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Let $M$ be an oriented null cobordant manifold.

Since $M$ is oriented its first Stiefel-Whitney class vanishes.

Since $M$ is null cobordant all of its Stiefel-Whitney numbers vanish.

Is it known if this implies that the second Stiefel-Whitney class vanishes so that $M$ admits a spin structure? Or are there known counterexamples?

psmears
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Flaccid Sheaf
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  • But how is the accepted answer consistent with the OP's "Since is null cobordant all of its Stiefel-Whitney numbers vanish " ? you pick a non-spin manifold with a second stiefel whitney class not vanish...? – Марина Marina S Aug 28 '21 at 18:03

1 Answers1

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No: Take your favourite orientable non-spin manifold $M$. Then $M\times S^1$ also does not admit a spin structure as the second stiefel whitney class does not vanish. But this manifold is nullbordant: It is the boundary of $M\times D^2$, where $D^2$ is the disc.

Thomas Rot
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