Can a smooth compact manifold be the boundary of a non-smoothable manifold? If so can any of its Stiefel-Whitney numbers be non-zero?

Thom's theorem says that a compact smooth manifold has zero Stiefel-Whitney numbers if and only if it is the boundary of a smooth manifold. One naturally asks whether there are cases where a compact smooth manifold is a boundary of a manifold that has no differential structure and whether in such a case it is possible for the manifold to have a non-zero Stiefel-Whitney number.

  • $\begingroup$ If $M$ is smooth, doesn't $\partial M$ have an induced smooth structure? $\endgroup$ Apr 17, 2019 at 17:16
  • $\begingroup$ Yes but that is a different question. $\endgroup$
    – Joe S
    Apr 17, 2019 at 17:19
  • $\begingroup$ Ah, I misread your question. $\endgroup$ Apr 17, 2019 at 17:22
  • $\begingroup$ I edited my answer to give a counterexample to a topological Thom's theorem. $\endgroup$ Sep 8, 2020 at 19:47

2 Answers 2


Let $M$ be the $E_8$ manifold with an open disc removed; note that $\partial M = S^3$ is a smooth manifold. If $M$ were smoothable, then its double would be as well, but the double of $M$ is just the connected sum of two copies of the $E_8$ manifold which is not smoothable by Donaldson's Theorem.

As Oscar Randall-Williams points out here, if two smooth manifolds are topologically cobordant, then they are smoothly cobordant. In particular, if a smooth manifold bounds a non-smoothable manifold, it also bounds a smooth manifold, and hence all of its Stiefel-Whitney numbers vanish by Thom's theorem.

  • $\begingroup$ OK. The question also was asks whether Thom's theorem applies only in the smooth category. That is can a smooth manifold that is the boundary os a non-smoothable manifold have non-zero Stiefel-Whitney numbers? $\endgroup$
    – Joe S
    Apr 18, 2019 at 12:20
  • $\begingroup$ I have edited my answer to address this question too. $\endgroup$ Apr 18, 2019 at 16:51
  • $\begingroup$ I understand that Stiefel-Whitney classes can be defined topologically via Poincare duality and the Steenrod algebra. But in the smooth category if a manifold N is is the boundary of a smooth manifold M then the pull back of the tangent bundle of M to N under the inclusion map is the tangent bundle of N plus a trivial line bundle. The theorem then follows by naturality of Stiefel-Whitney classes. In the topological category this still works except for one thing. Why is the pull back of TM equal to TN plus a trivial line bundle? Without this I do not see how the theorem. follows. $\endgroup$
    – Joe S
    Apr 23, 2019 at 12:00
  • $\begingroup$ @JoeS: In the topological category, $TM$ doesn't make any sense. You can only define the tangent bundle for smooth manifolds. $\endgroup$ Apr 25, 2019 at 15:32

You in fact can define Stiefel-Whitney classes directly for topological manifolds. Topological manifolds come with $\mathbb{R}^n$ bundles that intuitively correspond to continuously choosing a chart around the point homeomorphic to $\mathbb{R}^n$. Then we can take the Thom space of this and define Stiefel-Whitney classes in the usual way using the Thom Isomorphism and the Steenrod squares. And these clearly coincide with the definition using the tangent bundle in case our manifold is smoothable.

Now Thom's theorem in the topological unoriented case would say that a manifold is a boundary, if and only if, the Stiefel-Whitney numbers vanish. Now I do not know about the reverse implication, but the forward implication definitely does hold, so any such manifold (smooth or not) has its Stiefel-Whitney numbers vanish.

To remind you of the proof: the Stiefel-Whitney numbers of a manifold are the collection of numbers obtained by pushing the fundamental class to $H_n(BO(n))$ (coefficients always mod 2) via its tangent bundle and evaluating on the monomials in the Stiefel Whitney class of degree n.

If our manifold $M$ is the boundary of $W$, then the fundamental class of $M$ is the boundary of the fundamental class of $W$, and the tangent bundle of the pair $(W,M)$ is classified by a map $(W,M) \rightarrow (BO(n+1),BO(n))$. But evaluating a cohomology class on a boundary is the same as evaluating the coboundary on the original, so I may evaluate the coboundary of the monomial in the Stiefel-Whitney classes on the fundamental class of $W$. However, the Stiefel-Whitney classes $w_0,\dots,w_n$ all are the restrictions of the classes of the same name in $BO(n+1)$, so by exactness of the sequence for a pair, this evaluates to 0.

Now in the topological case, just replace $BO(n)$ with $BTop(n)$, the classifying space for $\mathbb{R}^n$ bundles and the same argument holds.


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