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Suppose $v$ is a $\mathbb{Z}_2$ cochain on a four dimensional spin manifold $M$, i.e. $v\in H^1(M, \mathbb{Z}_2)$. I am interested in evaluating the quantity $$\exp \bigg(i \frac{\pi}{2}\int_M \mathcal{P}_2 (v\cup v)\bigg)$$ where $\mathcal{P}_2$ is the Pontryagin square which maps $H^2(M, \mathbb{Z}_2)$ to $H^4(M, \mathbb{Z}_4)$.

My questions are:

Is the above quantity in general takes value $\pm 1$? I would think so because since $M$ is a spin manifold, any $\int_{M} \mathcal{P}_2 (v\cup v)$ is an even integer, so $\int_{M} \frac{\mathcal{P}_2 (v\cup v)}{2}\in \mathbb{Z}$ is an integer. Hence $\exp(i \pi \mathbb{Z})=\pm 1$ follows.

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If the above is correct, do we have the following expression? $$\int_{M} \mathcal{P}_2 (v\cup v)=2\int_{M} v\cup v\cup v\cup v\mod 4 $$

Thanks in advance!

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1) The quantitity $\int \mathcal P_2(v \smile v)$ is not an integer. It is an element of $\Bbb Z/4$. The reason it still makes sense to write $\exp(i \frac \pi 2 x)$ where $x \in \Bbb Z/4$ is that $\exp$ is $2\pi i$-periodic; if we choose any two representatives $n, n' \in \Bbb Z$ which reduce to $x$ modulo $4$, the term $i \frac \pi 2 n$ and $i \frac \pi 2 n'$ differ by a multiple of $2\pi i$. So the exponentials are the same.

2) If $M$ is an oriented, even-dimensional manifold, then the map $H^1(M;\Bbb Z/2) \to H^{2n}(M;\Bbb Z/2)$ given by $v \mapsto v^{2n}$, is identically zero. To see this, we will use properties of the first Steenrod square. (You may be more familiar with $\text{Sq}^1$ under the name of "Bockstein map $H^k(Y;\Bbb Z/2) \to H^{k+1}(Y;\Bbb Z/2)$".)

Using the Cartan formula, one may inductively identify $$\text{Sq}^1(v^k) = \begin{cases} v^{k+1} & k \text{ odd},\\ 0 & k \text{ even}. \end{cases}$$ In particular, we may identify $\text{Sq}^1(v^{2k-1}) = v^{2k}$. At the same time, there is a cohomology class on a manifold $M$, known as the first Wu class $v_1$, with the property that if $x \in H^{2n-1}(M;\Bbb Z/2)$, we have $$x \smile v_1 = \text{Sq}^1(x).$$

Applying this definition to $x = v^{2n-1}$ above, we see that if $v^{2n} \neq 0$, necessarily the Wu class $v_1 \neq 0$. At the same time, one of the other crucial properties of Wu classes are that, in particular, $v_1 = w_1$, the first Stiefel-Whitney class, which governs orientability. So if $v_1 \neq 0$, your manifold is not orientable. This proves the desired claim.

In particular, in your case, $v^4 = 0$.

3) For $x \in H^2(M;\Bbb Z/2)$, the Pontryagin square $\mathcal P_2(x)$ satisfies the property that $\mathcal P_2(x) \pmod 2 = x^2 \in H^4(M;\Bbb Z/2)$. In particular, $\mathcal P_2(v^2) \equiv v^4 \pmod 2$, and $v^4 = 0$. Therefore $\mathcal P_2(v^2)$ must be either $0$ or $2$ in $H^4(M;\Bbb Z/4) \cong \Bbb Z/4$. This is precisely the statement that $$\exp\left(i \frac{\pi}{2} \int \mathcal P_2(v^2)\right) \in \pm 1.$$

4) Your final expression does not make sense. $v^4$ lives in $H^4(M;\Bbb Z/2)$ and nowhere else. And if $M$ is orientable, we saw above that this class is zero.

Note that nowhere here did I use that $M$ is spin, only orientable.

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    $\begingroup$ There is probably a more elementary proof of (2). (I would try to use differential topology methods, for one.) But this is the first that came to mind. I didn't try to compute any examples of $\mathcal P_2(v^2)$ for spin manifolds. $\endgroup$ – user98602 Nov 11 '18 at 23:24
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    $\begingroup$ @user34104 I am not sure about $\text{Pin}^-$ structures. The standard example for me of something with $v^4 \neq 0$ is $\Bbb{RP}^4$, which is not Pin^- ($w_2 = 0$ but $w_1^2 \neq 0$). So maybe one should be inspired by this to try to prove that something with a Pin^- structures still have $v^4 = 0$. $\endgroup$ – user98602 Nov 12 '18 at 3:03
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    $\begingroup$ @user34104 Actually, here is a proof. The second Wu class $v_2$ is precisely $w_2 + w_1^2$, and $v_2 \smile x = x^2$ on a closed 4-manifold. So if $v^4 \neq 0$, then $v_2 \neq 0$. So you cannot even support a $\text{Pin}^-$ structure if $v^4 \neq 0$ on a 4-manifold. You can still have a $\text{Pin}^+$ structure, as in the case of $\Bbb{RP}^4$. $\endgroup$ – user98602 Nov 12 '18 at 3:06
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    $\begingroup$ On $\Bbb{RP}^4$, we have $v^4 = 1$. That is what inspired me to guess it was necessarily non-orientable, since I couldn't think of any orientable things with anywhere similar cohomology rings. $\endgroup$ – user98602 Nov 12 '18 at 3:07
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    $\begingroup$ I always forget the precise definition of $\mathcal P_2$... so I do not know whether it evalutes to $1$ or $3$ on $\mathcal P_2(v^2)$ for $\Bbb{RP}^4$. $\endgroup$ – user98602 Nov 12 '18 at 3:09

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