# A Multi-Task Learning Model for Super Resolution of Wireless Channel
Characteristics

Xiping Wang1, Zhao Zhang1, Danping He1, Ke Guan12, Dongliang Liu3, Jianwu
Dou3, and Bo Sun3 Corresponding Author: Danping He (e-mail:
hedanping@bjtu.edu.cn).  1State Key Laboratory of Rail Traffic Control and
Safety, Beijing Jiaotong University, 100044, Beijing, China  2Frontiers
Science Center for Smart High-speed Railway System, 100044, Beijing, China
3State Key Laboratory of Mobile Network and Mobile Multimedia Technology,
518055, Shenzhen, China

###### Abstract

Channel modeling has always been the core part in communication system design
and development, especially in 5G and 6G era. Traditional approaches like
stochastic channel modeling and ray-tracing (RT) based channel modeling depend
heavily on measurement data or simulation, which are usually expensive and
time consuming. In this paper, we propose a novel super resolution (SR) model
for generating channel characteristics data. The model is based on multi-task
learning (MTL) convolutional neural networks (CNN) with residual connection.
Experiments demonstrate that the proposed SR model could achieve excellent
performances in mean absolute error and standard deviation of error.
Advantages of the proposed model are demonstrated in comparisons with other
state-of-the-art deep learning models. Ablation study also proved the
necessity of multi-task learning and techniques in model design. The
contribution in this paper could be helpful in channel modeling, network
optimization, positioning and other wireless channel characteristics related
work by largely reducing workload of simulation or measurement.

###### Index Terms:

Wireless channel modeling, ray-tracing (RT), super resolution (SR), multi-task
learning (MTL), convolutional neural network (CNN)

##  I Introduction

Thanks to the fast advances in the fifth generation (5G) wireless
communication, our world is stepping into the era of Internet of Everything
(IoE)[1]. To enable ultra-low latency and high reliability communication
services in dense connected areas like urban districts, accurate wireless
channel model is a necessity. Correct knowledge of the propagation channel is
also critical for radio coverage estimation, network optimization and for many
related applications[2]. Naturally, channel modeling is seen as the foundation
for planning and optimizing communication and related systems[3].

Channel modeling is the process of characterizing the propagation principles
of radio waves in realistic environments, and provides insight theoretical
guidance for the design, deployment and optimization of communication systems.
Generally, stochastic channel modeling (SCM) and ray-tracing (RT) based
deterministic modeling are two main modeling approaches[4]. For SCM, channel
characteristics like path loss (PL), propagation condition (line of sight or
non line of sight), delay spread, angular spreads and Rician K-factor are
required to generate channel coefficients and thereafter to model the wireless
channel. Massive channel measurements in different environments, which are
usually time-consuming and expensive, must be conducted to obtain the
necessary channel characteristics data[5]. On the other side, RT based
modeling approach can generate accurate channel data if given precise
propagation environment and configuration, but at the cost of high
computational complexity and enormous calculation time[6]. Therefore, a fast
and reliable channel characteristics data generation method will effectively
address these limitations.

Figure 1: The overview of proposed MTL SR model for wireless channel
characteristics. ① RT simulation and dataset generation. ② The SR model
training process. ③ Evaluation.
Dataset1: Channel characteristics dataset

One of the promising solutions is machine learning (ML) method. The recent
surge of ML is revolutionizing almost every branch of science and technology,
including wireless channel modeling. The wireless channel is a time-varying
nonlinear system, which contains multi-dimensional information in time,
spatial and frequency domain. Machine learning has very powerful learning and
inferring capabilities. It can automatically learn from channel data so that
the structural relationship between data in complex environments can be
extracted to approximate nonlinear systems. Moreover, ML is very efficient in
mining information from high dimensional data, which can significantly
expedite data processing.

Current research in this area are still inadequate. Most of recent works
focused on predicting only one of the channel characteristics like PL (Radio
Coverage Prediction)[2]. The input is mostly restricted to building
information, satellite image[7], map[8] and just one of the characteristics.
Few works attempt to estimate several channel characteristics at the same
time. As for ML methods, traditional algorithms such as random forests (RF),
support vector machine (SVM) and K-nearest neighbors (KNN) as well as deep
learning (DL) models such as convolutional neural network (CNN)[8],
Transformer [9] and generative adversarial network (GAN) [10] are frequently
employed. Also, the estimation targets are usually restricted to one or two
characteristics. In general, few works have studied multi-task learning (MTL)
models for generating channel characteristics data with several
characteristics as input. The connection between different characteristics
awaits to be exploited.

In this paper, instead of prediction or estimation, we propose a super
resolution (SR) model for channel characteristics. The model is based on MTL
CNN with residual connection. Overview of our work is shown in Fig.1. Given 3D
models of urban areas and the corresponding EM parameters, CloudRT platform
outputs channel characteristics dataset and the dataset is used for SR model
training. Data from the dataset are degraded into low resolution data as
input. Original high resolution data are used as the ground truth. MTL loss is
employed to better balance multi tasks. We evaluate our proposed SR model by
ablation study and comparisons with other DL models. Specifically, we make the
following contributions:

  * •

In dense urban areas, RT simulation by self developed CloudRT was conducted
and channel characteristics dataset are constructed based on simulation
results.

  * •

A residual network based MTL SR model is proposed. Weighted masks are added in
loss function. Homoscedastic uncertainty is employed to balance the single
task losses during training. Residual connection and iterative up-and-down
technique are implemented in CNN blocks for better SR performances.

  * •

Training process and results of evaluation are provided. The proposed SR
approach generally performs better than other state-of-the-art DL models.
Comparing with baseline, the proposed approach could achieve very good SR
results in all channel characteristics targets and deteriorate significantly
less than baseline with larger scale factor. Ablation study proved that the
techniques in training and model design are necessary.

##  II Simulation and Dataset Construction

TABLE I: RT Simulation configuration Carrier frequency | 3.55 GHz
---|---
System bandwidth | 100 MHz
Frequency resolution | 1 MHz
Antenna | Omni-directional vertical polarization
Simulation range | 1 km${\times}$1 km
Tx location | 30 - 50 m above the ground
Rx location | 2 m above the ground

This section describes the procedures of RT simulation and dataset
construction. Self developed CloudRT[11][12] platform is used to generate
channel characteristics data.

RT approach is widely used to generate accurate channel characteristics data
in a specific environment. Based on 3D electronic map and electromagnetic (EM)
parameters provided by ITU-R P.1238-7, RT simulation is conducted in dense
urban areas on CloudRT platform. More than 100 simulation regions are manually
selected from urban areas of four major cities in China: Beijing, Shanghai,
Hangzhou and Xi’an. The simulation regions are 1km $\times$ 1km squares with
3D electronic map resolution of 200 $\times$ 200\. Transmitter (Tx) is located
on one of the high buildings near the center of simulation region, and a large
LOS area should be maintained. The receivers (Rx) are located 2 meters above
ground, uniformly distributed by distance of 5 meters on the horizontal plane.
Only receivers outside buildings are considered in simulation. Table I
summarizes the simulation configuration and details of it can be found in[13].

TABLE II: Channel characteristics dataset Characteristic | Normal Range | NaN Value
---|---|---
PL [dB] | [-200,0) | 200
$\textit{R}_{p}$ [dB] | (-30,0] | 100
DS [ns] | (0,500) | -100
$\phi$ [°] | [0,360) | -360
$\theta$ [°] | [0,180) | -180
LOS/NLOS | -1/0 | 1

$\textit{R}_{p}=\frac{\sum_{i=1}^{N_{r}}P_{i}-P_{0}}{\sum_{i=1}^{N_{r}}P_{i}}$
(1)

The channel characteristics dataset is constructed based on RT simulation
results. Definitions of PL, root mean squared (RMS) delay spread (DS), RMS
azimuth ($\phi$) angular spread of arrival, RMS elevation ($\theta$) angular
spread of arrival, line of sight (LOS) and non line of sight (NLOS) follows
the usual. We redefined Racian-K factor as multi-path power ratio
$\textit{R}_{p}$ in (1), where $P_{i}$ and $P_{0}$ is the power of ray $i$ and
ray of direct propagation. In LOS area, $\textit{R}_{p}$ is the ratio of power
of all rays except $P_{0}$ to the total power. In NLOS area, $\textit{R}_{p}$
is equal to zero. $\textit{R}_{p}$ is continuous in both LOS and NLOS areas
thus suitable for ML. The 6 characteristics are also SR targets in this paper.
We use PL, $\textit{R}_{p}$, DS, $\phi$, $\theta$, LOS/NLOS as the
abbreviations or symbols of the channel characteristics and SR targets, as
shown in TABLE II. Values which are far beyond ordinary thresholds in
communication systems are set as the minimum (PL, $\textit{R}_{p}$) or the
maximum (DS) of corresponding normal range. NaN value represents the channel
characteristics data of receivers which locate inside buildings. NaN value
should be void but set as real number out of normal range so that ML model is
able to distinguish. A data in channel characteristics dataset is a 200 by 200
tensor with 7 channel (including building heights). In total, 753 data were
generated and combined to construct the dataset. The input data are processed
by down-sampling by certain scale factor and up-sampling by interpolation so
that the shape remains the same.

##  III Methodology

###  III-A Problem Definitions and Terminologies

Super resolution is a notion of recovering high resolution (HR) data from the
low resolution (LR) data[14]. The LR data mostly originates from HR data with
a process of degradation:

$I_{LR}=\mathcal{D}\left(I_{HR};\delta\right)$ (2)

where $I_{LR}$, $I_{HR}$, $\mathcal{D}$, $\delta$ denote LR data, HR data,
degradation process which is usually unknown, and parameters of degradation
such as scaling factor. Usually $I_{LR}$, $I_{HR}$ are the input training data
and the ground truth correspondingly. SR can be described as the model of
recovering $I_{LR}$ to $I_{HR}$. The recovery (approximation) result is
denoted as $\hat{I}_{HR}$. $\mathcal{F}$ represents the recovery model with
parameters $\theta$. Take DL as the example, $\mathcal{F}$ is the neural
network while $\theta$ is the weights and other related parameters in the
neural network.

$\hat{I}_{HR}=\mathcal{F}\left(I_{LR};\theta\right)$ (3)

To this end, the objective of SR is as follows:

$\hat{\theta}=\mathop{\arg\min}\limits_{\theta}\mathcal{L}(\hat{I}_{HR},I_{HR})+\lambda\Phi(\theta)$
(4)

where $\mathcal{L}$ is the loss function between recovery result
$\hat{I}_{HR}$ and ground truth $I_{HR}$. $\Phi$ is the regularization term
with parameters $\theta$ and $\lambda$ is the weight factor. The best SR model
would minimize the loss to the least with respect to $\theta$.

###  III-B Residual Network based Multi-Task Learning Model

The proposed residual network based MTL model consists of two parts: backbone
part and fine-tune part, as in Fig. 2. The backbone part is to extract high-
dimensional features from input data while the fine-tune part focuses on each
of the tasks to achieve the best SR performances.

The CNN block contains two convolutional layers with activation function ReLU.
The size of data during convolution remains 200 $\times$ 200\. The number of
CNN blocks $N$ is set to be 3 in our work, but a larger number of blocks may
also achieve excellent SR performances based on our previous experiments. To
be noted, the numbers of channels in convolutional layers are set to be
iterative up-and-down, as shown in the illustration of CNN block. The
iterative up-and-down technique can filter irrelevant information in input
data. Residual connection is also necessary for good SR performances. Several
CNN blocks are concatenated with residual connection to construct the backbone
part. In fine-tune part, 6 lightweight models are designed for the
corresponding targets. Lightweight models are two layer CNN like in CNN block
but number of channels remain fixed. Note that for NaN/LOS/NLOS, outputs of
the fine-tune nets are the probabilities of corresponding propagation
conditions.

Figure 2: The overview of proposed MTL CNN model with residual connection

Figure 3: An illustration of $M_{gt}$ and $M_{na}$. Mask values are all set as
0.01. The background is electronic map of Tiananmen Square in Beijing, China.

###  III-C Loss Functions and Evaluation Metrics

As explained in Section II, only receivers outside the building are considered
in simulation. As a result, NaN values are given real number to guarantee that
the input data is a regular matrix or tensor. Also, there will always be some
ground truth values (elements) in the matrix of input data. However, both NaN
values and ground truth values are trivial in SR process. To keep the loss
function continuous and to help ML models concentrate on SR values of desired
positions, weighted loss is introduced as the mask of NaN values $M_{na}$ and
the mask of ground truth $M_{gt}$, which are illustrated in Fig. 3. Several
kinds of loss functions and evaluation metrics such as pixel loss, and content
loss are widely used in image SR tasks. Differently, only pixel loss is
accepted in channel characteristics SR task. According to previous
experiments, L1 (norm) loss is employed for training as it performs better
than L2 loss and peak signal noise ratio (PSNR).

There are 6 SR targets in this paper as mentioned in section II. For LOS/NLOS,
SR process is essentially to classify between LOS and NLOS area. For the rest
targets, the SR process are regression. Due to the difference between
classification and regression, the loss functions are categorized into two
scenarios where L1 norm are used for regression and cross entropy are used for
classification, as in (7) and (8). The loss function $\mathcal{L}_{m}$ in
training can be described as follows:

$\mathcal{L}_{m}(\hat{I},I)=\begin{cases}{}loss_{ce}(\hat{I},I),&\small{\text{if
target $m$ is LOS/NLOS}}\\\
loss_{l_{1}}(\hat{I},I),&\small{\text{others}}\end{cases}$ (5)

where $\hat{I}$ and $I$ are SR recovered data and ground truth. Before the
calculation of loss function, both $\hat{I}$ and $I$ should be weighted by
Hadamard Product with $M_{gt}$ and $M_{na}$:

$I_{weighted}=I_{original}\circ M_{na}\circ M_{gt}$ (6)

$n$, $h$, $w$, represents the number of values to be estimated (SR) and side
lengths of input data. $k$ stands for the class among LOS/NLOS/NaN in (8).

$loss_{l_{1}}(\hat{I},I)=\frac{n}{\left(hw\right)^{2}}\sum_{i,j}|\hat{I}_{i,j}-{I}_{i,j}|_{1}$
(7)

$loss_{ce}(\hat{I},I)=-\frac{n}{\left(hw\right)^{2}}\sum_{i,j}{I}_{i,j,k}\log{\hat{I}_{i,j,k}}$
(8)

During the pre-train stage, homoscedastic uncertainty[15] is employed to
balance the single-task losses as in (9). Not only the weights in neural
networks $W$ but also noise parameters $\boldsymbol{\sigma}$ are trainable and
updated through standard back propagation during training.

$\mathcal{L}_{MTL}(W,\boldsymbol{\sigma})=\sum_{m}\frac{\mathcal{L}_{m}}{2\sigma_{m}^{2}}+\sum_{m}log(\sigma_{m})$
(9)

The evaluation metric is much simpler than training loss. Basically, values of
receiver located in building areas (NaN values) and ground truth values are
not included in calculation. The metric for evaluating SR performance of
LOS/NLOS is classification accuracy. For the rest targets, mean absolute error
(MAE) and standard deviation of error (STDE) are regarded as the metrics.

##  IV Experiments

###  IV-A Training Configuration and Implementation

In this work, DL training are performed by PyTorch 1.10.2 on a work station
with 1 NVIDIA GeForce RTX 3090 GPU, Intel Core i9-9900K CPU and 32 GB DDR4
RAM. The training process are divided into two stage: pre-train stage and
fine-tune stage. In pre-train stage, (9) is used as the loss function for back
propagation and weights of the entire SR model are updated. Next, in fine-tune
stage, only weights of fine-tune part in the SR model are updated while
backbone part remains unchanged. The purpose of this design is to obtain a
general feature extractor (backbone part) for channel characteristics SR. By
doing so, it’s very fast and convenient to concatenate lightweight models for
specific tasks when comparing with DL models aiming at only one channel
characteristic. Data augmentation is employed to the training set for
enhancing diversity of the inputs. The training set is transformed by rotation
of 90, 180 and 270 degree as well as horizontally and vertically flipping. As
a result, the training dataset is incremented by 5 times.

The proposed SR model is trained for 100 epochs in both of two stages. The
learning rate is set as 0.00001. Adam optimizer is used for gradient descent.
Batchsize is set as 1. The channel characteristics dataset is randomly split
into training set and test set by ratio of 7:3. Experiments with scale factor
2, 4 and 8 were conducted.

(a) Path loss

(b) $R_{p}$

(c) RMS $\phi$ spread

(d) Classification accuracy of LOS/NLOS

(e) RMS delay spread

(f) RMS $\theta$ spread

Figure 4: MAE and classification accuracy of 6 SR targets during training
process.

###  IV-B Backbone Part

As mentioned above, backbone part of the proposed SR model is regarded as a
general feature extractor. The comparison of DL models is summarized in Table
III. Our proposed SR model could achieve 3.26 dB for MAE of PL with scale
factor 2. Under the same configuration, several kinds of DL models including
ResNet50, vision transformer (ViT)[16] and GAN-SR [17] were tested.
Performances of these state-of-the-art DL models are not satisfying. For
backbone part, ResNet50 is the best in tested DL models but the MAE of PL is
around 7-8 dB. After several experiments of ViT by changing the number of
transformer encoder and the number of heads in attention layer, the best
result of MAE is around 8 dB. Performances of GAN are much worse than CNN and
ViT. The best results of GAN is higher than 12 dB for MAE. Moreover, few
experiments show that the loss of generator and the output of discriminator
(probability that $\hat{I}_{HR}$ is real) converge at the end. Nash
equilibrium between generator and discriminator could hardly be reached. For
other targets, SR results of these DL models are also much worse than those of
the proposed model.

The reasons of why state-of-the-art DL models doesn’t work well are analyzed.
First, the data size of channel characteristics dataset is much smaller than
popular computer vision datasets like CIFAR-10, ImageNet and MINST. Second,
texture, style and smoothness of images should be considered in image SR but
not regarded as objectives in characteristics SR task. As a result, deeper
models with larger number of parameters are more liable to over-fitting. The
comparison of SR performances and model complexity are presented in Table III.
For the proposed model, both the number of parameters (Params) and floating
point of operations (FLOPs) are much smaller than other DL models.

TABLE III: Comparison of deep learning models | Proposed Model | ResNet50 | ViT | GANSR
---|---|---|---|---
Params / k | 4.25 | 8610 | 13040 | 28670
FLOPs / GMac | 0.34 | 2.58 | 10.33 | 12.42
MAE of PL / dB | 3.26 (best) | 7-8 | 7-8 |  $>$12

###  IV-C Performance of Proposed Model

MAE and classification accuracy of 6 SR targets during training process are
shown in Fig. 4. Clearly for the 6 targets, MAE and classification accuracy of
SR are decreasing during training and converge at the end. In pre-train stage
(1-100 epochs), MAE and classification accuracy decrease very quickly with
some minor jitters due to MTL loss. This indicates that the proposed model and
MTL loss is effective as a feature extractor to achieve fairly good SR
performances with multi inputs and tasks. In fine-tune stage (101-200 epochs),
few jitters are observed. MAE and classification accuracy decrease slowly to
converge at the end so that better SR performances could be achieved in fine-
tune stage. We also observed that given scale factor as 8, the classification
accuracy of LOS/NLOS jitters greatly. We will look into this part in future
research.

In general, the best SR results achieved by the proposed model is demonstrated
in Table IV and Table V. Compared with bilinear interpolation, both MAE and
STDE of the proposed model is far smaller. Moreover, SR performances of the
proposed model deteriorate significantly less than bilinear interpolation when
scale factor is relatively larger (4 or 8).

###  IV-D Ablation Study

Ablation study was conducted to investigate the effectiveness of MTL and
techniques in the proposed model. Cumulative SR performance gain of path loss
(MAE and STDE) in ablation study are demonstrated in Table VI. STL means the
proposed model for single task (PL) during training. MTL represents the
proposed model for multi-task learning but without residual connection and
iterative up-and-down technique. Apparently, SR of single task learning is
incomparable to MTL. RES Residual connection and iterative up-and-down
technique could reduce MAE and STDE for 20% on average. DA could also enhance
SR performance in some extent. It’s noticed that the enhancement of DA and RES
decreases much when scale factor are large (4 and 8).

TABLE IV: Super resolution performance (MAE) of proposed model and bilinear interpolation Scale Targets | PL | $R_{p}$ | DS | $\phi$ | $\theta$ |  | LOS/
---
NLOS
2 | Proposed | 3.26 | 0.89 | 6.84 | 5.34 | 0.84 | 98%
Bilinear | 16.58 | 7.21 | 13.46 | 16.60 | 12.11 | 85%
4 | Proposed | 4.29 | 1.28 | 9.99 | 8.55 | 1.24 | 95%
Bilinear | 26.76 | 11.74 | 20.41 | 26.53 | 19.90 | 72%
8 | Proposed | 5.81 | 1.98 | 14.09 | 12.53 | 1.78 | 87%
Bilinear | 36.77 | 16.51 | 27.91 | 36.92 | 27.97 | 65%
TABLE V: Super resolution performance (STDE) of proposed model and bilinear interpolation Scale Targets | PL | $R_{p}$ | DS | $\phi$ | $\theta$
---|---|---|---|---|---
2 | Proposed | 6.09 | 2.54 | 12.19 | 10.23 | 1.81
Bilinear | 31.69 | 13.19 | 18.18 | 27.57 | 24.39
4 | Proposed | 7.90 | 3.48 | 16.82 | 16.27 | 2.44
Bilinear | 43.24 | 18.18 | 24.94 | 38.41 | 33.75
8 | Proposed | 10.21 | 4.85 | 20.85 | 20.21 | 3.20
Bilinear | 51.15 | 21.82 | 30.23 | 46.20 | 40.42
TABLE VI: Cumulative super resolution performance gain of path loss

| MAE | STDE
---|---|---
scale=2 | scale=4 | scale=8 | scale=2 | scale=4 | scale=8
+DA | +36% | +26% | +18% | +26% | +14% | +8%
+RES | +28% | +20% | +13% | +20% | +12% | +7%
MTL | 0 | 0 | 0 | 0 | 0 | 0
STL | -149% | -170% | -124% | -167% | -221% | -205%

  * •

+DA: Add data augmentation to +RES.

+RES: Add residual connection and iterative up-and-down to MTL.

MTL: The proposed model without RES and DA in training.

STL: Single SR task (PL) learning using the proposed model

without RES and DA.

##  V Conclusion

In this paper, a novel residual network based MTL model is proposed for SR of
wireless channel characteristics. RT simulation was conducted and channel
characteristics dataset were constructed based on simulation results. Weighted
masks are introduced in loss function which can help better fit randomly
distributed building regions. A general MTL model with two stage training
methods are proposed. The proposed model could achieve SR results of PL with
MAE of 3.26 dB and 98% classification accuracy of LOS/NLOS areas given scale
factor as 2. It also outperforms other state-of-the-art DL models and the
reasons are discussed. The proposed model demonstrates huge advantages in
channel characteristics SR tasks especially when scale factor is relatively
large. Ablation study also proved the necessity of residual connection,
iterative up-and-down techinque and multi-task learning. In future, we will
continue the study of channel characteristics SR problem by refining proposed
MTL model on network structure and MTL loss function. Relations between upper
bounds of SR performances and scale factor will also be explored.

## Acknowledgment

This work is supported by the Fundamental Research Funds for the Central
Universities 2020JBZD005, National Science Foundation of China under Grant
61901029, Beijing Natural Science Foundation L212029, the State Key Laboratory
of Rail Traffic Control and Safety (Contract No. RCS2020ZZ005), Beijing
Jiaotong University, Ministry of Education of China under Grant (8091B032123),
ZTE Corporation and State Key Laboratory of Mobile Network and Mobile
Multimedia Technology.

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# Advances in the Geometry of the Ratio of Linear Points in the Desargues
Affine Plane Skew Field

Orgest ZAKA  Orgest ZAKA: Department of Mathematics-Informatics, Faculty of
Economy and Agribusiness, Agricultural University of Tirana, Tirana, Albania
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>and  James F. Peters
James F. PETERS: Department of Electrical & Computer Engineering, University
of Manitoba, WPG, MB, R3T 5V6, Canada and Department of Mathematics, Faculty
of Arts and Sciences, Adi̇yaman University, 02040 Adi̇yaman, Turkey
<EMAIL_ADDRESS>Dedicated to Girard Desargues and R.V. Banavar

###### Abstract.

This paper introduces advances in the geometry of the ratio of either two or
three points in a line in the Desargues affine plane, and we see this as a
ratio of elements of skew field which are constructed over a line in Desargues
affine plane. The results given here have a clean, geometric presentation
based Desargues affine plan axiomatics and definitions of addition and
multiplication of points on a line in this plane, and for skew field
properties. The results in this paper are: (1) study of properties for ratio
of two and three points, in a line on Desargues affine plane. Also, we discuss
the cases related to the ”line-skew field” characteristic, when it is two and
when it is different from two. (2) we have construct the maps for ratio
points-set, for two and three points, and have prove that, this maps are
bijections of the lines. (3) set of ratio points (for two and for three
points) with addition and multiplication of points, forms a skew fields, for
more, this skew fields are sub-skew fields of the ’line-skew field’ on
Desargues affine plane. (4) Every Dyck polygon containing co-linear ratio
vertices in the Desargues affine plane has a free group presentation.

###### Key words and phrases:

Co-linear, Ratio, Skew-Field, Desargues Affine Plane

###### 2010 Mathematics Subject Classification:

51A30; 51E15, 51N25, 30C20, 30F40

The research has been supported by the Natural Sciences & Engineering Research
Council of Canada (NSERC) discovery grant 185986, Instituto Nazionale di Alta
Matematica (INdAM) Francesco Severi, Gruppo Nazionale per le Strutture
Algebriche, Geometriche e Loro Applicazioni grant 9 920160 000362, n.prot U
2016/000036 and Scientific and Technological Research Council of Turkey
(TÜBİTAK) Scientific Human Resources Development (BIDEB) under grant no:
2221-1059B211301223.

##  1\. Introduction

The foundations for the study of the connections between axiomatic geometry
and algebraic structures were set forth by D. Hilbert [10]. And some classic
for this are, E. Artin [1], D.R. Huges and F.C. Piper [11], H. S. M Coxeter
[4]. Marcel Berger in [3], Robin Hartshorne in [8], etc. Even earlier, in we
works [20, 21, 7, 18, 19, 17, 24, 25, 23, 22] we have brought up quite a few
interesting facts about the association of algebraic structures with affine
planes and with ’Desargues affine planes’, and vice versa.

In this article we study ’ratio of 2-points’ and ’ratio of 3-points’ in a line
of the Desargues affine plane. Earlier, we have shown that on each line on
Desargues affine plane, we can construct a skew-field simply and
constructively, using simple elements of elementary geometry, and only the
basic axioms of Desargues affine plane (see [21], [7], [17], [24] ). In this
paper, we consider dilations and translations entirely in the Desargues affine
plane (see [20], [18], [17], [24]).

The novelty in this paper is that we achieve our results without the use
coordinates. Instead, we see the points of a line as elements of the skew
field which is constructed over this line and we make use of properties
enjoyed by transformations in Desargues affine planes such as parallel
projection, translations and dilations and is proved that the this-
transformations preserve the ratio of 2 and 3 points in a line. In addition,
we identify and present every Dyck polygons containing colinear points in the
Desargues affine plane as a free group, which make it possible to achieve a
concise view of complex geometric constructions.

##  2\. Preliminaries

This secion gives a brief introduction to addition and multiplication of co-
linear points in the Desargues affine plane and algebraic properties of skew
fields.

###  2.1. Desargues Affine Plane

Let $\mathcal{P}$ be a nonempty space, $\mathcal{L}$ a nonempty subset of
$\mathcal{P}$. The elements $p$ of $\mathcal{P}$ are points and an element
$\ell$ of $\mathcal{L}$ is a line.

######  Definition 1.

The incidence structure $\mathcal{A}=(\mathcal{P},\mathcal{L},\mathcal{I})$,
called affine plane, where satisfies the above axioms:

  1. 1o

For each points $\left\\{P,Q\right\\}\in\mathcal{P}$, there is exactly one
line $\ell\in\mathcal{L}$ such that $\left\\{P,Q\right\\}\in\ell$.

  2. 2o

For each point $P\in\mathcal{P},\ell\in\mathcal{L},P\not\in\ell$, there is
exactly one line $\ell^{\prime}\in\mathcal{L}$ such that $P\in\ell^{\prime}$
and $\ell\cap\ell^{\prime}=\emptyset$ (Playfair Parallel Axiom [14]). Put
another way, if the point $P\not\in\ell$, then there is a unique line
$\ell^{\prime}$ on $P$ missing $\ell$ [15].

  3. 3o

There is a 3-subset of points $\left\\{P,Q,R\right\\}\in\mathcal{P}$, which is
not a subset of any $\ell$ in the plane. Put another way, there exist three
non-collinear points $\mathcal{P}$ [15].

_Desargues’ Axiom, circa 1630_ [12, §3.9, pp. 60-61] [16]. Let
$A,B,C,A^{\prime},B^{\prime},C^{\prime}\in\mathcal{P}$ and let pairwise
distinct lines
$\ell^{AA_{1}},\ell^{BB^{\prime}},\ell^{CC^{\prime}},\ell^{AC},\ell^{A^{\prime}C^{\prime}}\in\mathcal{L}$
such that

$\displaystyle\ell^{AA_{1}}\parallel\ell^{BB^{\prime}}\parallel\ell^{CC^{\prime}}\
\mbox{(Fig.~{}\ref{fig:DesarguesAxiom}(a))}$ $\displaystyle\ \mbox{{or}}\
\ell^{AA_{1}}\cap\ell^{BB^{\prime}}\cap\ell^{CC^{\prime}}=P.\mbox{(Fig.~{}\ref{fig:DesarguesAxiom}(b)
)}$ $\displaystyle\mbox{and}\ \ell^{AB}\parallel\ell^{A^{\prime}B^{\prime}}\ $
$\displaystyle\ \mbox{and}\ \ell^{BC}\parallel\ell^{B^{\prime}C^{\prime}}.$
$\displaystyle
A,B\in\ell^{AB},A^{\prime}B^{\prime}\in\ell^{A^{\prime}B^{\prime}},$
$\displaystyle\ \mbox{and}\
B,C\in\ell^{BC},B^{\prime}C^{\prime}\in\ell^{B^{\prime}C^{\prime}}.$
$\displaystyle A\neq C,A^{\prime}\neq C^{\prime},$ $\displaystyle\ \mbox{and}\
\ell^{AB}\neq\ell^{A^{\prime}B^{\prime}},\ell^{BC}\neq\ell^{B^{\prime}C^{\prime}}.$

Figure 1. Desargues Axioms: (a) For parallel lines
$\ell^{AA^{\prime}}\parallel\ell^{BB^{\prime}}\parallel\ell^{CC^{\prime}}$;
(b) For lines which are cutting in a single point $P$,
$\ell^{AA^{\prime}}\cap\ell^{BB^{\prime}}\cap\ell^{CC^{\prime}}=P$.

Then $\boldsymbol{\ell^{AC}\parallel\ell^{A^{\prime}C^{\prime}}}$.
$\blacksquare$

######  Example 1.

In Euclidean plane, three vertexes $ABC$ and $A^{\prime}B^{\prime}C^{\prime}$,
are similar (in (a) are equivalent-triangle and in (b) are homothetical-
triangle) the parallel lines,
$\ell^{AC},\ell^{A^{\prime}C^{\prime}}\in\mathcal{L}$ in Desargues’ Axiom are
represented in Fig. 1. In other words, the side $AC$ of the triangle of
$\bigtriangleup ABC$ is parallel with the side $A^{\prime}C^{\prime}$ of the
triangle $\bigtriangleup A^{\prime}B^{\prime}C^{\prime}$, provided the
restrictions on the points and lines in Desargues’ Axiom are satisfied.
$\blacksquare$

A Desargues affine plane is an affine plane that satisfies Desargues’ Axiom.

######  Notation 1.

Three vertexes $ABC$ and $A^{\prime}B^{\prime}C^{\prime}$, which, fulfilling
the conditions of the Desargues Axiom, we call _’Desarguesian’_.

###  2.2. Addition and Multiplication of points in a line of Desargues affine
plane

Addition of points in a line in affine plane. In an Desargues affine plane
$\mathcal{A_{D}}=(\mathcal{P},\mathcal{L},\mathcal{I})$ we fix two different
points $O,I\in\mathcal{P},$ which, according to Axiom 1, determine a line
$\ell^{OI}\in\mathcal{L}.$ Let $A$ and $B$ be two arbitrary points of a line
$\ell^{OI}$. In plane $\mathcal{A_{D}}$ we choose a point $B_{1}$ not incident
with $\ell^{OI}$: $B_{1}\notin\ell^{OI}$ (we call the auxiliary point).
Construct line $\ell_{OI}^{B_{1}},$ which is only according to the Axiom 2.
Then construct line $\ell_{OB_{1}}^{A},$ which also is the only according to
the Axiom 2. Marking their intersection
$P_{1}=\ell_{OI}^{B_{1}}\cap\ell_{OB_{1}}^{A}.$ Finally construct line
$\ell_{BB_{1}}^{P_{1}}.$ For as much as $\ell^{BB_{1}}$ cuts the line
$\ell^{OI}$ in point $B$, then this line, parallel with $\ell^{BB_{1}}$, cuts
the line $\ell^{OI}$ in a single point $C$, this point we called the addition
of points $A$ with point $B$ (Figure 2 (a)).

Multiplication of points in a line in affine plane. Choose in the plane
$\mathcal{A_{D}}$ one point $B_{1}$ not incident with lines $\ell^{OI},$ and
construct the line $\ell^{IB_{1}}$. Construct the line $\ell_{IB_{1}}^{A},$
which is the only accoding to the Axiom 2 and cutting the line
$\ell^{OB_{1}}$. Marking their intersection with $P_{1}=\ell_{IB_{1}}^{A}\cap
OB_{1}.$ Finally, construct the line $\ell_{BB_{1}}^{P_{1}}.$ Since
$\ell^{BB_{1}}$ cuts the line $\ell^{OI}$ in a single point $B$, then this
line, parallel with $\ell^{BB_{1}}$, cuts the line $\ell^{OI}$ in one single
point $C$, this point we called the multiplication of points $A$ with point
$B$ (Figure 2 (b)).

Figure 2.  (a) Addition of points in a line in affine plane, (b)
Multiplication of points in a line in affine plane

The process of construct the points $C$ for adition and multiplication of
points in $\ell^{OI}-$line in affine plane, is presented in the tow algorithm
form

Addition Algorithm

Step.1:

$B_{1}\notin\ell^{OI}$

Step.2:

$\ell_{OI}^{B_{1}}\cap\ell_{OB_{1}}^{A}=P_{1}$

Step.3:

$\ell_{BB_{1}}^{P_{1}}\cap\ell^{OI}=C(=A+B)$

Multiplication Algorithm

Step.1:

$B_{1}\notin\ell^{OI}$

Step.2:

$\ell_{IB_{1}}^{A}\cap\ell^{OB_{1}}=P_{1}$

Step.3:

$\ell_{BB_{1}}^{P_{1}}\cap\ell^{OI}=C(=A\cdot B)$

In [17] and [7], we have prove that $(\ell^{OI},+,\cdot)$ is a skew field in
Desargues affine plane, and is field (commutative skew field) in the Papus
affine plane.

###  2.3. Some algebraic properties of Skew Fields

I n this section $K$ will denote a skew field [9] and $z[K]$ its center, where
is the set $K$ such that

$z[K]=\left\\{k\in K\quad|\quad ak=ka,\quad\forall a\in K\right\\}$

######  Proposition 1.

$z[K]$ is a commutative subfield of a skew field $K$.

###### Proof.

Show first, that $O,I\in z[K]$, because $O\cdot X=X\cdot O$ and $I\cdot
X=X\cdot I$, $\forall X\in K$ (for all point $X\in\ell^{OI}-$line), so $z[K]$
is non-empty.

If $A,B\in z[K]$ and $C\in K$, have

$(A-B)C=AC-BC=CA-CB=C(A-B),$

then

$\forall A,B\in z[K]\Rightarrow A-B\in z[K].$

let’s see now,

$(AB)C=A(BC)=A(CB)=(AC)B=(CA)B=C(AB),$

then, we have

$\forall A,B\in z[K]\Rightarrow AB\in z[K].$

If $A\in z[K]$, and $A\neq O$, then $A^{-1}\in z[K]$, truly, is clear that
$(A^{-1})^{-1}=A$, and let $B\in K$, then

$B^{-1}A^{-1}=(AB)^{-1}=(BA)^{-1}=A^{-1}B^{-1},$

thus, for $A\in z[K]$, and $A\neq O$, and $\forall B\in z[K]$, we have that,

$B^{-1}A^{-1}=A^{-1}B^{-1}\Rightarrow A^{-1}\in z[K]$

Thus, $z[K]$ is a commutative subring of $K$, then is a subfield of skew field
$K$. ∎

Let now $p\in K$ be a fixed element of the skew field $K$. We will denote
$z_{K}(p)$ the centralizer in $K$ of the element $p$, where is the set,

$z_{K}(p)=\left\\{k\in K|pk=kp,\right\\}.$

$z_{K}(p)$ is sub skew field of K, but, in general, it is not commutative.

Let $K$ be a skew field, $p\in K$, and let us denote by $[p_{K}]$ the
conjugacy class of $p$:

$[p_{K}]=\left\\{q^{-1}pq\quad|\quad q\in K\setminus\\{0\\}\right\\}$

If, $p\in z[K]$, for all $q\in K$ we have that $q^{-1}pq=p$

##  3\. Ratio of two and three Points in a line of Desargues affine plane

In this section, we will give the main result in this article. We will study
the ’ration of two points’ and ’ratio of three points’ on a line in Desargues
affine planes, we utilize a method that is naive and direct, in the abstract
and axiomatic sense of it, without requiring the concept of coordinates, and
metrics. We will defined the ’ratio of two points’ and ’ratio of three points’
as a point in the Desargues affine plane. For achieve our results, we do not
use coordinates, and follow simple techniques, of elementary geometry, for
proofs. We are mainly based on our previous work, on the close connection we
found between the lines, in Desargues affine planes and skew fields.

Mainly, we rely on our results regarding the addition and multiplication of
points on a line of the Desargues affine plane, and the fact that a line (set
of points), with addition and multiplication, forms a skew field (for more,
see [17], [21], [7], [19], [18], [20], [24], [25]).

We will select some transformations (some of them we are defining ourselves)
which are invariant regarding the ratio of points. As well on the properties
enjoyed by transformations in Desargues affine planes, like, parallel
projection, translations and dilations (see [18], [20], [24]), we prove that
this transformations preserving ’ratio of two points’ and ’ratio of three
points’ in a line on a Desargues affine plane. We use skew fields properties,
for our proofs.

###  3.1. Ratio of 2-Points in a line of Desargues affine plane

In this section we will study the ’ration’ of 2-points on a line in Desargues
affine planes. We will define ’ratio’ as a point in the Desargues affine
plane.

######  Definition 2.

Lets have two different points $A,B\in\ell^{OI}-$line, and $B\neq O$, in
Desargues affine plane. We define as ratio of this tow points, a point
$R\in\ell^{OI}$, such that,

$R=B^{-1}A$

we mark this, with,

$R=r(A:B)=B^{-1}A$

For a ’ratio-point’ $R\in\ell^{OI}$, and for point $B\neq O$ in line
$\ell^{OI}$, is a unique defined point, $A\in\ell^{OI}$, such that
$R=B^{-1}A=r(A:B)$.

Figure 3. Ilustrate the Ratio-Point, of 2-Points in a line of Desargues affine
plane $R=r(A:B)=B^{-1}A$.

######  Remark 1.

Our caution is maximum, since the points of a line in Desargues affine planes
are also elements of skew-fields built on it, and the commutative property for
multiplication does not hold! For this reason we have

$B^{-1}\cdot A\neq A\cdot B^{-1}.$

However, if we had defined the ratio of 2-points as,

$r(A:B)=A\cdot B^{-1},$

the results would be similar. However, we are keeping our definition!

######  Properties 1.

From Def.2, we derive a number of properties related to the position of the
points $A,B$ in $\ell^{OI}-$line in the Desargues affine plane.

  * •

If $A=B\neq O$, then $r(A:B)=r(A:A)=A^{-1}\cdot A=I$.

  * •

If $A=O$ and $B\neq O$, then $r(A:B)=r(O:B)=B^{-1}\cdot O=O$.

  * •

If $A=I$ and $B\neq O$, then $r(A:B)=r(O:I)=B^{-1}\cdot I=I$.

  * •

If $A=I$ and $B\neq I$, then $r(A:B)=r(I:I)=I^{-1}\cdot I=I$.

  * •

If $A\neq O$ and $B=O$, then $r(A:B)=r(A:O)=O^{-1}\cdot A=\infty$ (infinite
point).

######  Theorem 1.

If have two different points $A,B\in\ell^{OI}-$line, and $B\neq O$, in
Desargues affine plane, then,

$r^{-1}(A:B)=r(B:A)$

###### Proof.

From definition 2, we have that $r(A:B)=B^{-1}A$, we appreciate,

$\displaystyle r^{-1}(A:B)$ $\displaystyle=[r(B:A)]^{-1}$
$\displaystyle=[B^{-1}A]^{-1}$ $\displaystyle=A^{-1}[B^{-1}]^{-1}$
$\displaystyle=A^{-1}B\quad\text{(in skew fields
$(ab)^{-1}=b^{-1}a^{-1},\forall a,b$)}$ $\displaystyle=r(B:A)$

∎

######  Theorem 2.

For three collinear point $A,B,C$ and $C\neq O$, in $\ell^{OI}-$line, have,

$r(A+B:C)=r(A:C)+r(B:C)$

###### Proof.

From definition 2, we have that $r(A:B)=B^{-1}A$, we appreciate,

$\displaystyle r(A+B:C)$ $\displaystyle=C^{-1}(A+B)$
$\displaystyle=C^{-1}A+C^{-1}B)$ (from distribution property of skew fields
$K=(\ell^{OI},+,\cdot)$) $\displaystyle=r(A:C)+r(B:C)$

∎

######  Theorem 3.

For three collinear point $A,B,C$ and $C\neq O$, in $\ell^{OI}-$line, have,

  1. (1)

$r(A\cdot B:C)=r(A:C)\cdot B.$

  2. (2)

$r(A:B\cdot C)=C^{-1}r(A:C).$

$r(A\cdot B:C)=r(A:C)\cdot B$

###### Proof.

(1) From definition 2, we have that $r(A:B)=B^{-1}A$, we appreciate,

$\displaystyle r(A\cdot B:C)$ $\displaystyle=C^{-1}(A\cdot B)$
$\displaystyle=(C^{-1}A)\cdot B$ (from associative property of skew fields
$K=(\ell^{OI},+,\cdot)$) $\displaystyle=r(A:C)\cdot B.$

And for (2),

$\displaystyle r(A:B\cdot C)$ $\displaystyle=[BC]^{-1}A$
$\displaystyle=(C^{-1}B^{-1})\cdot A$ $\displaystyle=C^{-1}(B^{-1}\cdot A)$
(from associative property of skew fields $K=(\ell^{OI},+,\cdot)$)
$\displaystyle=C^{-1}r(A:C).$

∎

######  Theorem 4.

Let’s have the points $A,B\in\ell^{OI}-$line where $B\neq O$. Then have that,

$r(A:B)=r(B:A)\Leftrightarrow A=B.$

###### Proof.

Suppose that $r(A:B)=r(B:A)$, so $B^{-1}A=A^{-1}B$, but
$B^{-1}A=[A^{-1}B]^{-1}$, then we have,

$[A^{-1}B]^{-1}=A^{-1}B.$

But points, $A,B,A^{-1},B^{-1},A^{-1}B,B^{-1}A\in\ell^{OI}$, and
$(\ell^{OI},+,\cdot)$ is skew field. From skew field properties: If
$a^{-1}=a\Rightarrow a^{2}=1$, then element $a$ is _idempotent_ in skew field,
but, but, the idempotent elements in the skew field are alone $0$ and $1$
(see, e.g., [9]). So to us, we have that

$A^{-1}B-\text{Idempotent element of skew field $(\ell^{OI},+,\cdot)$,}$

hence,

$A^{-1}B=O\quad\text{or}\quad A^{-1}B=I$

so,

$A^{-1}B=O\Rightarrow B=O\quad\text{which contradicts}$

then, we have that,

$A^{-1}B=I\Rightarrow B=A.$

∎

We can say, now, that all two points in $\ell^{OI}-$line produce a point in
it, which we have called ’ratio of two points’. So, we can construct the map:

$r_{B}(X):\ell^{OI}\to\ell^{OI},$

such that,

$r_{B}(X)=r(X:B)=B^{-1}X,\forall X\in\ell^{OI}$

######  Theorem 5.

This ratio-map, $r_{B}:\ell^{OI}\to\ell^{OI}$ is a bijection in
$\ell^{OI}-$line in Desargues affine plane.

###### Proof.

Injective: If have,

$r_{B}(X)=r_{B}(Y)\Leftrightarrow B^{-1}X=B^{-1}Y$

then

$B^{-1}X-B^{-1}Y=O\Leftrightarrow B^{-1}(X-Y)=O.$

Let’s remember that these are the points of the skew fields $K$, and the skew
field has no ’divisor of zero’, for this reason we have,

$B^{-1}(X-Y)=O\Leftrightarrow B^{-1}=O\quad\text{or}\quad X-Y=O,$

but $B^{-1}\neq O$, then have that $X-Y=O\Leftrightarrow X=Y.$

Surjective: $\forall R\in\ell^{OI},\exists X\in\ell^{OI}$, such that
$r_{B}(X)=R$.
We know that, $r_{B}(X)=B^{-1}X=R\Leftrightarrow X=B\cdot R$, so,

$\forall R\in\ell^{OI},\exists X=B\cdot R\in\ell^{OI},\quad\text{i.e.,}\quad
r_{B}(X)=r_{B}(B\cdot R)=B^{-1}(BR)=(B^{-1}B)R=R.$

∎

For this map we have, $r_{B}(B)=r(B:B)=B^{-1}B=I$, $r_{B}(O)=r(O:B)=B^{-1}O=I$
and $r_{B}(I)=r(I:B)=B^{-1}I=B^{-1}$, and for all point $X\in\ell^{OI}$ have
$r_{B}(X)=r(X:B)$, which is the ratio of two points $X$ and $B$.

The map $r_{B}(X)=r(X:B)$ is a point in $\ell^{OI}$ line, for this reason,
addition of points (our algorithm for collecting points on a line of Desargues
affine plane), which we used earlier, is valid. We see that the set,

$\mathcal{R}_{2}=\\{r_{B}(X)|\forall X\in\ell^{OI},\\}$

It’s clear that $\mathcal{R}_{2}\subseteq\ell^{OI},$ and
$\mathcal{R}_{2}\neq\emptyset$, below we will prove that this set, is a sub-
skew field of $K=(\ell^{OI},+,\cdot)$. Let’s see now the following theorems,

######  Theorem 6.

Let’s have the point $B$ different for point $O$, in $\ell^{OI}-$line. The
ratio-maps-set $\mathcal{R}_{2}=\\{r_{B}(X)|\forall X\in\ell^{OI}\\}$ forms a
_Abelian_ (commutative) group with addition of points in $\ell^{OI}-$line.

###### Proof.

1. (Associativity) $\forall X,Y,Z\in\ell^{OI}$, we have that:

$\displaystyle\left[r_{B}(X)+r_{B}(Y)\right]+r_{B}(Z)$
$\displaystyle=[B^{-1}X+B^{-1}Y]+B^{-1}Z$ $\displaystyle=B^{-1}[X+Y]+B^{-1}Z$
$\displaystyle=B^{-1}[(X+Y)+Z]$ (from asociatiove properties of addition
groups $(\ell^{OI},+)$.) $\displaystyle=B^{-1}[X+(Y+Z)]$
$\displaystyle=B^{-1}X+B^{-1}(X+Y)$ $\displaystyle=B^{-1}X+[B^{-1}Y+B^{-1}Z]$
$\displaystyle=r_{B}(X)+[r_{B}(Y)+r_{B}(Z)],$

so,

$[r_{B}(X)+r_{B}(Y)]+r_{B}(Z)=r_{B}(X)+[r_{B}(Y)+r_{B}(Z)],\forall
X,Y,Z\in\ell^{OI}.$

2. (Commuativity) $\forall X,Y\in\ell^{OI}$, we have that:

$\displaystyle r_{B}(X)+r_{B}(Y)$ $\displaystyle=B^{-1}X+B^{-1}Y$
$\displaystyle=B^{-1}[X+Y]$ (from commutative property of addition groups
$(\ell^{OI},+)$.) $\displaystyle=B^{-1}[Y+X]$ $\displaystyle=B^{-1}Y+B^{-1}X$
$\displaystyle=r_{B}(Y)+r_{B}(X),$

so

$r_{B}(X)+r_{B}(Y)=r_{B}(Y)+r_{B}(X),\forall X,Y\in\ell^{OI}$

3. (’zero’) $\exists Z\in\ell^{OI},\forall X\in\ell^{OI}$, we have that:

$\displaystyle r_{B}(X)+r_{B}(Z)$ $\displaystyle=r_{B}(X)$ $\displaystyle
B^{-1}X+B^{-1}Z$ $\displaystyle=B^{-1}X$ $\displaystyle B^{-1}[X+Z]$
$\displaystyle=B^{-1}X$

so

$B^{-1}[X+Z]=B^{-1}(X)\Rightarrow X+Z=X\Leftrightarrow Z=O.$

Hence the ’zero’ element is $r_{B}(O)$.

4. (’opposite’) $\forall X\in\ell^{OI},\exists Y\in\ell^{OI}$, we have that:

$\displaystyle r_{B}(X)+r_{B}(Y)$ $\displaystyle=r_{B}(O)$ $\displaystyle
B^{-1}X+B^{-1}Y$ $\displaystyle=B^{-1}O$ $\displaystyle B^{-1}[X+Y]$
$\displaystyle=O$

so

$B^{-1}[X+Y]=B^{-1}O\Rightarrow X+Y=O\Leftrightarrow Y=-X.$

Hence the ’opposite’ element of $r_{B}(X)$ is $r_{B}(-X)$, $\forall
X\in\ell^{OI}$. ∎

######  Theorem 7.

Let’s have the point $B$ different for point $O$, in $\ell^{OI}-$line. The
ratio-maps-set $\\{r_{B}(X)|\forall X\in\ell^{OI}\\}$ forms a group with
’multiplication of points in $\ell^{OI}-$line.

###### Proof.

1. (Associativity) $\forall X,Y,Z\in\ell^{OI}$, we have that:

$[r_{B}(X)\cdot r_{B}(Y)]\cdot r_{B}(Z)=[B^{-1}X\cdot B^{-1}Y]\cdot B^{-1}Z$

2. (’unitary’) $\exists E\in\ell^{OI},\forall X\in\ell^{OI}$, we have that:

$\displaystyle r_{B}(X)\cdot r_{B}(E)$ $\displaystyle=r_{B}(X)$ $\displaystyle
B^{-1}X\cdot B^{-1}E=B^{-1}X$ $\displaystyle X\cdot B^{-1}E$ $\displaystyle=X$
$\displaystyle B^{-1}E$ $\displaystyle=X^{-1}X$ $\displaystyle B^{-1}E$
$\displaystyle=I$ $\displaystyle E$ $\displaystyle=B$

so unitary-element element is $r_{B}(B)$.

3. (’inverse’) $\forall X\in\ell^{OI},\exists Y\in\ell^{OI}$, we have that:

$\displaystyle r_{B}(X)\cdot r_{B}(Y)$ $\displaystyle=r_{B}(B)$ $\displaystyle
B^{-1}X\cdot B^{-1}Y$ $\displaystyle=I$ $\displaystyle X\cdot B^{-1}Y$
$\displaystyle=B$ $\displaystyle B^{-1}Y$ $\displaystyle=X^{-1}\cdot B$
$\displaystyle Y$ $\displaystyle=B\cdot X^{-1}\cdot B$

so

$r_{B}(Y)=B^{-1}Y=B^{-1}[B\cdot X^{-1}\cdot B]=X^{-1}\cdot B=r_{X}(B).$

Hence the ’inverse’ element of $r_{B}(X)$ is $r_{X}(B)$, $\forall
X\in\ell^{OI}$. ∎

######  Theorem 8.

In the ratio-maps-set $\\{r_{B}(X)|\forall X\in\ell^{OI}\\}$, for three
elements $r_{B}(X)$, $r_{B}(Y)$, $r_{B}(Z)$ of its, have true, that equations

  1. (1)

$r_{B}(X)\cdot[r_{B}(Y)+r_{B}(Z)]=r_{B}(X)\cdot r_{B}(Y)+r_{B}(X)\cdot
r_{B}(Z)$, and,

  2. (2)

$[r_{B}(X)+r_{B}(Y)]r_{B}(Z)=r_{B}(X)\cdot r_{B}(Z)+r_{B}(Y)\cdot r_{B}(Z)$

###### Proof.

Indeed $r_{B}(X)\cdot[r_{B}(Y)+r_{B}(Z)]=B^{-1}X[B^{-1}Y+B^{-1}Z]$, but the
’ratio’ points are points in $\ell^{OI}-$line, so they are also points of skew
fields $K=(\ell^{OI},+.\cdot)$, therefore, they enjoy the distribution
property, therefore

$B^{-1}X[B^{-1}Y+B^{-1}Z]=(B^{-1}X)(B^{-1}Y)+(B^{-1}X)(B^{-1}Z),$

so

$r_{B}(X)\cdot[r_{B}(Y)+r_{B}(Z)]=r_{B}(X)\cdot r_{B}(Y)+r_{B}(X)\cdot
r_{B}(Z).$

The other point (2) is proved in the same way. ∎

From the above three theorems 6, 7 and 8, have true, this,

######  Theorem 9.

The ratio-maps-set $\mathcal{R}_{2}=\\{r_{B}(X)|\forall X\in\ell^{OI}\\}$, for
a fixed point $B$ in $\ell^{OI}-$line, forms a skew-field with ’addition and
multiplication’ of points in $\ell^{OI}-$line.

This, skew field $(\mathcal{R}_{2},+,\cdot)$ is sub-skew field of the skew
field $(\ell^{OI},+,\cdot)$.

###  3.2. Ratio of 3-points in a line of Desargues affine plane

In this section we will study the ’ration’ of points on a line in Desargues
affine planes. We will define ’ratio’ as a point in the Desargues affine
plane.

The classical definition of the ratio (which is often called ”simple ratio”)
(see, e.g., [10], [2], [3]), which is given as a ratio of lengths, (in
coordinate geometry). So, for example, for three collinear points $A,B,C$,

$R=\frac{AC}{BC}$

where $AC$ is length of segment $[AB]$ and $BC$ is length of segment $[BC]$,
or in vectorial form: ’for three collinear points, $A,B,C$, a ’simple ratio’
called the ’scalar’ $\lambda$, such that,

$\vec{AC}=\lambda\vec{BC}$

Since we will not use coordinates and metrics, so we have only the axiomatic
of Desargues affine plane (our objects are only points and lines), we take as
definition the strictly algebraic, somewhat abstractly,

######  Notation 2.

For three points $A,B,C$ on a line $\ell^{OI}$ (collinear) in Desargues affine
plane, $B-C$ denotes the point $B+(-C)$ and $A-C$ denotes the point $A+(-C)$.
The operator ’-’ denotes the addition of point $B$ with an oposite point of
point $C$ [21], [7], [17]. ◼

######  Definition 3.

If $A,B,C$ are three points on a line $\ell^{OI}$ (collinear) in Desargues
affine plane, then we define their ratio to be a point $R\in\ell^{OI}$, such
that:

$(B-C)\cdot R=A-C,\quad\mbox{concisely}\quad R=(B-C)^{-1}(A-C)$

and we mark this with

$r(A,B;C)=(B-C)^{-1}(A-C)$

Figure 4. Ratio of 3-Points in a line of Desargues affine plane $R=r(A,B;C)$.

######  Example 2.

In Fig. 3, we have $R$ equal to the ratio of points $A$ and point $B$ in the
line $\ell^{OI}$ in the Desargues affine plane. Similarly, in Fig. 4, from
Def. 3, we have the ratio $R=r(A,B;C)$ of points $A,B,C$. ◼

######  Remark 2.

Similarly as in the ratio of two points, we can also define a ratio of three
points, as

$r(A,B;C)=(A-C)(B-C)^{-1},$

the results would be similar. Except in $\ell^{OI}-$line, in the Desargues
affine planes, the usual ratio points differ from ratios in Def. 3, since:

$(A-C)(B-C)^{-1}\neq(B-C)^{-1}(A-C).$

However, we are keeping our definition! ◼

In the [17], [21], [7], we prove that an line $\ell^{OI}$in Desargues affine
plane, with ’addition’ and ’multiplication’ of points in it, forms a Skew-
Field, so $(\ell^{OI},+,\cdot)-$is a skew field. For this reason we can define
it, ’ratio’ also for skew-fields. So, for a skew fields $K$, if we have
elements $a,b,c\in K$ then ratio we call the point $r=(b-c)^{-1}(a-c)$.
However, our work is more geometric, so we will follow the geometric
interpretation. But we will always keep in mind that: ’a line in Desargues
affine plane, is a skew-field’.

For a ’ratio-point’ $R\in\ell^{OI}$ and tow points $B,C\in\ell^{OI}$, is a
uniquely defined point $A\in\ell^{OI}$, such that $R=r(A,B;C)$.

######  Theorem 10.

For 3-points $A,B,C$ in a line $\ell^{OI}$ of Desargues affine plane, we have
that,

$r(-A,-B;-C)=r(A,B;C)$

###### Proof.

For proof of this, we will use the definition of ratio, and skew-fields
properties, so we have,

$\displaystyle r(-A,-B;-C)$ $\displaystyle=(-B-(-C))^{-1}(-A-(-C))$
$\displaystyle=(-B+C)^{-1}(-A+C)$ $\displaystyle=(-I[B-C])^{-1}[-I](A-C)$
$\displaystyle=(B-C)^{-1}[-I]^{-1}[-I](A-C)$
$\displaystyle=(B-C)^{-1}[-I][-I](A-C)$ $\displaystyle=(B-C)^{-1}(A-C)$
$\displaystyle=r(A,B;C)$

From skew fields properties we have that $(ab)^{-1}=b^{-1}a^{-1}$and $ab\neq
ba$, $[-I]^{-1}=-I$, and $[-I][-I]=I.$ ∎

######  Theorem 11.

For 3-points $A,B,C$ in a line $\ell^{OI}$ in the Desargues affine plane, have

$r^{-1}(A,B;C)=r(B,A;C)$

###### Proof.

For proof of this, we will use the definition of ratio, and skew-fields
properties, so we have,

$\displaystyle r^{-1}(A,B;C)$ $\displaystyle=[(B-C)^{-1}(A-C)]^{-1}$
$\displaystyle=(A-C)^{-1}[(B-C)^{-1}]^{-1}$ $\displaystyle=(A-C)^{-1}(B-C)$
$\displaystyle=r(B,A;C)$

∎

######  Properties 2.

From Def. 3, we derive a number of properties related to the position of the
points $A,B,C$ in $\ell^{OI}-$line in the Desargues affine plane.

  * •

If $A=B=O$ and $C\neq O$, have,

$r(A,B;C)=(O-C)^{-1}(O-C)=I.$

  * •

If $A=B=O$ and $C=I$, have,

$r(A,B;C)=(O-I)^{-1}(O-I)=I.$

  * •

If $A=B=I$ and $C=O$, have ,

$r(A,B;C)=(I-O)^{-1}(I-O)=I.$

  * •

If $A=B=I$ and $C\neq I$, have,

$r(A,B;C)=(I-C)^{-1}(I-C)=I.$

  * •

If $A=B=O$ and $C\neq O$ and $C\neq I$, have,

$r(A,B;C)=(O-C)^{-1}(O-C)=I.$

  * •

If $A=B=I$ and $C\neq O$ and $C\neq I$, have,

$r(A,B;C)=(I-C)^{-1}(I-C)=I.$

  * •

If ’$A=B\neq O$ and $A=B\neq I$’ and $C\neq A=B$, then

$r(A,B;C)=(A-C)^{-1}(A-C)=I.$

  * •

If $A=C$ and $B\neq A=C$, then $r(A,B;C)=O$.

  * •

If $B=C$ and $A\neq C$, then $r(A,B;C)-$is the infinite point, witch we mark
with ’$\infty$’. ◼

From Properties 2, we obtain the following result.

######  Theorem 12.

For 3-points in $\ell^{OI}-$line in the Desargues affine plane. if $A=B$ and
$C\neq A=B$, then $r(A,B;C)=I$, where $I$ is unitary point for ’multiplication
of points’ in $\ell^{OI}-$line

For three points $A,B,C$ such that $r(A,B;C)=R$, we have defined above,
$r(A,B;C)=(B-C)^{-1}(A-C)=R$. We can transform this expression into the form
used by Lüngerburg in [13], and have that,

$(B-C)^{-1}(A-C)=R\Rightarrow A-C=(B-C)R\Rightarrow A-C=BR-CR$ so

$A=BR+C-CR\Rightarrow A=BR+C(I-R)$

This form makes it easier for us to study the following cases, when, $B=C$ and
$A=C$! So,

  * •

If $B=C$, then have $A=BR+C(I-R)\Rightarrow A=BR+B(I-R)$, so

$A=BR+B-BR\Rightarrow\textbf{A=B}.$

So ’If $B=C$ then $B=C=A$’.

  * •

If $A=C$, then we have $A=BR+C(I-R)\Rightarrow A=BR+A(I-R)$, so,

$A=BR+A-AR\Rightarrow O=BR-AR\Rightarrow\textbf{B=A}.$

So ’If $A=C$ then $A=C=B$’.

Immediate from Properties 2, we obtain the following results.

######  Theorem 13.

For three points $A,B,C$ in a line $\ell^{OI}$ on Desargues affine plane,
i.e., $r(A,B;C)=R$,

  1. (1)

If $B=C$ then $B=C=A$.

  2. (2)

If $A=C$ then $A=C=B$.

######  Theorem 14.

If $A,B,C$, are three different points, and different from point $O$, in a
line $\ell^{OI}$ on Desargues affine plane, then

$r(A^{-1},B^{-1};C^{-1})=B[r(A,B;C)]A^{-1}.$

###### Proof.

Points $A,B,C$ and $A^{-1},B^{-1},C^{-1}$, are points of $\ell^{OI}-$line in
Desargues affine plane, so are and elements of the skew field
$K=(\ell^{OI},+,\cdot)$. First we prove that, for tow elements $X,Y$ in a skew
field $K$, we have that

$X^{-1}-Y^{-1}=Y^{-1}(Y-X)X^{-1}.$

Indeed

$\displaystyle Y^{-1}(Y-X)X^{-1}$ $\displaystyle=[Y^{-1}(Y-X)]X^{-1}$
$\displaystyle=(Y^{-1}Y-Y^{-1}X)X^{-1}$ $\displaystyle=(I-Y^{-1}X)X^{-1}$
$\displaystyle=IX^{-1}-Y^{-1}(XX^{-1})$ $\displaystyle=X^{-1}-Y^{-1}I$
$\displaystyle=X^{-1}-Y^{-1}.$

We use this result in the calculation of $r(A^{-1},B^{-1};C^{-1})$, we will
use the definition of ratio 3, and this skew-fields properties, so we have,

$\displaystyle r(A^{-1},B^{-1};C^{-1})$
$\displaystyle=(B^{-1}-C^{-1})^{-1}(A^{-1}-C^{-1})$
$\displaystyle=[C^{-1}(C-B)B^{-1}]^{-1}[C^{-1}(C-A)A^{-1}]$
$\displaystyle=[B(C-B)^{-1}C][C^{-1}(C-A)A^{-1}]$
$\displaystyle=B(C-B)^{-1}[CC^{-1}](C-A)A^{-1}$
$\displaystyle=B(C-B)^{-1}[CC^{-1}](C-A)A^{-1}$
$\displaystyle=B(C-B)^{-1}[I](C-A)A^{-1}$
$\displaystyle=B(C-B)^{-1}(C-A)A^{-1}$
$\displaystyle=B[(C-B)^{-1}(C-A)]A^{-1}$
$\displaystyle=B[([-I](B-C))^{-1}[-I](A-C)]A^{-1}$
$\displaystyle=B[(B-C)^{-1}[-I]^{-1}[-I](A-C)]A^{-1}$
$\displaystyle=B[(B-C)^{-1}[-I][-I](A-C)]A^{-1}$
$\displaystyle=B[(B-C)^{-1}(A-C)]A^{-1}$ $\displaystyle=B[r(B,A;C)]A^{-1}.$

From skew fields properties we have that $(abc)^{-1}=c^{-1}b^{-1}a^{-1}$and
$ab\neq ba$, $[-I]^{-1}=-I$, and $[-I][-I]=I.$ ∎

######  Corollary 1.

In the Pappus affine plane, for three point different from point $O$, in
$\ell^{OI}-$line, we have

$r(A^{-1},B^{-1};C^{-1})=r(A,B;C)\cdot r(B,A;O).$

###### Proof.

In theorem 14 we prove that,

$r(A^{-1},B^{-1};C^{-1})=B[r(A,B;C)]A^{-1}.$

We also know that the skew field, which constructed on a line in Pappus affine
plane, is commutative, so have

$r(A^{-1},B^{-1};C^{-1})=B[r(A,B;C)]A^{-1}=r(A,B;C)[A^{-1}B].$

We also know that $A=A-O$ and $B=B-O$, from construction of skew fields, then,

$A^{-1}B=(A-O)^{-1}(B-O)=r(B,A;O).$

Hence,

$r(A^{-1},B^{-1};C^{-1})=r(A,B;C)r(B,A;O).$

∎

We can say, now, that all three points in $\ell^{OI}-$line produce a point,
which we have called ’ratio’. So, we can construct the map:

$r_{BC}(X)=r(X,B;C)=(B-C)^{-1}(X-C),\forall X\in\ell^{OI}$

We call this map, ’equation of line which passes through the points $B$ and
$C$’, (in fact, this is the line $\ell^{OI}$, since $B,C\in\ell^{OI}$).

For this line we have, $r_{BC}(C)=r(C,B;C)=(B-C)^{-1}(C-C)=O$ and
$r_{BC}(B)=r(B,B;C)=(B-C)^{-1}(B-C)=I$, and for all point $X\in\ell^{OI}$ have
$r_{BC}(X)=r(X,B;C)$, which is the ratio point.

If we write the equation of the line $\ell^{OI}$, in the classical form, as
$M\cdot X+N$, for two fixed points $M,N\in\ell^{OI}$ and $\forall
X\in\ell^{OI}$. In fact, it is easy to find who the points $M$ and $N$ are, we
transform the expression,

$r_{BC}(X)=r(X,B;C)=(B-C)^{-1}(X-C)=(B-C)^{-1}\cdot X+[(B-C)^{-1}(-C)]$

so have points $M=(B-C)^{-1}$ and $N=(B-C)^{-1}(-C)$.

And since for the three points $A,B,C\in\ell^{OI}$ we have that ’lines’
$\ell^{CA}\equiv\ell^{CB}\equiv\ell^{OI}$, we have that,

$r_{CA}(X)=r(X,C;A)=(C-A)^{-1}(X-A),\forall X\in\ell^{OI}$

and

$r_{CB}(X)=r(X,C;B)=(C-B)^{-1}(X-B),\forall X\in\ell^{OI}$

also

$r_{CA}(X)=M\cdot X+N\quad\text{and}\quad r_{CB}(X)=M\cdot X+N.$

These equations are true and for points, $B$ and $A$, respectively, so

$r_{CA}(B)=r(B,C;A)=(C-A)^{-1}(B-A)=M\cdot B+N,$

and

$r_{CB}(A)=r(A,C;B)=(C-B)^{-1}(A-B)=M\cdot A+N.$

If we subtract side-by-side, we have,

$r_{CB}(A)-r_{CA}(B)=((C-B)^{-1}(A-B)-(C-A)^{-1}(B-A)=MA+N-MB-N,$

so

$(C-B)^{-1}(A-B)-(C-A)^{-1}(B-A)=M(A-B)$

thus

$(C-B)^{-1}(A-B)+(C-A)^{-1}(A-B)=M(A-B)$

since $B-A=[-I](A-B)$ and

$(C-A)^{-1}(B-A)=(C-A)^{-1}[-I](A-B)=[-I](C-A)^{-1}(A-B).$

The factor $A-B$ appears on both sides and we factorize this and have,

$[(C-B)^{-1}+(C-A)^{-1}](A-B)=M(A-B)$

Since the points $A$ and $B$, we have assumed them to be different, we have
that $A-B\neq O$, then, there exists the inverse $(A-B)^{-1}$. We multiply
from the right side by side by $(A-B)^{-1}$ and we have,

$[(C-B)^{-1}+(C-A)^{-1}](A-B)(A-B)^{-1}=M(A-B)(A-B)^{-1},$

hence,

$[(C-B)^{-1}+(C-A)^{-1}]=M.$

Now we can discuss a little about this value of point $M$, and ask whether the
point $M$ can be exactly the point $O$.

If $M=O$, we have that,

$(C-B)^{-1}+(C-A)^{-1}=O\Leftrightarrow(C-B)^{-1}=-(C-A)^{-1}\Leftrightarrow(C-B)=-(C-A)$

This is equivalent with,

$C+C=A+B\quad\text{or}\quad 2C=A+B.$

If characteristic of skew field $K=(\ell^{OI},+,\cdot)$ is different from $2$
($Char(K)\neq 2$), we have that, the point $C$ is _”midpoint”_ of points $A$
and $B$.

If characteristic of skew field $K=(\ell^{OI},+,\cdot)$ is $2$, ($Char(K)=2$)
then we have that, $A+B=O$ and $2A=O$, so $A+B=2A\Leftrightarrow A-B=O$, for
this we have $A=B$ which is against our assumption!

Therefore, the point $M=O$, if and only if characteristic of skew field
$K=(\ell^{OI},+,\cdot)$ is different from $2$, and if $A+B=C+C.$

For $M=O$, we have true the equation

$r_{CB}(A)=r(A,C;B)=(C-B)^{-1}(A-B)=N,\quad\text{and}\quad A+B=C+C,$

We substitute $A=C+C-B$ and have,

$\displaystyle N$ $\displaystyle=(C-B)^{-1}(C+C-B-B)$
$\displaystyle=(C-B)^{-1}[(C-B)+(C-B)]$
$\displaystyle=(C-B)^{-1}(C-B)+(C-B)^{-1}(C-B)$ $\displaystyle=I+I=2I.$

So, $N=I+I$. We also see the other equation,

$r_{CA}(B)=r(B,C;A)=(C-A)^{-1}(B-A)=N,\quad\text{and}\quad A+B=C+C,$

We substitute $B=C+C-A$ and have,

$\displaystyle N$ $\displaystyle=(C-A)^{-1}(B-A)$
$\displaystyle=(C-A)^{-1}(C+C-A-A)$ $\displaystyle=(C-A)^{-1}[(C-A)+(C-A)]$
$\displaystyle=(C-A)^{-1}(C-A)+(C-A)^{-1}(C-A)$ $\displaystyle=I+I.$

So, $N=I+I$.

In general, the form $MX+N$ is line equation for the line $\ell^{OI}$, we can
find exactly, who are the points $M$ and $N$, in $\ell^{OI}-$line. Since we
know that the points $O,I\in\ell^{OI}$.

Indeed, let us have, points $B,C\ell^{OI}$, we have the line equation for
$\ell^{BC}\equiv\ell^{OI}$,

$r_{BC}(X)=r(X,B;C)=(B-C)^{-1}(X-C),\forall X\in\ell^{OI}$

and on the other hand, we will have,

$r_{BC}(X)=MX+N,\forall X\in\ell^{OI}$

so,

$(B-C)^{-1}(X-C)=MX+N,\forall X\in\ell^{OI}$

Since $O\in\ell^{OI}$, we have true the equations,

$(B-C)^{-1}(O-C)=M\cdot O+N=N$

so the point $N$, is,

$N=-(B-C)^{-1}C\quad\text{or}\quad N=(C-B)^{-1}C$

Since $I\in\ell^{OI}$, we have true the equations,

$(B-C)^{-1}(I-C)=M\cdot I+N,$

replace $N$, found above, and we have

$(B-C)^{-1}(I-C)=M+N\Leftrightarrow M=(B-C)^{-1}(I-C)-[-(B-C)^{-1}C]$

So

$M=(B-C)^{-1}(I-C)+(B-C)^{-1}C=(B-C)^{-1}[(I-C)+C]$

Hence the point $M$, is,

$M=(B-C)^{-1}.$

So for two different-fixed point $B,C\in\ell^{OI}-$line we have the map,

$r_{BC}:\ell^{OI}\to\ell^{OI},$

such that,

$\forall X\in\ell^{OI},r_{BC}(X)=(B-C)^{-1}(X-C).$

######  Theorem 15.

This ratio-map, $r_{BC}:\ell^{OI}\to\ell^{OI}$ is a bijection in
$\ell^{OI}-$line in Desargues affine plane.

###### Proof.

Injective: If, have $r_{BC}(X)=r_{BC}(Y)$, then $X=Y$, truly,

$r_{BC}(X)=r_{BC}(Y)\Leftrightarrow(B-C)^{-1}(X-C)=(B-C)^{-1}(Y-C)$

so

$\displaystyle(B-C)^{-1}(X-C)-(B-C)^{-1}(Y-C)$ $\displaystyle=O$
$\displaystyle\Leftrightarrow(B-C)^{-1}[(X-C)-(Y-C)]$ $\displaystyle=O$
$\displaystyle\Leftrightarrow(B-C)^{-1}(X-Y)$ $\displaystyle=O$

the above factors are points of the line $\ell^{OI}$, so they are also
elements of skew fields $(\ell^{OI},+,\cdot)$, and we keep in mind that skew
field has no divisor of zero, therefore

$(B-C)^{-1}(X-Y)=O\Leftrightarrow X-Y=O\Leftrightarrow X=Y.$

Syrjective: $\forall R\in\ell^{OI},\exists X\in\ell^{OI}$, such that,
$R=r_{BC}(X)$.

From definition of ratio of three points, we have

$r_{BC}(X)=(B-C)^{-1}(X-C)=R\Rightarrow X-C=(B-C)R$

so

$X=(B-C)R+C.$

Hence, $\forall R\in\ell^{OI},\exists X=(B-C)R+C\in\ell^{OI}$, such that,
$R=r_{BC}[(B-C)R+C]$.

Points $B,C$ are fixed and different points in $\ell^{OI}$, and the point $R$
is in $\ell^{OI}$, we can construct in the line the points $B-C$, $(B-C)R$ and
$(B-C)R+C$, and have that,

$\displaystyle r_{BC}[(B-C)R+C]$ $\displaystyle=(B-C)^{-1}[(B-C)R+C-C]$
$\displaystyle=(B-C)^{-1}[(B-C)R]$ $\displaystyle=[(B-C)^{-1}(B-C)]R$
$\displaystyle=I\cdot R$ $\displaystyle=R.$

∎

We mark the set of maps-ratio-three-points, with,

$\mathcal{R}_{3}=\\{r_{BC}(X)|\forall X\in\ell^{OI}\\}$

It’s clear that $\mathcal{R}_{3}\subseteq\ell^{OI},$ and
$\mathcal{R}_{3}\neq\emptyset$, below we will prove that this set, is a sub-
skew field of $K=(\ell^{OI},+,\cdot)$. Let’s see now the following theorems,

######  Theorem 16.

Let’s have two differetn-fixet points $B,C\in\ell^{OI}$ and this points are
different for point $O$, in $\ell^{OI}-$line. The ratio-maps-set
$\mathcal{R}_{3}=\\{r_{BC}(X)|\forall X\in\ell^{OI}\\}$ forms a _Abelian_
(commutative) group with addition of points in $\ell^{OI}-$line.

###### Proof.

1. (Associativity) $\forall r_{BC}(X),r_{BC}(Y),r_{BC}(Z)\in\mathcal{R}_{3}$, we have three points $X,Y,Z\in\ell^{OI}$, and, we have that:

$\displaystyle\left[r_{BC}(X)+r_{BC}(Y)\right]+r_{BC}(Z)$
$\displaystyle=[(B-C)^{-1}(X-C)+(B-C)^{-1}(Y-C)]+(B-C)^{-1}(Z-C)$
$\displaystyle=(B-C)^{-1}\\{[(X-C)+(Y-C)]+(Z-C)\\}$ (from asociatiove
properties of addition groups $(\ell^{OI},+)$)
$\displaystyle=(B-C)^{-1}\\{(X-C)+[(Y-C)+(Z-C)]\\}$
$\displaystyle=(B-C)^{-1}(X-C)+(B-C)^{-1}[(Y-C)+(Z-C)]$
$\displaystyle=(B-C)^{-1}(X-C)+[(B-C)^{-1}(Y-C)+(B-C)^{-1}(Z-C)]$
$\displaystyle=r_{BC}(X)+[r_{BC}(Y)+r_{BC}(Z)],$

so,

$[r_{BC}(X)+r_{BC}(Y)]+r_{BC}(Z)=r_{BC}(X)+[r_{BC}(Y)+r_{BC}(Z)],\forall
X,Y,Z\in\ell^{OI}.$

2. (Commuativity) $\forall X,Y\in\ell^{OI}$, we have that:

$\displaystyle r_{BC}(X)+r_{BC}(Y)$
$\displaystyle=(B-C)^{-1}(X-C)+(B-C)^{-1}(Y-C)$
$\displaystyle=(B-C)^{-1}[(X-C)+(Y-C)]$ (from commutative property of addition
groups $(\ell^{OI},+)$) $\displaystyle=(B-C)^{-1}[(Y-C)+(X-C)]$
$\displaystyle=(B-C)^{-1}(Y-C)+(B-C)^{-1}(X-C)$
$\displaystyle=r_{BC}(Y)+r_{BC}(X),$

so

$r_{BC}(X)+r_{BC}(Y)=r_{BC}(Y)+r_{BC}(X),\forall X,Y\in\ell^{OI}$

3. (’zero’) $\exists Z\in\ell^{OI},\forall X\in\ell^{OI}$, we have that:

$\displaystyle r_{BC}(X)+r_{BC}(Z)$ $\displaystyle=r_{BC}(X)$
$\displaystyle(B-C)^{-1}(X-C)+(B-C)^{-1}(Z-C)$ $\displaystyle=(B-C)^{-1}(X-C)$
$\displaystyle(B-C)^{-1}[(X-C)+(Z-C)]$ $\displaystyle=B^{-1}(X-C)$

since $B\neq C$ then, $B-C\neq O$ then exists $(B-C)^{-1}$ and is different
from point $O$. So we have

$\displaystyle(B-C)^{-1}[(X-C)+(Z-C)]$ $\displaystyle=B^{-1}(X-C)$
$\displaystyle(B-C)^{-1}[(X-C)+(Z-C)]-B^{-1}(X-C)$ $\displaystyle=O$

so,

$(B-C)^{-1}[(X-C)+(Z-C)-(X-C)]=O\Leftrightarrow(B-C)^{-1}(Z-C)=O,$

we also know that these are points of skew-fields $(\ell^{OI},+,\cdot)$, and
skew fields do not have ’divisors of zero’, therefore,

$(B-C)^{-1}(Z-C)=O\Leftrightarrow Z-C=O\Leftrightarrow Z=C.$

Hence the ’zero’ element is $r_{BC}(C)$.

4. (’opposite’) $\forall X\in\ell^{OI},\exists\bar{X}\in\ell^{OI}$, we have that:

$\displaystyle r_{BC}(X)+r_{BC}(\bar{X})$ $\displaystyle=r_{BC}(C)$
$\displaystyle(B-C)^{-1}(X-C)+(B-C)^{-1}(\bar{X}-C)$
$\displaystyle=(B-C)^{-1}(C-C)$ $\displaystyle(B-C)^{-1}[(X-C)+(\bar{X}-C)]$
$\displaystyle=O$ $\displaystyle(B-C)^{-1}[X-C+\bar{X}-C]$ $\displaystyle=O$
$\displaystyle(B-C)^{-1}[X+\bar{X}-2C]$ $\displaystyle=O$

so

$X+\bar{X}-2C=O\Rightarrow\bar{X}=2C-X$

Hence the ’opposite’ element of $r_{BC}(X)$ is $r_{B}(2C-X)$, $\forall
X\in\ell^{OI}$.

∎

######  Theorem 17.

Let’s have tow different points $B,C$ and different for point $O$, in
$\ell^{OI}-$line. The ratio-maps-set $\\{r_{BC}(X)|\forall X\in\ell^{OI}\\}$
forms a group with ’multiplication of points in $\ell^{OI}-$line.

###### Proof.

1. (Associativity) $\forall X,Y,Z\in\ell^{OI}$, we have that:

$\displaystyle\left[r_{BC}(X)\cdot r_{BC}(Y)\right]\cdot r_{BC}(Z)$
$\displaystyle=[(B-C)^{-1}(X-C)\cdot(B-C)^{-1}(Y-C)]\cdot(B-C)^{-1}(Z-C)$ all
the factors are points of the line $\ell^{OI}$, they are also elements of skew
fields, therefore from the association property, we can move the brackets
$\displaystyle=(B-C)^{-1}(X-C)\cdot[(B-C)^{-1}(Y-C)\cdot(B-C)^{-1}(Z-C)]$
$\displaystyle=r_{BC}(X)\cdot[r_{BC}(Y)\cdot r_{BC}(Z)]$

2. (’unitary’) $\exists E\in\ell^{OI},\forall X\in\ell^{OI}$, we have that:

$\displaystyle r_{BC}(X)\cdot r_{BC}(E)$ $\displaystyle=r_{BC}(X)$
$\displaystyle(B-C)^{-1}(X-C)\cdot(B-C)^{-1}(E-C)$
$\displaystyle=(B-C)^{-1}(X-C)$ $\displaystyle(X-C)\cdot(B-C)^{-1}(E-C)$
$\displaystyle=(X-C)$ $\displaystyle(X-C)^{-1}(X-C)\cdot(B-C)^{-1}(E-C)$
$\displaystyle=(X-C)^{-1}(X-C)$ $\displaystyle(B-C)^{-1}(E-C)$
$\displaystyle=I$ $\displaystyle(E-C)$ $\displaystyle=I\cdot(B-C)$
$\displaystyle E-C$ $\displaystyle=B-C$ $\displaystyle E$ $\displaystyle=B$

so unitary-element element is $r_{BC}(B)$.

3. (’inverse’) $\forall X\in\ell^{OI},\exists Y\in\ell^{OI}$, we have that:

$\displaystyle r_{BC}(X)\cdot r_{BC}(Y)$ $\displaystyle=r_{BC}(B)$
$\displaystyle(B-C)^{-1}(X-C)\cdot(B-C)^{-1}(Y-C)$
$\displaystyle=(B-C)^{-1}(B-C)$ $\displaystyle(X-C)\cdot(B-C)^{-1}(Y-C)$
$\displaystyle=B-C$ $\displaystyle(B-C)^{-1}(Y-C)$
$\displaystyle=(X-C)^{-1}\cdot(B-C)$ $\displaystyle r_{BC}(Y)$
$\displaystyle=r_{XC}(B)$ or, distinguishing $Y$, we have $\displaystyle Y-C$
$\displaystyle=(B-C)\cdot(X-C)^{-1}\cdot(B-C)$ $\displaystyle Y$
$\displaystyle=(B-C)\cdot(X-C)^{-1}\cdot(B-C)+C$

Hence the ’inverse’ element of $r_{BC}(X)$ is $r_{XC}(B)$, $\forall
X\in\ell^{OI}$. ∎

######  Theorem 18.

In the ratio-maps-set $\\{r_{BC}(X)|\forall X\in\ell^{OI}\\}$, for three
elements $r_{BC}(X)$, $r_{BC}(Y)$, $r_{BC}(Z)$ of its, have true, that
equations

  1. (1)

$r_{BC}(X)\cdot[r_{BC}(Y)+r_{BC}(Z)]=r_{BC}(X)\cdot r_{BC}(Y)+r_{BC}(X)\cdot
r_{BC}(Z)$, and,

  2. (2)

$[r_{BC}(X)+r_{BC}(Y)]r_{BC}(Z)=r_{BC}(X)\cdot r_{BC}(Z)+r_{BC}(Y)\cdot
r_{BC}(Z)$

###### Proof.

Indeed

$\displaystyle r_{BC}(X)\cdot[r_{BC}(Y)+r_{BC}(Z)]$
$\displaystyle=(B-C)^{-1}(X-C)[(B-C)^{-1}(Y-C)+(B-C)^{-1}(Z-C)]$ all the
factors are points of the line $\ell^{OI}$, they are also elements of skew
fields, therefore from the distribution property, have
$\displaystyle=(B-C)^{-1}(X-C)\cdot(B-C)^{-1}(Y-C)$
$\displaystyle+(B-C)^{-1}(X-C)\cdot(B-C)^{-1}(Z-C)$
$\displaystyle=r_{BC}(X)\cdot r_{BC}(Y)+r_{BC}(X)\cdot r_{BC}(Z).$

Hence, $r_{BC}(X)\cdot[r_{BC}(Y)+r_{BC}(Z)]=r_{BC}(X)\cdot
r_{BC}(Y)+r_{BC}(X)\cdot r_{BC}(Z).$

The other point (2) is proved in the same way, so have

$\displaystyle\left[r_{BC}(X)+r_{BC}(Y)\right]\cdot r_{BC}(Z)]$
$\displaystyle=[(B-C)^{-1}(X-C)+(B-C)^{-1}(Y-C)]\cdot(B-C)^{-1}(Z-C)]$ all the
factors are points of the line $\ell^{OI}$, they are also elements of skew
fields, therefore from the distribution property, have
$\displaystyle=(B-C)^{-1}(X-C)\cdot(B-C)^{-1}(Z-C)$
$\displaystyle+(B-C)^{-1}(Y-C)\cdot(B-C)^{-1}(Z-C)$
$\displaystyle=r_{BC}(X)\cdot r_{BC}(Z)+r_{BC}(Y)\cdot r_{BC}(Z).$

Hence, $r_{BC}(X)+r_{BC}(Y)]\cdot r_{BC}(Z)]=r_{BC}(X)\cdot
r_{BC}(Z)+r_{BC}(Y)\cdot r_{BC}(Z).$

∎

From the above three theorems 16, 17 and 18, have true, this,

######  Theorem 19.

The ratio-maps-set $\mathcal{R}_{3}=\\{r_{BC}(X)|\forall X\in\ell^{OI}\\}$,
for a different fixed points $B,C$ in $\ell^{OI}-$line, forms a skew-field
with ’addition and multiplication’ of points in $\ell^{OI}-$line.

This, skew field $(\mathcal{R}_{3},+,\cdot)$ is sub-skew field of the skew
field $(\ell^{OI},+,\cdot)$.

##  4\. Free Group Presentation of Affine Plane Cycles

This section briefly introduces free groups resulting from cycles that share
colinear ratio points in simply-connected polygons in the Desargues affine
plane. Polygons are simply connected, provided the polygon vertices are in a
sequence of edges with no self-intersections.

Free groups were introduced by W. Dyck in 1882 as a natural biproduct of
simply connected polygons $P$ in which every vertex $a\in P$ can be reached by
traversing the edges between $a$ and a distinguished vertex $g\in P$ [5].

######  Definition 4.

Dyck Polygon.
A Dyck polygon is a simply-connected polygon in the Desargues affine plane. ◼

######  Example 3.

A simply-connected polygon that consists of three overlapping triangles with
colinear ratio points $A,B,C$ in common is shown in Fig. 5. This polygon
contains cycles $\mbox{cyc}E_{1},\mbox{cyc}E_{2},\mbox{cyc}E_{3}$:

  1. 1o

$\mbox{cyc}E_{1}=A\to A-C\to B_{3}\to A$.

  2. 2o

$\mbox{cyc}E_{2}=B\to B-C\to B_{4}\to B$.

  3. 3o

$\mbox{cyc}E_{3}=C\to O\to B_{1}\to B$. ◼

Figure 5. Polygon Cycles $\left\\{\mathop{\bigcup}\limits_{1\leq i\leq
3}\mbox{\bf cyc}E_{i}\right\\}\to\ \mbox{\bf free group}\ G(\beta,+)$

Notice that it is possible to start at any one of the three ratio points
$A,B,C$ and traverse a sequence of edges to reach any other vertex in the
polygon, e.g., if we start at $A$, we have

$\displaystyle A$ $\displaystyle\to B\to B_{4}.\ \mbox{or},$ $\displaystyle A$
$\displaystyle\to B\to C\to B_{1}.\ \mbox{or},$ $\displaystyle A$
$\displaystyle\to O\to B-C.\ \mbox{or},$ $\displaystyle A$ $\displaystyle\to
O\to B-C\to A-C.$

From Example 3, we see that the ratio points $A,B,C$ in the polygon (3
overlapping triangles) serve as distinguished vertices (called generators) so
that any other vertex in the polygon can be reached from a generator vertex by
traversing a sequence of edges.

######  Definition 5.

A free group $G(\beta,+)$ is nonempty finite set $G$ of $n$ elements equipped
with binary operation (typically written as a $+$) and a basis $\beta\subset
G$ so that every member $a\in G$ can be written as a linear combination the
generators $g\in\beta$, i.e., $a=\mathop{\sum}\limits_{g\in\beta}kg,k\
\mbox{mod}\ n.$ ◼

######  Lemma 1.

Every vertex in a Dyck polygon containing colinear ratio points can be reached
by a finite sequence of moves from a ratio point.

###### Proof.

Let $P$ be a Dyck polgon. From Def. 4, $P$ is simply-connected. Hence, if
vertices $g,v\in P$, then there is a finite sequence of $k$ moves to reach $v$
from $g$, denoted by $v=kg$. ∎

######  Theorem 20.

Every vertex in a Dyck polygon containing colinear ratio points is a linear
combination of the ratio points.

###### Proof.

Let $P$ be a Dyck polgon containing colinear ratio points $A,B,C$ and let
vertex $v\in P$. Let $k_{1},k_{2},k_{3}\in\mathbb{Z}$. From Lemma 1, we have

$\displaystyle B$ $\displaystyle=k_{1}A.$ $\displaystyle C$
$\displaystyle=k_{2}B+k_{1}A.$ $\displaystyle v$
$\displaystyle=k_{3}C+k_{2}B+k_{1}A$
$\displaystyle=\sum_{\begin{subarray}{c}k_{i}\in\mathbb{Z}\\\ g_{j}\in
P\end{subarray}}k_{i}g_{j}.$

∎

######  Definition 6.

Free group move operation +.
A Dyck free group denoted by $G(\beta,+)$ is defined in terms of a seqeunce of
moves from one vertex to another one in a polygon $G$ and basis $\beta$
containing collinear ratio points. From Theorem 20, every vertex in $G$ is a
linear combination of the elements of the basis $\beta$. For a Dyck free group
$G(\beta,+)$, the + operation is defined by a continuous map + such that

$\displaystyle\mathbb{Z}$ $\displaystyle=\mbox{integer}.$ $\displaystyle X$
$\displaystyle=\mbox{set of vertices in a polygon}.$ $\displaystyle
v,v^{\prime}$ $\displaystyle\in X.$ $\displaystyle G$
$\displaystyle=\left\\{g\in X\right\\}.$ $\displaystyle\beta$
$\displaystyle=\left\\{g\in P\right\\}.$ $\displaystyle g$
$\displaystyle\in\beta.$ $\displaystyle+:\mathbb{Z}\times
X\times\mathbb{Z}\times X$ $\displaystyle\to\mathbb{Z}\times X,\ \mbox{defined
by}$ $\displaystyle+(x,x^{\prime})$ $\displaystyle=x+x^{\prime}$
$\displaystyle=\sum_{\begin{subarray}{c}k_{i}\in\mathbb{Z}\\\
g_{j}\in\beta\end{subarray}}k_{i}g_{j}\ \mbox{from
Theorem~{}\ref{theorem:DyckSum}}.$ $\displaystyle+(v,0v^{\prime})$
$\displaystyle=\overbrace{kg+0g^{\prime}=kg+0=v\in
X.}^{\mbox{{\color[rgb]{0,0,0}\bf zero moves from $\boldsymbol{v}$$\in X$}}}$
$\displaystyle+(v,-v)$ $\displaystyle=\overbrace{kg-kg=(k-k)g=0g=g+0=v\in
X.}^{\mbox{{\color[rgb]{0,0,0}\bf move back to $\boldsymbol{v}$ from
$\boldsymbol{v}$}}}\mbox{\qquad{\color[rgb]{0.8,0.8,0.8}\SquareSteel}}$

######  Lemma 2.

Let $G(\beta,+)$ be a Dyck polygon $G$ with basis $\beta$ and equipped with
move +. The structure $G(\beta,+)$ is a free groupoid.

###### Proof.

From Theorem 20, $G$ has a basis $\beta\subset G$. Let $g\in\beta$. From Def.
6, + is a binary operation, i.e., every $+(v,v^{\prime})=kg+k^{\prime}g\in G$
for $v,v^{\prime}\in G$. Hence, $G(\beta,+)$ is a free groupoid. ∎

######  Lemma 3.

Let $G(\beta,+)$ be a free groupoid. Then $G(\beta,+)$ is a free semigroup.

###### Proof.

From Lemma 2, the structure $G(\beta,+)$ is a free groupoid on a Dyck polygon
$G$ with basis $\beta$ and operation $+$. Let $+(v,v^{\prime})=kg+k^{\prime}g$
for $v,v^{\prime}\in G,g\in\beta$. Then

$\displaystyle
v=kg,v^{\prime}=k^{\prime}g,v^{\prime\prime}=kk^{\prime\prime}g$
$\displaystyle\ \mbox{for}\ v,v^{\prime},v^{\prime\prime}\in G,g\in\beta$
$\displaystyle+((v,v^{\prime}),v^{\prime\prime})$
$\displaystyle=(kg+k^{\prime}g)+k^{\prime\prime}g$
$\displaystyle=kg+(k^{\prime}g+k^{\prime\prime}g)$
$\displaystyle=(v,(v^{\prime},v^{\prime\prime})).$

Consequently, the operation + is associative. Hence, $G(\beta,+)$ is a Dyck
free semigroup. ∎

######  Lemma 4.

Every Dyck free semigroup has an identity element.

###### Proof.

Immediate from Def. 6, since we can always write

$kg+0g=(k+0)g=v\in G,$

for every vertex $v\in G$. ∎

######  Notation 3.

Psudo-Identity element.
The psudo-identity element 0 in a Dyck free semigroup stems from integer 0,
since we can always write

$0g+kg=0+kg=kg=v\in G,0,k\in\mathbb{Z}.$

We also write

$0g=0+1g=v\in G.$

The notation $0+1g$ reads traversal from $v$ with zero length. ◼

######  Lemma 5.

Every member of a Dyck free semigroup has an inverse.

###### Proof.

Immediate from Def. 6, since we can always write

$kg-kg=(k-k)g=0v=v\in G$

for every vertex $v\in G$. ∎

######  Theorem 21.

$G(\beta,+)$ is a free Dyck group.

###### Proof.

From Lemma 3, the structure $G(\beta,+)$ is a free semigroup with basis
$\beta$. From Lemma 4, $G$ has an identity element, namely, $0v=v$. From Lemma
5, every member $v\in G$ has inverse, namely, $kg-kg=(k-k)g=0g=v$. Hence,
$G(\beta,+)$ is a free Dyck group. ∎

######  Theorem 22.

Every Dyck polygon containing colinear ratio vertices has a free group
presentation.

###### Proof.

Immediate from Theorem 21. ∎

From a Dyck free group presentation of a Desargues affine polygon containing
colinear, we obtain a concise means of identify the properties of colinear
ratio vertices, e.g.,

  1. Property-1o

Generator Colinear ratio vertices are generators in a Dyck free group.

  2. Property-2o

Traversal In a Dyck free group on a Dyck polygon, there is a traversal from
any ratio vertex to any other vertex in the polygon.

  3. Property-3o

Path Measure. Every vertex in a Dyck polygon with colinear ratio vertices can
be written concisely as a linear combination of the ratio vertices. Hence,
path has length $k_{1}+\cdots+k_{n}$ for
$v=\sum_{\begin{subarray}{c}k_{i}\in\mathbb{Z}\\\
g_{j}\in\beta\end{subarray}}k_{i}g_{j}^{\prime}$.

  4. Property-4o

Concise Presentation From Theorem 22, every Dyck polygon with colinear ratio
points has a free group presentation.

  5. Property-5o

Path From an application perspective, Theorem 22 is significant, since we can
then view a discrete Feynman path [6, p. xiv] (a trace of a trajectory of a
particle) having a geometric realization as a sequence of edges in a Dyck
polygon containing colinear ratio points.

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