Why are they NOT simply connected?

OK, it’s been so long since I last posted anything here.

http://math.columbia.edu/~shanbei/non%20simply%20connected

This is an expository notes for myself. In the beginning of this semester, I asked myself to give a talk on Johan’s paper in his seminar. It’s really fun, and I focused on the “easiest” case of it. Things are already very interesting at this stage. Hope you enjoy it as half (for otherwise you would be too excited to sleep today, hahaha) as I do.

My research somehow stucks, therefore I did this activity. Now I shall go back and drill on something else.

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记一个理想主义者–心中永远的坐标–黄兆镇 The Most dedicated and influential math teacher in Zhejiang Univ.

Brings me to the old days when I was an silly undergrad (am still silly though).

流星小屋

我非老黄的学生,因为老黄的学生都是好学生,都是“黄派中人”,而我只是上过老黄的课和讨论班的蹭课的学生。但在ZJU的岁月中谁给我的影响最大?不是老杨也不是老李,而是老黄。

一直想写点关于老黄的文字,一直拖到今日。在年初看到facebook中居然有老黄的专页,几个月后居然自己成为了专页的管理者,再一问老黄近况,发现老黄已经退休了,想来该写点东西了。

在写自己眼中的老黄之前,先转载别人眼中的老黄。自己也再读一遍。

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An ongoing write-up

Johan’s question seminar

Check out my ongoing write-up for an informal seminar. The theme of this seminar is that participants ask some little easy question which is somewhat fun and then we vote meet once a week to solve one or two of them. The reason to do so is to “make sure everyone get out of their bed once a week”.

I didn’t write down all the discussions of questions, due to my lack of attention, patience and time. Here are the questions of which I’ve written down some discussions:

(1) Given an étale map between varieties with constant number of fibers, is it necessarily a finite étale map?

(2) Is a quotient of locally ringed space by a finite group action always a locally ringed space?

(3) Why is there no section from PGL_n to GL_n?

(4) Is it true that any small perturbation of a diagonalizable matrix in GL_n(Z_p) with distinct eigenvalues diagonalizable? (Note that here diagonalization should happen within Z_p)

(5) How many lines of the 27 on a REAL cubic surface real (what are the possibilities)?

In the following weeks I will keep writing down further discussions of other fun little questions (hopefully), so check out the link above later~

A random thought

“I think a blessing/curse of our professions is that the lab is in our mind.” I don’t know who said that, but he’s damn right…

Today I want to talk about a wrong belief I held for the last 2 years. I thought that a variety X over k is projective if and only if each of it’s irreducible components is projective until very recently. I held the wrong belief because the following exercise in Hartshorne’s famous <Algebraic Geometry>: Let X be a scheme over k, a line bundle on X is ample if and only if the restriction on each irreducible components is ample. An problem would be that even on each irreducible component we have an ample line bundle we can not always patch them together (although we can always patch in 1-dimensional case). I came to the following (counter)example in my research.

Let’s take two pieces of P^2 blow up at one point, call them X and Y. There is a ordinary(a usual line not passing thru the point we blown up) line and an exceptional divisor on this blow up of P^2. Let’s identify the exceptional divisor on X with an ordinary line on Y  and identify the exceptional divisor on Y with an ordinary line on X. Call the result variety Z, which has two projective irreducible components. One can easily see that any two ample line bundles on X and Y can never be patched together (leave as an exercise to reader(s?)). Does anyone besides me had the same wrong belief before reading this post (Does anyone actually read this post?)? Or is it just me?

This example isn’t coming from nowhere, it is actually a reduction of some special non-archimedean Hopf surface. After a tedious computation one can show some reductions of some more general non-archimedean Hopf surfaces, whose irreducible components are blow up of Hirzebruch surfaces (toric surfaces), are also non-projective.

夜凉如水

“I have loved the stars too fondly to be fearful of the night.”

Let me recall one of my recent conversations with friends. So I was learning rigid geometry in the past month. One of fundamental things about rigid geometry was functional analysis over non-archimedean fields. In general, things would be nicer because of strong triangle inequality. However there is this weird thing that Huhn-Banach theorem may fail when working over some non-archimedean fields. One condition of ground field to make sure such a strange thing not happen was to ask the field to be spherically complete, which means a list of nested balls have nonempty intersection.

I was told that C_p is not spherically complete. If you think about it you will find it hard to believe. So suppose one was given nested balls B_i with radii r_i and (naturally?) suppose r_i goes to 0. Then if I pick an a_i from B_i, my claim is that a_i will have limit which is contained in the intersection of B_i (it’s easy to verify). So surprisingly (or maybe I’m the only one shocked by that?) the strange nested balls in C_p has radii bounded from below.

And indeed, let’s construct such a strange nested balls. One important lemma is the following:

Lemma: For every positive integer d, and every field K which is a finite extension of Q_p, there are only finitely many extension L/K of degree less than or equal to d. And let’s call the set of these fields to be F_d.

proof is easy, as every Galois extension of K is solvable. And cyclic extensions of K are understood in local class field theory…

Now let’s take B_1 to be the unit disc in C_p, and I want to construct the balls inductively. Suppose I have found B_i with the following property: r_i is bigger than 1/3+1/3i, and the intersection of B_i with O_L is empty where L runs over the set F_i. Then obviously one can always find a element a_{i+1} in B_i which is far away from all the integers of L in F_{i+1}, now let’s choose B_{i+1} to be the ball with center=a_{i+1} and radius 1/3+1/3(i+1). In this way we are given nested balls B_i and I claim the intersection is empty. Suppose there is an x inside all of B_i, then x is in O_{C_p} and will be at least 1/3 away from all the integers in finite extension of K which is a contradiction of the fact that union of such O_L’s is dense in O_{C_p}.

Anyhow, the amazing/strange fact to me is really that such a nested balls are forced to have a lower bound for the radii.

PS, I know that I promised to post once a month… Not to sure if I can do so, we’ll see.

World is full of fun.

Ok, I’ve noticed that there were a few people following this blog. So I felt obligated to post something on a regular phase, like once a month or something… I doubt if I’ll be that diligent…

Recently I’ve been spending my spare time surfing on the Mathoverflow, on which I don’t understand most questions and answers(well, as always…). But I’ve seen one interesting question yesterday and gave it some thoughts, the website of the question with a beautiful answer is at here.

We know that every curve can be embedded into 3-projective space. So the question is given a field k, whether there is a surface S over k such that every smooth curve over k can be embedded in S. The original question is asked for k=C, the complex numbers, and Olivier Benoist gave an beautiful answer to this question using following strategy. Let me (shamelessly copy?) sketch HIS approach: suppose there is such a complex surface S, then we consider the Hilbert schemes of genus g curves in S, because Hilbert schemes have only countably many component, one of them (say, H_g) will dominate M_g (the coarse moduli of genus g curves) which is of general type (proved by Mumford, relies on the hypothesis that we are over C). So H_g is not rational, hence the map from H_g to Pic^0(S) (sending a divisor to its associated line bundle) is not trivial, which means the Albanese map of S is not trivial. Then we would easily come to a contradiction. Sorry, it’s a very brief sketch…

So now we may ask ourselves this question, what if k is not C? The argument above would go wrong in two ways. Firstly, the reason that H_g would dominate M_g is because C is uncountable, so countably many proper subvarieties of M_g couldn’t cover whole M_g. But what if k is the algebraic closure of Q? That is to ask do we have a surface containin every curves defined over number field? This probably would be a very difficult and admittedly very interesting question. Secondly, we also have to consider the situation of characteristic p. If k is algebraic closure of F_p(t), then we don’t know if M_g is unirational or not (Ok, probably M_g is not unirational.)? At least I think most of people would guess it’s not (Well, let me speak for myself…). This is also an interesting question to ask. Long time ago the only technique we had for this kind of question was introduced by Mumford. Now I think we have another way to try, the thing below is just brainstorm…

Recently Professor Voisin has a new method called decomposition of diagonal to attack this kind of question (not unirationess but some other birational properties of a variety), so would her method help? Also one may ask her/himself whether given a general genus g curve we can put it in a nontrivial family over P^1 or not?

Anyway, I just realized there are still lots of open questions (as always?) out there. This world is full of challenges and fun, huh.

Hi there,

Outside of a dog, a book is a man’s best friend. Inside of a dog, it’s too dark to read.

I thought starting with a joke on the first paragraph in the first post of this blog would tell readers something about this blog. In the future, I will tell some of my recently understood math stuff in a sloppy way. I’m doing so mainly for fun and making sure I understood things correctly and less mainly for practicing my grammar (just kidding), so any comment on my mistakes (either grammatical or mathematical) will be appreciated.  I’ve noticed my website is a bit ugly, and probably it’s gonna keep being ugly…

Anyway, wish you a happy 2016.