# Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?

I'm looking for an elegant proof that any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$.

-
Kirby's book "Topology of 4-manifolds" or Gompf and Stipsicz's book "4-manifolds and Kirby Calculus" both have fairly elementary proofs of this. I believe one of the earliest proofs is in Thom's big paper on cobordism: R. Thom, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28, 17-86 (1954). –  Ryan Budney Apr 29 '11 at 3:20
Depending on what you mean by elegant there are also some more modern proofs that are closer to algorithmic questions, like Costantino and D.Thurston's work. –  Ryan Budney Apr 29 '11 at 3:22
I've taken the liberty of making the title more specific and replacing the dg.differential-geometry tag with a gt.geometric-topology one. Apologies if this changes the intent of the question, and please revert if that's the case. –  j.c. Apr 29 '11 at 12:26
A link to Costantino and Thurston's article: arxiv.org/abs/math/0506577 –  j.c. Apr 29 '11 at 13:27
@Ryan: Gompf and Stipsicz give brief descriptions of Rohlin's proof and Lickorish's proof, both described in the answers below. I haven't had a chance to look at Kirby's book yet. –  Dylan Thurston May 6 '11 at 9:54

I know of several different arguments. You can decide which one you think is most elegant...

1. Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Richard Kent's answer above.

2. Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.

3. Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.

4. I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4-manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3-sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.

-
Can you apply the argument 4 to a four-manifold $M^4$? Namely, map $M^4$ to $\mathbb R^3$ and apply the same method to construct a 5-manifold bounded by $M$. Of course, at some point you should need to do some topological blow-ups on $M^4$ to get zero-signature. –  Bruno Martelli Apr 30 '11 at 9:28
@Bruno: Yes, you can. You end up showing any 4-manifold is cobordant to a connect sum of $\mathbb{CP}^2$'s and $\overline{\mathbb{CP}}^2$'s. –  Dylan Thurston Apr 30 '11 at 9:43
I confess that I never understood the ideas of Thom's proof, certainly not well enough to see that it is fundamentally non-explicit. –  Greg Kuperberg May 1 '11 at 17:12
@Greg: I only understood Thom's proof well enough to see that it would be at least quite difficult to make it explicit, at least difficult enough that an explicit version might well be called a different proof. –  Dylan Thurston May 2 '11 at 3:39
Can you use your argument to bound the number of components of a link needed in a Kirby diagram? This gives some estimate on the minimal rank of H_2 of a simply-connected 4-manifold bounding the 3-manifold. –  Ian Agol May 5 '11 at 21:56

Maybe overkill, but elegant:

By a theorem of Hirsch, an oriented $3$-manifold $M$ embeds in the $5$-sphere (nonorientable case: Rohlin and Wall, independently). By Alexander duality, $M$ bounds a "Seifert $4$-manifold."

(Some references:

Hirsch, Immersions of almost parallelizable manifolds. Proc. Amer. Math. Soc. 12 1961 845–846.

Rohlin, The embedding of non-orientable three-manifolds into five-dimensional Euclidean space. Dokl. Akad. Nauk SSSR 160 1965 549–551.

Wall, All 3-manifolds imbed in 5-space. Bull. Amer. Math. Soc. 71 1965 564–567. )

-
That's quite interesting! And potentially related to the paper of Costantino and Thurston on complexity of 4-manifolds bounded by a 3-manifold. –  Greg Kuperberg Apr 29 '11 at 13:24
It's funny, I sorta thought this was how everybody thought about it. I guess having Cameron Gordon as an advisor colors your world view. –  Richard Kent Apr 29 '11 at 13:52
You can avoid Hirsch's theorem by using Whitney's immersion theorem of $M^n$ in $\mathbb R^{2n-1}$, which implies that every 3-manifold $M$ immerses in $\mathbb R^5$. Such an immersion can be perturbed so that self-intersections have dimension 3+3-5 = 1, i.e. are circles. You can then surger the manifold around the circles to get an embedding, and such a surgery can easily be realized by a cobordism. Then you use Alexander duality. This was Rohlin's original argument. –  Bruno Martelli Apr 29 '11 at 16:34
This is beautifully intuitive! But it's definitely massive overkill- each step is in itself more difficult that the fact being proven. Statement (1) is the easiest statement among: 1) Omega_3=0 2)Whitney immersion theorem 3)Hirsch's theorem 4) Alexander duality in this context. This is highlighted by Rourke's short elementary proof (which isn't half as conceptually satisfying for me, though). –  Daniel Moskovich Apr 29 '11 at 16:47
@Greg: It is related, though in a different direction; see my answer. @Daniel: I disagree that Whitney immersion is hard, and Hirsch's theorem is not necessary, as Bruno sketches. –  Dylan Thurston Apr 29 '11 at 23:37

MR0809959 (87f:57016) Rourke, Colin . A new proof that $\Omega_3$ is zero. J. London Math. Soc. (2) 31 (1985), no. 2, 373--376.

Edit: To summarize: Rourke's proof is short and elementary. Other proofs which I know involve either significant algebraic topology which is much harder than the theorem (Thom, or Rohlin), or lengthy calculations (Lickorish).
Any orientable 3-manifold has a Heegaard diagram $S(\mathbf{x},\mathbf{y})$, where $S$ is an orientable surface with two complete systems of curves $\mathbf{x}$ and $\mathbf{y}$ (a system of curves is complete if each curve it contains is simple and closed, its curves are pairwise disjoint, and their union does not separate $S$). A closed orientable 3-manifold $M(\mathbf{x},\mathbf{y})$ is obtained from $S(\mathbf{x},\mathbf{y})$ by attaching thickened 2-discs along $\mathbf{x}$ on $S\times {\{0\}}$ and along $\mathbf{y}$ on $S\times {\{1\}}$, and then filling in the resulting $S^2$ boundaries with 3-balls. The existence of a Heegaard diagram for any 3-manifold is a reformulation of the elementary fact that it has a handle decomposition.
Rourke's proof is by induction on the genus $g$ of $S$ and on the minimum intersection number $r$ of a curve in $\mathbf{x}$ with a curve in $\mathbf{y}$. Namely, he gives a straightforward combinatorial argument for why, if $r>1$, then there exists a third complete system of curves $\mathbf{z}$ on $S$ whose minimum pairwise linking with both $\mathbf{x}$ and $\mathbf{y}$ is less than $r$. Surgery of $M(\mathbf{x},\mathbf{y})$ around $\mathbf{z}$ gives the connect sum of $M(\mathbf{x},\mathbf{z})$ and $M(\mathbf{z},\mathbf{y})$. Finally, if $r\leq 2$, then you can chop one off the genus of $S$ pretty easily. And that's all there is to it, by induction.
Rourke's proof makes the fact that any closed orientable 3-manifold bounds a 4-manifold look like a stupidly easy combinatorics exercise. It's certainly my favourite proof of this theorem, although other ways of looking at the problem are not without their charm.

-
Nice induction proof. You can think of 3 handlebodies attached to the same surface $S$: if two pairs of them bound 4-manifolds $M_1, M_2$, then also the third pair does (the union of $M_1$ and $M_2$ along the third handlebody). –  Bruno Martelli Apr 30 '11 at 9:26