# S^3 (the most basic prime manifold) is prime

So a while ago, I was reading Hatcher’s notes on 3-manifolds. In there, he defines what it means for a manifold to be prime and states, casually, that the 3-sphere $S^3$ is prime. He later says that it follows immediately from Alexander’s Theorem as, and I quote: Every 2-sphere in $S^3$ bounds a 3-ball. And that’s it. Done.

Wait, what?!

Elsewhere, Hatcher expands his above statement: …every 2-sphere in $S^3$ bounds a ball on each side…[and h]ence $S^3$ is prime. Again, though, it isn’t accompanied by anything, and while this is clearly a trivial result, I just couldn’t see it for the longest time…I knew that it followed from a number of things, e.g. the fact that $S^3$ is the identity of the connected sum operation, that $S^3$ is irreducible (and that every irreducible manifold is prime), that one gets the trivial sum $M\# S^3=M$ by splitting along a 2-sphere $S$ in $M^3$ which bounds a 3-ball in $M$, etc. Even so, I didn’t want to leverage some enormous machinery to deduce the smallest of results and what I really wanted was for someone to tell me what I was missing. So I never stopped thinking about this, even after moving forward, until finally – it just clicked!

I figure other people who are as visualization-impaired as I may benefit from seeing this explained in greater depth, so in lieu of typing a blog post containing something new and attention-worthy, I figure I’d share this instead. Details after the break.

Throughout, I’m going to assume familiarity with the connected sum operation. From there, we begin with a definition:

Definition 1: A connected 3-manifold $M$ is said to be prime if writing $M=P\# Q$ implies that either $P=S^3$ or $Q=S^3$.

As Hatcher indicates, Alexander’s theorem will be a fundamental piece of the puzzle. I’ll state the version given in the first of the above-linked references here, deferring proof with the note that Hatcher’s proof is very readable (even if you’re totally unfamiliar with Morse theory).

Alexander’s Theorem: Every 2-sphere in $\mathbb{R}^3$ bounds a 3-ball.

A 1924 version of this result (again attributed to Alexander) says that, moreover, the result holds when $\mathbb{R}^3$ is replaced with $S^3$ and hence, every 2-sphere in $S^3$ bounds a 3-ball. This is the version we care about.

To prove that $S^3$ is prime, we’ll show that $S^3 = P\# Q$ is only possible if $P=Q=S^3$ and to do that, we’ll need the following observations/facts/lemmas/axioms/whatever…they’re the crux of the whole argument and they’re the things that totally escaped me.

Fact 2:

1. Every 2-sphere in $S^3$ bounds a ball on both sides;
2. Gluing two closed 3-balls along their 2-sphere boundaries via the identity map yields a 3-sphere.

To see the truth of the first fact, imagine $S^3$ as the union of $\mathbb{R}^3$ with a point $\{\infty\}$ at infinity. As above, a 2-sphere $S$ in $\mathbb{R}^3$ bounds a ball (in its interior) by Alexander’s theorem; moreover, the complement $\mathbb{R}^3\setminus S$ is homeomorphic to $\mathbb{R}^3\setminus\{pt\}$ (since we can homeomorphically shrink $S$ to be arbitrarily small). Replacing $\mathbb{R}^3$ with $S^3\cong\mathbb{R}^3\cup\{\infty\}$, we have that $S^3\setminus S\cong\mathbb{R}^3\cong B^3$ where $B^3$ is the open 3-ball.

The second fact can be seen in a couple different ways. In one dimension lower, note that a pair of closed 2-disks $D$ and $D'$ glued along their boundary circles $S$ and $S'$ by the identity map yield a 2-sphere $S^2$. In three dimensions, the argument is the same, and this can be seen to generalize to arbitrary finite dimensions using cell decompositions: Take cell decompositions of closed $n$-disks $D_1^n$ and $D_2^n$ consisting of one $n$-cell, one $(n-1)$-cell, and a 0-cell each, and glue them so that the boundary circles are identified, i.e. so that the resulting space $X$ consists of two $n$-cells, one $(n-1)$-cell, and one 0-cell. Appealing to Euler characteristic arguments, $\chi(X)=1\pm 1\mp 2\in\{0,2\}$, and there’s no topology, and homology stuff, and yadda yadda yadda and…. Hence $X\cong S^n$. 😆 😛 😉

Now, we formalize the point at hand.

Proposition 3: $S^3$ is prime.

Proof. Let $S$ be a 2-sphere in $S^3$ and let $S^3=P\# Q=P\#_S Q$ be the connected sum formed by splitting along $S$ (i.e., by removing a small open tubular neighborhood $N(S)$ of $S$ from $S^3$ and by defining $P$ and $Q$ to be the result of filling the boundary spheres corresponding to $S$ of the two components $P'$ and $Q'$ of $S^3|S=S^3-N(S)$ with a ball). The goal is to show that $P\cong Q\cong S^3$.

Now, $S$ bounds a ball in its interior (by Alexander) and in its exterior (by Fact 2.1 above) and so $S^3\setminus S$ is the disjoint union of two open 3-balls $B_{int}$ and $B_{ext}$: $S^3\setminus S = B_{int}\sqcup_S B_{ext}$. What’s more, the connected sum $P\# Q$ is formed by removing a tubular neighborhood $N(S)$ from $S^3$ and by gluing closed 3-balls back in the resulting boundary components, i.e. $P$ and $Q$ are both formed by gluing a closed 3-ball to another closed 3-ball along their 2-sphere boundaries: $P,Q \cong D^3 \cup_\partial D^3$ where $D^3$ denotes a closed 3-ball. By fact 2.2 above, it follows that both $P$ and $Q$ are homeomorphic to $S^3$: $P,Q\cong S^3$. $\square$

I’m not 100% sure which parts of this argument I struggled with the most but honestly, I never didn’t struggle with it. For me personally, it helped to draw the corresponding picture in one dimension lower:

• Take a 2-sphere $S$, form $S|S^1$ by deleting an (open) annular region from $S$ and look at the pair of resulting manifolds $M'$ and $N'$ (in this case, each are surfaces of genus 0 with one boundary component).
• Rearranging $M', N'$ and looking at how one would glue them to form the connected sum, it becomes obvious that (up to homeomorphism) it’s just a matter of capping the boundary of each of $D, D'$ with 2-disks then gluing the boundary circles $\partial D, \partial D'$ by the identity map.
• In particular, $S^2$ is being identified with the connected sum of two manifolds $M$ and $N$ which are each 2-spheres minus a (closed 2-)disk plus a (closed 2-)disk so that $M,N\cong S^2$ and so $S^2=S^2\# S^2$.

This is all very trivial, of course, but spending the extra time to ensure I wasn’t taking it for granted helped improved my ability to visualize things, I think. Even if not, maybe this (clumsy) (unpolished) (probably unsatisfactory) expounding will help someone fill in the gaps more quickly than I did!

Note: Though no exposition was entirely satisfactory for me, I found the proof of Corollary 2.2.3 of Schleimer’s lecture notes to be somewhat helpful, particularly in the realization of Fact 2.2.