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The Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Precisely, $$ \mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x. $$ The $\cup_1$ is a higher cup product. The $Sq^1 x= x \cup_1 x$.

  1. If I only consider the $\mathbb{Z}_2$ normal subclass as $$H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_2) \subset H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4),$$ then the $\mathbb{Z}_2$ generator I believe can be written as $$ x \cup x =x \cup (w_2(M)+w_1(M)^2) $$

  2. If I consider the full $\mathbb{Z}_4$ class as $H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4),$ then the $\mathbb{Z}_4$ generator, can it be written as

    $$ \mathcal{P}(x)= x \cup (w_2(M)+w_1(M)^2) + x \cup_1 2 Sq^1 x? $$ in general?

  3. Therefore, on the spin-manifold $w_2(M)=w_1(M)=0$, or the Pin$^-$-manifold, the $ (w_2(M)+w_1(M)^2)=0$, so the first term disappear completely?

So on the spin-manifold or the Pin$^-$-manifold, the $\mathbb{Z}_4$ class of $ \mathcal{P}(x)$ is reduced to a $\mathbb{Z}_2$ subclass generated by

$$ x \cup_1 Sq^1 x= \frac{1}{2}(\mathcal{P}(x) -x^2) ? $$ Is this precise?

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    $\begingroup$ This doesn't make sense to me. $w_i(M) \in H^i(M;\Bbb Z/2)$. It is not an element of the cohomology of $B\Bbb Z/2$. $\endgroup$
    – mme
    Oct 13, 2018 at 23:51
  • $\begingroup$ sorry I had typoes let me fix it $\endgroup$
    – annie marie cœur
    Oct 14, 2018 at 0:03
  • $\begingroup$ I meant Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $ \mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. I fix it, thanks - let me know your thoughts please $\endgroup$
    – annie marie cœur
    Oct 14, 2018 at 0:05
  • $\begingroup$ That doesn't seem to have been related to my comment. The Stiefel-Whitney classes of a manifold $M$ live in the cohomology of $M$, which never appears in your post. $\endgroup$
    – mme
    Oct 14, 2018 at 0:22
  • $\begingroup$ Thanks, but I mean we can pair it with the fundamental classes of the manifold $M$. Agree? $\endgroup$
    – annie marie cœur
    Oct 14, 2018 at 1:44

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