# Imitating in $\mathbb{R}^4$ the construction of tori in $\mathbb{R}^3$

This question belongs to a series of questions that keep arising in my mind from time to time: here and here are (in order) the two questions that I asked before.

The aim of the first question, let's say finding a similar "parametric definition" that was conformal for tori in $\mathbb{S}^3$ as that one in $\mathbb{R}^3$, was apparently too optimistic. Not even what I thought was a "parametric definition" in $\mathbb{R}^3$ seemed to be correct in the answer I got. Then I asked in the second question if it was possible to start with a closed curve and by rotating it construct a torus in $\mathbb{S}^3\subset\mathbb{R}^4$. I didn't get answers to this question.

At this point I'd like just to understand the most basic ideas. The parametric definition in $\mathbb{R}^3$ of a torus comes from a construction using a circle and rotating it. Suppose I start with a circle $$(x-a)^2+z^2=r^2,$$ for some fixed $a>r>0$.

Parametrizing this curve and multipliying the parametric vector by a matrix like $$\begin{pmatrix} \cos\phi & -\sin\phi & 0\\ \sin\phi & \cos\phi & 0\\ 0 & 0 & 1\end{pmatrix}$$ we rotate the curve around the $z$-axis (visually is easy to see that we construct in this way a torus in $\mathbb{R}^3$) and the usual parametric definition that one can find in the literature is obtained.

Now I'd like to imitate this idea but in $\mathbb{R}^4$. So I start with the same circle $(x-a)^2+z^2=r^2$, I take a parametrization $$x=r\cos\theta+a$$ $$z=r\sin\theta,$$ where $r\in(0,a)$ and $\theta\in[0,2\pi]$. Now I rotate it by doing $$\begin{pmatrix} \cos\phi & -\sin\phi & 0 & 0\\ \sin\phi & \cos\phi & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\cdot\begin{pmatrix}r\cos\theta+a\\ 0\\ r\sin\theta\\ 0\end{pmatrix}=\begin{pmatrix}\cos\phi(r\cos\theta+a)\\ \sin\phi(r\cos\theta+a)\\ r\sin\theta\\ 0\end{pmatrix}$$

This construction is exactly the same as that one in $\mathbb{R}^3$, which makes sense completely. Because of this I see that my simple question should be:

What is a torus (not topologically, that is clear) in $\mathbb{R}^4$? And once that's clear, what would be the analogous process to 'fit' a torus in $\mathbb{R}^4$? From there should be easy to understand how would it be in $\mathbb{S}^3\subset\mathbb{R}^4$.

• I'd say that without further context most people would take the definition of torus to be the topological one, so I'm not sure what kind of answer you're looking for when you ask "What is a torus in $\Bbb R^4$?" – Travis Mar 1 '18 at 12:27
• As for the second question, there is a standard embedding of the $2$-torus (i.e., $S^1 \times S^1$) in $\Bbb S^3 \subset \Bbb R^4$, namely $\{(x_1, y_1, x_2, y_2) \in \Bbb R^4 : x_1^2 + y_1^2 = \frac{1}{2} = x_2^2 + y_2^2 \}$, or better yet, $\{(z, w) \in \Bbb C^2 : |z| = \frac{1}{\sqrt{2}} = |w| \}$. This is a Clifford torus, which is distinguished among topological tori in $\Bbb R^4$ by having zero curvature. – Travis Mar 1 '18 at 12:40
• Again, I don't know precisely what you mean by torus in $\Bbb R^4$. An embedding of $S^1 \times S^1$? $S^1 \times S^1 \times S^1$? Both of these are possible, but these are both topological definitions. (NB it's possible to embed the first into $S^3$ but not the second.) – Travis Mar 1 '18 at 12:45
• No, there are many (topological) tori in $\Bbb R^3$---that is, topological embeddings $S^1 \times S^1 \to \Bbb R^3$, many (most) of which do not have simple explicit formulae. Just take your favorite torus with the usual parameterization and deform part of it a little without changing the topology. In order to get a nice result along the lines of "we can describe them all" for geometric objects, usually what we're describing all of needs to have stronger information than the topology alone entails. – Travis Mar 1 '18 at 14:03
• It depends on how you interpret the statement. It's correct to say that the parameterization (with specific fixed $a, r$) defines a torus in $\Bbb R^3$, but it's not true that every toplogically embedded torus in $\Bbb R^3$ can be parameterized that way for some $a, r$. – Travis Mar 1 '18 at 14:21

This construction can never produce a torus in $\mathbb{S}^{3}_{r}$ for any $r>0$. Since rotations with centre $0$ act on $\mathbb{S}^{3}_{r}$, the torus you construct will be contained in $\mathbb{S}^{3}_{r}$ if and only if the original circle is. The circle contains the points $(a-r,0,0,0)$ and $(a+r,0,0,0)$, which have different distance from the origin, so this is not the case.
(Notation: $\mathbb{S^{3}}_{r} = \{v \in \mathbb{R}^{4} : v \cdot v = r^2\}$)
Edit: case $a = 0$. in this case the point $(0,0,r,0)$ is fixed by the rotation, so the space you obtain is not homeomorphic to torus. In fact the space you obtain is homeomorphic to a $2$-sphere.
The standard way to embed a torus in $\mathbb{S}^{3} \subset \mathbb{C}^{2}$ is by the equaction $|z_{1}|^2 = |z_{2}|^{2} = \frac{1}{2}$ this is the so called Clifford torus. https://en.wikipedia.org/wiki/Clifford_torus.
• In the case of $r=1$ and $a=0$ the parametric vector clearly would be in $\mathbb{S}^3$, but firstly I'd like to actually understand why this does not (or maybe yes?) define a torus in $\mathbb{R}^4$... – Edu Mar 1 '18 at 12:17
• If this is the case you are interested in then why did you include the parameter $a$? Anyway, I will edit my answer to include this case – Nick L Mar 1 '18 at 12:18
• No. As the question illustrates I'm interested firstly on understanding 'what is a torus in $\mathbb{R}^4$', since the construction I'm using just gives me the same torus as in $\mathbb{R}^3$ but in a greater dimension... The parameter $a$ appears in the definition that one can usually find for a torus in $\mathbb{R}^3$ constructed by rotating the curve around an axis. Understanding how to fit a torus in $\mathbb{S}^3$ would come later. – Edu Mar 1 '18 at 12:24
• For $a \neq 0$ it is possible to obtain a torus in $\mathbb{R}^{4}$, I am not a aware of any way to "fit" this into the 3-sphere that would be geometric somehow. I think that Clifford Tori are definitely relevant to understanding the question "what is a torus in $\mathbb{R}^{4}$". Anyway, I hope this helped in some way. – Nick L Mar 1 '18 at 12:30