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

Question

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

Answer 1

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 Autumn 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.

Answer 2

Thom wrote two notes in the proceedings of the "Colloque de Topologie de Strasbourg", which was a topology seminar organized by Ehresmann at that time:

1. "Quelques propriétés des variétés-bords", Colloque de Topologie de Strasbourg, 1951, no. V. La Bibliothèque Nationale et Universitaire de Strasbourg, 1952.

2. "Sur les variétés cobordantes", Colloque de topologie et géométrie différentielle, Strasbourg, 1952, no. 7. La Bibliothèque Nationale et Universitaire de Strasbourg, 1953.

These two notes prefigure some of the results and some of the techniques that will appear slightly later in his 1954 paper "Quelques propriétés globales des variétés différentiables" and each of these notes contains a proof of the fact that $\Omega_3=0$. In fact, these two prior proofs are different from the demonstration that follows by specializing the results of his '54 paper, and which was alluded to in the above answers.

His '52 proof (in the 1953 proceedings) is close to Pontryagin's ideas relating framed cobordism groups to stable homotopy groups of spheres. Thom considers there the map $\psi_k:\pi_{n+k}(S^n) \to \Omega_k$ which consists in taking regular preimages of smooth approximations of maps $S^{n+k} \to S^n$. This map is additive and its image consists of the cobordism classes of those $k$-manifolds which can be embedded in $\mathbb{R}^{n+k}$ with trivial normal bundle. Since any closed orientable $3$-manifold is parallelizable, the map $\psi_3$ is surjective (for $n$ large enough) and it suffices to prove that $\psi_3$ vanishes on a generating set of the group $\pi_{n+3}(S^n)$. To conclude his argument, Thom mentions that this stable homotopy group is isomorphic to $\mathbb{Z}_{24}$ (which, I guess, he knew from Cartan or Serre) and, most importantly, that it is generated by the suspension of the quaternionic Hopf fibration with fiber $S^3= \partial D^4$.

In contrast with his '52 and '54 proofs, his '51 proof (in the 1952 proceedings) does not use any kind of Pontryagin-Thom constructions but, as Lickorish and Rourke will do some years later, he uses the fact that any closed oriented $3$-manifold can be presented by a Heegaard splitting. Let $H_g$ be the handlebody of genus $g$, let $\hbox{MCG}(\partial H_g)$ denote the mapping class group of its boundary and for any $f \in \hbox{MCG}(\partial H_g)$, set

$$ V_f := H_g \cup_f (-H_g). $$

Thom starts by observing that, if $V_f$ and $V_{f'}$ bound, then so does $V_{f \circ f'}$ (see Bruno's comment to Daniel's answer). Next, he recalls that Dehn found an explicit and finite system of generators $\{\tau_i\}$ for $\hbox{MCG}(\partial H_g)$ which consists of several Dehn twists: there are 2, 5 and 2g(g-1) such generators for $g=1$, $g=2$ and $g>2$ respectively. Thus, by the previous observation, it suffices to check that $V_{\tau_i}$ is a boundary for any $g\geq 1$ and for any $i$. For any $g>1$, it can be observed that each of the Dehn twists $\tau_i$ is performed along a curve which avoids a full (solid) handle of $H_g$: hence $V_{\tau_i}$ is the connected sum of $S^1 \times S^2$ with a $3$-manifold admitting a Heegaard splitting of genus $g-1$. Since such a connected sum can be realized in dimension $4$ by the attachement of a handle of index $1$, we are allowed to do an induction on the genus $g$. For $g=1$, the $3$-manifolds $V_{\tau_1}$ and $V_{\tau_2}$ are $S^3$ and $S^1 \times S^2$, which are obviously boundaries.

It seems that this early proof that $\Omega_3=0$ has been forgotten since then: this is probably due to the fact that the proceedings is not widely available and is written in French. Personally, I did not know it before I opened this proceedings in the Math' Library of Strasbourg a few months ago! Nonetheless, Haefliger mentions it in his survey paper of Thom's works (Publ. IHES 1988).

Answer 3

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. )

Answer 4

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.