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Let $w_1$ be the 1st Sitefel-Whitney classes of the tangent bundle of a 4-manifold $M$ ($M$ is non orientable). My question is

Is $$\exp\left(\frac{i\pi}{2}\int_{M_4} \mathcal{P}(w_1^2)\right)$$ well defined? Here $\mathcal{P}$ is the Pontryagin square.

It is known that for any $\mathbb{Z}_2$ cocycle $u$, the following quantity
$$\exp\left(\frac{i\pi}{2}\int_{M_4} \mathcal{P}(u)\right)$$ is well defined, in the sense that it does not depend on the lifting of $u$ from $\mathbb{Z}_2$ cocycle to $\mathbb{Z}_4$ cocycle. However, when we plug in $u=w_1^2$, it seems to be not well defined and depends on the lift of $w_1$, because $P(w_1^2)= w_1^2 \cup w_1^2+ 4 w_1^2 \cup_1 w_1^3 = w_1^2 \cup w_1^2\mod 4$. So there is a contradiction. How do we resolve this contradiction?

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    $\begingroup$ What does the notation $M_4$ mean? Also, do you have a reference for "It is known that..."? $\endgroup$
    – Mark Grant
    Oct 4 '19 at 8:50
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    $\begingroup$ @MarkGrant: I suspect $M_4 :=M$ to emphasize that $M$ is a 4-manifold. $\endgroup$
    – M.G.
    Oct 4 '19 at 16:05
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    $\begingroup$ I'm not sure I understand what the integration means. In particular, if M is non-orientable then there is neither a volume form or an integral fundamental class to evaluate a mod 4 cohomology class against. $\endgroup$
    – Mark Grant
    Oct 4 '19 at 20:15
  • $\begingroup$ @MarkGrant Yes, as M.G. explained, $M_4=M$. $\endgroup$
    – user34104
    Oct 5 '19 at 14:50
  • $\begingroup$ @MarkGrant This is precisely the problem posted in the question, i.e., when plug in $u=w_1^2$, the well-defined $\exp(i \pi/2 \int_M \mathcal{P}(u))$ becomes not well defined. So it seems to suggest that identifying $u$ with $w_1^2$ is somewhat problematic. But I do not know why. $\endgroup$
    – user34104
    Oct 5 '19 at 14:57
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I think the problem is just that the epxression $$ \operatorname{exp}\left(\frac{i\pi}{2} \int_{M} \mathcal{P}(u)\right) $$ is well-defined only when $M$ is oriented.

If $M$ is oriented, it has a fundamental class $[M]\in H_4(M;\mathbb{Z})$. The since $\mathcal{P}(u)\in H^4(M;\mathbb{Z}/4)$ is a mod $4$ cohomology class, we can make sense of the integration as evaluation against the fundamental class, $$ \int_M\mathcal{P}(u):=\langle\mathcal{P}(u),[M]\rangle \in \mathbb{Z}/4. $$ This gives an integer mod $4$, hence the complex exponential above gives a well-defined $4$-th root of unity.

If $M$ is non-orientable, however, it only has a mod $2$ fundamental class $[M]_2\in H_4(M;\mathbb{Z}/2)$, and so we can only define $$ \int_M\mathcal{P}(u):=\langle\mathcal{P}(u),[M]_2\rangle \in \mathbb{Z}/2. $$ as an integer mod $2$.

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