# Embedding the Riemann surface for $\sqrt z$ in $\mathbb{R}^3$

I've only ever seen the Riemann Surface for $\sqrt z$ described as below, with the $z$-axis typically being the $x$ or $y$ coordinate of $\sqrt z$. This embedding into $\mathbb{R}^3$ has intersection points however. For example, if our $z$-axis represents the $x$-coordinate, then both $i$ and $-i$ have the same $x$-coordinate ($0$), so they will be projected onto the same point in $\mathbb{R}^3$.

It seems however, that if we use the $z$-axis to represent the angle, then we can't have intersections (both roots of any number have a different angle). Am I missing something? In my mind, it seems like this picture would look like a small portion of the Riemann Surface for $\log z$, growing from $0$ to $2\pi$. However, this creates a boundary... and I'm really confused at this point. I hope some of you could follow this train of thought.

• This is unclear : "if our z-axis represents the x-coordinate, then both i and −i have the same x-coordinate (0)" Apr 29, 2017 at 22:21
• Say the $v$-axis is the vertical one at the center of your figure. This surface has two loops (or sheets) : one going on the top, the other one going down. The one going down has to intersect the other one because both are tighted to the $v$-axis. If you remove the "point $z=0$" then you will be able to avoid self-intersection. Apr 29, 2017 at 22:28
• See my comment to Andrew Hwang's answer for the video explaining my handwavy explanation. The $u$- and $v$-axes are the ones on the bottom horizontal plane. From what I understand, the vertical axis is usually a way to visualize the $z$ plane, so either its $x$ argument, $y$ argument, or a function of both. Apr 30, 2017 at 14:44

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$It sounds as if you're asking, "Can the Riemann surface of the square root be embedded into $\Reals^{3}$?", but the picture you seek is constrained by criteria you aren't quite sure how to articulate?

There is no two-valued function $f:\Cpx \to \Reals$ whose values at each non-zero point have the form $\{u, -u\}$ and whose graph in $(\Cpx \setminus\{0\}) \times \Reals \subset \Reals^{3}$ is a connected, embedded surface.

The "helicoidal" surface you envision has a boundary (as you say), so it's not an embedding of the Riemann surface of the square root.

There do (trivially!) exist embeddings of the Riemann surface of the square root into $\Reals^{3}$, but they're unlikely to satisfy because they aren't graph-like in the sense above: Instead, they're essentially graphs of the squaring function, or embeddings of the abstract Riemann surface, namely $\Cpx$, as a surface (e.g., a plane) in $\Reals^{3}$.

• In case it's of interest, here's an animation loop of the Riemann surface rotating in $\Cpx^{2} \simeq \Reals^{4}$. Note how the motion does look like rotation, but the entire plane containing the parabola is fixed. Apr 29, 2017 at 23:02
• That is exactly what I should have asked! Unfortunately, I couldn't follow most of your answer. Do you have a topology/embedding primer to recommend to help me understand the formalism behind what you are saying? Apr 30, 2017 at 14:40
• My question comes from this video: youtu.be/4MmSZrAlqKc?t=4m59s. Apr 30, 2017 at 14:40
• I don't have a primer in mind to suggest, but: The "graph" of a two-valued function is analogous to an ordinary graph, the set of points $(z, u)$ in $\Cpx \times \Reals$ such that $u = f(z)$ for some $z$ (and with two values for each $z$, which I'm taking to be $u$ and $-u$). Loosely, the set of non-zero complex numbers is a gasket; for the square root, walking once around the gasket causes the square root to pass from one sign/branch to the other. By the intermediate value theorem, the height has to pass through $0$, which causes the graph to self-intersect. Apr 30, 2017 at 15:22
• Starting to get it. How can we interpret the boundary in the helicoidal surface then? i.e. if it's not an embedding, then what is it...? Definitely injective, so then it must not be structure preserving. What structure exactly is it not preserving? Apr 30, 2017 at 15:45