# What does $S^1\times S^1 \times S^1$ look like?

The primary question is:

What does $S^1\times S^1 \times S^1$ look like? (assuming that it makes sense)

I have feeling that $S^1\times S^1 \times S^1$ sits in 4-D space because one envisions $S^1$ as being a circle on a plane and $S^1\times S^1$ being a torus in 3-D space. Almost an implication to a general rule. Which is probably why I'm having hard time imagining it.

Is there a general rule as implicated above?

It's natural to ask if $S^1\times (S^1 \times S^1)$ is the same as $(S^1\times S^1) \times S^1$?

• Do you see how you get the torus from the circle by adding on a circle at each point? Best I can do is ask you to do the same once more..
– M.B.
Commented Feb 7, 2014 at 20:41
• Is 'adding' a circle to each point on a torus the same as 'adding' a torus to each point on a circle? Commented Feb 7, 2014 at 20:55
• Yes. If you visualize it you will see it is the same thing - 'torus' and 'circle' are the same object in this case. Commented Feb 7, 2014 at 22:01

Another way to think about the $n$-torus is by identifying opposite faces of the $n$-hypercube. E.g.: You get $S^1$ by identifying the ends of the unit interval. You get $S^1\times S^1$ by identifying the opposite edges of the unit square. What you are interested in is what you get by gluing opposite faces of the unit cube.

You can find some nice illustrations here: http://www.geom.uiuc.edu/video/sos/materials/overview/

• Brilliant. For me, the best way to describe it. Commented Feb 7, 2014 at 21:00

A good way to visualize the 3-torus is in 3D slices, that rotate and scan the object at various angles.

Using the implicit definition of $T^3$ in Cartesian coordinates, where $R_1 > R_2 > R_3$ :

$$\left(\sqrt{\left(\sqrt{x^2 + y^2} -R_1\right)^2 + z^2} -R_2\right)^2 + w^2 = R_3^2$$

There are 3 distinct planes of rotation that will transform the two torus intercepts: $xz$ , $xw$ , and $zw$

The equation for rotating on plane $xz$ :

$$\left(\sqrt{\left(\sqrt{\left(x\cdot\cos(t)+z\cdot\sin(t)\right)^2 + y^2} -R_1\right)^2 + \left(x\cdot\sin(t)-z\cdot\cos(t)\right)^2} -R_2\right)^2 + w^2 = R_3^2$$

Rotating at origin by animating $0 < t < 2\pi$:

Holding at these angles while passing through the 3-plane:

The equation for rotating on plane $xw$ :

$$\left(\sqrt{\left(\sqrt{\left(x\cdot\cos(t)+w\cdot\sin(t)\right)^2 + y^2} -R_1\right)^2 + z^2} -R_2\right)^2 + \left(x\cdot\sin(t)-w\cdot\cos(t)\right)^2 = R_3^2$$

Rotating at origin:

Holding at these angles while passing through the 3-plane:

The equation for rotating on plane $zw$ :

$$\left(\sqrt{\left(\sqrt{x^2 + y^2} -R_1\right)^2 + \left(z\cdot\cos(t)+w\cdot\sin(t)\right)^2} -R_2\right)^2 + \left(z\cdot\sin(t)-w\cdot\cos(t)\right)^2 = R_3^2$$

Rotating at origin:

Holding at these angles while passing through the 3-plane:

• You know, I have a weakness for when shapes split into two (topological) torii and each half follows a toroidal path before merging back into a single shape again (fourth image). +1 from me; extremely entertaining. Commented Apr 22, 2016 at 3:33

A torus ($S^1\times S^1$) is just a circle rotated around an axis outside the circle. A 'hypertorus' (just my own word for it) is a torus rotated around an external axis, parallel to the $3$D space of the normal torus.

• I'm not sure I understand what you mean by rotation around an axis in 4D? A (Givens) rotation moves points in a plane, so in 4D it's set of fixed points is a 2D-plane rather than a line. Commented Feb 7, 2014 at 21:28
• @JyrkiLahtonen, I think you're right about that Commented Feb 7, 2014 at 22:17