Let $Y$ be an oriented manifold of dimension three, and let $X=Y\times S^1$. We have


$$H^2(X,\mathbb{Z}\_2)=H^2(Y,\mathbb{Z}\_2)\oplus H^1(Y,\mathbb{Z}\_2).$$


Pick an element $m\oplus n\in H^2(Y,\mathbb{Z}\_2)\oplus H^1(Y,\mathbb{Z}\_2)$, and consider $\mathfrak{P}(m\oplus n)$, where $\mathfrak{P}$ is the Pontryagin square operation. When evaluated on the fundamental class $[X]$, it is an integer modulo 4, and I believe it is even 


$$[X]\frown\mathfrak{P}(m\oplus n) =2k\_Y(m,n)$$


where $k\_Y(m,n)$ is defined modulo 2. I'd like to know more concrete explicit expression of $k\_Y$. 


(I think that it's of the form


$$k\_Y(m,n)=([Y]\frown(m\smile n)) + \mathfrak{Q}\_Y(n)$$


where $\mathfrak{Q}(n)$ is a quadratic form on $H^1(Y,\mathbb{Z}\_2)$. If it's true, is there a more explicit expression for $\mathfrak{Q}\_Y(n)$? The only thing I can think of is $[Y]\frown(n\smile\beta(n))$, where $\beta$ is the Bockstein. )



Pontryagin square on $Y\times S^1$ where $Y$ is three-dimensional
I think $[X] \frown \mathfrak{P}(m \oplus n) = 2\langle m \cdot n + n^3, [Y] \rangle$.


The class $m \oplus n$ is better thought of as $m \otimes 1 + n \otimes x$ under the Kunneth decomposition, where $x \in H^1(S^1;\mathbb{Z}/2)$ is the nontrivial element. Then the quadratic property of $\mathfrak{P}$ and naturality gives
$$\mathfrak{P}(m \otimes 1 + n \otimes x) = \mathfrak{P}(m) \otimes 1 + \mathfrak{P}(n \otimes x) + 2(m \smile n \otimes x).$$


Firstly, $\mathfrak{P}(m) = 0$ as $Y$ is 3-dimensional. 


Secondly, the fact that the suspension of the Pontrjagin square is the Postnikov square (and that the Postnikov square is not universally trivial, which bizarrely I can't find a reference for) means that
$$\mathfrak{P}(n \otimes x) = \bar{\mathfrak{P}}(n) = 2 n^3.$$


Remark: In an earlier version of this answer I had consulted



```
Nakaoka, Minoru
Note on cohomological operations. J. Inst. Polytech.
Osaka City Univ. Ser. A. Math. 4, (1953). 51–58.

```

which has the formula
$$\mathfrak{P}(n \otimes x) = \mathfrak{P}(n) \otimes \mathfrak{P}(x) + \bar{\mathfrak{P}}(n) \otimes \beta(Sq\_2(x)) + \beta(Sq\_2(n)) \otimes \bar{\mathfrak{P}}(n)$$
where $\bar{\mathfrak{P}}(-)$ is the Postnikov square (i.e. the operation given on cochains by $u \mapsto u \cup \delta u$), and $\beta$ is the Bockstein to $\mathbb{Z}/4$-cohomology. Each of $\mathfrak{P}(x)$ and $\bar{\mathfrak{P}}(x)$ must be trivial by degree reasons. If one interprets $Sq\_2(x)$ literally it also ought to be zero, but this is apparently wrong and it ought to be interpreted as $1$.


