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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?

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  • $\begingroup$ But how is the accepted answer consistent with the OP's "Since 𝑀 is null cobordant all of its Stiefel-Whitney numbers vanish " ? you pick a non-spin manifold 𝑀 with a second stiefel whitney class not vanish...? $\endgroup$
    – Марина Marina S
    Commented Aug 28, 2021 at 18:03

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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.

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  • $\begingroup$ But how is your example consistent with the OP's "Since 𝑀 is null cobordant all of its Stiefel-Whitney numbers vanish " ? you pick a non-spin manifold 𝑀 with a second stiefel whitney class not vanish...? $\endgroup$
    – Марина Marina S
    Commented Aug 28, 2021 at 18:03
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    $\begingroup$ Classes are not numbers $\endgroup$
    – Thomas Rot
    Commented Aug 28, 2021 at 18:25
  • $\begingroup$ yes, classes are not numbers. But they are related: how does a second stiefel whitney class relate to its stiefel whitney number? $\endgroup$
    – Марина Marina S
    Commented Aug 28, 2021 at 18:27

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