I'm completely aware of the triviality of this question, but for some reason, I can't visualize the argument.

In Hatcher's 3-manifold notes, the form of Alexander's theorem stating that Every embedded 2-sphere in $\mathbb{R}^3$ bounds an embedded 3-ball is given. Later, he uses this fact to conclude that $S^3$ is prime (recall that a 3-manifold $M$ is prime if, whenever $M=P\#Q$, either $P=S^3$ or $Q=S^3$ where here, $P\#Q$ denotes the connected sum of $P$ and $Q$), stating merely that every 2-sphere in S^3 bounds a 3-ball. I was hoping to visualize why this is true, and so far, I'm not having any luck; I'm hoping someone can help.

Things I've read:

  • From Hatcher's notes, it's mentioned that the trivial decomposition $M=M\# S^3$ is obtained by choosing the sphere $S$ (in the connected sum decomposition process) which bounds a ball in $M$. I realize that knowing this implies my result immediately, but it doesn't help me see the result.
  • Elsewhere, Hatcher states that the result follows from the fact that every 2-sphere in $S^3$ bounds a ball on each side. I've seen this justification elsewhere as well, but I can't visualize this one any easier.

I guess what I'm looking for is a direct proof of some kind. I tried to construct one as follows, but I'm stuck almost immediately after doing all the obvious things. Note that for a sphere $S$ in $M$, $M|S$ is Hatcher's notation for the manifold $M$ obtained by splitting along $S$, i.e. by removing an open tubular neighborhood $N(S)$ of $S$ from $M$.

Suppose that $S^3=P\#Q$. By definition of connected sum, there exists a sphere $S$ in $S^3$ such that $S^3|S$ has two components $P'$ and $Q'$ where $P$, respectively $Q$, is obtained from $P'$, respectively $Q'$, by filling in the boundary sphere corresponding to $S$ with a ball. By Alexander's theorem, $S$ bounds a 3-ball $B$ in $S^3$, so....

And that's it. I really have no idea how to proceed, and despite this being truly one of the most trivial things imaginable, I'm at a loss. Knowing why this is true for $S^3$ would be great, but knowing why the fact about general $M^3$ being trivially decomposed when $S$ bounds a ball would probably be much more helpful from a big picture perspective.

  • 1
    $\begingroup$ Here's what I think is happening. Saying that any (smooth) embedding of $S^3$ in $\mathbb{R}^3$ bounds a 3-ball is saying that the interior region of the embedded sphere is already a 3-ball. Thus any time you embed $S^2$ smoothly in $S^3$ (viewed as the one-point compactification of $\mathbb{R}^3$), it is automatically splitting $S^3$ into two 3-balls, one on each side. Thus the statement implies right away that $P$ and $Q$ are already 3-balls. $\endgroup$ – Ben Blum-Smith Feb 8 '15 at 0:53
  • $\begingroup$ @BenBlum-Smith - I understand what you're saying, but I still don't see why. Clearly, embedding $S^2$ smoothly into $\mathbb{R}^3$ (or thus into $S^3$) yields the interior region a 3-ball; I guess I'm confused as to why the exterior of an $S^2$ embedded into $S^3$ is necessarily a 3-ball? I agree that $S^3$ is the one-point compactification of $\mathbb{R}^3$...I don't know what I'm missing or why, but I'm definitely missing something. $\endgroup$ – cstover Feb 8 '15 at 3:12
  • 1
    $\begingroup$ Maybe this helps? The exterior region is homeomorphic to the interior region. You can see this by thinking about reflection in the unit sphere in $\mathbb{R}^3$. This is the map $\mathbb{R}^3\rightarrow\mathbb{R}^3$ obtained by taking a point of distance $r$ from the origin, and, keeping it on the same ray from the origin, relocating it to distance $1/r$. It is defined everywhere except at the origin. It fixes the points of the unit sphere. But if $\mathbb{R}^3$ is embedded in $S^3$, you can extend the map to the origin and it interchanges it with the point at infinity. $\endgroup$ – Ben Blum-Smith Feb 9 '15 at 14:25
  • 1
    $\begingroup$ This gives a smooth homomorphism between the interior ball and the exterior "ball" of the unit sphere. $\endgroup$ – Ben Blum-Smith Feb 9 '15 at 14:26
  • 1
    $\begingroup$ As Dan notes, you can also get some intuitive mileage by considering the analogy with the 2-d case. What I was describing also works in the plane - reflection in the unit circle. Viewing it as $\mathbb{C}$, it's the map $z\mapsto 1/\overline{z}$. $z\mapsto 1/z$ would also make the point. Viewed as a map on the Riemann sphere (aka $S^2$ seen as the one-point compactification of $\mathbb{C}$), it exchanges the interior disc with the exterior "disc" of the unit circle. $\endgroup$ – Ben Blum-Smith Feb 9 '15 at 14:29

Imagine an open ball $B^2$ in $\mathbb R^2$. If you consider the one point compactification of the complement, you get a closed ball. You could also consider an open ball in the one point compactification of $R^2$ which is $S^2$. Take the ball out and you'll see another ball.

The reason for this is that the boundary of the embedded ball, namely $S^1 = \partial B^2$, closes up at the other end, namely $\infty$.

This is how you can image the higher-dimensional analogue. The boundary of the ball, is also the boundary of another ball.

You could also imagine a bicollar of a codimension 1 sphere - it will close up at both ends.

Hope it helps your intuition.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.