Math sunday

I had a chill day thinking about math today without any pressure whatsoever. First I figured out, calculating inductively, that the order of GL_n(\mathbb{F}_p) is (p^n - 1)(p^n - p)(p^n - p^2)\cdots (p^n - p^{n-1}). You calculate the number of k-tuples of column vectors linear independent and from there derive p^k as the number of vectors that cannot be appended if linear independence is to be preserved. A Sylow p-group of that is the group of upper triangular matrices with ones on the diagonal, which has the order p^{n(n-1)/2} that we want.

I also find the proof of the first Sylow theorem much easier to understand now, the inspiration of it. I had always remembered that the Sylow p-group we are looking for can be the stabilizer subgroup of some set of p^k elements of the group where p^k divides the order of the group. By the pigeonhole principle, there can be no more than p^k elements in it. The part to prove that kept boggling my mind was the reverse inequality via orbits. It turns out that that can be viewed in a way that makes its logic feel much more natural than it did before, which like many a proof not understood, seems to spring out of the blue.

We wish to show that the number of times, letting p^r be the largest pth power dividing n, that the order of some orbit is divided by p is no more than r-k. To do that it suffices to show that the sum of the orders of the orbits, \binom{n}{p^k} is divided by p no more than that many times. To show that is very mechanical. Write out as m\displaystyle\prod_{j = 1}^{p^k-1} \frac{p^k m - j}{p^k - j} and divide out each element of the product on both the numerator and denominator by p to the number of times j divides it. With this, the denominator of the product is not a multiple of p, which means the number of times p divides the sum of the orders of the orbits is the number of times it divides m, which is r-k.

Following this, Brian Bi told me about this problem, starred in Artin, which means it was considered by the author to be difficult, that he was stuck on. To my great surprise, I managed to solve it under half an hour. The problem is:

Let H be a proper subgroup of a finite group G. Prove that the conjugate subgroups of H don’t cover G.

For this, I remembered the relation |G| = |N(H)||Cl(H)|, where Cl(H) denotes the number of conjugate subgroups of H, which is a special case of the orbit-stabilizer theorem, as conjugation is a group action after all. With this, given that |N(H)| \geq |H| and that conjugate subgroups share the identity, the union of them has less than |G| elements.

I remember Jonah Sinick’s once saying that finite group theory is one of the most g-loaded parts of math. I’m not sure what his rationale is for that exactly. I’ll say that I have a taste for finite group theory though I can’t say I’m a freak at it, unlike Aschbacher, but I guess I’m not bad at it either. Sure, it requires some form of pattern recognition and abstraction visualization that is not so loaded on the prior knowledge front. Brian Bi keeps telling me about how hard finite group theory is, relative to the continuous version of group theory, the Lie groups, which I know next to nothing about at present.

Oleg Olegovich, who told me today that he had proved “some generalization of something to semi-simple groups,” but needs a bit more to earn the label of Permanent Head Damage, suggested upon my asking him what he considers as good mathematics that I look into Arnold’s classic on classical mechanics, which was first to come to mind on his response of “stuff that is geometric and springs out of classical mechanics.” I found a PDF of it online and browsed through it but did not feel it was that tasteful, perhaps because I’m been a bit immersed lately in the number theoretic and abstract algebraic side of math that intersects not with physics, though I had before an inclination towards more physicsy math. I thought of possibly learning PDEs and some physics as a byproduct of it, but I’m also worried about lack of focus. Maybe eventually I can do that casually without having to try too hard as I have done lately for number theory. At least, I have not the right combination of brainpower and interest sufficient for that in my current state of mind.

一说起偏微分方程,想到此行有不少杰出的浙江裔学者,最典型的可以说是谷超豪。想起,华盛顿大学一位做非交换代数几何的教授,浙江裔也,的儿子,曾经说起他们回国时谷超豪,复旦的,如他父亲一样,逝世了,又半开玩笑言:“据说谷超豪被选为院士,是因为他曾经当过地下党。”记得看到杨振宁对谷超豪有极高的评价,大大出于谷超豪在杨七十年代访问复旦的促动下解决了一系列有关于杨-米尔斯理论的数学问题。之外,还有林芳华,陈贵强,都是非常有名气的这套数学的教授,也都是浙江人。我们都知道浙江人是中国的犹太人,昨天Brian Bi还在说”there are four times more Zhejiangnese than Jews.” 可惜我不是浙江人,所以成为数学家可能希望不大了。:(

Some speculations on the positive eugenics effects on the far right tail of intelligence of the Chinese population of the imperial examination system

Over the past few months, I had read casually on the imperial examination system (科举) out of curiosity. My knowledge of it, the system that very much defined pre-modern Chinese society, is still very limited and vague, but now I at least know what 进士 and 秀才 are, along with some classical Chinese, background indispensable for understanding that system. I hope, if time permits, to learn more about this over the next year, on the side.

It has occurred to me that the imperial examination system, while doing much to prevent China from developing modern science as the West had for cultural reasons, did select for intelligence at the far tail. The reason is simple. The tests, which were very g-loaded, conferred those who scored highly on them wealth, position, and status that enabled them to have more children, and those from families who scored highly married those from similar families. Over time, there emerged an elite subpopulation with very high base genotypic IQ, one that results in those born from such families to regress not to the overall Chinese mean but to the high mean of that subpopulation. This is consistent with the fact that in the 20th century and probably even today, a disproportionately high percentage of top scholars, scientists, engineers, and even revolutionaries and political leaders of Chinese descent can be traced back to those elite 科举 families, based on the many examples I have seen. I’ll not give specific examples for now; they can easily be found by anyone who reads Chinese.

I will conclude with a note that is likely to be very relevant. Brian Bi, about a year ago, made this following IQ map of China by province.

China_IQ_by_province

You can also view it here.

First of all, the data may not be very accurate; I’ll have to check on its source. But for now, let’s assume that it is. Then, what’s most noticeable is the high average of Zhejiang, consistent with the number of mathematical and scientific geniuses of Zhejiangnese ancestry relative to the number of those with ancestry of other provinces, adjusted for province population of course. Examples are numerous: Shiing-Shen ChernWu WenjunFeng KangYitang Zhang, etc for math. There is also, in another field, Qian Xuesen. Too many to name. Brian Bi and I have wondered the cause of this. It is plausible that the aforementioned effect was much more pronounced in Zhejiang than in other provinces in China. Of course, there is a probably substantial environmental effect here too. So I guess to satisfy this curiosity, I might study some Zhejiangnese history as well.

Aside from prominence in science, Zhejiangnese are stereotyped in China for being really entrepreneurial. They are now one of the most prosperous provinces in China, needless to say. They are, to put it simply, a super breed among Chinese, to my superficial view.