Slow to Speak

I've been pretty busy with my thesis stuff lately and just kinda dropped this blog thing. I figured that it was taking up way too much of my time, and well...it kinda was. But that was probably because I was spending a great deal of time writing up these fairly length posts. Hmm...maybe I should try just writing shorter posts and seeing if I'll get faster and faster over time?

To summarise the past few months: life's been pretty awesome. =P

Okay, now that the catching up is over and done with - let's do something mildly mathematical. Well, last week I gave a short talk for the Melbourne University Graduate Study Expo. I was meant to be saying a bit about what it's like to do a PhD, and I think that peeps enjoyed my talk but I'm pretty sure that it was less relevant to them. It was basically a room mainly composed of industry people who were thinking about doing further studies to increase their "human capital" - to use economist speak.

Anyways, here're the slides, and here's my talk:

The last conference that I went to was in Shanghai. For the very first talk, a Frenchman steps up to the podium as loud award-ceremony-orchestral music swells in the background. He's wearing a suit of dark grey, with a chain that I can only assume attaches to a fob watch. Instead of a necktie, he is sporting some unholy offspring of a cravat and a pussy bow. He walks up to the podium, leans into it and for the next hour, in his thick and impassioned French accent, he tells us - an auditorium full of 500 or so mathematicians - about the mathematics that he loved, the experiences he's had (like trying to find Boltzmann's grave, and the local grave-enthusiasts not knowing Boltzmann by name, but knowing his equation!), followed by a simple overview of the mathematics that he had invented. That man was Cedric Villani - a Fields medallist (it's like the Nobel prize of mathematics. Like many things in maths: it's lesser known, harder to get and comes with much MUCH less money).

That talk had everything that I love about maths: it's full of these wonderful and colourful characters, it's got the history, the passion and above all else, that talk showcased the beauty and creativity of mathematics.

Now, I've been asked to share with you my experiences doing a PhD in maths. And it's definitely been an adventure - my PhD has taken me to a month in Singapore, a month in America, a month in Germany and a month in China and a week in Istanbul. I mean, I wasn't just in front of a desk the whole time - I was on the alps, mountain biking...etc [something is omitted here for reasons about which I'd rather be cryptic].  BUT, the VAST majority of my time has been spent here in Melbourne, in my office, swivelling about on a chair in front of a Mac reading papers, frustrated by poor notation and typos, and checking my email every half an hour or so.

So, that's what it's like to do a PhD. And well, you've just heard my friend Jon (Kastelan, works at PWC) talk about what it's like to work in the industry and being typical of industry people, he gave you solid, practical advice. Well, I'm sitting in my ivory tower of academia and my questions are a lot more all-over-the-place, but I'll still ask them: what do YOU WANT in life?

I know that's a pretty big question, but a PhD is a pretty big decision - it's a fair few years of your life. For me, it was pretty simple. I love maths. But there are other reasons to do a PhD (and I'm not making a judgement call here): maybe you've heard that statisticians and mathematicians have some of the best jobs in the world in terms of pay and safety; maybe you've already worked in the industry for a few years and you find that you need some heavy mathematical machinery to solve the types of problems that you're dealing with; maybe you just like the sound of Doctor of Mathematics after your name. WHATEVER the reason, have a good think about if it's worth three or more years of reading and writing stuff that almost no-one-else in the world is going to read.

SO what if you don't know? Well, that's where a Masters comes in handy. There's a coursework component where you'll be "forced" to see different types of maths and figure out what you might like. And there's a thesis component where you'll get a preview of what it's like to do research and how to communicate it.

Of course, I'm not saying that you have to go and do a masters - if you're really not sure, and you've got a banking/engineering job line up, why not take it? I mean, getting a job outside of academia isn't a life sentence. I was just talking to a friend last week who's been working for a year or two and has decided that it sucks and he's going to come back and do a PhD next year.

Now, the other question I have for you is: CAN you do a PhD in maths?

As in, do you have the aptitude for it. If you don't know - again, a Masters is a good way for you to get a feel for it. And to be fair, I know that I've been pretty PhD-centric in my treatment of the Masters program, but it's a bit biased. You do get a lot out of a Masters - it gives you a very strong overview of the types of tools that we have at our disposal. Plus, it forces you to tighten up how you learn. There's no room for pride in mathematics - if you don't get something, you will have to ask - your supervisor will make sure of this.

Finally, is Melbourne the right place? For Masters? Ummm...YES. Melbourne's got a pretty well-rounded department, especially for its size. [Sth omitted here coz it's kinda cocky] As for PhD - it actually depends a lot more on what you're interested in. A much more important part of a PhDs is the specific project(s) that you want to work on; but if you do a Masters here, a lot of the people in the department are more than happy to help you find a good grad school/supervisor for your PhD.

So, I've shoved some details in your face, from Cedric Villani to sitting in front of a mac. And it's created this messy tapestry. And maths is all about sorting through these details and finding some underlying pattern. In your case --- these three questions [it makes more sense with the slides]. And I hope that, as you clarify your thoughts - the answer will reveal itself to you [it REALLY makes more sense with the slides]. =P

Statistics Quo

I loathe Chinese medicine, and I'm not entirely sure why. I wondered if it's simply because both my great-grandfather and my grandmother were both traditional Chinese medicinal herbalists (and acupuncturists to an extent) and I just like to antagonise my forebears. But then again, they were also both Christians and that's indelibly stuck with me?

I don't remember much of my great-grandfather, other than that he had this wooden reclining chair that I loved. He lived about a block away from me when I was a kid, and often I'd run over to his place and plonk myself atop his thin and fragile frame as children do, rollin' his tobacco as he gently rocked the chair. I guess that's probably why I like these types of chairs so much: there's something reassuringly crib-like about them.

They're both death now, but I vaguely recall that one of the last conversations that I had with my grandmother was about the day that my great-grandpa died. Apparently, when they gave me the news (*), I stood up in the cold fury of my then-boyish machismo and demanded that I wasn't going to let him die. I guess that ought to be the kind of story that I should tell if I ever become a doctor...well, a medical doctor, as opposed to a philosophical one (**).

And I know that rationally, given what I know, I probably shouldn't be hating Chinese medicine so much. I mean, it has the potential to actually be efficacious (both as placebo and drug). For example, the best antimalarial drug used world-wide has been employed by traditional Chinese medicinal practitioners to treat malaria for over 15 centuries (in fact, we're using it so much that malaria is beginning to acquire a tiny bit of resistance). Okay, that was somewhat of an extreme example. In particular because the history of how this drug came to be adopted is often paraded around as a prototypical demonstration of the disdain held by conventional scientists for all other paradigms of knowledge (***).

See, the thing about Chinese medicine is that its origins are probably statistically based...and now has evolved into a strange mish-mash of pseudo- and proto-science (****).

At the heart of much of Chinese medicine (at least historically) seems to be trial and error. Let's say that someone got sick in a village, and they hear from their neighbours that drinking the soup made from that tree near the town pond will help them get better. So, they willingly make this soup and two things happen: either they get better or they don't. Now, if the soup actually is an effective medicine, then the odds are that the person will probably get better and so this knowledge will probably be passed on. On the other hand, if it the soup isn't really effective, then it's a bit trickier because it's not just a case of the drug not working and this treatment being forgotten - because the drug could still function quite well as a placebo. Well, actually, the real story is even more complicated, some times the treatment could be useless but people just think that they've not boiled it properly or picked the wrong tree or something like that; and vice versa, you might think that you kinda just got lucky (especially if the soup tasted like and looked like water and you told your mum that she was crazy for thinking that your pernicious sickness could be cured by something so insubstantial).

But whatever the case, over a long period of time (and China has been practicing such herbal remedies for a fairly long period of time, and with a very large population) more than just a few truly effective drugs are bound to result from this. This is effectively a Darwinian method of doing pharmaceutical research.

Now, that all sounds nice and pragmatic, but unfortunately (or fortunately, depending on your beliefs), in the process of developing a scientific model for why and how these medicines function, the absence of an understanding of chemistry resulted in the current Chinese medicinal theory with its undeniable Taoist undertones.

Now I wish that, we can simply stand back in our scientific smugness and mock these prehistoric physicians. But I feel that I must concede that the underlying theory of Chinese medicine, given that it has been derived without the aid of any of the modern investigative tools of evidence-based medicine, is astonishingly functional and reasonable. I mean, okay...so the most basic idea is that all symptoms are classified as being either yin or yang in nature. Since yin is meant to be more moist in nature (it means shade), symptoms to do with a lack of yin include cold sweats, dry mouth...etc. And since yang is meant to be more heaty in nature (one of the characters for the Sun), a lack of yang is supposed to manifest in symptoms such as cold limbs (under room temperature), diarrhoea (*****) or having a pallid complexion (it's not quite as well defined as this, but it's just to give you an idea).

And the idea then is that when you get one of these symptoms you often have a lot of the others. For example: when I have the flu, I am likely to have a pallid complexion, I'm likely to feel "the chills" (hence cold limbs) and I might even have diarrhoea. Then, in treating for the chills (e.g.: chicken soup), I might be dealing with the root cause of the other symptoms too.

So, there is some merit to this classification. This yin/yang thing isn't a very well defined variable, but it's surprisingly difficult to disprove it without assuming some knowledge of human anatomy (even then, it's still difficult) - precisely because of this statistical grounding (and to give some perspective on how much human anatomy people understood back then, we didn't even figure out how our circulatory system worked until some time after the first millenia (okay, Galen came close, fair enough. But Galen was a Baus).

Now, you might think that modern evidence-based medicine would have weeded all of this out already, But not so! A lot of medical professionals in China (and out of China too) are being trained both in evidence-based (Western) medicine and traditional Chinese medicine. Now, a part of this is because many doctors are terrible statisticians, but I think that there's also some element of pragmatic dishonesty going on here. Some of the best Chinese medical practitioners are increasingly introducing new language to the theory of Chinese medicine, with concepts like heatiness-that-looks-like-coldness being used to account for (in my mind) the inadequacies of the yin/yang system. And there are definitely practical merits to doing this.

1. Many (esp. older) Chinese people are very vary of Western-based medicines, considering conventional drugs too "hard" or potent, and "hence" having many side-effects. (But they mentally ignore the fact that many Chinese herbal supplements also have drastic side-effects...including: death) In particular, they often function as excellent placebos.
2. There are a hella heap o' con-men who utilise Chinese medicine as their preferred method of robbery. So having an officially sanctioned form of Chinese medical practice helps to regulate this jazz.
3. It works. As in, in curing people of stuff.
4. It helps to preserve a significant aspect of the Chinese cultural heritage. And as the walls of Communism are slowly being eroded by the winds of economic growth, China is turning to things like nationalism to hope to maintain its political stability.

sigh. 

Well, well, well...this was unexpected. I think that I was originally just going to point out that stats is being used in medical research in a winnar way like:

Using stats yo'. Cancer treatment winnar. Chomp!

But I guess...yea...I really hate traditional Chinese medicine - especially because, I, of all people, am forced to occasionally defend it. Hm...and could it be that another reason that I dislike it is because it reminds me of religion? That is reminds me of the foolishness of what I preach and practice as a Christian?

Maybe that's a post for another day. =D


*: in no uncertain terms either. Hey man - we don't mince words in my family, and my parents have no qualms telling me that I was an accident =P.
**: strangely enough, the etymology of the word doctor comes from docere, which means teacher. Plus, a doctor used to be called a leech.
***: okay, so the mechanism by which Artemisinin works is really tricksy - it's kinda usually stable but becomes unstable upon contact with some part of the malaria parasite, which leads to it blowing up like a bomb. It was pretty surprising to all scientists that something like that could work. On the other hand, I do have to concede that most scientists are kind of closed minded (and with good historical reason, I guess?).
****: look, you don't need to lecture me on the irony of calling something so old a protoscience.
*****: stemming from the Chinese idea that cold/flus are caused by the wind getting into your stomach - of course, this is a surprisingly difficult hypothesis to disprove without modern medical equipment, since there is a big chance that you will catch a cold (during the right season) if you sleep in a poorly heated environment and your immune system isn't performing as well. Or to put it into statistician lingo: there's a high correlation between a drop in your core temperature (esp. during sleep) under "windy" conditions and the inefficacy of one's immunue response. SOOOO, the sensation of bloatedness often associated with diarrhoea conjures up the image of this wind rattling about one's stomach, hence the association with "coldness" - i.e.: the absence of yang.

Today I Learned: It's All Goodstein

I came back from China last Thursday and I've found it surprisingly difficult to restart this blog.

Maybe it's the thought that now that I've finished my semester with my Accelerated Maths tutorial class, I'll no longer have a real human audience for this writing? Or maybe it's just sloth. =P

So, I was away in China for a little over two weeks. It's summer there and suffocatingly hot and humid.

The first week or so I spent with my grandfathers: they live in the same apartment building, one floor away from each other. It was good to see that my (paternal) gramps was much healthier looking than the last time that I went back, although his attention span has grown somewhat evanescent. He would sit in his wicker chair, nude from the waist up, brows poised delicately on the fulcrum of furrowedom and unfurrowedness. The clacking of the rotating mechanical fan and his rhythmic laboured breathing lending his countenance an ethereal air. Except when we spoke, I was never sure if he was with me, or dreaming.

And the second week I spent in Shanghai. Yea, Shanghai's pretty chill, just not weatherwise. But let's start with the actual journey to China.

My flight was delayed by half an hour, so I sat in frustration (*) reading Harer and Penner's Combinatorics of Train Tracks. Nearing the original (undelayed) boarding time, I hear some commotion behind me and turn around to have a look: a Chinese girl had approached the metallic security door that you go through to get onto the plan. She was feeling...no...she was...groping every crevice of the door, until a few of the elderly aunties came up to her and told her that boarding had not yet commenced. She seemed content with that answer, strolled back to the waiting area, before swivelling about and running back to the gate, flailing at imaginary buttons that might magically open this door.

The buttons, they just weren't there.

And soon, the Chinese aunties came hobbling towards her once more, and so we saw the repetition of a few more cycles of this explanation/flailing-loop. Throughout this process, the airport staff are struggling to communicate with her, but to no avail. In between their attempts to talk to her, she makes a phone call to her dad (I found this out later), frantically crying out in fumbling, tumbling Chinese: "I'll listen to you", "The steel door is closed. It's closed" and "I want to go home; I want to go home; I JUST WANT TO GO HOME".

I just want to go home.

At this point the staff are asking for someone to help them translate, and I happily acquiesce, and we independently arrive at the (obvious) conclusion that she is probably a higher-functioning Asperger's syndrome-sufferer (**). Soon, security is called, and as they gently drag her from the security gate, I call out to her and begin explaining to her that I understand her desire to be on the plane since it's nearing 11. At this, she seemed relieved. I then ask if she received the firm instructions that she must be on the plane by 11 from a family member like her mother or her aunt (I didn't know that it was her father at this point) in the hopes that I could convince her to call them again and that I might then explain to them the actual situation (my guess is that she probably miscommunicated to the plane situation, and that perhaps they could give her new instructions). Maybe she was troubled at my line of questioning, maybe I just reminded her of her family and homesickness, but whatever the case she ran off.

Immediately, the security guards asked me to help talk to her, so I race around the crowd, get to her and by the time that I reach her she is in a state of hysteria. She is surrounded by a flock of chanting aunties, and has begun dry-sobbing. I offer her my hands (***), she takes them, and I begin anew my attempt to get her to call her father (by this time I've been informed by someone that it was her father).

She does (****) calm down.

But then I'm interrupted by a forceful middle-aged uncle who tells her that she must calm down and a middle-aged lady with an infant in her arm saying "look, my baby isn't afraid, so you shouldn't be scared either". And....and...I back away, entrusting them with the situation. And returning to my seat.

I..just walked away. =/

And then security took her away.

I mean, fair enough - security was probably always going to take her away; even if I'd managed to get her to contact her father, explain the situation, and have him relay to her new instructions regarding plane boarding, how might the airline staff trust her to behave sufficiently "normally" on the long flight back to China? A part of me keeps trying to find ways out - could her father have assigned me as a temporary "guardian" on this flight? Indeed, completely ridiculous "solutions" like these.

And after boarding, the empty seat next to mine on a mostly filled out plane did not help relieve me of these haunting thoughts: I kept wondering if this was her seat, the seat of that poor, crying, sobbing girl flailing herself against the hard, cold, metallic barrier: wanting to go home, but hindered by her own frets and incomprehension.

I guess, as a Christian, there's a parable here - there is, for me, a reminder that often we Christians, Jews and Muslims (worshippers of the same God) struggle and flail and cry and shout to "go home", but are stumbled by our own incomprehension. We do not see that Heaven/Paradise is to come to know, love and even befriend the God of the Jews, the Father of Abraham/Ibrahim, the Creator (*****) of the universe. And instead, miss the point.

So...erm...in remembrance of my inability to communicate with this girl...let's...let's try to communicate beyond the limitation of language. Let's try to do today's maths without words:

\(1=2^0.\)

1. \(\leadsto 3^0-1=0.\)

\(\Rightarrow G(1)=1.\)

 \(2=2^1.\)

1. \(\leadsto 3^1-1=2=2\cdot 3^0.\)
2. \(\leadsto 2\cdot 4^0-1=1=4^0.\)
3. \(\leadsto 5^0-1=0.\)

\(\Rightarrow G(2)=3.\)

\(3=2^1+2^0.\)

1. \(\leadsto 3^1+2^0-1=3^1.\)
2. \(\leadsto 4^1-1=3=3\cdot 4^0.\)
3. \(\leadsto 3\cdot 5^0-1=2=2\cdot 5^0.\)
4. \(\leadsto 2\cdot 6^0-1=1=6^0.\)
5. \(\leadsto 7^0-1=0.\)

\(\Rightarrow G(3)=5.\)

\(\Rightarrow G(4)=\ldots 7?\)

\(4=2^2.\)

1. \(\leadsto 3^3-1=28=2\cdot 3^2+2\cdot 3^1+2\cdot 3^0.\)
2. \(\leadsto 2\cdot 4^2+2\cdot 4^1+2\cdot 4^0-1=41=2\cdot 4^2+2\cdot 4^1+4^0.\)
3. \(\leadsto 2\cdot 5^2+2\cdot 5^1+5^0-1=60=2\cdot 5^2+2\cdot 5^1.\)
4. \(\leadsto 2\cdot 6^2+2\cdot 6^1-1=83=2\cdot 6^2+6^1+5\cdot 6^0.\)
5. \(\leadsto 2\cdot 7^2+7^1+5\cdot 7^0-1=109=2\cdot 7^2+7^1+4\cdot 7^0.\)
6. \(\leadsto 2\cdot 8^2+8^1+4\cdot 8^0-1=139=2\cdot 8^2+8^1+3\cdot 8^0.\)

\(\ldots\)

192. \(\leadsto 384^2+20\cdot 384-1=155136=384^2+19\cdot 384+383.\)

\(\ldots\)
\(\ldots\)
\(\vdots\)
\(\ldots\)

14. \(\leadsto 0.\)

\(\Rightarrow n=G(4)=3\cdot 2^{77}(2^{3\cdot 2^{77}}-1)+44>10^{100000}.\)

!!!

\(G(6)=???\Rightarrow\) $5. =)

Note that this is all made possible thanks to Goodstein's theorem, as in: it's not completely obvious that \(G(n)\) is finite for every positive integer \(n\). And I'm not the one paying out $5 for computing \(G(6)\), it's Ronald Graham.


*: My frustration did not stem from the delay of my flight, it was more because I was stuck on one of the steps in one of their proofs.
**: May be highly non-pc, but I will call it suffering, and I won't bother justifying myself in this footnote. =/
***: Let's not romanticise this in the most literal interpretation of "romanticise", she was no Rita Leeds.
****: At least...she does in my memories - which I don't completely trust, although I've written most of this down in a diary, so it should be fairly accurate.
*****: Whatever creation means.

I'm back!

I've finally returned from Shanghai this morning, and I must admit that I've had pretty interesting plane trips there and back, maybe I'll write about it when I've gotten back in the rhythm of things.
 
Oh and I totes coined a new interjection in the Mirror Symmetry seminar this arvo:

"maaaaaan, you're blowin' ma cheesecake!"

Cheesecake: blown.

Maths on Steroids: final?

Sorry guys: I tried, but I just couldn't think of a non-trite-sounding first sentence to this post, so I wrote a con-trite-sounding one instead. Okay, maybe I'm not that contrite...in fact, I'm pretty happy with how that sentence turned out. =P

There is perhaps, in the West, a slight cultural inclination towards the belief that teachers are somehow less. After all, it's often said that those who can, do. And those who can't, teach.

I can't be bothered refuting this claim, someone-else has already done it for me. But it does strike me as odd that even in a major segment of the teaching profession: tertiary education, this view is espoused. I took us about two hundred years after the adoption of the printing press in the West for scientific journals to happen (compare with erotic novels, which pretty much happened overnight).

There is no denying that this gadget has been instrumental (*) in the spreading of scientific ideas among academics. But one of the side-effects of the scientific journal has been a shift in the focus of academia. Traditionally, a professor's primary role is to profess that they know nothing about their field of expertise and then hammer you with all the details of how poorly they understand their material over the next few months. But some where along the line, academia started to focus less on teaching and more on the number of and quality of one's publications.

While I agree that there's probably a strong correlation between the two, I feel that it's tragic that some how research is touted as the primary objective of academia, and the noble act of teaching is...disdained.

...hmm...

I dunno, that was a bit of a random start to things. And...and I guess that I have been thinking a bit about teaching. Both of my parents were teachers in China, and they have a very pragmatic approach to teaching: make sure that the kids get good marks on their exams. On the other hand, I can't really name anyone at uni who would explicitly claim that this is the right way to do this; there is a hope that we can steer students far from the perils of an "exam-oriented" approach to study. And I feel... I feel like this is a reflection of an overly simplistic and naive view of human motivation. But...yea, I should go to sleep soon. I'mma fly off to China in a few hours and this is probably just some sort of procrastination.

Orritey! It's tiiime fooooor...

General Exam tips:

1. Get full marks.
2. Aim to get 120/100. When you fail to achieve your goals, you'll still have full marks.
3. Keep your hands warm. My personal preference is to bring a trashcan full of charcoal - this way, you can roast chestnuts while you get full marks! Failing that, I guess you could go buy one of those Japanese hand-warmer packs from Daiso?
4. Do the questions in order. I know that one of the standard pieces of advice that we're supposed to tell peeps is that you DON'T NEED TO do the questions in order. But, it's much harder for BOTH you and your exam marker to miss one of the questions this way. I think that the only reason that we're supposed to tell you that you don't need to do stuff in order is because sometimes people get stuck on one question and waste waaaaay too much time on it. So...
5. Don't waste waaaay too much time on one question. 
6. Do not leave an exam 2 hours early because you finished it and you and your friends "agreed" beforehand that you were all going to finish waay early and go out and party and stuff and then they end up not leaving 2 hours early and you get sunburnt/frostbite standing outside the exam hall and you realise much later that you forgot to do one of the questions because it was hella annoying and you were gonna leave it till the end sad face.
7. Write in English where necessary.

And hints to the questions that I gave you guys in the last post:

1. Duh. It's all Greek to me.
2. Just do some computations with monomials.
3. *Cough* the eigenvectors should be all the monomials...just make sure that you do some sample computations and define the right generalisation of \(T\).
4. I think that it's probably important to have two variables? Maybe???
5. The trick is to group things into sequences of two (non-overlapping) days.
6. *Cough* ...just...*cough*.
7. Diagonalise it!

Orrite. Own those exams guys! Own them like you paid fees to attend uni. =P

*: you can't really deny it because gadgets are pretty much instruments, right?

Puzzle Hunt 2013!

Hey guys, hey guys!!!

Shhhhh...

No maths tonight (*): it's Puzzle Hunt time!

*: I know, kinda sacrilegious. =)

Today I Learned: Generalised Pythagoras' Theorem

I had a hotpot dinner at church last Saturday. And speaking of hotpots - my friend broke her's (*). And no, that's neither a euphemism nor a metaphor. She was boiling water on the stove, forgot about it, only to come back to find a toasty dry pot roasting over blue flames. Quickly swiping the pot into the sink, she turned on the tap - only to find it shattered by the icy water.

This of course reminded me of my various childhood failures at drinking milk.

I used to drink about a litre of milk a day. I suppose that I should probably mention that China's experiments in pragmatic socialism had made this economically viable, but that's hardly the point. Instead, the point is that I was a champion of milk-drinking. So I was pretty shocked the first time that I tried milk here in Australia.

It was revolting.

It was all...creamy, and milky and not at all alike that watery malk produced by our proud proletariat Communist cows! Aussie milk: so gross.

So, to maintain my strong and healthy bones, I learned to take my milk with a big spoonful of sugar. Which again is slightly odd, because I was fine with taking medicine, in fact, I kind of liked it. But I'm getting distract when I should be distracting you with the story of that time that I mistook MSG for sugar - it certainly didn't help with my milk-loathing.

There was also one time when, home alone, I decided to microwave some milk during the commercial breaks (I can't remember the show, it was likely Sesame street or Play School), only to realise upon retrieving the glass that I'd forgotten to pour the actual milk into the glass (***). Shattered by my mindless folly, I took the glass to the sink and poured myself some water - only to find the glass shattered by this second act of mindless folly.

I've basically told you the same story twice: hot vessel + cold water = shattered pride. And this retelling of the same story but with minor details adjusted happens a lot in research mathematics. It is what we call generalising a result.

So let's talk about Pythagoras' theorem.

Oh wait, let's not - I've already done it here. Now there's an obvious generalisation to this result in higher dimensions. If we paraphrased/reinterpreted Pythagoras' theorem to say: the distance between two given points \((x_1,x_2)\) and \((y_1,y_2)\) on the plane \(\mathbb{R}^2\) is \begin{align}\sqrt{(y_1-x_1)^2+(y_2-x_2)^2},\end{align} then a fairly mundane generalisation would be that the distance between two points \((x_1,x_2,\ldots,x_n)\) and \((y_1,y_2,\ldots,y_n)\) on the plane \(\mathbb{R}^n\) is given by: \begin{align}\sqrt{(y_1-x_1)^2+(y_2-x_2)^2+\ldots+(y_n-x_n)^2}.\end{align}

But here's a more interesting generalisation that actually resembles the usual Pythagoras' theorem:

Take a square, slice off a small corner and you get a right-angled triangle. Take a cube, slice off a small corner and you get a right-angled tetrahedral. So take \(n\)-dimensional hypercube, slice off a small corner and you (should) get a right-angled \(n)-simplex (basically the generalisation of triangles and tetrahedrals)! Now, with a little imagination you should be able to see that this \(n\)-simplex has \(n+1\) "hyperfaces" - just like how a (\(2\)-dim) triangle has \(3\) sides or how a (\(3\)-dim) tetrahedral has \(4\) sides, You know how it is (****). And that one of these hyperfaces \(S_n\) is bigger than the rest \(S_0,\ldots,S_{n-1}\). Then we have following identity:\begin{align}vol(S_0)^2+\ldots+vol(S_{n-1})^2=vol(S_n)^2,\end{align} where \(vol(S_i)\) is the \((n-1)\)-dimensional volume of \(S_i\).

Awesome result, right? In particular, it says that the sum of the squares of the areas of the three smaller faces of a right-angled tetrahedral is equal to the square of the biggest face - i.e.: the sort of hypotenusoidal face.

To bad that the proof that I have for this isn't as cool as I want it to be. Although there are some nice ideas here, so very roughly outline it:

1. Show that a \(n\)-dimensional cone/pyramid with \((n-1)\)-dimensional base \(B\) has volume \(\frac{h}{n}vol(B)\), where \(h\) is the height of the apex (orthogonally) from the base. This can be done by considering about how the volume of \(B\) changes in the cross-sections of this cone parallel to the base, and then integrating these cross-sectional volumes along the height.
2. Conclude therefore that the volume of a right-angled simplex with vertices at \begin{align}(0,0,\ldots,0),(h_1,0,\ldots,0),(0,h_2,0,\ldots,0),\ldots,(0,\ldots,0,h_n)\end{align}is given by \(\frac{1}{n!}h_1\times h_2\times\ldots\times h_n\). And note that this may also be written as \(\frac{1}{n!}\) times the determinant of a matrix obtained by taking the non-origin vertex coordinates of our simplex as the rows of this matrix.
3. Combine this previous point with the fact that (high-dimensional versions of) shear mapping preserves volumes to get that the volume of a (general) \(n\)-simplex with one vertex at the origin, and the rest at \begin{align}(a_{1,1},a_{1,2},\ldots,a_{1,n}),(a_{2,1},a_{2,2},\ldots,a_{2,n}),\ldots,(a_{n,1},a_{n,2}\ldots,a_{n,n})\end{align} is given by \(\frac{1}{n!}\) times the determinant of the matrix \(A:=[a_{i,j}]\).
4. Now, embed this simplex in a one-dimension higher space, and in so doing + some shearing mappings, we get a formula for a \(n\)-simplex in \(\mathbb{R}^n\), which in term allows us to rearrage the vertex coordinate terms in this determinant into edge lengths. This is, to my current understanding, mainly (some impressive) algebraic manipulation, and may be found on page 124 and 125 of Sommerville's An Introduction to the Geometry of n Dimensions. It should be said that at this point, we're seeing that the square of the volume of our simplex is expressed completely in terms of the squares of the lengths of the edges of our simplex!
5. The proof of the last part that relates the volumes of all the hyperfaces of our simplex is also currently algebraic in nature, and may be found in the following paper.

Actually, in the process of writing this up, I've seen that Alvarez has an earlier proof of this result which seems much more elementary and uses induction. Maybe I should read it...

Maybe. =)

Update: yea, I read it, it's actually a much nicer proof. I mean, I don't get to make the cool observations about expressing volumes in terms of determinants and the shear mapping stuff, but it's a pretty chill induction proof.

*: I think that it was a hotpot pot, it might have been just a regular pot though. Who knows? She does, probably.

**: huh...it's curious now that I think about it...it's almost as if milk is actually an acquired taste, kinda like Vegemite?

***: on second thoughts, I could be lying here - I might have wanted water instead...huh, it was either water or milk...I'm leaning more towards water now.

****: this is an Arnold Schwarzenegger quote from Hercules in New York, but I just can't find footage of it on the interwebs. =(

Maths on Steroids: Tute 6

I have no idea why I thought that today was Yom Kippur - especially because I actually thought that it was Jewish New Year's Day: Rosh Hashanah. I suspect that a combination of illness, watching old Family Guy and South Park episodes and stealing a glance (over someone's shoulder on the train) on a Rabbi's chaplaincy report (*) must have caused my mind to mash all of these ingredients into some crazy-festive-kosher-holiday-cake-mix. And if that doesn't make sense, well good - it doesn't make any sense to me either. But let's all hope that the maths that I'm about to write does make sense.

Sooooo guys, remember this question from your assignment?

"Find an \(n_0\) such that for all \(n\geq n_0\), \(3^n>n^4\). Give a proof by induction."

Well, it's pretty painful for me to mark a proof by induction by induction in an induction in an induction (**), so I'm going to remind you guys that comparison trick that I showed you in the tutes...obviously, I'm only going to do the induction step, and I'll forgo the usual setting-up:

Assume that \(3^k>k^4\) is true, let's show that \(3^{k+1}>(k+1)^4.\)

Well, since \(k\geq 8\) - this follows from the base case holding for \(n_0=8\), we obtain the following sequence of inequalities:\begin{align}3^{k+1}> k^4+2k^4&\geq k^4+16k^3\\&=k^4+4k^3+12k^3\\&\geq k^4+4k^3+96k^2\\&>k^4+4k^3+6k^2+4k+1=(k+1)^4.\end{align} Note that I've tweaked it a tiny bit from what we did in class - just because \(k\geq 8\), doesn't mean that your inequalities always have to be of the form \(k^{m+1}\geq 8 k^{m}\). Speaking of tweaking stuff, Nick came up with a cute version of this comparison trick in his assignment. If you don't mind fractions, you can just straight off use that:\begin{align}k^4+4k^3+6k^2+4k+1\leq(1+\frac{4}{8}+\frac{6}{8^2}+\frac{4}{8^3}+\frac{1}{8^4})k^4=\frac{6561}{4096}k^4<3k^4<3^{k+1}.\end{align}

Oh, and one final comment before we go through the MATLAB graphing code from this week's tutorial: be aware of the flow of your logical arguments. I understand that sometimes your thinking as you progress through a proof is pragmatic, it's like: to get A, I need B; to get B, I need C...etc. But when it's written simply in the form: A, B, C, ... with no (backwards) implication arrows, I'm going to assume that you mean that A implies B and B implies C and so forth.

Orrite, MATLAB time! First things first, here's the code for the m-file that I used to generate the MATLAB plot during Tuesday's lab:

x=-3:0.1:3;
n=length(x);
y=[];
for k=1:n
    if x(k)<-2
        y(k)=0;
    elseif x(k)<0
        y(k)=x(k)^2;
    else
        y(k)=x(k)^3+2;
    end
end
plot(x,y,'*')

Ultimately, we want to produce a graph (of points) by the end of this exercise using MATLAB's "plot" function. Now, plot works by taking in two vectors of the same length \(x\) and \(y\) and using them respectively as the \(x\) and \(y\)-coordinates of the points that we wanna plot. And specifying '*' or 'x' or '+' or '-' in the third entry in plot(x,y,'*') tells MATLAB what type of points it should draw. If you just type plot(x,y), MATLAB will "join the dots" by default - try it out and see what happens!

Let's break it down, line by line:

1. We create a vector \(x=[-3, -2.9, -2.8,\ldots, 2.9, 3]\). This will constitute the \(x\)-values of the points that we'll draw.
2. Since we need to create a vector of \(y\)-values of the same length, I decided to create a variable \(n\) that keeps track of the length of \(x\).
3. We create a vector called \(y\) that at the moment is empty. As in, it's a length zero vector. This step is, I think, unnecessary in MATLAB, but I dunno, I see it around every so often - maybe it's important in another programming language? If anyone knows, tell me. =P
4. We start out for-loop. Note that \(1:n\) creates a vector of the form\[[1,2,3,\ldots,n-2,n-1,n],\]and for loop basically goes through this vector and runs everything inside of the for-loop (i.e.: lines 5~11) first for \(k=1\), then for \(k=2\), then for \(k=3\),...etc., until \(k=n\). 
5. This is the beginning of our if-loop inside of our for-loop. And the first condition is to check if the \(k\)th value of the vector \(x\) is less than -2. So recall: \(x(k)\) means the \(k\)th value of \(x\).
6. If the condition of \(x(k)<-2\) is satisfied, then we set the \(k\)th value of the vector \(y\) to be \(0\). Note that if \(y\) is shorter than length \(k\), this step implicitly extends the \(y\) vector to being length \(k\). Just to reiterate: the final outcome of this first portion of the if-loop is that if our \(x\) values are strictly less than \(-2\), our \(y\) is just \(0\). Which is precisely the first part of the hybrid function that we're meant to draw.
7. This second condition of our if-loop checks if the \(k\)th entry of \(x\) is less than \(0\). Note that since the first condition of the if-loop has already taken care of \(x(k)<-2\), this condition is really checking for when \(-2\leq x(k)<0\).
8. Given that the condition that \(-2\leq x(k)<0\) is fulfilled, we assign the \(k\)th entry of the \(y\) vector to be our \(x\) value squared. So, this is just saying that on the domain of \([-2,0)\), we're drawing the graph of \(f(x)=x^2\).
9. This last condition of our if-loop takes in everything-else in our domain (which in this case is \([-3,3]\) - I just picked this domain out of thin air, in the actual MATLAB exam, it should be given). So in this case, we'll end up dealing with \(0\leq x(k)\leq 3\).
10. Much like line 8, this is just saying that on the \([0,3]\) portion of our domain, the graph that we're drawing looks like \(f(x)=x^3+2\).
11. Ends our if-loop.
12. Ends our for-loop. Notice that you can always tell the start and ends of an if or a for loop using the natural indentations that MATLAB introduces.
13. Uses the plot-function native to MATLAB to draw our points on a plane.

Psst: at each step here we put a semi-colon ; at the end of every line where MATLAB does a computation. This stops the program from outputting something on the screen - otherwise you'd get clunky computations appearing in the program console. Try to remove one of those semi-colons and running the program if you wanna see what I mean. Also, the most common mistake that students make with the sort of problem in the MATLAB exam is to write \(x\) instead of \(x(k)\) and \(y\) instead of \(y(k)\) inside the loops. Always remember that \(x\) and \(y\) are vectors! Another common mistake is to get the conditions for the if-loop wrong, I mean: sometimes you need strict inequalities, sometimes you need non-strict ones.

So, I hope that all made sense, because I'm about to show you some other ways of drawing this graph. Now, recall from the first week that we can concatenate two row-vectors \(w1\) and \(w2\) by writing \([w1,w2]\) in MATLAB. Well, we could have used this to "extend" our \(y\) vector from a length \(0\) vector to a length \(n\) vector. And here's how: instead of writing \(y(k)=0\), \(y(k)=x(k)^2\) and \(y(k)=x(k)^3+2\), we could have respectively written \(y=[y,0]\), \(y=[y,x(k)^2]\) and \(y=[y,x(k)^3+2]\). Make sure that you understand why that works. Remember: it's MATLAB, if you wanna know how something works, just type it into the console and see what happens!

And there's an even "better" way of drawing these vectors without using if and for loops, which is very natural to MATLAB. It does have the slight disadvantage in that for the program that we wrote, we can increase the resolution of our graph by changing the parameter \(r\) in \(x=-3:r:3\) to be something smaller/finer. And the method that I'm about the describe is a tiny bit messier in this regard. But yea...the idea is that we can just produce the vector \(y\) "directly":

x1=-3:0.1:-2.1;
y1=0*x1;
x2=-2:0.1:-0.1;
y2=x2.^2;
x3=0:0.1:3;
y3=x3.^3+2;
x=[x1,x2,x3];
y=[y1,y2,y3];
plot(x,y,'+')

In fact, we could probably shorten all of that into just three lines, but I don't have MATLAB with me right now and I can't check if what I wanted to write down makes any sense. You guys should totes try to condense it to three lines though!

*: conclusion: Rabbis are the win! Respect ++. =P
**: having read that sentence, you now know how it feels.

Maths on Steroids: Tute 5

Arghghh...so cold.

I have basically zero cold tolerance. I wonder if it has something to do with my consistently running off to South East Asia during the last few Aussie winters. Maaan, those were good trips: got more than fifty mosquito bites on just one arm on the first night (in a malarial region, the guy I shared a room with and the two kids next door had malaria), spent hours cleaning toilets in a house of guys, nearly rode a motorbike off a cliff...etc, and yet still some of my happiest and most relaxing holidays (*) ever! =D

So yea, sometimes things aren't quite what you expect them to be - and today, I'mma write down one example of something that you mightn't have expected to be a vector space.

A vector space is essentially a whole bunch of vectors so that you may perform vector addition and scalar multiplication to your heart's content. So why are these two things important? Well, they're the basis of row-reduction, right? If you can add and subtract vectors and scalar multiply vectors then you can do row-reduction, which in turn means that you can solve linear equations...etc, and you basically get all of linear algebra.

We formalise this intuition with a semi-clunky definition that says that a vector space is a set \(V\) with a notion of addition \(\oplus\) and scalar multiplication \(\otimes\) that satisfy eight axioms. Most of the time, the addition and scalar multiplication is pretty obvious, but if I choose my vector space to be based on the set \(V:=\mathbb{R}^{+}=(0,\infty)\), then the usual addition and scalar multiplication will not suffice to make \(V:=\mathbb{R}^+\) a vector space.

But what if we took \begin{align}u\oplus v:=u\times v=uv\text{ and }\alpha\otimes v:=v^\alpha?\end{align} Turns out this will make \((V,\oplus,\otimes)\) a real vector space.

Let's check this for all eight axioms. We're going to assume through-out and \(u,v,w\) are arbitrary vectors in \(V\), and \(\alpha,\beta\) are arbitary real scalars in \(\mathbb{R}\).

1. Associativity of \(\oplus\): \begin{align}u\oplus(v\oplus w)=u\oplus (vw)=u(vw)=uvw=(uv)w=(uv)\oplus w=(u\oplus v)\oplus w.\end{align}
2. Commutativity of \(\oplus\): \begin{align}u\oplus v=uv=vu=v\oplus u.\end{align}
3. Additive identity element: this is actually slightly tricky. The identity (or zero) element in this vector space isn't the number \(0\) - which isn't even an element in \(V=\mathbb{R}^+\). It is in fact the number \(1\in V\)! And here's the proof that \(1\) is the zero-element (**):\begin{align}1\oplus v=1\times v=v.\end{align}
4. (Additive) Inverse element: this is also slightly tricky, but I guess if you're figured out the last point, this should make sense. Now, since \(v\in\mathbb{R}^+\), its reciprocal \(v^{-1}\) is also in \(V=\mathbb{R}^+\) and is indeed the inverse element of \(v\):\begin{align}v\oplus v^{-1}=v\times v^{-1}=1.\end{align}
5. Distributivity I:\begin{align}\alpha\otimes(u\oplus v)=\alpha\otimes (uv)=(uv)^{\alpha}=u^{\alpha}v^{\alpha}=(\alpha\otimes u)(\alpha\otimes v)=\alpha\otimes u\oplus\alpha\otimes v.\end{align}
6. Distributivity II:\begin{align}(\alpha+\beta)\otimes v=v^{\alpha+\beta}=v^{\alpha}v^{\beta}=(\alpha\otimes v)(\beta\otimes v)=\alpha\otimes v\oplus\beta\otimes v.\end{align}
7. Compatibility of scalar multiplication:\begin{align}\alpha\otimes(\beta\otimes v)=\alpha\otimes(v^{\beta})=(v^\beta)^{\alpha}=v^{\alpha\beta}=(\alpha\beta)\otimes v.\end{align}
8. Scalar identity element: in this case, the scalar identity element is \(1\in\mathbb{R}\) - note that we're distinguishing this from \(1\in V=\mathbb{R}^+\), which is totally a vector. =)\begin{align}1\otimes v=v^{1}=v.\end{align}

Oh good! We've checked all eight axioms and see that \((V,\oplus, \otimes)\) is indeed a vector space.

Now, if you really think about it, what's REALLY going on is that I've simply disguised the usual one-dimensional real vector space \((\mathbb{R},+,\times)\) as \((V,\oplus, \otimes)\) by exponentiating and then re-interpreting what addition and multiplication means once we've exponentiated (***). So here's an exercise:

Make \((0,1)\) into a vector space!

*: surprisingly more relaxing than spending a lot of time onsen-ing in Japan?
**: yes, it sounds stupid, but the truth often sounds stupid. =)
***:it doesn't matter which exponent you choose, so long as it's positive and not \(1\).

Monsters of the Pocket!

Awwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww...yeah!!!

Just found out about the new Pokemon series due to be released this year. So, why did they opt to call it Pokemon \(X\) and \(Y\) instead of naming it after increasingly obscure contrasting pairs of colours?...WHOCARESITSPOKEMONANDINEEDPUNCTUATION

Also, new Mewtwo look-a-like had better be named Mew-Z (as per the nomenclature of the Porygon evolution line). =P

Today I Learned: the Thue-Morse sequence

I did the 15 km run for the kids last Sunday. It took me about 80 minutes to run the whole thing, and I beat my frenemy Han Liang Gan by 1 second. So, yea, pretty much the best run ever! =)

Orrite - it's maths time! Now, let's say that you're only allowed to write "A"s and "B"s, no spaces and no punctuation. What's the length of a longest square-free word that you can write? That is: what is a longest word that you can write before you have the same sequence of letters appearing twice consecutively, such as in AA or ABAB?

Okay, orrite, orrite - that was a pretty silly question. The answer is obviously 3 (either ABA or BAB).

But what if I'd asked for the longest cube-free word instead? That is: subsequences like ABAABAABA are forbidden since it's ABA repeated thrice.

Well, the answer is \(\infty\).

And in fact, it's actually surprisingly easy to construct an example of an infinitely long cube-free word:

The Thue-Morse sequence.

1. Start with an initial sequence just given by: A.
2. Stick a copy of the sequence that you've got at the end, but with "A"s and "B"s switched.
3. Go back to step 2.

Specifically, applying those instructions, we grow the following sequence:
A\(\rightarrow\)AB\(\rightarrow\)ABBA\(\rightarrow\)ABBABAAB\(\rightarrow\)ABBABAABBAABABBA\(\rightarrow\)\(\ldots\)

So how can we prove this? Well, I'm certainly not going to write up a proof right now - I'm off to Easter camp tomorrow and there's much packing and stuff to be done! =)
 

Maths on Steroids: Tute 3

Guys from my tute! I lied to y'all on Tuesday (well, sorta lied)!

But first, a very short story.

There was a period, during the summer holidays, that my friends and I used to play 20 Questions (I did not know that it was made into a TV show) after lunch every day. And being the sound logicians that we are, we ask all the regular questions like:

• "Are you real?"
•  "Are you male?" and 
• "Do most of the letters constituting your name lie in the first half of the alphabet, where "most" is taken to mean "at least half""?

And I remember this one time that someone asked "Are you alive?", and this simple little question was met by a mask of furrow-browed incertitude and a very very hesitant "Yes, I think I'm alive".

We soon figured out that either he'd given us a really philosophical response, or - as was the case - we were dealing with Alexander Grothendieck.

And that's the end of my story. (Note: Grothendieck has since passed away. =/ )

Enter: vector resolutes vs. projections (a la our tutorial).

So, one of you asked me why we've renamed "vector resolutes" as "vector projections" and I made an offhanded remark about mathematicians loving the term "projection" partially due to Grothendieck. Well, that was at least half-false. =P  I don't really know why we like "projection" - although it is certainly a more conceptually meaningful word (*) than "resolute" - I was sorta hinting at the mathematical construction called "projectivisation" (usually just Proj) that's used like, a hella lot in algebraic geometry. And somewhere in the deep dark recesses of my mind, there's a small footnote that says that Grothendieck has a lot to do with projectivisations.

Anyhoo, let's talk about proving the following vector identity:\begin{align}\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}.\end{align}Now, you all got pretty close to bashing this out this identity as per the hint that was given in the yellow book, and I think that's probably the safest proof of this identity that I've seen. To my understanding, most of you will probably just expand out the left hand side, expand out the right hand side and go - O look, they're the same.

That's fine, but you should have a think about how to get straight from the left all the way to the right via factorisation near the end. Hint: you'll need to add and subtract the term \(a_ib_ic_i\) for each coordinate.

I'm going to explain an alternate way of proving this result, which - although algebraic - is slightly more geometric in feel. To begin with, let's talk about the standard unit vectors in \(\mathbb{R}^3\):\begin{align}\vec{e}_1=[1,0,0],\;\vec{e}_2=[0,1,0]\text{ and }\vec{e}_3=[0,0,1].\end{align}Now, these vectors are mutually perpendicular (ortho) + they're unit vectors (normal), and in maths, we call them (together) orthonormal vectors. And you guys can easily show that\begin{align}\vec{e}_1\times\vec{e}_2&=-\vec{e}_2\times\vec{e}_1=\vec{e}_3\\\vec{e}_2\times\vec{e}_3&=-\vec{e}_3\times\vec{e}_2=\vec{e}_1\\\vec{e}_3\times\vec{e}_1&=-\vec{e}_1\times\vec{e}_3=\vec{e}_2.\end{align}
Thanks to these cross-product identities, it's really really easy to verify our vector identity above for these three vectors, right? In fact, by the right-hand rule, if we think about what's geometrically happening to three generic orthogonal vectors \(\vec{u},\vec{v},\vec{w}=\vec{u}\times\vec{v}\), then we should certainly expect the following identities:\begin{align}\vec{u}\times\vec{v}&=-\vec{v}\times\vec{u}=\vec{w}\\\vec{v}\times\vec{w}&=-\vec{w}\times\vec{v}=\vec{u}\\\vec{w}\times\vec{u}&=-\vec{u}\times\vec{w}=\vec{v},\end{align}and we'd have no problems checking our result. So, wouldn't it be great if somehow the vectors \(\vec{a},\vec{b},\vec{c}\) were sorta close to being orthonormal-ish?

So here's a way to make that almost happen. Observe (i.e.: stare really hard) that the truthiness of our identity isn't affected by us multiplying any of the three vectors by a non-zero constant. So, what I mean is that if we know that: \begin{align}\vec{a}\times(\frac{\vec{b}}{|\vec{b}|}\times\vec{c})=(\vec{a}\cdot\vec{c})\frac{\vec{b}}{|\vec{b}|}-(\vec{a}\cdot\frac{\vec{b}}{|\vec{b}|})\vec{c}\end{align}
is true/false, then our desired identity must be true/false - this is of course assuming that \(\vec{b}\neq\vec{0}\), but then the identity is really easy to prove if that were the case. Therefore, we may assume without loss of generality (wlog) that \(\vec{b}\) is a unit vector.

Now, this scaling business means that we can make any of the vectors a unit vector (unless if it's the \(\vec{0}\) vector, but then the identity is really easy if any of them is the zero vector). But that's not quite good enough - I mean, orthogonality is sort of the more important bit. If only we could somehow replace \(\vec{c}\) with \[\vec{c}'=\vec{c}-(\vec{b}\cdot\vec{c})\vec{b},\]
which we know is orthogonal to \(\vec{b}\). Well, we can certainly do this on the left-hand side, because\[\vec{b}\times\vec{c}'=\vec{b}\times\vec{c}-(\vec{b}\cdot\vec{c})\vec{b}\times\vec{b}=\vec{b}\times\vec{c}.\]
And we can do so on the right hand side because\begin{align}&(\vec{a}\cdot(\vec{c}-(\vec{b}\cdot\vec{c})\vec{b}))\vec{b}-(\vec{a}\cdot\vec{b})(\vec{c}-(\vec{b}\cdot\vec{c})\vec{b})\\=&(\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})(\vec{a}\cdot\vec{b})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}+(\vec{a}\cdot\vec{b})(\vec{b}\cdot\vec{c})\vec{b}\\=&(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}.\end{align}So, we may assume wlog that \(\vec{c}\) is orthogonal to \(\vec{b}\). In fact, we can also assume that \(\vec{c}\) is normal by normalising \(\vec{c}'\), and again I should say something like if \(\vec{c}'\) is the zero vector, then everything is easy.

So, we've got orthonormal vectors \(\vec{b},\;\vec{c}\) and \(\vec{d}:=\vec{b}\times\vec{c}\), and if I re-position my \(x,y,z\)-axes in \(\mathbb{R}^3\) to line up with these three vectors, I can see that any vector in \(\mathbb{R}^3\) may be written in a unique way as a sum of these three vectors. In fact, you'll learn after the Easter break that this is what we call a basis for \(\mathbb{R}^3\). In particular, our vector \(\vec{a}\) may be written in the form:\[\vec{a}=\alpha_b\vec{b}+\alpha_c\vec{c}+\alpha_d\vec{d}.\]
Now, let's calculate the two sides of our desired identity and prove it!

On the left-hand side, we have:\begin{align}&\vec{a}\times(\vec{b}\times\vec{c})\\
&=(\alpha_b\vec{b}+\alpha_c\vec{c}+\alpha_d\vec{d})\times\vec{d}\\
&=-\alpha_b\vec{c}+\alpha_c\vec{b}.\end{align}Note that this makes use of that stuff we were saying about cross-products of orthonormal vectors at the start of this proof. And on the right-hand side, we have:\begin{align}(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}=\alpha_c\vec{b}-\alpha_b\vec{c},\end{align}
where we've made use of the orthogonality of the vectors \(\vec{b},\vec{c}\) and \(\vec{d}\).

Orrite, to be honest, that did seem kinda long as a proof, but I think that it's nicer than bashing it out directly because you get this additional understanding to do with scaling these vectors and making them orthogonal and jazz. But I have to warn you: I did make use of the right-hand rule and we had a physical-intuition based argument for why the cross products of any orthonormal vectors \(\vec{u},\vec{v},\vec{w}\) should be as I described. And for that reason alone, if this question crops up in an assignment or (God-forbid?) the exam, just bash it out. It's actually not so painful the second time you do it - I think that my second time took about 2~3 minutes?

So...ugh...happy Easter guys!

(*): I'm thinking of movie projectors here, but I think that might actually be kinda backwards, because the peeps who first invented and made projectors might actually be semi-familiar with the word "projection".

Maths on Steroids (Accelerated Mathematics): Tutes 1 and 2

Orrite guys, some of you were asking me about the last tute question in the MATLAB prac sheet today, so I'll try to walk you through how to do the "simple programming" task there. But first, let me tell you a little story.

First week of my first year at uni, I learned a little something called row reduction. Naturally, the first thing that I did was to ignore the lecturer (sorry Paul (*)), and fiddle around with some matrices. And, here's what I did: \begin{align}&\left[\begin{array}{rrr|l} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right] \begin{array}{l} R_1\rightarrow R_1-R_2 \\ R_2\rightarrow R_2-R_3 \\ R_3\rightarrow R_3-R_1\end{array} \sim \left[\begin{array} {rrr|l} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ -1 & 0 & 1 & 0 \end{array} \right]. \end{align} So, that was pretty innocuous right? I mean, all I did was to be a fully-sick efficient guy and do a few row-reductions together; what could possibly go wrong?

And yet, something is very wrong.

I mean, I started off with this augmented matrix that had a unique solution of \(x=y=z=2\) and winded up with a system of equations that has \(\infty\) many solutions (you can check this by getting the row echelon form of this latter matrix)! Being a totally cocky first year, I was all like...oh man oh man oh man oh man, I broke maths.

Soooo, for any one in my tute who's reading this - see if you can figure out what I screwed up, and see if you can clearly articulate your idea/s to me before/during our next tute. =)

Onto this week's Lab sheet exercise 4!

You played around with the following rowswap function today in MATLAB:

function [B,E]=rowswap(A,x,y)
    % this function performs a row swap operation and creates the
    % associated elementary matrix.
    temp=A(x,:);
    A(x,:)=A(y,:);
    A(y,:)=temp;
    E=eye(size(A,1));
    temp=E(x,:);
    E(x,:)=E(y,:);
    E(y,:)=temp;
    B=A;

And exercise 4 requires you to tweak this function so that this becomes a column-swapping function instead. So let's first try to understand these lines.

1. The very first line:

"function [\(B,E\)]=rowswap(\(A,x,y\))" 

has always been the most inscrutable one to me. Now, I am not programmer and I can only tell you what this practically means. Basically, it tells Matlab that this m-file is going to be a function called rowswap. And that this rowswap function takes in 3 inputs: \(A, x, y\) and is capable of spitting out two things: \(B\) and \(E\). We saw from the lab that \(A\) is a matrix (in our case, a \(4\times 4\) matrix) and we tell the function which rows we want to swap by specifying \(x\) and \(y\).

2,3. The next two lines have a percentage sign out the front and are commented out. That is, MATLAB totes ignores these two lines.

4. In the fourth line, we create a new variable called temp (the name isn't important, I could have called it werewolfram but that'd take more letters to type) and we assign it to be \(A(x,:)\). So let's remind ourselves a little something that we learned last week.

Now, when we run this program, \(x\) is some number that we enter, like \(x=1\). Last week we saw that in MATLAB,
typing \(A(1,1)\) gives you the \((1,1)\)-th entry of the matrix \(A\),
typing \(A(1,2)\) gives you the \((1,2)\)-th entry of the matrix \(A\),
typing \(A(1,3)\) gives you the \((1,3)\)-th entry of the matrix \(A\), and so forth.

We also saw that the colon \(:\) has to do with producing vectors of integers. For example, typing \(1:9\) into MATLAB gives you the vector: \begin{align}[1\, 2\, 3\, 4\, 5\, 6\, 7\, 8\, 9],\end{align}
so it's perhaps not surprising that if you type in \(A(x,:)\), you should get you get the \(x=1\)th row of the matrix \(A\): \begin{align}[A(1,1)\, A(1,2)\, \ldots A(1,n)].\end{align} In short: the fourth line sets temp to be the \(x\)th row of \(A\).

5. If you followed the above point, it should be clear that the fifth line sets \(A(x,:)\) to be \(A(y,:)\), that is: we're replacing the \(x\)th row of \(A\) with the \(y\)th row of \(A\). So, this achieves half of our goal, which is to switch rows \(x\) and \(y\).

6. Now, obviously we can't just replace the \(y\)th row of \(A\) with the (current) \(x\)th row of \(A\), because we've just replaced the \(x\)th row of \(A\) in the last step. But thankfully we've got a backup of the original \(x\)th row of \(A\) stored as temp. And so the sixth line is just to replace the \(y\)th row of \(A\) with temp.

7-10. I'm not going to spell out the specific details of each step because you guys are all smart enough to figure this out. But the aim of these steps is to create a matrix called \(E\) that corresponds to the elementary matrix that does a row swap between rows \(x\) and \(y\) when you multiply a matrix on the left by \(E\). Oh, and if you're thrown off by what \(size(A,1)\) does, go to a lab, type in a (preferably non-square) matrix \(A\) and type \(size(A)\) and \(size(A,1)\) and \(size(A,2)\) and see what you get.

11. In this last step, we set \(B\) to be the row-swapped matrix \(A\) that we've produced in lines 4-6, and this is one of the two matrices that the rowswap function outputs (we mentioned this in line 1).

Now, you've been asked to turn this whole thing into a column-swapping function, well, before I tell you how to do it, you should really have a think about how you might do this. In fact, at the end of my second year, I decided to learn some topology, so I went to the library and found a nice looking book and started reading the definitions. Being the typical maths book, it's got a section of definitions, then some examples (actually, I might be wrong about this, it might have been one of those rare books that gives motivating examples before definitions) followed by some theorems. And I just recall reading the statement of every theorem and then proving/trying to prove the result myself before reading their proof.

And you learn so much from doing this. Because often you'll find that perhaps your proof is more intuitive, but their's is cleaner. In fact, if you find that their proof is really unintuitive and seemingly belaboured, then that's especially useful: it tells you that either you've understood some of the concepts wrongly, or that their proof actually generalises much more easily in a more general framework than the particular result that's given.

So...that was enough time for you to have figured out how to do it yourself, right?

The idea is to just go through every line, and switch the roles of the row-coordinates and column-coordinates. Specifically, replace \(A(x,:)\) with \(A(:,x)\), and similarly for the matrix \(E\). Also, take care with what's going on with the \(size(\cdot)\) function - it shouldn't make a difference for a square matrix, but it will for a non-square matrix.

We finish this section with a few small remarks:

1. Every line of the code ends with a semi-colon. Is this necessary? Maybe you can try to run a version of the code where this semi-colon is taken away. Does the code still work? And if it still works, then what's changed?

2. Also, having now understood this code, you should realise that it seems to be a bit inefficient. In the sense that you should be able to do the whole thing without actually writing lines 4-6, because in lines 7-10 you produce an elementary matrix whose role is to switch rows \(x\) and \(y\). In fact, the following code should suffice:

function [B,E]=rowswap(A,x,y)
    E=eye(size(A,1));
    temp=E(x,:);
    E(x,:)=E(y,:);
    E(y,:)=temp;
    B=E*A;

So, why isn't it written this way? Maybe it's easier for you guys to figure out what's going on in the code without having to deal with the \(size\) function straight up? Maybe the guy who wrote the sheet didn't think of this? Whatever the case, I think that the original code is possibly more...computer-friendly. And I mean this in the sense that the original algorithm requires computers to do fewer computation. But yea, whatevs...that's more of a random remark. =)

Allllso...some additional shameless advertising (that I'm doing outside of my duties as your tutor, much as writing up this stuff is outside of my duties as your tutor :P): I'm helping to start up a new Bible study thingy on Tuesday arvos (especially for peeps who can speak Chinese, I guess? Coz some of the students who are currently in it are a lot more comfortable with Mandarin), so if you're interested you can always contact me?

(*) you guys are thinking of the wrong Paul (**), I meant Paul Pearce,
(**) I hope that you guys are thinking of the wrong Ron Paul. =P

MoMath!

It's time for MoMath - that is, the Museum of Mathematics.

I first heard about MoMath from a pretty cool looking flyer. You know, the kind with some demurely (dare I say...indiesque?) coloured triangles arranged in the shape of a nautilus and stuff - like, an actual well-designed flyer, as opposed to the really nerdy stuff that we mathematicians produce most of the time. And that's all I know about MoMath.

Except the following puzzle given at the opening ceremony of this museum:

Eight people are seated around a round table (the table seats 8 precisely). They all get up at the same time (ignore physics for a moment here) and each person either moves to their left or right by one chair or stays where they are, before sitting down again (miraculously avoiding two people trying to get into the same chair). In how many distinct ways is this possible?

This was actually a question posed on the NPR Sunday Puzzles last week (hence my later than usual post, I was a tad paranoid that I'd get it wrong and waited for the official answer to be revealed :P), and the guy on the program who got it right said that he took some time over a few days bashing out all the solutions. Let's see if we can at least do this a tiny bit more efficiently.

Let's first label the positions going around the table slightly irregularly: let's order them 1, 3, 5, 7, 8, 6, 4 and 2 without loss of generality. Now, given a (legal) re-seating, we can roll-call these guys in order from 1 to 8, and just take note of the number \(\sigma\) just prior to first person who's changed their seat. The idea is that \(\sigma\) partitions our set of legal re-seatings and helps us organise our counting of legal re-seatings.

For example, if this number is 8, then everyone ends up where they started - clearly there's only one way to do this. So:

when \(\sigma=8\), there's precisely \(1\) re-seating.

Next, if \(\sigma=7\), then clearly 7 stays where s/he was seated, and since 8 can't go anywhere-else, s/he's forced to stay where s/he is - which isn't allowed, because then we'd have that \(\sigma=8\). Soo:

when \(\sigma=7\), there're precisely \(0\) re-seatings.

Now, for \(\sigma=6\), 6 stays and 7 has to move, and it's forced to switch with 8. Sooo:

when \(\sigma=6\), there's precisely \(1\) re-seating.

For \(\sigma=5\), 5 stays and 6 has to move, and it's forced to switch with 7. And by switch, I mean that 7 goes to sit where 6 started from - so...actually switch. Poor 8 is forced to stay. Soooo:

when \(\sigma=5\), there's precisely \(1\) re-seating.

For \(\sigma=4\), 4 stays and 5 has to move, and it's forced to switch with 6. We have 7 and 8 left, who're free to either switch or to stay put. Sooooo:

when \(\sigma=4\), there're precisely \(2\) re-seatings.

For \(\sigma=3\), 3 stays and 4 has to move, and it's forced to switch with 5. We have 6, 7 and 8 left, who're free to do whatever. And we could just count this as before, but if you think about it, the number of such arrangements is the same as the number of legal re-seatings where 4 and 5 are forced to stay where they started instead of switching. That is, the count for \(\sigma=3\) is the same as the the total of the legal re-seatings for when \(\sigma\geq5\). Soooooo:

when \(\sigma=3\), there're precisely \(3\) re-seatings.

For \(\sigma=2\), 2 stays and 3 has to switch with 4. So, by the same logic as for the previous count, the total number of legal seatings may be obtained by summing the legal re-seatings for \(\sigma\geq4\). Sooooooo:

when \(\sigma=2\), there're precisely \(5\) re-seatings.

For \(\sigma=1\), 1 stays and 2 switches with 3. Rinse and repeat. Soooooooo:

when \(\sigma=1\), there're precisely \(8\) re-seatings.

At this point, let's pause for a while, we've got the sequence of numbers:\begin{align}1,0,1,1,2,3,5,8\end{align} does it ring a bell? Why yes, it's (pretty much) the Fibonacci sequence. And indeed this behaviour will manifest itself should we wish to do the same computations for more people seated around the table. But for now, let's get back to the problem. So, we might mindlessly now expect that for \(\sigma=0\), that we'd end up with \(13\) re-seatings, but that it's slightly wrong. So, let's walk through our logic.

For \(\sigma=0\), we know that 1 must switch with someone. However, unlike before, it isn't forced to switch with 2, but may switch with 8 as well. Which means that we should double the \(13\) re-seatings to correct for this extra freedom. But, \(26\) is still under-counting the number of legal reseatings for \(sigma=0\) because there are 2 additional legal re-seatings which arise when everyone just moves either one seat to their left or one seat to their right. We miss these 2 possibilities in our count because we only considered the cases when 1 switches with another dude, and not when the other dude (either 2 or 8) just keeps shufflin' along.

Sooooooooo...

when \(\sigma=0\), there're precisely \(28\) re-seatings.

Making a total of \(1+0+1+1+2+3+5+8+28=49\) legal re-seatings.

So, umm...homework. Show that the the number of legal re-seatings for \(n\) people is precisely the coefficient of \(x^n\) when we expand \begin{align}\frac{1-x+3x^3-2x^4-3x^5}{1-2x+x^3}\end{align} as an infinite series around \(x=0\).

Okay, I admit that wasn't the most straight-forward proof that something was Fibonacci-esque or whatevs, but umm...just...deal with it? I guess?

Mirror Symmetry!

Wednesday last week, my friends and I decided that we're going to take a stab at learning mirror symmetry this semester. We decided this based on a few talks that we gave to each other about geometric Langlands, cobordismy stuff, mirror symmetry and enumerative geometry. In the course of preparing for my talk about mirror symmetry, I stumbled across this in a paper:

"In this section I will attempt to outline the mirror symmetry principle in mathematical terms, and describe some of the mathematical evidence for it. I apologize to physicists for my misrepresentations of their ideas, and I apologize to mathematicians for the vagueness of my explanations."

-David R. Morrison.

Frickin' awesome.

I suppose that I should really write up something here that gives a flavour of (homological) mirror symmetry but I really don't understand this nearly well enough yet. I mean, okay, sure - (a small part of the mathematical side of) the story revolves around two (families of) spaces \(X\) and \(X^{\vee}\), one of which is symplectic and one of which is complex, that demonstrate dual Hodge structures dimensions.

Much of the current mathematical perspective on homological mirror symmetry is about the equivalence of the derived Fukaya category on \(X\) and the derived category of coherent sheaves on \(X^{\vee}\). Although sitting through Motohiko Mulase's (*) talk on Friday, you can see that he really thinks of mirror symmetry as...the Laplace transform.

Again, this is something that I'd love to be able to explain here, but I just don't get it. I mean, an example of this phenomenon is applying the Laplace transform to the cut and join relation governing certain characteristic classes on the (DM-compactified) moduli space of curves to obtain certain recursion relations (which, awesomely enough, prove Witten's conjecture and the \(\lambda_g\)-conjecture). And it's infuriating, because I spoke to him for a fair while after his wonderfully engaging and understandable talk, and yet here I am...still not quite catching how he thinks of mirror symmetry.

I guess I'mma just have to find him on Tuesday, admit that I'm a total idiot and ask him to explain it in greater detail (assuming that I can find him =P).

It's good that there's no room for pride in maths. =)

By the way #1: Cyberdyne systems from the Terminator vs Cyberdyne Inc. of robot legs fame, seriously Skynet? You couldn't have been a teeny bit subtler?

By the way #2: And here are roughly 450 million reasons to learn your maths right.

*: the guy is just...so...effervescent, especially for a uni professor! Totes likeable, and an excellent speaker. It's going to take a great deal of commitment to hate him. =)

Today I Learned: The Rational Circle

I think that tonight is meant to be the Lantern Festival or whatever. But let's not get distracted - let's talk about:

The Rational Circle.

Huh...I like the sound of that.

It vaguely conjures up the image of a clandestine cult, perhaps akin to the Pythagorean Brotherhood (*), encircling an altar of smooth grey slate; each crimson robed member up-raising a dragon's fang of chalk in fearful exultation, only lowering these totems for the transcription of mystical symbols of power.

In short: a typical group of mathematicians sitting around a blackboard.

For the past year, a couple of friends and I have been going through Deligne's proof of the Weil conjectures - a finite fields version of the Riemann hypothesis. We've done Weil I, and we've gone through parts of Weil II and now (most of us) have Weil-fatigue (okay, sure, it should probably be Deligne-fatigue).

So we're currently looking at four potential new topics, one of which will be selected for this coming year: geometric Langlands, Cobordismy stuff (in particular, Madsen Weiss), (Homological) Mirror Symmetry and Topological Recursion (i.e.: Eynard-Orantin invariants and stuff). Oh, and I guess there was a fifth suggestion - we thought about sitting down and proving the Volume Conjecture, but decided that it wasn't exciting enough. =P

And in amidst all the discussion about the merits of each of these topics we started (as you do, assuming that your life is like American television) to reminisce all the good times that we had in the Weil conjecture series. Like that very first session when Norman Do, for whatever reason, gave us a pretty sweet description of the rational points of the unit circle in a passing remark. A description so sweet that we immediately distracted ourselves for a minute or so proving it.

Claim: the rational points on the unit circle \(\mathbb{S}^1\), i.e. the coordinates are fractions (of integers), may all be written in the form:\begin{align}(\frac{u^2-v^2}{u^2+v^2},\frac{2uv}{u^2+v^2}),\,u,v\in\mathbb{Z}.\end{align}
Huzzah.

In fact, we can easily see that any point of such a form is a rational circle point, unless if both \(u\) and \(v\) are zero. But I didn't really wanna type all that in the claim because MathJax makes words that you type in mathmode look funny.

Whatevs man, let's just prove this jazz.

To begin with, by the unit circle \(\mathbb{S}^1\), I mean the set:\begin{align}\mathbb{S}^1:=\{(x,y)\in\mathbb{R}^2\,:\,x^2+y^2=1\}.\end{align}
Now, consider the following map from \(\mathbb{S}^1\) to the real projective line \(\mathbb{R}\mathrm{P}^1\) given by:\begin{align}\psi:\mathbb{S}^1\rightarrow\mathbb{R}\mathrm{P}^1,\,(x,y)\mapsto[y:1-x].\end{align}
Heeeey, whaddya know - this is precisely the stereographic projection map of \(\mathbb{S}^1\) from \((1,0)\) onto the \(y\)-axis (plus an extra point at \(\infty\) for the projection of \((1,0)\) itself). So \(\psi\) is totally a bijection.

Now, it's clear that \(\psi(x,y)=[y:1-x]\) can be written in the form \([u:v]\in\mathbb{Z}\mathrm{P}^1\subset\mathbb{R}\mathrm{P}^1\) if \((x,y)\) is a rational circle point. In fact, for any \([u:v]\in\mathbb{Z}\mathrm{P}^1\), the inverse point \(\psi^{-1}[u:v]\) is a rational circle point!

Let's check this by bashing stuff.

So we're looking for some \(x,y)\) such that \([u:v]=[y:1-x]\). Therefore, we have that \(u(1-x)=vy\) and shoving this into \(x^2+y^2=1\) yields that:\begin{align}(u^2+v^2)x^2-2u^2x+u^2-v^2=0.\end{align}This quadratic yields the solutions that:\begin{align}x=\frac{u^2\pm v^2}{u^2+v^2}=1,\frac{u^2-v^2}{u^2+v^2}.\end{align} The case that \(x=1\) is covered by the latter (via \([u:v]=[1:0]\) ) and hence may be excluded.

Repeating the same calculations for \(y\), we get that\begin{align}y=0,\frac{2uv}{u^2+v^2}.\end{align}As before, the case that \(y=0\) is covered by the latter case (via \([u:v]=[\pm1:0]\) ) and may be excluded.

Oh look, not only did we show that the stereographic projection map \(\psi\) gives us a bijection between the rational circle points and \(\mathbb{Z}\mathrm{P}^1\), but we also proved our initial claim.

-*Insert obligatory proof-ending blank square*

Going back to my story: I think that the reason that Norm brought up this example was to demonstrate to us the following guiding principle: rational solutions to polynomials are all about number theory.

As an example: we've just characterised the rational solutions to the polynomial \(x^2+y^2=1\) as being of the form:\begin{align}(\frac{u^2-v^2}{u^2+v^2},\frac{2uv}{u^2+v^2}).\end{align}And anyone's who's played around with number theory will immediately recognise these as being related to the general form for any Pythagorean triad:\begin{align}(u^2-v^2,2uv,u^2+v^2).\end{align}Which, by the way, is hardly surprising once you think about rescaling Pythagorean triangles so that its hypotenuse is length \(1\).

What might be surprising, is that we started off with a rant about the Pythagorean Brotherhood, wandered about and wound up back at Pythagoras - effectively tracing out a big circle of thought, a rational circle, if you will. =P

Since it's pretty hard to finish on a circle, let's shoot off on a tangent. This characterisation of the rational points of the unit circle also comes as a consequence of Hilbert's Theorem 90 which is a pretty early example of how number theory may be phrased in terms of (and resolved via) cohomology theories.

Sometimes I worry about this.

Sometimes I worry one day we'll show that pretty much everything in maths may be phrased in terms of cohomologies...I mean, cohomology is one of the most straight forward ways that we can extract invariants from stuff...and to be honest, we usually can't do very much in maths apart from the most obvious things. So, who knows...may be my worries are...semi-quasi-pseudo-meta-justifiable?

Who knows, who cares: we thought in a big loop and went off on a perpendicular, thereby forming a Q - that is: Q for quitting.

Let's quit here.

I am become tired and (more) incoherent (than usual).

*: which, despite its name, did include women.

I'mma \(f(n)\) maths you up: 3

To: mX Funny Valentine,

I'mma just give you a moment to appreciate that jazzy, punny salutation.

Awww yeah!

Okay, fair enough, Valentine's day was like, three days ago - but hey, I've been busy, didn't get a chance to write this up. But when the Valentine's day issue of the mX (free newspaper in Melbourne) comes out with a front page title like this gem:

"Only 599 eligible bachelors in town."

you've gotta write about it...right?

So, this claim is supposedly based upon surveys done by Roy Morgan Research (although I kinda (*) wonder if the mX just yoinked it from NewsMaker (**)). The idea is that they compiled a list of "must-haves" in an eligible dude, in particular:

• financial stability: 55%
• is totes a cook(ie monster): 57%
• bought deodorant in the last 6 months: 57%
• is a Mr. Fix It-type (is that a real type?): 40%
• believes in helping peeps: 80%
• believes that expensive restaurants are worth it: 35%
• read a book in the last three months: 56%
• has acceptable BMI (somehow, that seems worse than saying not too fat, not too thin): 52%
• like weekends away: 51%
• like to know what's going on in the world (really? not everyone's into hermitty artists?): 67%
• recycles everything: 84%
• don't have substance abuse issues or dandruff (yes, I know, this is a pretty weird combo): 87%

and just multiplied all of those probabilities and the population of Melbourne.

Homework: work our roughly what the population of Melbourne is.

Needless to say, this is probably very bad maths. This calculation assumes independence, which is totally suss in this case. Like, I'm fairly sure that guys who know what's going on in the world, and read books are going to be just a tad more financial stable - so that's good news for any conservationists who are worried about eligible bachelor populations. On the other hand, I do worry that people who are good cooks might be less inclined to go to expensive restaurants...but maybe I'm wrong on that, maybe good cooks can actually appreciate the amount of work that goes into creating a dish at The Fat Duck.

Whatever the case, I declare this rant over. =P

Maaaan, bad use of statistics, that's just \(f(n)\) mathed up.

*: kinda in the sense that I'm almost certain that it's true.
**: in the mX's defence, that website is all about making news - and who better to distribute it than a newspaper?

Today I Learned: On De-Klein-ing Lie's - Part 2

Update: Part 0 Part 1.

Roughly a year ago, I waltzed through the streets of Munich.

Everything was wonderful - I'd recently befriended an American couple and a African+German couple (they all very kindly let me crash at their places for weeks) and life was a whimsical concoction of Japanese conversations with strangers (I'm not Japanese), Bible studies, scaling European architecture (in town and on the alps), office meetings with my supervisor at the Ludwig Maximilian University (he was technically holidaying) and asking God to provide for all of my accommodations (I'm currently fairly well-off, but I asked it for good reason, and He did provide for a whole month - including a free week at a hotel in Istanbul, so I'd better hold up my end of the bargain). Even the weather was wonderful: cold, lippy, but not so chilling as to be penetrating.

The only thing that really bothered me was...why did they love Bayesian Statistics so much? I mean, the signs were everywhere. Above libraries, on opera posters, on random landmarks were scribed these...these impenetrable and insufferable letters: "Bayerische Staats". They haunted me. Statistics had made her mark upon the capital of Bavaria.

Yes. Bavaria. I only realised (much to my humiliation) after I'd landed back in Melbourne that I'd mistranslated Bayerische Staats, that is: the Bavarian State.

It was a good trip, I even managed to miss my flight back to Australia because I was doing some maths with my supervisor and scored having my first Thanksgiving turkey - with the American couple I mentioned earlier. Life (*) is awesome.

Pity that I hadn't also visited the second largest city in Bavaria - neighbour to a little university/Siemens town called Erlangen.

A while back (to all AMSI peeps who're still reading this, after that meandering introduction, yes, I know that your summer school is nearly over and your exams are coming up - hey, I'm a slow writer), I decided to write up a few things that I've learned from reading one of Klein's lectures on Lie's line-sphere correspondence. This is part 2: the Erlangen program.

A slightly modern understanding of the Erlangen program is that it is the study of the invariants of a group \(G\) acting transitively on a homogeneous space \(X\). There are key words here: groups, transitive, spaces and invariants. Let's consider an example.

Example 1.

The group \(SO(n)\) made up of orthogonal matrices with determinant \(1\) may be thought of as matrices whose column (and rows) constitute an orthonormal basis for \(\mathbb{R}^n\). This means that:

• any vector gets mapped to a vector of the same length;
• angles between vectors are preserved.

The only transformations of \(\mathbb{R}^n\) which can do this are rotations and reflections - we discount the latter option in this case because the determinant of a matrix in \(SO(n)\) is \(1\) as, opposed to \(-1\).


Now, because orthogonal matrices must preserve vector lengths, they cannot act transitively on all of \(\mathbb{R}^n\), so our space needs to be something smaller - like the unit sphere (any radius will do):\[\mathbb{S}^{n-1}:=\{(x_1,\ldots,x_{n})\,:\,x_1^2+\ldots+x_{n}^2=1\}.\]

The transitivity of the action of \(SO(n)\) on \(\mathbb{S}^{n-1}\) means that for any two points \(x,y\in \mathbb{S}^{n-1}\) there are a whole bunch of matrices taking \(x\) to \(y\), and either by post- or pre-multiplication by a matrix, it's clear that this set of matrices is in bijection with the group of matrices fixing \(x\) (or fixing \(y\)). Take \[x=(0,0,\ldots,0,1)\] and we see that this group of point-fixing matrices needs to be \(x\) in the bottom row and \(x^{t}\) in the last column, so the group of point-fixing matrices is precisely \(SO(n-1)\). Which of course means that: \[\mathbb{S}^{n-1}=SO(n)/SO(n-1).\] And this is what we mean by a homogeneous space: a group quotiented by a sufficiently nice (possibly non-normal) subgroup.

So, we've got our group, we've got our space - what are these invariants?

Well, we know that \(SO(n)\) is distance preserving on \(\mathbb{R}^n\), and knowing this it's not too hard to show that orthogonal matrices also preserve distances on \(\mathbb{S}^{n-1}\). So there you go: distances are preserved. In addition, the angle between two intersecting curves on \(\mathbb{S}^{n-1}\) are also preserved - you can show this by thinking about the tangent lines of these two curves at the relevant point of intersection (an exercise!). So, distances are preserved, angles are preserved, surely this means that spherical polygons must be preserved (insert higher dimensional generalisations of polygons where appropriate).

In fact, we're seeing that these invariants are pretty much what we'd expect to care about as geometers: spherical lengths, angles, spherical shapes, spherical volumes...etc.

In fact, it's as if:

the Erlangen program is an attempt to unify geometry in a cohesive language.

Let's list a few more examples:

Example 2.

Group: \(O(n)\) - the orthonormal matrices.
Space: \(\mathbb{S}^{n-1}\)
Invariants: spherical lengths, angles, spherical shapes, spherical volumes...etc.

Wait, isn't this the same example? Well, almost. The only difference being that \(O(n)\) includes not just rotations but reflection matrices too. And so there is one invariant that is retained for \(SO(n)\) that isn't for \(O(n)\): orientation (mirror-image-ness). So in our first example, we were able to distinguish a triangle on \(\mathbb{S}^2\) and its mirror image, whereas the two would be regarded as being identical in this second example.

Example 3.

Group: the group of linear functions look-a-likes on \(\mathbb{R^n}\) of the form: \[x\mapsto Ax+b,\]
where \(A\in O(n)\) is an orthogonal matrix and  \(b\in\mathbb{R^n}\) is just another vector.
Space: \(\mathbb{R}^n\).
Invariants: lengths, angles, shapes, volumes...etc.

This might look unfamiliar, but it's really just Euclidean geometry in disguise. Our space is \(\mathbb{R}\) and our symmetries are any combination of translations and rotations and reflections - precisely the symmetries that we'd think of in primary school.

Example 4.

Group: the special linear group \(SL_2(\mathbb{R})\) of \(2\times2\) matrices of determinant \(1\).
Space: the hyperbolic upper half plane \(\mathbb{H}\) given by \[\mathbb{H}:=\{z=x+yi\in\mathbb{C}\,:\,\mathrm{Im}(z)=y>0\},\]
and the action of the group is given by:\begin{align}\left(\begin{array}{c c}a & b \\ c & d \end{array}\right)\cdot z:=\frac{az+b}{cz+d}.\end{align}
Invariants: angles, orientation, hyperbolic length, hyperbolic polygons, hyperbolic area...etc.


It's sometimes said that polygons in in spherical, Euclidean and hyperbolic geometry are respectively fat, boring and thin (okay, I sorta lied: I haven't heard anyone other than high school kids call Euclidean triangles "boring"). As in, the sides of spherical triangles are curved slightly outwards, the sides of a Euclidean triangle are straight and the sides of a hyperbolic triangle are curved slightly inwards. In fact, this simple phenomenon is a manifestation of a deeper invariant of these geometries called the Gaussian curvature. In particular, the curvature of a sphere is positive, for the plane it's zero and for the hyperbolic plane it's negative, and you can think of curvature either in terms of how triangles curve or how much "space" you squeeze in at each "point".

Also, it's worth pointing out that the fact that the hyperbolic plane is a homogeneous space does make it extremely rigid. I'm just mentioning that as an alternate handwavy explanation of the rigidity of hyperbolic spaces to any individuals who might have asked me about this in the past.

It's also worth pointing out that I've provided a presentation of the hyperbolic plane that does not generalise very well to general hyperbolic spaces (and instead generalises more nicely to the Siegel upper half space). Go check out the Wikipedia article for homogeneous spaces to see how to do it for hyperbolic spaces.

Recently, I stumbled across this discussion on MathOverflow about the Erlangen program and what precisely it states. And I think that one of take home message that I got was that Erlangen program is that curious combination of beautiful logic and deep intuition - in parallel with Klein's self-assessment as a mathematician (perhaps not so colourfully phrased). It's like Monet's impressionism: colour, fire, light, captured by masterful brush strokes. Its triumph is that of intuition - we see once and once again the her indelible fingerprints in geometry. In Riemannian geometry, she is present at the grand level by uniformization and local models and present at the microscopic level via the Levi-Civita connection. Which in turn leads us to Cartan geometries and Cartan connections. In symplectic geometry, Lie's contact-transformations seem also to be grounded in this philosophy. The details are somewhat foggy to me, perhaps like a Seurat, but the forms are there.

So...um...Felix Klein, more than just the guy who came up with the Klein bottle.

*: Secretly, I wanted to write that God is awesome, but I chickened out. =P