# Pontryagin square and $\frac{1}{2}(\mathcal{P}(x) -x^2) =x \cup_1 Sq^1 x$

The Pontryagin square, maps $$x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$$ to $$\mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$$. Precisely, $$\mathcal{P}(x)= x \cup x+ x \cup_1 2 Sq^1 x.$$ The $$\cup_1$$ is a higher cup product. The $$Sq^1 x= x \cup_1 x$$. It shall be true that $$\mathcal{P}(x) \mod 2= x \cup x.$$

Question 1: If $$x \cup x =0 \mod 2$$, and if $$x \cup_1 Sq^1 x =0 \mod 2$$, is it true that $$\mathcal{P}(x) =0 \mod 4?$$

• If not, please provide some counter examples.

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Question 2: $$\mathcal{P}(x)$$ is a well-defined invariant for the cobordism $$\Omega^4_{SO}(B^2 \mathbb{Z}_2)=\mathbb{Z}_4$$. Is $$\frac{1}{2}(\mathcal{P}(x) -x^2) \mod 2 = x \cup_1 Sq^1 x \mod 2$$ a well-defined invariant of the cobordism $$\Omega^4_{Spin}(B^2 \mathbb{Z}_2)=\mathbb{Z}_2$$?

Question 3: Please provide the correct way to write the cobordism generator of $$\Omega^4_{Spin}(B^2 \mathbb{Z}_2)=\mathbb{Z}_2.$$

• I meant Pontryagin square, maps $x \in H^2({B}^2\mathbb{Z}_2,\mathbb{Z}_2)$ to $\mathcal{P}(x) \in H^4({B}^2\mathbb{Z}_2,\mathbb{Z}_4)$. Typo fixed Oct 14 '18 at 0:06
• I'd like to get a little clarification because usually the cup-1 is only defined on the cochain level, and doesn't give an operation on cohomology groups, e.g. $x \cup_1 Sq^1 x$ usually doesn't have boundary zero. Are you asking about the cohomology level or the cochain level? Oct 15 '18 at 8:30
• $x$ should be a cocycle defined above. It can be written as on a 2-simplex as the identity map $x∈H^2(B^2ℤ_2,ℤ_2)$. Oct 15 '18 at 15:37
• Then the full $P(x)=x∪x+x∪_12Sq^1x$ is also a 4-cocycle. Oct 15 '18 at 15:38
• Other things I am not so sure -- so the comments answers from you are encouraged very much. Oct 15 '18 at 15:39

(1) No, not for any reasonable interpretation of your condition "$$x \cup_1 Sq^1 x = 0$$". Consider $$M= S^2 \times S^2$$, let $$y \in H^2(M; \mathbb{Z})$$ be the sum of the two obvious generators, and let $$x \in H^2(M; \mathbb{Z}_2)$$ be the mod 2 reduction of $$y$$. Then $$P(x)$$ is the mod 4 reduction of $$y^2 = 2$$, but certainly $$x^2 = 0$$ mod 2 and $$Sq^1 x = 0$$.

(2) No, I don't think either side of your equation is a well-defined element of $$H^4(M; \mathbb{Z}_2)$$ in general. Rather, Wu's theorem implies that if $$M$$ a closed spin 4-manifold and $$x \in H^2(M;\mathbb{Z}_2)$$ then $$x^2 = Sq^2x = w_2 x = 0$$ (mod 2), so since the mod 2 reduction of $$P(x)$$ equals $$x^2$$ we get that $$P(x) \in 2\mathbb{Z}_4$$. Hence $$\frac12 P(x)$$ can be interpreted as a well-defined homomorphism $$\Omega_4^{Spin}(B^2\mathbb{Z}_2) \to \mathbb{Z}_2$$.

(3) Taking $$M = S^2 \times S^2$$ and $$x$$ as in (1) shows that $$\frac12 P(x)$$ is non-trivial.

• +1 Thanks, this is useful, but I like to have (3) written as precise 4d topological invariant term like $P(x)$ as for cobordism group generators. Thanks! What you wrote is for the bordism group generators, which is good but not enough Oct 17 '18 at 17:32
• I think you mean it is $(1/2) P(x)$? Oct 17 '18 at 17:42
• When you wrote " Wu's theorem implies that if $M$ a closed spin 4-manifold and $x \in H^2(M;\mathbb{Z}_2)$ then $x^2 = Sq^2x = w_2 x = 0$," did you mean that $$x^2 = Sq^2x = w_2 x = 0 \mod 2$$ or mod 4? Oct 17 '18 at 18:34
• I mean $x^2 = 0$ mod 2, so hence $P(x)$ is an even element of $\mathbb{Z}_4$ (I've made a clarifying edit). Oct 17 '18 at 19:05
• @annieheart In the following comment, I hope. Wu constructs, for a closed manifold $M$, characteristic classes $\nu_i \in H^i(M;\Bbb Z/2)$. These satisfy two properties: 1) for $x \in H^{n-i}(M;\Bbb Z/2)$, we have $\nu_i \cup x = \text{Sq}^i x$; 2) $w_k(M) = \sum_{i=0}^{\lfloor k/2 \rfloor} \text{Sq}^i \nu_{k-i}$. Two axioms of Steenrod squares are that the operation $\text{Sq}^0 = 1$ and if $x \in H^k(M;\Bbb Z/2)$, then $\text{Sq}^k x = x^2$. Now $M$ is spin, so $w_1 = w_2 = 0$. Then (2) immediately implies that $\nu_1 = \nu_2 = 0$. Now $0 = \nu_2 \cup x = \text{Sq}^2 x = x^2$ as desired.
– mme
Oct 18 '18 at 3:29