June 15th: Singularities

This post is based upon Richard Borcherds’ video on Singularities. I will write out the key concepts that I have learned from this video in my own words (without referring to my notes) and some questions that I have.

We begin by looking at removable singularities. How do we understand such singularities? I like how Borcherds said these singularities arise from our own laziness to write the function properly (or maybe because we intentionally want to create such a phenomenon), i.e. the singularity can be easily corrected by representing the function in another way. I interpreted this type of singularities as ‘man-made’.

Example: $f(z) = \frac{sin(z)}{z}$ .

It has a removable singularity at the origin. Key: we look at its taylor series. Notice some important properties about $f(z)$ near the origin: it is holomorphic at any point that is not the origin and it is also bounded.

$f(z) = \frac{z + z^3/3! + z^5/5! + ...}{z} = 1 + z^2/e! + z^4/5! + ...$

As $z \rightarrow 0$ , $f(z) \rightarrow 1$ . If we write $f(z)$ in its taylor series, it really has no singularities.

Borcherds listed a criterion for functions with removable singularity:

If f is holomorphic for $0 < |z| < \epsilon$ and f is bounded, then the singularity is removable.

How do we prove this? We show that f can be rewritten in a nicer form.

Define $g(w) = \frac{1}{2i\pi} \int_C \frac{f(z)}{z - w}dz$ , where C is a closed curve around the origin.

$g(z)$ is holomorphic for $|w| < \epsilon$ because $\frac{f(z)}{z - w}$ is holomorphic on C (since C is away from the origin). Something I am not too clear about is what happens if $w \in C$ ? Is C fixed? From the way Borcherds described it in the video, it seems C can be moved around, so that we can avoid w being on C (the integral will not change since $f(z)$ is holomorphic in the region away from the origin). It follows that g(0) is well-defined.

Claim: $g(w) = f(w)$ for any $0 < |w| < \epsilon$ .

In order for this to work, we would need to move C across the origin so that it encloses w. Since f is bounded in the neighbourhood around the origin, the integral of $\frac{f(z)}{z-w}$ around a circle enclosing the origin is controlled by the radius, so as the radius tends to the origin, the integral also tends to 0. My interpretation: we don’t have to think of it as moving C across the origin, we can really deform C so that it looks like a closed curve around w:

Hence, it follows that f can be expressed g, and g is defined at the origin.

Now, what happens when the value of f near a singularity becomes unbounded? i.e. as we approach the origin in any direction, the function tends towards infinity? We get a pole (if we look at the modulus of the function, we really see a pole). Borcherds said that poles are really quite nice, because if we look at $\frac{1}{f}$ , then it is holomorphic at that point. My takeaway from this is that when we study meromorphic, we really need to study both $f$ and $\frac{1}{f}$ at once, i.e. near the poles of $f$ , we study the behaviour of $\frac{1}{f}$ and vice versa.

Near a pole, $f$ will locally look like $\frac{1}{z^n}$ for some $n \in \mathbb{N}$ .

So what happens if we have a pole-looking singularity at which f cannot be approximated by a $\frac{1}{z^n}$ ? Suppose the taylor series is of the form $f(z) \sum_{n \in \mathbb{Z}} a_n z^n$ . What’s the difference between this and the poles we saw before? The difference occurs when we approaching this singular point. For poles, it doesn’t matter the direction we are approaching, we are all tending towards a single value (i.e. infinity). But with this type of singularities, the direction in which approach it matters.

Take for instance $e^{-frac{1}{z^2}}$ . When restricted to the real line:

It is bounded and smooth near the origin. So we might think that the origin is a removable singularity. But no! If we approach the origin along the imaginary axis, we would be tending towards infinity.

Notice that as we move towards the origin in various directions, any complex value would be achieved infinitely many times. This is true for any essential singularity (this is the difference between poles and this type of singularity). So suppose there exists $a \in \mathbb{C}$ such that only occurs finitely many times. Then, there exists a neighbourhood around the origin such that $f \neq a$ for all points in that neighbourhood. This means that $|f(z) - a| \geq \delta$ (for some $\delta > 0$ ) in that neighbourhood. Why does such $\delta$ exist? It seems obvious isn’t it because if such $\delta$ does not exist, then we would have $f \rightarrow a$ inside that region (so in order for this to be obvious, I think we need to assume that f does not have any removable singularities in this region).

This means that $|\frac{1}{f(z) - a}|$ is bounded, and so $\frac{1}{f(z) - a}$ is holomorphic, i.e.

$\frac{1}{f(z) - a} = g(z)$ , where $g(z)$ is holomorphic at the origin.

Key: we can split f into a holomorphic part and a not-so-nice part, i.e. $f(z) = \sum_{n \in \mathbb{Z}_{-}} a_n z^n + \sum_{n \in \mathbb{N}} a_n z^n$ .

June 15th: Singularities

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