I am currently reading Kervaire-Milnor's paper "Groups of Homotopy Spheres I", Annals of Mathematics, and I am trying to prove (or disprove) the following result. The more elementary the proof, the better.

If two smooth manifolds are homeomorphic, then their stable tangent bundles (i.e. the Whitney sum of the tangent bundle with the trivial line bundle) are vector bundle isomorphic.

I am trying to prove this as an intermediate step to give an alternative proof for KM's Theorem 3.1: Every homotopy sphere is $s$-parallelizable.


2 Answers 2


The result you are hoping for is in fact false.

In section 9 of Microbundles: Part I, Milnor constructs an open set $U \subset \mathbb{R}^m$. With its standard smooth structure, the (stable) tangent bundle of $U\times\mathbb{R}^k \subset \mathbb{R}^{m+k}$ is trivial, while in Corollary 9.3, Milnor shows that it admits a smooth structure for which the tangent bundle has a non-zero Pontryagin class. As Pontryagin classes are stable, the stable tangent bundle of the latter manifold is not trivial, and hence not isomorphic to the stable tangent bundle of $U\times\mathbb{R}^k$ with its standard smooth structure.

Milnor, John W., Microbundles, Topology 3, Suppl. 1, 53-80 (1964). ZBL0124.38404.

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    $\begingroup$ Is this true for compact manifolds though? I think it should follow from the fact the Spivak normal bundle is unique. (I also see the claim that any homotopy equivalence of manifolds can be covered by fiberwise isomorphisms of normal bundles, which I believe implies it). $\endgroup$ Sep 9, 2020 at 20:04
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    $\begingroup$ @ConnorMalin: I am not sure if it holds for compact manifolds, but I would be interested to know one way or the other. $\endgroup$ Sep 9, 2020 at 20:41
  • $\begingroup$ If one can embed the mapping cylinder of a (smooth map) between compact manifolds, then I think it should be true. This will need some type of relative version of the Whitney embedding theorem, which I don't know to be true. $\endgroup$ Sep 9, 2020 at 22:22
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    $\begingroup$ @ConnorMalin: It seems like this answer to your question provides an example of closed smooth manifolds which are homeomorphic but do not have stably equivalent tangent bundles. $\endgroup$ Nov 28, 2020 at 18:31
  • $\begingroup$ Yes its nice to have a concrete example. It definitely felt like it should be true from fundamental results of smoothing theory (apparently not when I made those comments though), but for the easiest examples, spheres and tori, it fails. $\endgroup$ Nov 28, 2020 at 19:10

Let me add something to Michael Albanese's great answer to see this question in a broader context.

Novikov proved that the rational Pontryagin classes are homeomorphism invariants (in fact, this was one of the achievements for which he received the Fields medal in 1970). The integral Pontryagin classes, however, are not invariant under homeomorphism, see Chapter 4.4 of "The Novikov Conjecture" by Kreck and Lück.

Some polynomials in the Pontryagin classes are even homotopy invariant: for instance, $p_1$ of a closed oriented $4$-manifold $M$ agrees (by Hirzebruch's signature theorem) with $3\sigma(M)$ times the fundamental class in cohomology (where $\sigma(M)$ denotes the signature of $M$), which is invariant under homotopy equivalence.

The famous Novikov conjecture asks whether certain so-called "higher signatures" are also invariant under homotopy equivalence. It is one of the most important open questions in topology.

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    $\begingroup$ I was under the impression that Novikov proved that the rational Pontryagin classes are homeomorphism invariants of a closed, orientable, smooth manifold. Can the closedness and orientability hypotheses be removed? It should be noted that the non-zero Pontryagin class I refer to in my answer is torsion. $\endgroup$ Aug 7, 2020 at 10:15
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    $\begingroup$ I am not sure what exactly Novikov proved originally, but I think these extra conditions are not necessary for the statement to hold. I suggest having a look at the "epilogue" of Milnor-Stasheff, which explicity mentions Novikov's result. There are also other proofs using different methods: arxiv.org/pdf/0901.0819.pdf $\endgroup$ Aug 7, 2020 at 13:52

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