Math Blog

Morphism Summary Chart

tomcircle's avatarMath 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.

image

5. Monomorphism 单同态 = Injective + Homomorphism
image

6. Epimorphism 满同态 = Surjective + Homomorphism

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

tomcircle's avatarMath 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.Universal Property of Quotient

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).

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.

fig2

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 ? 

ChefCouscous's avatarMath 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

ChefCouscous's avatarMath 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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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

tomcircle's avatarMath 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.

  • 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

ChefCouscous's avatarMath 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

ChefCouscous's avatarMath 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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Analysis: 97 marks not enough for Higher Chinese cut-off point for Pri 1 pupils

Quite tough to be a primary school kid nowadays, even 97 marks is not enough to be admitted for Higher Chinese classes.

From experience, the main underlying reasons behind this scenario could be:

  • Due to intensive tuition starting from preschool, students enter primary 1 already knowing primary 3 syllabus, so everyone is scoring 100/100. So top 25% percentile mark becomes 99/100.
  • Lack of manpower (Chinese teachers). It is well-known that Singaporeans are not very interested in general in pursuing the career of Mother Tongue teacher (look at the cut-off points of Chinese studies in universities). So only enough manpower for limited number of Higher Chinese classes.
  • Kiasu principals / HODs who want to “quality-control” those taking Higher Chinese to boost the distinction rate of the cohort (a common but unethical tactic to improve the cohort’s performance in national exams is to force those who are not doing well to drop the subject)
  • Lastly, it is not known if 97 is the overall mark, or just one of the marks in the continual assessment. It is possible to score 97 in one test, but the average can be much lower.

This is quite a serious issue as Chinese is no longer a minor/unimportant subject, like in the past it was. In fact, under the new PSLE scoring system, Chinese is one of the major game-changing core components, a severe Achilles’ heel for those in English-speaking families. Getting proficient in Chinese from an early age is a must for the new PSLE system, so no doubt many parents are anxious about Higher Chinese.

http://www.straitstimes.com/singapore/education/how-can-97-marks-be-not-good-enough

Parents of some children in a well-known primary school have complained about the selection process for Higher Chinese.

St Hilda’s Primary pupils are routed into Higher Chinese classes in Primary 2 based on continual assessment test results in Primary 1.

What upset the parents was that pupils who scored as high as 97 marks in Chinese last year were told that they had failed to make the cut for Higher Chinese.

Read more at: http://www.straitstimes.com/singapore/education/how-can-97-marks-be-not-good-enough

Functors, Homotopy Sets and Groups

Functors
Definition:
A functor F from a category \mathscr{C} to a category \mathscr{D} is a function which
– For each object X\in\mathscr{C}, we have an object F(X)\in\mathscr{D}.
– For each f\in\hom_\mathscr{C}(X,Y), we have a morphism \displaystyle F(f)\in\hom_\mathscr{D}(F(X),F(Y)).

Furthermore, F is required to satisfy the two axioms:
– For each object X\in\mathscr{C}, we have F(1_X)=1_{F(X)}. That is, F maps the identity morphism on X to the identity morphism on F(X).

– For f\in\hom_{\mathscr{C}}(X,Y), g\in\hom_\mathscr{C}(Y,Z) we have \displaystyle F(g\circ f)=F(g)\circ F(f)\in\hom_\mathscr{D}(F(X),F(Z)). That is, functors must preserve composition of morphisms.

Definition:
A cofunctor (also called contravariant functor) F^* from a category \mathscr{C} to a category \mathscr{D} is a function which
– For each object X\in\mathscr{C}, we have an object F^*(X)\in\mathscr{D}.
– For each f\in\hom_\mathscr{C}(X,Y) we have a morphism \displaystyle F^*(f)\in\hom_\mathscr{D}(F^*(Y),F^*(X)) satisfying the two axioms:
– For each object X\in\mathscr{C} we have F^*(1_X)=1_{F^*(X)}. That is, F^* preserves identity morphisms.
– For each f\in\hom_\mathscr{C}(X,Y) and g\in\hom_\mathscr{C}(Y,Z) we have \displaystyle F^*(g\circ f)=F^*(f)\circ F^*(g)\in\hom_\mathscr{D}(F^*(Z),F^*(X)). Note that cofunctors reverse the direction of composition.

Example

Given a fixed pointed space (K,k_0)\in\mathscr{PT}, we define a functor \displaystyle F_K:\mathscr{PT}\to\mathscr{PS} as follows: for each (X,x_0)\in\mathscr{PT} we assign F_K(X,x_0)=[K,k_0; X,x_0]\in\mathscr{PS}. Given f: (X,x_0)\to (Y,y_0) in \hom((X,x_0),(Y,y_0)) we define F_K(f)\in\hom([K,k_0; X,x_0],[K,k_0;Y,y_0]) by \displaystyle F_k(f)[g]=[f\circ g]\in[K,k_0; Y,y_0] for every [g]\in [K,k_0; X,x_0].

We can check the two axioms:
– F_k(1_X)[g]=[1_X\circ g]=[g] for every [g]\in[K,k_0; X, x_0].
– For f\in\hom((X,x_0),(Y,y_0)), h\in\hom((Y,y_0),(Z,z_0)) we have \displaystyle F_K(h\circ f)[g]=[h\circ f\circ g]=F_K(h)\circ F_K(f)[g]\in[K,k_0; Z,z_0] for every [g]\in[K,k_0; X,x_0].

Similarly, we can define a cofunctor F_K^* by taking F_K^*(X,x_0)=[X,x_0; K,k_0] and for f:(X,x_0)\to (Y,y_0) in \hom((X,x_0),(Y,y_0)) we define \displaystyle F_K(f)[g]=[g\circ f]\in[X,x_0; K,k_0] for every [g]\in[Y,y_0; K,k_0].

Note that if f\simeq f' rel x_0, then F_K(f)=F_K(f') and similarly F_K^*(f)=F_K^*(f'). Therefore F_K (resp.\ F_K^*) can also be regarded as defining a functor (resp.\ cofunctor) \mathscr{PT}'\to\mathscr{PS}.

Homotopy Sets and Groups
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 1:
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 2:
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.

H2 Maths Tuition by Ex-RI, NUS 1st Class Honours (Mathematics)

Junior College H2 Maths Tuition

About Tutor (Mr Wu): https://mathtuition88.com/singapore-math-tutor/
– Raffles Alumni
– NUS 1st Class Honours in Mathematics

Experience: More than 10 years experience, has taught students from RJC, NJC, ACJC and many other JCs.

Personality: Friendly, patient and good at explaining complicated concepts in a simple manner

Email: mathtuition88@gmail.com

Areas teaching (West / Central Singapore):

  • Clementi
  • Jurong East
  • Buona Vista
  • West Coast
  • Dover
  • Central Areas like Bishan/Toa Payoh/Marymount (near MRT)

viXra vs arXiv

viXra (http://vixra.org/) is the cousin of arXiv (http://arxiv.org/) which are electronic archives where researchers can submit their research before being published on a journal.

The difference is that viXra allows anyone to submit their article, whereas arXiv requires an academic affiliation to recommend before submitting. There are pros and cons to viXra, the pros being freedom of submission open to everyone on the world. The cons is that, naturally, there may be more crackpots who submit nonsense.

There are, however, some serious papers on viXra.

After submitting, the viXra admin will send an email something like this:

Thank you for your submission request to viXra.org

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If you need to replace the document you should use the replacement form and enter the viXra number
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Featured book:

An Introduction to the Theory of Numbers

An Introduction to the Theory of Numbers by G. H. Hardy and E. M. Wright is found on the reading list of virtually all elementary number theory courses and is widely regarded as the primary and classic text in elementary number theory. Developed under the guidance of D. R. Heath-Brown, this Sixth Edition of An Introduction to the Theory of Numbers has been extensively revised and updated to guide today’s students through the key milestones and developments in number theory.

Prof ST Yau’s 邱成桐 Talk to Chinese Youth on Math Education 

ChefCouscous's avatarMath Online Tom Circle

Prof ST Yau邱成桐, Chinese/HK Harvard Math Dean, is the only 2 Mathematicians in history (the other person is Prof Pierre Deligne of Belgium) who won ALL 3 top math prizes: Fields Medal (at 27, proving Calabi Conjecture), Crafoord Prize(1994),Wolf Prize(2010).

Key Takeaways:

1. On Math Education:
◇ Compulsary Math training for reasoning skill applicable in Economy, Law, Medicine, etc.
◇ Study Math Tip: read the new topic notes 1 day before the lecture, then after it do the problems.
◇ Read Math topics even though you do not understand in first round, re-read few more times, then few days / months / years / decades later you will digest them. (做学问的程序).
◇ Do not consult students in WHAT to teach, because they don’t know what to learn.
◇ Love of Math beauty is the “pull-factor” for motivating students’ interest in Math.
◇ Parental…

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Algebraic Topology: Fundamental Group

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.

Example: Linear Homotopies
Any two paths f_0 and f_1 in \mathbb{R}^n having the same endpoints x_0 and x_1 are homotopic via the homotopy \displaystyle f_t(s)=(1-t)f_0(s)+tf_1(s).

Simply-connected
A space is called simply-connected if it is path-connected and has trivial fundamental group.

A space X is simply-connected iff there is a unique homotopy class of paths connecting any two parts in X.
Path-connectedness is the existence of paths connecting every pair of points, so we need to be concerned only with the uniqueness of connecting paths.

(\implies) Suppose \pi_1(X)=0. If f and g are two paths from x_0 to x_1, then f\simeq f\cdot \bar{g}\cdot g\simeq g since the loops \bar{g}\cdot g and f\cdot\bar{g} are each homotopic to constant loops, due to \pi_1(X,x_0)=0.

(\impliedby) Conversely, if there is only one homotopy class of paths connecting a basepoint x_0 to itself, then all loops at x_0 are homotopic to the constant loop and \pi_1(X,x_0)=0.

\pi_1(X\times Y) is isomorphic to \pi_1(X)\times \pi_1(Y) if X and Y are path-connected.
A basic property of the product topology is that a map f:Z\to X\times Y is continuous iff the maps g:Z\to X and h:Z\to Y defined by f(z)=(g(z),h(z)) are both continuous.

Hence a loop f in X\times Y based at (x_0,y_0) is equivalent to a pair of loops g in X and h in Y based at x_0 and y_0 respectively.

Similarly, a homotopy f_t of a loop in X\times Y is equivalent to a pair of homotopies g_t and h_t of the corresponding loops in X and Y.

Thus we obtain a bijection \pi_1(X\times Y, (x_0,y_0))\approx \pi_1(X,x_0)\times \pi_1(Y,y_0), [f]\mapsto([g],[h]). This is clearly a group homomorphism, and hence an isomorphism.

Note: The condition that X and Y are path-connected implies that \pi_1(X,x_0)=\pi_1(X), \pi_1(Y,y_0)=\pi_1(Y),\pi_1(X\times Y,(x_0,y_0))=\pi_1(X\times Y).

Chinese Remainder Theorem

Any short-cut method ? Yes, by L.C.M…

ChefCouscous's avatarMath Online Tom Circle

How to formulate this problem in CRT ?

Hint: Sunday = 7 , Interval 2 days = mod 2, …

Let d = week days {1, 2, 3, 4, 5, 6, 7} for {Monday (Prof M), tuesday (Prof t), Wednesday (Prof W), Thursday (Prof T), Friday (Prof F), saturday (Prof s), Sunday (Prof S)}

d : 1 2 3 4 5 6 [7] 1 2 3 4 5 6 [7] 1 2
M: m 0 m 0 m 0 [m] ==> fell on 1st sunday
t: - t 0 0 t 0 [0 ] t 0 0 t 0 0 [t ] ==> fell on 2nd sunday
W: – - w 0 0 0 [w] 0 0 0 w 0 0 [0] ==> fell on 1st sunday
T: - - - T T T [T] ==> fell on 1st sunday (TRIVIAL CASE!)
F: - - - - f 0…

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Some Math Connotations Demystified 数学内涵解密

ChefCouscous's avatarMath Online Tom Circle

This Taiwanese Math Prof is very approachable in clarifying the doubts in an unconventional way different from the arcane textbook definitions. Below are his few key tips to breakthrough the “mystified”concepts :

1. “Dual Space“(对偶空间) : it is the evaluation of a “Vector Space”.

Example: A student studies few subjects {Math, Physics, English, Chemistry…}, these subjects form a “Subject Vector Space” (V), if we associate the subjects with weightages (加权) , say, Math 4, Physics 3, English 2, Chemistry 1, the “Weightage Dual Space” of V will be W= {4, 3, 2, 1}.

2. Vector: beyond the meaning of a physical vector with direction and value, it extends to any “object” which can be manipulated (抵消) by the 4 operations “+, – , x, / ” in a FieldF = {R or Z2 …}.

Eg. $latex alpha_{1}.v_{1} + alpha_{2}.v_{2} + alpha_{3}.v_{3}, forall alpha_{j} in…

View original post 111 more words

Homology: Why Boundary of Boundary = 0 ?

ChefCouscous's avatarMath Online Tom Circle

This equation puzzles most people. WHY ?
$latex boxed {{delta}^2 = 0 { ?}}&fg=aa0000&s=3 $

It is analogous to the Vector Algebra:
Let the boundary of {A, B} =
$latex delta (A,B) = overrightarrow{AB }$

$latex overrightarrow{AB } + overrightarrow {BA} =overrightarrow{AB } – overrightarrow {AB} = vec 0 $

Source: http://mathoverflow.net/questions/640/what-is-cohomology-and-how-does-a-beginner-gain-intuition-about-it

Note: Co-homology: (上)同调

Euclid Geometry & Homology:

Isabell Darcy Lecture: cohomology

View original post

Existence and properties of normal closure

If E is an algebraic extension field of K, then there exists an extension field F of E (called the normal closure of E over K) such that
(i) F is normal over K;
(ii) no proper subfield of F containing E is normal over K;
(iii) if E is separable over K, then F is Galois over K;
(iv) [F:K] is finite if and only if [E:K] is finite.

The field F is uniquely determined up to an E-isomorphism.

Proof:
(i) Let X=\{u_i\mid i\in I\} be a basis of E over K and let f_i\in K[x] be the minimal polynomial of u_i. If F is a splitting field of S=\{f_i\mid i\in I\} over E, then F=E(Y), where Y\supseteq X is the set of roots of the f_i. Then F=K(X)(Y)=K(Y) so F is also a splitting field of S over K, hence F is normal over K as it is the splitting field of a family of polynomials in K[x].

(iii) If E is separable over K, then each f_i is separable. Therefore F is Galois over K as it is a splitting field over K of a set of separable polynomials in K[x].

(iv) If [E:K] is finite, then so is X and hence S. Say S=\{f_1,\dots,f_n\}. Then F=E(Y), where Y is the set of roots of the f_i. Then F is finitely generated and algebraic, thus a finite extension. So [F:K] is finite.

(ii) A subfield F_0 of F that contains E necessarily contains the root u_i of f_i\in S for every i. If F_0 is normal over K (so that each f_i splits in F_0 by definition), then F\subset F_0 (since F is the splitting field) and hence F=F_0.

Finally let F_1 be another extension field of E with properties (i) and (ii). Since F_1 is normal over K and contains each u_i, F_1 must contain a splitting field F_2 of S over K with E\subset F_2. F_2 is normal over K (splitting field over K of family of polynomials in K[x]), hence F_2=F_1 by (ii).

Therefore both F and F_1 are splitting fields of S over K and hence of S over E: If F=K(Y) (where Y is set of roots of f_i) then F\subseteq E(Y) since E(Y) contains K and Y. Since Y\supseteq X, so K(Y) contains E=K(X) and Y, hence F=E(Y). Hence the identity map on E extends to an E-isomorphism F\cong F_1.

Happy New Year to Readers of Mathtuition88.com

Wishing all readers of Mathtuition88.com a happy new year, and may 2017 bring you peace and joy in your life.

No matter which stage of life you are in (student/career/parent/retiree), here is my sincere wishes that you will achieve your goals in 2017, and more importantly be happy in the process.

Best wishes,
Mathtuition88.com

Cours Raisonnements (Logics) , Ensembles ( Sets), Applications (Mappings)

ChefCouscous's avatarMath Online Tom Circle

This is an excellent quick revision of the French Baccalaureat Math during the first month of French university. (Unfortunately common A-level Math syllabus lacks such rigourous Math foundation.)

Most non-rigourous high-school students / teachers abuse the use of :

“=> ” , “<=>” .

Prove by “Reductio par Absudum” 反证法 (by Contradiction) is a clever mathematical logic :

$latex boxed {(A => B) <=> (non B => non A)} &s=3$

Famous Examples: 1) Prove $latex sqrt 2 $ is irrational ; 2) There are infinite prime numbers (both by Greek mathematician Euclid 3,000 years ago)

The young teacher showed the techniques of proving Mapping:

Surjective (On-to) – best understood in Chinese 满射 (Full Mapping)

Injective (1-to-1) 单射

Bijective (On-to & 1-to-1) 双射

He used an analogy of (the Set of) red Indians shooting (the Set of bisons 野牛):

All bisons are shot by arrows from1 or more Indians. (Surjective…

View original post 81 more words

A finitely generated torsion-free module A over a PID R is free

A finitely generated torsion-free module A over a PID R is free.
Proof
(Hungerford 221)

If A=0, then A is free of rank 0. Now assume A\neq 0. Let X be a finite set of nonzero generators of A. If x\in X, then rx=0 (r\in R) if and only if r=0 since A is torsion-free.

Consequently, there is a nonempty subset S=\{x_1,\dots,x_k\} of X that is maximal with respect to the property: \displaystyle r_1x_1+\dots+r_kx_k=0\ (r_i\in R) \implies r_i=0\ \text{for all}\ i.

The submodule F generated by S is clearly a free R-module with basis S. If y\in X-S, then by maximality there exist r_y,r_1,\dots,r_k\in R, not all zero, such that r_yy+r_1x_1+\dots+r_kx_k=0. Then r_yy=-\sum_{i=1}^kr_ix_i\in F. Furthermore r_y\neq 0 since otherwise r_i=0 for every i.

Since X is finite, there exists a nonzero r\in R (namely r=\prod_{y\in X-S}r_y) such that rX=\{rx\mid x\in X\} is contained in F:

If y_i\in X-S, then ry=r_{y_1}\dots r_{y_n}y_i\in F since r_{y_i}y_i\in F. If x\in S, then clearly rx\in F since F is generated by S.

Therefore, rA=\{ra\mid a\in A\}\subset F. The map f:A\to A given by a\mapsto ra is an R-module homomorphism with image rA. Since A is torsion-free \ker f=0, hence A\cong rA\subset F. Since a submodule of a free module over a PID is free, this proves A is free.

Tensor is a right exact functor Elementary Proof

This is a relatively elementary proof (compared to others out there) of the fact that tensor is a right exact functor. Proof is taken from Hungerford, and reworded slightly. The key prerequisites needed are the universal property of quotient and of tensor product.

Statement

If A\xrightarrow{f}B\xrightarrow{g}C\to 0 is an exact sequence of left modules over a ring R and D is a right R-module, then \displaystyle D\otimes_R A\xrightarrow{1_D\otimes f}D\otimes_R B\xrightarrow{1_D\otimes g}D\otimes_R C\to 0 is an exact sequence of abelian groups. An analogous statement holds for an exact sequence in the first variable.

Proof

(Hungerford 210)

We split our proof into 3 parts: (i) \text{Im}(1_D\otimes g)=D\otimes_R C; (ii) \text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g); and (iii) \text{Ker}(1_D\otimes g)\subseteq\text{Im}(1_D\otimes f).

(i) Since g is an epimorphism by hypothesis every generator d\otimes c of D\otimes_R C is of the form d\otimes g(b)=(1_D\otimes g)(d\otimes b) for some b\in B. Thus \text{Im}(1_D\otimes g) contains all generators of D\otimes_R C, hence \text{Im}(1_D\otimes g)=D\otimes_R C.

(ii) Since \text{Ker} g=\text{Im} f we have gf=0 and \displaystyle (1_D\otimes g)(1_D\otimes f)=1_D\otimes gf=1_D\otimes 0=0, hence \text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g).

(iii) Let \pi:D\otimes_R B\to(D\otimes_R B)/\text{Im}(1_D\otimes f) be the canonical epimorphism. From (ii), \text{Im}(1_D\otimes f)\subseteq\text{Ker}(1_D\otimes g) so (by universal property of quotient Theorem 1.7) there is a homomorphism \alpha:(D\otimes_R B)/\text{Im}(1_D\otimes f)\to D\otimes_R C such that \displaystyle \alpha(\pi(d\otimes b))=(1_D\otimes g)(d\otimes b)=d\otimes g(b). We shall show that \alpha is an isomorphism. Then \text{Ker}(1_D\otimes g)=\text{Im}(1_D\otimes f).

We show first that the map \beta:D\times C\to(D\otimes_R B)/\text{Im}(1_D\otimes f) given by (d,c)\mapsto\pi(d\otimes b), where g(b)=c, is independent of the choice of b. Note that there is at least one such b since g is an epimorphism. If g(b')=c, then g(b-b')=0 and b-b'\in\text{Ker} g=\text{Im} f, hence b-b'=f(a) for some a\in A. Since d\otimes f(a)\in\text{Im}(1_D\otimes f) and \pi(d\otimes f(a))=0, we have
\begin{aligned} \pi(d\otimes b)&=\pi(d\otimes(b'+f(a))\\ &=\pi(d\otimes b'+d\otimes f(a))\\ &=\pi(d\otimes b')+\pi(d\otimes f(a))\\ &=\pi(d\otimes b'). \end{aligned}

Therefore \beta is well-defined.

Verify that \beta is middle linear:
\begin{aligned} \beta(d_1+d_2,c)&=\pi((d_1+d_2)\otimes b)\qquad\text{where }g(b)=c\\ &=\pi(d_1\otimes b+d_2\otimes b)\\ &=\pi(d_1\otimes b)+\pi(d_2\otimes b)\\ &=\beta(d_1,c)+\beta(d_2,c). \end{aligned}

\begin{aligned} \beta(d,c_1+c_2)&=\pi(d\otimes(b_1+b_2))\qquad\text{where }g(b_i)=c_i\\ &=\pi(d\otimes b_1+d\otimes b_2)\\ &=\pi(d\otimes b_1)+\pi(d\otimes b_2)\\ &=\beta(d,c_1)+\beta(d,c_2). \end{aligned}

Let r\in R.
\begin{aligned} \beta(dr,c)&=\pi(dr\otimes b)\qquad\text{where }g(b)=c\\ &=\pi(d\otimes rb)\\ &=\beta(d,rc)\qquad\text{where }g(rb)=rg(b)=rc. \end{aligned}

By universal property of tensor product there exists a unique homomorphism \bar{\beta}:D\otimes_R C\to(D\otimes_R B)/\text{Im}(1_D\otimes f) such that \bar{\beta}(d\otimes c)=\beta(d,c)=\pi(d\otimes b), where g(b)=c.

Therefore, for any generator d\otimes c of D\otimes_R C, \displaystyle \alpha\bar{\beta}(d\otimes c)=\alpha\pi(d\otimes b)=d\otimes g(b)=d\otimes c, hence \alpha\bar{\beta} is the identity map.

Similarly
\begin{aligned} \bar{\beta}\alpha(d\otimes b+\text{Im}(1_D\otimes f))&=\bar{\beta}\alpha\pi(d\otimes b)\\ &=\bar{\beta}(d\otimes g(b))\\ &=\pi(d\otimes b)\\ &=d\otimes b+\text{Im}(1_D\otimes F) \end{aligned}
so \bar{\beta}\alpha is the identity so that \alpha is an isomorphism.

Category Theory in Computing Languages

ChefCouscous's avatarMath Online Tom Circle

​Is there any connection between category theory and the way computer languages work?by Thorsten Altenkirch

Yes, lots.

Just one example: a function with 2 inputs from A and B and results from C would have the type A x B -> C but in functional languages like Haskell we are using A -> (B -> C), i.e. a function that returns a function. This “currying” is exactly a the categorical definition of a cartesian closed category as one where Hom(AxB,C) is isomorphic to Hom(A,B -> C) and in this false you can replace Hom(X,Y) with X -> Y.

It is well known that effects in functional programming can be modelled by monads which is a concept from category theory. Nowadays a weaker structure called applicative functors has become very popular – needless to say also a concept from Category Theory.

Not all languages are functional (yet) but…

View original post 24 more words

Note on Finitely Generated Abelian Groups

We state and prove a sufficient condition for finitely generated Abelian Groups to be the direct product of its generators, and state a counterexample to the conclusion when the condition is not satisfied.

Theorem

Let G be an abelian group and G=\langle g_1,\dots, g_n\rangle.

Suppose the generators g_1,\dots,g_n are linearly independent over \mathbb{Z}, that is, whenever c_1g_1+\dots+c_ng_n=0 for some integers c_i\in\mathbb{Z}, we have c_1=\dots=c_n=0.

(Here we are using additive notation for (G,+), where the identity of G is written as 0, the inverse of g is written as -g).

Then \displaystyle G\cong\langle g_1\rangle\times\dots\times\langle g_n\rangle.

Proof

Define the following map \psi:\langle g_1\rangle\times\dots\times\langle g_n\rangle\to\langle g_1,\dots,g_n\rangle by \displaystyle \psi((c_1g_1,\dots,c_ng_n))=c_1g_1+\dots+c_ng_n.

We can check that \psi is a group homomorphism.

We have that \psi is surjective since any element x\in\langle g_1,\dots,g_n\rangle is by definition a combination of finitely many elements of the generating set and their inverses. Since G is abelian, x=c_1g_1+\dots+c_ng_n for some c_i\in\mathbb{Z}.

Also, \psi is injective since if c_1g_1+\dots+c_ng_n=0, then all the coefficients c_i are zero (by the linear independence condition). Thus \ker\psi is trivial.

Hence \psi is an isomorphism.

Remark

Note that without the linear independence condition, the conclusion may not be true. Consider G=\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5 which is abelian with order 30. Consider g_1=(1,1,0), g_2=(0,1,1).

We can see that G=\langle g_1,g_2\rangle, by observing that 3g_1=(1,0,0), 4g_1=(0,1,0), 2g_1+g_2=(0,0,1). However \langle g_1\rangle\times\langle g_2\rangle=\mathbb{Z}_6\times\mathbb{Z}_{15} has order 90. Thus \langle g_1,g_2\rangle\not\cong\langle g_1\rangle\times\langle g_2\rangle.

Locally Lipschitz implies Lipschitz on Compact Set Proof

Assume \phi is locally Lipschitz on \mathbb{R}^n, that is, for any x\in \mathbb{R}^n, there exists \delta, L>0 (depending on x) such that |\phi(z)-\phi(y)|\leq L|z-y| for all z,y\in B_\delta(x)=\{t\in\mathbb{R}^n: |x-t|<\delta\}.

Then, for any compact set K\subset\mathbb{R}^n, there exists a constant M>0 (depending on K) such that |\phi(x)-\phi(y)|\leq M|x-y| for all x,y\in K. That is, \phi is Lipschitz on K.

Proof

Suppose to the contrary \phi is not Lipschitz on K, so that for all M>0, there exists x,y\in K such that \displaystyle \frac{|\phi(x)-\phi(y)|}{|x-y|}>M.

Then there exists two sequences x_n, y_n\in K such that \displaystyle \frac{|\phi(x_n)-\phi(y_n)|}{|x_n-y_n|}\to\infty.

Since \phi is locally Lipschitz implies \phi is continuous, so \phi is bounded on K by Extreme Value Theorem. Hence |x_n-y_n|\to 0.

By sequential compactness of K, there exists a convergent subsequence x_{n_k}\to x, and thus y_{n_k}\to x.

Then for any L>0, there exists k such that x_{n_k},y_{n_k}\in B_\delta(x) but \displaystyle \frac{|\phi(x_{n_k})-\phi(y_{n_k})|}{|x_{n_k}-y_{n_k}|}>L which contradicts that \phi is locally Lipschitz.

Gauss Lemma Proof

There are two related results that are commonly called “Gauss Lemma”. The first is that the product of primitive polynomial is still primitive. The second result is that a primitive polynomial is irreducible over a UFD (Unique Factorization Domain) D, if and only if it is irreducible over its quotient field.

Gauss Lemma: Product of primitive polynomials is primitive

If D is a unique factorization domain and f,g\in D[x], then C(fg)=C(f)C(g). In particular, the product of primitive polynomials is primitive.

Proof

(Hungerford pg 163)

Write f=C(f)f_1 and g=C(g)g_1 with f_1, g_1 primitive. Consequently \displaystyle C(fg)=C(C(f)f_1C(g)g_1)\sim C(f)C(g)C(f_1g_1).

Hence it suffices to prove that f_1g_1 is primitive, that is, C(f_1g_1) is a unit. If f_1=\sum_{i=0}^n a_ix^i and g_1=\sum_{j=0}^m b_jx^j, then f_1g_1=\sum_{k=0}^{m+n}c_kx^k with c_k=\sum_{i+j=k}a_ib_j.

If f_1g_1 is not primitive, then there exists an irreducible element p in D such that p\mid c_k for all k. Since C(f_1) is a unit p\nmid C(f_1), hence there is a least integer s such that \displaystyle p\mid a_i\ \text{for}\ i<s\ \text{and}\ p\nmid a_s.

Similarly there is a least integer t such that \displaystyle p\mid b_j\ \text{for}\ j<t\ \text{and}\ p\nmid b_t.

Since p divides \displaystyle c_{s+t}=a_0b_{s+t}+\dots+a_{s-1}b_{t+1}+a_sb_t+a_{s+1}b_{t-1}+\dots+a_{s-t}b_0, p must divide a_sb_t. Since every irreducible element in D (UFD) is prime, p\mid a_s or p\mid b_t. This is a contradiction. Therefore f_1g_1 is primitive.

Primitive polynomials are associates in D[x] iff they are associates in F[x]

Let D be a unique factorization domain with quotient field F and let f and g be primitive polynomials in D[x]. Then f and g are associates in D[x] if and only if they are associates in F[x].

Proof

(\impliedby) If f and g are associates in the integral domain F[x], then f=gu for some unit u\in F[x]. Since the units in F[x] are nonzero constants, so u\in F, hence u=b/c with b,c\in D and c\neq 0. Thus cf=bg.

Since C(f) and C(g) are units in D, \displaystyle c\sim cC(f)\sim C(cf)=C(bg)\sim bC(g)\sim b.

Therefore, b=cv for some unit v\in D and cf=bg=vcg. Consequently, f=vg (since c\neq 0), hence f and g are associates in D[x].

(\implies) Clear, since if f=gu for some u\in D[x]\subseteq F[x], then f and g are associates in F[x].

Primitive f is irreducible in D[x] iff f is irreducible in F[x]

Let D be a UFD with quotient field F and f a primitive polynomial of positive degree in D[x]. Then f is irreducible in D[x] if and only if f is irreducible in F[x].

Proof

(\implies) Suppose f is irreducible in D[x] and f=gh with g,h\in F[x] and \deg g\geq 1, \deg h\geq 1. Then g=\sum_{i=0}^n(a_i/b_i)x^i and h=\sum_{j=0}^m(c_j/d_j)x^j with a_i, b_i, c_j, d_j\in D and b_i\neq 0, d_j\neq 0.

Let b=b_0b_1\dots b_n and for each i let \displaystyle b_i^*=b_0b_1\dots b_{i-1}b_{i+1}\dots b_n. If g_1=\sum_{i=0}^n a_ib_i^* x^i\in D[x] (clear denominators of g by multiplying by product of denominators), then g_1=ag_2 with a=C(g_1), g_2\in D[x] and g_2 primitive.

Verify that g=(1_D/b)g_1=(a/b)g_2 and \deg g=\deg g_2. Similarly h=(c/d)h_2 with c,d\in D, h_2\in D[x], h_2 primitive and \deg h=\deg h_2. Consequently, f=gh=(a/b)(c/d)g_2h_2, hence bdf=acg_2h_2. Since f is primitive by hypothesis and g_2h_2 is primitive by Gauss Lemma, \displaystyle bd\sim bdC(f)\sim C(bdf)=C(acg_2h_2)\sim acC(g_2h_2)\sim ac.

This means bd and ac are associates in D. Thus ubd=ac for some unit u\in D. So f=ug_2h_2, hence f and g_2h_2 are associates in D[x]. Consequently f is reducible in D[x] (since f=ug_2h_2), which is a contradiction. Therefore, f is irreducible in F[x].

(\impliedby) Conversely if f is irreducible in F[x] and f=gh with g,h\in D[x], then one of g, h (say g) is a unit in F[x] and thus a (nonzero) constant. Thus C(f)=gC(h). Since f is primitive, g must be a unit in D and hence in D[x]. Thus f is irreducible in D[x].

Animation: Linear Algebra 

ChefCouscous's avatarMath Online Tom Circle

Abstract Vector Spaces ​向量空间

Eigenvalues & Eigenvectors (valeurs propres et vecteurs propres) 特征值/特征向量

[ Note: “Eigen-” is German for Characteristic 特征.]

The Essence of Determinant (*): (行列式)

(*) Determinant was invented by the ancient Chinese Algebraists 李冶 / 朱世杰 /秦九韶 in 13th century (金 / 南宋 / 元) in《天元术》.The Japanese “和算” mathematician 关孝和 spread it further to Europe before the German mathematician Leibniz named it the “Determinant” in 18th century.

[NOTE] 金庸 武侠小说 《神雕侠女》里 元朝初年的 黄蓉 破解 大理国王妃 瑛姑 苦思不解的 “行列式”, 大概是求 eigenvalues & eigenvectors ? 🙂

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Cours Diagonalisation 1/2. Cours Maths spé

ChefCouscous's avatarMath Online Tom Circle

Maths Spéciales (2nd Year French Engineering University Math ”Linear Algebra”): The French way of treating Matrices is very general and abstract. Advantage is it studies Matrices at a theoretical high-level, disadvantage is it ignores on applications.

This young French prof explained in 30 mins at great length of what is simply the Characteristic Polynomial Equation:

$latex boxed {det (A – lambda.I) = 0 }&fg=aa0000&s=3$

Compare it with the more practical (but less theoretical – 知其然而不知其所以然) American teaching below from the famous MIT Prof Gilbert Strang for the same Diagonalization of Matrices:

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Free Math Notes by AMS

Just learnt from Professor Terence Tao’s blog that there is a new series of free math notes by the American Mathematical Society: http://www.ams.org/open-math-notes.

Many of the notes there are of exceptionally high quality (check out “A singular mathematical promenade”, by Étienne Ghys).

Welcome to AMS Open Math Notes, a repository of freely downloadable mathematical works in progress hosted by the American Mathematical Society as a service to researchers, teachers and students.

These draft works include course notes, textbooks, and research expositions in progress. They have not been published elsewhere, and, as works in progress, are subject to significant revision.

Visitors are encouraged to download and use these materials as teaching and research aids, and to send constructive comments and suggestions to the authors.

Non-trivial submodules of direct sum of simple modules

Suppose M_1 and M_2 are two non-isomorphic simple, nonzero R-modules.

Determine all non-trivial submodules of M_1\oplus M_2.

Let N be a non-trivial submodule of M_1\oplus M_2. Note that \displaystyle \{0\}\subset M_1\subset M_1\oplus M_2 is a composition series. By Jordan-Holder theorem, all composition series are equivalent and have the same length. Hence \displaystyle \{0\}\subset N\subset M_1\oplus M_2 must be a composition series too.

Thus N\cong M_1 or M_2. In particular N is simple.

Let \pi_1: M_1\oplus M_2\to M_1 and \pi_2:M_1\oplus M_2\to M_2 be the canonical projections. Note that \pi_1(N) is a submodule of M_1, so \pi_1(N)\cong 0 or M_1. Similarly, \pi_2(N)\cong 0 or M_2.

By Schur’s Lemma \pi_1|_N: N\to \pi_1(N) and \pi_2|_N: N\to\pi_2(N) are either 0 or isomorphisms.

They cannot be both zero since N is non-zero. They cannot be both isomorphisms either, as that would imply M_1\cong\pi_1(N)\cong\pi_2(N)\cong M_2.

Hence, exactly one of \pi_1, \pi_2 are zero. So N=M_1\oplus\{0\} or \{0\}\oplus M_2.

Recommended Educational Toys

Does your child complain that science is “boring”? This may be because science is often taught in a boring manner. The solution may be to supplement teaching with hands-on experiments that develop the inner curiosity of the child.

For parents looking to buy an educational toy for their child, here are two recommendations:

Educational Insights GeoSafari Micropro 48-Piece Microscope Set

For its price, it is one of the most affordable microscopes around. Suitable as a starter microscope for children interested in doing experiments. Suitable for upper primary (Grade 5/6) onwards.

For more serious/experienced students, they can consider AmScope B120C-E1 Siedentopf Binocular Compound Microscope, 40X-2500X Magnification, LED Illumination, Abbe Condenser, Two-Layer Mechanical Stage, 1.3MP Camera and Software Windows XP/Vista/7/8/10 which is probably even better than the microscope in your secondary school / junior college. It can be connected to the computer for deeper analysis.

Snap Circuits Jr. SC-100 Electronics Discovery Kit

Electricity is one of the greatest inventions in the past century, and also a key component of the science syllabus from primary all the way to university. Learn more about circuits in this amazing toy. Suitable for lower primary onwards (Grade 1-3).

Kids with tuition fare worse?

Article: http://www.straitstimes.com/opinion/kids-with-tuition-fare-worse

Those who read the news, either online or in print, would probably have seen this article: “Kids with tuition fare worse”.

In the article, it is claimed that: “In fact, children who received tuition actually scored about 0.256 standard deviations lower on their tests than those who did not (standard deviation is a measure of how spread out test scores are from the average).”

The headline is actually quite misleading, causing people to think that tuition causes worse performance. One needs to read the final part of the article: “The first is that students who receive tuition choose to receive it precisely because they are not doing well in school. In other words, weak performance may be what is driving students to enrol for tuition.”

The correct way to measure the effect of tuition is via a “before and after” experiment. Scores of students before and after enrolling in tuition should be compared to truly see if tuition has any effect. Many tuition centers are already doing this, it is not a rocket science experiment.

Without the “before and after” comparison, the research is meaningless. It is like saying, “People who see a medical doctor frequently have poorer health.”, it is true, but obviously one cannot conclude that medical doctors cause poor health!

Lastly, the research is analysing PISA data (Programme for International Student Assessment). Clearly, there is no tuition centre tutoring PISA, which is significantly different from the ordinary curriculum (I was a PISA grader). As tuition is highly specialized, it is true that tuition can have close to zero effect on PISA scores. It is like PSLE / O Level Math tuition has close to no effect on Math Olympiad scores; even if it is both “Math”, it is possible to score full marks in PSLE / O Level Math but zero marks in Olympiad Math!

Commutator subgroup G’ is the unique smallest normal subgroup N such that G/N is abelian.

Commutator subgroup G' is the unique smallest normal subgroup N such that G/N is abelian.

If G is a group, then G' is a normal subgroup of G and G/G' is abelian. If N is a normal subgroup of G, then G/N is abelian iff N contains G'.

Proof

Let f:G\to G be any automorphism. Then \displaystyle f(aba^{-1}b^{-1})=f(a)f(b)f(a)^{-1}f(b)^{-1}\in G'.

It follows that f(G')\leq G'. In particular, if f is the automorphism given by conjugation by a\in G, then aG'a^{-1}=f(G')\leq G', so G'\unlhd G.

Since (ab)(ba)^{-1}=aba^{-1}b^{-1}\in G', abG'=baG' and hence G/G' is abelian.

(\implies) If G/N is abelian, then abN=baN for all a,b\in G. Hence ab(ba)^{-1}=aba^{-1}b^{-1}\in N. Therefore, N contains all commutators and G'\leq N.

(\impliedby) If G'\subseteq N, then ab(ba)^{-1}=aba^{-1}b^{-1}\in G'\subseteq N. Thus abN=baN for all a,b\in G. Hence G/N is abelian.

Ascending Central Series and Nilpotent Groups

Ascending Central Series of G

Let G be a group. The center C(G) of G is a normal subgroup. Let C_2(G) be the inverse image of C(G/C(G)) under the canonical projection G\to G/C(G). By Correspondence Theorem, C_2(G) is normal in G and contains C(G).

Continue this process by defining inductively: C_1(G)=C(G) and C_i(G) is the inverse image of C(G/C_{i-1}(G)) under the canonical projection G\to G/C_{i-1}(G).

Thus we obtain a sequence of normal subgroups of G, called the ascending central series of G: \displaystyle \langle e\rangle<C_1(G)<C_2(G)<\cdots.

Nilpotent Group

A group G is nilpotent if C_n(G)=G for some n.

Abelian Group is Nilpotent

Every abelian group G is nilpotent since G=C(G)=C_1(G).

Every finite p-group is nilpotent (Proof)

G and all its nontrivial quotients are p-groups, and therefore have non-trivial centers.

Hence if G\neq C_i(G), then G/C_i(G) is a p-group, and C(G/C_i(G)) is non-trivial. Thus C_{i+1}(G), the inverse image of C(G/C_i(G)) under \pi:G\to G/C_i(G), strictly contains C_i(G).

Since G is finite, C_n(G) must be G for some n.

Singaporean Student Wins US$250K Scholarship

Congratulations to Ms See for the win. The amount of the scholarship is no joke, it is the price of a HDB flat (a home in Singapore).

If you are interested to take part, here are the details: The Breakthrough Junior Challenge is an annual competition for students, ages 13-18, to share their passion for math and science with the world! In partnership with the Khan Academy, each student submits a video that explains a challenging and important concept or theory in mathematics, life sciences, or physics. The winner receives a $250,000 college scholarship. The winning student’s teacher and school also benefit: $50,000 for the teacher and a state-of-the-art $100,000 science lab for the school. Learn more at https://breakthroughjuniorchallenge.org.

AsianScientist (Dec. 6, 2016) – Ms Deanna See, a 17-year-old studying at Raffles Institution in Singapore, was one of two students to win the Breakthrough Junior Challenge, a global science video competition that is part of the 2017 Breakthrough Prizes. See’s fun and quirky five-minute video, titled ‘Superbugs! And Our Race Against Resistance,’ explains the phenomenon of antibiotic resistance with the use of Lego figurines. She takes the viewer through the evolution and genetics behind how superbugs amass their formidable defenses against various antibiotics. Watch See’s video below:

Read more from Asian Scientist Magazine at: http://www.asianscientist.com/2016/12/topnews/deanna-see-breakthrough-junior-challenge-singapore/

The other finalist, Antonella Masini from Peru