So I know the fact that the join of $S^1$ and $S^1$ is homeomorphic to the 3-sphere, but I'm having trouble "seeing" this. I'd prefer something that appeals to geometric intuition, but more formal abstract arguments are welcome as well! I want to invoke the unit quaternions here, but I can't think of a particularly simple way to do it.

If this question is too "soft" I sincerely apologize.

  • $\begingroup$ @AgustíRoig Methinks in general, S^m joined to S^n is homeomorphic to S^(m+n+1)... $\endgroup$ – AsinglePANCAKE Sep 20 '12 at 18:39
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    $\begingroup$ The join in question is not $\vee$ but $*$: see this (One calls $\vee$ a wedge in this context, and dies a little every time one types \wedge to get a smash...) $\endgroup$ – Mariano Suárez-Álvarez Sep 20 '12 at 18:45
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    $\begingroup$ @AgustíRoig Methinks you're wedging and not joining... $\endgroup$ – AsinglePANCAKE Sep 20 '12 at 18:46
  • $\begingroup$ Ok, I mistook the "join" for the "wedge". Sorry. $\endgroup$ – d.t. Sep 20 '12 at 18:51
  • $\begingroup$ Thanks for your explanations, Asingle and Mariano. $\endgroup$ – d.t. Sep 20 '12 at 18:52

We discussed here a little while ago the fact that $S^3$ can be described as the result of gluing two solid tori along their boundaries.

To get a geometric reason for the homeomorphism $S^1*S^1\cong S^3$ you can look at how the two constructions are related: notice that two two tori have two central $S^1$s inside them...

(Alternatively: try to see what $S^1*[0,1]$ is, notice that you can describe $S^1$ as two copies of $[0,1]$ identified along the boundaries, and see how you can use this description on one of the two factors of $S^1*S^1$.)

(Another alternative: if you construct $S^1*S^1$ by doing identifications on $S^1\times[0,1]\times S^1$, cut the latter space in two parts $S^1\times[0,1/2]\times S^1$ and $S^1\times[1/2,1]\times S^1$, look at what the identifications do on each half, and then glue back the two parts)

  • $\begingroup$ I could add pictures... but that would ruin half of the fun for the OP, so I won't :-) $\endgroup$ – Mariano Suárez-Álvarez Sep 20 '12 at 19:01
  • $\begingroup$ But I like pictures! $\endgroup$ – AsinglePANCAKE Sep 21 '12 at 4:16
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    $\begingroup$ That is precisely why you have to learn to make them yourself! :-) $\endgroup$ – Mariano Suárez-Álvarez Sep 21 '12 at 4:26

It's better to think of $S^3$ as sitting inside $\mathbb{C}^2$ rather than $\mathbb{H}$ to understand $S^3=S^1\ast S^1$.

In $\mathbb{C}^2$, there are two "coordinate axes" $\mathbb{C}\times\{0\}$ and $\{0\}\times\mathbb{C}$ each containing a circle $S^1$, and the sphere $S^3$ is "filled out" by adjoining arcs $\{\cos(\theta)u+\sin(\theta)v\mid 0\le\theta\le\frac{\pi}{2}\}$ between points $u$ in the first circle and points $v$ in the second circle. Many things come from this description:

  • $ \{\cos(\theta)u+\sin(\theta)v\mid \theta=0\}=S^1\times\{0\} $
  • $ \{\cos(\theta)u+\sin(\theta)v\mid \theta=\frac{\pi}{2}\}=\{0\}\times S^1$
  • $ \{\cos(\theta)u+\sin(\theta)v\mid \theta=\frac{\pi}{4} \}=$ flat Clifford torus
  • $ \{\cos(\theta)u+\sin(\theta)v\mid 0\le\theta\le\frac{\pi}{4}\} =$ solid torus
  • $ \{\cos(\theta)u+\sin(\theta)v\mid \frac{\pi}{4}\le\theta\le\frac{\pi}{2}\} =$ solid torus

Evidently the two solid tori share a boundary, the Clifford torus. (Above, $u,v$ are not fixed.)

The arc-filling description also applies to $S^1=S^0\ast S^0$ (one $S^0$ are the East/West poles, the other are the North/South poles) and $S^2=S^1\ast S^0$ (the $S^0$ are the North/South poles, and $S^1$ is the equator; arcs are parts of longitudes).

It is possible to visualize the arc-filling description of $S^3$ somewhat by using stereographic projection $S^3\to\mathbb{R}^3\cup\{\infty\}$. Every line in $\mathbb{R}^3$ is a "generalized circle" that goes through $\infty$, wrapping around the other side. The two circles in $S^3=S^1\ast S^1$ project to one line through the origin and a unit circle around it. To get the arcs between the line and the circle, consider the half-plane cross-sections bounded by the line (they intersect the circle around it in a single point).

$\hskip 1in$ cross-section

The arcs between the point and the line comprise all semicircles with a diameter on the line and which intersect the point. If you fix $\theta$ and let $u,v$ vary you get tori in $S^3$, and their cross sections in the half-planes are in purple above.

Exercise: A solid torus is a circle's worth of disks. The above description implies the complement of a solid torus is itself a solid torus in $\mathbb{R}^3\cup\{\infty\}$. How is this so?

Quaternions help with the alternative description $S^3=S^0\ast S^2$. The $S^0=\{\pm1\}$ is in the real axis, the $S^2$ are the unit vectors in $\mathbb{R}^3$ (the subspace of imaginary quaternions), and the "arc-filling" is implicit in the fact unit quaternions all have a unique polar form $\exp(\theta\mathbf{u})$ where $\mathbf{u}$ is a unit vector (so, a sqrt of $-1$) and $0\le\theta\le\frac{\pi}{2}$.

If you stereographically project that description, the $S^0$ gives you the origin and the point at infinity, the $S^2$ is the usual unit sphere, and the arcs make up all the lines ("generalized circles") through the origin.

(I edited a picture from this webpage instead of making my own.)


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