4
$\begingroup$

I am working thru Hatcher on my own and having some problems with what he calls the "standard decomposition" of the 3-sphere into two solid tori. Specifically,

(1) Why is the boundary of a 4-disk equal to the boundary of the product of two 2-disks?
More importantly, is this a particular example of a more general result for decomposing boundaries of complex shapes into boundaries of products of simpler spaces?

(2) I cannot follow his geometric argument, especially concerning the second torus. Is there an alternative explanation that any one would care to share?

(I realize this question overlaps some earlier questions, but I don't believe any earlier posts answer part (1) of my question.)

$\endgroup$
1
  • 2
    $\begingroup$ $D^4 \cong [0, 1]^4 \cong [0, 1]^2 \times [0,1]^2 \cong D^2 \times D^2$. As $D^4$ and $D^2 \times D^2$ are homeomorphic, they have homeomorphic boundaries. $\endgroup$ Feb 29, 2016 at 16:23

1 Answer 1

8
$\begingroup$

You may read [this post] 1, but let me give you a geometric interpretation of decomposing 3-sphere into two solid tori.

The idea is to regard $S^3$ as $\mathbb R^3$ plus a infinity point, and embed one solid torus $\textbf{T}$ into $\mathbb R^3$, and try to think why $(\mathbb R^3\cup {\infty})- \textbf{T}$ corresponds to the other solid torus.

Let's say $\textbf{T}$ is bounded by the torus $$x(u,v)=(3+\cos v)\cos u,\\ y(u,v)=(3+\cos v)\sin u,\\ z(u,v)=\sin v$$

Let me explain the decomposition in the following pictures.

decomposition of three sphere into two solid tori

The first picture is some disks attached on $\textbf{T}$, and the second picture show how these disks gives you another solid torus.

As you see in the picture, the boundary of $\textbf{T}$ is a torus drawn in black, and the green circle is its equator. Now pick any meridional circle, say the red loop on torus, you can have a $D^2$ with boundary attached with the circle, e.g., the surface A.

Now, the key point is to regard the upper space $\mathbb R_+^3-\textbf{T}$, i.e, the complement of $\textbf{T}$ with coordinate $z>0$, as slices of disks $(0,1)\times D^2$, with $\{0\}\times D^2$ corresponds to the disk B, and $\{1\}\times D^2$ corresponds to disk D at the equator (we haven't define this disk yet). And you can do the same thing downside for the lower space $\mathbb R_-^3 - \textbf{T}$, which gives you $(-1,0)\times D^2$, with point $\{-1\}\times D^2$ corresponds to disk at equator. Now, Attach them together at point $\{0\}\times D^2$ with disk B in blue color, we get $(-1,1)\times D^2$, and the remaining space is just $\{(x,y,0)|x^2+y^2>9\}\cup \{\infty\}$, but this is just an open disk removed from $S^2$, which is exactly $D^2$, and this gives you the last disk drawn in green color, which corresponds to $\{1\}\times D^2=\{-1\}\times D^2$. (At this point, you should think topologically why the green disk fit in place the second picture!)

Finally, you get a $S^1\times D^2$, which is exactly another solid torus.

$\endgroup$
1
  • $\begingroup$ Thanks, I can now at least see the correspondence of points in both spaces, although the geometry is slippery, for me. $\endgroup$
    – PossumP
    Mar 2, 2016 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .