Let $n$ be an even integer and $X$ a manifold. Given a cohomology class $B \in H^k(X,\mathbb{Z}_n)$, the Pontryagin square is a class $\mathfrak{P}(B)\in H^{2k}(X,\mathbb{Z}_{2n})$. Is it true that if $X$ is a spin manifold then $\mathfrak{P}(B)$ is an even class? In this case it should make sense to consider the class $\frac{\mathfrak{P}(B)}{2}\in H^{2k}(X,\mathbb{Z}_n)$. What is the most general class of manifolds where this makes sense? Could you provide an example where this does not make sense?
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$\begingroup$ If $\mathfrak{P}(B)$ is an even class, then wouldn't $\frac{\mathfrak{P}(B)}{2} \in H^{2k}(X, \mathbb{Z}_{\color{red}{2n}})$? $\endgroup$– Michael AlbaneseMay 23 at 13:48
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If $B \in H^2(X;\mathbb{Z}_2)$ satisfies $B^2 \neq 0$, then $\mathfrak{P}(B) \in H^4(X; \mathbb{Z}_4)$ is not even as $\mathfrak{P}(B) \equiv B^2 \bmod 2$. Such classes can exist on spin manifolds, for example, take $B$ to be the generator of $H^2(\mathbb{CP}^3; \mathbb{Z}_2)$.
If $X$ is a closed $2k$-dimensional manifold and $B \in H^k(X; \mathbb{Z}_2)$, then $B^2 = \nu_k\cup B$ where $\nu_k$ is the $k^{\text{th}}$ Wu class. So if $\nu_k = 0$, then $B^2 = 0$, so $\mathfrak{P}(B) \equiv 0 \bmod 2$, i.e. $\mathfrak{P}(B)$ is even for every $B \in H^k(X; \mathbb{Z}_2)$. When $X$ is a closed four-manifold, then the relevant Wu class is $\nu_2 = w_1^2 + w_2$. In particular, if $X$ is spin (or more generally, pin$^-$), we see that $\mathfrak{P}(B)$ is even for every $B \in H^2(X; \mathbb{Z}_2)$.
answered May 23 at 13:43
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$\begingroup$ Thank you, indeed I was looking at the case of $H^2$ for four-manifolds. This clarifies how it works in general. About the question I asked on the reduction to $\mathbb{Z}_{n}$ (that you mentioned in the comment), I would say that an even class can be consistently divided by $2$, and the result is well definite modulo $n$. Differently said you have the SES $\mathbb{Z}_n\rightarrow \mathbb{Z}_{2n}\rightarrow \mathbb{Z}_2$ where the right map is $\text{mod}(2)$ and even elements are in the kernel of this map, so they are lifted to elements in $\mathbb{Z}_n$. $\endgroup$ May 23 at 15:26
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$\begingroup$ Sorry, I am assuming your answer does hold for any $\mathbb{Z}_n$ with $n$ even, not only for $n=2$. Am I right? Also, can you suggest a reference where the story you described is explained? (I think it is textbook material at the end, but I am a physicist...) $\endgroup$ May 23 at 18:31
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$\begingroup$ I don't know how things work with $\mathbb{Z}_n$ coefficients. One key fact about $\mathbb{Z}_2$ coefficients is that $B \mapsto B^2$ is linear, so it can be given by the cup product with a fixed class; this is not true for other coefficients. There is an analogue of the Wu class for odd primes though, see this comment. Regarding a reference, I have some notes on Steenrod Squares and Wu classes that you may find helpful. $\endgroup$ May 23 at 20:25
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$\begingroup$ Also, what do you mean by the Pontryagin square with $\mathbb{Z}_n$ coefficients? I am familiar with the case where $n$ is a power of $2$, and when $n$ is a power of an odd prime, there is a Pontryagin $p^{\text{th}}$ power. There is a generalisation to arbitrary coefficients, but I'm not sure that's what you mean. See this page for details. $\endgroup$ May 26 at 20:46