# Pontryagin square of first Stiefel-Whitney class

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?

• What does the notation $M_4$ mean? Also, do you have a reference for "It is known that..."? Oct 4 '19 at 8:50
• @MarkGrant: I suspect $M_4 :=M$ to emphasize that $M$ is a 4-manifold.
– M.G.
Oct 4 '19 at 16:05
• 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. Oct 4 '19 at 20:15
• @MarkGrant Yes, as M.G. explained, $M_4=M$. Oct 5 '19 at 14:50
• @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. Oct 5 '19 at 14:57

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 $$2$$.