Comparing Binomial Tree, Monte Carlo Simulation And Finite

In recent years, numerical methods for valuing options such as binomial tree models, Monte Carlo simulation and finite difference methods are use for a wide range of financial purposes. This paper illustrates and compares the three numerical methods. On one hand, it provided general description of the three methods separately involved their definitions, merits and drawbacks and determinants of each method. On the other hand, this paper makes a concrete comparison in valuing options between the three numerical methods. Overall, the three numerical methods have proven to be valuable and efficient methods to value options.

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Introduction
In recent years, option valuation methods are very important in the theory of finance and increased wildly in the practice field. The various approaches on option prices valuation included binomial tree models, Monte Carlo simulation and finite difference methods. Binomial models are suggested by Cox, Ross and Rubinstein (1979). Boyle (1977) firstly discussed Monte Carlo simulation and then it has been used by both Johnson and Shanno(1985) and Hull and White(1987) to value options when it is a stochastic process. Finite difference methods are discussed by Schwartz (1977), Brennan and Schwartz (1979), and Courtadon (1982) (Hull and White, 1988). This essay aims to provide a comparison and contrast among the three numerical methods mentioned above. All these numerical methods focus on the objectives of both calculation accuracy and speed. The only way for any given method to achieve better accuracy and speed is to calculate with many times (Hull and White, 1988). For one thing, this essay provides general description about binomial trees, Monte Carlo simulation and finite difference methods and defines benefits and drawbacks of each method. For another thing, it makes contrast on the valuation option prices involved American and European options.
Binomial tree models
Hull and White (1988) provide a general description about binomial trees. They concluded that” Binomial model is a particular case of a more general set of multivariate multinomial models”. All multivariate multinomial models are characteristics as lattice approaches such as binomial and trinomial lattice models(Hull and White, 1988).And the binomial trees, a valuation option approach, which involved separating option into a large number of small time intervals of length Δt. The assumption of this method is that the asset price changed from its initial value to two new values, both upward and downward movement, Su and Sd separately. The probability of an upward movement was indicated as p, while the probability of a downward movement is 1-p and the parameter u, d, p are used to value option prices. (Hull, 2008)
The binomial model focused on option replication. For the binomial trees, the only way to reproduce the payoff of an option is to trade a portfolio involved the stock and the risk-free asset. Within other lattice approaches, involved the trinomial tree model, do not admit option replication(Figlewski&Gao, 1999).However, the fair value of option can be valued under the basic assumptions of option pricing which is the world is risk-neutral. (Hull, 2008)
In this case, the fair value can be valued simply by computing the expected values within the risk neutral distribution and discounting at the risk-free interest rate (Hull, 2008).When the world is risk-neutral, any approximation procedure which is based on a probability distribution and rough risk neutral distribution and make convergence to its limit, can be used to value options prices properly. Therefore, it is necessary to use trinomial tree model even a more complex structure without lack of the ability to calculating unique option payoffs (Figlewski&Gao, 1999).
What is also worth mentioning about the application of binomial tree is that there exists known payouts involved dividends (Hull and White, 1988). Dividend policy was based on the principle that the stock maintains a constant yield on each ex-dividend date which was denoted by δ (Cox et.al, 1979)
Essentially, binomial and trinomial models are powerful, intuitive methods to value both American and European option. Moreover, it also provides asymptotically exact approximation based on Black-Scholes assumptions (Figlewski&Gao, 1999).
Consider the efficiency and accuracy of this method, the binomial method is more efficient and accurate when there are a small number of options values without dividends. However it lacks of efficient in a situation where effects of cash dividends should be analysed. Actually, the fixed dividend yield generated an improper hedge ratio despite that the assumption of fixed dividend yield is an efficient and accurate approximation. Furthermore, the binomial tree models are inefficient in valuing American options compared with European option. And it is less efficient and accurate than finite difference methods for multiple options valuation. This is because it has a conditional starting point (Geske&Shastri, 1985).
Monte Carlo simulation
Monte Carlo simulation is a useful numerical method to implement for various kinds of purposes of finance such as securities valuation. For the valuation of option, Monte Carlo simulation use risk-neutral measure (Hull, 2008). For example, a call option is a security whose expected payoffs depend on not only one basic security. The value of a derivative security can be obtained by discounted the expected payoff in the risk-neutral world at the riskless rate (Boyle, et.al, 1997).
Boyle et.al (1997) stated that “this approach comprises several steps in the following. Firstly, Simulate sample paths of the underlying state variables (e.g., underlying asset prices and interest rates) over the relevant time horizon. Stimulate these according to the risk-neutral measure. Secondly, evaluate the discounted cash flows of a security on each sample path, as determined by the structure of the security in question. Thirdly, average the discounted cash flows over sample paths”
There is a tendency that high-dimensional integral is becoming more and more necessary to evaluate in the derivative security. Monte Carlo simulation is widely used in the option valuation due to the increases of high dimension (Ibanez &Zapatero, 2004). Regarding the integral of the function f(x) over the d-dimensional unit hypercube, the simple Monte Carlo estimate of the integral is equivalent to the average value of the function f over n random points from the unit hypercube. When n tends to be infinite, this estimate converges to the true value of the integral. Furthermore, the distinct advantage of this method compared with other numerical approaches is that the error convergence rate is independent dimension. In addition, the function f should be square integrable and this is the only restriction which is relative and slight ((Boyle, et.al, 1997).
Monte Carlo simulation is simple, flexible. It can be easily modified to adapt different processes which involved governing stock returns. Moreover, compared other methods, it has distinct merit in some specific circumstances. Essentially Monte Carlo simulation can be used when the process of generating future stock value movement determined the final stock value. This process mentioned above is created on a computer and aims to generate a series of stock price trajectories which is used to obtain the evaluation of option. In addition, the standard deviation also can be used simultaneously in order to make sure the accuracy of the results (Boyle, 1977).
However, there are some disadvantages of this method. In recent years, some new techniques were developed so as to overcome the disadvantages. One key drawback is that it is wasteful to calculate many times and difficult to control situations when there are early exercise opportunities (Hull, 2008). Different variances reduction techniques involved control variate approach and antithetic variate method are used to solve these problems. Furthermore, deterministic sequences also known as low-discrepancy sequences or quasi-random sequences are used to accelerate the valuation of multi-dimensional integrals, (Boyle, et.al, 1997).
Quasi-Monte Carlo methods are suggested as a new approach to supplement Monte Carlo simulation. It uses deterministic sequences rather than random sequences. These sequences are used to obtain convergence with known error bounds¼ˆJoy¼Œet.al. 1996¼‰
Until recently, Monte Carlo simulation has not been used in American options. The key problem is that payoff depends on some sources of uncertainty. The optimal exercise frontier of American options is uncertain (Barraquand &Martineau, 1995).
Finite difference methods
Hull (2008) provides a general description of finite difference methods. He concluded that “finite difference methods value a derivative by solving the differential equation that the derivative satisfies.” Finite difference methods are classified into two ways those are implicit and explicit finite difference method. The former approach is related the value of option at time t+Δt to three alternative values at time t, while the latter one is related the value of option at time t to three alternative values at time t+Δt (Hull& White, 1990).
The explicit finite difference method is equivalent to a trinomial lattice approach. Compared with the two finite difference methods, the distinct advantage of explicit finite difference method is that it has fewer boundary conditions than the implicit way. For instance, to implement implicit method, considering the price of a derivative security S, it is vital to specify boundary conditions for the derivative security whether minimising or maximising price. By contrast, the explicit method, regarded as a trinomial lattice approach, does not need specific boundary conditions (Hull& White, 1990).
There are two alternative problems of partial differential equations. The first, known as boundary value problems where a wide range of boundary conditions must be specified, the second, known as initial value problems where only a fraction of valuation required to be specified. There is a fact that most option valuation problems are initial value problems. The explicit finite difference method is the most appropriate method to solve initial value problems because implicit finite method used extra boundary condition which was produced errors (Hull& White, 1990).
Furthermore, consider the efficiency and accuracy of valuing option, the explicit finite difference method, with logarithmical transformation, is more efficient than the implicit method. This is because it does not need the solution solved a series of simultaneous equations (Geske&Shastri, 1985).
In addition, for the finite difference method and jump process, the simple explicit difference approximation is harmonized with a three-point jump process, while the more complex implicit difference approximation corresponds a generalized jump process which is based on that the value of derivative security will jump to infinite future values, not just three points(Brennan&Schwartz, 1978)
Finite difference approach can be used in the same situation as binomial tree approach. They can control American and European option and cannot easily used when the payoff of an option depends on the past history of the state variable. Furthermore, finite difference methods can be used in the situation where there are some state variable¼ˆHull 2008). However, the binomial tree method is more intuitive and easily implemented than the finite difference methods. Therefore, financial economists tend to use binomial tree methods when there are a small number of option values. In contrast, finite difference methods are frequently used and more efficient in a situation where there are a large number of option values (Geske&Shastri, 1985).
The comparison between the three methods
Overall, compared with the three numerical methods of valuing option, Monte Carlo simulation should be seen as a supplement methods for the binomial tree models and finite difference methods. This is because the increase of a variety of complexity in financial instruments (Boyle, 1977). Furthermore, binomial and finite difference methods are implemented with low dimension of problems and standard dynamics, while Monte Carlo simulation is the proper methods to solve high dimension problems and stochastic parameters (Ibanez &Zapatero, 2004)
The binomial tree models and finite difference methods are classified as backward methods and can easily handle early exercise opportunities. On the contrary, Monte Carlo simulation is a looking forward method and may be opposed with backward induction (Ibanez &Zapatero, 2004)
For the two similar methods, finite difference approach is equivalent to a trinomial lattice method. They are both useful for American and European options and tend not to be used in a situation where the options’ payoff depends on the past history of state variables. However, there also are some differences between them. Binomial tree methods can be used to calculate a small number of values of options, while finite difference methods can be used and more efficient and accurate when there exit a large number of option values. In addition, binomial tree models are more intuitive and readily completed than the finite difference methods
Monte Carlo simulation is a powerful and flexible method to value various options. In principle, Monte Carlo simulation is calculated a multi-dimension integral and this is becoming an attractiveness compared other numerical methods. It can be used to solve the problem of high dimension. The drawbacks should not be neglected. The computation with many times and cannot easily handle the situation where there are early exercise opportunities. Based the traditional Monte Carlo simulation, a new approach was developed, known as Quasi-Monte Carlo methods to improve the efficiency of Monte Carlo method. The basic theorem is to use deterministic number rather than random.
However, it has not been used in valuing American options due to the optimal exercise frontier is uncertain. One way to value American option is to achieve combination of Monte Carlo simulation and dynamic programming (Ibanez &Zapatero, 2004)
Conclusion
To sum up, with the complexity of numerical computation, numerical methods are wildly used to value derivative security. This paper provided general description and specific comparison between the three numerical methods mentioned above. Binomial tree models, known as lattice approach, are a powerful and intuitive tool to value both American and European option with and without dividend. When there are a small number of option values, binomial method is more efficient and accurate. On the contrary, it is inefficient in a situation where effects of cash dividend should be analysed.
Finite difference method can be seen as the trinomial lattice approach. They are used with the problems of low dimension and have been regarded as efficient and accurate methods to value American and European options. Compared with binomial tree models, finite difference methods is more efficient and accurate when practicers computing a large number of values of options.
Monte Carlo simulation can be seen as a supplement tool for the two methods mentioned above to value options. It can be used with high dimensional problems whereas other two methods are used with low dimensional problems. The flows of Monte Carlo simulation are that it consumes time for calculating and cannot readily handle the situation where there are early exercise opportunities. In this case, Quasi-Monte Carlo methods based on traditional Monte Carlo simulation utilise deterministic sequences known as quasi-random sequences. These sequences provide an opportunity to acquire convergence with known error bounds.
Referenc:
Barraquand¼ŒJ.& Martineau, D. (1995)”Numerical Valuation of High Dimensional Multivariate American Securities “The Journal of Financial and Quantitative Analysis, Vol. 30, No. 3 pp. 383 -405
Boyle, P.P., “Option: A Monte Carlo Approach,” Journal of Financial Economics, Volume:4, pp: 323-338
Boyle, P. Broadie, M. and Glasserman,P.(1997) “Monte Carlo methods for security pricing,” Journal of Economic Dynamics and Control, Volume 21, Issues 8-9,29,pp:1267-1321
Brennan, M.J. & Schwartz, E.S., (1978)”Finite Difference Methods and Jump Processes Arising in the Pricing of Contingent Claims: A Synthesis,” The Journal of Financial and Quantitative Analysis, Vol. 13, No. 3 pp. 461 -474
Cox, J.C., Ross, S.A. and Rubinstein. M.(1979) “Option pricing: A simplified approach,” Journal of Financial Economics, Volume 7, Issue 3, pp: 229-263
Figlewski,S.&Gao,B.(1999)”The adaptive mesh model: a new approach to efficient option pricing,” Journal of Financial Economics, Volume 53, Issue 3, pp: 313-351
Geske,R. &Shastri, K.(1985) “Valuation by Approximation: A Comparison of Alternative Option Valuation Techniques,” The Journal of Financial and Quantitative Analysis, Vol. 20, No. 1 pp. 45- 71
Hull, J.(2008) Option, Futures, and Other Derivatives,7th edition, Upper Saddle River: Pearson Prentice Hall
Hull, J, &White, A.(1988) “The Use of the Control Variate Technique in Option Pricing,” Journal of Financial and Quantitative Analysis. Vol. 23, Issue. 3; p. 237-251
Hull, J, &White, A. (1990) “Valuing Derivative Securities Using the Explicit Finite Difference Method,”Journal of Financial and Quantitative Analysis. Vol. 25, No. 1; pp: 87-100
Ibanez, A. &Zapatero, F. (2004) “Monte Carlo Valuation of American Options through Computation of the Optimal Exercise Frontier,” Journal of Financial and Quantitative analysis Vol.39, No. 2, pp: 253-275
Joy, C., Boyle, P.P. and Tan, K.S.(1996)” Quasi-Monte Carlo Methods in Numerical Finance,” Management Science.Vol.42, No.6,pp:926-938
 

Finite and Non-Finite Verb

Finite and Non-Finite Verb 
Learning Objectives
I will learn :

how to distinguish between Finite and Non-Finite Verbs
Infinitives, Participles and Gerunds

Introduction
Read the sentences given below:
I am writing.
You are writing.
They were writing
The verbs are different in the three sentences. This is because the verb is controlled by the number, person and tense of the subject. Therefore they are finite verbs.

A finite verb is controlled by the number of the subject. If the subject is singular, the verb is singular. If the subject is plural the verb is plural

Example : The boy runs fast. (subject is singular)
The boys run fast. (subject is plural)

A finite verb is controlled by the person.

Example : I go to the gymnasium on Sunday. (I – First Person)
He goes to the gymnasium on Sunday. (He – Third Person)

A finite verb is controlled by the tense. It can be in the past, present or future tense.

Example : She lives in Kolkata (Present Tense)
She lived in Kolkata (Past Tense)
Try Your Hand 1( H3)
Pick out the finite verbs in the following sentences:

She kept the plates in the sink.
Looking at both sides, he crossed the street.
The curtain is flying in the wind.
Many of us will sign the petition.
They are eager to join our club.
It was raining heavily when I left.
Rolling stones gather no moss.
I have requested him to come.
I enjoy reading books.
Alighting from the aircraft she switched her mobile phone on.

Now read the sentences given below:
I enjoy listening to music.
You enjoy listening to music.
They enjoy listening to music.
The verb ‘listening’ is not controlled by the number, person and tense of the subject. Therefore it is a non-finite verb.
Take a look at a few more sentences:
I like to swim every evening.
He likes to swim every evening
They liked to swim every evening.
The verb ‘swim’ remains unchanged whatever be the person, number and tense of the subject.
On reading these sentences we see that some non-finite verbs end with -ing and some have to before them.
Try Your Hand 2 (H3)
Read the sentences given below and state whether the highlighted verbs are finite or non-finite:

He was drawing a picture in his notebook.
He wanted to draw a cartoon.
Keeping her bag on the floor, she ran out.
They wanted to meet the head teacher.
I polish the silver ear-rings regularly.
The weeping woman appealed for help.
She had spoken the truth, but we had found it hard to believe her.

Non -finite verbs are of three kinds :
1 Infinitives 2.Participles 3. Gerunds
In this lesson we shall take up the Infinitive.
Read the given sentences:
I like to sing
She tried to help the old man.
‘To sing’ and ‘to find’ are infinitives.
The infinitive is the base form of the verb and it often has ‘to’ before it.
Try Your Hand 3 (H3)
Pick out the infinitives in the given sentences:

To find fault is easy.
I did not want to scare him.
The best course of action would be to take leave.
She phoned to speak to my father.
In order to get a job he left college.
He had no choice but to travel to Chennai.
The young man rose to address the gathering.

After certain verbs, such as bid, let, make, need, dare, see, hear, the infinitive is used without ‘to’
Example : Let us go for a picnic.
She saw her brother win the race.
He makes me read the lesson aloud.
Try Your Hand 4 (H3)
Pick out the infinitives in the given sentences:

They need not cook today.
He bid me sing a song.
Let them draw whatever they want.
The children saw her buy food.
She is making him revise the lesson.
They heard her speak to the policeman on duty.
I dared not look into the room.

Infinitives can be used to join sentences. Look at the examples given below.
Ravi spoke the truth.
He was not afraid to do so.
These two sentences can be combined to form one sentence:
Ravi was not afraid to speak the truth.
We see that one of the main verbs ‘spoke’ is changed into the infinitive ‘to speak’ and used to combine the sentences.
Take a look at another example.
He remained inside the burning house. He wanted to rescue all the children.
This pair of sentences will change to :
He remained inside the burning house to rescue all the children.
We see that an infinitive ‘to rescue’ which was already present in one of the sentences has been used to combine the sentences.
Try Your Hand 5 (H3)
Combine each pair of sentences by using an infinitive.

We started running. We wanted to reach the station on time.
Every school has a Principal. He decides how the school will function.
The old man gave his servant one thousand rupees. He wanted to reward him for his faithful service.
He is determined to attain a high post. He works day and night for that purpose.
She collects old cook books from various parts of India. This is her hobby.
They must write a letter of apology. That is the only way to avoid punishment.
She goes to the hospital every day. She is always willing to look after needy patients.

Quick Recap

Finite verbs are controlled by the number, person and tense of the subject.
Non-finite verbs are not controlled by the number, person and tense of the subject
There are three kinds of non-finite verbs : Infinitives, Participles and Gerunds.
Participles and Gerunds may be similar in form as both end with –ing.
Infinitives are usually preceded by ‘to’. However there are infinitives that are not preceded by ‘to’.
Infinitives can be used to combine sentences

Brush Up Your Grammar (H2)
Revision 1(H3)
In the following sentences pick out the finite verbs and the non-finite words. The first sentence has been done for you :

Hearing the knock on the door, she ran to open it.

Non- finite : Hearing Finite: ran

I want you to keep the keys in the drawer.
She saw the fishermen casting their nets.
Rina stopped at the shop to buy something.
They enjoy performing in front of a live audience.
She read the instructions on how to bake a cake.
I heard the man give instructions to the taxi driver.
Her sparkling eyes revealed her excitement.
He kept himself busy by teaching children.
Please permit me to order the rest of the books

Revision 2 (H3)
Fill in the blanks with appropriate non-finite verbs:
I decided _____ out of the hotel and take a walk. Suddenly a number of monkeys surrounded me and began ____ my purse away. ____ here and there for help, I spotted a banana seller who was setting up his stall. I waved to him and bade him _____ towards me. _____ about two dozen bananas he hurried forwards. I saw him ____ the bananas on the ground. Then he began ____ their attention by making clicking sounds. _________ my bag the monkeys rushed towards him. I made my escape but came back later______ him and ______ for the bananas.
Revision 3 (H3)
Work in pairs. One of you will ask the question the other will give the answer with the help of the clues in brackets. The first one has been done for you :
Why did you go out? (buy some fruits)
I went out to buy some fruits

Why are you standing here? (greet the Chief Guest)
Why will Rajni go to Chennai? (visit her aunt)
Why did you sell your car? (buy a car)
Why did you travel by plane? (save time)
Why did you go to the bazaar? (buy grocery)
Why did you spend an extra hour in school? (practise football)
Why are you going to College Street? (buy second hand books)

Revision 4(H3)
Complete the sentences using the correct non-finite form of the verb given in brackets:

I am sorry for ________ (speak) rudely to you.
It gives me great pleasure ________ (inform) you that you have been selected.
I want to start now _______ (complete) the project on time.
I shall continue ______ (run) early in the morning.
It was rude of him ______ (scream) at his mother.
I must stop ______ (give) advice to every one.
I almost came under the wheels of the _____ (speed) bus

Try This 1
Make a list of 4 things that you enjoy doing

Looking at the starry sky
__________________
_________________-
_________________
_________________

Now, make a list of things you must remember to do in the coming week

To cover my text books
__________
__________
__________
__________

Try This 2
Use non-finites to write a dialogue between two friends about a three day camping trip that both will be going for. You could begin like this :
Seema : I am going for the school camping trip.
Wendy : So will I. In fact I saw you write your name on the list.
Seema : I have decided to take my transistor. Carrying it will be no problem.
 

Finite Groups of Isometries in the Euclidean Space

Finite Groups of Isometries

Abstract

The aim of this report is to find the finite groups of isometries in the Euclidean space
Rn
. We shall specifically be considering finite groups of the special orthogonal group
SO(3)
, which is a natural subgroup of the isometries in the Euclidean space. We shall assemble a set of definitions and theorems related to group theory, Euclidean geometry and spherical geometry to help gain an understanding of the finite groups of
SO(3)
.

Introduction

This paper will explore the finite groups of isometries, specifically the special orthogonal group
SO(3,R)
which is a natural subgroup of
Isom(Rn)
. We will understand that all these finite groups are isomorphic to either a cyclic group, a dihedral group, or one of the groups of a Platonic solid.

 As well as the aid of definitions and theorems leading up to the finite groups of
SO(3)
, we shall study topics, such as polygons in the Euclidean plane and spherical triangles, which will be beneficial in gaining a larger insight to the subject. The main concept that we shall be investigating is the rotation groups of these finite subgroups.

1 Euclidean Geometry

1.1 Euclidean Space
R
1 refers to the real line which is all real numbers from least to greatest.
R
2 is the plane, where points are represented as ordered pairs:
(x1,x2)
.
R
n, which is the n-dimensional Euclidean space, is the space of n-tuples of real numbers:
(x1,x2,…,xn)
.

In this report, the Euclidean space
R
n shall be used, equipped with the standard Euclidean inner-product
( , ).
The inner-product is defined by

 
(x,y
) =
∑i=1nxiyi
.

The Euclidean norm on
R
n is defined by
∥x∥ =(x,x)

and the distance function
d
  is defined by
dx,y= ∥x–y∥
.

Definition 1.1 A metric space is a set
X
equipped with a metric
d
, namely
d: X × X → R
, satisfying the following conditions:

dP,Q≥0 
with
dP,Q=0 
if and only if
P=Q
.

dP,Q= d(Q,P)

dP,Q+ dQ,R= d(P,R)

for any points
P, Q, R
.

The Euclidean distance
d
function is an example of a metric as it also satisfies the above conditions.

Lemma 1.2 A map
f: X → Y
of metric spaces is continuous if and only if, under
f
, the inverse image of every open subset of
Y
is open in
X
.

A homeomorphism between given metric spaces
(X,dX)
and
Y,dY
is a continuous map with a continuous inverse.

A topological equivalence between the spaces is when the open sets in two spaces correspond under the bijection; the two spaces are then considered homeomorphic.

1.2 Isometries

Definition 1.3 Let
X
and
Y
be metric spaces with metrics
dX
and
dY
. An isometry    
f:(X, dX) → (Y, dY)
is a distance-preserving transformation between metric spaces and is assumed to be bijective.

i.e.
dYfx1,fx2= dXx1,x2   ∀ x1, x2 ∈X.
 

Isometries are homeomorphisms since the second condition implies that an isometry and its inverse are continuous. A symmetry of the space is an isometry of a metric space to itself.
Isom(X)
denotes the isometry group or the symmetry group, which are the isometries of a metric space
X
to itself that form a group under composition of maps.

An isometry is a transformation in which the original figure is congruent to its image. Reflections, rotations and translations are isometries.

 

Definition 1.4 A group
G
is a set of elements with a binary operation
x,y∈G⟼x,y∈G

called multiplication, satisfying three axioms:

xyz=xyz∀x,y,z∈G
,                        

xe=ex=x  ∀x,e∈G
,

There exists an inverse
x–1∈G
such that
xx–1=x–1x=e  ∀x∈G
.

Definition 1.5 A group
G
is isomorphic to a group
G‘
if there is a bijection
ϕ
from
G
to
G‘
such that
ϕxy=ϕ(x)ϕ(y)
.

 

Definition 1.6 A group
G
acts on a set
X
if there is a map
G × X → X
;
g,x↦g⋅x
, such that

e⋅x=x
for the identity
e
of
G
and any point
x∈X

g⋅h⋅x=(gh)⋅x 
for
g,h ∈G
and any point
x∈X
.

If for all
x,y ∈X
, there exists
g ∈G
with
gx=y
then the action of
G
is transitive.

For the case of the Euclidean space
Rn
, with its standard inner-product
( , )
and distance function
d
, the isometry group
Isom(Rn)
 acts transitively on
Rn
since any translation of
Rn
is an isometry. A rigid motion is sometimes used to refer to an isometry of
Rn
.

Theorem 1.7 An isometry
f:Rn → Rn
is of the form
f(x)=Ax+b
, for some orthogonal matrix
A
and vector
 b∈Rn
.

Lemma 1.8 Given points
P ≠ Q in Rn
, there exists a hyperplane
H
, consisting of the points of
Rn
which are equidistant from P and Q, for which the reflection
RH
swaps the points P and Q.

 

Theorem 1.9 Any isometry of
Rn
can be written as the composite of at most
(n+1)
reflections.

 

 

1.3 The group
O(3,R)

The orthogonal group, denoted
O(n)=O(n,R)
, is a natural subgroup
of Isom(Rn)
which consists of those isometries that are fixed at the origin. These can therefore be written as a composite of at most n reflections. It is the group of
n × n
orthogonal matrices.
On≔{A∈Mn×nR:ATA=AAT=I}

is a group with respect to matrix multiplication
X,Y⟼XY
.

If
A ∈ O(n)
, then
detA detAt= det(A)2 = 1,

and so
detA=1
or
detA=–1
.

The special orthogonal group, denoted
SO(n)
,is the subgroup of
O(n)
which consists of elements with
detA=1.

Direct isometries of
Rn
are the isometries of
Rn
of the form
f(x)=Ax+b
, for some
A∈SO(n)
and
b∈Rn
. They can be expressed as a product of an even number of reflections.

Suppose that
A∈O(3)
. First consider the case where
A∈SO(3)
, so
detA=1
. Then
det⁡A–I=det(At–I)=detA(At–I)=det⁡(I–A)
⇒ det⁡A–I=0, 

i.e.
+1
is an eigenvalue.

Therefore, there exists an eigenvector
v1
such that
Av1=v1
.
W=v1⊥
is set to be the orthogonal complement to the space spanned by
v1
. Then                                         
Aw,v1=Aw,Av1=w,v1=0
if
w∈W
. Thus
A(W)⊂W
and
A|W
is a rotation of the two-dimensional space
W
, since it is an isometry of
W
fixing the origin and has determinant
1
. If
{v1,v2}
is an orthonormal basis for
W
, the matrix
1000cos⁡θ–sin⁡θ0sin⁡θcos⁡θ

represents the action of
A
on
R3
with respect to the orthonormal basis
{v1,v2,v3}
.

This is just rotation about the axis spanned by
v1
through an angle
θ
. It may be expressed as a product of two reflections.

Now suppose
detA=–1
.

Using the previous result, there exists an orthonormal basis with respect to which
–A
is a rotation of the above form, and so
A
takes the form
–1000cos⁡ϕ–sin⁡ϕ0sin⁡ϕcos⁡ϕ

With
ϕ=θ+π
. Such a matrix
A
represents a rotated reflection, rotating through an angle
ϕ
about a given axis and then reflecting in the plane orthogonal to the axis. In the special case
ϕ=0
,
A
is a pure reflection. The general rotated reflection may be expressed as a product of three reflections.

 

1.4 Curves and their lengths

Definition 1.10 A curve (or path)
γ
in a metric space
(X,d) 
is a continuous function                   
γ :[a,b]→X,
for some real closed interval
[a,b]
.

If a continuous path can join any two points of
X
, a metric is called path connected. Both connectedness and path connectedness are topological properties, in that they do not change under homeomorphisms. If X is path connected, then it is connected.

 

Definition 1.11 We consider dissections
D: a = t0  t1  …  tN = b

of
[a,b]
, with
N
arbitrary, for a curve
 γ : [a,b] → X
on a metric space
(X,d)
.

We set
Pi= γ (ti)
and
SD:= ∑ d(Pi,Pi+1).

The length
Ɩ
of
γ
is defined by
Ɩ = sup SD

if this is finite.

For curves in
Rn
, this is illustrated below:

A straight-line segment is any curve linking the two endpoints which achieves this minimum length in the Euclidean space.

There are curves
 γ  : [a,b] → R2
which fail to have finite length but for sufficiently nice curves, this does not apply. A curve of finite length may connect any two points if
X
denotes a path connected open subset or
Rn
.

A metric space
(X,d)
is called a length space if
d(P,Q) = 
inf
{
length
( γ )
:     
γ
  a curve joining
P
to
Q
},

for any two points
P,Q
of
X
.

The metric is sometimes called intrinsic metric.

We can identify a metric
d
on
X
, defining
d(P,Q)
to be the infimum of lengths of curves joining the two points, if we start from a metric space
(X,d0)
that satisfies the property that any two points may be joined by a curve of finite length. This is a metric, and
(X,d)
is then a length space.

Proposition 1.12 If
 γ  : [a,b] → R3
is continuously differentiable, then
length  γ  = ∫ ‖  γ‘(t) ‖ dt
,

where the integrand is the Euclidean norm of the vector
 γ‘ t∈R3.

 

1.5 Completeness and compactness

Completeness and compactness are another two recognised conditions on metric spaces.

Definition 1.13 A sequence
x1,x2,… 
of points in a metric space
(X,d)
is called a Cauchy sequence if, for any
ε > 0
there exists an integer
N
such that if
m,n ≥ N
then
d(xm,xn)  ε.

A metric space (X, d) in which every Cauchy sequence
(xn)
converges to an element of X is called complete. This means that a point
x ∈ X
  such that
d(xn,x)→0
as
n→∞
. These limits are unique.

The real line is complete since real Cauchy sequences converge. The Euclidean space
Rn
is also complete when this is applied to the coordinates of points in
Rn
. A subset
X
of
Rn
will be complete if and only if it is closed.

Definition 1.14 Let
X
be a metric space with metric d. If every open cover of X contains a finite subcover,
X
is compact.

An open cover of
X
is a collection
{Ui}i ∈I
of open sets if every
x∈X
belongs to at least one of the
Ui
, with
i∈I
. If the index
I
is finite, then an open cover is finite.

Compactness is a property that establishes the notion of a subset of Euclidean space being closed and bounded. A subset being closed means to contain all its limit points. A subset being bounded means to have all its points lie within some fixed distance of each other. If every sequence in a X has a convergent subsequence, then a metric space
(X,d)
is called sequentially compact.

Lemma 1.15 A continuous function
f: X → R
on a compact metric space
(X,d)
is uniformly continuous.

i.e. given
ε > 0
there exists
δ > 0
such that if
d(x,y)  δ
, then
|f(x)–f(y)|  ε
.

 

Lemma 1.16 If
Y
is a closed subset of a compact metric space
X
, then
Y
is compact.

Since
X
is a closed subset of some closed box
Rn
, we infer that any closed and bounded subset
X
of
Rn
is compact.

 

Lemma 1.17 If
f: X → Y
is a continuous surjective map of metric spaces, with
X
compact, then so is
Y
.

 

1.6 Polygons in the Euclidean Plane

Euclidean polygons in
R2
will be considered as the ‘inside’ of a simple closed polygon curve.

Definition 1.18 For a metric space, a curve
γ: [a,b]→X
is called closed if
γ(a)=γ(b)
. It is called simple if, for
t1 t2
, we have
γ(t1)≠γ(t2)
, except for
t1=a
and
t2=b
, when the curve is closed.

Proposition 1.19 Let
γ: [a,b] → R2
be a simple closed polygonal curve, with
C⊂R2
denoting the image
γ([a,b]).
Then
R2C
has at most two path connected components.

Given a set
A⊂C*=C{0},
a continuous function
h: A→R
such that
h(z)
is an argument of
z
for all
z∈A
, is a continuous branch of the argument on
A
.

A continuous branch of the argument exists on
A
if and only if a continuous branch of the logarithm exists.

i.e. a continuous function
g: A→R
such that exp
g(z)=z
for all
z∈A
.

For a curve
γ: a,b→C*
; a continuous branch of the argument for
γ
is a continuous function
θ: [a,b]→R
such that
θ(t)
is an argument for
γ(t)
for all
t∈[a,b].
Continuous branches of the argument of curves in
C
* always exist, unlike continuous branches of the argument for subsets. The use of continuity of the curve can show that they exist locally on
[a,b]
. Then, a continuous function overall of
[a,b]
can be achieved using the compactness of
[a,b]
.

For a closed curve
γ: [a,b]→C*
, the winding number of
γ
about the origin, is any continuous branch of the argument
θ
for
γ
. This is denoted
n(γ,0)
and is defined
n(γ,0) =  θ(b) – θ(a)2π
.

Given a point
w
not on a closed curve
γ: [a,b]→C=R2
,  the integer
n(γ,w):=n(γ–w,0)
defines the winding number of
γ
about
w
, where
γ–w
is the curve whose value at
t∈[a,b]
is
γ(t)–w
. The integer
n(γ,w)
describes how many times the curve
γ
‘winds around
w‘
.

Elementary properties of the winding number of a closed curve
γ
:

The winding number does not change when reparametrising
γ
or changing the starting point on the curve. However, if
–γ
denotes the curve
γ
travelled in the opposite direction i.e.
(–γ)(t)= γ(b–(b–a)t),
then for any
w
not on the curve,

n((–γ),w)=–n(γ,w).

We have
n(γ,w)=0
for the constant curve
γ
.

n(γ,w)=0
if a subset
A⊂C*
contains the curve
γ–w
on which a continuous branch of the argument can be defined. Therefore, if a closed ball

contains
γ
, then
n(γ,w)=0
for all
w∉B̅
.

The winding number
n(γ,w)
is a constant on each path connected component of the complement of
C:= γa,b
, as a function of
w
.

If
γ1,γ2: [0,1] → C
are two closed curves with
γ10=γ11=γ20=γ2(1)
, we can form the link
γ=γ1*γ2: [0,1]→C
, defined by

γt=       γ1t   for 0 ≤ t ≤ 1

                              
γ2(t–1)   for 1 ≤ t ≤ 2.

Then, for
w
not in the image of
γ1*γ2
, we have
n(γ1*γ2,w) = n(γ1,w) + n(γ2,w).

Definition 1.20
C
is compact for a simple closed polygonal curve with image
C⊂R2
, and hence bounded. Therefore, some closed ball

contains
C
. One of the two components of
R2/C
contains the complement of

since any two points in the complement of

may be joined by a path and hence is unbounded, whilst the other component of
R2/C
is contained in

, and hence is bounded. The closure of the bounded component will be a closed polygon in
R2
or a Euclidean polygon. This consists of the bounded component together with
C
. Since a Euclidean polygon is closed and bounded in
R2
, it is also compact.

 

1.21 Exercise The rotation group for a cube centred at the origin in
R3
is isomorphic to
S4
, considering the permutation group of the four diagonals.

Proof A cube has 4 diagonals and any rotation induces a permutation of these diagonals. However, we cannot assume different rotations correspond to different rotations.

We need to show all 24 permutations of the diagonals come from rotations.

Two perpendicular axes where
90°
rotations give the permutations
α=(1 2 3 4)
and
β=(1 4 3 2)
can be seen by numbering the diagonals as 1,2,3 and 4. These make an 8-element subgroup
{ε,α,α2,α3,β2,β2α,β2α2,β2α3}
and the 3-element subgroup
{ε,αβ,αβ2}
.

Thus, the rotations make all 24 permutations since
lcm8,3=24=|S4|
.

2 Spherical Geometry

2.1 Introduction

Let
S=S2
denote a unit sphere in
R3
with centre
O=0
.

The intersection of
S
with a plane through the origin is a great circle on
S
. This is the spherical lines on
S
.

S

 

 

 

Definition 2.1 The distance
d(P,Q)
between
P
and
Q
on
S
is defined to be the length of the shorter of the two segments
PQ
along the great circle. This is
π
if
P
and
Q
are on opposite sides.
d(P,Q)
is the angle between
P=OP⃗
and
Q=OQ⃗
, and hence is just
cos–1(P,Q)
, where
(P,Q)=P·Q
is the Euclidean inner-product on
R3
.

 

 

 

 

 

 

2.2 Spherical Triangles

Definition 2.2 A spherical triangle
ABC
on
S
is defined by its vertices
A,B,C∈S
, and sides
AB, BC 
and
 AC
, where these are spherical line segments on
S
of length
 π
.

S2

The triangle
ABC
is the region of the sphere with area
 2π
enclosed by these sides.

Setting
A=OA⃗, B=OB⃗ 
and
 C=OC⃗
,
c=cos–1(A·B)
gives the length of the side
AB
. For the lengths
a,b
of the sides
BC
and
CA
, similar formulae are used.

The unit normals to the planes
OBC, OAC, OBA
  are set by denoting the cross-product of vectors in
R2
by
×
;
n1 = C × B / sin a
n2= A × C / sin b
n3 = B × A / sin c
.

Given a spherical triangle
∆ABC
, the polar triangle
∆A‘B‘C‘
is the triangle with
A
a pole of
B‘C‘
on the same side as
A‘
,
B
a pole of
A‘C‘
on the same side as
B‘
, and
C
a pole of
A‘B‘
on the same side as
C‘
.

Theorem 2.3 If
∆A‘B‘C‘
is the polar triangle to
∆ABC
, then
∆ABC
is the polar triangle to
∆A‘B‘C‘
.

Theorem 2.4 If
∆A‘B‘C‘
is the polar triangle to
∆ABC
, then
∠A + B‘C‘ = π.

 

Theorem 2.5 (Spherical cosine formula)
sin(a) sin(b) cos(γ) = cos(c) – cos(a) cos(b).

 

Corollary 2.6 (Spherical Pythagoras theorem) 

When
γ=π2
,
cos(c)=cos(a) cos(b).

Theorem 2.7 (Spherical sine formula)
sin⁡(a)sin⁡(α)=sin⁡(b)sin⁡(β)=sin⁡(c)sin⁡(γ).

Corollary 2.8 (Triangle inequality)

For
P,Q,R ∈S2
,
d(P,Q) + d(Q,R) ≥ d(P,R)

with equality if and only if
Q
is on the line segment
PR
.

Proposition 2.9 (Second cosine formula)
sin(α) sin(β) cos(c) =cos(γ) – cos(α) cos(β).

 

2.3 Curves on the sphere

The restriction to
S
of the Euclidean metric on
R3
and the spherical distance metric are two natural metrics defined on the sphere.

Proposition 2.10 These two concepts of length coincide, given a curve
γ
on
S
joining points
P,Q 
on
 S
.

Proposition 2.11 Given a curve
 γ  
on
 S
joining points

and
 Q
, we have
Ɩ=
length
 γ  ≥ d(P,Q).
In addition, the image of
 γ 
is the spherical line segment
PQ 
on
 S
if
Ɩ=d(P,Q)
.

A spherical line segment is a curve 
γ
   of minimum length joining
 P 
and
 Q
. So

length
 γ  |[0,1] = d(P, γ (t)),

for all
t
. Therefore, the parameterisation is monotonic since
d(P, γ (t))
is strictly increasing as a function of
t
.

 

2.4 Finite Groups of Isometries

Definition 2.12 Let
X={1,2,…,n}
be a finite set. The symmetric group
Sn
is the set of all permutations of
X
. The order of
Sn
is
Sn=n!=1∙2∙…∙n
.

Definition 2.13 The alternating group
An
is the set of all even permutations in
Sn
. The order of group
An
is
An=|Sn|2=n!2
.

Definition 2.14 The dihedral group
Dn
is the symmetry group of a regular polygon with
n
sides.

Definition 2.15 The cyclic group
Cn
, with
n
elements, is a group that is generated by combining a single element of the group multiple times.

A matrix in
O(3,R)
determines an isometry of
R3
which fixes the origin. Such a matrix preserves both the lengths of vectors and angles between vectors since it preserves the standard inner-product.

Any isometry
f: S2→S2
may be extended to a map
g: R3→R3
fixing the origin, which for non-zero
x
is defined by
g(x):= ‖x‖ f(x/‖x‖).

With the standard inner-product
( , )
on
R3
,
(g(x),g(y))=(x,y)
for any
x,y∈R3
. For
x,y
non-zero, this follows since
(g(x),g(y)) = ‖x‖ ‖y‖ (f(x/‖x‖),f(y/‖y‖))
= ‖x‖ ‖y‖ (x/‖x‖,y/‖y‖) = (x,y).

From this we infer that
g
is an isometry of
R3
which fixes the origin and is given by a matrix in
O(3).
  Therefore,
Isom(S2)
is naturally acknowledged with the group
O(3, R)
.

The restriction to
S2
of the isometry
RH
of
R3
, the reflection of
R3
in the hyperplane
H
is defined as the reflection of
S2
in a spherical line
Ɩ
. Therefore, three such reflections are the most any element of
Isom(S2)
can be composite of. Isometries that are just rotations of
S2
and are the composite of two reflections are an index two subgroup of
Isom(S2)
corresponding to the subgroup
SO(3)⊂O(3)
. The group
O(3)
is isomorphic to
SO(3)×C2
, since any element of
O(3)
is of the form
±A
, with
A∈SO(3)
.

Any finite subgroup
G
of
Isom(R3)
has a fixed point in
R3
,
1|G| ∑g(0)∈R3,

and corresponds to a finite subgroup of
Isom(S2)
. Since any finite subgroup of
Isom(R2)
has a fixed point, it is either a cyclic or dihedral group.

We consider the group of rotations
SO(3)
. All finite subgroups of
SO(3)
are isomorphic to either the cyclic group, the dihedral group, or one of the groups of a Platonic solid. There are five platonic solids: the icosahedron, the dodecahedron, the tetrahedron, the octahedron and the cube.

Copies of a cyclic group
Cn
are contained in
SO(3) 
by considering rotations of
S2
about the
z
-axis through angles which are multiples of
 2π/n
. We generate a new subgroup of
SO(3)
by also including the rotation of
S2
about the
 x
-axis through an angle
π
which is isomorphic to the group of symmetries
D2n
of the regular
n–
gon for
 n > 2
. We have the special case
D4=C2×C2 
when
n=2
.

However, corresponding to the rotation groups of the regular solids, there are further finite subgroups of
SO3
. The tetrahedron has rotation group
A4
, the cube has rotation group
S4
and the octahedron is dual to the cube. Dual solids are solids that can be constructed from other solids; their faces and vertices can be interchanged. The dodecahedron and the icosahedron are also dual solids and have rotation group
A5
.

Proposition 2.16 The finite subgroups of
SO(3)
are of isomorphism types
Cn
for
n ≥ 1
,
D2n
for
n ≥ 2
,
A4, S4, A5
, the last three being the rotation groups arising from the regular solids.

Since
–I∈O(3)  SO(3),
H=C2× G
is a subgroup of
O(3)
of twice the order if
G
is a finite subgroup of
SO(3)
, with elements
±A
for
A∈G
.

The reason why extra finite groups do not occur for either the Euclidean or hyperbolic cases but does occur for the sphere is because we can consider the subgroup of isometries
G
generated by the reflections in the sides of the triangle, if we have a spherical triangle
Δ
with angles
π/p, π/q 
and
 π/r 
with
r≥q≥p≥2
.

The tessellation of
S2
is by the images of Δ under the elements of
G
by the theory of reflection groups. This means that the spherical triangles
g(
Δ
)
for
g∈G
covers
S2
and that any two such images have disjoint interiors. A special type of geodesic triangulation for which all triangles are congruent is developed by such a tessellated
S2
. Therefore, the reflection group
G
is finite.

From Gauss-Bonnet Theorem, the area of Δ is
π(1/p + 1/q + 1/r–1)
, and hence
1/p + 1/q + 1/r > 1
.

The only solutions are:

G
has order 4n, 24, 48 and 120 in these cases. This is implied from the tessellation of
S2
by the images of Δ under
G
. It is then clear that
G
is
C2× D2n
in the first case, and it is the full symmetry group of the tetrahedron, cube and dodecahedron in the remaining cases.

 

2.5 Gauss-Bonnet and Spherical Polygons

The statement that angles of a Euclidean triangle add up to
π
is the Euclidean version of Gauss-Bonnet.

Proposition 2.17 If Δ is a spherical triangle with angles
α,β,γ
, its area is
(α+β+γ)– π
.

For a spherical triangle,
α+β+γ > π
. We obtain the Euclidean case;
α+β+γ = π
in the limit as area
Δ → 0
.

We can subdivide the triangle, whose sides have length less than
 π
, into smaller ones if one of the sides of the spherical triangle has length
≥ π
. The area of the original triangle is still
α+β+γ+π–2π = α+β+γ–π

when applying Gauss-Bonnet to the two smaller triangles and adding.

The Gauss-Bonnet can be extended to spherical polygons on
S2
. Consider a simple closed polygonal curve
C
on
S2
, where spherical line segments are the segments of
C
. Suppose that the north pole does not lie on
C
. We consider a simple closed curve in
C
the image
┌ 
of
C
under stereographic projection. Stereographic projection is a mapping that projects a sphere onto a plane. 
 
Arcs of certain circles or segments of certain lines are the segments of

. A bounded and an unbounded component are contained by the complement of

  in
C
. Therefore, two path connected components are also contained in the complement of
C
in
S2
. Each component corresponds to the bounded component in the image of a stereographic projection. A spherical polygon is determined by the information of the polygonal curve
C
and a choice of a connected component of its complement in
S2
.

A subset
A
of
S2
is called convex if there is a unique spherical line segment of minimum length joining

to
 Q
, for any points
P,Q∈A
and this line segment is contained in
A
.

 

Theorem 2.18 If
∏⊂S2
is a spherical
n
-gon, contained in some open hemisphere, with interior angles
α1,…,αn
, its area is
α1+…+αn– (n–2) π.

 

2.6 Möbius Geometry

Möbius transformations on the extended complex plane
C∞=C U {∞}
is closely related to spherical geometry, with a coordinate
ϛ
. The stereographic projection map
π: S2→C∞
,

defined geometrically by the diagram below provides this connection.

The point of intersection of the line through
N
and
P
with
C
is
π(P)
, where the plane
z=0
identifies
C
, and where we define
πN:=∞
;
π
is a bijection.

Using the geometry of similar triangles, an explicit formula for
π
can be formed;
π(x,y,z) =x+iy1–z 

since in the diagram below
rR=1–z1
and so
R=r1–z
.

 

 

 

Lemma 2.19 If
π‘: S2→ C∞
denotes the stereographic projection from the south pole, then
π‘(P) = 1 / π(P)̅

for any
P∈S2.

The map
π‘∘π–1 :C∞→C∞
is just inversion in the unit circle,
ϛ↦1/ϛ̅
.

If
P=(x,y,z)∈S2
, then
πP=ϛ== x + iy1–z
.

The antipodal point
–P = (–x,–y–z)
has
π–P=–x+iy1+z
and so
πPπ–P̅=–x2+y21–z2=–1. 

Therefore
π(–P) = – 1 / π(P).̅

The group
G
, of Möbius transformations, is acting on
C∞
.
A
defines a Möbius transformation on
C∞
by
ς↦aς+bcς+d

if
A=abcd∈GL(2,C)
.
λA
defines the same Möbius transformation for any
λ∈C*=C{0}
.

Conversely, if
A1,A2
define the same Möbius transformation, then the identity transformation is identified by
A2–1A1
. This simplifies that
A2–1A1= λI
for some
λ∈C*,
and hence that
A1=λA2
. Therefore
G = PGL(2,C) := GL(2,C) / C*
,

identifying elements of
GL(2,C)
attains the group on the right, which are non-zero multiples of each other.

If
det A1=1=det A2
and
A1=λA2
, then
λ2=1
, and so
λ=±1
. Therefore
G = PSL(2,C) := SL(2,C) / {±1}
,

where identifying elements of
SL(2,C)
which differ only by a sign attains the group on the right. The quotient map
 SL(2,C) → G
is a surjective group homomorphism which is 2-1.
SL(2,C)
is a double cover of
G
.

Elementary facts about Möbius transformation

The group
G
of Möbius transformations is generated by elements of the form

z↦z + a   for a∈C

z↦az        for a∈C* =C  {0}

z ↦1/z. 

Any circle/straight line in
C
is of the form

azz̅ – w̅z – wz̅ + c = 0,

for
a,c∈R
,
w ∈C
such that
|w|2> ac
, and therefore is determined by an indefinite hermitian
2 × 2
matrix   
aww̅c.

Möbius transformations send circles/straight lines to circles/straight lines.

There exists a unique Möbius transformation
T
such that  

T(z1)=0, T(z2)=1, T(z3)=∞
,
Tz=z–z1z–z3z2–z3z2–z1,

given distinct points
z1,z2,z3∈C∞
.

The image of
z4
under the unique map
T
defined above in iv. is defined by the cross-ratio
[z1,z2,z3,z4]
of distinct points of
C∞
.

There exists a unique Möbius transformation
T
sending
R(z1),R(z2)
and
R(z3)
to
0,1 
and
 ∞
, given distinct points
z1,z2,z3,z4
and a Möbius transformation
R
. The composite
TR
is therefore the unique Möbius transformation sending
z1,z2
and
z3
to
0,1 
and
 ∞
. Our definition of cross-ratio then implies that
[Rz1,Rz2,Rz3,Rz4]
= T(Rz4) = (TR) z4=
z1,z2,z3,z4.

 

2.7 The double cover of
SO(3)

We have an index two subgroup of the full isometry group
O(3)
, the rotations
SO(3)
on
S2
. The section aims to show that the group
SO(3)
is established isomorphically with the group
PSU(2)
by the stereographic projection map
π
. There is a surjective homomorphism of groups
SU(2)→SO(3)
, which is
2–1
map.

 

Theorem 2.20 Every rotation of
S2
corresponds to a Möbius transformation of
C∞
in
PSU2
via the map π.

Theorem 2.21 The group of rotations
SO(3) 
acting on
S2
corresponds isomorphically with the subgroup
PSU(2)=SU(2)/{±1}
of Möbius transformations acting on
C∞
­.

Corollary 2.22 The isometries of
S2
which are not rotations correspond under stereographic projection precisely to the transformations of
C∞
of the form
z↦az̅–bb̅z̅+a̅

with
|a|2+|b|2=1.

There exists a 2-1 map
SU(2)→PSU(2)≅SO(3).

This map is usually produced using quaternions.

This is the reason why a non-closed path of transformations in
SU(2) 
going from

to
–I
exists, corresponding to a closed path in
SO(3)
starting and ending at
100010001
.

Since
SU(2)
consist of matrices of the form
a–bb̅a̅
  

with
|a|2+|b| 2=1
, geometrically it is
S3⊂R4.

There are finite subgroups of
SU(2)
of double the order corresponding to finite subgroups of
SO(3)
, specifically cyclic, dihedral and the rotation groups of the tetrahedron, cube and dodecahedron.

2.8 Circles on
S2

We consider the locus of points on
S2
, whose spherical distance from
P
is
ρ
, given an arbitrary point
P
on
S2
and
0 ≤ p ≤ π
. In spherical geometry, this is what is meant by a circle.

To ensure the point
P
is always at the north pole, we may rotate the sphere, as shown below:

Therefore, the circle is also a Euclidean circle of radius
sin(ρ)
and that it is the intersection of a plane with
S2
. Conversely, a plane cuts out a circle if its intersection with
S2
consists of more than one point. Great circles correspond to the planes passing through the origin. The area of such a circle is calculated by
2π1–cos⁡ρ=4π sin2ρ2,

which, from the Euclidean case, is always less than the area
πρ2.
For small
ρ
this may be expanded as
πρ21–112ρ2+Oρ4.

2.23 Exercise Two spherical triangles
∆1,∆2
on a sphere
S2
are said to be congruent if there is an isometry of
S2
that takes
∆1
to
∆2
.
∆1,∆2
are congruent if and only if they have equal angles.

Proof Let
∆ABC
and
∆DEF
have
∠A=∠D
etc and let
∆A‘B‘C‘
and
∆D‘E‘F‘
be the polar triangles. By theorem 2.18,
B‘ C‘ = π – ∠A = π – ∠D =E‘F‘

and so on. So, by the three sides,
∆A‘B‘C‘
is congruent to
∆D‘E‘F‘
which means that they have the same angles. Now theorem 2.17 implies that
∆ABC
and
∆DEF
are the polar triangles of
∆A‘B‘C‘
and
∆D‘E‘F‘
. Thus, with roles reversed, theorem 2.18 can be applied to get
BC = π – ∠A 0 = π – ∠D 0 = EF

and so on. Therefore, the original triangles are congruent.

Conclusion

In conclusion, in this report we have discussed isometries and the group
O(3,R)
, including the special orthogonal group
SO(3)
. As well as exploring related concepts within Euclidean geometry and spherical geometry, we have analysed the finite groups of
SO(3)
and classified their symmetry groups by considering their rotational symmetry.

We also checked two examples: one which aided to understand the rotational symmetry of a cube, which is one of the finite subgroups of
SO(3)
and one which helped us understand the congruence of spherical triangles under certain circumstances.

References

Wilson, P. M. H. (2007). Curved spaces: from classical geometries to elementary differential geometry. Cambridge University Press.

Armstrong, M. A. (2013). Groups and symmetry. Springer Science & Business Media.

 

Finite Element Analysis of a Rocker Arm

Finite Element Analysis of a Rocker Arm

 

Component Description

 

1.1.            Component Function

The rocker arm is an oscillating,  two-arm lever that provides a means of actuating the valves in the combustion chamber of an internal combustion engine. It translates the radial motion of the profile of the cam lobe through a fulcrum into linear motion for opening and closing the intake and exhaust valves.  It also provides a means of multiplying the lift ratio.

During operation, the rocker arm experience stresses and undergo deflection. Severe rocker arm deflection causes inefficient engine performance, and often results in metal fatigue leading to increased wear and friction in the valve train and eventually engine failure.

Figure 1: Diesel engine valve train showing rocker arm (1)

1.2.            Component Geometry

The rocker arm is a two-arm lever that pivots about a fulcrum. One end is connected to the push rod which rests over cams on the camshaft, while the other acts on the spring-loaded valve stem and pivoted on the rocker shaft.

Figure 2:(From left) 3-D model of the rocker arm and the 2-D geometry

 

1.3.            Service Loading

The rocker arm is subjected to compressive load at the fulcrum on the rocker shaft during the opening and closing of the valves.  The forces acting at the valve end (FE) include; the gas back pressure on the valve, the spring force and the force due to valve acceleration. There is also the load at the cam end, (FC) which is transmitted to the rocker arm through the push rod.

Figure 3: Free Body Diagram of Rocker arm depicting the location of the acting forces: FE and FC

Methodology (154)

         Two-Dimensional Idealization

In a rocker arm, the constraint at the pivot hole and the applied loads at both ends, all act in a plane, parallel to the cross-section plane of the rocker arm.

To estimate the stress concentration of the geometry, it is idealized as a shell and shaped in the profile of the planar cross section of the rocker arm under the assumption that the rocker arm is completely solid with only the holes drilled through as seen in Figure 4. The two-dimensional idealization makes it easier to deal with and the main stresses are obtained with reasonable accuracy.

Figure 4: 2D geometrical Idealization using Autodesk Inventor

2.2.            Loading, Boundary Conditions and Constraints

The rocker arm has two major loads applied on either end respectively:

Force due to exhaust valve loading; FE

Force due to cam action through the push rod; FC

The main forces will be calculated based on the Diesel Engine specifications below.

Type of Engine

Turbocharged 6-cylinder, Diesel Engine, 2523CCm2DiCR Engine.

Number of Cylinders

6

Engine capacity

2523 CC

Maximum Engine Power

46.3 kW @ 3200 rpm

Maximum Torque

195 Nm @ 1440 – 2200 rpm.

Table 1: Diesel Engine Specification (2)

Mass of the valve mv

0.09kg

Diameter of the valve head dv

40mm

Lift of the valve   h

9.4mm

Cylinder pressure Pc

0.4N/mm2

Maximum suction pressure Ps

0.02N/mm2

Diameter of fulcrum pin d1

22mm

Diameter of boss D1

34mm

Rocker arm ratio

1.64

Engine Speed

3200 RPM

Angle action of cam (Ɵ)

110o

Spring rate k

23 N/mm

Spring Preload P1

249.5N

Weight of associates parts with valve w

0.882 N

Acceleration of valve a

1550 m/s2

Table 2: Input data for calculation of applied load on rocker arm (2)

Total load on valve:
Pt=Pg+w=π4×d12×Pc+w=502.4N

Initial spring force:
Fs
=  
π4×d12×Pc–w
= 24.249 N

Force due to valve acceleration with valve weight:
Fa=m×a–w=140.38N
  

Maximum load on the rocker arm for exhaust valve:
Fe=P+Fs+Fa+(P1+k×h)=1133.87N

Load on push rod side of the rocker arm:
 Fc=Fe×Rocker arm ratio=1859.54N

A pinned constraint is applied on the pivoting hole in all degrees of freedom.

Finite Element Representation

         Modelling and Justification of Shell type

The material for the rocker arm; carbon steel is assumed to be homogeneous and isotropic with details in Table 3.

Young’s Modulus (GPa)

200

Poisson’s ratio

0.29

Density (kg/m3)

7850

Yield Strength (MPa)

415

Ultimate Strength (MPa)

585

Table 3: Carbon Steel Material Properties

Plane stress idealization is assumed because the thickness in the z-direction is smaller than the length in the x-y plane. Given the 2D geometry, it is idealized as a plane stress model and shell elements must be used. The shape of the element can either be triangular or quadrilateral and the element order is either linear or parabolic.

Element Size

Maximum von-Mises Stress

(MPa)

(Triangular element)

Maximum von-Mises Stress (MPa)

(Quadrilateral element)

Linear

Parabolic

Linear

Parabolic

1.0mm

97.51

140.7

141.9

167.8

0.8mm

109.2

149.0

137.6

167.9

0.6mm

118.9

162.3

143.6

176.5

0.4mm

144.9

182.3

170.5

179.4

0.2mm

162.6

185.6

178.6

185.4

Table 4: Comparison of Element type and order

The element type chosen for the meshing is the second order quadrilateral element. This was due to the accuracy and consistency of the von Mises stress values obtained when the Force FE is applied on the rocker arm in comparison to other element types. Therefore, it has proven to be the best for this geometry.

To accurately represent the pinned pivot constraint, a pinned, rigid body connector was applied at the centre point and connected to the pivot hole so all the nodes at the hole can better represent the motion.  For each loading scenario, one hole must have a fixed constraint the other has the load applied for the solution to be solved. This can be seen in Figure 5 and Figure 6.

Figure 5: Mesh diagram for FC loading scenario

Figure 6: Mesh diagram for FE loading scenario

Contour Plots

After the finite element model was solved, the maximum Von-Mises stress was recorded. The finite element contour plot for each loading scenario can be seen in Figure 7 and Figure 8.

Figure 7: FE loading scenario

Figure 8: FC loading scenario

When Force FC was applied, it predicted the largest maximum von-Mises stress compared to when Force FE was applied. Hence, FC is the worst-case loading scenario and the stress obtained from this scenario will be evaluated.

4.1.            Justification of results parameter

The von Mises stress is chosen as the main result parameter to be evaluated. This is required to evaluate the failure rate of the rocker arm, since it is made of a ductile material and is largely likely to fail as a result of shearing action. The von Mises stress yield criterion predicts failure for a ductile material, when the von Mises stress becomes equal to the yield strength.

The contour plot predicts the maximum von Mises stress value as 203.3MPa, occurring at the neck of the rocker arm.  This value is within the Yield strength for carbon steel of 415MPa.

Mesh Convergence

In order to explore the effects of mesh density on the prediction from the finite element model, several finite element models were generated, varying the number of elements from 39 (coarsest mesh) to 12069 (finest mesh) from element sizes ranging from 0.2mm to 20.5 mm. The von Mises stress from the worst-case scenario were calculated and compared, to study the influence of mesh density on the stresses predicted, as shown below.

Number of elements

39

84

500

1019

2364

3045

3970

5368

9189

12069

Von Mises stress (MPa)

107.4

122.5

189.6

195.6

205.6

203.1

205.7

204.5

204.4

203.3

Table 5: Effect of mesh density

From Figure 9, the stress converges to the maximum possible value of 203.3 MPa.

Conclusion

In this case, them model produced acceptable results. The maximum stress was 203.3 MPa and was localized mostly and the neck of the rocker arm which is where the stress was expected to be most concentrated. The convergence of the results shows that the result is not significantly dependent on the choice of discretization and is an indication of the validity of the accuracy of the prediction. Also, the result obtained under both loading conditions, indicating the same location (the neck of the rocker arm) as where the maximum von Mises stress is predicted and where failure is likely to occur provides further validation of the result. However, the component was assumed to be completely solid except for the holes. In reality, the 3D geometry (Figure 2) is a better representation of the component and the aforementioned assumption is a relatively poor one. Apart from the applied loads, the component would be subjected to thermal stresses. Furthermore, though plane stress was assumed, the thickness of 19 mm was relatively large compared to the length of 50.8 mm. To validate the model, plane strain assumptions may be made during the next analysis in which a displacement-controlled test can be used to identify the strain that causes failure.

References

1.) Diesel Engine Valve Train. Nuclear Power Training. [Online] [Cited: November 14th, 2018 at 21:34.] http://nuclearpowertraining.tpub.com/h1018v1/css/Figure-10-Diesel-Engine-Valve-Train-31.htm.

2.) DESIGN AND STATIC STRUCTURAL ANALYSIS OF ROCKER ARM IN I.C. Bacha, Sachin et al. s.l. : INTERNATIONAL JOURNAL OF ENGINEERING SCIENCES & RESEARCH, 2018, INTERNATIONAL JOURNAL OF ENGINEERING SCIENCES & RESEARCH, p. 9.