Wall has calculated enough about the cobordism ring of oriented smooth manifolds that we know that two oriented smooth manifolds are oriented cobordant if and only if they have the same Stiefel--Whitney and Pontrjagin numbers.

Novikov has shown that rational Pontrjagin classes may be defined for topological manifolds; thus smooth manifolds which are topologically cobordant have equal Pontrjagin numbers. It is also easy to see that they have the same Stiefel--Whitney numbers (for this they only need to be Poincare cobordant).

It follows that smooth manifolds which are topologically cobordant are in fact smoothly cobordant. Is there a direct geometric proof of this fact?

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    $\begingroup$ Is this analogous, or equivalent through Thom-Pontrjagin construction, to asking whether or not within the homotopy class of a continuous map between two smooth manifolds, there is a choice of a smooth map, which vaguely I would imagine would follow from Weierstrass theorem?! $\endgroup$
    – user51223
    Jun 12, 2015 at 12:05
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    $\begingroup$ No: it is not the quality of the map that affects the outcome of the Pontrjagin--Thom construction, but the quality of the bundle. $\endgroup$ Jun 23, 2015 at 14:35
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    $\begingroup$ What about the unoriented case? Is there a direct proof that if a smooth manifold bounds a topological manifold (or even a Poincare complex) then it bounds a smooth manifold? $\endgroup$ Mar 3, 2017 at 20:47
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    $\begingroup$ I can see that the Pontryagin classes are defined for non-smooth manifolds (more generally for any oriented topological $R^n$-bundle) because the map $BSO\to BSTOP$ is known to be a rational homotopy equivalence. But why Stiefel-Whitney numbers (or classes?) can be defined for non-smooth manifolds? $\endgroup$
    – Victor
    Jan 4, 2018 at 1:58
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    $\begingroup$ @Victor: on a closed smooth manifold, the Wu formula describes how to recover the Stiefel-Whitney classes from Poincaré duality and the Steenrod module structure of mod 2 cohomology. The Steenrod module structure and Poincaré duality are present on closed topological manifolds, so one can use them in the same way to define Stiefel-Whitney classes. Then Stiefel-Whitney numbers can be obtained by evaluating on the fundamental class as usual. $\endgroup$ Jun 26, 2018 at 19:53

1 Answer 1


Is there a direct geometric argument to show that two smooth manifolds that are topologically bordant are actually smoothly bordant? This is the question to be addressed. The response below is a guide to a more satisfying situation....

In dimension three there is Likorish's argument using Dehn twists that oriented three manifolds bound. There is also Rochlin's geometric proof that in dimension four the complete bordism invariant is the signature circa 1950 [Oleg Viro at Stonybrook, his student, has these early references....] Rochlin was the student of Pontryagin. Thom suggested that he deserved the 1958 Fields Medal less than these predecessors.

Above that dimension, the only purely geometric step, to my knowledge, is the Pontryagin–Thom construction (1953 Thom, 1942 Pontryagin)reducing bordism questions to homotopy questions.

Serre's 1950 thesis opened the way to compute homotopy groups and the rest is the history alluded to in the question. Nevertheless.....

There is the theory of surgery or what was known as "constructive cobordisms" which one can learn from Milnor–Kervaire (Annals 1963) or Browder's book "Simply connected surgery theory" or Novikov's papers from early 60's, or Wall's book on "Non simply connected surgery"

There is also obstruction theory as explained in Steenrod's book "Topology of fibre bundles" circa late 50's.

With these three tools in hand one can try to do something step by step.

In my Princeton thesis January 1966 "Triangulating homotopy equivalences" I tried this for the problem of constructing a PL homotopy bordism of a homotopy equivalence to a PL homeomorphism when it was impossible to compute the torsion aspects in the smooth case and when the torsion aspects of the difference between PL bundles and smooth bundles was also unknown.

By a stroke of luck the two unknown torsion groups canceled each other and an obstruction theory for the question with explicit coefficients in $0$ $\mathbb{Z}/2$ $0$ $\mathbb{Z}$ $0$ $\mathbb{Z}/2$ $0$ $\mathbb{Z}$.... emerged. [At the thesis defense Steenrod asked how these explicit obstructions obstructions could be calculated. The question was unexpected but a very good one. They were like any obstruction theory explicitly ill-determined but in this case the effective obstructions could be determined using bordism theory itself.]

Takeaway, there may be a tweaked version of the challenge being discussed which has a step by step construction which is geometric and constructive yet still obstructed but now by more explicit groups related to the signature and Kervaire–Arf invariant of surgery & one might do something that is part geometric and part algebraic but still rather concrete. I would try to first lift the challenge to trying to lift a PL bordism to a smooth bordism because the relative homotopy group between topological and PL is one $\mathbb{Z}/2$ which is described by a signature divided by $8$ and reduced mod $2$. Novikov's theorem/information expanded by et al comprises this information.

(BTW The sad news appeared today that Serge Novikov is very ill.)

Then directly attack lifting a PL bordism to a smooth bordism. Now the elementary geometric argument of Thom from Colloque Topologie Mexico City 1957 suffices to understand the rational Pontryagin classes by geometry. And BTW Novikov's argument is built on this argument of Thom and both are geometric plus a dollop of Serre which is easy.

Conclusion: There is a chance to have a geometric understanding of this question using just basic algebraic topology/geometric information about manifolds with a dash of serre algebraic topology/ algebra .

Dennis Sullivan Friday December 13th, 2019.


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