The Shape of Space

Just came across this book: The Shape of Space (Chapman & Hall/CRC Pure and Applied Mathematics). It is a very unique book, in the sense that it is aimed at high school students, but even a undergraduate or graduate student can benefit from it. It has a lot of diagrams, that are missing in most textbooks, presumably because it takes a lot of effort to draw a mathematical (3D) diagram.

It will be useful to students who want to learn more about topology. This book can be read casually, it is not like a textbook, yet it has substantial mathematical content.

Example of an illustration in the book:

illustration topology

CW Approximation

A weak homotopy equivalence is a map $f:X\to Y$ that induces isomorphisms $\pi_n(X,x_0)\to\pi_n(Y,f(x_o))$ for all $n\geq 0$ and all choices of basepoint $x_0$ .

In other words, Whitehead’s theorem says that a weak homotopy equivalence between CW complexes is a homotopy equivalence. Just to recap, a map $f:X\to Y$ is said to be a homotopy equivalence if there exists a map $g:Y\to X$ such that $fg\cong id_Y$ and $gf\cong id_X$ . The spaces $X$ and $Y$ are called homotopy equivalent.

It turns out that for any space $X$ there exists a CW complex $Z$ and a weak homotopy equivalence $f:Z\to X$ . This map $f:Z\to X$ is called a CW approximation to $X$ .

Excision for Homotopy Groups

According to Hatcher (Chapter 4.2), the main difficulty of computing homotopy groups (versus homology groups) is the failure of the excision property. However, under certain conditions, excision does hold for homotopy groups:

Theorem (4.23): Let $X$ be a CW complex decomposed as the union of subcomplexes $A$ and $B$ with nonempty connected intersection $C=A\cap B$ . If $(A,C)$ is m-connected and $(B,C)$ is n-connected, $m,n\geq 0$ , then the map $\pi_i(A,C)\to\pi_i(X,B)$ induced by inclusion is an isomorphism for $i<m+n$ and a surjection for $i=m+n$ .

Miscellaneous Definitions

Suspension: Let $X$ be a space. The suspension $SX$ is the quotient of $X\times I$ obtained by collapsing $X\times\{0\}$ to one point and $X\times\{1\}$ to another point.

The definition of suspension is similar to that of the cone in the following way. The cone $CX$ is the union of all line segments joining points of $X$ to one external vertex. The suspension $SX$ is the union of all line segments joining points of $X$ to two external vertices.

The classical example is $X=S^n$ , when $SX=S^{n+1}$ with the two “suspension points” at the north and south poles of $S^{n+1}$ , the points $(0,\dots,0,\pm 1)$ .

Here are some graphical sketches of the case where $X$ is the 0-sphere and the 1 sphere respectively.

Linear Fractional Transformation (Mobius Transformation)

The transformation $w=\frac{az+b}{cz+d}$ , with $ad-bc\neq 0$ , and $a,b,c,d$ are complex constants, is called a linear fractional transformation, or Mobius transformation.

One key property of linear fractional transformations is that it transforms circles and lines into circles and lines.

Let us find the linear fractional transformation that maps the points $z_1=2$ , $z_2=i$ , $z_3=-2$ onto the points $w_1=1$ , $w_2=i$ , $w_3=-1$ . (Question taken from Complex Variables and Applications (Brown and Churchill))

Solution: $w=\frac{3z+2i}{iz+6}$

What we have to do is basically solve the three simultaneous equations arising from $w=\frac{az+b}{cz+d}$ , namely $1=\frac{2a+b}{2c+d}$ , $i=\frac{ia+b}{ic+d}$ and $-1=\frac{-2a+b}{-2c+d}$ .

Eventually we can have all the variables in terms of $c$ : $a=-3ic$ , $b=2c$ , $d=-6ic$ . Substituting back into the Mobius Transformation gives us the answer.

Donate to Singapore Charity @ Giving.sg

URL: https://www.giving.sg/

Just to introduce this website to readers who haven’t heard of it. Donations above $50 are tax deductible, and it also features ways to volunteer for the charity organisations. Do check it out!

Q: What is Giving.sg? (taken from their FAQ page)

A: Giving.sg is Singapore’s very own one-stop portal for empowering all of us on our Giving Journey, whether we are looking to help local non-profit organisations (NPOs) by giving our TIME, by general volunteering; our TALENT, by skills volunteering; or [our] TREASURE, by donations.

Giving.sg brings together its predecessors sggives.org and sgcares.org, both of which have helped raise over S$51 million for more than 350 [local] non-profits, as well as seen over 40,000 volunteers generously gift their time to these [local] non-profits.

News article featuring Giving.sg: http://www.straitstimes.com/singapore/online-platform-among-initiatives-to-promote-giving-in-singapore

Analytic Isomorphism (Complex Analysis)

An analytic isomorphism $f:U\to V$ (where $U$ and $V$ are open sets) is an analytic function such that there exists another analytic function $g:V\to U$ , satisfying $f\circ g=id_V$ and $g\circ f=id_U$ .

Interesting Blog Post on Mathematical Conversations

Source: http://www.theliberatedmathematician.com/2015/12/why-i-do-not-talk-about-math/

A honest opinion on the nature of mathematical conversations, by this blog post author Piper Harron. (Also see our previous blog post on her interesting PhD Thesis) Very interesting read, for those who are in the mathematical community.

Secondary Chinese Tuition

For those interested in Secondary Level (O Level) Chinese Tuition, do check out http://chinesetuition88.com/.

The tutor (Ms Gao) will be taking in a few more students in 2016 for Chinese Tuition. Do check out http://chinesetuition88.com/ for more info!

(1/2)! = (√π)/2

Math Online Tom Circle

Richard Feynman (Nobel Physicist) proved it in high school using a funny Calculus: “Differentiating under Integral” — is it legitimate to do so ? Of course it is by “The Fundamental Theorem of Calculus”

Note: We were thought in high school the “HOW” of calculating (such as integration and differentiation), but not the “WHY” (the Theorem behind). Richard Feynman was unique in exploring the WHY since high school, it helped later he was assigned by President Reagan to investigate the 1986 ‘Challenger’ disaster ?

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Cellular Approximation for Pairs

(This is Example 4.11 in Hatcher’s book).

Cellular Approximation for Pairs: Every map $f:(X,A)\to (Y,B)$ of CW pairs can be deformed through maps $(X,A)\to (Y,B)$ to a cellular map $g:(X,A)\to (Y,B)$ .

What “map of CW pairs” mean, is that $f$ is a map from $X$ to $Y$ , and the image of $A\subseteq X$ under $f$ is contained in $B$ . CW pair $(X,A)$ means that $X$ is a cell complex, and $A$ is a subcomplex.

First, we use the ordinary Cellular Approximation Theorem to deform the restriction $f:A\to B$ to be cellular. We then use the Homotopy Extension Property to extend this to a homotopy of $f$ on all of $X$ . Then, use Cellular Approximation Theorem again to deform the resulting map to be cellular staying stationary on $A$ .

We use this to prove a corollary: A CW pair $(X,A)$ is n-connected if all the cells in $X-A$ have dimension greater than $n$ . In particular the pair $(X,X^n)$ is n-connected, hence the inclusion $X^n\hookrightarrow X$ induces isomorphisms on $\pi_i$ for $i<n$ and a surjection on $\pi_n$ .

First we note that being n-connected means that the space is non-empty, path-connected, and the first n homotopy groups are trivial, i.e. $\pi_i(X)\cong 0$ for $1\leq i\leq n$ .

Proof: First, we apply cellular approximation to maps $(D^i,\partial D^i)\to (X,A)$ with $i\leq n$ , thus the map is homotopic to a cellular map of pairs $g$ . Since all the cells in $X-A$ have dimension greater than $n$ , the n-skeleton of $X$ must be inside $A$ . Therefore $g$ is homotopic to a map whose image is in $A$ , and thus it is 0 in the relative homotopy group $\pi_i(X,A)$ . This proves that the CW pair $(X,A)$ is n-connected. Note that 0-connected means path-connected.

Consider the long exact sequence of the pair $(X,X^n)$ :

$\dots\to\pi_n(X^n,x_0)\xrightarrow{i_*}\pi_n(X,x_0)\xrightarrow{j_*}\pi_n(X,X^n,x_0)\xrightarrow{\partial}\pi_{n-1}(X^n,x_0)\to\dots\to\pi_0(X,x_0)$

Since it is an exact sequence, the image of any map equals the kernel of the next. Thus, $\text{Im}(i_*)=\ker j_*=\pi_n(X,x_0)$ (since $\pi_n(X,X^n,x_0)=0$ ). Thus $i_*$ is surjective. Since $\pi_n(X,X^n,x_0)=0$ , the later terms in the long exact sequence are also 0, thus, the inclusion $X^n\hookrightarrow X$ induces isomorphisms on $\pi_i$ for $i<n$ , since the first n homotopy groups all vanish.

Piper Harron discusses her artistic and wonderful math Ph.D. thesis

This is the most “unique” PhD thesis I have ever seen. Very special, and humorous to read, and coming from the most elite institution Princeton, under the guidance of Fields Medalist Manjul Bhargava.

mathbabe

Piper Harron is a mathematician who is very happy to be here, and yes, is having a great time, despite the fact that she is standing alone awkwardly by the food table hoping nobody will talk to her.

Piper, would you care to write a mathbabe post describing your thesis, and yourself, and anything else you’d care to mention?

When Cathy (Cathy? mathbabe?) asked if I would like to write a mathbabe post describing my thesis, and myself, and anything else I’d care to mention, I said “sure!” because that is objectively the right answer. I then immediately plunged into despair.

Describe my thesis? My thesis is this thing that was initially going to be a grenade launched at my ex-prison, for better or for worse, and instead turned into some kind of positive seed bomb where flowers have sprouted beside the foundations I thought I wanted to crumble…

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2015 in review

The WordPress.com stats helper monkeys prepared a 2015 annual report for this blog.

Here’s an excerpt:

The Louvre Museum has 8.5 million visitors per year. This blog was viewed about 170,000 times in 2015. If it were an exhibit at the Louvre Museum, it would take about 7 days for that many people to see it.

Click here to see the complete report.

Rouche’s Theorem and Applications

This blog post is on Rouche’s Theorem and some applications, namely counting the number of zeroes in an annulus, and the fundamental theorem of algebra.

Rouche’s Theorem: Let $f(z)$ , $g(z)$ be holomorphic inside and on a simple closed contour $K$ , such that $|g(z)|<|f(z)|$ on $K$ . Then $f$ and $f+g$ have the same number of zeroes (counting multiplicities) inside $K$ .

Rouche’s Theorem is useful for scenarios like this: Determine the number of zeroes, counting multiplicities, of the polynomial $f(z)=2z^5-6z^2-z+1=0$ in the annulus $1\leq |z|\leq 2$ .

Solution:

Let $K_1$ be the unit circle $|z|=1$ . We have

$\begin{aligned}|2z^5-z+1|&\leq |2z^5|+|z|+|1|\\ &=2+1+1\\ &=4\\ &<6\\ &=|-6z^2| \end{aligned}$

on $K_1$ .

Since $-6z^2$ has 2 zeroes in $K_1$ , therefore $f$ has 2 zeroes inside $K_1$ , by Rouche’s Theorem.

Let $K_2$ be the circle $|z|=2$

$\begin{aligned} |-6z^2-z+1|&\leq |-6z^2|+|-z|+|1|\\ &=6(2^2)+2+1\\ &=27\\ &<64\\ &=|2z^5| \end{aligned}$

on $K_2$ . Therefore $f$ has 5 zeroes inside $K_2$ .

Therefore $f$ has 5-2=3 zeroes inside the annulus.

We do a computer check using Wolfram Alpha (http://www.wolframalpha.com/input/?i=2z%5E5-6z%5E2-z%2B1%3D0). The moduli of the five roots are (to 3 significant figures): 0.489, 0.335, 1.46, 1.45, 1.45. This confirms that 3 of the zeroes are in the given annulus.

Fundamental Theorem of Algebra Using Rouche’s Theorem

Rouche’s Theorem provides a rather short proof of the Fundamental Theorem of Algebra: Every degree n polynomial with complex coefficients has exactly n roots, counting multiplicities.

Proof: Let $f(z)=a_0+a_1z+a_2z^2+\dots+a_nz^n$ . Chose $R\gg 1$ sufficiently large so that on the circle $|z|=R$ ,

$\begin{aligned} |a_0+a_1z+a_2z^2+\dots+a_{n-1}z^{n-1}|&\leq|a_0|+|a_1|R+|a_2|R^2+\dots+|a_{n-1}|R^{n-1}\\ &<(\sum_{i=0}^{n-1}|a_i|)R^{n-1}\\ &<|a_n|R^n\\ &=|a_nz^n| \end{aligned}$

Since $a_nz^n$ has $n$ roots inside the circle, $f$ also has $n$ roots in the circle, by Rouche’s Theorem. Since $R$ can be arbitrarily large, this proves the Fundamental Theorem of Algebra.

Star Wars Ball Droid: How is the head attached to the body?

Just watched Star Wars: The Force Awakens, here is my review on it. Overall a good movie, enjoyed watching it. The storyline and lightsaber duels are a bit weak in my opinion. How Rey, an untrained person holding a lightsaber for the first time, managed to defeat Kylo Ren with his crossguard lightsaber remains a mystery to me. My favorite episode remains Episode 1: The Phantom Menace.

Many mysteries remain unanswered, like the identities of Rey and Snoke. Looking forward to the next episode.

Something I find very interesting is the Ball Droid BB-8. Something even more interesting about the droid is that it is not CGI effects, it is a real prop. How the head of BB-8 is being attached to the body seems to be via strong magnets.

The toy-version of BB-8 is being sold on Amazon, a possible gift idea for those who are Star Wars fans. Sphero BB-8 App-Enabled Droid

Culture, Intellect & Attitude

Math Online Tom Circle

Culture, Intellect & Attitude Most Important
Let f: {A,B…Y,Z} -> {1,2… 25,26}
f(AB}=f(A)+f(B)=1+2=3
=> f homomorphism

f(LUCK)= f(L)+f(U)+f(C)+f(K)= 12+21+3+11 =47
f(LOVE)=54
f(KNOWLEDGE)=96
f(LEADERSHIP)=97
f(HARDWORK)=98

f(CULTURE)=100
f(INTELLECT)=100
f(ATTITUDE)=100
=> Last 3 most important in life!

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Galois Group (Example)

This post is about the Galois group of $K$ over $\mathbb{Q}$ , where $K$ is the splitting field of $f(x)=x^p-2$ , where $p$ is an odd prime.

First we show that the polynomial $f(x)=x^p-2$ is irreducible over $\mathbb{Q}$ . This follows immediately by Eisenstein’s Criterion, since $2\mid (-2)$ , $2\nmid 1$ and $2^2\nmid (-2)$ .

Next, we show that the splitting field $K$ of $f(x)$ in $\mathbb{C}$ is $Q(\sqrt[p]{2},\omega)$ , where $\omega=e^{2\pi i/p}$ is a primitive p-th root of unity. The roots of $f$ are $\sqrt[p]2, \sqrt[p]2\omega, \sqrt[p]2\omega^2, \dots, \sqrt[p]2\omega^{p-1}$ .

The splitting field $K$ contains $\sqrt[p]2$ and $\omega=\frac{\sqrt[p]2\omega^2}{\sqrt[p]2\omega}$ . Thus $\mathbb{Q}(\sqrt[p]2,\omega)\subseteq K$ .

On the other hand, $\mathbb{Q}(\sqrt[p]2,\omega)$ contains all the roots of $f$ , hence $f$ splits in $\mathbb{Q}(\sqrt[p]2, \omega)$ . Thus $K\subseteq\mathbb{Q}(\sqrt[p]2,\omega)$ , since $K$ is the smallest field that contains $\mathbb{Q}$ and all the roots of $f$ . All in all, we have that the splitting field $K=\mathbb{Q}(\sqrt[p]2, \omega)$ .

The next part involves determining the Galois group of $K$ over $\mathbb{Q}$ . We have $|Gal(K/\mathbb{Q})|=[K:\mathbb{Q}]$ . Since $[\mathbb{Q}(\sqrt[p]2):\mathbb{Q}]=p$ (minimal polynomial $x^p-2$ ), and $[\mathbb{Q}(\omega):\mathbb{Q}]=p-1$ (minimal polynomial the cyclotomic polynomial $1+x+x^2+\dots+x^{p-1}$ ), thus $|Gal(K/\mathbb{Q})|=p(p-1)$ . Here we have used the lemma that suppose $[F(\alpha):F]=m$ and $[F(\beta):F]=n$ with $\gcd(m,n)=1$ , then $[F(\alpha,\beta):F]=mn$ .

What the Galois group does is it permutes the roots of $f$ . Let $\sigma$ be an element of the Galois group. $\sigma(\sqrt[p]2)$ can possibly be $\sqrt[p]2, \sqrt[p]\omega, \dots, \sqrt[p]2\omega^{p-1}$ , a total of $p$ choices. Similarly, $\sigma(\omega)=\omega, \omega^2, \dots,\omega^{p-1}$ , a total of $p-1$ choices. All these total up to $p(p-1)$ elements, which is exactly the size of the Galois group.

The above Galois group $Gal(K/\mathbb{Q})$ is described by how its elements act on the generators. For a more concrete representation, we can actually prove that the Galois group above is isomorphic to the group of matrices $\begin{pmatrix}a&b\\0&1\end{pmatrix}$ , where $a,b\in\mathbb{F}_p$ , $a\neq 0$ . We denote the group of matrices as $M$ .

To show the isomorphism, we define a map $\phi: Gal(K/\mathbb{Q})\to M$ , mapping $\sigma_{a,b}$ to $\begin{pmatrix}a&b\\0&1\end{pmatrix}$ .

Notation: $\sigma_{a,b}$ is defined on the generators as follows, $\sigma_{a,b}(\sqrt[p]2)=\sqrt[p]2\omega^b$ , $\sigma_{a,b}(\omega)=\omega^a$ .

We can clearly see that the map $\phi$ is bijective. To see it is a homomorphism, we compute $\phi(\sigma_{a,b}\circ\sigma_{c,d})=\begin{pmatrix}ac&a+bd\\0&1\end{pmatrix}=\phi(\sigma_{a,b})\phi(\sigma_{c,d})$ .

SAT Lessons free on Khan Academy

Math Online Tom Circle

You need to take SAT during JC2 year to apply for USA universities, a good SAT score is more important than GCE A-level.

https://www.khanacademy.org/sat

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Cellular Approximation Theorem and Homotopy Groups of Spheres

First we will state another theorem, Whitehead’s Theorem: If a map $f:X\to Y$ between connected CW complexes induces isomorphisms $f_*:\pi_n(X)\to\pi_n(Y)$ for all $n$ , then $f$ is a homotopy equivalence. If $f$ is the inclusion of a subcomplex $X\to Y$ , we have an even stronger conclusion: $X$ is a deformation retract of $Y$ .

The main theorem discussed in this post is the Cellular Approximation Theorem: Every map $f:X\to Y$ of CW complexes is homotopic to a cellular map. If $f$ is already cellular on a subcomplex $A\subset X$ , the homotopy may be taken to be stationary on $A$ . This theorem can be viewed as the CW complex analogue of the Simplicial Approximation Theorem.

Corollary: If $n<k$ , then $\pi_n(S^k)=0$ .

Proof: Consider $S^n$ and $S^k$ with their canonical CW-structure, with one 0-cell each, and with one n-cell for $S^n$ and one k-cell for $S^k$ . Let $[f]\in\pi_n(S^k)$ , where $f:S^n\to S^k$ is a base-point preserving map. By the Cellular Approximation Theorem, $f$ is homotopic to a cellular map $g$ , where cells map to cells of same or lower dimension.

Since $n<k$ , the n-cell $S^n$ can only map to the 0-cell in $S^k$ . The 0-cell in $S^n$ (the basepoint) is also mapped to the 0-cell in $S^k$ . Thus $g$ is the constant map, hence $\pi_n(S^k)=0$ .

Arzela-Ascoli Theorem and Applications

The Arzela-Ascoli Theorem is a rather formidable-sounding theorem that gives a necessary and sufficient condition for a sequence of real-valued continuous functions on a closed and bounded interval to have a uniformly convergent subsequence.

Statement: Let $(f_n)$ be a uniformly bounded and equicontinuous sequence of real-valued continuous functions defined on a closed and bounded interval $[a,b]$ . Then there exists a subsequence $(f_{n_k})$ that converges uniformly.

The converse of the Arzela-Ascoli Theorem is also true, in the sense that if every subsequence of $(f_n)$ has a uniformly convergent subsequence, then $(f_n)$ is uniformly bounded and equicontinuous.

Explanation of terms used: A sequence $(f_n)$ of functions on $[a,b]$ is uniformly bounded if there is a number $M$ such that $|f_n(x)|\leq M$ for all $f_n$ and all $x\in [a,b]$ . The sequence is equicontinous if, for all $\epsilon>0$ , there exists $\delta>0$ such that $|f_n(x)-f_n(y)|<\epsilon$ whenever $|x-y|<\delta$ for all functions $f_n$ in the sequence. The key point here is that a single $\delta$ (depending solely on $\epsilon$ ) works for the entire family of functions.

Application

Let $g:[0,1]\times [0,1]\to [0,1]$ be a continuous function and let $\{f_n\}$ be a sequence of functions such that $f_n(x)=\begin{cases}0,&0\leq x\leq 1/n\\ \int_0^{x-\frac{1}{n}}g(t,f_n(t))\ dt,&1/n\leq x\leq 1\end{cases}$

Prove that there exists a continuous function $f:[0,1]\to\mathbb{R}$ such that $f(x)=\int_0^x g(t,f(t))\ dt$ for all $x\in [0,1]$ .

The idea is to use Arzela-Ascoli Theorem. Hence, we need to show that $(f_n)$ is uniformly bounded and equicontinuous.

We have

$\begin{aligned}|f_n(x)|&\leq |\int_0^{x-\frac{1}{n}} 1\ dt|\\ &=|x-\frac{1}{n}|\\ &\leq |x|+|\frac{1}{n}|\\ &\leq 1+1\\ &=2 \end{aligned}$

This shows that the sequence is uniformly bounded.

If $0\leq x\leq 1/n$ ,

$\begin{aligned}|f_n(x)-f_n(y)|&=|0-f_n(y)|\\ &=|\int_0^{y-\frac{1}{n}} g(t,f_n(t))\ dt|\\ &\leq |\int_0^{y-\frac{1}{n}} 1\ dt|\\ &=|y-\frac{1}{n}|\\ &\leq |y-x| \end{aligned}$

Similarly if $0\leq y\leq 1/n$ , $|f_n(x)-f_n(y)|\leq |x-y|$ .

If $1/n\leq x\leq 1$ and $1/n\leq y\leq 1$ ,

$\begin{aligned}|f_n(x)-f_n(y)|&=|\int_0^{x-1/n} g(t,f_n(t))\ dt-\int_0^{y-1/n}g(t,f_n(t))\ dt|\\ &=|\int_{y-1/n}^{x-1/n}g(t,f_n(t))\ dt|\\ &\leq |\int_{y-1/n}^{x-1/n} 1\ dt|\\ &=|(x-1/n)-(y-1/n)|\\ &=|x-y| \end{aligned}$

Therefore we may choose $\delta=\epsilon$ , then whenever $|x-y|<\delta$ , $|f_n(x)-f_n(y)|\leq |x-y|<\epsilon$ . Thus the sequence is indeed equicontinuous.

By Arzela-Ascoli Theorem, there exists a subsequence $(f_{n_k})$ that is uniformly convergent.

$f_{n_k}(x)\to f(x)=\int_0^x g(t,f(t))\ dt$ .

By the Uniform Limit Theorem, $f:[0,1]\to\mathbb{R}$ is continuous since each $f_n$ is continuous.

Merry Christmas

Wishing all readers a Merry Christmas and Happy New Year!

For parents looking for an ideal Christmas gift for their child, do consider buying an enrichment book from Recommend Math Books. As a quote goes, “A book is a gift you can open again and again.” – Garrison Keillor

Some other excellent educational books for Christmas gifts are:

2nd Isomorphism Theorem (Lattice Diagram)

Math Online Tom Circle

I found this “lattice diagram” only in an old Chinese Abstract Algebra Textbook, never seen before in any American/UK or in French textbooks . Share here with the students who would find difficulty remembering the 3 useful Isomorphism Theorems.

Reference: 2nd Isomorphism Theorem (“Diamond Theorem”)

Let G be a group. Let H be a subgroup of G, and let N be a normal subgroup of G. Then:

1. The product HN is a subgroup of G,
The intersection H ∩ N is a normal subgroup of H, and

2. The 2 quotient groups
(HN) / N and
H / (H∩ N)
are isomorphic.

It is easy to remember using the green diagram below: (similarly can be drawn for 1st & 3rd Isomorphism)

This 2nd isomorphism theorem has been called the “diamond theorem” due to the shape of the resulting subgroup lattice with HN at the top, H∩ N…

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3rd Isomorphism Theorem

Math Online Tom Circle

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Quotient Ring of the Gaussian Integers is Finite

The Gaussian Integers $\mathbb{Z}[i]$ are the set of complex numbers of the form $a+bi$ , with $a,b$ integers. Originally discovered and studied by Gauss, the Gaussian Integers are useful in number theory, for instance they can be used to prove that a prime is expressible as a sum of two squares iff it is congruent to 1 modulo 4.

This blog post will prove that every (proper) quotient ring of the Gaussian Integers is finite. I.e. if $I$ is any nonzero ideal in $\mathbb{Z}[i]$ , then $\mathbb{Z}[i]/I$ is finite.

We will need to use the fact that $\mathbb{Z}[i]$ is an Euclidean domain, and thus also a Principal Ideal Domain (PID).

Thus $I=(\alpha)$ for some nonzero $\alpha\in\mathbb{Z}[i]$ . Let $\beta\in\mathbb{Z}[i]$ .

By the division algorithm, $\beta=\alpha q+r$ with $r=0$ or $N(r)<N(\alpha)$ . We also note that $\beta+I=r+I$ .

Thus,

$\begin{aligned}\mathbb{Z}[i]/I&=\{\beta+I\mid\beta\in\mathbb{Z}[i]\}\\ &=\{r+I\mid r\in\mathbb{Z}[i],N(r)<N(\alpha)\} \end{aligned}$ .

Since there are only finitely many elements $r\in\mathbb{Z}[i]$ with $N(r)<N(\alpha)$ , thus $\mathbb{Z}[i]/I$ is finite.

Behavior of Homotopy Groups with respect to Products

This blog post is on the behavior of homotopy groups with respect to products. Proposition 4.2 of Hatcher:

For a product $\prod_\alpha X_\alpha$ of an arbitrary collection of path-connected spaces $X_\alpha$ there are isomorphisms $\pi_n(\prod_\alpha X_\alpha)\cong\prod_\alpha \pi_n(X_\alpha)$ for all $n$ .

The proof given in Hatcher is a short one: A map $f:Y\to \prod_\alpha X_\alpha$ is the same thing as a collection of maps $f_\alpha: Y\to X_\alpha$ . Taking $Y$ to be $S^n$ and $S^n\times I$ gives the result.

A possible alternative proof is to first prove that $\pi_n(X_1\times X_2)\cong\pi_n(X_1)\times\pi_n(X_2)$ , which is the result for a product of two spaces. The general result then follows by induction.

We construct a map $\psi:\pi_n(X_1\times X_2)\to\pi_n(X_1)\times\pi_n(X_2)$ , $\psi([f])=([f_1],[f_2])$ .

Notation: $f:S^n\to X_1\times X_2$ , $f_1=p_1\circ f:S^n\to X_1$ , $f_2=p_2\circ f:S^n\to X_2$ where $p_i:X_1\times X_2\to X_i$ are the projection maps.

We can show that $\psi ([f]+[g])=\psi([f])+\psi([g])$ , thus $\psi$ is a homomorphism.

We can also show that $\psi$ is bijective by constructing an explicit inverse, namely $\phi:\pi_n(X_1)\times\pi_n(X_2)\to\pi_n(X_1\times X_2)$ , $\phi([g_1],[g_2])=[g]$ where $g:S^n\to X_1\times X_2$ , $g(x)=(g_1(x),g_2(x))$ .

Thus $\psi$ is an isomorphism.

Graph of measurable function is measurable (and has measure zero)

Let $f$ be a finite real valued measurable function on a measurable set $E\subseteq\mathbb{R}$ . Show that the set $\{(x,f(x)):x\in E\}$ is measurable.

We define $\Gamma(f,E):=\{(x,f(x)):x\in E\}$ . This is popularly known as the graph of a function. Without loss of generality, we may assume that $f$ is nonnegative. This is because we can write $f=f^+ - f^-$ , where we split the function into two nonnegative parts.

The proof here can also be found in Wheedon’s Analysis book, Chapter 5.

The strategy for proving this question is to approximate the graph of the function with arbitrarily thin rectangular strips. Let $\epsilon>0$ . Define $E_k=\{x\in E\mid \epsilon k\leq f(x)<\epsilon (k+1)\}$ , $k=0,1,2,\dots$ .

Also, $\Gamma(f,E)=\cup\Gamma(f,E_k)$ , where $\Gamma(f,E_k)$ are disjoint.

$\begin{aligned}|\Gamma(f,E)|_e&\leq\sum_{k=1}^\infty|\Gamma(f,E_k)|_e\\ &\leq\epsilon(\sum_{k=1}^\infty|E_k|)\\ &=\epsilon|E| \end{aligned}$

If $|E|<\infty$ , we can conclude $|\Gamma(f,E)|_e=0$ and thus $\Gamma(f,E)$ is measurable (and has measure zero).

If $|E|=\infty$ , we partition $E$ into countable union of sets $F_k$ each with finite measure. By the same analysis, each $\Gamma(f,F_k)$ is measurable (and has measure zero). Thus $\Gamma(f,E)=\bigcup_{k=1}^\infty\Gamma(f,F_k)$ is a countable union of measurable sets and thus is measurable (has measure zero).

LaTeX to WordPress Converter

Just created a LaTeX to WordPress Converter: http://mathtuition88.blogspot.sg/2015/12/latex-to-wordpress-converter.html

Currently it is a very basic converter, just changes “$abc$” to “$ latex abc$”. To change back from WordPress to LaTeX, a simple text editor will do the job, with replace “$ latex ” with “$”.

Test code:

LaTeX: From the above inequality $|z^n|>|a_1z^{n-1}+\ldots+a_n|$ we can conclude that the polynomial $p_t(z)=z^n+t(a_1z^{n-1}+\ldots+a_n)$ has no roots on the circle $|z|=r$ when $0\leq t\leq 1$.

WordPress: From the above inequality $|z^n|>|a_1z^{n-1}+\ldots+a_n|$ we can conclude that the polynomial $p_t(z)=z^n+t(a_1z^{n-1}+\ldots+a_n)$ has no roots on the circle $|z|=r$ when $0\leq t\leq 1$ .

Covering space projection induces isomorphisms

Proposition 4.1 (from Hatcher): A covering space projection $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ induces isomorphisms $p_*:\pi_n(\tilde{X},\tilde{x}_0)\to\pi_n(X,x_0)$ for all $n\geq 2$ .

We will elaborate more on this proposition in this blog post. Basically, we will need to show that $p_*$ is a homomorphism and also bijective (surjective and injective).

Homomorphism

$p_*([f]):=[pf]$

$p_*([f]+[g])=[p(f+g)]$

$p(f+g)(s_1,s_2,\dots,s_n)=\begin{cases}pf(2s_1,s_2,\dots,s_n)&s_1\in[0,\frac 12]\\ pg(2s_1-1,s_2,\dots,s_n)&s_1\in[\frac 12,1] \end{cases}$

$p_*[f]+p_*[g]=[pf]+[pg]$

$(pf+pg)(s_1,s_2,\dots,s_n)= \begin{cases}pf(2s_1,s_2,\dots,s_n)&s_1\in[0,\frac 12]\\ pg(2s_1-1,s_2,\dots,s_n)&s_1\in[\frac 12,1] \end{cases}$ , which we can see is the same.

Thus, $p_*$ is a homomorphism.

Surjective

For surjectivity, we need to use a certain Proposition 1.33: Suppose given a covering space $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ and a map $f:(Y,y_0)\to (X,x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $\tilde{f}:(Y,y_0)\to (\tilde{X},\tilde{x}_0)$ of $f$ exists iff $f_*(\pi_1(Y,y_0))\subset p_*(\pi_1(\tilde{X},\tilde{x}_0))$ .

Let $[f]\in\pi_n(X,x_0)$ , where $f:(S_n,s_0)\to(X,x_0), n\geq 2$ . Since $S^n$ is simply connected for $n\geq 2$ , $\pi_1(S_n,s_0)=0$ . Thus $f_*(\pi_1(S_n,s_0))=0\subset p_*(\pi_1(\tilde{X},\tilde{x}_0))$ . By Proposition 1.33, a lift $\tilde{f}:(S_n,s_0)\to (\tilde{X},\tilde{x}_0)$ of $f$ exists, where $p\tilde{f}=f$ .

i.e. we have $\boxed{p_*[\tilde{f}]=[p\tilde{f}]=[f]}$ . Hence $p_*$ is surjective.

Injective

Let $[\tilde{f}_0]\in\ker p_*$ , where $\tilde{f}_0:I^n\to \tilde{X}$ with a homotopy $f_t:I^n\to X$ of $f_0=p\tilde{f}_0$ to the trivial loop $f_1$ .

By the covering homotopy property (homotopy lifting property), there exists a unique homotopy $\tilde{f}_t:I^n\to \tilde{X}$ of $\tilde{f}_0$ that lifts $f_t$ , i.e. $p\tilde{f}_t=f_t$ . There is a lifted homotopy of loops $\tilde{f}_t$ starting with $\tilde{f}_0$ and ending with a constant loop. Hence $[\tilde{f}_0]=0$ in $\pi_n(\tilde{X},\tilde{x}_0)$ and thus $p_*$ is injective.

Rock-Paper-Scissors 石头 – 剪刀 – 布

Math Online Tom Circle

A Chinese Mathematician Figured Out How To Always Win At Rock-Paper-Scissors – (Business Insider)

This is “Game Theory” demonstrating the Nash Equilibrium.
Very good to understand the “Kia-Soo” (Singlish means: 惊(怕)输 “afraid to lose”) syndrome of Singaporeans.

To win this game and beat the “kia-soo” mentality — 反其道而行 Adopt the reverse way of the opposition’s anticipated kia-soo way 🙂

Key points:
(1). Sequence : “R- P -S” or (中文习惯) “石头 – 剪刀 – 布”;
(2). Winner tends to stay same way in next move;
(3). Loser likely to switch to the next step in the Sequence (1).

Reflection:
In business,
(2) is where big conglomerates like IBM , HP, Sony, Microsoft etc lose because they stay put with the same strategy (Corporate Data Center, Sell thru Channel distributors with mark-up, CD/DVD music… ), and products (Mainframes, Servers, PC, CRT-TV, Packaged software…) which brought them to success but never…

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Star Wars Math Song

I like the first one about Physics, and the last one about Math!

Curiously, they used Darth Maul’s Theme for Physics, and Darth Vader’s Theme for Math. 🙂

How to calculate Homology Groups (Klein Bottle)

This post will be a guide on how to calculate Homology Groups, focusing on the example of the Klein Bottle. Homology groups can be quite difficult to grasp (it took me quite a while to understand it). Hope this post will help readers to get the idea of Homology. Our reference book will be Hatcher’s Algebraic Topology (Chapter 2: Homology). I will elaborate further on the Hatcher’s excellent exposition on Homology.

This is also Exercise 5 in Chapter 2, Section 2.1 of Hatcher.

The first step to compute Homology Groups is to construct a $\Delta$ -complex of the Klein Bottle.

klein bottle

One thing to note for $\Delta$ -complexes, is that the vertices cannot be ordered cyclically, as that would violate one of the requirements which is to preserve the order of the vertices.

The key formula for Homology is: $\boxed{H_n=\ker\partial_n/\text{Im}\ \partial_{n+1}}$ .

We have $\ker\partial_0=\langle v\rangle$ , the free group generated by the vertex $v$ , because there is only one vertex!

Next, we have $\partial_1(a)=\partial_1(b)=\partial_1(c)=v-v=0$ . Thus $\text{Im}\ \partial_1=0$ .

Therefore $H_0=\ker\partial_0/\text{Im}\ \partial_1=\langle v\rangle /0\cong\mathbb{Z}$ .

Next, we have $\ker\partial_1=\langle a,b,c\rangle$ . $\partial_2U=a+b-c$ , $\partial_2L=c+a-b$ . To learn more about calculating $\partial_2$ , check out the diagram on page 105 of Hatcher.

We then have $\text{Im}\ \partial_2=\langle a+b-c, c+a-b\rangle=\langle a+b-c, 2a\rangle$ , where we got $2a$ from adding the two previous generators $(a+b-c)+(c+a-b)$ .

Thus $H_1=\ker\partial_1/\text{Im}\ \partial_2=\langle a,b,c\rangle/\langle a+b-c, 2a\rangle=\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}$ .

To intuitively understand the above working, we need to use the idea that elements in the quotient are “zero”. Hence $a+b-c=0$ , implies that $c=a+b$ , thus $c$ can be expressed as a linear combination of $a, b$ , thus is not a generator of $H_1$ . $2a=0$ implies that $a+a=0$ , which gives us the $\mathbb{Z}/2\mathbb{Z}$ part.

Finally we note that $\ker\partial_2=0$ , and also for $n\geq 3$ , $\ker\partial_n=0$ since there are no simplices of dimension greater than or equal to 3. Thus, the second homology group onwards are all zero.

In conclusion, we have $H_k(K)=\begin{cases}\mathbb{Z}&k=0\\ \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}&k=1\\ 0&\text{otherwise} \end{cases}$

Mean Value Theorem for Higher Dimensions

Let $f$ be differentiable on a connected set $E\subseteq \mathbb{R}^n$ , then for any $x,y\in E$ , there exists $z\in E$ such that $f(x)-f(y)=\nabla f(z)\cdot (x-y)$ .

Proof: The trick is to use the Mean Value Theorem for 1 dimension via the following construction:

Define $g:[0,1]\to\mathbb{R}$ , $g(t)=f(tx+(1-t)y)$ . By the Mean Value Theorem for one variable, there exists $c\in (0,1)$ such that $g'(c)=\frac{g(1)-g(0)}{1-0}$ , i.e.

$\nabla f(cx+(1-c)y)\cdot (x-y)=f(x)-f(y)$ . Here we are using the chain rule for multivariable calculus to get: $g'(c)=\nabla f(cx+(1-c)y)\cdot (x-y)$ .

Let $z=cx+(1-c)y$ , then $\nabla f(z)\cdot (x-y)=f(x)-f(y)$ as required.

Function of Bounded Variation that is not continuous

This is a basic example of a function of bounded variation on [0,1] but not continuous on [0,1].

The key Theorem regarding functions of bounded variation is Jordan’s Theorem: A function is of bounded variation on the closed bounded interval [a,b] iff it is the difference of two increasing functions on [a,b].

Consider $g(x)=\begin{cases}0&\text{if}\ 0\leq x<1\\ 1&\text{if}\ x=1 \end{cases}$

$h(x)\equiv 0$

Both $g$ and $h$ are increasing functions on [0,1]. Thus by Jordan’s Theorem, $f(x)=g(x)-h(x)=g(x)$ is a function of bounded variation, but it is certainly not continuous on [0,1]!

Please help to do Survey

URL: https://career-test.com/s/sgamb?reid=210

The results of this Personality Test is quite surprisingly accurate, do give it a try to see if you are a Careerist, Entrepreneur, Harmonizer, Idealist, Hunter, Internationalist or Leader?

Do try out this Free Career Guidance Personality Test at https://career-test.com/s/sgamb?reid=210 while it is still available!

Benefits of doing the (Free) Career Test:

Get familiar with the top companies in Singapore
There are seven distinct career types based on career preferences, goals and personality. Get to know yours! (My result: Harmonizer)
Take part in the annual Universum survey and win prizes!

Chilli Crab

Bought a Mud Crab from Sheng Shiong at $6. The live ones were even cheaper, at $4 each. Much cheaper than ordering crab outside at restaurants, where they would at least cost $30.

My wife then cooked the crab in the “Chilli Crab” style. Yummy!

List of Fundamental Group, Homology Group (integral), and Covering Spaces

Just to compile a list of Fundamental groups, Homology Groups, and Covering Spaces for common spaces like the Circle, n-sphere ( $S^n$ ), torus ( $T$ ), real projective plane ( $\mathbb{R}P^2$ ), and the Klein bottle ( $K$ ).

Fundamental Group

Circle: $\pi_1(S^1)=\mathbb{Z}$

n-Sphere: $\pi_1(S^n)=0$ , for $n>1$

n-Torus: $\pi_1(T^n)=\mathbb{Z}^n$ (Here n-Torus refers to the n-dimensional torus, not the Torus with n holes)

$\pi_1(T^2)=\mathbb{Z}^2$ (usual torus with one hole in 2 dimensions)

Real projective plane: $\pi_1(\mathbb{R}P^2)=\mathbb{Z}_2$

Klein bottle $K$ : $\pi_1(K)=(\mathbb{Z}\amalg\mathbb{Z})/\langle aba^{-1}b\rangle$

Homology Group (Integral)

$H_0(S^1)=H_1(S^1)=\mathbb{Z}$ . Higher homology groups are zero.

$H_k(S^n)=\begin{cases}\mathbb{Z}&k=0,n\\ 0&\text{otherwise} \end{cases}$

$H_k(T)=\begin{cases}\mathbb{Z}\ \ \ &k=0,2\\ \mathbb{Z}\times\mathbb{Z}\ \ \ &k=1\\ 0\ \ \ &\text{otherwise} \end{cases}$

$H_k(\mathbb{R}P^2)=\begin{cases}\mathbb{Z}\ \ \ &k=0\\ \mathbb{Z}_2\ \ \ &k=1\\ 0\ \ \ &\text{otherwise} \end{cases}$

Klein bottle, $K$ : $H_k(K)=\begin{cases}\mathbb{Z}&k=0\\ \mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})&k=1\\ 0&\text{otherwise} \end{cases}$

Covering Spaces

A universal cover of a connected topological space $X$ is a simply connected space $Y$ with a map $f:Y\to X$ that is a covering map. Since there are many covering spaces, we will list the universal cover instead.

$\mathbb{R}$ is the universal cover of the unit circle $S^1$

$S^n$ is its own universal cover for $n>1$ . (General result: If $X$ is simply connected, i.e. has a trivial fundamental group, then it is its own universal cover.)

$\mathbb{R}^2$ is the universal cover of $T$ .

$S^2$ is universal cover of real projective plane $RP^2$ .

$\mathbb{R}^2$ is universal cover of Klein bottle $K$ .

Every non-empty open set in R is disjoint union of countable collection of open intervals

Question: Prove that every non-empty open set in $\mathbb{R}$ is the disjoint union of a countable collection of open intervals.

The key things to prove are the disjointness and the countability of such open intervals. Otherwise, if disjointness and countability are not required, we may simply take a small open interval centered at each point in the open set, and their union will be the open set.

Elementary Proof: Let $U$ be a non-empty open set in $\mathbb{R}$ .

Let $x\in U$ . There exists an open interval $I\subseteq U$ containing $x$ . Let $I_x$ be the maximal open interval in $U$ containing $x$ , i.e. for any open interval $I\subseteq U$ containing $x$ , $I\subseteq I_x$ . (The existence of $I_x$ is guaranteed, we can take it to be the union of all open intervals $I\subseteq U$ containing $x$ .)

We note that such maximal intervals are equal or disjoint: Suppose $I_x\cap I_y\neq\emptyset$ and $I_x\neq I_y$ then $I_x\cup I_y$ is an open interval in $U$ containing $x$ , contradicting the maximality of $I_x$ .

Each of the maximal open intervals contain a rational number, thus we may write $\displaystyle U=\bigcup_{q\in U\cap\mathbb{Q}}I_q$ . Upon discarding the “repeated” intervals in the union above, we get that $U$ is the disjoint union of a countable collection of open intervals.

There are many other good proofs of this found here (http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv), though some can be quite deep for this simple result.

7 Cardinal Rules in Life

(See more at: http://www.lifehack.org/articles/lifestyle/7-cardinal-rules-life-everyone-should-know-about.html)

Useful Theorem in Introductory Ring Theory

Something interesting I realised in my studies in Math is that certain theorems are more “useful” than others. Certain theorems’ sole purpose seem to be an intermediate step to prove another theorem and are never used again. Other theorems seem to be so useful and their usage is everywhere.

One of the most “useful” theorems in basic Ring theory is the following:

Let $R$ be a commutative ring with 1 and $I$ an ideal of $R$ . Then

(i) $I$ is prime iff $R/I$ is an integral domain.

(ii) $I$ is maximal iff $R/I$ is a field.

With this theorem, the following question is solved effortlessly:

Let $R$ be a commutative ring with 1 and let $I$ and $J$ be ideals of $R$ such that $I\subseteq J$ .

(i) Show that $J$ is a prime ideal of $R$ iff $J/I$ is a prime ideal of $R/I$ .

(ii) Show that $J$ is a maximal ideal of $R$ iff $J/I$ is a maximal ideal of $R/I$ .

Sketch of Proof of (i):

$J$ is a prime ideal of $R$ iff $R/J$ is an integral domain. ( $R/J\cong \frac{R/I}{J/I}$ by the Third Isomorphism Theorem. ) $\iff$ $J/I$ is a prime ideal of $R/I$ .

(ii) is proved similarly.

ICM2014 (Seoul): Zhang YiTang

Math Online Tom Circle

ICM2014 (VideoSeries LC7):
Yitang Zhang

A conversation between Yitang Zhang and Daniel Goldston — who is one of the team GPY:

Simplified Prime Gap Video Explanations:

Video 1 – https://youtu.be/vkMXdShDdtY

Video 2 (extra) – https://youtu.be/D4_sNKoO-RA

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Live your Dream (Motivational Video) 夢想．激勵人心的演說

Just to share a very inspiring motivational video from YouTube. Not sure which movie it is from. (any readers know, please comment below as I would be interested)

Highly suitable for students (and their parents) who have just completed their PSLE, whether their PSLE 2015 results are good or not, it is now a good time to reflect on their dreams and the next step to take in the next year 2016.

“Proof School” For Math Kids

Math Online Tom Circle

To overcome the USA general school low Math standard, there are some elite schools to groom the gifted students such as this “Proof School”:

San-Fancisco new “Proof School” for Math kids
The Math Syllabus : Topology, Number Theory, …
IT: Python programming

Proof School Website
http://www.proofschool.org/#we-love-math

What is Proof School?

https://www.quora.com/What-is-Proof-School/answer/Alon-Amit?srid=oZzP&share=50c583a6

Math Syllabus:

http://www.proofschool.org/course-descriptions-1#math-courses

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Fundamental Group of Torus (van Kampen method)

There are various methods of computing fundamental groups, for example one method using maximal trees of a simplicial complex (considered a slow method). There is one “trick” using van Kampen’s Theorem that makes it relatively fast to compute the fundamental group.

This “trick” doesn’t seem to be explicitly written in books, I had to search online to learn about it.

Fundamental Group of Torus

First we let $U$ and $V$ be open subsets of the torus (denoted as $X$ )as shown in the diagram below. $U$ is an open disk, while $V$ is the entire space with a small punctured hole. We are using the fundamental polygon representation of the torus. This trick can work for many spaces, not just the torus.

$U$ is contractible, thus $\pi_1(U)=0$ . $U\cap V$ has $S^1$ as a deformation retract, thus $\pi_1(U\cap V)=\mathbb{Z}$ . We note that $X=U\cup V$ and $U\cap V$ is path-connected. These are the necessary conditions to apply van Kampen’s Theorem.

Then, by Seifert-van Kampen Theorem, $\displaystyle\boxed{\pi_1(X)=\pi_1(U)\coprod_{\pi_1(U\cap V)}\pi_1(V)}$ , the free product of $\pi_1(U)$ and $\pi_1(V)$ with amalgamation.

Let $h$ be the generator in $U\cap V$ . We have $j_{1*}(h)=1$ and $j_{2*}(h)=aba^{-1}b^{-1}$ . ( $j_1:U\cap V\to U$ and $j_2:U\cap V\to V$ are the inclusions. )

Therefore

$\begin{aligned} \pi_1(X)&=\langle a,b\mid aba^{-1}b^{-1}=1\rangle\\ &=\langle a,b\mid ab=ba\rangle\\ &\cong\mathbb{Z}\times\mathbb{Z} \end{aligned}$

vankampen_torus

Free Math Newsletter

Interpolation Technique in Analysis

Question: Let $f$ belong to both $L^{p_1}$ and $L^{p_2}$ , with $1\leq p_1<p_2<\infty$ . Show that $f\in L^p$ for all $p_1\leq p\leq p_2$ .

There is a pretty neat trick to do this question, known as the “interpolation technique”. The proof is as follows.

For $p_1<p<p_2$ , there exists $0<\alpha<1$ such that $\displaystyle\boxed{p=\alpha p_1+(1-\alpha)p_2}$ . This is the key “interpolation step”. Once we have this, everything flows smoothly with the help of Holder’s inequality.

$\displaystyle\begin{aligned} \int |f|^p\ d\mu&=\int (|f|^{\alpha p_1}\cdot |f|^{(1-\alpha)p_2})\ d\mu\\ &\leq\||f|^{\alpha p_1}\|_\frac{1}{\alpha}\||f|^{(1-\alpha)p_2}\|_\frac{1}{1-\alpha}\\ &=(\int |f|^{p_1}\ d\mu)^\alpha\cdot (\int |f|^{p_2}\ d\mu)^{1-\alpha}\\ &<\infty \end{aligned}$

Thus $f\in L^p$ .

Note that the magical thing about the interpolation technique is that $p=\frac{1}{\alpha}$ and $q=\frac{1}{1-\alpha}$ are Holder conjugates, since $\frac{1}{p}+\frac{1}{q}=1$ is easily verified.

Undergraduate Math Books

“latex path not specified” WordPress LaTeX Bug

Recently, there is a “latex path not specified” WordPress LaTeX bug, it is very weird. Some LaTeX expressions will get rendered and some will not. Will have to postpone my math blogging till it is fixed. Worst case scenario is I have to abandon this blog and move to Blogger (http://mathtuition88.blogspot.com) if the issue remains unfixed.

Testing: $x$ , $y$ , $z$ , $e^{i\pi}+1=0$ .

Hope this bug gets fixed soon. If anyone knows the solution to solve this bug, please inform me in the comments below!

Note: Thanks to Professor Terence Tao who has replied in the comments below and shown us a link where there is ongoing discussion about the highly mysterious “latex path not specified” issue.

Interesting Measure and Integration Question

Let $(\Omega,\mathcal{A},\mu)$ be a measure space. Let $f\in L^p$ and $\epsilon>0$ . Prove that there exists a set $E\in\mathcal{A}$ with $\mu(E)<\infty$ , such that $\int_{E^c} |f|^p<\epsilon$ .

Solution:

The solution strategy is to use simple functions (common tactic for measure theory questions).

Let $0\leq\phi\leq |f|^p$ be a simple function such that $\int_\Omega (|f|^p-\phi)\ d\mu<\epsilon$ .

Consider the set $E=\{\phi>0\}$ . Note that $\int_\Omega \phi\ d\mu\leq\int_\Omega |f|^p\ d\mu<\infty$ . Hence each nonzero value of $\phi$ can only be on a set of finite measure. Since $\phi$ has only finitely many values, $\mu(E)<\infty$ .

Then,

$\begin{aligned} \int_{E^c}|f|^p\ d\mu&=\int_{E^c} (|f|^p-\phi)\ d\mu +\int_{E^c}\phi\ d\mu\\ &\leq \int_\Omega (|f|^p-\phi)\ d\mu+0\\ &<\epsilon \end{aligned}$

Students in Singapore going for ‘ad-hoc’ tuition in specific topics

http://www.straitstimes.com/singapore/education/students-going-for-ad-hoc-tuition?&utm_source=google_gmail&utm_medium=social-media&utm_campaign=addtoany

Instead of committing to regular tuition sessions throughout the year, some students who need academic help are choosing to attend such classes on an “ad-hoc” basis.. Read more at straitstimes.com.

SAJC Retain Rate (JC 1)

SAJC Retain / Retention Rate (Estimate)

Just heard from a reliable source (cousin who is in the school) that SAJC’s tentative retention rate for 2015 is around 10%. On average, for a class of 25, around 2 or 3 are retained, after the Promos (Promotional Exams) in JC 1.

This is an estimate, intended to give information to those seeking it, hope it helps. By today’ s standards, 10% retention rate is considered “moderate”, considering official statistics from MOE shows that “The two JCs with the highest retention rates at JC1 averaged around 15% over the past three years.”

Side note: Some of those “retained” in SAJC are given a second chance to take another exam, upon passing they can be promoted. Hence the actual retain rate will be less than 10%, which is considered quite ok (compared to other JCs).

Success consists of going from failure to failure without loss of enthusiasm.

Winston Churchill

For those who are retained, do not despair, check out my motivational page for motivational quotes and stories.

Actually JC life is difficult for students, they have to wake up at 6am everyday, and go home at around 6-7 pm or later (due to CCA). After reaching home, it is just the beginning and they have to revise / do homework / go for tuition. It is much tougher than even the typical adult’s job of 8-5pm work. And JC students have to repeat the schedule daily for two years. The problem is that too much stuff is being crammed into two years.

Apparently, the retain rate / retention rate of JCs is a source of concern for many. Some official statistics has been released by MOE. The statistics given are “over the last three years, approximately 6% of first year JC students in each cohort failed some subjects in their promotional exams and were retained.” “The two JCs with the highest retention rates at JC1 averaged around 15% over the past three years.”

As a student who has gone through the system, rumours of JCs like MJC having 50% retain rate (most likely exaggerated, but having some basis of truth, since there is no smoke without fire) do cause some concern. Currently the JC system works by setting extremely tough internal exams, including promos and prelims (compared to the A levels), such that a D or E in the prelims in top JCs (e.g. RI/HCI/NJC) is very likely equivalent to an A in the eventual A levels. This works for some students to spur them to study harder, but may be overly demoralising for many students. For retention rate, common sense and logic would tell that a high retention rate would boost the school’s eventual A level results (one extra year of study is a lot), however that is at the expense of the student spending one extra year in JC. Since the retention rate is entirely up to the school’s decision (i.e. not regulated by MOE), each JC has different retain rate.

Students choosing a JC should check out their retention rate from reliable seniors / relatives / teachers (there is no official source released online for individual retention rate for JCs).

For students looking for tuition, do check out a highly recommended tuition agency.

Video on Computing Homology Groups

Just watched this again. I can truely say that this is the most clear explanation of Homology on the entire internet!

Mathtuition88

This video follows after the previous video on Simplicial Complexes.

If you are looking for quality Math textbooks to study from (including Linear Algebra and Calculus, the two most popular Math courses), check out my page on Recommended Math Books for students!

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