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A famous result by Thom states that Oriented Bordism classes are determined by characteristic numbers; specifically, two closed manifolds are orientedly bordant if and only if they have the same Stiefel-Whitney and Pontryagin numbers (I'll just talk about Stiefel-Whitney for brevity). An immediate consequence is that if $M$ is a closed manifold which has a non-vanishing Stiefel-Whitney number involving $w_k$ for some $k$, then $w_k(N)\neq 0$ for any $N$ which is bordant to $M$; in other words this non-vanishing characteristic number provides a "bordism reason" for why a characteristic class should be non-zero.

My question concerns the converse. Given an $M$, suppose that for some $k$ every Steifel-Whitney number involving $w_k$ vanishes, so that there is "no bordism reason" for the class to be non-vanishing: is it then possible to find an $N$ which is bordant to $M$ and has $w_k(N)=0$? If so, is it possible to simultaneously eliminate all classes which have no bordism reason to be non-zero?

This seems like something which maybe shouldn't be expected since characteristic classes often provide obstructions to doing surgery, and two manifolds are orientedly bordant exactly when they differ by a finite sequence of surgeries. On the other hand the manifold $\mathbb{CP}^n\#\overline{\mathbb{CP}^n}$ is null-bordant even though it has many non-vanishing characteristic classes for $n>1$ (half the time it isn't even spin).

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    $\begingroup$ Have you checked whether some set of standard generators of the cobordism ring yield such representatives? Eg the Dold manifolds, in Erzeugende der Thomschen Algebra 𝔑. Math. Z. 65 (1956), 25--35. See also map.mpim-bonn.mpg.de/Unoriented_bordism $\endgroup$
    – Danny Ruberman
    Commented Oct 31, 2016 at 12:48

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This is not a complete answer, but rather two specific instances where the answer is yes.

  1. If every Stiefel-Whitney number of $M$ involving $w_1$ vanishes, then $M$ is cobordant to an orientable manifold $N$. See Proposition 4 of Wall, C. T. C., Determination of the cobordism ring, Ann. Math. (2) 72, 292-311 (1960). ZBL0097.38801.

  2. If every Stiefel-Whitney number of $M$ involving $w_1$ and $w_2$ vanishes, then $M$ is cobordant to a spin manifold $N$. See Corollary 2.4 of Anderson, D. W.; Brown, Edgar H. jun.; Peterson, Franklin P., The structure of the Spin cobordism ring, Ann. Math. (2) 86, 271-298 (1967). ZBL0156.21605.

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I suggest to consider $\mathbb CP^4 \# (\mathbb CP^2 \times \mathbb CP^2)$. Then all SW-numbers involving $w_4$ vanish, but on the other hand we have $\text{Sq}^2w_4 = w_2w_4 + w_6$ in $H^{\ast}(BSO;\mathbb F_2)$, so for any manifold with $w_4 = 0$ we also have $w_6 = 0$. But $\mathbb CP^4 \# (\mathbb CP^2 \times \mathbb CP^2)$ has nontrivial SW-number $w_2w_6$, so no manifold cobordant to it can have vanishing $w_6$.

EDIT: As pointed out in a comment below, this is wrong.

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    $\begingroup$ I did the computation but I got a different result: I found that $w_2w_6$ vanished for both $\mathbb{C}P^4$ and $\mathbb{C}P^2 \times \mathbb{C}P^2$, but that $w_4^2$ is non-zero for the connected sum. $\endgroup$
    – William
    Commented Nov 1, 2016 at 13:02
  • $\begingroup$ But I think it's maybe true that there are too many relations between characteristic classes that it might be difficult to phrase this problem accurately $\endgroup$
    – William
    Commented Nov 1, 2016 at 13:10
  • $\begingroup$ You are right, I had an error in my calculation. For some reason I thought $\binom{5}{2}$ would be odd and hence $w_4(\mathbb CP^4) \neq 0$... but I am feeling a similar example might actually work. I will continue to think about it. $\endgroup$
    – Jens Reinhold
    Commented Nov 1, 2016 at 13:19

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