# Mazur Manifold – Class Lecture

Let G = < A | R > be a presentation of a group.

Tietze Transformations (finite versions) are ways that a group presentation can be altered without changing the group.

$T_I$ – add a relator that is a consequence of the other relators.

example: $aba^{-1}b^{-1} , aba$ implies $a^2b$

$T_{II}$ – add a new generator while simultaneously adding a relator that expresses that new generator as a word in the existing generators.

example: $$ transformed into $$

$T_{I}^{-1}, T_{II}^{-1}$ – reverses of the above moves.

Fact: Finite presentation < A | R > and  < B | S > represent isomorphic groups if one may be obtained from the other via a finite sequence of Tietze Transformations.

Note: Also true for non-finitely presented groups if we generalize the Tietze Transformation.

Goal: To construct a compact contractible 4 – manifold, that is not homeomorphic to a $B^4$.

We will construct a Mazur 4-manifold. We begin with a 4 -dimensional solid torus $S^2 \times B^3$.

We will attach a single 2-handle which kills the fundamental group but does not yield a 4-ball.

Clearly $\partial (S^1 \times B^3 ) = S^1 \times S^2$

Begin by creating a picture of $S^2 \times S^2$ by doing a single 0 – Dehn surgery on the punctured unknot.

Next, we attach a 2 – handle whose attaching sphere (i.e. circle) lies in the complement of that unknot and hence in $S^1 \times S^2$.

That circle is pictured.

For meridional curve in the corresponding integral Dehn surgery, choose a simple closed curve that stays parallel to $\Gamma$ in the given picture.

$\pi_1 (S^3 - \Gamma \cup \Theta ) = < x_1 , ... , x_9 | r_1 , ... , r_9 >$

$\pi_1(S^1 \times S^2 - \Gamma ) = < x_1 , ..., x_9 | r_1, ... r_9 , x_5 x_2^{-1}x_1^{-1} >$

$\pi_1 ( \partial M^4 ) = $

Perform Tietze Transformations

Let $\beta = x_7 , \lambda = x_2 , \alpha = \beta \lambda$

Perform Tietze Transformations

$\pi_1 ( \partial M^4 ) \equiv < \alpha , \beta | \beta^5 = \alpha^7 , \beta^4 = \alpha^2 \beta \alpha^2 >$

Goal: To show that this group is non-trivial.

First, we will introduce an additional relator $\beta^5 = 1$

The resulting group is the quotient group $\pi_1 (\partial M^4 ) / << \beta^5 >>$. If it is non-trivial then so is $\pi_1 ( \partial M^4 )$.

We now have $< \alpha, \beta | \alpha^7 , \beta^5, \beta^4 = \alpha^2 \beta \alpha^2 >$

# Genus Two Symmetry

Thurston 1.2.6 (a)

Julian Rosen created a photo and answer:

Imagine three lines of longitude on a unit sphere (embedded in ℝ3), at angles 0, 2π/3, and 4π/3. Let G be the union of those three lines of longitude. Then G has three-fold rotational symmetry and the set of points in ℝ3 that are distance exactly 14 from G form a genus 2 surface.

1.2.6 (b)

# Thurston Chapter 1 Doubts

• Problem 1.1.1(a) Stereographic Projection is the homeo?
• (b), (c) ??
• What are the ‘hidden symmetries’ of a torus? Is it a rotation of points along the vertical circles?
• Regular Hexagon to torus: Is this a twisted torus?
• 1.1.2 (c), (d)
• Are there two tori? Before identification and after identification? As they have different curvatures.
• Torus =  Plane / Group of which isometries? (Integer translations?)
• Why is the hyperbolic plane homeomorphic to $\mathbb{R}^2$. Is it because we want the Topological Space $\mathbb{R}^2$ to have a geometry (a metric) that defies the parallel axiom?
• Are we trying to backtrack to the metric from geodesics and isometries?
• What is an angle in the hyperbolic plane? Usually, we understand ‘angle’ as a measure of rotation. How do we imagine angle in the Hyperbolic Plane?
• Mechanical Linkage – skipped in the first run
• Page 13: The driving idea is that hyperbolic reflections should preserve distances, that is they should be isometries. This is enough to pin down the metric up to a constant (why?)
• Double check the simple argument (Page 13)
• Page 13: Now apply hyperbolic reflection to the whole picture (about which circle?)
• Page 14: The Euclidean and Hyperbolic Lengths should be proportional to the first order. Why? Is this because inversions are conformal?
• Page 14: The Poincare model is conformal because the Euclidean and Hyperbolic angles are equal.
• Problem 1.2.6 (b), (c), (d)
• Problem 1.3.2
• Page 19: Euler Number = Number of even cells – Number of odd cells (is this just an extension of V- E + F)
• Page 20: Each face receives the net charge from an open interval along its boundary. What open interval?
• Does all differentiable surfaces have differentiable triangulation?
• Page 22: How to achieve transverse triangulation so that ‘the direction of each edge and of the field in the star of v, measured in these coordinates, changes by no more than $\epsilon$.

# একটি সূত্রের জন্ম

(ফরাসি গণিতজ্ঞ, সেড্রিক ভিলানি, ‘থিয়োরেম ভিভান্ত’ গ্রন্থ রচনা করেন ২০০৮ সালে। বইটি খানিকটা ডায়রি লেখার ঢঙে রচিত। মূল ফরাসি থেকে বাংলায় অংশ বিশেষ অনুবাদ করছি স্রেফ খেলার ছলে। কোনো বাণিজ্যিক উদ্দেশ্য নেই।)।

লিয়ঁ, ২৩শে মার্চ, ২০০৮

রবিবার দুপুর ১টা। প্রায় জনহীন গবেষণাগারে দু’জন ব্যস্ত গণিতজ্ঞ ছাড়া বোধকরি আর কেউ নেই। লিয়ঁর এই গবেষণাকেন্দ্রে আমি গত আট বছর ধরে আছি। ইকোল নরম্যাল সুপিরিয়রের তিন তলার এই ঘর আমাদের যাবতীয় নিবিড় চিন্তনের সূতিকাগার।

এই ঘরে একটা বেশ মস্ত আরামকেদারা আছে। সেখানে বসে আমি মাকড়ষার মত আঙুল ছড়িয়ে তাল ঠুকছিলাম টেবিলে। ঠিক যেমন আমার পিয়ানো শিক্ষক আমায় শিখিয়েছিলেন।

আমার বাম দিকে, টেবিলে, একটা কম্পিউটার রাখা। ডানদিকে, একটা মস্ত ক্যাবিনেট। সেখানে কয়েকশো গণিত ও পদার্থবিদ্যার বই ঠাসা। আমার ঠিক পেছনে, টানা বই-এর তাক। সেখানে হাজার হাজার পাতার প্রবন্ধ, আদ্যিকালের গবেষণাপত্রের ফোটোকপি রাখা। এসব সেই সময়ের সংগ্রহ যখন আমার বেতন নিতান্ত অল্প ছিল। পয়সা খরচ করে বই-তেষ্টা মেটানো তখন কল্পনাতীত। এছাড়া তাকে সারদিয়ে অনেকগুলি খসড়া, হাতে লেখা ক্লাসনোট, সেমিনার নোট রাখা। কথ অগুনতি ঘণ্টা যে আমি গবেষণাপত্র পাঠ শুনে কাটিয়েছি তার ইয়ত্তা নেই। এগুলি তারই সাক্ষ্য দিচ্ছে।

সামনের ডেস্কে বসে আছে গ্যাসপার্ড। গ্যাসপার্ড আমার ল্যাপটপের নাম। প্রবাদপ্রতীম গণিতজ্ঞ গ্যাসপার্ড মঞ্জের নামে রাখা। ডেস্কের ওপর বেশ কিছু কাগজ রাখা। সেখানে দিন দুনিয়ার আট মুলুক থেকে আনা অজস্র গাণিতিক আঁকিবুকি কিলবিল করছে।

সামনের দেওয়াল জুড়ে মস্ত হোয়াইট বোর্ড। আমার সহচর ক্লেমেন্ট মোহুট একটা মার্কার পেন নিয়ে, বোর্ডের সামনে দাঁড়িয়ে। তার চোখে বিদ্যুৎ খেলছে যেন।

# Motivation

Locally finite: Each vertex is attached to only finitely many simplices.

Why locally finite: To make sure it is a CW complex. Notice that the closure-finite criteria require each cell of a CW complex to meet only finitely many other cells.

Hence we do not have a situation like this:

Suppose K is any finite subcomplex of X.

We are interested in the number of infinite connected components of X – K. Call that number n(K).

Example:

Consider the Real line (with usual triangulation). If we remove one vertex (shown in the picture), then n(I) = 2 (as there are two open rays or two infinite connected components).

From $\mathbb{R}^n, n \ge 2$ onward, if you remove a finite subcomplex K, that can be enclosed within a (large enough) closed ball. Hence there can only one infinite connected component of $\mathbb{R}^n - K , n \ge 2$

This number n(K) can be different for different K.  Take for example the infinite binary tree.

Removing the top vertex, the number of infinite connected components are 2. But removing up to level 2, we will have 4 infinite connected components. And as we go down the levels, n(K) will grow.

So, we try with all possible such finite subcomplexes (possibly infinitely many of them) and check each time how many infinite connected components of X – K is there.

This is precisely what leads us to ends of X.

Ends of (X) =  E(X) = sup n(K)

Example: $E(\mathbb{R}) = 2, E(\mathbb{R^2}) = 1, E(T) = \infty, \textrm{T = infinite binary tree}$

Star of K: If K is a (finite) subcomplex of a simplicial complex X, then st(K) or star of K is defined to be the interior of all simplices which meet the vertices in K.

Since K is locally finite st(K) is an open finite subset of K.

If two point $v_1, v_2$ in X – st(K) can be connected by a path in X – st(K) then they can be connected via an edge path in X – st(K). Hence they are in a path component of X – K. Hence it is sufficient to look into the 1-skeleton of X – K to find out n(K).

# Cohomology

Recall that given a simplicial complex, one can create Chain complexes and Co-Chain Complexes.

A chain complex, $C_0 (X)$ of vertices is simply the set of formal sums of vertices in X. In other words, if $v_1, v_2, v_3$ are vertices in X then $v_1 + v_2 - v_3$ is an element of $C_0 (X)$. In fact it has all such finite formal sums.

Once the chain complex is defined, one may quickly define co-chain complexes. They are formal sums of homomorphisms from $C_0(X) \to G$ where G is a suitable group. Usually $G = \mathbb{Z}_2, \textrm {or} G = \mathbb{Z}$. It depends on what we are trying to achieve.

One may think of each member of 0th cochain complex as an assignment of values to each vertex in X.

In the above picture, we have assigned 0’s and 1’s to each vertex of the finite simplicial complex. This is one member of $C^0(X)$. Each such assignment is nothing but a function from $C_0(X) \to \mathbb{Z}_2$.

We collect all such functions and their formal sum. They constitute $C^0(X)$.

In the present context, we may think of $\mathbb{Z}_2$ as {off, on}. Whenever we pick a finite subcomplex K in the simplicial complex X, we are indirectly assigning 1 to each vertex in K and 0 to all the vertices in X – K. That assignment is indeed a function from the set of vertices to {0, 1} hence a member of the cochain complex.

Formally we say that the function has finite support; that is it is non-zero at only finitely many points and zero everywhere else.

Thus the member of $C^0(X)$ which have finite support, are our algebraic tools for ‘picking’ finite subcomplexes K from simplicial complex X. In fact, each such cochain corresponds to one such ‘pick’.

Let $C_f^0 (X)$ denote the subset of $C^0 (X)$ with finite support.

Notice that similarly we can define $C_f^1 (X)$. They are maps from (formal sums of) edges of X to $\mathbb{Z}_2$ which have 1 assigned to finitely many edges (and 0 elsewhere).

$C_f^0 (X)$ maps inside $C_f^1 (X)$ under the coboundary map.

Why? Suppose $\phi \in C_f^0 (X)$ . Then $\delta \phi (v_1, v_2) = \phi (v_2) - \phi (v_1)$ (Recall that the coboundary map assigns ‘difference’ of values at the vertices. This can be intuited as a change in height function: value at each edge is the difference of height at its vertices).

Clearly $\phi (v_2) - \phi (v_1)$ can be non zero if and only if one of them is 0 and the other one is 1. But as $\phi \in C_f^0(X)$ has finite support, it gives nonzero output only at finitely many vertices hence $\phi (v_2) - \phi (v_1)$ can be non zero only at finitely many edges.

# ‘Ends’ is the dimension of Cohomology

Recall that cohomology groups of a space X are the homology groups of Cochain Complex.

In more simpler terms, consider the following Cochain complex:

$... \overset{\delta_2} \Leftarrow C^1(X) \overset{\delta_1} \Leftarrow C^0(X) \overset{\delta_0} \Leftarrow 0$

Then $H^0 (X) = \frac{ker (\delta_1)}{im (\delta_0)} = ker (\delta_1)$

Intuition: Image of the lower level difference map contains lower level ‘difference’ data. By quotienting that out, we may focus only on higher level differences.

We previously constructed $C_f^0 (X)$ that singled out the finite subcomplexes. (K’s). Next, we wish to turn our attention toward X – K. Hence it makes sense to ‘quotient out’ $C_f^0 (X)$ from $C^0 (X)$. Then we will be left out with those functions which assign 1 to infinitely many vertices.

Define $C_e^0 (X) = \frac{C^0 (X)}{C_f^0 (X)}$

Similarly define $C_e^1 (X) = \frac{C^1 (X)}{C_f^1 (X)}$

Intuition: Imagine as ends. Functions in $C_e^0 (X)$ flow 1’s upto ends (as 1’s appear infinitely many times at the vertices).

Claim: $dim_{\mathbb{Z}_2} H_e^0 (X) = E(X)$

It is not hard to see why this could be true. Afterall what is $latex H_e^0 (X)$ ? It is the collection of all those $\phi \in C_e^0 (X)$ that maps to 0 of $latex C_e^1 (X)$.

More explicitly, these are 0-1assignments on vertices, which have 1 at infinitely many vertices, but non-zero differences at finitely many edges.

Example:

Again consider the real line. Assign 1 at all vertices leftward from U. Assign 0 W onward (to the right).

Note that this is a member of (an equivalence class of) $C^0_e (X)$ as it has 1 at infinitely many vertices. Call it $\phi$ .

But after applying the coboundary map $\delta$ to $\phi$, we get 0 at all edges except on UV.  Clearly $\delta \phi$ has finite support, hence is a member of $C^1_f (X)$. Therefore it is in the equivalence class of 0 in $C^1_e (X)$ (as we have quotiented out all edge-functions with finite support).

Note that U and V are the crossover points in the above example. When we talk about the $ker \delta_1$, we are seeking vertex-functions in $C^0_e (X)$ we have finitely many crossover points (but infinitely many points with 1 assigned).

Why? Because if we cut at the crossover, we have an infinite component at least at one side of it (after all the infinitely many 1’s need to accommodated somewhere).

The recipe for finding the finite crossover but infinitely flowing vertex functions:

• Look at all vertex-functions that map to edge-functions with finite support ($\delta^{-1} {C^1_f(X) }$)
• Quotient out those which are 1 (on) at finitely many vertices (they do not flow to the ends)
• Final result $\frac{\delta^{-1} {C^1_f(X) }}{C^0_f(X)}$

The rest is easy!

Each path component of X – K must bear the same value at each vertex. Hence we can create linearly independent vertex functions corresponding to each component of X – K.

But we have taken care of all possible K as we considered all cochains.

Hence the number of linearly independent finite crossover but infinitely flowing vertex maps provide the Ends of X.

Good News: This data is cohomological. Hence it is independent of the triangulation of the space. (It is a deep theorem of algebraic topology that cohomology groups are independent of triangulation; infact can be achieved without triangulation).

# Which manifold is this?

This is the first exercise from Thurston’s Three Dimensional Geometry and Topology Vol. 1.

Which manifold is this?

It is like an old trick. Try following the lines. There are actually 6 loops (circles) in this maze.

Here is a color coded picture of it.

# Direction of a vector field

Let ${ f: S^n \rightarrow S^n }$ be a map of degree zero. Show that there exists points ${ x, y \in S^n }$ with ${ f(x) = x }$ and ${ f(y) = - y}$. Use this to show that if F is a continuous vector field defined on the unit ball ${ D^n }$ in ${ R^n }$ such that ${ F(x) \neq 0 }$ for all x, then there exits a point in ${ \partial D^n }$ where F points radially outward and another point in ${\partial D^n }$ where F points radially inward.

Solution:

Suppose g is the antipodal map.

If ${ f(x) \neq x }$ for all ${x \in S^n }$ then ${ f \sim g }$ . This implies ${deg (f) = deg(g) = (-1)^{n+1} \neq 0 }$ hence giving us a contradiction.

Similarly if ${ f(y) \neq - y \forall y \in S^n }$ then ${ g \circ f (y) \neq y }$ for all ${ y \in S^n }$ . This implies ${ g \circ f }$ is homotopic to the antipodal map. But then ${ deg (g \circ f) = (-1)^{n+1} }$ . This is impossible as ${ deg (g \circ f ) = deg (g) \cdot deg (f) = 0 }$ . Hence we again have a contradiction.

Now define ${ \bar{F} (x) = \frac{F(x)}{||F(x)||} , x\in S^{n-1} }$ . It is clearly a continuous vector field from ${ S^{n-1} \rightarrow S^{n-1} }$ . (It is well defined as F(x) never vanishes. Also restriction of F to ${S^{n-1} }$ is continuous as F is continuous. Finally as F is continuous, composing it with absolute value function is continuous. ${||F|| }$ is continuous and F is continuous implies their ratio is continuous).

Note that ${\bar{F} = S^{n-1} \hookrightarrow D^n \rightarrow S^{n-1} }$ where the first map is inclusion ${i }$ and the second map is ${ \frac{F}{||F||}}$ . Since ${ H_n (D^n ) = 0 }$ (as n-disk is contractible), therefore ${ \left ( i \circ \frac{F}{||F||} \right)_* = 0}$ implying ${ \bar{F}_* }$ is of zero degree.

As ${ \bar{F} }$ is of degree zero therefore there is a point ${ x \in S^{n-1} }$ such that ${ \bar{F}(x) = x }$. Hence ${ \frac{F(x)}{||F(x)||} = x \Rightarrow F(x) = c \cdot x }$. This F points radially outward at this point (c is positive constant).

Similarly there is a point ${ y \in S^{n-1} }$ such that ${ \bar{F}(y) = -y }$. This is the point where F points radially inward.

# Round robin tournament

Problem : Suppose there are ${k}$ teams playing a round robin tournament; that is, each team plays against all the other teams and no game ends in a draw.Suppose the ${i^{th}}$ team loses ${l_{i}}$ games and wins ${w_{i}}$ games.Show that

${{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}$ = ${{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}$

Solution : Each team plays exactly one match against each other team.

Consider the expression $\displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) }$

Since each team plays exactly k-1 matches and no match ends in a draw, hence number of wins plus numbers of loses of a particular team is k-1 (that is the number of matches it has played). In other words $l_i + w_i = k-1$ for all i (from 1 to k).

Hence

$\displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} }$
$\displaystyle{= \sum_{i=1}^{k}(l_i + w_i)(l_i - w_i) }$
$\displaystyle{= \sum_{i=1}^{k}(k-1)(l_i - w_i) }$
$\displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) }$

But $\displaystyle{ \sum_{i=1}^{k} l_i = \sum_{i=1}^{k} w_i }$ (as total number of loses = total number of matches = total number of wins; as each match results in a win or lose of some one)

Hence $\displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i - \sum_{i=1}^{k} w_i\right) = (k-1) \times 0 = 0 }$

Therefore $\displaystyle{\sum_{i=1}^{k} l_i^{2} - {w_i^2} = 0 }$ implying ${{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}$ = ${{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}$

Proved.

# Number Theory 1 Teaching Schedule

This document is useful for current students. It contains teaching schedule for Number Theory 1.

## Overview:

Number Theory 1 is an introductory module. It is useful for beginner math olympiad aspirants (preparing for AMC, AIME, ARML, Duke Math Meet etc.)

• Number systems
• Prime numbers
• Arithmetic and geometric sequences
• Mathematical Induction
• Divisibility techniques
• Arithmetic of remainders
• Modular Arithmetic and Gauss’s theory
• Equivalence Relations
• Mathematical games

## Specifications

• Each session (day) is 2 hours long.
• It is followed by a homework assignment.
• Apart from regular theoretical work and problem-solving, each section consists of mathematical games
• Books:
• Challenges and Thrills of Pre-College Mathematics
• Mathematics can be Fun by Yakov Perelman
• Excursion Into Mathematics
• Mathematical Circles, Russian Experience by Fomin

## Sessions

### Session 1

• Formula for nth odd number and nth even number
• Sum of first n odd numbers and their visual treatment
• Arithmetic Progression

### Session 2 and 3

• Arithmetic Progression’s description (nth number)
• Gauss’s method for summing arithmetic progression (rewriting a finite sum in reverse order).
• Sum of n terms of an arithmetic sequence
• Geometric sequence
• Sum of nth term of a geometric sequence

### Session 4 and 5

• Mathematical induction
• Strong form of induction

### Session 6 and 7

• Divisibility and prime numbers
• Fundamental Theorem of Arithmetic

### Session 8

• Types of numbers
• Well ordering principles
• Irrationality of square root of 2 (and primes)

### Session 9 and 10

• Modular Arithmetic – similarity and differences with equality
• Notion of equivalence relation