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  1. The Arf invariant is a nonsingular quadratic form over a field of characteristic 2.

The form that I looked at was: $$ S(q)=|H^1(M^2,\mathbb{Z}_2)|^{-1/2} \sum_{x\in H^1(M^2,\mathbb{Z}_2)} \exp[\pi \;i\; q(x)] $$ see Kirby-Taylor, Pin structures on low-dimensional manifolds.

The $M^2$ is an oriented 2 dimensional manifold with spin structures that has $\mathbb{Z}_2$ valued quadratic forms on $H_1(M^2,\mathbb{Z}_2)$, which obeys $$q(x + y) = q(x) + x ∩ y + q(y) \mod 2,$$ here $x ∩ y$ denotes the $\mathbb{Z}_2$ intersection pairing. The bordism invariant is the Arf invariant.

  1. Stiefel-Whitney class is $\mathbb{Z}_2$-characteristic class associated to real vector bundles. For example, we can take the $w_i(TM)$ for the tangent bundle of the base manifold $M$.

Question:

  • Is there some precise way to pair the Arf invariant $S(q)$ (or any variation of Arf) as a $\mathbb{Z}_2$-cohomology class $\text{arf}$ with Stiefel-Whitney class $w_i(TM)$? How would one write and define them formally as a topological invariant? Like $$ \text{arf} \cup w_{d-2}(TM)? $$ where the $M$ is a $d$-dimensional manifold? How do one cup product this two objects formally?

  • Must this $\text{arf} \cup w_{d-2}(TM)$ be a $\mathbb{Z}_2$-characteristic class?

  • What will be a nontrivial manifold generator $M$ for such an invariant $\text{arf} \cup w_{d-2}(TM)$?

  • How to view the Poincare dual PD for the 2 and $d-2$ submanifolds from $\text{arf}$ or $w_{d-2}(TM)$?

This is a variation of the problem from a previous Math.SE post which receives 0 feedback.

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    $\begingroup$ One problem with this is that the Arf invariant is either $1$ or $-1$, whereas Stiefel-Whitney classes are an element of mod 2 cohomology, in which $1$ and $-1$ are equal. So these things just look different to me. $\endgroup$ Oct 1, 2018 at 22:11
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    $\begingroup$ sorry, let us map the Arf invariant 1 or −1 to 0 and 1 mod 2 cohomology. Is that possible? $\endgroup$
    – wonderich
    Oct 1, 2018 at 22:17

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