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Let $E=\mathbb C/\Lambda$ be an elliptic curve, and let $D\subset E$ be a very small disc.
($D$ is round for the usual flat metric on $E$)

By the main result of [1], there exists a holomorphic immersion $f:E\setminus D \to \mathbb C$.
The image $f(\partial D)$ is a closed curve in $\mathbb C$ that self-intersects a bunch of times.

What is the shape of $f(\partial D)$?

I understand that such a curve is not unique. What I want is a qualitative description of an example of such a curve. A drawing would be great.

Example: When $D\subset E$ is a "rather big" disk, the curve ⌘ is the image of $\partial D$ under an immersion $E\setminus D \to \mathbb C\mathbb P^1$ (points in the central square have two preimages; points in the four lobes have zero preimages). But that only works when $D$ is a rather big compared to the size of $E$, and I don't know how to modify this curve as the size of $D$ tends to zero.


[1]: Gunning, R. C., Narasimhan, R., Immersion of open Riemann surfaces. Math. Ann. 174, 103–108 (1967).

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  • $\begingroup$ Neighborhood of the origin of $\mathbb{C}$ ? Take $U = \{s\in \mathbb{C}, s > r\}$ and glue $U / \{ r e^{i t}= \infty e^{i t}\}$ to obtain a torus. $\endgroup$ – reuns Nov 29 '17 at 13:10
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    $\begingroup$ What do you mean by "shape"? This curve is not unique, what information about it you need? $\endgroup$ – Alexandre Eremenko Nov 29 '17 at 13:20
  • $\begingroup$ @Alexandre Eremenko. That's right. The curve is not unique. I just want an example of what it can look like. @ reuns: The words "neighborhood of the origin" were confusing so I removed them. $\endgroup$ – André Henriques Nov 29 '17 at 14:22
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    $\begingroup$ @Andre Henriques: I still don't understand the question: "what does it look like"? Can you be more precise? Do you want an example, or you want some general property of these curves? $\endgroup$ – Alexandre Eremenko Nov 29 '17 at 15:04
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    $\begingroup$ @Adam P. Goucher : The Weierstrass P-function is defined on the complement of the point $0\in E$, and has zero derivative at three other points. So it's not an immersion. Immersions have everywhere non-zero derivative on their domain of definition. $\endgroup$ – André Henriques Nov 29 '17 at 17:30
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I don't know if this is the kind of answer which will satisfy. Write $f$ for the function on $\mathbb{C} - \Lambda$ and $z$ for the coordinate on $\mathbb{C}$. Write $D^{\ast}$ for the punctured disc.

Forstneric and Ohsawa show we can take $f$ to be of finite order as $z \to 0$, meaning $|f| < \exp(z^{-N})$ for some $N$. Put $f' = g$, so $g$ is a nowhere vanishing function on $D^{\ast}$ and also of finite order. Let $k$ be the winding number of $g'$ as we travel around $0$, so $z^{-k} g$ has a logarithm on $D^{\ast}$, say $g = z^k e^{h(z)}$. Then $|h| < z^{-N}$ so $h$ is meromorphic. Say $h(z) = c z^{-n} + \cdots$.

Undoing the changes of variable, $f = \int z^k \exp(c z^{-n} + \cdots)$.

This integral has different behavior on the sectors where $c z^{-n}$ lies in the left or the right half plane. If we fix $\theta \in (0, \pi/2)$ and consider the $n$ wedges where $\arg(c)-n \arg(z) \in (-\theta, \theta)$, then we have an asymptotic series starting $- \tfrac{z^{k+2n+1}}{c (n+1)} \exp(c z^{-n})$. So, for $z = r e^{i \phi}$ with $\phi$ ranging through each of these wedges, $f$ swings around a circle of large radius, roughly $r^{-n}$ times.

Meanwhile, if $\arg(c)-n \arg(z) \in (\pi-\theta, \pi+\theta)$, then $f'$ is very small, so $f$ stays very near some limiting value. I don't know whether it is the same limiting value in the $n$ different wedges, I don't see why it should be. It circles that value roughly $r^{-n}$ times. However, I am not claiming that I understand the behavior near the $2n$ rays where $\arg(c)-n \arg(z) \in \{0, \pi \}$ well enough to subtract off these numbers of criclings from each other.

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  • $\begingroup$ Very nice answer! Can any value of $\ell$ occur? Or do you think that $\ell$ is always equal to some fixed number? $\endgroup$ – André Henriques Nov 30 '17 at 10:17
  • $\begingroup$ I would expect it can be arbitrary since it is a lower order term than $N$. But I'll think about it a little. $\endgroup$ – David E Speyer Nov 30 '17 at 12:37
  • $\begingroup$ My asymptotic analysis was broken: When $\mathrm{Re}(c z^{-n})$ is negative, I picked out the wrong dominant term. Will try to fix. $\endgroup$ – David E Speyer Nov 30 '17 at 13:50
  • $\begingroup$ Fixed now (10 hours ago, actually). $\endgroup$ – David E Speyer Dec 1 '17 at 1:34
  • $\begingroup$ I'm guessing that one can arrange to have $n=3$. Namely, that one can arrange for there to be exactly three limiting values ( where the curve is roughly stationary, and circles around a given point in the complex plane). This $n=3$ should be related to the fact that the genus is $1$ (for general genus, I'm guessing that one can arrange for $n=3g$). Is this something that you can see from your analysis? $\endgroup$ – André Henriques Dec 1 '17 at 9:29
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Based on David Speyer's answer, I got what I want, namely some intuition about what these closed paths look like. It's a bit hard to draw them, so I include an inline description.


Let $X:=3i$, $Y:=0$, $Z:=-3i$.
Given $T\in[0,\infty)$, I'll define a closed path in $\mathbb C$, which I'll call $\gamma_T$.

Let $R:=5e^T$ and $r:=e^{-T}$.
Let $C_X$ be the circle of radius $r$ centred around $X$.
Let $C_{Y}$ be the circle of radius $r$ centred around $Y$.
Let $C_{Z}$ be the circle of radius $r$ centred around $Z$.
Let $C_\infty$ be the circle of radius $R$ centred around $0$.

The path $\gamma_T$ is as follows:

• Start at $Re^{-iT}$
• straight line to $X+re^{-iT}$
• run along $C_X$ for an angle of $4\pi+2T$, until the point $X+re^{iT}$
• straight line to $Re^{iT}$
• run along $C_\infty$ for an angle of $-2T$, until the point $Re^{-iT}$
• straight line to $Y+re^{-iT}$
• run along $C_Y$ for an angle of $4\pi+2T$, until the point $Y+re^{iT}$
• straight line to $Re^{iT}$
• run along $C_\infty$ for an angle of $-2T$, until the point $Re^{-iT}$
• straight line to $Z+re^{-iT}$
• run along $C_Z$ for an angle of $4\pi+2T$, until the point $Z+re^{iT}$
• straight line to $Re^{iT}$
• run along $C_\infty$ for an angle of $-4\pi-2T$, until the point $Re^{-iT}$

Remark: The path $\gamma_0$ is easy to visualise. After deforming it a tiny bit, it becomes isotopic to the image of the curve ⌘ under the map $z\mapsto (z-a)^{-1}$, where $a$ is the middle point of one of the lobes.


I believe that there exists an elliptic curve $E=\mathbb C/\Lambda$ and an immersion $$f:E\setminus\{0\}\to\mathbb C$$ such that $\forall T\in [0,\infty)$, the path $\gamma_T$ is the image under $f$ of a small counterclockwise loop (not quite round, but of bounded distortion) of radius $\simeq T^{-1/3}$ around the puncture.

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