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.


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