Mathtuition88

Homology Group of some Common Spaces

Homology of Circle
$\displaystyle H_n(S^1)=\begin{cases}\mathbb{Z}&\text{for}\ n=0,1\\ 0&\text{for}\ n\geq 2. \end{cases}$

Homology of Torus
$\displaystyle H_n(T)=\begin{cases}\mathbb{Z}\oplus\mathbb{Z}&\text{for}\ n=1\\ \mathbb{Z}&\text{for}\ n=0, 2\\ 0&\text{for}\ n\geq 3. \end{cases}$

Homology of Real Projective Plane
$\displaystyle H_n(\mathbb{R}P^2)=\begin{cases} \mathbb{Z}&\text{for}\ n=0\\ \mathbb{Z}_2&\text{for}\ n=1\\ 0&\text{for}\ n\geq 2. \end{cases}$

Homology of Klein Bottle
$\displaystyle H_n(K)=\begin{cases} \mathbb{Z}&\text{for}\ n=0\\ \mathbb{Z}_2\oplus\mathbb{Z}&\text{for}\ n=1\\ 0&\text{for}\ n\geq 2. \end{cases}$

Also see How to calculate Homology Groups (Klein Bottle).

北大高等代数 Beijing University Advanced Algebra

Math Online Tom Circle

辛弃疾的《青玉案·元夕》：“…众里寻他千百度；蓦然回首，那人却在灯火阑珊处。” –表达出了我的一种 (网上)意外相逢的喜悦，又表现出对心中(名师)的追求。

2011 年北京大学教授丘维声教授被邀给清华大学物理系(大学一年级) 讲一学期课 : (Advanced Algebra) 高等代数, aka 抽象代数 (Abstract Algebra)。

丘维声（1945年2月－）生于福建省龙岩市[1]，中国数学家、教育家。16岁时以全国高考状元的成绩考入北京大学，1978年3月至今担任北京大学数学科学学院教授，多年坚持讲授数学专业基础课程[2]。截至2013年，共著有包括《高等代数（上册、下册）》、《简明线性代数》两本国家级规划教材在内的40部著述[3]。于1993-97年的一系列文章中逐步解决了n=3pr情形的乘子猜想，并取得了一系列进展[2]。

———————

72岁的丘教授学问渊博, 善于启发, 尤其有别于欧美的”因抽象而抽象”教法, 他独特地提倡用”直觉” (Intuition) – 几何概念, 日常生活例子 (数学本来就是源于生活)- 来吸收高深数学的概念 (见:数学思维法), 谆谆教导, 像古代无私倾囊相授的名师。

全部 151 (小时) 讲课。如果没时间, 建议看第1&第2课 Overview 。

http://www.bilibili.com/mobile/video/av7336544.html?from=groupmessage

第一课: 导言 : n 维方程组 – 矩阵 (Matrix)-n 维向量空间 (Vector Space) – 线性空间 (Linear Space)

第二课:

上表 (左右对称):

双线性函数 (Bi-linear functions) / 线性映射 (Linear Map)

线性空间 + 度量 norm =>

Euclidean Space (R) => (正交 orthogonal , 对称 symmetric)变换
酉空间 Unitary Space (C)… =>酉变换, Hermite变换

近代代数 (Modern Math since 19CE Galois): 从研究结构 (环域群) 开始: Polynomial Ring, Algebraic structures (Ring, Field, Group).

第三课: 简化行阶梯形矩阵 Reduced Row Echelon Matrix

第四课: 例子 (无解)

第五课: 证明无解/唯一解/无穷解
[几何直觉]: 任何2线 1) 向交(唯一解) ; 2) 平行 (无解) ; 3) 重叠 (无穷解)

n次方程組的解也只有3个情况:

无解:

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中国数学考研 Graduate Math Exams

Math Online Tom Circle

中国”考研”究生:

考题难, 重视理论基础, 不是技巧。计算量大, 时间(3小时)不够。

国家 “及格” 底线 : 58~ 90分 (总分 : 150 分) – 根据理工 / 经管系 , 不同重点大学, 底线各异。

http://www.bilibili.com/mobile/video/av2261356.html

[例子] $latex p (x) = a + bx+cx^{2}+dx^{3}$

$latex p(x) – tan x sim x^{3}, text { when } x to 0$

Find a, b, c, d ?

[Solution] :

1. Don’t use l’Hôpital Rule for $latex displaystyle lim frac {f}{g}$

2. Apply Taylor expansion :

$latex tan x = x + frac {1}{3}x^{3} + o (x^{3})$
$latex p (x) – tan x = a + (b -1)x + (c – frac {1}{3})x^{3} + o (x^{3})$

$latex p(x) – tan x sim x^{3}, text { when } x to 0 $

$latex iff boxed {a=0, b=1, c=frac {4}{3}}&fg=aa0000&s=2$

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Barry Mazur – Harvard Lecture on Primes and the Riemann Hypothesis for High School Students

Math Online Tom Circle

Prelude:

Harvard Lecture:

The Key toopen this secret …

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Summary of Persistent Homology

We summarize the work so far and relate it to previous results. Our input is a filtered complex $K$ and we wish to find its $k$ th homology $H_k$ . In each dimension the homology of complex $K^i$ becomes a vector space over a field, described fully by its rank $\beta_k^i$ . (Over a field $F$ , $H_k$ is a $F$ -module which is a vector space.)

We need to choose compatible bases across the filtration (compatible bases for $H_k^i$ and $H_k^{i+p}$ ) in order to compute persistent homology for the entire filtration. Hence, we form the persistence module $\mathscr{M}$ corresponding to $K$ , which is a direct sum of these vector spaces ( $\alpha(\mathscr{M})=\bigoplus M^i$ ). By the structure theorem, a basis exists for this module that provides compatible bases for all the vector spaces.

Specifically, each $\mathcal{P}$ -interval $(i,j)$ describes a basis element for the homology vector spaces starting at time $i$ until time $j-1$ . This element is a $k$ -cycle $e$ that is completed at time $i$ , forming a new homology class. It also remains non-bounding until time $j$ , at which time it joins the boundary group $B_k^j$ .

A natural question is to ask when $e+B_k^l$ is a basis element for the persistent groups $H_k^{l,p}$ . Recall the equation $\displaystyle H_k^{i,p}=Z_k^i/(B_k^{i+p}\cap Z_k^i).$ Since $e\notin B_k^l$ for all $l<j$ , hence $e\notin B_k^{l+p}$ for $l+p<j$ . The three inequalities $\displaystyle l+p<j,\ l\geq i,\ p\geq 0$ define a triangular region in the index-persistence plane, as shown in Figure below.

The triangular region gives us the values for which the $k$ -cycle $e$ is a basis element for $H_k^{l,p}$ . This is known as the $k$ -triangle Lemma:

Let $\mathcal{T}$ be the set of triangles defined by $\mathcal{P}$ -intervals for the $k$ -dimensional persistence module. The rank $\beta_k^{l,p}$ of $H_k^{l,p}$ is the number of triangles in $\mathcal{T}$ containing the point $(l,p)$ .

Hence, computing persistent homology over a field is equivalent to finding the corresponding set of $\mathcal{P}$ -intervals.

Source: “Computing Persistent Homology” by Zomorodian and Carlsson

Part 4 群的线性表示的结构

Math Online Tom Circle

不变子空间: Invariant Sub-space

第一课 : Direct Sum 直和 $latex oplus$of Representations

直和 = $latex {oplus}&fg=aa0000&s=3$

第二课: 群表示可约 Reducible Representation

Analogy :
Prime number decomposition
Irreducible Polynomial

外直和 : $latex { dot{ +} }&fg=aa0000&s=3$

$latex boxed { displaystyle phi_{1} dot {+} phi_{2} = tilde {phi_{1}} oplus tilde {phi_{2}}}&fg=aa0000&s=3$

* 第三课: 完全可约表示 Completely Reducible Representation

完全表示是可完全分解为 不可约表示的一种表示。

完全可约表示 => 其子表示也完全可约。
不可约 一定是完全可约的!
一次表示一定是不可约的!
[Analogy: Polynomial degree 1 (x + 1) is irreducible. ]

註: (*) 深奥课, 可以越过直接跳到结果。(证明待以后复习)。

集合证明: 交(和)⊇和(交)
如果也是⊆ , 则交(和) =和(交)
Ref 2 《高代》 Pg 250 命题 1

$latex boxed {U cap (U_{1} oplus W) supseteq (U cap U_{1} ) oplus (U cap W)}&fg=aa0000&s=3$
Also,
$latex U cap (U_{1} oplus W) subseteq (U cap U_{1} ) oplus (U cap W)$
Then,
$latex boxed {U…

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Part 3 (b) 群的线性表示和例

Math Online Tom Circle

第七课 Group Action群作用

$latex x_{i} in Omega = big{0.x_{1},0.x_{2},…, 0.x_{i-1},1.x_{i}, 0.x_{i+1}, ….0.x_{n} big}$

…

第11课:Cyclic Group (循环群) Representation , Dihedron 二面体

$latex begin{pmatrix}
0 & 0 & 1
1 & 0 &…
0 & 1 &…
end{pmatrix} = P (a) $3 阶 Cyclic Group (循环群) Representation

$latex boxed{ Bigr|D_{n} Bigr| = 2n }&fg=aa0000&s=3$

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苏联老师Arnold 如何教中小学抽象”群”

Math Online Tom Circle

Download eBook (PDF) here:

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Part 3 群的线性表示和例

Math Online Tom Circle

[Part 1 引言 : 温习]
[Part 2 群的基础概念 : 温习]

北大: 丘维声

Part 1 & 2 : 本科班 (Undergraduate) 数学温习

Part 3 开始: 研究班 (Graduate) 数学

第一课 群表示 Group Representation

Φ: Group homomorphism 群同态
V: Linear Space 线性空间 (K 域上 Over Field K) => 表示空间

有限 V => deg (Φ) : 次数 / 维数

无限 V => 无限维

$latex boxed {text {Group Representation : }(phi, V)}&fg=aa0000&s=3$

群表示: 通过研究 1)Φ 同态 2) 像 = 线性空间3)Φ核 = Normal Subgroup => 了解群

KerΦ = {e} =>Φinjective =>ΦFaithful 忠实表示

KerΦ = G =>Φ平凡表示 (全部G 都映射到零, 平凡)

若平方表示Φ 是一次的 ( 即V 是 1 维) => 主表示 (或单位表示)

$latex boxed {GL(V) cong GL_{n} (K)}&fg=aa0000&s=2$ 可逆矩阵

$latex boxed { Phi : G to GL_{n} (K)}&fg=aa0000&s=2 $ G…

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Cast Iron Pan Singapore Review

Recently bought a cast iron pan/skillet for home cooking. Cast iron is an ancient technology that has several benefits over the more modern non-stick technology. It is supposed to be cheap (just US$15 in America Lodge L8SK3 Cast Iron Skillet, Pre-Seasoned, 10.25-inch), but in Singapore it is quite expensive probably due to import fees.

I bought the mid-range USA brand Lodge 10.25-inch skillet (around $60 SGD). It can be found in Qoo10:

Where to buy Cast Iron Pan/Pot/Skillet Singapore

Lodge Pre-Seasoned 8-inch Cast Iron Skillet: http://www.qoo10.sg/su/412118339/Q100000595

Lodge Pre-Seasoned Cast Iron Skillet 10.25-inch: http://www.qoo10.sg/su/412118386/Q100000595

Lodge Cast Iron Skillet (Midrange, affordable)

The high-end brands include Le Creuset, Staub. These are very expensive (at least $100 SGD).

Le Creuset Skillet (High-end, Super Expensive)

Benefits of Cast Iron Cookware vs Non-stick:

Non-stick Teflon, even with extreme care, tends to flake off and end up in food. Also it is released as fumes during cooking. It has dubious, unknown effects on humans, but is scientifically proven to be toxic to birds and rats.
Adds iron supplementation to cooked food. Iron is essential for human health to make hemoglobin in blood.
Technically lasts forever, as it is very durable. Save cost in the long run, as you don’t have to keep replacing the pan.

Also, other benefits include:

Works with induction cookers. (Iron is magnetic.)
It is moderately non-stick, almost as non-stick as Teflon. If anything sticks, just boil with water. Also, the more you use and season it, the more non-stick it becomes.

Downsides include: Heavy weight, needs seasoning (wipe dry and coat with oil) after cooking otherwise it can rust.

The third popular alternative, Aluminum pans, are definitely not good as it may be linked to Alzheimer’s and dementia.

Check out my wife’s food blog on making Cheese Zucchini Patties using the new cast iron skillet.
Also: Garlic Butter Lobster using Cast Iron Skillet

Part 2: 群表示论的基本概念和Abel群的表示

Math Online Tom Circle

[引言 : Part 1 温习]

第一课:映射(f) 集合A,B

$latex f: A to B$
$latex f: a mapsto b , a in A, b in B$
$latex f(A) = { f(a) | a in A } subseteq B$ (f的值域， Im f)

A : 象域 domain:
B : 陪域 co-domain: 唯一
满射 Surjective，单射 Injective ，双射 Bijective

第二课: 线性空间，线性变化，同态

Projection 投影 $latex P_{U} implies $ 线性变化

$latex V = U oplus W$ , W non-unique

$latex V = U oplus U^{perp}$

北大丘维声的 “群论” List of All Videos:http://www.youtube.com/playlist?list=PLwzFfIxhEkcxvU7-c8rPBbPLHUeacPIpa

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Pure Mathematicians versus Applied Mathematicians

Math Online Tom Circle

“A pure mathematician, when stuck on the problem under study, often decides to narrow the problem further and so avoid the obstruction. An applied mathematician interprets being stuck as an indication that it is time to learn more mathematics and find better tools”

— Distinguished differential geometer EugenioCalabi

Ref:

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Inspirational Scientist: Dan Shechtman

Source: https://www.theguardian.com/science/2013/jan/06/dan-shechtman-nobel-prize-chemistry-interview

To stand your ground in the face of relentless criticism from a double Nobel prize-winning scientist takes a lot of guts. For engineer and materials scientist Dan Shechtman, however, years of self-belief in the face of the eminent Linus Pauling‘s criticisms led him to the ultimate accolade: his own Nobel prize.

The atoms in a solid material are arranged in an orderly fashion and that order is usually periodic and will have a particular rotational symmetry. A square arrangement, for example, has four-fold rotational symmetry – turn the atoms through 90 degrees and it will look the same. Do this four times and you get back to its start point. Three-fold symmetry means an arrangement can be turned through 120 degrees and it will look the same. There is also one-fold symmetry (turn through 360 degrees), two-fold (turn through 180 degrees) and six-fold symmetry (turn through 60 degrees). Five-fold symmetry is not allowed in periodic crystals and nothing beyond six, purely for geometric reasons.

Shechtman’s results were so out of the ordinary that, even after he had checked his findings several times, it took two years for his work to get published in a peer-reviewed journal. Once it appeared, he says, “all hell broke loose”.

Many scientists thought that Shechtman had not been careful enough in his experiments and that he had simply made a mistake. “The bad reaction was the head of my laboratory, who came to my office one day and, smiling sheepishly, put a book on x-ray diffraction on my desk and said, ‘Danny, please read this book and you will understand that what you are saying cannot be.’ And I told him, you know, I don’t need to read this book, I teach at the Technion, and I know this book, and I’m telling you my material is not in the book.

“He came back a couple of days later and said to me, ‘Danny, you are a disgrace to my group. I cannot be with you in the same group.’ So I left the group and found another group that adopted a scientific orphan.”

He says that the experience was not as traumatic as it sounded. Scientists around the world had quickly replicated Shechtman’s discovery and, in 1992, the International Union of Crystallography accepted that quasi-periodic materials must exist and altered its definition of what a crystal is from “a substance in which the constituent atoms, molecules or ions are packed in a regularly ordered, repeating three-dimensional pattern” to the broader “any solid having an essentially discrete diffraction diagram”.

That should have been the end of the story were it not for Linus Pauling, a two-time Nobel laureate, once for chemistry and a second time for peace. Shechtman explains that at a science conference in front of an audience of hundreds Pauling claimed, “Danny Shechtman is talking nonsense, there are no quasi-crystals, just quasi-scientists.”

Pauling told everyone who would listen that Shechtman had made a mistake. He proposed his own explanations for the observed five-fold symmetry and stuck to his guns, despite repeated rebuttals. “Everything he did was wrong and wrong and wrong and wrong; eventually, he couldn’t publish his papers and they were rejected before they were published,” says Shechtman. “But he was very insistent, was very sure of himself when he spoke; he was a flamboyant speaker.”

数学是什么 ? What is Mathematics?

Math Online Tom Circle

北京大学：丘维声教授

第1讲数学的思维方式

3000 年前希腊，巴比伦，中国，印度， 10世纪阿拉伯， 16世纪欧洲文艺复兴数学 – 经典数学

1830 年数学的革命 – 近代数学: 法国天才少年伽瓦罗 (Evariste Galois 1811 – 1832)

观察 (Observe): 客观现象
$latex downarrow$
抽象 (Abstraction) : 概念, 建立模型 (Model)
$latex downarrow$
探索 (Explore): 自觉 (Intuition), 解剖 , 类比(Analogy), 归纳 (Induction), 联想, 推理 (Deduction) 等…
$latex downarrow$
猜测 (Conjecture) : eg. Riemann Conjecture (unsolved)
$latex downarrow$
论证 (Prove): 只能用公理 (Axioms)(以知的共识), 定义 (概念), 已经证明的定理 (Theorems), 进行逻辑推理并计算.
$latex downarrow$
揭示 (Reveal): 事物的内在规律 (井然有序)

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2016 Nobel-Prize Winning Physics Explained Through Pastry

Math Online Tom Circle

2016 Nobel Prize Physics is Mathematics (Topology) applied in SuperConductor and SuperFluid to explain the Phase Transitions and Phase matters.

Phase matters: Solid, Liquid, Gas

Phase Transition: Solid -> Liquid -> Gas

Superconductor below Tc (critical temperature) : zero resistance.

Superfluid below Tc : zero viscosity.

Reason explained by Mathematics : Topological invariance increased step-wise.

Eg. Disk (0 hole), Circle (1 hole), Donut (2 holes), Coffee Cup (2 holes)… XYZ (n holes). [n increased by steps from 0, 1, 2, 3… ]

We say donut and coffee cup are homeomorphic (同胚) because they have the same topological invariant 拓扑不变量(2 holes).

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Structure Theorem for finitely generated (graded) modules over a PID

If $R$ is a PID, then every finitely generated module $M$ over $R$ is isomorphic to a direct sum of cyclic $R$ -modules. That is, there is a unique decreasing sequence of proper ideals $(d_1)\supseteq(d_2)\supseteq\dots\supseteq(d_m)$ such that $\displaystyle M\cong R^\beta\oplus\left(\bigoplus_{i=1}^m R/(d_i)\right)$ where $d_i\in R$ , and $\beta\in\mathbb{Z}$ .

Similarly, every graded module $M$ over a graded PID $R$ decomposes uniquely into the form $\displaystyle M\cong\left(\bigoplus_{i=1}^n\Sigma^{\alpha_i}R\right)\oplus\left(\bigoplus_{j=1}^m\Sigma^{\gamma_j}R/(d_j)\right)$ where $d_j\in R$ are homogenous elements such that $(d_1)\supseteq(d_2)\supseteq\dots\supseteq(d_m)$ , $\alpha_i, \gamma_j\in\mathbb{Z}$ , and $\Sigma^\alpha$ denotes an $\alpha$ -shift upward in grading.

Secondary Level Chinese Tuition

Looking for O Level / IP / JC Chinese Tuition?

Ms Gao specializes in teaching secondary level chinese (CL/HCL) tuition in Singapore. Ms Gao has taught students from various schools, including RI (Raffles Institution IP Programme).

Teaches West / Central Area: E.g. Clementi, Jurong East, Bukit Timah, Dover, Bishan, Marymount

Email: chinesetuition88@gmail.com
Website: http://chinesetuition88.com

Persistence Interval

Next, we want to parametrize the isomorphism classes of the $F[t]$ -modules by suitable objects.

A $\mathcal{P}$ -interval is an ordered pair $(i,j)$ with $0\leq i<j\in\mathbb{Z}^\infty=\mathbb{Z}\cup\{+\infty\}$ .

We may associate a graded $F[t]$ -module to a set $\mathcal{S}$ of $\mathcal{P}$ -intervals via a bijection $Q$ . We define $\displaystyle Q(i,j)=\Sigma^i F[t]/(t^{j-i})$ for a $\mathcal{P}$ -interval $(i,j)$ . When $j=+\infty$ , we have $Q(i,+\infty)=\Sigma^iF[t]$ .

For a set of $\mathcal{P}$ -intervals $\mathcal{S}=\{(i_1,j_1),(i_2,j_2),\dots,(i_n,j_n)\}$ , we define $\displaystyle Q(\mathcal{S})=\bigoplus_{k=1}^n Q(i_k, j_k).$

We may now restate the correspondence as follows.

The correspondence $\mathcal{S}\to Q(\mathcal{S})$ defines a bijection between the finite sets of $\mathcal{P}$ -intervals and the finitely generated graded modules over the graded ring $F[t]$ .

Hence, the isomorphism classes of persistence modules of finite type over $F$ are in bijective correspondence with the finite sets of $\mathcal{P}$ -intervals.

The Map of Mathematics (YouTube)

A nicely done video on how the various branches of mathematics fit together. It is amazing that he has managed to list all the major branches on one page.

Also see: Beautiful Map of Mathematics.

Homogenous / Graded Ideal

Let $A=\bigoplus_{i=0}^\infty A_i$ be a graded ring. An ideal $I\subset A$ is homogenous (also called graded) if for every element $x\in I$ , its homogenous components also belong to $I$ .

An ideal in a graded ring is homogenous if and only if it is a graded submodule. The intersections of a homogenous ideal $I$ with the $A_i$ are called the homogenous parts of $I$ . A homogenous ideal $I$ is the direct sum of its homogenous parts, that is, $\displaystyle I=\bigoplus_{i=0}^\infty (I\cap A_i).$

Donate to help Stray Dogs in Singapore

URL: https://give.asia/movement/run_for_exclusively_mongrels

3 Singaporeans – Dr Gan, A Dentist, Dr Herman, A Doctor, and Mr Ariffin, a Law Undergraduate will be taking on the Borneo Ultra Trail Marathon on Feb 18th 2017 to raise 30k for Exclusively Mongrels Ltd; a welfare group set up for Mongrels in Singapore. (https://www.facebook.com/exclusivelymongrels/)

Do support them in their cause, if you can. And share this story so as to spread the word (maintenance and upkeep of the dogs can be a huge cost). Mongrels are actually highly intelligent, and can be more healthy and robust as compared to pedigrees, which may have hereditary diseases. For example, the popular Golden Retriever breed is prone to hip dysplasia.

A story told by Dr Gan summarizes everything — The state and welfare of stray dogs in Singapore, supposedly a first-world country, is actually worse than jungle dogs in Borneo. The Orang Asli, primitive junglers in Sabah, apparently treat dogs better than the average layperson in Singapore:

When Dr Gan, an EM member, was running through the trails of Sabah in Oct 2016, he stumbled upon a stray dog.

Being an avid dog lover and the proud father to three rescued Mongrels, he had to stop in his tracks. He fed the dog and it even ran alongside him for a mile or two. Further along the route, he encountered more stray dogs too.

All of the stray dogs he encountered seemed well-fed and were very approachable. They all displayed no aggression, despite being in the middle of a jungle. To Dr Gan, this was a tell-tale sign that the Orang Asli, who lived in villages in these jungles, took care of the dogs by feeding them. The fact that these Orang Aslis were living in harmony with these strays was indeed very commendable in his eyes.

These thoughts stuck with him throughout the run, and on the journey home too.

He couldn’t help but compare the Orang Asli’s hospitality to how a Singaporean layperson would react upon encountering a stray dog. More often than not, even in the absence of aggressive behaviour, a Singaporean who sees a stray dog would view it as no more than a pest and would either chase it away or even, call the authorities. As it so often is when the latter option is exercised, the authorities would have a hard time rehoming the dog and EM has to step in to ‘bail’ the dog out before the authorities euthanize it.

It is strange, he remarked, how the Orang Asli from the jungle can treat these strays with reverence while many Singaporeans would report a stray to the authorities without the slightest hesitation.

“Would the situation end up the same way if, instead of a stray mongrel, there was a stray pedigree dog?”

Armed with the notion that more needs to be done not just for these dogs but also to empower and educate the general public in Singapore about the plight of these strays and what can be done to help them, he then called on his two running buddies to undertake this journey with him.

It was going to be a journey that united his two passions – running and dogs; a journey back to the jungles where he first encountered the strays; back to where he first witnessed the hospitality of the Orang Asli; back to where where the spark was first ignited. He, and his Team, hope to bash through the jungles of Borneo, all in the hopes of blazing a new trail for Mongrels back home, in Singapore.

Water cuts through rock, not because of its strength, but because of its persistence.

群表示论引言 Introduction to Group Representation

Math Online Tom Circle

北京大学数学系丘维声教授

引言: 基本数学强化班 — 深入浅出介绍

群表示论是什么?
有何用 ?

第一课:环Ring
丘教授不愧是大师, 也和一些良师一样, 认同 “数”的(代数)结构先从“环” (Ring)开始教起, 再域, 后群 : 美国/法国/英国都从 “群”(Group)开始, 然后 “环”, “域” (Field) , 是错误的教法, 好比先穿鞋后穿袜, 本末倒置!

精彩的”环” (Ring) 引出 6 条 axioms 公理:

4条 ” + ” 法:

Commutative 交换律, Associative 结合律, Neutral element ” 0″ 零元, Inverse (-) 逆元

2 条 “x ” 法: (exclude ”1″ Unit, WHY ?)

Associative 结合律, Distributive (wrt “+”) 分配律

如果:

环 + 交换 = 交换环 (Commutative Ring)

环 + 单位 ‘1’ =单位环 (Unit Ring)

第二课: 域 Field

星期: 子集的划分 Partitions

$latex mathbb {Z} _7 =
{ bar {0} , bar {1} , bar {2} , bar {3} , bar {4} , bar {5} , bar {6} } $

模m剩余类 : Mod m
$latex mathbb {Z} _ m =
{ bar {0} , bar {1} , bar…

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Smooth/Differentiable Manifold

Smooth Manifold
A smooth manifold is a pair $(M,\mathcal{A})$ , where $M$ is a topological manifold and $\mathcal{A}$ is a smooth structure on $M$ .

Topological Manifold
A topological $n$ -manifold $M$ is a topological space such that:
1) $M$ is Hausdorff: For every distinct pair of points $p,q\in M$ , there are disjoint open subsets $U,V\subset M$ such that $p\in U$ and $q\in V$ .
2) $M$ is second countable: There exists a countable basis for the topology of $M$ .
3) $M$ is locally Euclidean of dimension $n$ : Every point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$ . For each $p\in M$ , there exists:
– an open set $U\subset M$ containing $p$ ;
– an open set $\widetilde{U}\subset\mathbb{R}^n$ ; and
– a homeomorphism $\varphi: U\to\widetilde{U}$ .

Smooth structure
A smooth structure $\mathcal{A}$ on a topological $n$ -manifold $M$ is a maximal smooth atlas.

Smooth Atlas
$\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in J}$ is called a smooth atlas if $M=\bigcup_{\alpha\in J}U_\alpha$ and for any two charts $(U,\varphi)$ , $(V,\psi)$ in $\mathcal{A}$ (such that $U\cap V\neq\emptyset$ ), the transition map $\displaystyle \psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V)$ is a diffeomorphism.

Source:
Introduction to Smooth Manifolds (Graduate Texts in Mathematics, Vol. 218) by John Lee

Differentiable Manifolds (Modern Birkhäuser Classics) by Lawrence Conlon

These two books are highly recommended books for Differentiable Manifolds. John Lee’s book has almost become the standard book. Its style is similar to Hatcher’s Algebraic Topology, it can be wordy but it has detailed description and explanation of the ideas, so it is good for those learning the material for the first time.

Lawrence Conlon’s book is more concise, and has specialized chapters that link to Algebraic Topology.

Persistence module and Graded Module

We show that the persistent homology of a filtered simplicial complex is the standard homology of a particular graded module over a polynomial ring.

First we review some definitions.

A graded ring is a ring $R=\bigoplus_i R_i$ (a direct sum of abelian groups $R_i$ ) such that $R_iR_j\subset R_{i+j}$ for all $i$ , $j$ .

A graded ring $R$ is called non-negatively graded if $R_n=0$ for all $n\leq 0$ . Elements of any factor $R_n$ of the decomposition are called homogenous elements of degree $n$ .

Polynomial ring with standard grading:
We may grade the polynomial ring $R[t]$ non-negatively with the standard grading $R_n=Rt^n$ for all $n\geq 0$ .

Graded module:
A graded module is a left module $M$ over a graded ring $R$ such that $M=\bigoplus_i M_i$ and $R_iM_j\subseteq M_{i+j}$ .

Let $R$ be a commutative ring with unity. Let $\mathscr{M}=\{M^i,\varphi^i\}_{i\geq 0}$ be a persistence module over $R$ .

We now equip $R[t]$ with the standard grading and define a graded module over $R[t]$ by $\displaystyle \alpha(\mathscr{M})=\bigoplus_{i=0}^\infty M^i$ where the $R$ -module structure is the sum of the structures on the individual components. That is, for all $r\in R$ , $\displaystyle r\cdot (m^0,m^1,m^2,\dots)=(rm^0,rm^1,rm^2,\dots).$

The action of $t$ is given by $\displaystyle t\cdot (m^0, m^1, m^2,\dots)=(0,\varphi^0(m^0),\varphi^1(m^1),\varphi^2(m^2),\dots).$
That is, $t$ shifts elements of the module up in the gradation.

Source: “Computing Persistent Homology” by Zomorodian and Carlsson.

GEP Selection Test Review and Experience

The following is a parent’s review and experience of the GEP Selection Test (2016). Original text (in Chinese) at: http://mp.weixin.qq.com/s/xQpLynFWpZ6QNpI_vlw4cw

Interested readers may also want to check out Recommended Books for GEP Selection Test.

Translation:

One day in September 2016 afternoon, read the son of the third son as usual time to go home, after the door looked calmly handed me a letter ~ OMG! A letter from the MOE to inform the son passed the GEP first round Examination, will be held on October 18 to participate in the second round of selection.

The son of the school in Singapore ranked 100 +, the third grade a total of seven classes, a total of about 280 students, he is in the best class. According to him, almost all the classmates participated in the first round of examinations, only through the eight individuals, including him. Later learned that, in fact, the school also 8 individuals to participate in the second round of their selection. Due to the small number of schools will not send people to pick up. Examination place in a subway station, never been to the school. The original quiet life, because to send test and upset, and finally have the opportunity to close feeling the legendary GEP.

A. Parents around the campus export was packed, looking at the eagerly a pair of eyes, I immediately think of China’s college entrance examination. Originally even sent too lazy to send his son to the exam, that is only an examination only, did not expect her husband told me to pick up the road, I began to excitement.

B. Carefully observed the son of the school to take the exam students, are not usually learn top-notch, but not usually take the scholarship. Such as the son of English is poor, but also through the first round.

Further, GEP study focus on learning with the usual very different. Also confirmed the rivers and lakes in the legendary: GEP will try to reduce the impact of language on the selection, so that truly talented children to stand out, and as much as possible without interference. Nevertheless, English is actually bad or affected. I asked the four students, all of the questions are difficult to answer the most difficult IQ, and the son of English that is better than IQ difficult, but there are several IQ questions did not understand, because the word does not know, of. In this case,

C. There are eight children in the class reference, thought that there will be a few other classes, did not think the day before the collection know that their school also their 8 classes. In fact, before the class this year, his son was assigned to other classes of students, there are several aspects of the results are good. Why the last one did not pass the GEP first round?

I think the first is the environment, in improving class, the teacher will be strict a lot of the other classes are not necessarily. Son is after almost a year, only to adapt to such a fast-paced and strict requirements.

Second, the amount of information provided is different. I remember the beginning of the beginning of his son’s class soon, on a large number of additional courses, including Mathematical Olympiad, Science Olympiad, Chinese writing, the second foreign language (Malay), plus a day CCA and school normal plus lesson. . .

Never had a tutorial managed son plus a lesson, home every day at least 4 points, and sometimes 6 points, as well as the violin and Chinese Orchestra, once tired and round and round all day shouting hungry. Home do not want to do anything, followed by his brother to play, to think of homework to do quickly, the next day and get up.

After six months, tired not, but the results plummeted. I have wanted his son not to learn these extra lessons, and his son said that these classes only their classes have, and other classes will not notice the information plus lesson, or learn it!

It now appears that the school had great efforts to catch them this class, the son is still helpful, and sometimes really forced a force, hold on, or there will be harvest. At least the son did not spend extra effort to improve classes, but also an improvement! This also fully shows that folklore, the small three-class is how important and tragic. I also know it!

From the test finished out of the children’s face, you can guess the state of the exam!

D. Elite is the elite schools, such as the son of this little-known school, a school had only a few people in the first round. The elite is the school charter to pick up, as well as teachers to accompany. Because the reference is really many people, a car also sat down, opened a few.

Nanyang Primary School is said to have 120 reference. People usually test and this test is almost, not just like to play like a try test chant. In this case,

E. When the son, met a lot of acquaintances. Parents who have children’s kindergarten students, parents who have attended the parents’ meeting, parents who have written classes, parents who have Chinese orchestra help, parents who have neighbors playmates, friends who have friends with God, and my fellow villagers and husband colleagues Even though the children in different schools, but the emphasis on education, parents, will eventually meet ~ to wait for the child to test this way to meet, quite special.

F. From the parents of the ratio can be inferred: the Chinese to the absolute high rate of reference, a small amount of Indian, a small amount of Malay, did not see Europe and the United States. Chinese like to test, but also good at the test, really reflected most vividly. After my visual, the number of boys more than girls. Take the son school, for example, 8 people have only 1 girl. I guess half of half a far cry. After all, his son son school class first, almost the girls occupied. Impression in the class last year, single scholarship, only the son of a boy.

G. GEP ultimately can be admitted to the rare, most parents are holding try to see the idea of the problem, let the children participate in, do not need all the energy on the GEP, but no need to focus on depletion in the primary three. Son of a classmate did not apply for GEP, heard there are not admitted to the second round, and some even admitted to the elite do not read.

Have seen a documentary article, his children’s classmates, the results are very good score is also high, can enter the first-class university, but eventually chose to read poly, because that read enough, never want to read!

Summary

Although most of the parents of the GEP rush, often the results are unsatisfactory. If the child has the ability to have a high degree of quality into the GEP selected elite, of course, is very good!

But if it is to further test, in order to further fight, one year or even several years earlier to the child overweight, premature energy consumption in reading this matter, the child’s desire to pursue knowledge and innovation, personal opinion, for the long And a variety of life, it is not worth!

I am a student of English in the workplace, said her daughter through the GEP test class children to go, now mixed very general.

Postscript

Participating in GEP is a good experience. No matter what the outcome, are worth a try Oh!

In addition, the son of GEP in the second round of the examination notice, the accident received three years to transfer the success of the phone in the fourth grade to go home only 5 minutes away from the school, and is directly assigned to the best classes to This ended his last three years, 5-15 minutes a day, take a 15-minute bus, but also to go some way to learn the experience.

Attached: GEP introduction of Singapore

GEP History

In 1984, the Ministry of Education of Singapore launched the Gifted Class, which aims to foster gifted students and give full play to their talents so as to better serve the community in the future.

The nine schools that provide talent education are: Anglo-Chinese School (Primary), Catholic High School (Primary), Henry Park Primary School, Nan Hua Primary School (Nan Hua Primary School) ), Nanyang Primary School, Raffles Girls’ Primary School, Rosyth School, St. Hilda’s Primary School and Tao Primary School. Nan School).

GEP screening process

All primary school students have the opportunity to participate in the first round of screening tests, voluntary, not mandatory. The first round of screening tests, including English and mathematics, usually in late August each year (the specific test location and time to the Ministry of Education notice).

In the first round, only 5% of students will be selected to participate in the second round of the selection test (usually the examination time in mid-October each year). Usually only 1% of the students will be selected last year, from the fourth grade, more than 9 schools to enter the genius classes.

Genius classes differ from ordinary students in their curricula.

On the basis of the general curriculum, intensive classes will be arranged for the Gifted students to explore and expand the capacity of gifted students to stimulate their more personalized and profound learning.

(Text: Tao Ying)

Persistence module and Finite type

A persistence module $\mathcal{M}=\{M^i,\varphi^i\}_{i\geq 0}$ is a family of $R$ -modules $M^i$ , together with homomorphisms $\varphi^i: M^i\to M^{i+1}$ .

For example, the homology of a persistence complex is a persistence module, where $\varphi^i$ maps a homology class to the one that contains it.

A persistence complex $\{C_*^i, f^i\}$ (resp.\ persistence module $\{M^i, \varphi^i\}$ ) is of finite type if each component complex (resp.\ module) is a finitely generated $R$ -module, and if the maps $f^i$ (resp.\ $\varphi^i$ ) are isomorphisms for $i\geq m$ for some integer $m$ .

If $K$ is a finite filtered simplicial complex, then it generates a persistence complex $\mathscr{C}$ of finite type, whose homology is a persistence module $\mathcal{M}$ of finite type.

To Live Your Best Life, Do Mathematics

This article is a very good read. 100% Recommended to anyone interested in math.

The ancient Greeks argued that the best life was filled with beauty, truth, justice, play and love. The mathematician Francis Su knows just where to find them.

Source: https://www.quantamagazine.org/20170202-math-and-the-best-life-francis-su-interview/

Math conferences don’t usually feature standing ovations, but Francis Su received one last month in Atlanta. Su, a mathematician at Harvey Mudd College in California and the outgoing president of the Mathematical Association of America (MAA), delivered an emotional farewell address at the Joint Mathematics Meetings of the MAA and the American Mathematical Society in which he challenged the mathematical community to be more inclusive.

Su opened his talk with the story of Christopher, an inmate serving a long sentence for armed robbery who had begun to teach himself math from textbooks he had ordered. After seven years in prison, during which he studied algebra, trigonometry, geometry and calculus, he wrote to Su asking for advice on how to continue his work. After Su told this story, he asked the packed ballroom at the Marriott Marquis, his voice breaking: “When you think of who does mathematics, do you think of Christopher?”

Su grew up in Texas, the son of Chinese parents, in a town that was predominantly white and Latino. He spoke of trying hard to “act white” as a kid. He went to college at the University of Texas, Austin, then to graduate school at Harvard University. In 2015 he became the first person of color to lead the MAA. In his talk he framed mathematics as a pursuit uniquely suited to the achievement of human flourishing, a concept the ancient Greeks called eudaimonia, or a life composed of all the highest goods. Su talked of five basic human desires that are met through the pursuit of mathematics: play, beauty, truth, justice and love.

If mathematics is a medium for human flourishing, it stands to reason that everyone should have a chance to participate in it. But in his talk Su identified what he views as structural barriers in the mathematical community that dictate who gets the opportunity to succeed in the field — from the requirements attached to graduate school admissions to implicit assumptions about who looks the part of a budding mathematician.

When Su finished his talk, the audience rose to its feet and applauded, and many of his fellow mathematicians came up to him afterward to say he had made them cry. A few hours later Quanta Magazine sat down with Su in a quiet room on a lower level of the hotel and asked him why he feels so moved by the experiences of people who find themselves pushed away from math. An edited and condensed version of that conversation and a follow-up conversation follows.

Homotopy for Maps vs Paths

Homotopy (of maps)

A homotopy is a family of maps $f_t: X\to Y$ , $t\in I$ , such that the associated map $F:X\times I\to Y$ given by $F(x,t)=f_t(x)$ is continuous. Two maps $f_0, f_1:X\to Y$ are called homotopic, denoted $f_0\simeq f_1$ , if there exists a homotopy $f_t$ connecting them.

Homotopy of paths

A homotopy of paths in a space $X$ is a family $f_t: I\to X$ , $0\leq t\leq 1$ , such that

(i) The endpoints $f_t(0)=x_0$ and $f_t(1)=x_1$ are independent of $t$ .
(ii) The associated map $F:I\times I\to X$ defined by $F(s,t)=f_t(s)$ is continuous.

When two paths $f_0$ and $f_1$ are connected in this way by a homotopy $f_t$ , they are said to be homotopic. The notation for this is $f_0\simeq f_1$ .

The above two definitions are related, since a path is a special kind of map $f: I\to X$ .

NOUVEAU : découvrez l’appli mobile d’Optimal Sup Spé !

A free new mobile apps on French Math (Classe Prepa) for engineering undergraduate 1st & 2nd years. Very high standard!

Math Online Tom Circle

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【区别：代数拓扑 (Algebraic Topology) 微分拓扑 (Differential Topology ) 微分几何 ( Differential Geometry ) 代数几何 (Algebraic Geometry ) 交换代数 (Commutative Algebra ) 微分流形 (Differential Manifold )

Sheaf (束) originated from Algebraic Geometry, but applied in other areas eg. Algebraic Topology.

Math Online Tom Circle

【区别：代数拓扑 (Algebraic Topology) 微分拓扑 (Differential Topology ) 微分几何 ( Differential Geometry ) 代数几何 (Algebraic Grometry ) 交换代数 (Commutative Algebra ) 微分流形 (Differential Manifold ) ？】月如歌：并不能理解什么叫做楼主所说的配对。我简要谈下我对于上述所列名词的理解。… http://www.zhihu.com/question/23848852/answer/26771912 （分享自知乎网）

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Morphism Summary Chart

Math Online Tom Circle

The more common morphisms are:

1. Homomorphism (Similarity between 2 different structures) 同态
Analogy: Similar triangles of 2 different triangles.

2. Isomorphism (Sameness between 2 different structures) 同构
Analogy: Congruence of 2 different triangles

Example: 2 objects are identical up to an isomorphism.

3. Endomorphism (Similar structure of self) = {Self + Homomorphism} 自同态
Analogy: A triangle and its image in a magnifying glass.

4. Automorphism (Sameness structure of self) = {Self + Isomorphism} 自同构
Analogy: A triangle and its image in a mirror; or
A triangle and its rotated (clock-wise or anti-clock-wise), or reflected (flip-over) self.

5. Monomorphism 单同态 = Injective + Homomorphism

6. Epimorphism 满同态 = Surjective + Homomorphism

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Isomorphism = Congruence, Homomorphism = Similar

Math Online Tom Circle

New Math <=> Old Math

1. Isomorphism of Groups (or any structures)
<=> Congruence Triangles
(Faithful Representation)

2. Homomorphism of Groups (or any structures)
<=> Similar Triangles
(unFaithful Representation)

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Homomorphism History

Math Online Tom Circle

1830 Group Homomorphism
(1831 Galois)

1870 Field Homomorphism
(1870 Camile Jordan Group Isomorphism)
(1870 Dedekind: Automorphism Groups of Field)

1920 Ring Homomorphism
(1927 Noether)

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Quora: Galois Field Automorphism for 15/16 year-old kids

Math Online Tom Circle

3 common Fields: $latex mathbb{R, Q, C}$ with 4 operations : {+ – × ÷}

Automorphism = “self” isomorphism (Analogy: look into mirror of yourself, image is you <=> Automorphism of yourself).

The trivial Field Automorphism of : $latex mathbb{R, Q}$ is none other than Identity Automorphism (mirror image of itself).

Best example for Field Automorphism : : $latex mathbb{C}$ and its conjugate. (a+ib) conjugate with (a-ib)

Field automorphisms using terms a 15/16/ year oldwould understand? by David Joyce

What interesting results are there regardingautomorphisms of fields? by Henning Breede

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Universal Property of Quotient Groups (Hungerford)

If $f:G\to H$ is a homomorphism and $N$ is a normal subgroup of $G$ contained in the kernel of $f$ , then $f$ “factors through” the quotient $G/N$ uniquely.

This can be used to prove the following proposition:
A chain map $f_\bullet$ between chain complexes $(A_\bullet, \partial_{A, \bullet})$ and $(B_\bullet, \partial_{B,\bullet})$ induces homomorphisms between the homology groups of the two complexes.

Proof:
The relation $\partial f=f\partial$ implies that $f$ takes cycles to cycles since $\partial\alpha=0$ implies $\partial(f\alpha)=f(\partial\alpha)=0$ . Also $f$ takes boundaries to boundaries since $f(\partial\beta)=\partial(f\beta)$ . Hence $f_\bullet$ induces a homomorphism $(f_\bullet)_*: H_\bullet (A_\bullet)\to H_\bullet (B_\bullet)$ , by universal property of quotient groups.

For $\beta\in\text{Im} \partial_{A,n+1}$ , we have $\pi_{B,n}f_n(\beta)=\text{Im}\partial_{B,n+1}$ . Therefore $\text{Im}\partial_{A,n+1}\subseteq\ker(\pi_{B,n}\circ f_n)$ .

Some Homology Definitions

Chain Complex
A sequence of homomorphisms of abelian groups $\displaystyle \dots\to C_{n+1}\xrightarrow{\partial_{n+1}}C_n\xrightarrow{\partial_n}C_{n-1}\to\dots\to C_1\xrightarrow{\partial_1}C_0\xrightarrow{\partial_0}0$ with $\partial_n\partial_{n+1}=0$ for each $n$ .

$n$ th Homology Group
$H_n=\text{Ker}\,\partial_n/\text{Im}\,\partial_{n+1}$

$\Delta_n(X)$
$\Delta_n(X)$ is the free abelian group with basis the open $n$ -simplices $e_\alpha^n$ of $X$ .

$n$ -chains
Elements of $\Delta_n(X)$ , called $n$ -chains, can be written as finite formal sums $\sum_\alpha n_\alpha e_\alpha^n$ with coefficients $n_\alpha\in\mathbb{Z}$ .

RP^n Projective n-space

Define an equivalence relation on $S^n\subset\mathbb{R}^{n+1}$ by writing $v\sim w$ if and only if $v=\pm w$ . The quotient space $P^n=S^n/\sim$ is called projective $n$ -space. (This is one of the ways that we defined the projective plane $P^2$ .) The canonical projection $\pi: S^n\to P^n$ is just $\pi(v)=\{\pm v\}$ . Define $U_i\subset P^n$ , $1\leq i\leq n+1$ , by setting $\displaystyle U_i=\{\pi(x^1, \dots, x^{n+1})\mid x^i\neq 0\}.$

Prove
1) $U_i$ is open in $P^n$ .
2) $\{U_1, \dots, U_{n+1}\}$ covers $P^n$ .
3) There is a homeomorphism $\varphi_i: U_i\to\mathbb{R}^n$ .
4) $P^n$ is compact, connected, and Hausdorff, hence is an $n$ -manifold.

Proof:
1) $\pi^{-1}U_i=\{(x^1, \dots, x^{n+1})\mid x^i\neq 0\}$ is open in $S^n$ , so $U_i$ is open in $P^n$ .
2) Let $y=\pi(x^1,\dots, x^{n+1})\in P^n$ . Then since $(x^1,\dots, x^{n+1})\neq(0,\dots,0)$ , so $y\in\bigcup_{i=1}^{n+1}U_i$ . Hence $P^n\subset\bigcup_{i=1}^{n+1}U_i$ .
3) Consider $A=\{(x^1,\dots, x^{n+1})\mid x^i+1\}\cong\mathbb{R}^n$ . Define $\displaystyle \varphi_i(\pi(x^1,\dots, x^{n+1}))=(\frac{x^1}{\|x^i\|},\dots,\frac{x^{i-1}}{\|x^i\|},1,\dots,\frac{x^{n+1}}{\|x^i\|})$ for $x^i>0$ . If $x^i<0$ , then $\varphi_i(\pi(x^1,\dots, x^{n+1}))=\varphi_i(\pi(-x^1,\dots, -x^{n+1}))$ . Then $\varphi_i$ is well-defined.

$\displaystyle \varphi_i^{-1}(x^1,\dots,1,\dots,x^{n+1})=\pi(\frac{x^1}{\|v\|},\dots,\frac{1}{\|v\|},\dots,\frac{x^{n+1}}{\|v\|}),$ where $v=(x^1,\dots, 1,\dots, x^{n+1})$ . Both $\varphi_i$ and $\varphi_i^{-1}$ are continuous, so $\varphi_i: U_i\to A$ is a homeomorphism.
4) Since $S^n$ is compact and connected, so is $P^n=S^n/\sim$ . $P^n$ is a CW-complex with one cell in each dimension, i.e.\ $P^n=\bigcup_{i=0}^n e^n$ . Since CW-complexes are Hausdorff, so is $P^n$ .

Introduction to Persistent Homology (Cech and Vietoris-Rips complex)

Motivation
Data is commonly represented as an unordered sequence of points in the Euclidean space $\mathbb{R}^n$ . The global `shape’ of the data may provide important information about the underlying phenomena of the data.

For data points in $\mathbb{R}^2$ , determining the global structure is not difficult, but for data in higher dimensions, a planar projection can be hard to decipher.
From point cloud data to simplicial complexes
To convert a collection of points $\{x_\alpha\}$ in a metric space into a global object, one can use the points as the vertices of a graph whose edges are determined by proximity (vertices within some chosen distance $\epsilon$ ). Then, one completes the graph to a simplicial complex. Two of the most natural methods for doing so are as follows:

Given a set of points $\{x_\alpha\}$ in Euclidean space $\mathbb{R}^n$ , the Cech complex (also known as the nerve), $\mathcal{C}_\epsilon$ , is the abstract simplicial complex where a set of $k+1$ vertices spans a $k$ -simplex whenever the $k+1$ corresponding closed $\epsilon/2$ -ball neighborhoods have nonempty intersection.

Given a set of points $\{x_\alpha\}$ in Euclidean space $\mathbb{R}^n$ , the Vietoris-Rips complex, $\mathcal{R}_\epsilon$ , is the abstract simplicial complex where a set $S$ of $k+1$ vertices spans a $k$ -simplex whenever the distance between any pair of points in $S$ is at most $\epsilon$ .

Top left: A fixed set of points. Top right: Closed balls of radius $\epsilon/2$ centered at the points. Bottom left: Cech complex has the homotopy type of the $\epsilon/2$ cover ( $S^1\vee S^1\vee S^1$ ) Bottom right: Vietoris-Rips complex has a different homotopy type ( $S^1\vee S^2$ ). Image from R. Ghrist, 2008, Barcodes: The Persistent Topology of Data.

Does Abstract Math belong to Elementary Math ?

Math Online Tom Circle

Yes.

Most pedagogy mistake made in Abstract Algebra teaching is in the wrong order (by historical chronological sequence of discovery):

[X ] Group -> Ring -> Field

It would be better, conceptual wise, to reverse the teaching order as:

Field -> Ring -> Group

or better still as (the author thinks):

Ring -> Field -> Group

Reason 1: Ring is the Integers, most familiar to 8~ 10-year-old kids in primary school arithmetic class involving only 3 operations: ” + – x”.
Reason 2: Field is the Real numbers familiar in calculators involving 4 operations: ” + – × ÷”, 1 extra division operation to Ring.
Reason 3: Group is “Symmetry”, although mistakenly viewed as ONLY 1 operation, but not as easily understandable like Ring and Field, because group operation can be non-numeric such as “rotation” of triangles, “permutation” of roots of equation, “composition” of functions, etc. The only familiar Group…

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In Search for Radical Roots of Polynomial Equations of degree n > 1

Math Online Tom Circle

Take note: Find roots 根 to solve polynomial 多项式方程式equations, but find solution解to solve algebraic equations代数方程式.

Radical : (LatinRadix = root): $latex sqrt [n]{x} $

Quadratic equation (二次方程式) 有 “根式” 解:[最早发现者 : Babylon 和三国时期的吴国数学家赵爽]

$latex {a.x^{2} + b.x + c = 0}&fg=aa0000&s=3$

$latex boxed{x= frac{-b pm sqrt{b^{2}-4ac} }{2a}}&fg=aa0000$

Cubic Equation: 16 CE Italians del Ferro, Tartaglia & Cardano
$latex {a.x^{3} = p.x + q }&fg=0000aa&s=3$

Cardano Formula (1545 《Ars Magna》):
$latex boxed {x = sqrt [3]{frac {q}{2} + sqrt{{ (frac {q}{2})}^{2} – { (frac {p}{3})}^{3}}}
+ sqrt [3]{frac {q}{2} -sqrt{ { (frac {q}{2})}^{2} – { (frac {p}{3})}^{3}}}}&fg=0000aa$

Quartic Equation: by Cardano’s student Ferrari
$latex {a.x^{4} + b.x^{3} + c.x^{2} + d.x + e = 0}&fg=00aa00&s=3$

Quintic Equation:
$latex {a.x^{5} + b.x^{4} + c.x^{3} + d.x^{2} + e.x + f = 0}&s=3$

No radical solution (Unsolvability) was suspected by Ruffini (1799)…

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Why call the Algebraic Structure Z a “Ring” ?

Equivalence of C^infinity atlases

Equivalence of $C^\infty$ atlases is an equivalence relation. Each $C^\infty$ atlas on $M$ is equivalent to a unique maximal $C^\infty$ atlas on $M$ .

Proof:

Reflexive: If $A$ is a $C^\infty$ atlas, then $A\cup A=A$ is also a $C^\infty$ atlas.

Symmetry: Let $A$ and $B$ be two $C^\infty$ atlases such that $A\cup B$ is also a $C^\infty$ atlas. Then certainly $B\cup A$ is also a $C^\infty$ atlas.

Transitivity: Let $A, B, C$ be $C^\infty$ atlases, such that $A\cup B$ and $B\cup C$ are both $C^\infty$ atlases.

Notation:
$\begin{aligned} A&=\{(U_\alpha,\varphi_\alpha)\}\\ B&=\{(V_\beta, \psi_\beta)\}\\ C&=\{(W_\gamma, f_\gamma)\}. \end{aligned}$

Then $\displaystyle \varphi_\alpha\circ f_\gamma^{-1}=\varphi_\alpha\circ\psi_\beta^{-1}\circ\psi_\beta\circ f_\gamma^{-1}: f_\gamma(U_\alpha\cap W_\gamma)\to\varphi_\alpha(U_\alpha\cap W_\gamma)$ is a diffeomorphism since both $\varphi_\alpha\circ\psi_\beta^{-1}$ and $\psi_\beta\circ f_\gamma^{-1}$ are diffeomorphisms due to $A\cup B$ and $B\cup C$ being $C^\infty$ atlases. Also, $M=\bigcup U_\alpha$ , $M=\bigcup W_\gamma$ implies $M=(\bigcup U_\alpha)\cup(\bigcup W_\gamma)$ so $A\cup C$ is also a $C^\infty$ atlas.

Let $A$ be a $C^\infty$ atlas on $M$ . Define $B$ to be the union of all $C^\infty$ atlases equivalent to $A$ . Then $B\sim A$ . If $B'\sim A$ , then $B'\subseteq B$ , so that $B$ is the unique maximal $C^\infty$ atlas equivalent to $A$ .

代数拓扑 Algebraic Topology

Math Online Tom Circle

Excellent Advanced Math Lecture Series (Part 1 to 3) by齊震宇老師

（2012.09.10) Part I:

History: 1900 H. Poincaré invented Topologyfrom Euler Characteristic (V -E + R = 2)

Motivation of Algebraic Topology: Find Invariants[1]of various topological spaces (in higher dimension). 求拓扑空间的“不变量” eg.

AbelianFundamental Group(so as to manipulate by + -);

Vector Space (to + – , × ÷ by multiplier Field scalars);

Ring (to + x), etc.

then apply algebra (Linear Algebra, Matrices) with computer to compute these invariants (homology, co-homology, etc).

A topological space can be formed by a “Big Data” Point Set, e.g. genes, tumors, drugs, images, graphics, etc. By finding (co)- / homology – hence the intuitive Chinese term (上) /同调 [2] – is to find “holes” in the Big Data in the 10,000 (e.g.) dimensional space the hidden information (co-relationship, patterns, etc).
Note: [1]…

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Natural Equivalence relating Suspension and Loop Space

Theorem:
If $(X,x_0)$ , $(Y,y_0)$ , $(Z,z_0)\in\mathscr{PT}$ , $X$ , $Z$ Hausdorff and $Z$ locally compact, then there is a natural equivalence $\displaystyle A: [Z\wedge X, *; Y,y_0]\to [X, x_0; (Y,y_0)^{(Z,z_0)}, f_0]$ defined by $A[f]=[\hat{f}]$ , where if $f:Z\wedge X\to Y$ is a map then $\hat{f}: X\to Y^Z$ is given by $(\hat{f}(x))(z)=f[z,x]$ .

We need the following two propositions in order to prove the theorem.
Proposition
\label{prop13}
The exponential function $E: Y^{Z\times X}\to (Y^Z)^X$ induces a continuous function $\displaystyle E: (Y,y_0)^{(Z\times X, Z\vee X)}\to ((Y,y_0)^{(Z,z_0)}, f_0)^{(X,x_0)}$ which is a homeomorphism if $Z$ and $X$ are Hausdorff and $Z$ is locally compact\footnote{every point of $Z$ has a compact neighborhood}.

Proposition
\label{prop8}
If $\alpha$ is an equivalence relation on a topological space $X$ and $F:X\times I\to Y$ is a homotopy such that each stage $F_t$ factors through $X/\alpha$ , i.e.\ $x\alpha x'\implies F_t(x)=F_t(x')$ , then $F$ induces a homotopy $F':(X/\alpha)\times I\to Y$ such that $F'\circ (p_\alpha\times 1)=F$ .

Proof of Theorem
i) $A$ is surjective: Let $f': (X,x_0)\to ((Y,y_0)^{(Z,z_0)},f_0)$ . From Proposition \ref{prop13} we have that $E: (Y,y_0)^{(Z\times X, Z\vee X)}\to ((Y,y_0)^{(Z,z_0)},f_0)^{(X,x_0)}$ is a homeomorphism. Hence the function $\bar{f}: (Z\times X, Z\vee X)\to (Y,y_0)$ defined by $\bar{f}(z,x)=(f'(x))(z)$ is continuous since $(Ef'(x))(z)=f'(z,x)$ and thus $\bar{f}=E^{-1}f'$ . By the universal property of the quotient, $\bar{f}$ defines a map $f:(Z\wedge X, *)\to (Y,y_0)$ such that $f[z,x]=\bar{f}(z,x)=(f'(x))(z)$ . Thus $\hat{f}=f'$ , so that $A[f]=[f']$ .

ii) $A$ is injective: Suppose $f,g: (Z\wedge X, *)\to (Y,y_0)$ are two maps such that $A[f]=A[g]$ , i.e.\ $\hat{f}\simeq\hat{g}$ . Let $H': X\times I\to (Y,y_0)^{(Z,z_0)}$ be the homotopy rel $x_0$ . By Proposition \ref{prop13} the function $\bar{H}: Z\times X\times I\to Y$ defined by $\bar{H}(z,x,t)=(H'(x,t))(z)$ is continuous. This is because $\bar{H}(z,x,t)=(E\bar{H}(x,t))(z)$ so that $E\bar{H}=H'$ , thus $\bar{H}=E^{-1}H'$ where $E$ is a homeomorphism. For each $t\in I$ we have $\bar{H}((Z\vee X)\times\{t\})=y_0$ . This is because if $(z,x)\in Z\vee X$ , then $z=z_0$ or $x=x_0$ . If $z=z_0$ , then $(H'(x,t))(z_0)=y_0$ . If $x=x_0$ , $(H'(x_0,t))(z)=y_0$ as $H'$ is the homotopy rel $x_0$ . Then by Proposition \ref{prop8} there is a homotopy $H:(Z\wedge X)\times I\to Y$ rel $*$ such that $H([z,x],t)=\bar{H}(z,x,t)=(H'(x,t))(z)$ . Thus $H_0([z,x])=(H_0'(x))(z)=(\hat{f}(x))(z)=f[z,x]$ and similarly $H_1([z,x])=(H_1'(x))(z)=(\hat{g}(x))(z)=g[z,x]$ . Thus $[f]=[g]$ via the homotopy $H$ .

Loop space
If $(Y,y_0)\in\mathscr{PT}$ , we define the loop space $(\Omega Y, \omega_0)\in\mathscr{PT}$ of $Y$ to be the function space $\displaystyle \Omega Y=(Y,y_0)^{(S^1,s_0)}$ with the constant loop $\omega_0$ ( $\omega_0(s)=y_0$ for all $s\in S^1$ ) as base point.

Suspension
If $(X,x_0)\in\mathscr{PT}$ , we define the suspension $(SX,*)\in\mathscr{PT}$ of $X$ to be the smash product $(S^1\wedge X, *)$ of $X$ with the 1-sphere.

Corollary (Natural Equivalence relating $SX$ and $\Omega Y$ )
If $(X,x_0)$ , $(Y,y_0)\in\mathscr{PT}$ and $X$ is Hausdorff, then there is a natural equivalence $\displaystyle A: [SX, *; Y,y_0]\to [X, x_0; \Omega Y, \omega_0].$

Russian Math Education

Math Online Tom Circle

In the world of Math education there are 3 big schools (门派) — in which the author had the good fortune to study under 3 different Math pedagogies:

“武当派” French (German) -> “少林派” Russian (China) -> “华山派” UK (USA).

( ) : derivative of its parent school. eg. China derived from Russian school in 1960s by Hua Luogeng.

Note:
武当派 : 内功, 以柔尅刚, 四两拨千斤 <=> “Soft” Math, Abstract, Theoretical, Generalized.

少林派: 拳脚硬功夫 <=> “Hard” Math, algorithmic.

华山派: 剑法轻灵 <=> Applied, Astute, Computer-aided.

The 3 schools’ pioneering grand masters (掌门人) since 16th century till 21st century, in between the 19th century (during the French Revolution) Modern Math (近代数学) is the critical milestone, the other (现代数学) is WW2 : –

France: Descartes / Fermat / Pascal (17 CE : Analytical Geometry, Number Theory, Probability), Cauchy / Lagrange / Fourier /Galois (19 CE, Modern Math : Analysis, Abstract Algebra),

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Fundamental Group of S^n is trivial if n>=2

$\pi_1(S^n)=0$ if $n\geq 2$
We need the following lemma:

If a space $X$ is the union of a collection of path-connected open sets $A_\alpha$ each containing the basepoint $x_0\in X$ and if each intersection $A_\alpha\cap A_\beta$ is path-connected, then every loop in $X$ at $x_0$ is homotopic to a product of loops each of which is contained in a single $A_\alpha$ .

Proof:
Take $A_1$ and $A_2$ to be the complements of two antipodal points in $S^n$ . Then $S^n=A_1\cup A_2$ is the union of two open sets $A_1$ and $A_2$ , each homeomorphic to $\mathbb{R}^n$ such that $A_1\cap A_2$ is homeomorphic to $S^{n-1}\times\mathbb{R}$ .

Choose a basepoint $x_0$ in $A_1\cap A_2$ . If $n\geq 2$ then $A_1\cap A_2$ is path-connected. By the lemma, every loop in $S^n$ based at $x_0$ is homotopic to a product of loops in $A_1$ or $A_2$ . Both $\pi_1(A_1)$ and $\pi_1(A_2)$ are zero since $A_1$ and $A_2$ are homeomorphic to $\mathbb{R}^n$ . Hence every loop in $S^n$ is nullhomotopic.

Tangent Space is Vector Space

Prove that the operation of linear combination, as in Definition 2.2.7, makes $T_p(U)$ into an $n$ -dimensional vector space over $\mathbb{R}$ . The zero vector is the infinitesimal curve represented by the constant $p$ . If $\langle s\rangle_p\in T_p(U)$ , then $-\langle s\rangle_p=\langle s^-\rangle_p$ where $s^-(t)=s(-t)$ , defined for all sufficiently small values of $t$ .

Proof:

We verify the axioms of a vector space.

Multiplicative axioms:
* $1\langle s_1\rangle_p=\langle 1s_1+0-(1+0-1)p\rangle_p=\langle s_1\rangle_p$
* $(ab)\langle s_1\rangle_p=\langle abs_1-(ab-1)p\rangle_p$
$\begin{aligned} a(b\langle s_1\rangle_p)&=a\langle bs_1-(b-1)p\rangle_p\\ &=\langle abs_1-(ab-a)p-(a-1)p\rangle_p\\ &=\langle abs_1-(ab-1)p\rangle_p\\ &=(ab)\langle s_1\rangle_p \end{aligned}$

Additive Axioms:
* $\langle s_1\rangle_p+\langle s_2\rangle_p=\langle s_2\rangle_p+\langle s_1\rangle_p=\langle s_1+s_2-p\rangle_p$
* $\begin{aligned} (\langle s_1\rangle_p+\langle s_2\rangle_p)+\langle s_3\rangle_p&=\langle s_1+s_2-p\rangle_p+\langle s_3\rangle_p\\ &=\langle s_1+s_2-p+s_3-p\rangle_p\\ &=\langle s_1+s_2+s_3-2p\rangle_p \end{aligned}$
$\begin{aligned} \langle s_1\rangle_p+(\langle s_2\rangle_p+\langle s_3\rangle_p)&=\langle s_1\rangle_p+\langle s_2+s_3-p\rangle_p\\ &=\langle s_1+s_2+s_3-2p\rangle_p \end{aligned}$
* $\langle s\rangle_p+\langle s^-\rangle_p=\langle s+s^- -p\rangle_p$

$\frac{d}{dt}f(s(t)+s(-t)-p)|_{t=0}=0=\frac{d}{dt}f(p)|_{t=0}$

Hence $\langle s\rangle_p+\langle s^-\rangle_p=\langle p\rangle_p$ .
* $\langle s_1\rangle_p+\langle p\rangle_p=\langle s_1+p-p\rangle_p=\langle s_1\rangle_p$

Distributive Axioms:
* $\begin{aligned} a(\langle s_1\rangle_p+\langle s_2\rangle_p)&=a\langle s_1+s_2-p\rangle_p\\ &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p \end{aligned}$
$\begin{aligned} a\langle s_1\rangle_p+a\langle s_2\rangle_p&=\langle as_1-(a-1)p\rangle_p+\langle as_2-(a-1)p\rangle_p\\ &=\langle a(s_1+s_2)-2(a-1)p-p\rangle_p\\ &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p\\ &=a(\langle s_1\rangle_p+\langle s_2\rangle_p) \end{aligned}$
* $(a+b)\langle s_1\rangle_p=\langle (a+b)s_1-(a+b-1)p\rangle_p$

$a\langle s_1\rangle_p+b\langle s_1\rangle_p=\langle as_1+bs_1-(a+b-1)p\rangle_p=(a+b)\langle s_1\rangle_p$

Hence $T_p(U)$ is a vector space over $\mathbb{R}$ . Since $U\subseteq\mathbb{R}^n$ , $T_p(U)$ is $n$ -dimensional.

群论的哲学 Philosophical Group Theory

Math Online Tom Circle

在一个群体里, 每个会员互动中存在一种”运作” (binary operation)关系, 并遵守以下4个原则:

1) 肥水不流外人田: 任何互动的结果要回归群体。(Closure) = C

2) 互动不分前后次序 (Associative) = A

(a.*b)*c = a*(b*c)

3) 群体有个”中立” 核心 (Neutral / Identity) = N (记号: e)

4) 和而不同: 每个人的意见都容许存在反面的意见 “逆元” (Inverse) = I (记号: a 的逆元 = $latex a^{-1}$)

Agree to disagree = Neutral

$latex a*a^{-1} = e $

具有这四个性质的群体才是

群体的 “美 : “对称”

如果没有 (3)&(4): 半群

如果没有 (4) 反对者: 么半群
以上是 Group (群 ) 数学的定义: “CAN I”

CA = Semi-Group 半群

CAN = Monoid 么半群

群是 19岁Evariste Galois 在法国革命时牢狱中发明的, 解决 300年来 Quintic Equations (5次以上的方程式) 没有 “有理数” 的解 (rational roots)。19世纪的 Modern Math (Abstract Algebra) 从此诞生, 群用来解释自然科学(物理, 化学, 生物)里 “对称”现象。Nobel Physicists (1958) 杨振宁/李政道用群来证明物理弱力 (Weak Force) 粒子(Particles) 的不对称 (Assymetry )。

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Balance Quote

Never let success go to your head, and never let failure go to your heart.