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 is prime. He later says that it follows immediately from Alexander’s Theorem as, and I quote: Every 2-sphere in bounds a 3-ball. And that’s it. Done.

Wait, what?!

Elsewhere, Hatcher expands his above statement: …every 2-sphere in bounds a ball on each side…[and h]ence 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 is the identity of the connected sum operation, that is irreducible (and that every irreducible manifold is prime), that one gets the trivial sum by splitting along a 2-sphere in which bounds a 3-ball in , 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 is said to be prime if writing implies that either or .

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 bounds a 3-ball.

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

To prove that is prime, we’ll show that is only possible if 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 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 as the union of with a point at infinity. As above, a 2-sphere in bounds a ball (in its interior) by Alexander’s theorem; moreover, the complement is homeomorphic to (since we can homeomorphically shrink to be arbitrarily small). Replacing with , we have that where 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 and glued along their boundary circles and by the identity map yield a 2-sphere . 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 -disks and consisting of one -cell, one -cell, and a 0-cell each, and glue them so that the boundary circles are identified, i.e. so that the resulting space consists of two -cells, one -cell, and one 0-cell. Appealing to Euler characteristic arguments, , and there’s no topology, and homology stuff, and yadda yadda yadda and…. Hence . 😆 😛 😉

Now, we formalize the point at hand.

Proposition 3: is prime.

Proof. Let be a 2-sphere in and let be the connected sum formed by splitting along (i.e., by removing a small open tubular neighborhood of from and by defining and to be the result of filling the boundary spheres corresponding to of the two components and of with a ball). The goal is to show that .

Now, bounds a ball in its interior (by Alexander) and in its exterior (by Fact 2.1 above) and so  is the disjoint union of two open 3-balls and : . What’s more, the connected sum is formed by removing a tubular neighborhood from and by gluing closed 3-balls back in the resulting boundary components, i.e. and are both formed by gluing a closed 3-ball to another closed 3-ball along their 2-sphere boundaries: where denotes a closed 3-ball. By fact 2.2 above, it follows that both and are homeomorphic to : .

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 , form by deleting an (open) annular region from and look at the pair of resulting manifolds and (in this case, each are surfaces of genus 0 with one boundary component).
• Rearranging 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 with 2-disks then gluing the boundary circles by the identity map.
• In particular, is being identified with the connected sum of two manifolds and which are each 2-spheres minus a (closed 2-)disk plus a (closed 2-)disk so that and so .

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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