# Computing Pontryagin Square

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.

$$~$$

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

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.

• 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.
– user98602
Nov 11, 2018 at 23:24
• @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 $v^4 = 0$.
– user98602
Nov 12, 2018 at 3:03
• @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 $\Bbb{RP}^4$.
– user98602
Nov 12, 2018 at 3:06
• 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.
– user98602
Nov 12, 2018 at 3:07
• 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 $\Bbb{RP}^4$.
– user98602
Nov 12, 2018 at 3:09