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For $\mathbb{Z}_2$ cohomology classes, we have a very useful Wu formula: In $d$-dimensional manifold and for a $n$-cocycle in $x_n \in H^n(M^d; \mathbb{Z}_2)$, we have $Sq^{d-n}(x_n)=u_{d-n}\cup x_n$, where $u_m$ is the $m^{th}$ Wu class, and $Sq^k$ is the Steenrod square.

I wonder if there is a similar formula for integral cohomology classes? For example, if there is a cohomological operation $P^4$, such that $P^4(y_{d-4}) = p_1 \cup y_{d-4}$ where $p_1$ is the first Pontryagin class and $y_n \in H^n(M^d; \mathbb{Z})$?

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    $\begingroup$ The issue seems to involve integral cohomology operations. But you might be able to get something to work at primes $p\ne 2$ for oriented manifolds: there are Steenrod power operations $P^i$ on $\mathbb F_p$-cohomology, and they're linear, so on an oriented $d$-manifold $M$, Poincaré duality guarantees each $P^i\colon H^{d-2i(p+1)}(M;\mathbb F_p)\to H^d(M;\mathbb F_p)$ is represented by cup product with some cohomology class $u_{2i(p+1)}$. $\endgroup$ – Arun Debray Apr 27 '17 at 2:28
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I do not think such a cohomology operation can exist, as it would descend to a rational cohomology class in $H^{n+4}(K(\mathbb Z,n);\mathbb Q)$. But for any odd $n$ and also for all even $n >4$, this cohomology group is zero! This in turn gives that $P$ is trivial after rationalizing, so the formula you want could never be true.

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The Wu classes are, or at least can be, defined by $Sq^{d-n}(x_n) = u_{d-n}\cup x_n$. The interesting part is not that the Wu classes exist, it is that they are related to Stiefel--Whitney classes, which is not at all a priori. Indeed, this implies that SW classes are homotopy-invariant objects (since $Sq^i$ are), which is not at all god-given. Most characteristic classes are not homotopy-invariant, and so will not participate in a Wu-type formula in terms of cohomology operations.

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