Two-Dimensional Geometries and Topologies, generated by a special

inner product for Complex-Type Numbers

A brief report on work performed in the topic for evaluation process

by

Yellanki Abhinav

(08MA2013)

Under the guidance of

Prof. Debapriya Biswas

Department of Mathematics

Indian Institute of Technology, Kharagpur

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CERTIFICATE

This is to certify that the M.Sc. project entitled Two-Dimensional and higher

Dimensional Geometries, generated by a special inner product for Complex-Type

Numbers submitted by Yellanki Abhinav (08MA2013) to the Department of

Mathematics in partial fulfillment of the requirements for the degree of Master of

Science is a bonafide record of the work carried out by him under my supervision

and guidance during the academic year 2012-2013.

Professor Debapriya Biswas Department of Mathematics

Indian Institute of Technology, Kharagpur

Kharagpur 721302

India

Date:

Signature:

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Acknowledgement

I declare all that this report is my own work. And I would like to express my

sincere gratitude to Professor Debapriya Biswas for providing me with the

opportunity to work in this project under her supervision. Her able guidance and

constant inspiration was a major source of motivation to help me carry out this

semester project.

Abhinav Yellanki

(08MA2013)

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TABLE OF CONTENTS:

Sl.no Topic Page No

1. Abstract……………………………………………………………………………………………. 5

2. Introduction……………………………………………………………………………………… 6

3. Preliminaries…………………………………………………………………………………….. 6

4. General complex type number systems ……………………………………………. 7

5. Geometries induced on 𝑹𝟐by 𝒈 ………………………………………………………. 8

6. Topologies induced on 𝑹𝟐by 𝒈 ……………………………………………………….. 16

7. Futurescope ……………………………………………………………………………………. 18

References………………………………………………………………………………………........... 18

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1. Abstract

In non relativistic classical mechanics we use Euclidean space[1][2], as time is treated as

independent of state of motion of the observer and the object. which won’t work in relativistic

mechanics since they usually have a time dimension. Theory of Relativity is a theory of Spacetime.

Spacetime is Mathematical model that combines space and time into a single continuum. It is usually

interpreted as 1or 2or 3 spatial dimensions along with a time dimension which isn’t spatial. Spacetime

can be viewed as the consequence of Albert Einstein’s “theory of special relativity”[1] It is explicitly

proposed mathematically by the mathematician Hermann Minkowski in a essay, building on and

extending Einsteins’s works.

In his work on special theory of relativity [3], G.L Naber provides a mathematically rigorous way

to understand the special theory of relativity showing the physical significance of mathematics. His model

is based on a special indefinite inner product.

Paul Fjelstad has worked on this field extending Naber’s work by showing that inner product and

topologies from Naber’s work can be generated by the hyperbolic complex numbers[4]. He studied two

dimensional geometries and physics generated in a similar manner to Naber for more general Complex

Type Numbers 𝐶𝑞 = { 𝑧 = 𝑥 + 𝑞𝑦; 𝑥,𝑦 ∈ 𝑹, 𝑞 ɇ 𝑹, 𝑞2 = 𝐴 + 𝑞 2𝐵 𝑤𝑕𝑒𝑟𝑒 𝐴,𝐵 ∈ 𝑹 } by defining a

more general inner product. His work starts with remark that all the general Complex type numbers 𝐶𝑞 are

ring isomorphic to one of the three types of familiar complex type numbers(elliptic, hyperbolic and dual).

Later he proves these complex type number systems are as a group of homeomorphisms to each other on

the topologies induced by the more general inner product he constructed.

Paul Fjelstad’s paper contain a number of propositions which are drawn from the similarities with

the Naber’s work on Minkowski space (𝑴 with three spacial dimensions and a time dimension). In this

paper the main focus is on understanding, verifying and proving the various propositions, lemmas and

theorems in the works mentioned above.

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2. Introduction

G.L Naber model is based on a special indefinite inner product 𝑔1:𝑴 x 𝑴 → 𝑹 [3] and defined as

𝑔1 𝑣,𝑤 = 𝑣1𝑤1 + 𝑣2𝑤2 + 𝑣3𝑤3 − 𝑣4𝑤4

where 𝑣,𝑤 ∈ 𝑅4 and 𝑣1 ,𝑣2 ,𝑣3 , 𝑣4 ,𝑤1 ,𝑤2 ,𝑤3 ,𝑤4 ∈ 𝑅

𝑔1𝑖s of index ‘1’ and the associated Lorentz group of transformations in 𝑅4. Index indicate

negative definiteness of the inner product it is otherwise the dimension of subspace of a given vector

space for which 𝑔(𝑣, 𝑣) = 0 where 𝑣 ∈ 𝑉𝑒𝑐𝑡𝑜𝑟 𝑆𝑝𝑎𝑐𝑒 on which 𝑔 is defined. His book also presents

some collection of unique work in topologies of the 4-D geometry he worked upon.

Paul Fjelstad defined inner product for more general Complex Type Numbers[4] 𝐶𝑞 = { 𝑧 = 𝑥 +

𝑞𝑦; 𝑥, 𝑦 ∈ 𝑹, 𝑞 ɇ 𝑹, 𝑞2 = 𝐴 + 𝑞 2𝐵 𝑤𝑕𝑒𝑟𝑒 𝐴,𝐵 ∈ 𝑹 } as 𝑔: 𝐶𝑞𝐴,𝐵 𝑥 𝐶𝑞

𝐴,𝐵 → 𝑹 and 𝑔 𝑣,𝑤 =

𝑣𝑤 ′ + 𝑤𝑣 ′ /2 Where 𝑧 ∈ 𝐶𝑞 , 𝑧 = 𝑥 + 𝑞𝑦 and the conjugate is defined as 𝑧′ = 𝑥 + 2𝐵𝑦 − 𝑞𝑦. Which can be

explicitly written as

𝑔 𝑣,𝑤 = 𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 − 𝐴𝑣2𝑤2 (1)

Where 𝑣,𝑤 ∈ 𝐶𝑞𝐴,𝐵

and 𝑣1 ,𝑣2 ,𝑤1 ,𝑤2 ∈ 𝑹

Which is also of Index ‘1’ and gives Naber inner product when B=0 and A=1. Likewise most of

the geometrical properties induced by inner product are also similar to that of the properties in Naber’s

works.

3. Preliminaries Indefinite Inner product:

This type of inner product satisfies all properties of standard inner product except the positive

definiteness property.

i.e., g(x,x) ≥ 0 is not always true.

Index of Inner product:

This is defined as the number of basis vectors of a finite dimensional vector space V with orthonormal

basis is of the form {e1, e2, … en}, for which g(ei, ei) ≤ 0 where i ϵ {1,2,..n}.

Null vector, Null Cone, Null wordline:

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𝑥 ∈ 𝑅2 will be called null (or light like) if 𝑔′(𝑥, 𝑥) = 𝑂. The null (light) cone 𝐶𝑁(𝑥0) at 𝑥0 ∈ 𝑅2 is

defined by 𝐶𝑁(𝑥0) = {𝑥 ∈ 𝑅2; 𝑄′(𝑥 − 𝑥0) = 0}, where 𝑄’(𝑥) = 𝑔’(𝑥, 𝑥). For 𝑥 ∈ 𝐶𝑁 𝑥0 ,𝑥 ≠ 𝑥0

the null wordline (of light ray) containing 𝑥0 and 𝑥 is defined by

𝑹𝒙𝟎,𝒙 = {𝒚 = 𝒙𝟎 + 𝒕(𝒙 − 𝒙𝟎); 𝒕 ∈ 𝑹}

Spacelike, Timelike:

If 𝑄’(𝑥) > 0, 𝑥 will be called spacelike and if 𝑄′(𝑥) < 𝑂, x will be called timelike.

Collection of all timelike vectors be 𝝉.

Time cone:

For each timelike 𝑥0 we define the time cone 𝐶𝑇(𝑥0), future time cone 𝐶𝑇+(𝑥0) and past time cone 𝐶𝑇

− 𝑥0

at 𝑥0 by

𝐶𝑇(𝑥0) = {𝑥 ∈ 𝑅2; 𝑄′ 𝑥 − 𝑥0 < 0}

𝐶𝑇+ 𝑥0 = 𝐶𝑇 𝑥0 ∩ 𝜏+, 𝐶𝑇

− 𝑥0 = 𝐶𝑇 𝑥0 ∩ 𝜏−

Duration of 𝑥:

For any 𝑥 ∈ 𝜏 one defines the duration of 𝑥 by 𝝉∗ (x) = −𝑸′(𝒙) .

Timelike straight line:

A subset of 𝑅2of the form 𝑥0 + 𝑡 𝑥 − 𝑥0 ; 𝑡 ∈ 𝑅 where 𝑄′(𝑥 − 𝑥0) < 0, is called timelike straight

line in 𝑅2.

4. General complex type number systems

A general complex type number system forms an Algebraic ring and can be written in the form

𝑧 = 𝑥 + 𝑞𝑦 where 𝑥,𝑦 𝜖 𝑅 and 𝑞 ɇ 𝑅 Satisfies the equation 𝑞2 = 𝐴 + 𝑞 2𝐵 with fixed𝐴,𝐵 𝜖 𝑹. In

general all the complex type number systems can be represented by Cq = {z = x + qy; x, y ϵ R }. All the

number systems of Cq are ring isomorphic with one of the three types of familiar complex type numbers

1. System of Complex numbers, if 𝐵2 + 𝐴 < 0

2. System of Dual numbers, if 𝐵2 + 𝐴 = 0

3. System of hyperbolic Complex numbers, if 𝐵2 + 𝐴 > 0

To prove the ring isomorphism, we can construct a function which establishes the ring isomorphism

present. In the case of B2 + A < 0 we can use the function ∅ : 𝐶𝑞𝐴,𝐵 → 𝐶𝑖

1,0 defined as

∅ 𝑥 + 𝑞𝑦 = 𝑥 + 𝐵𝑦 + 𝑖 𝐴 + 𝐵2

Where 𝑖2 = 1, 𝑞2 = 𝐴 + 𝑞(2𝐵) and 𝐵2 + 𝐴 > 0

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So inverse function will be

∅−1 𝑥 + 𝑖𝑦 = 𝑥 −𝐵𝑦

𝐴 + 𝐵2 + 𝑞 𝐴 + 𝐵2 𝑦

As the inverse exists ∅ is onto…..(1)

Consider 𝑧1 , 𝑧2 ∈ 𝐶𝑞𝐴,𝐵

where 𝑧1 = 𝑥1 + 𝑦1 𝑎𝑛𝑑 𝑧2 = 𝑥2 + 𝑦2

Now let us check for one-one

Let ∅ 𝑧1 = ∅ 𝑧2 ∅ 𝑥1 + 𝑞𝑦1 = ∅ 𝑥2 + 𝑞𝑦2

𝑥1 + 𝐵𝑦1 + 𝑖 𝐴 + 𝐵2 𝑦1 = 𝑥2 + 𝐵𝑦2 + 𝑖 𝐴 + 𝐵2 𝑦2

As real and imaginary parts have to be equal

𝐴 + 𝐵2 𝑦1 = 𝐴 + 𝐵2 𝑦2 & 𝑥1 + 𝐵𝑦1 = 𝑥2 + 𝐵𝑦2

𝑦1 = 𝑦2 & 𝑥1 = 𝑥2 Which implies 𝑥1 + 𝑞𝑦1 = 𝑥2 + 𝑞𝑦2 or 𝑧1 = 𝑧2 and ∅ is one-one ….(2)

From hence (1) & (2) ∅ is bijective.

Now ∅(𝑧1 +𝑧2) = ∅( (𝑥1 + 𝑥2) + 𝑞(𝑦1 + 𝑦2))

= (𝑥1 + 𝑥2) + 𝐵 𝑦1 + 𝑦2 + 𝑖 𝐴 + 𝐵2 𝑦1 + 𝑦2

= 𝑥1 + 𝐵𝑦1 + 𝑖 𝐴 + 𝐵2 𝑦1 + 𝑥2 + 𝐵𝑦2 + 𝑖 𝐴 + 𝐵2 𝑦2 ∅(𝑧1 +𝑧2) = ∅(𝑧1) + ∅(𝑧2) ∅(𝑧1 𝑧2) = ∅((𝑥1 + 𝑞𝑦1) 𝑥2 + 𝑞𝑦2 )

= ∅((𝑥1𝑥2 + 𝐴𝑦1𝑦2 ) + 𝑞 𝑥1𝑦2 + 𝑦1𝑥2 + 2𝐵𝑦1𝑦2 ) = 𝑥1𝑥2 + 𝐴𝑦1𝑦2 + 𝑥1𝑦2 + 𝑦1𝑥2 + 2𝐵𝑦1𝑦2

+ 𝑖 𝐴 + 𝐵2 𝑥1𝑦2 + 𝑦1𝑥2 + 2𝐵𝑦1𝑦2

= 𝑥1 + 𝐵𝑦1 + 𝑖 𝐴 + 𝐵2 𝑦1 . 𝑥2 + 𝐵𝑦2 + 𝑖 𝐴 + 𝐵2 𝑦2

= ∅ 𝑥1 + 𝑞𝑦1 .∅ 𝑥2 + 𝑞𝑦2 = ∅(𝑧1) . ∅(𝑧2) hence ∅ is ring isomorphism Likewise we can use the same function for establishing ring isomorphism in the remaining two cases too.

5. Geometry Induced on 𝑹𝟐 by 𝒈

Few important properties of 𝑔 given by (1)

Lemma:

(i) For all 𝐴,𝐵 ∈ 𝑅 the function 𝑔 in (i) is bilinear and symmetric.

(ii) 𝑔 is nondegenerate (i.e. 𝑔(𝑣,𝑤) = 𝑂, ∀ 𝑤 ∈ 𝑅2 implies v = (0, 0)) if and only if 𝐵2 + 𝐴 ≠ 𝑂

(iii) If 𝐵2 + 𝐴 > 0 then 𝑔 is of index 1, if 𝐵2 + 𝐴 < 0 then 𝑔 is of index 0.

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Lemma (iii) implies that indefiniteness of inner product in the systems of complex type numbers

is exclusively due to number systems which are ring isomorphic with hyperbolic complex numbers. And

other two types of number systems are usual positive definite and follow all geometrical properties of

definite inner product spaces.

So, we study only systems of complex type numbers satisfying 𝐵2 + 𝐴 > 0. For clarity let the

inner product 𝑔 in (1) be denoted by 𝑔′. The above lemma permits us to extend all the results of Naber’s

book to number systems for which 𝐵2 + 𝐴 > 0. Few of the important results are mentioned below

Theorem (1): Two nonzero null vectors 𝑥,𝑦 ∈ 𝑅2 are 𝑔’-orthogonal iff they are parallel (i.e. 𝑖𝑓𝑓 ∃ 𝑡 ∈

𝑅 𝑠𝑢𝑐𝑕 𝑡𝑕𝑎𝑡 𝑥 = 𝑡𝑦).

Proof: (=>)

Let 𝑥 be 𝑥1 ,𝑥2 and 𝑦 be 𝑦1 ,𝑦2

Given 𝑥,𝑦 are null vectors ,

𝑖. 𝑒. , 𝑥12 + 2𝐵𝑥1𝑥2 − 𝐴𝑥2

2 = 0

𝑦12 + 2𝐵𝑦1𝑦2 − 𝐴𝑦2

2 = 0

𝑥1 + 𝐵𝑥2 2 = (𝐴 + 𝐵2)𝑥2

2

𝑦1 + 𝐵𝑦2 2 = (𝐴 + 𝐵2)𝑦2

2

𝑥1 + 𝐵𝑥2 𝑦1 + 𝐵𝑦2 = (𝐴 + 𝐵2)𝑥2𝑦2

𝑔′ 𝑥,𝑦 = 0

Hence 𝑥,𝑦 are parallel.

Proof :(<=)

Given 𝑥, 𝑦 are parallel

𝑖. 𝑒. , 𝑥 = 𝑡𝑦 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑡 ∈ 𝑅

𝑔′ 𝑥, 𝑦 = 𝑔′ 𝑡𝑦,𝑦

= 𝑡𝑔′ 𝑦,𝑦

= 0 (𝑠𝑖𝑛𝑐𝑒 𝑦 𝑖𝑠 𝑛𝑢𝑙𝑙 𝑣𝑒𝑐𝑡𝑜𝑟)

Hence 𝑥,𝑦 are orthogonal

Theorem 2: If 𝑥 ≠ 𝑥𝑜 and 𝑄′(𝑥 − 𝑥𝑜) = 0 then

𝑹𝒙𝟎,𝒙 = 𝐶𝑁(𝑥) ∩ 𝐶𝑁(𝑥0)

Proof:

Consider a element 𝑦 =𝑥0 + 𝑡 𝑥 – 𝑥0 ; of 𝑹𝒙𝟎,𝒙

𝑦- 𝑥0 = 𝑡 𝑥 – 𝑥0

𝑄′(𝑦 -𝑥0) = 𝑄′(𝑡 𝑥 – 𝑥0 )

= 𝑡2(𝑄′ 𝑥 – 𝑥0 )

= 𝑡2 0 ( given 𝑄′ 𝑥 – 𝑥𝑜 = 0)

𝑄′(𝑦 -𝑥0) = 0

So, 𝑦 belongs to 𝐶𝑁(𝑥0)

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Now let us prove 𝑹𝒙𝟎,𝒙 = 𝑹𝒙,𝒙𝟎 then we can say 𝑦 also belongs to 𝐶𝑁(𝑥)

we know 𝑅𝑥0,𝑥 = { 𝑥0 + 𝑡(𝑥 − 𝑥0); 𝑡 ∈ 𝑅}

= { 𝑡𝑥 − (𝑡 − 1) 𝑥0); 𝑡 ∈ 𝑅}

= { 𝑥 + (𝑡 − 1)𝑥 − (𝑡 − 1) 𝑥0); 𝑡 ∈ 𝑅}

= { 𝑥 + (𝑡 − 1)(𝑥 − 𝑥0)); 𝑡 ∈ 𝑅}

= { 𝑥0 + 𝑘(𝑥 − 𝑥0); 𝑘 ∈ 𝑅}

𝑹𝒙𝟎,𝒙 = 𝑹𝒙,𝒙𝟎

So, 𝑦 belongs to 𝐶𝑁(𝑥)

Hence 𝑹𝒙𝟎,𝒙 ⊆ 𝐶𝑁(𝑥) ∩ 𝐶𝑁(𝑥0)

Now we have to prove 𝐶𝑁(𝑥) ∩ 𝐶𝑁 𝑥0 ⊆ 𝑹𝒙𝟎,𝒙

Consider a element 𝑦 ∈ 𝐶𝑁(𝑥) ∩ 𝐶𝑁 𝑥0 𝑦 − 𝑥,𝑦 − 𝑥0 𝑎𝑟𝑒 𝑛𝑢𝑙𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠

And we already now 𝑥 − 𝑥0 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑎 𝑛𝑢𝑙𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 (given 𝑄′ 𝑥 – 𝑥0 = 0)

we know 𝑦 − 𝑥0 = 𝑦 − 𝑥 − 𝑥 – 𝑥0 𝑄′ 𝑦 − 𝑥0 = 𝑄′( 𝑦 − 𝑥 − 𝑥 – 𝑥0 )

= 𝑄′( 𝑦 − 𝑥 + 𝑥0 − 𝑥 )

𝑄′ 𝑦 − 𝑥0 = 𝑄′ 𝑦 − 𝑥 + 𝑄′ 𝑥0 − 𝑥 + 2𝑔 𝑦 − 𝑥, 𝑥0 − 𝑥

𝑔 𝑦 − 𝑥, 𝑥0 − 𝑥 = 0 (since Q’ 𝑦 − 𝑥0 = 𝑄′ 𝑦 − 𝑥 = 𝑄′ 𝑥0 − 𝑥 = 0)

𝑦 − 𝑥, 𝑥0 − 𝑥 orthogonal

Case(i): 𝑦 = 𝑥 then 𝑦 ∈ 𝑹𝒙𝟎,𝒙

Case(ii): 𝑦 ≠ 𝑥 then 𝑦 − 𝑥 must be parallel with 𝑥 – 𝑥0

i.e., 𝑦 − 𝑥 = 𝑡 𝑥0 − 𝑥 𝑦 = 𝑥 + 𝑡(𝑥0 − 𝑥)

𝑦 ∈ 𝑹𝒙𝟎,𝒙

So, 𝐶𝑁(𝑥) ∩ 𝐶𝑁 𝑥0 ⊆ 𝑹𝒙𝟎,𝒙

Hence 𝑹𝒙𝟎,𝒙 = 𝐶𝑁 𝑥 ∩ 𝐶𝑁 𝑥0 is proved

Theorem (3): Suppose that 𝑣 = (𝑣1 ,𝑣2) is timelike and 𝑤 = (𝑤1 ,𝑤2)is either timelike or null and

nonzero. Then either

(i) 𝑣2𝑤2> 𝑂, in which case 𝑔′(𝑣,𝑤) < 𝑂,

or

(ii) 𝑣2𝑤2< 𝑂, in which case 𝑔′ 𝑣,𝑤 > 𝑂.

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Proof:

𝑣 𝑖𝑠 𝑡𝑖𝑚𝑒𝑙𝑖𝑘𝑒 => 𝑣12 + 2𝐵 𝑣1𝑣2 < 𝐴𝑣2

2

Add 𝐵2𝑣22 on both the sides

𝑣12 + 2𝐵 𝑣1𝑣2 + 𝐵2𝑣2

2 < 𝐴𝑣22 + 𝐵2𝑣2

2

𝑣1 + 𝐵𝑣2 2 < 𝐴𝑣2

2 + 𝐵2𝑣22 (𝑖)

𝑤 𝑖𝑠 𝑡𝑖𝑚𝑒𝑙𝑖𝑘𝑒𝑜𝑟 𝑛𝑢𝑙𝑙 => 𝑤12 + 2𝐵 𝑤1𝑤2 ≤ 𝐴𝑤2

2

Add 𝐵2𝑤22 on both the sides

𝑤12 + 2𝐵 𝑤1𝑤2 + 𝐵2𝑤2

2 ≤ 𝐴𝑤22 + 𝐵2𝑤2

2

𝑤1 + 𝐵𝑤2 2 ≤ 𝐴𝑤2

2 + 𝐵2𝑤22 (𝑖𝑖)

Multiplying (i) and (ii)

𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 2 < 𝐴 + 𝐵2 2𝑣2

2𝑤22

𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 < 𝐴 + 𝐵2 𝑣2𝑤2 (𝑖𝑖𝑖)

if 𝑣2𝑤2 > 0 by equation (iii) & 𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 ≤ 𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2

𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 < 𝐴 + 𝐵2 𝑣2𝑤2

𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 + 𝐵2𝑣2𝑤2 < 𝐴𝑣2𝑤2 + 𝐵2𝑣2𝑤2

𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 − 𝐴𝑣2𝑤2 < 0

𝑔′ 𝑣,𝑤 < 0

If 𝑣2𝑤2 < 0 by equ (iii) & − 𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 ≤ 𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2

− 𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 < 𝐴 + 𝐵2 𝑣2𝑤2

𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 > 𝐴 + 𝐵2 𝑣2𝑤2

𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 + 𝐵2𝑣2𝑤2 > 𝐴𝑣2𝑤2 + 𝐵2𝑣2𝑤2

𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 − 𝐴𝑣2𝑤2 > 0

𝑔′ 𝑣,𝑤 > 0

Hence Proved.

Corollary: A nonzero vector in 𝑅2 is 𝑔′-orthogonal to a timelike vector, then it must be spacelike.

Let us define a relation ~ on 𝜏 as follows, 𝑖𝑓 𝑣,𝑤 ∈ 𝜏 , then v~𝜏 w iff 𝑔′(𝑣,𝑤) < 𝑂 .

Proof: It is directly implied from above theorem

Theorem (4): The relation, ~ is an equivalence on 𝜏, and 𝜏 is the union of two disjoint sets denoted by

𝜏+and 𝜏−, which are cones.

Proof:

Reflexive: 𝑄′ 𝑣 < 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑣 ∈ 𝜏

𝑣 ~ 𝑣 ∀ 𝑣 ∈ 𝜏

Symmetric: 𝑣 ~ 𝑤 => 𝑔′ 𝑣,𝑤 < 0 => 𝑔′ 𝑤, 𝑣 < 0 => 𝑤 ~ 𝑣

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Transitive: let 𝑣~𝑤 𝑎𝑛𝑑 𝑤~𝑧

𝑔′ 𝑣,𝑤 < 0 𝑎𝑛𝑑 𝑔′ 𝑤, 𝑧 < 0

𝑣2𝑤2 > 0 𝑎𝑛𝑑 𝑤2𝑧2 > 0 (from theorem 3)

𝑣2𝑧2 > 0 ( for both 𝑤2 > 0 𝑎𝑛𝑑 𝑤2 < 0 𝑖𝑡 𝑕𝑜𝑙𝑑𝑠)

𝑔′ 𝑣, 𝑧 < 0

𝑣 ~ 𝑧

Hence ′~′ is equivalence relation on 𝝉

𝜏+ = {𝑣 = (𝑣1 , 𝑣2) |𝑣 ∈ 𝜏 𝑎𝑛𝑑 𝑣2 > 0}

𝜏− = {𝑣 = (𝑣1 , 𝑣2) |𝑣 ∈ 𝜏 𝑎𝑛𝑑 𝑣2 < 0}

To prove 𝜏+ and 𝜏− are cones we have to show that 𝑣,𝑤 ∈ 𝜏+ (𝜏−) then for any positive real number 𝑟,

𝑟𝑣 , 𝑣 + 𝑤 ∈ 𝜏+ (𝜏−)

We know 𝑟𝑣2 > 0 => 𝑟𝑣 ∈ 𝜏+

Let us consider the case of 𝑣 + 𝑤, we know 𝑣1 + 𝐵𝑣2 2 < 𝐴 + 𝐵2 𝑣2

2 & 𝑤1 + 𝐵𝑤2 2 ≤ (𝐴 + 𝐵2)𝑤2

2

𝑣1 + 𝐵𝑣2 𝑤1 + 𝐵𝑤2 < 𝐴 + 𝐵2 𝑣2𝑤2

𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 + 𝐵2𝑣2𝑤2 < 𝐴𝑣2𝑤2 + 𝐵2𝑣2𝑤2

𝑣1𝑤1 + 𝐵 𝑣1𝑤2 + 𝑣2𝑤1 − 𝐴𝑣2𝑤2 < 0

𝑔′ 𝑣,𝑤 < 0 & 𝑣 + 𝑤 ∈ 𝜏+

Hence 𝜏+, 𝜏− are both disjoint cones

Theorem (5): (Reversed Schwartz Inequality) If 𝑣 and 𝑤 are timelike vectors in 𝑅2 then

𝑔′ 𝑣,𝑤 2 ≥ [𝑔′(𝑣, 𝑣)𝑔′(𝑤,𝑤)] ,

and equality holds if and only if 𝑣 and 𝑤 are linearly dependent.

Proof: Construct a vector 𝑢

𝑢 = 𝑎𝑣 − 𝑏𝑤 where 𝑎 = 𝑔 𝑣,𝑤 and 𝑏 = 𝑔(𝑣, 𝑣)

consider 𝑔 𝑢, 𝑣 = 𝑔 𝑎𝑣 − 𝑏𝑤, 𝑣

= 𝑎𝑔 𝑣, 𝑣 − 𝑏𝑔(𝑤, 𝑣)

= 𝑔 𝑣,𝑤 𝑔 𝑣, 𝑣 − 𝑔(𝑣, 𝑣)𝑔(𝑤, 𝑣)

= 0

𝑢, 𝑣 are orthogonal

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𝑢 is spacelike or zero vector (from theorem 3)

𝑔 𝑢,𝑢 ≥ 0

𝑔 𝑎𝑣 − 𝑏𝑤,𝑎𝑣 − 𝑏𝑤 > 0

𝑎2𝑔 𝑣, 𝑣 + 𝑏2𝑔 𝑤,𝑤 − 2𝑎𝑏𝑔 𝑣,𝑤 ≥ 0

𝑔 𝑣,𝑤 2𝑔 𝑣, 𝑣 + 𝑔 𝑣, 𝑣 2𝑔 𝑤,𝑤 ≥ 2 𝑔 𝑣,𝑤 2𝑔 𝑣, 𝑣)

2𝑔 𝑣,𝑤 2 ≥ 𝑔 𝑣,𝑤 2 + 𝑔 𝑣, 𝑣 𝑔 𝑤,𝑤 (cancelling 𝑔 𝑣, 𝑣 and since 𝑔 𝑣, 𝑣) < 0

𝑔 𝑣,𝑤 2 ≥ 𝑔 𝑣, 𝑣 𝑔 𝑤,𝑤

Equality holds when 𝑢 = 0

𝑎𝑣 − 𝑏𝑤 = 0

𝑎𝑣 = 𝑏𝑤

𝑣,𝑤 are linear

Hence proved

Theorem (6): (Reversed Triangle Inequality) Let 𝑣 and 𝑤 be timelike with 𝑔′(𝑣,𝑤) < 𝑂. Then

𝜏∗ (𝑣 + 𝑤) ≥ 𝜏∗(𝑣) + 𝜏∗(𝑤),

and equality holds if and only if 𝑣 and 𝑤 are linearly dependent.

Proof: By theorem 5 𝑔′ 𝑣,𝑤 2 ≥ 𝑔′ 𝑣, 𝑣 𝑔′ 𝑤,𝑤 ≥ [−𝑔′(𝑣, 𝑣)][−𝑔′(𝑤,𝑤)]

So, 𝑔′ 𝑣,𝑤 ≥ −𝑔′(𝑣, 𝑣) −𝑔′(𝑤,𝑤)

𝑔′ 𝑣,𝑤 ≤ − −𝑔′(𝑣, 𝑣) −𝑔′(𝑤,𝑤) (given 𝑔′(𝑣,𝑤) < 𝑂)

𝑔′ 𝑣,𝑤 ≤ − −𝑄′(𝑣) −𝑄′(𝑤) (i)

𝜏∗ 𝑣 + 𝑤 = −𝑄′(𝑣 + 𝑤)

𝜏∗ 𝑣 + 𝑤 = −(𝑄′ 𝑣 + 𝑄′ 𝑣 + 2𝑔′ 𝑣,𝑤 )

𝜏∗ 𝑣 + 𝑤 ≥ −𝑄′ 𝑣 2

+ −𝑄′ 𝑤 2

+ 2 −𝑄′(𝑣) −𝑄′(𝑤) (𝑓𝑟𝑜𝑚 𝑖 )

𝜏∗ 𝑣 + 𝑤 ≥ 𝜏∗ 𝑣 2+ 𝜏∗ 𝑤 2 + 2𝜏∗ 𝑣 𝜏∗ 𝑤

𝜏∗ 𝑣 + 𝑤 ≥ 𝜏∗(𝑣) + 𝜏∗(𝑤)

Since we used Reversed Schwartz Inequality theorem in which equality holds only when 𝑣 ,𝑤 are linear

it’s the same in this result also.

Lemma: Let 𝛼: 𝐴,𝐵 → 𝑚 be a smooth timelike and future directed curve. Then for any 𝑡0 in 𝐴,𝐵

there exist 𝜀 > 0 such that (𝑡0 − 𝜀 , 𝑡0 + 𝜀 ) is contained in 𝐴,𝐵 , 𝛼(𝑡) is in past cone at 𝛼(𝑡0) for every

𝑡 in (𝑡0 − 𝜀 , 𝑡0) and is in future cone at for every 𝑡 in (𝑡0 , 𝑡0 + 𝜀 ).

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Proof: We prove that there exists a 𝜀1 > 0 such that 𝛼 𝑡 is in 𝐶𝑇+(𝛼 𝑡0 ) for each 𝑡 in (𝑡0 , 𝑡0 + 𝜀1 ). The

argument to produce and 𝜀2 > 0 with 𝛼 𝑡 is in 𝐶𝑇−(𝛼 𝑡0 ) for each 𝑡 in (𝑡0 − 𝜀2 , 𝑡0 ) is similar. Taking 𝜀

be the smallest of 𝜀1 and 𝜀2 proves the lemma

If no such 𝜀1 exists then one can produce a sequence 𝑡1 > 𝑡2 > 𝑡3 … > 𝑡0 in 𝑡0 ,𝐵 such that

lim𝑛→∞ 𝑡𝑛 = 𝑡0 and such that one of the following is true

1) 𝑄′(𝛼 𝑡𝑛 − 𝛼 𝑡0 ) ≥ 0 for all 𝑛 (i.e., 𝛼 𝑡𝑛 − 𝛼 𝑡0 is spacelike )

2) 𝑄′(𝛼 𝑡𝑛 − 𝛼 𝑡0 ) < 0 but 𝛼 𝑡𝑛 − 𝛼 𝑡0 is past directed for every 𝑛

Consider case(1), On contrary let us assume that such a sequence doesn’t exist,

i.e., 𝑄′ 𝛼 𝑡𝑛 −𝛼 𝑡0

𝑡𝑛− 𝑡0 ≥ 0 for all 𝑛

𝑄′ 𝑥1 𝑡𝑛 −𝑥

1 𝑡0

𝑡𝑛− 𝑡0 ,𝑥2 𝑡𝑛 −𝑥

2 𝑡0

𝑡𝑛− 𝑡0 ≥ 0

Thus,

lim𝑛→∞ 𝑄′ 𝑥1 𝑡𝑛 −𝑥

1 𝑡0

𝑡𝑛− 𝑡0 ,𝑥2 𝑡𝑛 −𝑥

2 𝑡0

𝑡𝑛− 𝑡0 ≥ 0

𝑄′ 𝑑𝑥1 𝑡0

𝑑𝑡 ,𝑑𝑥2 𝑡0

𝑑𝑡 ≥ 0

𝑄′ 𝛼′ 𝑡0 ≥ 0

‘𝛼’ is spacelike or null which is contrary to the fact ‘𝛼’ is timelike.

Now consider Case(2) :

If 𝑄′(𝛼 𝑡𝑛 − 𝛼 𝑡0 ) < 0 but 𝛼 𝑡𝑛 − 𝛼 𝑡0 is past directed

by definition of 𝛼′ 𝑡0 is future directed and from theorem(3) for all 𝑛

𝑔′(𝛼 𝑡𝑛 − 𝛼 𝑡0 , 𝛼′ 𝑡0 ) > 0

lim𝑛→∞

𝑔′ 𝛼 𝑡𝑛 − 𝛼 𝑡0

𝑡𝑛 − 𝑡0 ,𝛼′ 𝑡0 > 0

𝑔′ lim𝑛→∞

𝛼 𝑡𝑛 − 𝛼 𝑡0

𝑡𝑛 − 𝑡0 ,𝛼′ 𝑡0 > 0

𝑔′ 𝛼′ 𝑡0 ,𝛼′ 𝑡0 > 0

𝑄′ 𝛼′ 𝑡0 > 0

Which is again contrary to the fact 𝛼 is timelike. Hence the lemma is true.

15

Theorem (7): let 𝑝 and 𝑞 be in 𝑹𝟐 then 𝑝 − 𝑞 is timelike and future directed if and only if there exists a

smooth future directed timelike curve 𝛼: 𝑎, 𝑏 → 𝑹𝟐 such that 𝛼 𝑎 = 𝑞 and 𝛼 𝑏 = 𝑝.

Proof: The necessity is clear since always there exists a straight line(curve) joining 𝑝 and 𝑞. i.e., there

exists a curve 𝛼: 𝑎, 𝑏 → 𝑹𝟐 such that 𝛼 𝑡 = 𝑝 𝑡−𝑎

𝑏−𝑎 + 𝑞

(𝑏−𝑡)

(𝑏−𝑎)

To prove sufficiency we take 𝛼 as smooth future directed timelike extension of 𝛼 to some interval 𝐴,𝐵

containing [a,b]. By the above lemma there exists an 𝜀1 > 0, such that

𝑎,𝑎 + 𝜀1 ⊆ 𝐴,𝐵 & 𝛼 𝑡 ∈ 𝐶𝑇+ 𝛼 𝑡0 ∀ 𝑡 ∈ 𝑎,𝑎 + 𝜀1

Let 𝑡0 be supremum of all such 𝜀1. Since 𝑏 < 𝐵 it will be sufficient to show 𝑡0 = 𝐵 for this let us assume

the contrary that 𝐴 < 𝑡0 < 𝐵. According to the above lemma,

∃𝜀 > 0 such that 𝑡0 − 𝜀, 𝑡0 + 𝜀 ⊆ 𝐴,𝐵 for 𝑡 𝑖𝑛 𝑡0 − 𝜀 , 𝑡0 and 𝛼 𝑡 ∈ 𝐶𝑇− 𝛼 𝑡0 and for 𝑡 in

𝑡0 , 𝑡0 + 𝜀 and 𝛼 𝑡 ∈ 𝐶𝑇+ 𝛼 𝑡0 . Now if 𝛼 𝑡0 is in 𝐶𝑇

+ 𝑞 then for any 𝑡 in 𝑡0 , 𝑡0 + 𝜀 which implies

𝛼 𝑡0 − 𝑞 + 𝛼 𝑡 − 𝛼 𝑡0 = 𝛼 𝑡 − 𝑞

Which is timelike and future directed contradicting the definition of 𝑡0 . And if 𝛼 𝑡0 is outside Null

cone at ′𝑞′ then 𝑡0 − 𝜀, 𝑡0 is disjoint from future cone at ′𝑞′ , which implies𝑡0 = 𝐵 which proves the

theorem.

Definition: If 𝛼: 𝑰 → 𝑹𝟐 , the proper time function on 𝑰 = 𝑎, 𝑏 is defined by

𝜏∗ 𝑡 = [𝑔′(𝛼′ 𝑢 ,𝛼′ 𝑢 |]1/2 𝑑𝑢𝑏

𝑎

, 𝑡 ∈ 𝑰

For 𝑥, 𝑥0 ∈ 𝑹𝟐 with 𝑄′ 𝑥 − 𝑥0 > 0, the proper spatial separation defined by 𝑆∗ 𝑥 − 𝑥0 =

𝑄′ 𝑥 − 𝑥0 .

Theorem (8) : Let 𝑥, 𝑥0 , 𝑥1 ∈ 𝑹𝟐 for which 𝑥1 − 𝑥0, 𝑥1 − 𝑥 are spacelike and 𝑥 − 𝑥0 , 𝑥1 − 𝑥 are

orthogonal then ,

𝑆∗2 𝑥1 − 𝑥0 = 𝑆∗

2 𝑥1 − 𝑥 – 𝜏2 𝑥 − 𝑥0 .

Proof: 𝑥1 − 𝑥0 can be written as 𝑥1 − 𝑥 + 𝑥 − 𝑥0 . So,

𝑆∗2 𝑥1 − 𝑥0 = 𝑆∗

2 𝑥1 − 𝑥 + 𝑥 − 𝑥0

= 𝑄′ 𝑥 − 𝑥0 + 𝑄′ 𝑥1 − 𝑥 − 2𝑔′(𝑥1 − 𝑥, 𝑥 − 𝑥0)

As 𝑥 − 𝑥0 , 𝑥1 − 𝑥 are orthogonal

16

= 𝑄′ 𝑥 − 𝑥0 + 𝑄′ 𝑥1 − 𝑥

= 𝑄′ 𝑥 − 𝑥0 − (−𝑄′ 𝑥1 − 𝑥 )

= 𝑆∗2 𝑥1 − 𝑥 + 𝑥 − 𝑥0

6.Topologies induced on 𝑹𝟐 by g

We can consider the Topologies generated by the following Bases of topology

𝑩− = 𝐷− 𝑥; 𝜀 ; 𝑥 ∈ 𝑹𝟐 , 𝜀 > 0 ∪ ∅ & 𝑩+ = 𝐷+ 𝑥; 𝜀 ; 𝑥 ∈ 𝑹𝟐 , 𝜀 > 0 ∪ ∅

Where

𝐷− 𝑥; 𝜀 = 𝑦 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥,𝑦 < 𝜀 𝑎𝑛𝑑 𝑔 𝑦 − 𝑥,𝑦 − 𝑥 < 0 ∪ 𝑥

𝐷+ 𝑥; 𝜀 = 𝑦 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥,𝑦 < 𝜀 𝑎𝑛𝑑 𝑔 𝑦 − 𝑥,𝑦 − 𝑥 > 0 ∪ 𝑥

Let us denote the topologies generated by 𝑩− and 𝑩+, with 𝑇−and 𝑇+, respectively.

Concerning 𝑔 we have three possibilities:

(i) 𝐵2 + 𝐴 > 0; (ii) 𝐵2 + 𝐴 < 0;

(iii) 𝐵2 + 𝐴 = 0

Case (i). Firstly we have

Lemma: 𝑩− and 𝑩+ are bases for certain topologies

Proof: Let us consider for example B_ (the case for /3+ is similar). We have to check the following two

properties:

a) 𝑹𝟐 = 𝐵𝑩 ∈𝑩− ;

b) ∀ 𝑩1,𝑩2 ∈ 𝑩−,𝑩1 ∩ 𝑩2 can be written as union of elements of 𝑩−.

For a) it is obvious since for every 𝑥 ∈ 𝑹𝟐 ∃ 𝐷− 𝑥; 𝜀 ∈ 𝑩− such that 𝑥 ∈ 𝐷− 𝑥; 𝜀

In order to prove b) let 𝑩1 = 𝐷− 𝑥1; 𝜀1 𝑎𝑛𝑑 𝑩2 = 𝐷− 𝑥2; 𝜀2 . If 𝑥1 = 𝑥2 then (b) is obvious and the

case of 𝑥1 ≠ 𝑥2 𝑩1 ∩𝑩2 = ∅ is also trivial.

So let us consider the case 𝑩1 ∩𝑩2 ≠ ∅. It is easy to see that 𝑩1 = 𝑃𝑥1 ∪ {𝑥1 } , 𝑩2 = 𝑃𝑥2

∪ {𝑥2 }

Where 𝑃𝑥𝑖 ={ 𝑥 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥, 𝑥𝑖 < 𝜀𝑖} ∩ {𝑥 ∈ 𝑹𝟐 ;𝑔′ 𝑥 − 𝑥𝑖 ,𝑥 − 𝑥𝑖 < 0} for 𝑖 = 1,2 which are

obviously open subsets in the Euclidean topology on 𝑹𝟐. It follows,

𝑩1 ∩ 𝑩2 = 𝑃𝑥1 ∩ 𝑃𝑥2

∪ {𝑥1} ∩ 𝑃𝑥2 ∪ 𝑃𝑥1

∩ {𝑥2}

Let 𝑥 ∈ 𝑃𝑥1 ∩ 𝑃𝑥2

is a non empty openset, there exists 𝑟𝑥 > 0 such that

𝐶 = { 𝑥 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥,𝑦 < 𝑟𝑥} ⊂ 𝑃𝑥1 ∩ 𝑃𝑥2

Denoting

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𝐴 = { 𝑦 ∈ 𝑹𝟐 ;𝑔′ 𝑥 − 𝑥𝑖 ,𝑥 − 𝑥𝑖 <0}

And

𝐷− 𝑥; 𝑟𝑥 = 𝐴 ∩ 𝐶 ∪ 𝑥 ⊂ 𝑃𝑥1 ∩ 𝑃𝑥2

It easily follows

𝑃𝑥1 ∩ 𝑃𝑥2

= 𝐷− 𝑥; 𝑟𝑥

𝑥∈𝑃𝑥1 ∩𝑃𝑥2

Now if 𝑥1 ∉ 𝑃𝑥2 𝑎𝑛𝑑 𝑥2 ∉ 𝑃𝑥1

then

𝑩1 ∩ 𝑩2 = 𝑃𝑥1 ∩ 𝑃𝑥2

= 𝐷− 𝑥; 𝑟𝑥

𝑥∈𝑃𝑥1 ∩𝑃𝑥2

If for example 𝑥1 ∈ 𝑃𝑥2 and 𝑥2 ∉ 𝑃𝑥1

then because 𝑃𝑥1 is open there exists 𝑟1 < 𝜀1sufficiently small such

that

𝐶1 = {𝑥 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥,𝑦 < 𝑟𝑥} ⊂ 𝑃𝑥2

This immediately implies

𝐷− 𝑥1; 𝑟1 = 𝐶1 ∩ {𝑥 ∈ 𝑹𝟐 ;𝑔′ 𝑥−𝑥𝑖 ,𝑥−𝑥𝑖 < 0} ⊂ 𝑃𝑥2

And

𝐷− 𝑥1; 𝑟1 \{𝑥1} ⊂ 𝑃𝑥1

That is 𝐷− 𝑥1; 𝑟1 \{𝑥1} ⊂ 𝑃𝑥1 ∩ 𝑃𝑥2

. Consequently,

𝑩1 ∩ 𝑩2 = 𝑃𝑥1 ∩ 𝑃𝑥2

∪ {𝑥1} = 𝐷− 𝑥; 𝑟𝑥

𝑥∈𝑃𝑥1 ∩𝑃𝑥2

∪ 𝐷− 𝑥1; 𝑟1

The cases 𝑥1 ∉ 𝑃𝑥2 and 𝑥2 ∈ 𝑃𝑥1

and 𝑥1 ∈ 𝑃𝑥2 and 𝑥2 ∈ 𝑃𝑥1

are similar hence 𝑩− is base

Similar argument can be extended to prove 𝑩+ is base of certain topology which proves lemma.

Lemma: The closure of 𝑀𝑃 of 𝑁εP (𝑥) is 𝐶𝑙𝐸 𝑁ε

P 𝑥 − (𝑏𝑑𝑦𝐸 𝑁εP 𝑥 ∩ 𝑏𝑑𝑦𝐸 𝑥 )

Proof: Since 𝑃-Topology is finer than 𝐸-Topology 𝐶𝑙𝑃 𝐴 ⊆ 𝐶𝑙𝐸 𝐴 for any set 𝐴 𝑜𝑓 𝑀. Moreover the

points in 𝑏𝑑𝑦𝐸 𝑁εP 𝑥 ∩ 𝑏𝑑𝑦𝐸 𝑥 are not in 𝐶𝑙𝑃 𝑁ε

P 𝑥 since if ‘𝑦’ is such a point then 𝑁ε/2P 𝑦

doesn’t intersect 𝑁εP 𝑦 (the null cone at ‘𝑦’ is tangent to the surface of the Euclidean ball 𝑁ε

E 𝑥

Thus, 𝐶𝑙𝑃 𝑁εP 𝑥 ⊆ 𝐶𝑙𝐸 𝑁ε

P 𝑥 − (𝑏𝑑𝑦𝐸 𝑁εP 𝑥 ∩ 𝑏𝑑𝑦𝐸 𝑥 ). Reverse containment is clear since if

‘𝑦’ is in the set on right hand side, every 𝑁δP 𝑦 intersects 𝑁ε

P 𝑥 .

Theorem (9): (a) If B2 + A > 0 then the topologies 𝑇−and 𝑇+ generated by the bases 𝑩− 𝑎𝑛𝑑 𝑩+

respectively, are finer than the Euclidean topology, are Hausdorff, are connected, but not regular, and not

locally compact.

Proof:

i) 𝑇−and 𝑇+ are finer: Let 𝑇𝐸be Euclidean topology with base

𝑩𝐸 = {𝐷𝐸 𝑥; 𝑟 ; 𝑥 ∈ 𝑹𝟐 , 𝑟 > 0 }

𝑤𝑕𝑒𝑟𝑒 𝐷𝐸 𝑥; 𝑟 = {𝑦 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥,𝑦 < 𝑟}

Take any element of base 𝑩𝐸 say 𝐷𝐸 𝑦; 𝑟𝑦 . For any 𝑥 ∈ 𝐷𝐸 𝑦; 𝑟𝑦 ∃ 𝑩1 ∈ 𝑩− such that

𝑏 = 𝐷− 𝑥; 𝜀 where 0 < 𝜀 < 𝑟𝑦 − 𝑑𝐸 𝑦; 𝑥 . Hence any 𝐷𝐸 𝑦; 𝑟𝑦 is also open in 𝑇−.

Conversely take any open set in base 𝑩− of 𝑇− say 𝐷− 𝑥; 𝜀 . Clearly there doesn’t exist any open

set in 𝑩𝐸 which contains ‘𝑥’ and is contained in 𝐷− 𝑥; 𝜀 .

18

Hence 𝑇− is finer topology than 𝑇𝐸 . With the similar argument we can say 𝑇+ is finer than 𝑇𝐸

which proves the theorem.

ii) Consider Topology 𝑇− any two points 𝑥 = (𝑥1 ,𝑥2 ) and 𝑦 = (𝑦1 ,𝑦2 )

Let 𝜀 = 𝑑𝐸 𝑥,𝑦 , take 𝑟𝑥 , 𝑟𝑦 < 𝜀/3 𝐷− 𝑥; 𝑟𝑥 and 𝐷− 𝑦; 𝑟𝑦 are disjoint open sets in 𝑇− .

Hence (𝑹𝟐 ,𝑇−) is hausdorff. Similarly (𝑹𝟐 ,𝑇+) is also Hausdorff.

iii) Claim:𝑹𝟐 𝑎𝑛𝑑 ∅ are the only two open and closed sets in (𝑹𝟐 ,𝑇−).

Take any open set in (say A). we can write

𝐴 = 𝐷− 𝑥𝑖 ; 𝑟𝑖

𝑖∈𝑁

𝐴 = [ 𝑦 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥𝑖 ,𝑦 < 𝑟𝑖 𝑎𝑛𝑑 𝑔 𝑦 − 𝑥𝑖 ,𝑦 − 𝑥𝑖 < 0 ∪

𝑖∈𝑁

{𝑥𝑖}]

And

𝐴 = [ 𝑦 ∈ 𝑹𝟐 ;𝑑𝐸 𝑥𝑖 ,𝑦 ≤ 𝑟𝑖 𝑎𝑛𝑑 𝑔 𝑦 − 𝑥𝑖 ,𝑦 − 𝑥𝑖 ≤ 0

𝑖∈𝑁

]

≠ 𝐴

Hence (𝑹𝟐 ,𝑇−) is connected. Similarly (𝑹𝟐 ,𝑇+) is also connected.

iv) From the lemma from before it can be said that 𝐶𝑙𝑃 𝑁δP 𝑥 ⊄ 𝑁ε

P 𝑥 for any value of ε. Hence

(𝑹𝟐 ,𝑇−) is not regular. Similarly, (𝑹𝟐 ,𝑇+) is also not regular.

v) Since 𝑃- topology is finer than 𝐸- topology, any 𝑃- compact is 𝐸- compact. And no 𝐶𝑙𝑃 𝑁εP 𝑥

is compact (since not even 𝐸- closed) we find that no point in (𝑹𝟐 ,𝑇−) has a compact

neighborhood. In particular (𝑹𝟐 ,𝑇−) is also not locally compact. Similarly, (𝑹𝟐 ,𝑇+) is also not

locally compact.

7.Future scope:

My project is to understand the works of Naber and Paul thoroughly and work upon the

challenges in this explorable field of mathematics. The above study comprised of Geometrical properties

Topologies induced by the inner product’ 𝑔’ on different 2 dimensional systems. And similarly we can

study other characteristics like trigonometry and physics generated by these complex type numbers which

are extensively used in Relativity and extend the results in this study to the higher dimensional vector

spaces.

References: 1. Einstein Albert, "Space–Time", Encyclopedia Britannica (1926), 13th ed

2. Feynman, Richard Phillips; Morínigo, Fernando B.; Wagner, William; Pines, David; Hatfield, Brian

(2002). “Feynman Lectures on Gravitation”. West view Press.

3. Naber G.L., “The Geometry of Minskowski Spacetime . An Introduction to Mathematics of

Special Theory of Relativity “ Springer- verlag, Newyork (1992)

4. Paul Flejstad, “Two-Dimensional Geometries, Topologies, Trigonometry and Physics generated

by Complex-Type Numbers” (2001).

5. Stephen Willard, “General Topology”, Addison Wisley (1970)