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I took a course in complex variables and I remember reading something about Riemann surfaces on Wikipedia. There are even some examples in this page:

enter image description here

I am curious about the following: How do we actually plot them? The definition in there doesn't look like anything I could translate into points $(x,y,z)$. I've been trying to guess for a while how to plot the Riemann surface for $f(z)=z^{\frac{1}{2}}$ but with no success. I think we need to compute the branches for $f(z)=z^{\frac{1}{2}}$ and we can do that with the following formula:

$$f_k(z)=|z|^{\frac{1}{n} } \left( \cos\left(\frac{\arg z + 2 k \pi}{n}\right) + i\sin\left(\frac{\arg z + 2 k \pi}{n}\right) \right) \quad k=0,1,2,..,n-1 $$

So in our case, we have $f_0$ and $f_1$ with $n=2$. Now I tried to plot the following sets of points:

$$(a,b,\arg f_0(a+bi))\qquad (a,b,\arg f_1(a+bi))$$

And I got this:

$\quad \quad \quad \quad \quad \quad \quad $enter image description here

Which kinda looks like it can be cut and glued like the figure I gave previously. Is it correct?

EDIT:

I had some progress on it, I read on Agarwal's Introduction to Complex Analysis:

enter image description here

enter image description here

I tried to do as he describes and am plotting the function $z=w^2$, with the restrictions he gave in the text. By plotting the pairs $(\Re(z),\Im(z),\arg(z))$ I obtained this:

enter image description here

I don't know if this is the correct result but it still doesn't look like the one I found on Wikipedia:

enter image description here

Red Banana
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    Mathematica has a function for this: https://resources.wolframcloud.com/FunctionRepository/resources/RiemannSurfacePlot3D – David G. Stork Jun 09 '22 at 04:34
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    @DavidG.Stork I know, but I want to be able to plot them by hand. – Red Banana Jun 09 '22 at 04:36
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    By hand? Why would anyone want to make a complex plot by hand? I wouldn't even plot a sine wave by hand. – David G. Stork Jun 09 '22 at 04:38
  • @DavidG.Stork I meant that I want to code it on Mathematica and hence, I need to know how to construct the points. – Red Banana Jun 09 '22 at 04:43
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    It's pretty simple really. Take a few sheet of paper for each branch, now connect them along the branch cut where the values are same. The orientation you connect the sheets by depends on value function takes. You can check out the book Road to Reality By Roger penrose, it's shown it being done for log(z). Also this https://www.youtube.com/watch?v=4MmSZrAlqKc&list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF&index=13 – tryst with freedom Jun 11 '22 at 16:58
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    The wikipedia-ish images can be created by graphing a complex power function and projecting it to three-space. For instance, the last diagram can be made with something like$$f(r,\theta)=(r^2\cos(2\theta),r^2\sin(2\theta), r\cos\theta, r\sin\theta);$$then discarding one of the last two coordinates. Similar tricks work for $\arcsin$ and $\log$. – Andrew D. Hwang Jun 11 '22 at 19:06
  • @AndrewD.Hwang It doesn't seems to be working. See here. What did I do wrong? – Red Banana Jun 12 '22 at 06:38
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    You have cosine in both of the first two components. :) – Andrew D. Hwang Jun 12 '22 at 11:33
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    @AndrewD.Hwang Oh, a silly mistake. It works now! But It's not clear to me how you converted the function $w=z^{1/2}$ to $f(r,\theta)=(r^2\cos(2\theta),r^2\sin(2\theta), r\cos\theta, r\sin\theta)$. – Red Banana Jun 12 '22 at 12:38

1 Answers1

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Modulo details about domains and images, if we view a function $f$ as a "maps-to" relation $y = f(x)$, the same set of points defines the inverse function $x = f^{-1}(y)$. More correctly, this works if and only if $f:X \to Y$ is a bijection. In that situation, each $x$ in $X$ corresponds to precisely one $y$ in $Y$, there exists an inverse mapping $f^{-1}:Y \to X$, and for all $x$ in $X$ and all $y$ in $Y$, $y = f(x)$ if and only if $x = f^{-1}(y)$.

This all holds for complex-valued functions. For example, the set of points $(z, w)$ in $\mathbf{C}^{2}$ satisfying $w = z^{2}$ represents the squaring function $f(z) = z^{2}$. Formally, the same set of points represents the complex square root. To make this correct we must restrict to an open set $Z$ in the $z$-plane where $f$ is injective, and take $W = f(Z)$ to be the image. But if we do not restrict, the "complex parabola" with equation $w = z^{2}$ may be viewed as the "Riemann surface of the square root" by interpreting $w$ as the independent variable.

If we write $z = re^{i\theta}$ in polar form, the graph of the squaring map may be viewed as the set of points $$ (z, w) = (z, z^{2}) = (re^{i\theta}, r^{2}e^{2i\theta}) \leftrightarrow (r\cos\theta, r\sin\theta, r^{2}\cos(2\theta), r^{2}\sin(2\theta)). $$ Up to permutation of coordinates in real four-space (swapping $z$ and $w$), this parametrizes the Riemann surface of the square root. (The calculations to do this in Cartesian, writing $z = x + iy$ and $w = u + iv$, are left as a pleasant exercise. The resulting plot, the Wikipedia diagram at the top of the question, is nice in its own way!)

This viewpoint sheds light on branch cuts and how sheets connect across branch cuts. A branch of square root or a sheet of the Riemann surface is a non-empty open subset $U$ of the graph $\{(z, w) : w = z^{2}\}$ that "passes the vertical line test in $w$" in the sense that the projection $(z, w) \to w$ is injective when restricted to $U$. A standard choice is to fix a ray from $0$ in the $w$-plane, remove the real parabola sitting over this ray when we project to the $w$-plane, and to take $U$ to be one of the two connected components.

We can play the same game with the complex logarithm: Here, the graph $w = \exp z$ may be parametrized by $$ (z, w) = (x, y, e^{x}\cos y, e^{x}\sin y). $$ The Riemann surface of $\log$ as pictured at the top of the post is obtained by discarding the real part of $z$. If we restrict $y$ to an interval $(\theta_{0}, \theta_{0} + 2\pi)$ (i.e., we restrict $z$ to a horizontal band of height $2\pi$) we obtain a branch of logarithm. Alternatively, we can visualize the complex logarithm by the parametrization $$ (x, y) \mapsto (e^{x}\cos y, e^{x}\sin y, x + y). $$ This depicts both the "parking garage" behavior of branches and the "log-ish" singularity near $0$.

The Riemann surface of log

Andrew D. Hwang
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    Really nice answer! Thank you very much! BTW: What book did you use to learn it this way? Most books I've seen describe the Riemann surface in a super abstract way, what you wrote is really neat and understandable. – Red Banana Jun 13 '22 at 10:23
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    I first learned complex analysis from Ahlfors' book, but arrived at the picture described here only gradually over the years. The square root in polar form looks like $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\theta/2}$; I remember trying to sketch this by hand as a student, and struggling particularly to understand the behavior near $0$. Over time, the realization about inverse functions sank in. <> Your question reminds me about a course handout on the complex square root I thought to submit to an expository journal a couple of years ago, but events intervened. So, thank you! – Andrew D. Hwang Jun 13 '22 at 11:25