Pontryagin square on $Y\times S^1$ where $Y$ is three-dimensional

Question

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. )

Answer 1

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 $1$.