A celebrated theorem of Rohlin states the following

An oriented closed 4-manifold $M^4$ bounds an oriented 5-manifold if and only if the signature of $M^4$ is zero.

Simple homological arguments based on Lefschetz duality show that the vanishing of the signature is a necessary condition. Showing that it is also sufficient is however harder.

I know two proofs of this fact. Each is a variation of Rohlin's proof of the simpler 3-dimensional case, which says

An oriented closed 3-manifold $M^3$ bounds an oriented 4-manifold.

I was wondering if someone knows a more elementary proof, for instance based on Kirby calculus. The two proofs I know start as follows.

  1. By Whitney's theorem we can embed any closed oriented n-manifold $M^n$ in $\mathbb R^{2n}$ and we can immerse it in $\mathbb R^{2n-1}$. The immersion self-intersects into circles, and by accurately surgerying $M^n$ we can eliminate these self-intersections. Surgerying changes $M^n$ via a $(n+1)$-dimensional cobordism, hence we can suppose that $M^n$ itself embeds in $\mathbb R^{2n-1}$.
  2. As for knots in 3-space, any codimension-2 closed oriented manifold $M^n \subset \mathbb R^{n+2}$ bounds an oriented "Seifert" $(n+1)$-manifold $W^{n+1}$.

When $n=3$ these two facts imply that every closed oriented 3-manifold bounds an oriented 4-manifold. When $n=4$ we only obain that every closed oriented 4-manifold is cobordant to a codimension-3 embedded $M^4 \subset \mathbb R^7$ and more work has to be done.

  • In his original proof, Rohlin shows that up to blowing up $M^4$ in some points (i.e. making connected sums with $\pm\mathbb {CP}^2$) we can suppose that $M^4$ bounds a 5-cycle in $\mathbb R^7$, which can be subsequently smoothed to an oriented 5-manifold (blow-ups are needed in both steps!). This proof is explained in A la recherche de la topologie perdue.
  • In Kirby's book The topology of 4-manifolds, he proves that up to cobordism the 4-manifold $M^4$ can be immersed in $\mathbb R^6$. Such an immersion has double and triple points, like a surface in $\mathbb R^3$. Triple points have signs. He proves a nice theorem which says that the number of triple points counted with sign equals (up to a factor) the first Pontryagin number, which in turns equals (up to a factor 3) the signature thanks to the Hirzebruch formula! Therefore if $M$ has signature zero we can pair double points with opposite signs and destroy them by surgery. Finally we obtain an embedded cobordant 4-manifold $M^4 \subset \mathbb R^6$. Now codimension is two and there is a "Seifert" 5-manifold bounding $M^4$.

Finally, here is my question:

Do you know any other proof different from these ones? For instance, a proof which does not use embeddings in Euclidean space?

References are of course welcome.

  • 5
    $\begingroup$ Rene Thom, "Quelques propriétés globales des variétés différentiables", Commentarii Mathematici Helvetici 28, page 17–86?? In this well known and fields-medal-winning paper, Thom computes the unoriented cobordism completely, and also a large chunk of oriented cobordism, including the statement that $\Omega^{SO}_{4}=\mathbb{Z}$, which immediately implies the result you are asking for. But Thom's construction relies on embedding into euclidean space as well. $\endgroup$ – Johannes Ebert Nov 14 '10 at 13:26
  • $\begingroup$ Thank you very much, I though Thom only considered the non-oriented case. I am curious to see which techniques he used. $\endgroup$ – Bruno Martelli Nov 15 '10 at 17:39

It depends on what you consider elementary. Gompf-Stipcisz has something like this: Morse theory gives a handle decomposition. Surger circles (the trace gives a bordism) to kill the 1-handles (this introduces 2-handles). Turn upside down and kill the 3-handles. 2 handles are attached along a framed link whose surgery gives an $S^3$ since the 4-manifold is closed. Kirby calculus says your diagram is Kirby move equivalent to the empty framed link. So do the Kirby moves, but every time you blow up a +1 also blow up a -1 (off in a corner) and note that $CP^2 \# -CP^2$ bounds (it's the boundary of a $D^3$ bundle over $S^2$.). Handle slides are diffeos of the 4-manifold. Don't blow down, just move extra $\pm 1$ unknots aside. When you are done, your picture is a unlink with framings $\pm 1$, the same number of each since the signature is zero. This shows your manifold is bordant to a connected sum of $CP^2\# -CP^2$.

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    $\begingroup$ How can one prove Kirby calculus (namely, that two diagrams of the same 3-manifold are Kirby move equivalent) without using Rohlin's theorem? Gompf-Stipcisz' proof of Kirby calculus (in page 161) makes use of Rohlin's theorem, which is stated without proof in page 341. $\endgroup$ – Bruno Martelli Nov 13 '10 at 20:50
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    $\begingroup$ Oh, you're looking for a non circular proof! I don't know. Incidentally, on that page they attribute this fact to Thom, not Rohlin. $\endgroup$ – Paul Nov 13 '10 at 22:57
  • $\begingroup$ @Bruno: Kirby's first published proof just used Cerf's theorem on generic 1-parameter families of smooth functions connecting two Morse functions on a manifold, didn't it? $\endgroup$ – Ryan Budney Nov 14 '10 at 6:52
  • $\begingroup$ @Ryan: Kirby seems to use Rohlin's theorem at the beginning of page 37 of his paper. As in Gomps-Stipcisz, given two Kirby diagrams of the same 3-manifold he builds a closed 4-manifold by gluing the two resulting 4-dimensional handlebodies, he makes some blow-ups in order to kill the signature and he then uses that the resulting 4-manifold bounds a 5-manifold. Then he applies Cerf's theory to this 5-manifold, as far as I can understand from a quick reading. $\endgroup$ – Bruno Martelli Nov 14 '10 at 11:54
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    $\begingroup$ The MCG proofs of Kirby's theorem don't use Rokhlin's theorem. See my answer here: mathoverflow.net/questions/16848/proofs-of-kirbys-theorem $\endgroup$ – Daniel Moskovich Nov 14 '10 at 12:39

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