# Using Alexander's Theorem to show that the sphere $S^3$ is a prime manifold

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.

• 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.

• 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. – Ben Blum-Smith Feb 8 '15 at 0:53
• @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. – cstover Feb 8 '15 at 3:12
• 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. – Ben Blum-Smith Feb 9 '15 at 14:25
• This gives a smooth homomorphism between the interior ball and the exterior "ball" of the unit sphere. – Ben Blum-Smith Feb 9 '15 at 14:26
• 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. – 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 $\infty$.