## Big Picard theorem

I’ve been asked to prove the Big Picard theorem, assuming the fundamental normality test. Assuming the latter, it is a very short proof, and I could half-ass with that. I don’t like writing up stuff that I don’t actually understand for the sake of doing so. There’s little point, and if I’m going to actually write up a proof of it, I’ll do so for real, which means that I go over the fundamental normality test in its entirety.

## On grad school, science, academia, and also a problem on Riemann surfaces

I like mathematics a ton and I am not bad at it. In fact, I am probably better than many math graduate students at math, though surely, they will have more knowledge than I do in some respects, or maybe even not that, because frankly, the American undergrad math major curriculum is often rather pathetic, well maybe largely because the students kind of suck. In some sense, you have to be pretty clueless to be majoring in just pure math if you’re not a real outlier at it, enough to have a chance at a serious academic career. Of course, math professors won’t say this. So we have now an excess of people who really shouldn’t be in science (because they much lack the technical power or an at least reasonable scientific taste/discernment, or more often both) adding noise to the job market. On this, Katz in his infamous Don’t Become a Scientist piece writes:

## Elliptic functions

I am writing this as a way to go through in detail the section on elliptic functions in Schlag’s book.

Proposition 4.14.  Let $\Lambda = \{m\omega_1 + n\omega_2 | m,n \in \mathbb{Z}\}$ and set $\Lambda^* = \Lambda \setminus \{0\}$. For any integer $n \geq 3$, the series Continue reading “Elliptic functions”

## Construction of Riemann surfaces as quotients

There is a theorem in Chapter 4 Section 5 of Schlag’s complex analysis text. I went through it a month ago, but only half understood it, and it is my hope that passing through it again, this time with writeup, will finally shed light, after having studied in detail some typical examples of such Riemann surfaces, especially tori, the conformal equivalence classes of which can be represented by the fundamental region of the modular group, which arise from quotienting out by lattices on the complex plane, as well as Fuchsian groups.

In the text, the theorem is stated as follows.

Theorem 4.12.  Let $\Omega \subset \mathbb{C}_{\infty}$ and $G < \mathrm{Aut}(\mathbb{C}_{\infty})$ with the property that

• $g(\Omega) \subset \Omega$ for all $g \in G$,
• for all $g \in G, g \neq \mathrm{id}$, all fixed points of $g$ in $\mathbb{C}_{\infty}$ lie outside of $\Omega$,
• for all $K \subset \Omega$ compact, the cardinality of $\{g \in G | g(K) \cap K \neq \phi\}$ is finite.

Under these assumptions, the natural projection $\pi : \Omega \to \Omega / G$ is a covering map which turns $\Omega/G$ canonically onto a Riemann surface.

The properties essentially say that the we have a Fuchsian group $G$ acting on $\Omega \subset \mathbb{C}_{\infty}$ without fixed points, excepting the identity. To show that quotient space is a Riemann surface, we need to construct charts. For this, notice that without fixed points, there is for all $z \in \Omega$, a small pre-compact open neighborhood of $z$ denoted by $K_z \subset \Omega$, so that

$g(\overline{K_z} \cap \overline{K_z}) = \emptyset \qquad \forall g \in G, g \neq \mathrm{id}$.

So, in $K_z$ no two elements are twice represented, which mean the projection $\pi : K_z \to K_z$ is the identity, and therefore we can use the $K_z$s as charts. The $g$s as Mobius transformations are open maps which take the $K_z$s to open sets. In other words, $\pi^{-1}(K_z) = \bigcup_{g \in G} g^{-1}(K_z)$ with pairwise disjoint open sets $g^{-1}(K_z)$. From this, the $K_z$s are open sets in the quotient topology. In this scheme, the $g$s are the transition maps.

Finally, we verify that this topology is Hausdorff. Suppose $\pi(z_1) \neq \pi(z_2)$ and define for all $n \geq 1$,

$A_n = \left\{z \in \Omega | |z-z_1| < \frac{r}{n}\right\} \subset \Omega$

$B_n = \left\{z \in \Omega | |z-z_2| < \frac{r}{n}\right\} \subset \Omega$

where $r > 0$ is sufficiently small. Define $K = \overline{A_1} \cup \overline{B_1}$ and suppose that $\pi(A_n) \cap \pi(B_n) \neq \emptyset$ for all $n \geq 1$. Then for some $a_n \in A_n$ and $g_n \in G$ we have

$g_n(a_n) \in B_n \qquad \forall n \geq 1$.

Since $g_n(K) \cap K$ has finite cardinality, there are only finitely many possibilities for $g_n$ and one of them therefore occurs infinitely often. Pass to the limit $n \to \infty$ and we have $g(z_1) = z_2$ or $\pi(z_1) = \pi(z_2)$, a contradiction.

## Variants of the Schwarz lemma

Take some self map on the unit disk $\mathbb{D}$, $f$. If $f(0) = 0$, $g(z) = f(z) / z$ has a removable singularity at $0$. On $|z| = r$, $|g(z)| \leq 1 / r$, and with the maximum principle on $r \to 1$, we derive $|f(z)| \leq |z|$ everywhere. In particular, if $|f(z)| = |z|$ anywhere, constancy by the maximum principle tells us that $f(z) = \lambda z$, where $|\lambda| = 1$. $g$ with the removable singularity removed has $g(0) = f'(0)$, so again, by the maximum principle, $|f'(0)| = 1$ means $g$ is a constant of modulus $1$. Moreover, if $f$ is not an automorphism, we cannot have $|f(z)| = |z|$ anywhere, so in that case, $|f'(0)| < 1$.

## Cauchy’s integral formula in complex analysis

I took a graduate course in complex analysis a while ago as an undergraduate. However, I did not actually understand it well at all, to which is a testament that much of the knowledge vanished very quickly. It pleases me though now following some intellectual maturation, after relearning certain theorems, they seem to stick more permanently, with the main ideas behind the proof more easily understandably clear than mind-disorienting, the latter of which was experienced by me too much in my early days. Shall I say it that before I must have been on drugs of something, because the way about which I approached certain things was frankly quite weird, and in retrospect, I was in many ways an animal-like creature trapped within the confines of an addled consciousness oblivious and uninhibited. Almost certainly never again will I experience anything like that. Now, I can only mentally rationalize the conscious experience of a mentally inferior creature but such cannot be experienced for real. It is almost like how an evangelical cannot imagine what it is like not to believe in God, and even goes as far as to contempt the pagan. Exaltation, exhilaration was concomitant with the leap of consciousness till it not long after established its normalcy.

## Weierstrass products

Long time ago when I was a clueless kid about the finish 10th grade of high school, I first learned about Euler’s determination of $\zeta(2) = \frac{\pi^2}{6}$. The technique he used was of course factorization of $\sin z / z$ via its infinitely many roots to

$\displaystyle\prod_{n=1}^{\infty} \left(1 - \frac{z}{n\pi}\right)\left(1 + \frac{z}{n\pi}\right) = \displaystyle\prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2\pi^2}\right)$.

## Riemann mapping theorem

I am going to make an effort to understand the proof of the Riemann mapping theorem, which states that there exists a conformal map from any simply connected region that is not the entire plane to the unit disk. I learned of its significance that its combination with the Poisson integral formula can be used to solve basically any Dirichlet problem where the region in question in simply connected.

## More mathematical struggles

Math is hard. It wrecks my self-esteem, and at times, it makes me feel an utter loser, who simply isn’t smart enough, who is a league if not multiple away from the big name mathematicians, who come up with much if not most of the most original results in mathematics. There are times when the formalism within the mathematics looks, perhaps superficially out of lack of perception no the part of its viewer, so excruciatingly complex and dry, and that one is inclined to simply go: this is too hard, give up. I’ve felt that, and I think just about everyone, no matter how smart, has, to some extent. Over time, I’ve come to realize that the dirty details tend to be a natural product of a few main ideas behind the proof, and once such ideas as grasped, every detail can easily be seen to have its rightful place within the entire construction. There was a time when I felt demoralized or slightly baffled upon seeing this answer of Ron Maimon that can totally come across as intellectually too presumptuous, from a guy too smart who never had to struggle like all us ordinary folks, from a guy who takes for granted as routine what is a slog for most, without being metacognitively aware enough to appreciate that he is of a totally different beast. In this, stood out the following quote:

You need to learn to “unpack” proofs into the construction that is involved, to know what the proof is saying really. It is no good to memorize the proof, you need to understand the construction, and this will motivate the proof.

## An unpacking of Hurwitz’s theorem in complex analysis

Let’s first state it.

Theorem (Hurwitz’s theorem). Suppose $\{f_k(z)\}$ is a sequence of analytic functions on a domain $D$ that converges normally on $D$ to $f(z)$, and suppose that $f(z)$ has a zero of order $N$ at $z_0$. Then for every small enough $\rho > 0$, there is $k$ large such that $f_k(z)$ has exactly $N$ zeros in the disk $\{|z - z_0| < \rho\}$, counting multiplicity, and these zeros converge to $z_0$ as $k \to \infty$.

As a refresher, normal convergence on $D$ is convergence uniformly on every closed disk contained by it. We know that the argument principle comes in handy for counting zeros within a domain. That means

The number of zeros in $|z - z_0| < \rho$, $\rho$ arbitrarily small, goes to the number of zeros inside the same circle of $f$, provided that

$\frac{1}{2\pi i}\int_{|z - z_0| = \rho} \frac{f'_k(z)}{f_k(z)}dz \longrightarrow \frac{1}{2\pi i}\int_{|z - z_0| = \rho} \frac{f'(z)}{f(z)}dz$.

To show that boils down to a few technicalities. First of all, let $\rho > 0$ be sufficiently small that the closed disk $\{|z - z_0| \leq \rho\}$ is contained in $D$, with $f(z) \neq 0$ inside it everywhere except for $z_0$. Since $f_k(z)$ converges to $f(z)$ uniformly inside that closed disk, $f_k(z)$ is not zero on its boundary, the domain integrated over, for sufficiently large $k$. Further, since $f_k \to f$ uniformly, so does $f'_k / f_k \to f' / f$, so we have condition such that convergence is preserved on application of integral to the elements of the sequence and to its convergent value. With $\rho$ arbitrarily small, the zeros of $f_k(z)$ must accumulate at $z_0$.