplify the Sim2Real transfer procedure. For this purpose, preprocessing
is performed to convert the transmitted voltage signal into a binary contact
signal according to a selected threshold $\theta_{th}$ before being used as
input to the control.

IssacGym simulator [4] was used for the training of our tactile based robot
system. The simulation settings for each task are visualized in Fig. DexTouch:
Learning to Seek and Manipulate Objects  with Tactile Dexterity (left). A
multi-degree-of-freedom robotic arm-hand system was implemented in the
simulation, identical to a real robot. Sixteen virtual contact sensors are
configured in the simulation and $\|F\|$ calculated from the net contact force
$F=[F_{x},F_{y},F_{z}]$ measured for each step is used as the contact force.
Then, using a selected threshold $\tilde{\theta}_{th}=0.01$ $\mathrm{N}$ the
binary contact signal is calculated in the same way as in real-world
environments. The control frequency was set to 10 Hz in both real-world
environments and simulations.

Figure 2: The object sets used in grasping object task. The objects in red box
are the unseen objects.

###  III-B Task Description

We study dexterity using a robot’s sense of touch through various types of
object manipulation tasks aimed at exploring and manipulating objects in an
invisible environment. In this paper, we mainly focus on three distinct types
of benchmark tasks that are relevant to our daily lives as illustrated in Fig
1. (i) Grasping Object. In this task, inspired by human behavior of retrieving
items from a high, invisible shelf, the robot is required to utilize its sense
of touch to seek and grasp objects randomly positioned on the table and then
bring them to the target point without dropping them. Furthermore, the policy
should demonstrate that the robot can also manipulate multiple objects with
different shapes. Fifteen objects were selected from the YCB dataset [32], as
depicted in the Fig 2, and experiments were conducted on seven of them in a
real-world environment. Three of these are unseen objects. (ii) Door Opening.
In this task, the robot needs to locate a door handle that is positioned
randomly about the x and y-axes, rotate the door handle, and pull it to open
the door. In an invisible environment, the robot shoud use its sense of touch
to locate door handles and learn complex movements to open the door. (iii)
Valve Rotating. The purpose of this task is to locate the position of a valve
located randomly on a plane and rotate the valve clockwise. Compared to
objects in other tasks, the size of the valve is small, requiring more
sophisticated exploration ability of the robot.

##  IV Learning Tactile Manipulation

###  IV-A Problem Formulation

Our tasks are formulated as a Markov Decision Process (MDP)
$\mathcal{M}=(\mathcal{S},\mathcal{A},\mathcal{R},\mathcal{P})$. Here,
$\mathcal{S}$ is the state space, $\mathcal{A}$ is the action space,
$\mathcal{R}$ is the reward function, and $\mathcal{P}$ is the transition
dynamics. The objective of robot agent is to find optimal policy $\pi$ to
maximize the $\gamma$ discounted return $\sum_{t=0}^{T}\gamma^{t}r_{t}$. The
agent observes state $s_{t}$ at each time step $t$. Then, it take action
$a_{t}=\pi(s_{t})$ calculated by policy $\pi$ and receive a reward
$r_{t}=\mathcal{R}(s_{t},a_{t},s_{t+1})$. During this process, the agent
doesn’t know $\mathcal{R}$ and $\mathcal{P}$. An episode terminates when the
agent exceeds the maximum number of steps $T$ or achieves the goal of task or
reset conditions are achieved.

####  IV-A1 State

The state of the system consists of the joint position
$q_{t}\in\mathbb{R}^{22}$, joint velocity $\dot{q}_{t}\in\mathbb{R}^{22}$ of
6-DoF UR-5e arm and 16-DoF Allegro hand, the binary contact signal
$o_{t}\in\\{0,1\\}^{16}$, 7-Dof palm pose, 3-Dof linear and 3-Dof angular
velocity $p_{plam}\in\mathbb{R}^{13}$, 4-fingertip positions relative to robot
palm $p_{tips}\in\mathbb{R}^{12}$, and task information $I$ for each task. $I$
consists of the goal position $(goal_{x},goal_{y},goal_{z})$ and random range
$(range_{x},range_{y})$ $I\in\mathbb{R}^{5}$ in Grasping Object, random range
$I\in\mathbb{R}^{2}$ in Door Opening, random range $I\in\mathbb{R}^{2}$ in
Valve Rotating.

####  IV-A2 Action

At each step, the action calculated by the policy network is a normalized
vector $a_{t}\in\mathbb{R}^{22}$ that consists of two parts, 6-DoF for robot
arm and 16-DoF for robot hand. For the robot arm, the first 6-D vector was
utilized as a position control command for each joint of the robot arm. In the
case of the robot hand, a joint position controller is employed to command the
position target of its 16 finger joint angles. Then, both the hand and arm are
controlled by PD controllers.

Figure 3: Visualization of the reward for each task. Yellow arrows represent
reward $r_{reach}$ based on the distance between each fingertip and the target
object. Blue arrows represent reward $r_{execute}$ based on manipulative
actions to perform each task. The total reward function is the sum of the two
rewards.

###  IV-B Reward Design

For a successful application of reinforcement learning, the density of rewards
should be enough to facilitate exploration by the agent. At the same time, the
agent should be focused on achieving the goal without being distracted from
taking actions that deviate from the final goal. To this end, a reward
function is proposed that decomposes the task into two phases: reaching the
target object and executing the purposeful manipulation task, as illustrated
in Fig 3. Additionally, inspired by [33], a reward function is designed such
that the agent can only receive rewards for movements toward the goal.

Reaching the target object. To encourage the robot to get closer to the target
object, the reach reward $r_{reach}$ was designed as follows:

$r_{reach}=\underset{finger}{\sum}\alpha_{reach}\times\max(d_{closest}-d,0)$
(1)

This reward is common to all three tasks. Here, $d$ and $d_{closest}$ are both
the distance between each fingertip and the target object. The target object
is the object on the table in Grasping Object, the door handle in Door
Opening, the center position of valve in Valve Rotating. $d$ is the current
distance, and $d_{closest}$ is the closest distance achieved during actions so
far. At the beginning of the episode, set $d_{closest}=d$. $\alpha_{reach}$ is
a relative reward weight.

Executing the manipulation. Since the target manipulation varies for each
task, a manipulation reward $r_{execute}$ was designed according to each task.
First, $r_{execute}$ for Grasping Object task is designed as follows:

$r_{execute}=(1-\mathds{1}_{picked})\times\alpha_{pick}\times
h_{obj}+r_{picked}\\\
+\mathds{1}_{picked}\times\alpha_{goal}\times\max(\tilde{d}_{closest}-\tilde{d},0)$
(2)

In equation 2, the component $r_{execute}$ rewards the agent for picking up an
object, lifting it off the table, and bringing it to the target point. Here,
$\mathds{1}_{picked}$ is an indicator function that activates when the height
of the object $h_{obj}$ relative to the table becomes 1 above the preset
threshold of 10 $\mathrm{cm}$. At this moment the agent receives an additional
bonus reward $r_{picked}$. After lifting an object, the agent does not receive
a reward based on the object’s height. That is, the agent then focuses on
moving the object to the target point. Similar to equation 1, $\tilde{d}$ and
$\tilde{d}_{closest}$ represent the distance between the object and the target
point, where $\tilde{d}$ is the current distance and $\tilde{d}_{closest}$ is
the closest distance achieved during attempt so far. $\alpha_{pick}$ and
$\alpha_{goal}$ are relative reward weights.

For Door Opening task, $r_{execute}$ is designed as follows:

$\displaystyle r_{execute}$
$\displaystyle=(1-\mathds{1}_{rotated})\times\alpha_{rot}\times\max(\phi-\phi_{max},0)$
$\displaystyle+\mathds{1}_{rotated}\times\alpha_{open}\times\max(\psi-\psi_{max},0)$
$\displaystyle+r_{rotated}+r_{opened}$ (3)

In equation 3, the component $r_{execute}$ rewards the agent for actions such
as rotating the door handle and opening the door. Similar to equation 2,
$\mathds{1}_{picked}$ is an indicator function that becomes 1 when the
rotation angle of the door handle exceeds the preset threshold of 1.047
$\mathrm{rad}$ (60 $\mathrm{deg}$). After rotating the door handle beyond the
threshold, the agent can only receive a reward for the action of opening the
door. Therefore, the agent subsequently focuses on the action of opening the
door. $\phi$ and $\phi_{max}$ are the door handle angles, $\phi$ is the
current door handle angle, and $\phi_{max}$ is the maximum door handle angle
achieved during attempt. Similarly, $\psi$ and $\psi_{max}$ are the door
angles, $\psi$ is the current door angle, and $\psi_{max}$ is the maximum door
angle achieved. Agents receive bonus rewards when they complete each sub-task.
When the door handle is rotated beyond the threshold, reward $r_{rotated}$ is
received, and when the door is opened beyond a preset threshold of 0.873
$\mathrm{rad}$ (50 $\mathrm{deg}$), reward $r_{opened}$ is received.
$\alpha_{rot}$ and $\alpha_{open}$ are relative reward weights.

In case of Valve Rotating task, $r_{execute}$ is designed as follows:

$\displaystyle r_{execute}$
$\displaystyle=\alpha_{rot}\times\max(\theta-\theta_{max},0)+r_{success}$ (4)

In equation 4, the component $r_{execute}$ rewards the agent for the action of
rotating the valve. It is similar to the reward for rotating the door handle
in the previous task; however, the difference is that there is no threshold.
Therefore, the more the agent rotates the valve, the greater the reward.
However, when the valve is rotated more than 135 $\mathrm{deg}$, the agent
receives a success bonus. $\alpha_{rot}$ is a relative reward weight.

Finally, to prevent unstable jerky movements of the robot, we add a simple
adjustable penalty containing the L1 norm of the velocity of each joint to
promote smooth movement of the robot agent.

###  IV-C Training Procedure

The policy is trained using experiences simulated in IsaacGym [4], a highly
parallelized GPU-accelerated physics engine. Both the policy and value
networks consist of 3 multi-layer perceptrons (MLPs) with sizes 512, 256, and
128, respectively, and ELU [34] as the activation function. To train the
control policy, the Proximal Policy Optimization (PPO) [35] algorithm is
employed with the following hyper-parameters: clipping $\epsilon=0.2$,
discount factor $\gamma=0.99$ and 0.016 KL threshold. To reduce the difficulty
of learning, asymmetric observation is applied to policy and value networks
[12]. Specifically, the value network includes privileged information such as
the exact pose of the target object, linear and angular velocities, distance
from the robot hand, and physical parameters, and includes information for
calculating rewards such as indicator functions depending on the task.

For the IsaacGym simulation, 4096 parallel environments were used in
$dt=0.01667$ $\mathrm{s}$ with 2 simulation sub-steps. The action generated by
the policy network is executed over 6 steps, corresponding to a control
frequency of 10Hz in real-world.

##  V Experiments

Figure 4: Training process depending on the attachment and sensitivity of the
tactile sensor. The results are averaged on 3 seeds, and the shaded area
indicates the standard deviation. The x-axis is the environment steps. The
y-axis of the upper row is the success rate for each task. The y-axis of the
lower row is the episodic return. Results showed that without tactile sensors,
the task was not successful, and higher sensitivity in the tactile sensor
corresponded to better performance.  TABLE I:  Performance of various methods
on each manipulation task in simulation. The results are averaged over 3
policies trained on 3 seeds.

Method | Grasping Object | Door Opening | Valve Rotation
---|---|---|---
Success rate | Reward | Success rate | Reward | Success rate | Reward
LQ-Sensor | 0.44±0.16 | 699.70±202.58 | 0.45±0.20 | 753.57±213.49 | 0.68±0.14 | 1664.11±312.18
WO-Sensor | 0.0±0.0 | 260.03±12.51 | 0.0±0.0 | 162.23±11.44 | 0.02±0.01 | 442.45±67.84
WO-PInfo | 0.19±0.12 | 405.36±168.77 | 0.17±0.09 | 460.73±137.89 | 0.36±0.13 | 867.57±184.57
Ours | 0.78±0.08 | 972.78±74.58 | 0.73±0.11 | 1060.71±71.27 | 0.89±0.06 | 2199.29±72.17

In this part, our DexTouch system was compared to several baselines in both
the simulation and real-environments. Experiments are conducted on the
proposed tasks, which include grasping objects, opening doors, and rotating
valves as defined in the method. Experiments and analyses are mainly conducted
on the following perspectives: (i) For every task, the task success rate of
the proposed system on randomly positioned target objects is evaluated. The
success rate of each task is used as an evaluation indicator, and comparative
analysis with baselines is performed. In particular, an ablate study is
conducted on how tactile information affects the search of objects in an
environment without vision. (ii) For every task, transfer performance to the
real world is evaluated. As in simulation, the success rate of each task is
evaluated for objects randomly located within a random range. In particular,
we analyze the ability to search for randomly located target objects. Overall,
each task is evaluated based on its success rate and episodic returns. The RL
policy is trained with three different random seeds for each experiment and
transferred to the real world.

###  V-A Baselines

In experiments, our method is mainly compared with the following baselines.

  * 1)

WO-Sensor (Without-Sensor). This PPO policy is learned without any tactile
information from the robot. It should use only the robot’s proprioception to
search target object and learn the interactions required depending on the
task.

  * 2)

LQ-Sensor (Low Quality-Sensor). The threshold of the tactile sensor to detect
touch was set to 0.3 $\mathrm{N}$. In other words, by reducing the sensor’s
sensitivity, a lower-quality tactile sensor is employed to train the PPO
policy. The robot can still be capable of sensing touch, but it requires
greater force when making contact with the object.

  * 3)

WO-PInfo (Without-Privileged Information). This PPO policy is learned without
privileged information such as the exact pose of the target object, linear and
angular velocities, distance from the robot hand, and physical parameters, and
includes information for calculating rewards such as indicator functions
depending on the task.

  * 4)

DA-Sensor (Deactivation-Sensor). Training is conducted following the same
procedure as the proposed method, with the only difference being the
deactivation of the tactile sensor during evaluation. This policy is used for
ablation purposes and to test the extent to which tactile information is used
when applied in a real-world environment.

Figure 5: Visualization of the results of performing each task from a top view
in the real-world experiment. The red arrow indicates the difference in the
position of the target object. Frames illustrating a series of operations
demonstrate that the policy can search randomly located objects and then
manipulate them according to the objectives.

###  V-B Ablation of tactile sensors during policy training

The Fig 4 shows the success rate and episodic return of each task according to
the training process of the proposed method and baselines and Table I shows
the learning results of each method. The findings are highlighted as follows:

(i) In environments without visual information, tactile information is crucial
for object interaction. WO-Sensor, which does not have tactile information,
learns by leveraging the location information known in the initial training
stage (e.g., position of the table in the case of the grasping object task).
However, after that, it fails to learn interactions that involve manipulating
the target object to succeed in the task. In other words, robot can approach
an initial location given by a task, such as a table, but is unable to learn
the behavior of detecting and interacting with objects. These results show
that it is difficult to detect the position of an object during interaction
using a robot’s proprioception alone, and that tactile sensors play an
important role in object detection.

(ii) High-quality tactile sensors enhance the success rates of dexterous
manipulation tasks. As the sensitivity of the tactile sensor decreases, more
force is required for object contact detection. Phenomena such as objects
falling or being thrown have been observed during this process. For example,
in the Grasping Object task, there were instances of objects falling under the
table and resulting in failure. In the Door Opening task, the object was
occasionally lost from the hand due to excessive manipulation of the door
handle. While this allowed for the rotation of the door handle, it posed
challenges in pulling the handle to open the door. Additionally, in the Valve
Rotating task, failures were observed, including excessive force leading to
the valve rotating in the opposite direction or contact with the floor by the
hand. These findings indicate that high-sensitivity tactile sensors are
beneficial for the reliable manipulation of objects.

Furthermore, the findings indicate that high-quality tactile sensors are
beneficial not only for detecting objects through touch but also for
subsequent object manipulation. Table 1 shows the training results of each
method. The high standard deviation of LQ-Sensor indicates a significant
likelihood of task failure during object manipulation. While performing a
task, excessive force may cause objects to slip out of hand or cause
unintended interactions, which may result in task failure. The policy should
consider these movements and manipulate objects with appropriate force during
interactions. The proposed method can perform the task reliably and with a
high success rate compared to LQ-Sensor. This suggests that high-quality
tactile sensors can help infer information about the dynamic movement of
objects and the sophisticated hand movements needed to manipulate them for the
purpose.

###  V-C Real-world evaluation with different sensing capabilities

It is confirmed that the proposed method can achieve superior performance
under ideal simulation settings. In this part, the trained policy is then
transferred to a real robot to ensure that it is equally useful as the
previously demonstrated tactile sensors. In particular, the analysis focuses
on the transferability of the capability to locate randomly positioned target
objects. In the real-world experiment, the identical environment used in the
simulation is recreated, and the trained policy is implemented to control the
robot. The results are shown in Table II.

For the LQ-Sensor, it was observed that more force was applied to the object
to detect object contact, similar to the simulation. Consequently, during the
grasping object task, instances were observed where the object fell off the
table or was inadvertently released from the hand. In the case of DA-Sensor,
it had the lowest success rate. From the initial stage to the reaching stage,
it shows a similar trajectory to the proposed method. This suggests that even
though the robot performs the action of searching for an object, the inability
to utilize the touch sensor prevents it from observing the object, rendering
subsequent manipulation actions challenging.

TABLE II:  Performance of different methods on each manipulation task in real-
world. The results are averaged over 3 policies trained on 3 seeds.

Method | Grasping Object | Door Opening | Valve Rotation
---|---|---|---
LQ-Sensor | 0.32±0.24 | 0.39±0.21 | 0.43±0.25
DA-Sensor | 0.11±0.09 | 0.13±0.07 | 0.16±0.05
Ours |  | Seen: 0.72±0.15
---
Unseen : 0.51±0.27
0.70±0.18 | 0.78±0.13

Fig 5 illustrates the process of manipulating the target object from a top
view. As depicted in the initial stage in Fig 5, target objects are randomly
positioned for the same task. In the reaching stage, the robot searches for
the target object using a tactile sensor as it approaches the object. In all
tasks, the initial reaching process was observed to have similar trajectories.
This suggests that the policy recognizes a random area given for each task and
attempts to search a target object within this area. When an object is
detected within a random area, the robot performs manipulation according to
each task. It is demonstrated that successful manipulation is possible even in
real environments despite the different positions of the objects.

In the case of the grasping object task, it was observed that the point at
which the object was lifted differed depending on when the touch occurred.
This demonstrates that the policy doesn’t merely learn a specific path for
grasping the object but utilizes tactile sensors to locate the object and
subsequently conducts manipulation actions accordingly. In the door opening
task, fingers were observed to bend inward during the reaching stage to search
the door handle. This seems to be an action taken to avoid contact with an
obstacle, such as the door board, rather than the intended target object,
which is the door handle. This observation suggests that the policy tends to
act to avoid external obstacles to concentrate on the target object and
proceed with the search. In the case of the valve rotation task, unlike other
tasks, it was observed that the robot attempted to search and manipulate the
valve using its fingertips. For the LQ-Sensor, the task often failed due to
excessive force applied to the valve during this process. Hence, it is crucial
to employ a high-sensitivity tactile sensor and manipulate it with the
appropriate force. Therefore, these results suggest that the high sensitivity
of tactile sensors is advantageous for dexterous manipulation.

The results demonstrated that policies trained in simulation can be
transferred to the real environment. Even in a real environment, the policy
can seek target objects located randomly within a random area and manipulate
them according to the task. Therefore, it shows that object search and
manipulation using tactile information can be effectively applied in real
environments even without visual information.

##  VI Conclusion

In this paper, a new multi-finger manipulation system is proposed that
searches and manipulates objects through touch without relying on visual
information. Tasks that mimic daily life were defined, and a reinforcement
learning framework was presented to learn complex multi-degree-of-freedom
movements of the robot arm and hand for each task. The proposed system
demonstrated that tactile information can be used to seek and manipulate
randomly located objects. The results show that the ability to utilize tactile
information to seek and manipulate objects is also applicable to robots. This
suggests that objects can be manipulated without relying on visual
information, and proves that tactile information is useful for object
manipulation in environments where visual information is absent. Future
research directions are focused on applying sensors that can provide various
tactile information, such as 3-axis force sensors, and expanding the system to
study generalizability.

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Mathematics Department, Imperial College, London SW7 2AZ<EMAIL_ADDRESS>
Mathematics Department, London School of Economics, Houghton Street, London
WC2A 2AE<EMAIL_ADDRESS>
##  7 Appendix (arXiv only):

To lighten the burden of the main text above, the four sections here serve as
a repository: first for some further examples, second for an independent
derivation of the radiality formula of Theorem 5.2 in the Euclidean case that
illuminates the role of spectral conditions, third for the fulfilment of
earlier promised referrals to some lengthier, straightforward, albeit
necessary, arguments, and lastly to reproduce Sasane’s proof of his Theorem S.

###  7.1 Further Examples

Example 7.1. Take $\mathbb{A=R}^{2}$; then the $\mathbb{A}$-differentiable
solutions are of the form

$S(x)=\mathbf{1}+\rho x=(1+\rho_{1}x_{1},1+\rho_{2}x_{2}).$

Here provided $\rho$ is invertible,
$t1_{\mathbb{A}}=S((t-1)/\rho_{1},(t-1)/\rho_{2}).$

We consider the alternative Fréchet differentiable (but not
$\mathbb{A}$-differentiable) solution given by Th.3.1(ii) (with $d=2$ and
trivial partition)

$\displaystyle S(x)$ $\displaystyle:$
$\displaystyle=(1+\sigma_{1}x_{1}+\sigma_{2}x_{2},1+\sigma_{1}x_{1}+\sigma_{2}x_{2})\in\langle(1,1)\rangle=\langle
1_{\mathbb{A}}\rangle:$ $\displaystyle\gamma(u)$ $\displaystyle=$
$\displaystyle DS(0)u=\left[\begin{array}[]{cc}\sigma_{1}&\sigma_{2}\\\
\sigma_{1}&\sigma_{2}\end{array}\right]u=\left[\begin{array}[]{c}\tau(u)\\\
\tau(u)\end{array}\right]\text{ for
}\tau(u):=\sigma_{1}u_{1}+\sigma_{2}u_{2},$ $\displaystyle\mathcal{N}$
$\displaystyle:$
$\displaystyle=\langle(\sigma_{2},-\sigma_{1})\rangle=\\{u:\tau(u)=0\\}.$

So $e^{\gamma(u)}-1=(e^{\tau(u)}-1,e^{\tau(u)}-1)$ is non-invertible for
$\tau(u)=0,$ implying $N$ is homogeneous on $\mathcal{N}$ and so linear, as we
shall see.

$\rho:=S(\mathbf{1})-\mathbf{1}=(\sigma_{1}+\sigma_{2},\sigma_{1}+\sigma_{2})=(\sigma_{1}+\sigma_{2})(1,1).$

Adjusting $\mathbf{1}+\rho x$ into agreement with $S(x)$ gives

$N(x):=(\sigma_{2}(x_{2}-x_{1}),\sigma_{1}(x_{1}-x_{2}))=(x_{2}-x_{1})(\sigma_{2},-\sigma_{1})\in\mathcal{N},$

which is indeed linear. Then
ran$N=\langle(\sigma_{2},-\sigma_{1})\rangle=\mathcal{N}$, $\ker
N=\langle(1,1)\rangle$ and

$\mathcal{N}=\ker(S^{\prime}(0))=\\{(x_{1},x_{2}):\sigma_{1}x_{1}+\sigma_{2}x_{2}=0\\}=\langle(\sigma_{2},-\sigma_{1})\rangle.$

Tilting here is :

$T(u)=u(e^{\gamma(u)}-1)/\gamma(u)=u\left(\frac{e^{(\sigma_{1}u_{1}+\sigma_{2}u_{2})}-1}{\sigma_{1}u_{1}+\sigma_{2}u_{2}},\frac{e^{(\sigma_{1}u_{1}+\sigma_{2}u_{2})}-1}{\sigma_{1}u_{1}+\sigma_{2}u_{2}}\right)\text{
or }u\text{ if }\tau(u)=0.$

Example 7.2. As in Example 7.1, take $\mathbb{A=R}^{3}$; then the
$\mathbb{A}$-differentiable solutions are of the form

$S(x)=1+\rho x=(1+\rho_{1}x_{1},1+\rho_{2}x_{2},1+\rho_{3}x_{3}).$

We consider the alternative (Fréchet differentiable) solution given by Th.
3.3(iv):

Here, provided $(\sigma_{1},\sigma_{2})\neq 0$ and $\sigma_{3}\neq 0$, if
$(x_{1},x_{2})$ solves $\sigma_{1}x_{1}+\sigma_{2}x_{2}=(t-1)$, then
$t1_{\mathbb{A}}=S(x_{1},x_{2},(t-1)/\sigma_{3}),$ in which case

$(\mathrm{ran}S)u\supseteq\langle 1_{\mathbb{A}}\rangle u=\langle u\rangle.$
$\displaystyle S(x)$ $\displaystyle:$
$\displaystyle=(1+\sigma_{1}x_{1}+\sigma_{2}x_{2},1+\sigma_{1}x_{1}+\sigma_{2}x_{2},1+\sigma_{3}x_{3}):$
$\displaystyle\gamma(u)$ $\displaystyle=$ $\displaystyle
DS(0)u=\left[\begin{array}[]{ccc}\sigma_{1}&\sigma_{2}&0\\\
\sigma_{1}&\sigma_{2}&0\\\
0&0&\sigma_{3}\end{array}\right]u=\left[\begin{array}[]{c}\sigma_{1}u_{1}+\sigma_{2}u_{2}\\\
\sigma_{1}u_{1}+\sigma_{2}u_{2}\\\ \sigma_{3}u_{3}\end{array}\right],$
$\displaystyle(e^{\gamma(u)}-1)/\gamma(u)$ $\displaystyle=$
$\displaystyle\left((e^{\tau(u)}-1)/\tau(u),(e^{\tau(u)}-1)/\tau(u),\frac{e^{\sigma_{3}u_{3}}-1}{\sigma_{3}u_{3}}\right),$
$\displaystyle\mathcal{N}$ $\displaystyle:$
$\displaystyle=\\{(x_{1},x_{2},0):\sigma_{1}x_{1}+\sigma_{2}x_{2}=0\\}=\langle(\sigma_{2},-\sigma_{1},0)\rangle,$
$\displaystyle\rho$ $\displaystyle:$
$\displaystyle=S(\mathbf{1})-\mathbf{1}=(\sigma_{1}+\sigma_{2},\sigma_{1}+\sigma_{2},\sigma_{3}).$

Adjusting $\mathbf{1}+\rho x$ into agreement with $S(x)$ gives

$N(x):=(\sigma_{2}(x_{2}-x_{1}),\sigma_{1}(x_{1}-x_{2}),0)=(x_{2}-x_{1})(\sigma_{2},-\sigma_{1},0)\in\mathcal{N},$

which is linear; then ran$N=\mathcal{N}$, $\ker
N=\langle(1,1,0),(0,0,1)\rangle.$

Example 7.3. (Non-linear adjustor for a degenerate $S$). In the previous
examples the solution function $S$ had an underlying linear form. This is not
so for $\mathbb{A=R}^{2}=\mathbb{G}_{S}^{\ast}(\mathbb{R}^{2}\mathbb{)}$, when
$S(x_{1},x_{2}):=(1,e^{x_{1}}).$ Note that
$e(x)=(0,e^{x_{1}}-1-x_{1})=(0,\frac{1}{2}x_{1}^{2}+...)$ and $e(x\circ
x)=(0,2x_{1}^{2}+...).$ Here

$e(x)/||x||^{2}=\frac{1}{2}\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}^{2}}\rightarrow\text{depends
on }x_{2}/x_{1}$

with variable limit as $x\rightarrow 0.$

$S(x)(0,u)=(0,e^{x_{1}}u):\qquad(\mathrm{ran}S)(0,u)=\langle(0,u)\rangle.$
$\displaystyle S(x_{1},x_{2})$ $\displaystyle:$
$\displaystyle=(1,e^{x_{1}})=\mathbf{1+(}0,e^{x_{1}}-1):$
$\displaystyle\mathcal{N}$ $\displaystyle:$
$\displaystyle=\\{x:S(x)=\mathbf{1}\\}=\\{x:x_{1}=0\\}=\langle e_{2}\rangle,$
$\displaystyle\gamma(u)$ $\displaystyle=$ $\displaystyle
DS(0)u=\left[\begin{array}[]{cc}0&0\\\
1&0\end{array}\right]u=\left[\begin{array}[]{c}0\\\
u_{1}\end{array}\right]=0\text{ iff }u_{1}=0.$ $\displaystyle\rho$
$\displaystyle=$ $\displaystyle S(\mathbf{1})-\mathbf{1}=(0,e-1):\qquad\rho
x=(0,(e-1)x_{2}):$ $\displaystyle N(x_{1},x_{2})$ $\displaystyle:$
$\displaystyle=S(x_{1},x_{2})-\mathbf{1}-\rho
x=(0,(e^{x_{1}}-1)-(e-1)x_{2})\in\mathcal{N}.$

Applying the L’Hospital convention:

$\displaystyle\mathrm{spec}(\gamma(u))$ $\displaystyle=$
$\displaystyle\\{0,u_{1}\\}:\qquad\lambda_{u}(t)=(t,\frac{e^{tu_{1}}-1}{e^{u_{1}}-1}),\text{
or }(t,t)\text{ if }u_{1}=0:$ $\displaystyle N(te_{2})$ $\displaystyle=$
$\displaystyle t\mathbf{1}N(e_{2})=tN(e_{2})\qquad\text{ (for }u_{1}=0);$

but for $u_{1}\neq 0:$

$\displaystyle N((t,e^{t}-1)e_{1})$ $\displaystyle=$
$\displaystyle(t,\frac{e^{t}-1}{e-1})N((1,e^{1}-1)e_{1}):$ $\displaystyle
N(te_{1})$ $\displaystyle=$ $\displaystyle(t,\frac{e^{t}-1}{e-1})N(e_{1}):$
$\displaystyle N(t(1,0))$ $\displaystyle=$
$\displaystyle(0,e^{t}-1)=\frac{e^{t}-1}{e-1}(0,e-1)=\lambda(t)N(1,0).$

Here

$S(x)(0,u)=(0,e^{x_{1}}u):\qquad(\mathrm{ran}S)(0,u)=\langle(0,u)\rangle.$

and tilting is given by

$T(u):=u(e^{\gamma(u)}-1)/\gamma(u)=u(1,\frac{e^{u_{1}}-1}{u_{1}})\text{ or
}u\text{ if }u_{1}=0.$

Example 7.4. $S(x)=(1,1,e^{x_{1}+x_{2}})$. Here
$\mathcal{N}=\\{x:x_{1}+x_{2}=0\\},$ which is 2-dimensional. As in Cor. 4.4:

$\displaystyle S(x)(u_{1},u_{2},u_{3})$ $\displaystyle=$
$\displaystyle(u_{1},u_{2},e^{x_{1}+x_{2}}u_{3}):\qquad u\in(\mathrm{ran}S)u:$
$\displaystyle\qquad(-u)+(\mathrm{ran}S)u$ $\displaystyle\supseteq$
$\displaystyle\\{(0,0,(e^{t}-1)u_{3}):t\in\mathbb{R}\\};$
$\displaystyle(\mathrm{ran}S)\mathcal{N}$ $\displaystyle=$
$\displaystyle\mathcal{N}:\qquad\mathcal{N}\supseteq\\{(0,0,\lambda
u_{3}):\lambda>-1\\}\text{ for }u\in\mathcal{N}\text{ and so
}\mathcal{H}\neq\emptyset.$ $\displaystyle\\{u$ $\displaystyle:$
$\displaystyle(\mathrm{ran}S)u=\\{u\\}\\}=\\{u:u_{3}=0\\}\text{ orthogonal to
}(0,0,1).$ $\displaystyle DS(0)u$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{ccc}0&0&0\\\ 0&0&0\\\
1&1&0\end{array}\right]u=\left[\begin{array}[]{c}0\\\ 0\\\
u_{1}+u_{2}\end{array}\right]=0\text{ iff }u_{1}+u_{2}=0,$
$\displaystyle\lambda_{u}(t)$ $\displaystyle=$
$\displaystyle\left(t,t,\frac{e^{t(u_{1}+u_{2})}-1}{e^{(u_{1}+u_{2})}-1}\right)\text{
or }(t,t,t)\text{ if }u_{1}+u_{2}=0.$

So $\gamma(e_{1}-e_{2})=\gamma(e_{1})-\gamma(e_{2})=0$ and $\gamma(e_{3})=0,$
confirming that $\mathcal{N=\langle}e_{1}-e_{2},e_{3}\mathcal{\rangle}$.

So tilting is given by

$T(u)=u(e^{\gamma(u)}-1)/\gamma(u)=u(1,1,\frac{e^{(u_{1}+u_{2})}-1}{u_{1}+u_{2}})\text{
or }u\text{ if }u_{1}+u_{2}=0.$

We may check this directly:

$\displaystyle\rho$ $\displaystyle=$ $\displaystyle
S(\mathbf{1})-\mathbf{1}=(0,0,e^{2}-1):\qquad\rho x=(e^{2}-1)x_{3}:$
$\displaystyle N(x)$ $\displaystyle=$
$\displaystyle(0,0,(e^{x_{1}+x_{2}}-1)-(e^{2}-1)x_{3}).$

Here

$\displaystyle N(tx)$ $\displaystyle=$
$\displaystyle(0,0,(e^{tx_{1}+tx_{2}}-1)-(e^{2}-1)tx_{3}),$ $\displaystyle
N(tx_{1},-tx_{1},tx_{3})$ $\displaystyle=$
$\displaystyle(0,0,-(e^{2}-1)tx_{3})=tN(x_{1},-x_{1},x_{3}):$ $\displaystyle
N(tx_{1},-tx_{1},0)$ $\displaystyle=$ $\displaystyle tN(x_{1},-x_{1},0)\text{
and }N(te_{3})=tN(e_{3}),$

yielding the two directions along which $N$ is homogenous:

$N(t(1,-1,0)=tN(1,-1,0),\text{ and }N(te_{3})=tN(e_{3}).$

Remark. Along the natural base directions (with $\lambda$ subscript $i$ for
$e_{i}$):

$\lambda_{1}(t)=\lambda_{2}(t)=(t,t,(e^{t}-1)/(e-1))\qquad\lambda_{3}(t)=(t,t,t).$

So

$\displaystyle N(u(e^{t\gamma(u)}-1)/\gamma(u))$ $\displaystyle=$
$\displaystyle\lambda_{u}(t)N(u(e^{\gamma(u)}-1]/\gamma(u)):$ $\displaystyle
N(e_{i}(t,t,(e^{t}-1))$ $\displaystyle=$
$\displaystyle(t,t,(e^{t}-1)/(e-1))N(e_{i}(1,1,(e-1)):\qquad(i=1,2)$
$\displaystyle N(te_{i})$ $\displaystyle=$
$\displaystyle(t,t,(e^{t}-1)/(e-1))N(e_{i})\qquad(i=1,2),$ $\displaystyle
N(te_{3})$ $\displaystyle=$ $\displaystyle t(1,1,1)N(e_{3})=tN(e_{3}).$

###  7.2 Radiality formula in $\mathbb{R}^{d}$

We give a direct proof of Theorem 4.2 when $\mathbb{A}=\mathbb{R}^{d}$. We
begin with two Lemmas. In $\mathbb{R}^{d}$ we will write $x_{i}$ or $(x)_{i}$
for the $i$-th component of $x;$ of course $x_{i}=xe_{i}$ where the $e_{i}$
are the natural base vectors. We write $\gamma(u)=S_{u}^{\prime}(0).$ Recall
that the spectrum of $x$ [Rud, 10.10] is defined by

$\mathrm{spec}(x):=\\{\lambda:(\lambda
1_{\mathbb{A}}-x)\notin\mathbb{A}^{-1}\\}.$

So in $\mathbb{A=R}^{d},$ $\mathrm{spec}(\gamma(u))$ comprises the components
$\gamma(u)_{i}.$

Lemma 7.1. In $\mathbb{R}^{d}$, $e^{\gamma(u)}-1_{\mathbb{A}}$ is invertible
iff $0\notin\mathrm{spec}(\gamma(u)).$

Proof. $\gamma(u)$ has real components, so $\mathrm{spec}(\gamma(d))\subseteq$
$\mathbb{R}$. So by [Rud, 10.28], $e^{\gamma(u)}-1_{A}$ is invertible iff
$e^{\lambda}-1\neq 0$ for all $\lambda\in\mathrm{spec}(\gamma(d))\subseteq$
$\mathbb{R}$ iff $0\notin\mathrm{spec}(\gamma(d)).$ $\square$

Lemma 7.2. In $\mathbb{R}^{d}$, if $S_{u}^{\prime}(0)_{i}=0$ for some $i$ and
$u=e_{j}$, then $S(te_{j})_{i}=1$ for all $t\in\mathbb{R}$, i.e. $\langle
e_{j}\rangle\subseteq\\{x:S(x)_{i}=1\\}.$ In particular, if
$S_{u}^{\prime}(0)_{i}=0,$ for $u=e_{j}$ and for all $i,$ then $S(te_{j})=1$
for all $t\in\mathbb{R}$, i.e. $\langle
e_{j}\rangle\subseteq\mathcal{N}=\\{x:S(x)=1\\}.$

Proof. Fix $i.$ Working in direction $e_{j}$, $(GS)$ implies that for
$s,t\in\mathbb{R}$

$S(se_{j}+tS(se_{j})_{j}e_{j})_{i}=(S(se_{j}))_{i}(S(te_{j}))_{i}.$

This is a pexiderized form of the real-valued univariate version of $(GS)$
(see e.g. [Jab]). Since $t\mapsto S(te_{j})_{j}$ is continuous for all $j,$ it
follows that, for some $\sigma_{ij}\in\mathbb{R}$,

$S(te_{j})_{j}=S(te_{j})_{i}=1+t\sigma_{ij}\qquad(t\in\mathbb{R}).$

Then for $u=e_{j}$

$S(tu)_{i}-S(0)_{i}=t\sigma_{ij}:\qquad\sigma_{ij}=S_{u}^{\prime}(0)_{i}=0.$

So $S(te_{j})_{j}=S(te_{j})_{i}=1+t\sigma_{ij}=1.$ $\square$

Theorem 7.1Euclidean. For $S$ Fréchet differentiable, the solution $(N,S)$ to
the Goldie equation in $\mathbb{R}^{d},$

$N(x+S(x)y)=N(x)+S(x)N(y),$

satisfies, for $u$ any natural base vector, the radiality formulae below (with
the L’Hospital convention); that is, with $\gamma(u)=S_{u}^{\prime}(0)$ and

$\lambda_{u}(t):=[e^{t\gamma(u)}-1]/[e^{\gamma(u)}-1]:\qquad\lambda_{u}(t)_{i}=[e^{t\gamma(u)_{i}}-1]/[e^{\gamma(u)_{i}}-1]\text{,}$
$N(u(e^{t\gamma(u)}-1)/\gamma(u))=\lambda_{u}(t)N(u(e^{\gamma(u)}-1)/\gamma(u)).$

In particular, if $\gamma(u)_{i}=0$ for all $i,$ then

$N(tu)=t\mathbf{1}N(u)=tN(u).$

Proof. Referring again to the polynomials $\wp_{n}(x)=1+x+...+x^{n-1}$ and
rational polynomials $[\wp_{m}/\wp_{n}](x),$ [BinO4] Theorem 2 gives with $N$
here for $K$ there that for any $u$

$N(\wp_{m}(S(u/n))u/n)=[\wp_{m}/\wp_{n}](S(u/n))N(\wp_{n}(S(u/n))u/n).$

By Gateaux differentiability, write

$S(u/n)-S(0)=S_{u}^{\prime}(0)(1/n)+\varepsilon_{n}(u),$

with $n\varepsilon_{n}(u)\rightarrow 0.$ Fix $u=e_{j}.$ We work with odd
integers $n$ below and take $\gamma(u):=S_{u}^{\prime}(0)$ as above and
$\xi_{n}:=S(u/n).$

Case 1. $\gamma(u)_{i}\neq 0$ for all $i.$ Provided $(\xi_{n})_{i}\neq 1$ for
infinitely many (odd) $n:$

$\displaystyle[\wp_{m}/\wp_{n}](\xi_{n})_{i}$ $\displaystyle=$
$\displaystyle[(S(u/n)_{i})^{m}-1]/[(S(u/n)_{i})^{n}-1]$ $\displaystyle=$
$\displaystyle[(1+S_{u}^{\prime}(0)_{i}/n+\varepsilon_{n}(u))^{nt(n)}-1]/[(1+S_{u}^{\prime}(0)_{i}/n+\varepsilon_{n}(u))^{n}-1]$
$\displaystyle\rightarrow$
$\displaystyle[e^{t\gamma(u)_{i}}-1]/[e^{\gamma(u)_{i}}-1],$

as $n\rightarrow\infty$ with $m(n)/n\rightarrow t\in\mathbb{R}.$ If for some
$i,$ $(\xi_{n})_{i}=1$ for all large (odd) $n$, then

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
S(u/n)_{i}-S(0)_{i}=S_{u}^{\prime}(0)_{i}(1/n)+\varepsilon_{n}(u):$
$\displaystyle S_{u}^{\prime}(0)_{i}$ $\displaystyle=$
$\displaystyle-n\varepsilon_{n}(u)\rightarrow 0:\qquad\gamma(u)_{i}=0,$

a contradiction to this case.

Case 2. For some $i,S_{u}^{\prime}(0)_{i}=0.$ Here $u=e_{j},$ so by Lemma 2
$(\xi_{n})_{i}=S(e_{j}/n)_{i}=1$ for all $n;$ then, again as $m/n\rightarrow
t$ and as $\wp_{n}(1)=n,$

$[\wp_{m}/\wp_{n}](\xi_{n})_{i}=\frac{m(n)}{n}\rightarrow t.$

So we may interpret the earlier displayed formula using the L’Hospital
convention. $\square$

###  7.3 Expanded arguments

Below we make good on the promise to set out certain routine arguments.

####  7.3.1 Proof of Prop 1.1

(i) $\mathbb{G}_{S}^{\ast}(\mathbb{A})$ is closed under the operation
$\circ_{S}$ since

$S(a\circ_{S}b)=S(a)S(b),$

so that $S(a)S(b)$ is invertible when $S(a)$ and $S(b)$ are invertible. Note
that for $a\in\mathbb{G}_{S}(\mathbb{A})$

$a=a\circ_{S}b=a+S(a)b\text{ iff }b=0.$

Also for $a\in\mathbb{G}_{S},$ as $S(a)$ is invertible,

$S(a)=S(a\circ_{S}0)=S(a)S(0):\quad S(0)=1_{\mathbb{A}}.$

Thus the neutral element for $\circ_{S},$ i.e. $0$ is in
$\mathbb{G}_{S}^{\ast}$. The $\circ_{S}$-inverse of $a$ is $b=-aS(a)^{-1}$. As
in [Jav], the operation is associative:

$\displaystyle(a\circ_{S}b)\circ_{S}c$ $\displaystyle=$
$\displaystyle(a+S(a)b)+S(a+S(a)b)c=a+S(a)b+S(a)S(b)c,$ $\displaystyle
a\circ_{S}(b\circ_{S}c)$ $\displaystyle=$ $\displaystyle
a+S(a)[b+S(b)c]=a+S(a)b+S(a)S(b)c.$

As the elements of $\mathbb{A}_{1}$ are invertible,
$\mathbb{G}_{S}:=S^{-1}(\mathbb{A}_{1})\subseteq\mathbb{G}_{S}^{\ast}$. As
$S(0)=1_{\mathbb{A}},$ $0\in\mathbb{G}_{S},$ and furthermore for
$a,b\in\mathbb{G}_{S}:$ as $S(a),S(b)\in\mathbb{A}_{1},$

$S(a\circ_{S}b)=S(a)S(b)\in\mathbb{A}_{1}$

(as $\mathbb{A}_{1}$ is a multiplicative group), so
$a\circ_{S}b\in\mathbb{G}_{S},$ and also $a_{S}^{-1}\in\mathbb{G}_{S}$ because

$S(a_{S}^{-1})=S(a)^{-1}\in\mathbb{A}_{1},$

since

$1_{\mathbb{A}}=S(0)=S(a\circ_{S}a_{S}^{-1})=S(a)S(a_{S}^{-1}).$

So $\mathbb{G}_{S}$ is a subgroup of $\mathbb{G}_{S}^{\ast}.$

(ii) Evidently, for $a,b\in\mathcal{N}$

$a+b=a+S(a)b=a\circ_{S}b.$

So if $a,b\in\mathcal{N},$ then $a+b\in\mathcal{N}$ and $-a\in\mathcal{N},$
since

$\displaystyle S(a+b)$ $\displaystyle=$ $\displaystyle
S(a+S(a)b)=S(a)S(b)=1_{\mathbb{A}},$ $\displaystyle 1_{\mathbb{A}}$
$\displaystyle=$ $\displaystyle S(0)=S(a-a)=S(a-S(a)a)=S(a)S(-a)=S(-a).$

Also, for $a\in\mathcal{N},$ $-a=-aS(-a)=-aS(a)^{-1}$ is the
$\circ_{S}$-inverse of $a,$ so $\mathcal{N}$ is both an additive subgroup of
$\mathbb{A}$ and a subgroup of $\mathbb{G}_{S}^{\ast}.$

####  7.3.2 Proof of Th. 2.2

Below we use the notation $a\circ b:=a+bS(a)$ on
$\mathbb{G}_{S}:=(a\in\mathbb{A}:S(a)\in\mathbb{A}^{-1}\\}.$ Differentiation
of $(GS)$ w.r.t. to $b$ gives

$S^{\prime}(a+bS(a))S(a)=S(a)S^{\prime}(b);$

now take $a=b_{S}^{-1}$ so $S(a)^{-1}=S(b)$ to obtain the similarity
relations:

$\displaystyle S(a)^{-1}S^{\prime}(0)S(a)$ $\displaystyle=$ $\displaystyle
S^{\prime}(b):$ $\displaystyle S(b)\gamma S(b)^{-1}$ $\displaystyle=$
$\displaystyle S^{\prime}(b).$

Below we repeatedly write

$S(a+h)=S(a)+S^{\prime}(a)h+o(h).$

We consider $(GS)$ after $a$ and $b$ are equally incremented by $h:$

$\displaystyle S((a+h)+(b+h)(S(a)+S^{\prime}(a)h+o(h))$ $\displaystyle=$
$\displaystyle S((a+bS(a))+h+bS^{\prime}(a)h+hS(a)+o(h))$ $\displaystyle=$
$\displaystyle S(a\circ b)+S^{\prime}(a\circ
b)[h+bS^{\prime}(a)h+S(a)h+o(h)]+o(h)$ $\displaystyle=$
$\displaystyle(S(a)+S^{\prime}(a)h+o(h))(S(b)+S^{\prime}(b)h+o(h))$
$\displaystyle=$ $\displaystyle
S(a)S(b)+S(a)S^{\prime}(b)h+S(b)S^{\prime}(a)h+o(h).$

Comparison of the two sided gives to within $o(h)$

$S^{\prime}(a\circ
b)[1+bS^{\prime}(a)+S(a)]=S(a)S^{\prime}(b)+S(b)S^{\prime}(a).$

Applying the similarity relations gives

$\displaystyle S(a\circ b)\gamma S(a\circ b)^{-1}[1+bS^{\prime}(a)+S(a)]$
$\displaystyle=$ $\displaystyle S(a)S(b)\gamma S(b)^{-1}+S(b)S(a)\gamma
S(a)^{-1}.$

Cancelling $S(a)S(b)$ on the left-hand side$:$

$\gamma S(a)^{-1}S(b)^{-1}[1_{\mathbb{A}}+bS(a)\gamma S(a)^{-1}+S(a)]=\gamma
S(b)^{-1}+\gamma S(a)^{-1}.$

Absorbing $S(a)^{-1}$ on the left-hand side:

$\gamma S(b)^{-1}[S(a)^{-1}+b\gamma S(a)^{-1}+1_{\mathbb{A}}]=\gamma
S(b)^{-1}+\gamma S(a)^{-1}.$

Cancelling the term $\gamma S(b)^{-1}$ appearing on each side:

$\gamma[S(b)^{-1}S(a)^{-1}+S(b)^{-1}b\gamma S(a)^{-1}]=\gamma S(a)^{-1}.$

Cancelling $S(a)^{-1}$ on the right on both sides gives:

$\gamma[S(b_{s}^{-1})-b_{s}^{-1}\gamma]=\gamma,$

with $b_{s}^{-1}$ the $\circ_{S}$-inverse of $b.$ Put $c=b_{s}^{-1};$ then, on
rearranging,

$\gamma(S(c)h)=\gamma(h)+\gamma(c\gamma h).$ ($\ast^{\prime}$)

All the steps above are reversible. We now improve on the last equation.

Differentiating $(\ast^{\prime})$ with respect to $c$ in direction $k:$

$\gamma(S^{\prime}(c)(k)h)=\gamma(k\gamma h).$

Using the similarity relations gives

$\gamma(S(c)\gamma S(c)^{-1}(k)h)=\gamma(k\gamma h),$

which for $a=S(c)$ yields the claim $(\ast\ast\ast)$ and for $c=0$ yields the
claim $(\ast\ast).$

Writing $S(c)k$ for $k$ in the last equation gives

$\gamma(S(c)\gamma(k)h)=\gamma(S(c)k\gamma h).$ ($\sharp$)

It now follows from $(\ast^{\prime})$ and $(\ast\ast)\ $that

$\gamma(S(c)h)=\gamma(h)+\gamma(c\gamma
h)=\gamma(h)+\gamma(h\gamma(c))=\gamma(h(1+\gamma(c)).$

That is, $(\ast)\ $holds. Evidently $(\ast)$ and $(\ast\ast)$ yield
$(\ast^{\prime}),$ and so the conjuction of $(\ast)$ and $(\ast\ast)$ yields
$(GS).$

It is immediate from $(\ast)$ that

$\gamma(S(c))=\gamma(1_{\mathbb{A}}+\gamma(c)).$

Furthermore, $(\ast\ast)$ with $k=1_{\mathbb{A}}$ yields

$\gamma(\gamma(h))=\gamma(\gamma(1_{\mathbb{A}})h).$

Differentiating $(\natural)$ with respect to $c$ in direction $u$ and setting
$c=0$ gives

$\gamma(\gamma(u)\gamma(k)h)=\gamma(\gamma(u)k\gamma(h)).$

Taking $k=1_{\mathbb{A}}$ yields

$\displaystyle\gamma(\gamma(1_{\mathbb{A}})\gamma(u)h)$ $\displaystyle=$
$\displaystyle\gamma(\gamma(u)\gamma(h)):$
$\displaystyle\gamma(\gamma(\gamma(u)h)$ $\displaystyle=$
$\displaystyle\gamma(\gamma(u)\gamma(h)).\qquad\qquad\qquad\square$

####  7.3.3 Proof of Prop. 2.2.

Assume the weaker homogeneity property and suppose for some $n$ that

$\gamma(x\gamma(x)^{n})=\gamma(x)^{n+1},$

which is valid for $n=0;$ the inductive step is provided by

$\gamma(x\gamma(x)^{n+1})=\gamma((x\gamma(x)^{n}).\gamma(x))=\gamma(x\gamma(x)^{n}).\gamma(x)=\gamma(x)^{n+1}\gamma(x).$

Conversely, the case of $(\times)$ for $k$ implies for $y=x\gamma(x)^{k}$ that

$\displaystyle\gamma(y\gamma(x)$ $\displaystyle=$
$\displaystyle\gamma(x\gamma(x)^{k}\gamma(x))=\gamma(x\gamma(x)^{k+1})=\gamma(x)^{k+1}\gamma(x)$
$\displaystyle=$
$\displaystyle\gamma(x\gamma(x)^{k})\gamma(x)=\gamma(y)\gamma(x).\qquad\qquad\qquad\qquad\square$

####  7.3.4 Proof of Proposition 2.3.

We first prove the case $k=1$ of $(\times),$ which case also establishes the
converse. By linearity of $\gamma,$ it is enough to prove this identity for
any $u$ of norm $1.$ Fix $u$ of norm 1, and take $x=tu$ with
$t=||x||\rightarrow 0$. Substitution into $GS$ leads to

$\gamma(x\gamma(x))-\gamma(x)^{2}+\gamma(xe(x))-2e(x)\gamma(x)=2e(x)-e(x\circ
x).$ (†)

Since $e(x)=o(x),$

$||\gamma(xe(x))||/||x||^{2}=||\gamma(u\cdot
e(x)/||x||)||\leq||\gamma||.||u||.(||e(x)/||x||)\rightarrow 0.$

Furthermore,

$||x\circ x||/||x||=||2u+tu\gamma(u)+ue(tu)||\rightarrow 2,$

and so RHS of (†) gives

$[2e(x)-e(x\circ x)]/||x||^{2}=2e(x)/||x||^{2}-\frac{||x\circ
x||^{2}}{||x||^{2}}\cdot e(x\circ x)/||x\circ x||^{2}\rightarrow 0.$

Hence

$\gamma(u\gamma(u))=\gamma(u)^{2},$

and holds for all $u.$ This last identity implies the converse: from
$({\dagger}),$

$\gamma(u\gamma(u))-\gamma(u)^{2}+\gamma(ue(tu)/t)-2\gamma(u)e(tu)/t=\frac{2e(x)}{t^{2}}(1_{\mathbb{A}}-\frac{e(x\circ
x)}{2e(x)}).$

Passage to the limit yields $e(x)/||x||^{2}\rightarrow 0.$

To deduce the cases $k\geq 2$ of $(\times),$ it again suffices to establish
them for $u$ of norm $1$. Comparing the two sides of $S(x\circ y)=S(x)S(y)$
yields:

$\displaystyle\gamma(y\gamma(x))-\gamma(x)\gamma(y)+\gamma(ye(x))-e(x)\gamma(y)-e(y)\gamma(x)$
$\displaystyle=$ $\displaystyle e(x)+e(y)-e(x\circ y).$

With $x=tu$ and $y=x\gamma(x)^{k}=t^{k+1}u\gamma(u)^{k},$ the analogous
estimates of ratios against $t^{2}$ are similar and easier because the power
of $t$ in $y$ (strictly) exceeds $2$ and $e(x)=o(x).$ Note that $||x\circ
y||/t\rightarrow 1.$ $\square$

####  7.3.5 Proof of Th. 3.2 (Exhaustivity)

Put $s(x,y)=(s_{1}(x,y),s_{2}(x,y))=(\sigma(x,y),\tau(x,y));$ evaluating
components on right- and left-hand sides, the two of the right are

$RHS_{1}=\sigma(a_{1},a_{2})\sigma(b_{1},b_{2}):\qquad
RHS_{2}=\tau(a_{1},a_{2})\tau(b_{1},b_{2}).$

Likewise the first component on the left is

$LHS_{1}=\sigma(a+s(a)b)=\sigma(a_{1}+\sigma(a)b_{1},a_{2}+\tau(a)b_{2}).$

So

$\sigma(a_{1}+\sigma(a)b_{1},a_{2}+\tau(a)b_{2})=\sigma(a_{1},a_{2})\sigma(b_{1},b_{2}).$
($S1$)

Similarly,

$\tau(a_{1}+\sigma(a)b_{1},a_{2}+\tau(a)b_{2})=\tau(a_{1},a_{2})\tau(b_{1},b_{2}).$
($S2$)

In $(S1),$ taking $a_{2}=b_{2}=0$ gives

$\sigma(a_{1}+\sigma(a_{1},0)b_{1},0)=\sigma(a_{1},0)\sigma(b_{1},0).$

So $\varphi(x):=\sigma(x,0)$ solves the standard $(GS)$ equation

$\varphi(x+y\varphi(x))=\varphi(x)\varphi(y),$

and so for some $\sigma_{1}$

$\sigma(x,0)=1+\sigma_{1}x.$

Likewise, working with $(S2)$ with $a_{1}=b_{1}=0$ gives

$\displaystyle\tau(0,a_{2}+\tau(0,a_{2})b_{2})$ $\displaystyle=$
$\displaystyle\tau(0,a_{2})\tau(0,b_{2}):$ $\displaystyle\tau(0,y)$
$\displaystyle=$ $\displaystyle 1+\tau_{2}y.$

Now taking $a_{2}=b_{2}=0$ in $(S2)$ gives

$\tau(a_{1}+\sigma(a_{1},0)b_{1},0)=\tau(a_{1},0)\tau(b_{1},0).$

So $\varphi(x):=\tau(x,0)$ and $h(x):=\sigma(x,0)$ solve the pexiderized
version of the (scalar) $(GS):$

$\varphi(x+h(x)y)=\varphi(x)\varphi(y),$

for which see [Jab, Cor. 1(iv)]. Here there are two possibilities

$\displaystyle h$ $\displaystyle=$ $\displaystyle 1\text{ and
}\varphi=e^{\gamma x},$ $\displaystyle h$ $\displaystyle=$ $\displaystyle
1+cx\text{ and }\varphi=(1+cx)^{\gamma}.$

That is,

$\displaystyle\sigma(x,0)$ $\displaystyle=$ $\displaystyle
1+\sigma_{1}x=1\text{ and }\tau(x,0)=e^{\gamma x},\text{ i.e. }\sigma_{1}=0.$
$\displaystyle\sigma(x,0)$ $\displaystyle=$ $\displaystyle
1+\sigma_{1}x=1+cx\text{ and }\tau(x,0)=(1+cx)^{\gamma}.$

The latter case should be interpreted as either $\gamma\neq 0$ and
$\tau_{1}=c=\sigma_{1},$ or $\gamma=0$ and so $\tau_{1}=0.$

Also taking $a_{1}=b_{1}=0$ in $(S1)$ gives

$\sigma(0,a_{2}+\tau(0,a_{2})b_{2})=\sigma(0,a_{2})\sigma(0,b_{2}).$

So $\varphi(x):=\sigma(0,y)$ and $h(x):=\tau(0,y)$ solve the pexiderized
version

$\varphi(x+h(x)y)=\varphi(x)\varphi(y).$

Here again there are two possibilities:

$\displaystyle\tau(0,y)$ $\displaystyle=$ $\displaystyle 1+\tau_{2}y=1\text{
and }\sigma(0,y)=e^{\gamma y},\text{i.e. }\tau_{2}=0;$
$\displaystyle\tau(0,y)$ $\displaystyle=$ $\displaystyle
1+\tau_{2}y=1+cy\text{ and }\sigma(0,y)=(1+cy)^{\gamma},\text{ i.e.
}\sigma(0,y)=(1+\tau_{2}y)^{\gamma}.$

In summary,

$\sigma(x,0)=1+\sigma_{1}x,\qquad\tau(0,y)=1+\tau_{2}y,$ (summary)

and the following ‘side’ conditions hold.

If $\tau_{2}\neq 0,$ then $\sigma_{2}:=\tau_{2}$ and
$\sigma(0,y)=(1+\sigma_{2}y)^{\gamma};$ otherwise $\sigma(0,y)=e^{\gamma y}.$

If $\sigma_{1}\neq 0,$ then $\tau_{1}:=\sigma_{1}$ and
$\tau(x,0)=(1+\tau_{1}x)^{\delta};$ otherwise $\tau(x,0)=e^{\delta x}.$

Finally in $(S1)$ and in $(S2)$ take $a_{2}=b_{1}=0.$ Then

$\sigma(a_{1},\tau(a_{1},0)b_{2})=\sigma(a_{1},0)\sigma(0,b_{2})\qquad\tau(a_{1},\tau(a_{1},0)b_{2})=\tau(a_{1},0)\tau(0,b_{2}).$

Thus

$\displaystyle\tau(x,y)$ $\displaystyle=$
$\displaystyle\tau(x,0)(1+\tau_{2}y/\tau(x,0))\text{\quad(taking
}y=\tau(x,0)b_{2}\text{)}$ $\displaystyle=$
$\displaystyle\tau(x,0)+\tau_{2}y=\left\\{\begin{array}[]{cc}(1+\tau_{1}x)^{\delta}+\tau_{2}y&\text{if
}\tau_{1}=\sigma_{1}\neq 0\\\ e^{\delta x}+\tau_{2}y&\text{if
}\sigma_{1}=0.\end{array}\right.$

Similarly

$\displaystyle\sigma(a_{1},\tau(a_{1},0)b_{2})$ $\displaystyle=$
$\displaystyle\sigma(a_{1},0)\sigma(0,b_{2})$
$\displaystyle\sigma(x,\tau(x,0)b_{2})$ $\displaystyle=$
$\displaystyle(1+\sigma_{1}x)\sigma(0,y/\tau(x,0))\text{ (take
}y=\tau(x,0)b_{2}\text{)}$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{cc}(1+\sigma_{1}x)(1+\sigma_{2}y/\tau(x,0))^{\gamma}&\text{if
}\sigma_{2}=\tau_{2}\neq 0\\\ (1+\sigma_{1}x)e^{\gamma y/\tau(x,0)}&\text{if
}\sigma_{2}=0\end{array}\right.,$

$=\left\\{\begin{array}[]{cc}((1+\sigma_{1}x)^{1/\gamma}+\sigma_{2}y(1+\sigma_{1}x)^{[1/\gamma-\delta]})^{\gamma}&\text{if
}\tau_{1}=\sigma_{1}\neq 0\text{ \& }\sigma_{2}=\tau_{2}\neq 0\\\
(1+\sigma_{2}ye^{-\delta x})^{\gamma}&\text{if }\sigma_{1}=0\text{ \&
}\sigma_{2}=\tau_{2}\neq 0\\\ e^{\gamma ye^{-\delta x}}&\text{if
}\sigma_{1}=\sigma_{2}=0\\\ (1+\sigma_{1}x)e^{\gamma
y(1+\sigma_{1}x)^{-\delta}}&\text{if }\tau_{1}=\sigma_{1}\neq 0\text{ \&
}\sigma_{2}=0.\end{array}\right..$

We examine all the possible cases. The first and second special cases below
emerge as the only viable ones, i.e. leading to a solution. The remaining
general cases turn out to be impossible on asymptotic growth grounds.

First Special Case: Take $\sigma_{1}\neq 0\neq\sigma_{2}$ and
$\gamma=\delta=1;$ then $\sigma(x,0)=\tau(x,0)=1+\sigma_{1}x$ and
$\sigma(0,y)=\tau(0,y)=1+\sigma_{2}y$ with $:$

$\displaystyle\sigma(a_{1},(1+\sigma_{1}a_{1})b_{2})$ $\displaystyle=$
$\displaystyle(1+\sigma_{1}a_{1})(1+\sigma_{2}b_{2})\qquad\text{(}z-b_{2}:=\sigma_{1}a_{1}b_{2}\text{),}$
$\displaystyle\sigma(a_{1},z)$ $\displaystyle=$ $\displaystyle
1+\sigma_{1}a_{1}+\sigma_{2}b_{2}+\sigma_{1}\sigma_{2}a_{1}b_{2}$
$\displaystyle=$ $\displaystyle
1+\sigma_{1}a_{1}+\sigma_{2}b_{2}+\sigma_{2}[z-b_{2}]$ $\displaystyle=$
$\displaystyle 1+\sigma_{1}a_{1}+\sigma_{2}z.$

We verify that this formula with $\sigma=\tau$ satisfies $(GS).$ It will
suffice to check the first component:

$\displaystyle LHS_{1}$ $\displaystyle=$
$\displaystyle\sigma(a_{1}+\sigma(a)b_{1},a_{2}+\tau(a)b_{2})$
$\displaystyle=$ $\displaystyle
1+\rho_{1}(a_{1}+(1+\rho_{1}a_{1}+\rho_{2}a_{2})b_{1})+\rho_{2}(a_{2}+(1+\rho_{1}a_{1}+\rho_{2}a_{2})b_{2})$
$\displaystyle=$ $\displaystyle
1+\rho_{1}a_{1}+\rho_{1}b_{1}+\rho_{1}^{2}a_{1}b_{1}+\rho_{1}\rho_{2}a_{2}b_{1}$
$\displaystyle+\rho_{2}a_{2}+\rho_{2}b_{2}+\rho_{2}\rho_{1}a_{1}b_{2}+\rho_{2}^{2}a_{2}b_{2}$
$\displaystyle=$ $\displaystyle
1+\rho_{1}a_{1}+\rho_{2}a_{2}+\rho_{1}b_{1}+\rho_{2}b_{2}+(\rho_{1}a_{1}+\rho_{2}a_{2})(\rho_{1}b_{1}+\rho_{2}b_{2}).$

This agrees with

$\displaystyle RHS_{1}$ $\displaystyle=$
$\displaystyle(1+\rho_{1}a_{1}+\rho_{2}a_{2})(1+\rho_{1}b_{1}+\rho_{2}b_{2})$
$\displaystyle=$ $\displaystyle
1+\rho_{1}a_{1}+\rho_{2}a_{2}+\rho_{1}b_{1}+\rho_{2}b_{2}$
$\displaystyle+(\rho_{1}a_{1}+\rho_{2}a_{2})(\rho_{1}b_{1}+\rho_{2}b_{2}).$

The calculation of the second component is exactly the same in these
circumstances.

Second Special Case. Take $\tau_{2}\neq 0,$ and $\gamma=0,$ equivalently
$\sigma_{2}=0.$ Then $\sigma(x,y)=1+\sigma_{1}x.$ Take $\delta=0,$ then
$\tau(x,0)=1,$ equivalently $\tau_{1}=0.$ Then $\tau(x,y)=1+\tau_{2}y.$ We
check the proposed solution:

$s(x_{1},x_{2})=(1+\sigma_{1}x_{1},1+\tau_{2}x_{2}).$

The first component on the left is

$\displaystyle LHS_{1}$ $\displaystyle=$
$\displaystyle\sigma(a_{1}+\sigma(a)b_{1},a_{2}+\tau(a)b_{2})$
$\displaystyle=$ $\displaystyle
1+\sigma_{1}(a_{1}+b_{1}+\sigma_{1}a_{1}b_{1})=1+\sigma_{1}a_{1}+\sigma_{1}b_{1}+\sigma_{1}^{2}a_{1}b_{1},$

and this matches

$\displaystyle RHS_{1}$ $\displaystyle=$
$\displaystyle\sigma(a_{1},a_{2})\sigma(b_{1},b_{2})$ $\displaystyle=$
$\displaystyle(1+\sigma_{1}a_{1})(1+\sigma_{1}b_{1})$ $\displaystyle=$
$\displaystyle 1+\sigma_{1}a_{1}+\sigma_{1}b_{1}+\sigma_{1}^{2}a_{1}b_{1}.$

The calculation of the second component (with $\tau$) is similar.

It turns out that these special cases are the only possible ones. So it now
remains to eliminate the remaining (ostensibly ‘general’) cases.

General Cases (non-viable).

We first compute $\sigma,$ according to the side condition below $($summary$)$
above, treating it ‘disjunctively’ as Cases A and B.

Case A (If $\tau_{2}\neq 0$ then $\sigma_{2}=\tau_{2}$ and
$\sigma(0,y)=(1+\sigma_{2}y)^{\gamma}.).$ Here

$\sigma(\sigma(0,a_{2})b_{1},a_{2})=\sigma(0,a_{2})\sigma(b_{1},0).$

Substitution under this case yields

$\displaystyle\sigma((1+\sigma_{2}y)^{\gamma}b_{1},y)$ $\displaystyle=$
$\displaystyle(1+\sigma_{2}y)^{\gamma}(1+\sigma_{1}b_{1})$
$\displaystyle\sigma(x,y)$ $\displaystyle=$
$\displaystyle\sigma_{1}x+(1+\sigma_{2}y)^{\gamma}\text{ with
}\sigma_{2}=\tau_{2}\neq 0.$

Case B ($\sigma(0,y)=e^{\gamma y}).$ Substituting as in Case A

$\displaystyle\sigma(e^{\gamma y}b_{1},y)$ $\displaystyle=$ $\displaystyle
e^{\gamma y}(1+\sigma_{1}b_{1})=e^{\gamma y}+\sigma_{1}e^{\gamma y}b_{1}$
$\displaystyle\sigma(x,y)$ $\displaystyle=$ $\displaystyle e^{\gamma
y}(1+\sigma_{1}b_{1})=\sigma_{1}x+e^{\gamma y}\text{ with }\tau_{2}=0.$

Next we compute $\tau$ analogously to $\sigma$, again ‘disjunctively’.

Case A (If $\sigma_{1}\neq 0$ then $\sigma_{1}=\tau_{1}$ and
$\tau(x,0)=(1+\sigma_{1}x)^{\delta}).$ Substituting under this case into

$\tau(a_{1},\tau(a_{1},0)b_{2})=\tau(a_{1},0)\tau(0,b_{2})$

yields

$\displaystyle\tau(x,(1+\sigma_{1}x)^{\delta}b_{2})$ $\displaystyle=$
$\displaystyle(1+\sigma_{1}x)^{\delta}(1+\tau_{2}b_{2})$
$\displaystyle\tau(x,y)$ $\displaystyle=$
$\displaystyle(1+\sigma_{1}x)^{\delta}+\tau_{2}b_{2}(1+\sigma_{1}x)^{\delta}$
$\displaystyle=$ $\displaystyle(1+\tau_{1}x)^{\delta}+\tau_{2}y\text{ with
}\tau_{1}:=\sigma_{1}\neq 0.$

Case B ($\tau(x,0)=e^{\delta x}).$ Substituting as in the preceeding Case A
yields:

$\displaystyle\tau(x,e^{\delta x}b_{2})$ $\displaystyle=$
$\displaystyle\tau(x,0)\tau(0,b_{2})=e^{\delta x}(1+\tau_{2}b_{2})$
$\displaystyle\tau(x,y)$ $\displaystyle=$ $\displaystyle e^{\delta
x}+\tau_{2}b_{2}e^{\delta x}$ $\displaystyle=$ $\displaystyle e^{\delta
x}+\tau_{2}y\text{ with }\tau_{1}=\sigma_{1}=0.$

In summary, we have the following possibilities for $\sigma$ and $\tau.$

$\sigma(x_{1},x_{2})=\left\\{\begin{array}[]{cc}\sigma_{1}x_{1}+(1+\sigma_{2}x_{2})^{\gamma}&\text{with
}\sigma_{2}=\tau_{2}\neq 0,\\\ \sigma_{1}x_{1}+e^{\gamma x_{2}}&\text{with
}\sigma_{2}=\tau_{2}=0,\end{array}\right.$
$\tau(x_{1},x_{2})=\left\\{\begin{array}[]{cc}(1+\tau_{1}x_{1})^{\delta}+\tau_{2}x_{2}&\text{with
}\tau_{1}:=\sigma_{1}\neq 0,\\\ e^{\delta x_{1}}+\tau_{2}x_{2}&\text{ with
}\tau_{1}=\sigma_{1}=0.\end{array}\right.$

We now rule out all four possible pairings of $\sigma$ and $\tau$ by their
asymptotic behaviour for large values of the arguments on both sides of the
first component equation $(S1).$ This last asserts that

$\sigma(a_{1}+\sigma(a)b_{1},a_{2}+\tau(a)b_{2})=\sigma(a_{1},a_{2})\sigma(b_{1},b_{2}).$

Case 1. Consider pairing the first choices available to $\sigma$ and $\tau.$

$\sigma(x)=\sigma_{1}x_{1}+(1+\sigma_{2}x_{2})^{\gamma},\tau(x)=(1+\tau_{1}x_{1})^{\delta}+\tau_{2}x_{2}.$

On the LHS of $(S1)$ the two arguments for $\sigma$ are

$x_{1}:=a_{1}+b_{1}[\sigma_{1}a_{1}+(1+\sigma_{2}a_{2})^{\gamma}],\quad
x_{2}:=a_{2}+b_{2}(1+\tau_{1}a_{1})^{\delta}+\tau_{2}a_{2}b_{2}.$

So

$\displaystyle LHS$ $\displaystyle=$
$\displaystyle\sigma_{1}x_{1}+(1+\sigma_{2}x_{2})^{\gamma}$ $\displaystyle=$
$\displaystyle\sigma_{1}a_{1}+\sigma_{1}\sigma_{1}a_{1}b_{1}+\sigma_{1}b_{1}(1+\sigma_{2}a_{2})^{\gamma}$
$\displaystyle+(1+\sigma_{2}a_{2}+\sigma_{2}\tau_{2}a_{2}b_{2}+\sigma_{2}b_{2}(1+\tau_{1}a_{1})^{\delta})^{\gamma}.$

But

$\displaystyle RHS$ $\displaystyle=$
$\displaystyle[\sigma_{1}a_{1}+(1+\sigma_{2}a_{2})^{\gamma}][\sigma_{1}b_{1}+(1+\sigma_{2}b_{2})^{\gamma}]$
$\displaystyle=$
$\displaystyle\sigma_{1}a_{1}\sigma_{1}b_{1}+\sigma_{1}a_{1}(1+\sigma_{2}b_{2})^{\gamma}+\sigma_{1}b_{1}(1+\sigma_{2}a_{2})^{\gamma}$
$\displaystyle+(1+\sigma_{2}a_{2})^{\gamma}(1+\sigma_{2}b_{2})^{\gamma}.$

Equating sides and setting $b_{1}=a_{2}=0$ gives

$\sigma_{1}a_{1}+(1+\sigma_{2}b_{2}(1+\tau_{1}a_{1})^{\delta})^{\gamma}=\sigma_{1}a_{1}(1+\sigma_{2}b_{2})^{\gamma}+(1+\sigma_{2}b_{2})^{\gamma}.$

Letting $a_{1},b_{2}\rightarrow\infty$ yields

$\sigma_{1}a_{1}+(\sigma_{2}b_{2})^{\gamma}(1+\tau_{1}a_{1})^{\delta+\gamma}\sim\sigma_{1}a_{1}(\sigma_{2}b_{2})^{\gamma}+(\sigma_{2}b_{2})^{\gamma},$

a contradiction unless $\sigma_{1}=\tau_{1}=0$. This yields
$\sigma=(1+\sigma_{2}x_{2})^{\gamma}$ and $\tau=1+\tau_{2}x_{2}$ in the
univariate format. Clearly $(S2)$ is satisfied, whereas $(S1)$ requires that

$[1+\sigma_{2}(x_{2}+y_{2}(1+\tau_{2}x_{2}))]^{\gamma}=[(1+\sigma_{2}x_{2})(1+\sigma_{2}y_{2})]^{\gamma}.$

The case $\gamma=0$ gives the ‘independent’ format with $\sigma=1$, as does
$\sigma_{2}=0$. Otherwise one has $\sigma_{2}=\tau_{2},$ a univariate type.

Case 2. Consider now pairing second choices, so that

$\sigma=\sigma_{1}x_{1}+e^{\gamma x_{2}},\quad\tau=e^{\delta
x_{1}}+\tau_{2}x_{2}.$

Here the arguments on the left of $(S1)$ are

$x_{1}=a_{1}+b_{1}[\sigma_{1}a_{1}+e^{\gamma a_{2}}],\quad
x_{2}=a_{2}+b_{2}[e^{\delta a_{1}}+\tau_{2}a_{2}].$

So

$LHS=\sigma_{1}x_{1}+e^{\gamma
x_{2}}=\sigma_{1}a_{1}+\sigma_{1}b_{1}\sigma_{1}a_{1}+\sigma_{1}b_{1}e^{\gamma
a_{2}}+\exp[\gamma a_{2}+\gamma b_{2}[e^{\delta a_{1}}+\tau_{2}a_{2}]].$

But

$RHS=[\sigma_{1}a_{1}+e^{\gamma a_{2}}][\sigma_{1}b_{1}+e^{\gamma
b_{2}}]=\sigma_{1}a_{1}\sigma_{1}b_{1}+\sigma_{1}b_{1}e^{\gamma
a_{2}}+\sigma_{1}a_{1}e^{\gamma b_{2}}+e^{\gamma(a_{2}+b_{2})}.$

Again the asymptotic behaviours of both sides do not match unless
$\sigma_{1}=\gamma=0$. when both sides reduce to 1. So here $\sigma=1$ and so
by $(S2)$

$\displaystyle e^{\delta(x_{1}+y_{1})}+\tau_{2}(x_{2}+y_{2}(e^{\delta
x_{1}}+\tau_{2}x_{2}))$ $\displaystyle=$ $\displaystyle(e^{\delta
x_{1}}+\tau_{2}x_{2})(e^{\delta y_{1}}+\tau_{2}y_{2})$ $\displaystyle
e^{\delta(x_{1}+y_{1})}+\tau_{2}x_{2}+\tau_{2}y_{2}e^{\delta
x_{1}}+\tau_{2}x_{2}\tau_{2}y_{2}$ $\displaystyle=$ $\displaystyle
e^{\delta(x_{1}+y_{1})}+\tau_{2}x_{2}e^{\delta y_{1}}+\tau_{2}y_{2}e^{\delta
x_{1}}+\tau_{2}x_{2}\tau_{2}y_{2}$ $\displaystyle\tau_{2}x_{2}$
$\displaystyle=$ $\displaystyle\tau_{2}x_{2}e^{\delta y_{1}}$

leading to the solutions $\sigma=1,\tau=1+\tau_{2}x_{2}$ covered by the
‘independent’ format and finally the ‘univariate’ type

$\sigma=1,\tau=e^{\delta x_{1}}.$

The remaining Cases 3 and 4 (cross-choices) are similar. $\square$

####  7.3.6 Th. 3.3 – Stone-Weierstrass argument

Using the Structure Theorem of the Euclidean case of
$\mathbb{G}_{S}^{\ast}(\mathbb{R}^{d})$ established in §3, we may now describe
$S$ by reference to $C(T)$ as follows.

Denote by $\delta$ and $\delta_{T}$ the Dirac mass function respectively for
$[0,1]^{2}$ and for $T^{2}$ so that $\delta(t)(s)=1$ iff $t=s;$ then by Th.
3.1, there are numbers $\sigma_{ij}$ with

$\sigma_{T}(x_{T})(t_{i})=\Sigma_{j}\sigma_{ij}x_{T}(t_{j})\text{ so that
}\sigma_{i}{}_{j}=\sigma_{T}(t_{i})_{j}=S(\delta_{T}(t_{j}))(t_{i})-1.$

The corresponding ‘generator’ will be represented here by
$\rho_{T}(x_{T})\in\mathbb{R}^{n+1}=C(T)$ with

$\rho_{T}(x_{T})(t_{i}):=\Sigma_{j}\sigma_{ij}x_{T}(t_{j})=\Sigma_{i=0}^{n}\sigma_{T}(x_{T})(t_{i})\rightarrow
S(x^{T})(t_{i}),$

with the limit here again under refinement of subdivisions.

Define the partition $\mathcal{P}_{T}$ of $T$ to comprise all the distinct
sets $I_{T}(t)$ for $t\in T$ with

$I_{T}(t):=\\{t_{i}\in T:(\forall x)S(x)(t_{i})=S(x)(t)\\}.$

Then for $s,t\in I_{T}(t)$

$S(\delta(t_{j}))(s)=S(\delta(t_{j}))(t)\qquad(j=0,1,...,n).$

Here taking limits under refinement of subdivisions $T$ yields

$I_{T}(t)\rightarrow K_{t}.$

For $s,t\in I\in\mathcal{P}_{T}$, as $S(\delta(t_{j}))(t)=0$ iff
$S(\delta(t_{j}))(s)=0,$ we may define

$J_{T}(I):=\\{t_{j}\in T:S(\delta(t_{j}))(t)\neq 0,t\in I\\}.$

Then for $x=x_{T}$

$\sigma_{I}(x):=\rho_{T}(e_{I}\cdot x)e_{I}\text{ for }e_{I}:=\Sigma_{t\in
J(I)}\delta_{T}(t)\text{ and }I\in\mathcal{P}_{T}.$

Here $x\mapsto e_{I}\cdot x$ is the projection from $\mathbb{R}^{n+1}$ onto
the span of $\\{\delta_{T}(t):t\in J(I)\\}.$

Taking limits over subdivisions $T$ under refinement with inclusion of $t$
yields

$\rho_{T}(x_{T})(t)\rightarrow\rho(x)(t);$

this limit generator gives a continuous linear map $\rho:C[0,1]\rightarrow
C[0,1]$. For $s\in[0,1],$ put $\rho_{s}(x):=\rho(x)(s),$ a continuous linear
functional. Then

$\\{t:\rho_{s}=\rho_{t}\\}=K_{s}.$

Let $e_{K}$ denote the map $x\mapsto e_{K}\cdot x$ projecting from $C[0,1]$
onto $C(K)$. Now define $\sigma_{K}:C(K)\rightarrow\mathbb{R}$ for
$K_{t}=K\in\mathcal{P}$ by

$\sigma_{K}(x):=\rho_{t}(e_{K}\cdot x),$

thereby completing the analysis of the action of $S$. $\square$

####  7.3.7 Proof of Lemma 4.1

With restrictions on $a,b,c,z$ as above, using $(GS)$ and invertibility of
$S,$

$\displaystyle S(c+(a-b))$ $\displaystyle=$ $\displaystyle
S(a+(c-b)S(a)S(b)^{-1})=S(a)S((c-b)S(b)^{-1})$ $\displaystyle=$ $\displaystyle
S(b)S((c-b)S(b)^{-1})=S(b+S(b)(c-b)S(b)^{-1})=S(c).$

Since

$S(z+S(z)a)=S(z)S(a)=S(z)S(b)=S(z+S(z)b)\in\mathbb{A}^{-1},$

replacing in (i) $a$ by $z+S(z)a$ and $b$ by $z+S(z)b$ yields (ii). In
particular, for any $z\in\mathbb{G}_{S}^{\ast}$

$1_{\mathbb{A}}=S(0)=S(S(z)a)$

(take $b=c=0),$ giving
$S(\mathbb{G}_{S}^{\ast})\mathcal{N}\subseteq\mathcal{N}$ and so also the last
assertion, as $S(0)\mathcal{N}=\mathcal{N}$. $\square$

####  7.3.8 Proof of Corollary 4.4

Note first that since
$\mathbb{C}_{1}\mathbb{=C}\backslash\\{0\\}=\mathbb{C}^{-1},$
$\mathbb{G}_{S}^{\ast}(\mathbb{C})=\mathbb{G}_{S}(\mathbb{C}).$ It is known
[Bar] and reproved in Prop. 4.2 that in the case $\mathbb{A=C}$ any continuous
solution $S$ of $(GS)$ is either canonical, i.e. of the form $S(z)=1+\rho z$
so that $\mathbb{G}_{S}(\mathbb{C})=\mathbb{G}_{\rho}(\mathbb{C}),$ or takes
the ‘non-canonical’ form
$S(z):=1+a\mathop{\mathrm{R}e}(z)+b\mathop{\mathrm{I}m}(z)$ with $a,b$ real
constants. Here if $a=b=0,$ this leads to $\mathrm{ran}S=\\{1\\};$ then $1$ is
an isolated point, and the preceeding result cannot be applied, although its
conclusion still holds with $c=0.$ The alternative is that
$\mathrm{ran}S=\langle 1\rangle.$ This presents two possibilities:

(i) $S(z)=1+az$ for $z\in\langle 1\rangle$ with $a\neq 0,$ yielding a Popa
subgroup $\mathbb{G}_{\alpha}^{\ast}(\langle
1\rangle)=\mathbb{G}_{\alpha}^{\ast}(\mathbb{R})$ with $\alpha=1/a;$

(ii) $S(z)=1-ibz$ for $z\in\langle i\rangle$ with $b\neq 0,$ yielding a Popa
subgroup $\mathbb{G}_{\beta}^{\ast}(\langle
i\rangle)\approx\mathbb{G}_{|\beta|}^{\ast}(\mathbb{R})$ with $\beta=1/(-ib).$

These two separate restrictions of $S,$ both in ‘canonical’ form, correspond
to $S((z-1)/a)=z$ for $z\in\langle 1\rangle$ (with $c=1/a),$ and
$S((iz-1)/(-ib))=iz$ for $z\in\langle i\rangle$ (with $c=i/b).$

Write $z_{1}=u+iv$ and $z_{2}=x+iy;$ then, since $S(z_{1})=1+au+bv,$

$z_{1}\circ_{S}z_{2}=(u+iv)+(x+iy)+(1+au+bv)(x+iy).$

This corresponds to a Popa operation $\circ_{\sigma}$on the set
$\mathbb{G}_{\alpha}^{\ast}(\mathbb{R})\times\mathbb{G}_{\text{\mbox{$|$}}\beta|}^{\ast}(\mathbb{R})$
with $\sigma(u,v):=1+au+bv$ and

$(u,v)\circ_{\sigma}(x,y)=(u+x+\sigma(u,v)x,v+y+\sigma(u,v)y).$

We shall identified this in §3 as
$\mathbb{G}_{\sigma}^{\ast}(\mathbb{R}^{2})$. $\square$

####  7.3.9 Example 5.1 (Standardized Tilting in $\mathbb{C}$)

Writing $\omega=x+iy,$ the real and imaginary parts give:

$\displaystyle e^{x}\cos y$ $\displaystyle=$ $\displaystyle 1+x,\qquad
e^{x}\sin y=y:$ $\displaystyle y^{2}$ $\displaystyle=$ $\displaystyle
y^{2}(x):=e^{2x}-(1+x)^{2}>0,\text{ for }x>0.$

For $x\geq 0,$ $y(x)$ is monotonic and unbounded, whereas

$x\mapsto e^{-x}y(x)=\sqrt{1-(e^{-x}(1+x))^{2}}$

increases on $[0,\infty)$ strictly from $0$ to $1,$ yielding solutions in $x$
to the equation

$\sin y(x)=e^{-x}y(x),$

one within each consecutive interval in which $\sin y(x)$ traces the interval
$[-1,1]$ with $x$ near the solution point of $y(x)=\pi/2$ mod
$2\pi\mathbb{Z}$. It follows that, in alternate intervals where $\cos y(x)>0,$

$e^{x}\cos y(x)=\sqrt{e^{2x}-y(x)^{2}}=1+x,$

thus satisfying $(ST)$.

Remark. The function $y(x)$ is also defined in a maximal interval $[-\xi,0],$
with $\xi$ satisfying $e^{-2\xi}=(1-\xi)^{2},$ equivalently $e^{-\xi}=\xi-1.$
From here $\xi>1,$ and in fact

$\xi=1.27846...;$

it follows that $y(-\xi)=0.$ (Evidently $y(-1)=1.)$ So $\sin
y(-\xi)=e^{\xi}y(-\xi)=0;$ however,

$e^{-\xi}\cos y(-\xi)=e^{-\xi}=\xi-1\neq 1-\xi,$

since $\xi\neq 1.$

####  7.3.10 Proof of Prop. 5.4

Using orthogonality,

$\displaystyle RHS_{GS}$ $\displaystyle=$
$\displaystyle(1_{\mathbb{A}}+\Sigma_{i}\sigma(e_{i}x)e_{i})(1_{\mathbb{A}}+\Sigma_{j}\sigma(e_{i}y)e_{j})$
$\displaystyle=$ $\displaystyle
1_{\mathbb{A}}+\Sigma_{i}\sigma(e_{i}x)e_{i}+\Sigma_{j}\sigma(e_{i}y)e_{j}$
$\displaystyle+\Sigma_{i}\sigma(e_{i}x)\sigma(e_{i}y)e_{i}.$

Noting that

$x+[1_{\mathbb{A}}+\Sigma_{i}\sigma(e_{i}x)e_{i}]y=x+y+\Sigma_{i}\sigma(e_{i}x)e_{i}y,$

we compute, using orthogonality, that

$\displaystyle\sigma(e_{j}[x+y+\Sigma_{i}\sigma(e_{i}x)e_{i}y])$
$\displaystyle=$
$\displaystyle\sigma(e_{j}x+e_{j}y+\Sigma_{i}\sigma(e_{i}x)e_{j}e_{i}y)$
$\displaystyle=$
$\displaystyle\sigma(e_{j}x)+\sigma(e_{j}y)+\sigma[\sigma(e_{j}x)e_{j}y]$
$\displaystyle=$
$\displaystyle\sigma(e_{j}x)+\sigma(e_{j}y)+\sigma(e_{j}x)\sigma[e_{j}y].$

So

$\displaystyle LHS_{GS}$ $\displaystyle=$ $\displaystyle
1_{\mathbb{A}}+\Sigma_{j}\sigma(e_{j}[x+y+\Sigma_{i}\sigma(e_{i}x)e_{i}y])e_{j}$
$\displaystyle=$ $\displaystyle
1_{\mathbb{A}}+\Sigma_{j}\sigma(e_{j}x)e_{j}+\sigma(e_{j}y)e_{j}+\sigma(e_{j}x)\sigma(e_{j}y)e_{j},$

and the two sides match. $\square$

####  7.3.11 Proof of Lemma 5.5

W.l.o.g. $||e^{-a}||<1.$ Then $||e^{a}||>1,$ for otherwise $||e^{a}||\leq 1,$
leading to

$1=||e^{-a}e^{a}||\leq||e^{-a}||\cdot||e^{a}||<1,$

a contradiction. Furthermore, for $n\in\mathbb{Z}$,

$||e^{na}||=||e^{(n+1)a}e^{-a}||\leq||e^{-a}||\cdot||e^{(n+1)a}||<||e^{(n+1)a}||.$

The sequence $\\{||e^{na}||\\}_{n\in\mathbb{N}}$ is thus monotonically
increasing. If $||e^{na}||\rightarrow c$ for some finite $c>0,$ then, by the
preceding inequality

$c\leq||e^{-a}||c<c,$

also a contradiction. So $||e^{na}||$ is unbounded. Similarly,
$\\{||e^{-na}||\\}_{n\in\mathbb{N}}$ is monotonically decreasing, this time
with limit $0.$

As for the final statement, provided $||e^{\gamma(u)}||\neq 1$,

$T(\pm us)=u(e^{\pm s\gamma(u)}-1)/\gamma(u)$

is unbounded as $s\rightarrow\infty$ in one of the directions $\pm u$.
$\square$

####  7.3.12 Proof of Th. 6.1 (Wołodźko-Javor Theorem)

Given $S,$ take $\mathcal{N}:=\\{u\in\mathbb{G}_{S}^{\ast}:S(u)=1_{A}\\}$,
$\Lambda:=S\mathcal{(}\mathbb{G}_{S}^{\ast}),$ and choose any right inverse
$W$ with $S(W(a))\equiv a.$ Then use Lemma 2.1(i) and (ii).

For the reverse direction, as $\mathcal{N}$ is a subgroup, we may work
$\mathop{\mathrm{m}od}\mathcal{N},$ indicating this now with
$\equiv_{\mathcal{N}}.$ First note that $S$ is well-defined. For if
$W(\lambda_{1})=W(\lambda),$ then, taking
$\lambda_{2}=\lambda\lambda_{1}^{-1},$ so that
$\lambda=\lambda_{1}\lambda_{2}$ by (iii),

$\lambda_{1}W(\lambda_{2})=\lambda_{1}W(\lambda_{2})+[W(\lambda_{1})-W(\lambda_{1}\lambda_{2})]\equiv_{\mathcal{N}}0.$

So $\lambda_{1}W(\lambda_{2})\in\mathcal{N}$, or
$W(\lambda_{2})\in\lambda_{1}^{-1}\mathcal{N}\subseteq\Lambda\mathcal{N=N}$;
so by (ii), $\lambda_{2}=1,$ i.e. $\lambda=\lambda_{1},$ as required.

We check that (iv) satisfies $(GS)$ with $x=x_{1}$ and $y=x_{2}.$

If $S(x_{1})=0,$ then $(GS)$ holds trivially.

If $S(x_{1})\neq 0$ and $S(x_{1}+S(x_{1})x_{2})\neq 0,$ then pick
$\lambda_{1}$ and $\lambda$ with $W(\lambda_{1})=x_{1}$ and $W(\lambda)\equiv
x_{1}+S(x_{1})x_{2}.$ Take $\lambda_{2}=\lambda\lambda_{1}^{-1};$ then

$\displaystyle W(\lambda_{1}\lambda_{2})$ $\displaystyle=$ $\displaystyle
W(\lambda)\equiv_{\mathcal{N}}x_{1}+S(x_{1})x_{2}=W(\lambda_{1})+\lambda_{1}x_{2}:$
$\displaystyle\lambda_{1}x_{2}$ $\displaystyle\equiv$
${}_{\mathcal{N}}W(\lambda_{1}\lambda_{2})-W(\lambda_{1})\equiv_{\mathcal{N}}\lambda_{1}W(\lambda_{2}):\qquad
x_{2}\equiv_{\mathcal{N}}W(\lambda_{2}).$

So here $S(x_{2})=\lambda_{2}\neq 0$, i.e. passing to the contrapositive: if
$S(x_{2})=0,$ then $S(x_{1}+S(x_{1})x_{2})=0$ and $(GS)$ holds.

Now consider $x_{i}$ with both $S(x_{i})\neq 0.$ Write
$x_{i}\equiv_{\mathcal{N}}W(\lambda_{i}).$ Then

$\displaystyle x_{1}+S(x_{1})x_{2}$ $\displaystyle\equiv$
${}_{\mathcal{N}}W(\lambda_{1})+\lambda_{1}W(\lambda_{2})\equiv_{\mathcal{N}}W(\lambda_{1}\lambda_{2}):$
$\displaystyle S(x_{1}+S(x_{1})x_{2})$ $\displaystyle=$
$\displaystyle\lambda_{1}\lambda_{2}=S(x_{1})S(x_{2}).$

This completes the check that $(GS)$ holds. $\square$

###  7.4 Proof of Theorem S

We are grateful to Amol Sasane for the following proof.

For fixed $\theta\in M_{\mathbb{A}}$, the maximal ideal space of $\mathbb{A}$
(viewed as comprising characters in $\mathcal{L}(\mathbb{A},\mathbb{C})),$
define for $\zeta\in\mathcal{D}_{u}$

$\displaystyle\varphi_{\theta u}(\zeta)$ $\displaystyle:$
$\displaystyle=\theta(f(\zeta\gamma(u))),$ $\displaystyle\psi_{\theta
u}(\zeta)$ $\displaystyle:$
$\displaystyle=\theta(\gamma(f(\zeta\gamma(u))\cdot u/\gamma(u)));$

both are holomorphic on $\mathcal{D}_{u}$, since $\theta$ is linear.
Furthermore, by the assumed identity,

$\varphi_{\theta u}(\zeta)=\psi_{\theta
u}(\zeta)\qquad\hfill(\zeta\in\Sigma).$

So, by Riemann’s Uniqueness (Identity) theorem [Gam2,V.7], [Rem, Ch. 8], also
$\varphi_{\theta u}=\psi_{\theta u}$ on $\mathcal{D}_{u}.$ As $\theta$ was
arbitrary, this may be restated using Gelfand transforms as

$f(\zeta\gamma(u)\;\widehat{}=\gamma(f(\zeta\gamma(u)\cdot
u/\gamma(u))\;\widehat{}\;.$

For $\mathbb{A}$ semisimple, the Gelfand transform is injective and so

$f(\zeta\gamma(u))=\gamma(f(\zeta\gamma(u)\cdot
u/\gamma(u))\qquad\hfill(\zeta\in\mathcal{D}_{u}),$

thus extending an identity from $\Sigma$ to $\mathcal{D}_{u}.$ $\square$