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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. )

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up vote 2 down vote accepted

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 $1$.

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Thanks; where can I have a look at the list of $H^i(K(\mathbb{Z}/2,1);\mathbb{Z}/4)$ and $H^i(K(\mathbb{Z},1);\mathbb{Z}/4)$? – Yuji Tachikawa Aug 21 '13 at 12:16
Also, I'm a bit confused. Take $Y=\mathbb{RP}^3$, and consider a flat $SO(3)$ bundle over $Y\times S^1$, such that the holonomy $a$ around the generator of $\pi_1(\mathbb{RP}^3)$ and the holonomy $b$ around $S^1$ are given by $a=diag(-1,-1,1)$ and $b=diag(-1,1,-1)$ respectively. This bundle has $p_1=0$, and $w_2=\alpha^2+\alpha x$ where $\alpha$ is the generator of $H^1(\mathbb{RP}^3,\mathbb{Z}/2)$. Then $\mathfrak{P}(w_2)=\alpha^3 x$ according to your formula, but there's a general formula saying $p_1=\mathfrak{P}(w_2)$ mod 4, i.i.r.c. Where did I make a mistake? – Yuji Tachikawa Aug 21 '13 at 12:25
You probably didn't, and I probably did. Let me think about it. – Oscar Randal-Williams Aug 21 '13 at 12:56
I had been led astray by a strange formula. I think its right now (and fits with your example). – Oscar Randal-Williams Aug 21 '13 at 14:55
Thanks, but $\mathfrak{P}(n\otimes x)=2n^3$ doesn't seem to satisfy $\mathfrak{P}((n+n')\otimes x)=\mathfrak{P}(n\otimes x)+\mathfrak{P}(n'\otimes x)$ as I don't see $n^2 n'+ n' n^2$ to vanish in general... – Yuji Tachikawa Aug 21 '13 at 15:15

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