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.

- 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 $\mathbb R^{2n-1}$.
- 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.