Let A be a simple R-algebra, then M_n(A) is simple

We let $A$ be a simple $R$ -algebra. Then $M_n(A)$ , the n by n matrix algebra over $A$ is simple. In particular if $D$ is a division algebra, then $M_n(D)$ is simple.

Reference: Associative Algebras (Graduate Texts in Mathematics)

We will split our proof into 2 Lemmas.

Lemma 1: If $I\lhd A$ , then $M_n(I)\lhd M_n(A)$ .

Proof: Let $X=(x_{ij}), Y=(y_{ij})\in M_n(I), Z=(z_{ij})\in M_n(A)$ .

$(X-Y)_{ij}=x_{ij}-y_{ij}\in I$ . Therefore $X-Y\in M_n(I)$ .

$(XY)_{ij}=\sum_{k=1}^n x_{ik}y_{kj}\in I$ , thus $XY\in M_n(I)$ . Similarly $YX\in M_n(I)$ . Thus $M_n(I)$ is indeed an ideal of $M_n(A)$ .

Lemma 2 (Tricky part, takes some time to digest): If $J\lhd M_n(A)$ , then $J=M_n(I)$ for some $I\lhd A$ .

A technical thing we need to know is the $E_{ij}$ matrix, which is a n by n matrix with 1 in the (i,j) entry, 0 elsewhere. A key property is that $E_{ij}E_{kl}=E_{il}$ if $j=k$ , and 0 otherwise (zero matrix).

The key idea here is that $I$ can be taken to be $I=\{r\in A\mid rE_{11}\in J\}$ . We can check that $I$ is indeed an ideal. We need to know that it is possible to premultiply and postmultiply a given matrix by “elementary matrices” such that any entry of the given matrix is “shifted” to the (1,1) entry. An example will be $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}d&0\\0&0\end{pmatrix}$ .

Let us begin the proof to first show $J\subseteq M_n(I)$ . Let $X=(x_{ij})\in J$ . We write $X=\sum_{i,j}^n x_{ij}E_{ij}$ . For any $1\leq p,q\leq n$ ,

$\begin{aligned}E_{1p}XE_{q1}&=E_{1p}(\sum x_{ij}E_{ij})E_{q1}\\ &=(\sum x_{pj}E_{1j})E_{q1}\\ &=x_{pq}E_{11}\in J \end{aligned}$

Therefore $x_{pq}\in I$ . Therefore $X\in M_n(I)$ . What we are actually doing here is first pick an arbitrary matrix $X$ in $J$ . Then, we do the “shifting” process to show that any (p,q) entry in $X$ can be “shifted” to the (1,1) entry, thus it is inside our ideal $I$ . Since any arbitrary (p,q) entry in $X$ is inside the ideal $I$ , we conclude that $X$ is an element of $M_n(I)$ .

Next part is to show $M_n(I)\subseteq J$ . Let $X=(x_{ij})\in M_n(I)$ . We have $x_{ij}E_{11}\in J$ for all $i,j$ . We compute that

$\begin{aligned}E_{i1}(x_{ij}E_{11})E_{1j}&=x_{ij}E_{i1}E_{1j}\\ &=x_{ij}E_{ij}\in J \end{aligned}$

Therefore $X=\sum x_{ij}E_{ij}\in J$ . We will illustrate what we are doing here by an example in the 2 by 2 case. Let us have $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ . Since $X$ is in $M_n(I)$ , what we have, by definition of $I$ , is that $\begin{pmatrix}a&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}b&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}c&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}d&0\\0&0\end{pmatrix}$ are all in $J$ . We then proceed to “shift” them all into their correct positions: $\begin{pmatrix}a&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}0&b\\0&0\end{pmatrix}$ , $\begin{pmatrix}0&0\\c&0\end{pmatrix}$ , $\begin{pmatrix}0&0\\0&d\end{pmatrix}$ , all of which are still in the ideal $J$ . $X$ is the sum of all of them, thus also in $J$ .

Final Conclusion

Since $A$ is simple, its ideals are 0 and itself. Thus, the ideals of $M_n(A)$ will be also the 0 matrix and itself, and thus is simple. A division algebra $D$ is simple (Any ideal containing a nonzero element $x$ , when multiplied by its inverse will give 1. Thus the ideal will contain 1 and thus the entire ring $D$ ), thus $M_n(D)$ is also simple.

1000 Posts

Screen Shot 2016-01-12 at 8.26.51 PM

Finally, reached 1000 posts!

Better To Try And Fail Than Never To Try At All – Poem by William F. O’Brien

Source: http://www.poemhunter.com/poem/better-to-try-and-fail-than-never-to-try-at-all/

Some say risk nothing, try only for the sure thing,
Others say nothing gambled nothing gained,
Go all out for your dream.
Life can be lived either way, but for me,
I’d rather try and fail, than never try at all, you see.

Some say “Don’t ever fall in love,
Play the game of life wide open,
Burn your candle at both ends.”
But I say “No! It’s better to have loved and lost,
Than never to have loved at all, my friend.”

When many moons have gone by,
And you are alone with your dreams of yesteryear,
All your memories will bring you cheer.
You’ll be satisfied, succeed or fail, win or lose,
Knowing the right path you did choose.

William F. O’Brien

Recommended Functional Analysis Book (Graduate Level)

Functional Analysis is a subject that combines Linear Algebra with Analysis. I researched online, and it seems one of the best Functional Analysis Book for Graduate level is Functional Analysis by Peter Lax. This book is ideal for a second course in Functional Analysis. For a first course in Functional Analysis, I would recommend Kreyszig, which is listed on my Recommended Undergraduate Math Books page.

This is Theorem 5 in the book: Let $X$ be a linear space over the reals.

The following 9 properties hold:

The empty set is convex.
A subset consisting of a single point is convex.
Every linear subspace of $X$ is convex.
The sum of two convex subsets is convex.
If $K$ is convex, so is $-K$ .
The intersection of an arbitrary collection of convex sets is convex.
Let $\{K_j\}$ be a collection of convex subsets that is totally ordered by inclusion. Then their union $\cup K_j$ is convex.
The image of a convex set under a linear map is convex.
The inverse image of a convex set under a linear map is convex.

Brief sketch of proofs:

We give a brief sketch of the idea behind the proofs.

We are using the definition of convex as follows: $X$ is a linear space over the reals; a subset $K$ of $X$ is called convex if, whenever $x$ and $y$ belong to $K$ , all points of the form $ax+(1-a)y$ , $0\leq a\leq 1$ also belong to $K$ .

Property 1 is vacuously true.

Property 2 is true because of $ax+(1-a)x\equiv x$ .

Property 3 is true because $ax+(1-a)y$ is a linear combination and is thus in the linear subspace.

Property 4) Let $C_1$ and $C_2$ be the two convex subsets. Let $x_1+y_1$ and $x_2+y_2$ be points in $C_1+C_2=\{x+y:x\in C_1, y\in C_2\}$

$\begin{aligned} a(x_1+y_1)+(1-a)(x_2+y_2)&=ax_1+ay_1+x_2+y_2-ax_2-ay_2\\ &=[ax_1+(1-a)x_2]+[ay_1+(1-a)y_2]\\ &\in C_1+C_2 \end{aligned}$

Property 5) We just need to know that $-K=\{-x:x\in K\}$ and this algebraic observation: $a(-x)+(1-a)(-y)=-(ax+(1-a)y)$ .

Property 6) Let $x,y\in\bigcap_{i\in I}C_i$ . $ax+(1-a)y\in C_i$ for all $i\in I$ , thus $ax+(1-a)y\in\bigcap_{i\in I}C_i$ .

Property 7) Let $x,y\in\bigcup K_j$ , where $K_i\subseteq K_{i+1}$ . Let $x\in K_n$ , $y\in K_m$ , then either $K_n\subseteq K_m$ or $K_m\subseteq K_n$ . If $K_n\subseteq K_m$ , $ax+(1-a)y\subseteq K_m\subseteq \bigcup K_j$ . Similarly for the other case $K_m\subseteq K_n$ .

Property 8) Observe that $af(x)+(1-a)f(y)=f(ax+(1-a)y)\in f(K)$ .

Property 9) The only tricky thing about this part is that we cannot assume that the inverse $f^{-1}$ exists. We can only talk about the pre-image.

Let $w,z\in f^{-1}(K)$ . $f(w)\in K$ and $f(z)\in K$ .

We have $f(aw+(1-a)z)=af(w)+(1-a)f(z)\in K$ .

Thus $aw+(1-a)z\in f^{-1}(K)$ .

The End!

Image of Vertical Strip under Inversion

This is a slight generalisation of Example 3 in Churchill’s Complex Variables and Applications.

Consider the infinite vertical strip $c_1<x<c_2$ , under the transformation $w=1/z$ , where $c_1, c_2$ are of the same sign.

When $x=c_1$ , note that by arguments similar to earlier analysis, we have $x=\frac{u}{u^2+v^2}$ .

$u^2+v^2=\frac{1}{c_1}u$ , upon completing the square, becomes

$(u-\frac{1}{2c_1})^2+v^2=(\frac{1}{2c_1})^2$ — Circle 1

Similarly the line $x=c_2$ is transformed into:

$(u-\frac{1}{2c_2})^2+v^2=(\frac{1}{2c_2})^2$ — Circle 2

Note that as $x$ gets larger, the radius of the resultant circle gets smaller.

Thus, the resultant image is the (open) region between the two circles.

The converse holds too, i.e. the region between two circles will be mapped back to the vertical strip by inversion.

Example

Find an analytic isomorphism from the open region between the two circles $|z|=1$ and $|z-\frac{1}{2}|=\frac{1}{2}$ to the vertical strip $0<\text{Re}(z)<1$ .

First, we “shift” the circles to the left by 1 unit via the map $w=z-1$ . Now we are in the situation of our above analysis, thus inversion $w=1/z$ maps the region to the vertical strip $-1<x<-1/2$ . The map $w=-z$ (reflection), followed by $w=z-\frac{1}{2}$ , finally finishing up with a scaling of factor 2 brings us to the vertical strip desired.

Composition of all the above functions leads us to the desired transformation $f(z)=2(\frac{-1}{z-1}-\frac 12)=\frac{-1-z}{z-1}$ to the very nice strip $0<\text{Re}(x)<1$ .

We will mention an analytic isomorphism of this vertical strip to the upper half plane $\text{Im} z>0$ in a subsequent blog post.

Ultimate Tic-Tac-Toe

Math with Bad Drawings

Updated 7/16/2013 – See Original Here

Once at a picnic, I saw mathematicians crowding around the last game I would have expected: Tic-tac-toe.

As you may have discovered yourself, tic-tac-toe is terminally dull. There’s no room for creativity or insight. Good players always tie. Games inevitably go something like this:

But the mathematicians at the picnic played a more sophisticated version. In each square of their tic-tac-toe board, they’d drawn a smaller board:

As I watched, the basic rules emerged quickly.

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Clash of Clans Math – Using Linear Inequalities to Save Elixir

This is quite a nice video on Clash of Clans and Math. Only got 70 views so far, but it is definitely a well prepared video.

Another video:

Farming in Clash of Clans has just gotten harder, due to nerf on Town Hall not granting shield, and very powerful defenses for TH10 and TH11. Watch these videos and you may gain some idea on a better way to farm.

Definition of Tensor Product: M Tensor N

Let $R$ be a commutative ring with 1 and let $M$ and $N$ be $R$ -modules. This blog post will be about what is the $R$ -module $M\otimes_R N$ . The source of the material will primarily come from Abstract Algebra, 3rd Edition (by Dummit and Foote) which is a highly recommended Algebra book for undergraduates.

Motivation

The tensor product is a construction that, roughly speaking, allows us to take “products” $mn$ of elements $m\in M$ and $n\in N$ .

Definition

(Following Dummit; there is another equivalent definition using “universal property”, see Wikipedia)

$M\otimes_R N$ is the quotient of the free $\mathbb{Z}$ -module over $M\times N$ (also called module of the formal linear combinations of elements of $M\times N$ ) by the subgroup generated by elements of the form:

$(m_1+m_2,n)-(m_1,n)-(m_2,n)$

$(m,n_1+n_2)-(m,n_1)-(m,n_2)$

$(mr,n)-(m,rn)$

The outcome of the above definition is that the following nice properties hold:

$(m_1+m_2)\otimes n=m_1\otimes n+m_2\otimes n$

$m\otimes (n_1+n_2)=m\otimes n_1+m\otimes n_2$

$mr\otimes n=m\otimes rn$

Recent News about MOE (Ministry of Education)

The following are some interesting news on Singapore’s MOE Ministry of Education. Readers interested may want to check them out:

Motivational: Kung Fu Panda 3 Song

The new Kung Fu Panda 3 song lyrics is out!

Very motivational lyrics. The title itself <Try> is very motivational.

“Only those who dare to fail greatly can ever achieve greatly.” ― Robert F. Kennedy

The full lyrics are as follows (includes some Chinese):

Try – 派伟俊/周杰伦
(《功夫熊猫3》电影全球主题曲)
中文词：方文山
英文词：冼佩瑾
曲：派伟俊
小派：You always have to do something
Just to show the world that you exist
So you try
You hope they’ll see
If on this brand new day you’ll look
On the bright side of the same old street
You will see
What you deserve
Jay：Let’s go
我说几华里我送别了过去
他们说人生的结局非常的戏剧
塞外羌笛孤城马蹄
在武侠的世界里谁与谁来为敌
合：La la la la la la la la la
黄沙里用竹笔写下的字叫勇气
Jay：You just have to try
To be who you are
And you ought to fly
Step into the light
小派：And soon you will find
Be yourself
Somewhere deep inside
There’s a universe right there waiting to be unlocked
The key lies in looking into yourself
Jay：Oh Try try try try
Just do what is right
You’ll fly so high
Let go of the brakes
Be who you are
Be yourself ’cause your power is on
合： When you believe in what you’ve got
You know you’re perfect just be who you are
小派：So they don’t see what you’re made of
But I like you and I know they’re wrong
Now it’s time
To show them what you got
Let the blue skies cheer you on
Embrace the wind we’ll ride along
You’re perfect when you’re who you are
Jay：这世界有些事有些人凭感觉
别管他旌旗密布遍野狼烟霜雪
那故事在穿越而我也在翻页
一行行做好准备敏锐而直接
合：La la la la la la la la la
爱不灭真实的一切废话全收回
小派：You just have to try
To be who you are
Jay：And you ought to fly
Step into the light
小派：And soon you will find
Be yourself
Somewhere deep inside
There’s a universe right there waiting to be unlocked
The key lies in looking into yourself
Jay：Oh Try try try try
Just do what is right
You’ll fly so high
Let go of the brakes
合： Be who you are
Be yourself ’cause your power is on
When you believe in what you’ve got
You know you’re perfect just be who you are
小派：You just have to try
To be who you are
Jay：And you ought to fly
Step into the light
小派：And soon you will find
Be yourself
Somewhere deep inside
There’s a universe right there waiting to be unlocked
The key lies in looking into yourself
Jay：Oh Try try try try
Just do what is right
You’ll fly so high
Let go of the brakes

合： Be who you are
Be yourself ’cause your power is on
When you believe in what you’ve got

Chinese Tuition

Chinese Tuition: http://chinesetuition88.com/

Chinese Tuition Singapore

新加坡华文补习老师

Tutor: Ms Gao (高老师）

Ms Gao is a patient tutor, and also effectively bilingual in both Chinese and English.

A native speaker of Mandarin, she speaks clearly with perfect accent and pronunciation. She is also well-versed in Chinese history, idioms and proverbs.

Ms Gao is able to teach Chinese at the Primary and Secondary school level. She will teach in an exam-oriented style, but will also try her best to make the lesson interesting for the student.

Ms Gao graduated from Huaqiao University, which is founded by late Chinese premier Zhou Enlai.

Contact:

Email: chinesetuition88@gmail.com

Website: http://chinesetuition88.com/

(Preferably looking for students staying in the West side of Singapore, e.g. Clementi / Dover / Jurong East / Boon Lay / Queenstown)

Image of Infinite Strip under Transformation w=1/z

Q: Find the image of the infinite strip $0<y<1/(2c)$ under the transformation $w=1/z$ .

(This question is taken from Complex Variables and Applications (Brown and Churchill))

Answer: $u^2+(v+c)^2>c^2$ , $v<0$ .

Solution:

We are using the standard notation $w=u+iv$ , $z=x+iy$ . From the equation $\displaystyle z=\frac{1}{w}=\frac{1}{u+iv}=\frac{u-iv}{u^2+v^2}$ , we can conclude that $\displaystyle y=\frac{-v}{u^2+v^2}$ .

With some algebra and completing the square, one can obtain $u^2+(v+\frac{1}{2y})^2=(\frac{1}{2y})^2$ , which is the equation of a circle centered at $(0,-\frac{1}{2y})$ with radius $1/2y$ . When $y=1/(2c)$ , the equation of the circle is $u^2+(v+c)^2=c^2$ . As $y$ gets smaller (closer to zero), the radius of the circle becomes larger, while still remaining tangent to the horizontal axis.

Thus, the image of the strip is $u^2+(v+c)^2>c^2$ , $v<0$ .

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
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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.