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 – annie heart Oct 14 at 0:06
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    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? – Tyler Lawson Oct 15 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)$. – annie heart Oct 15 at 15:37
  • Then the full $P(x)=x∪x+x∪_12Sq^1x$ is also a 4-cocycle. – annie heart Oct 15 at 15:38
  • Other things I am not so sure -- so the comments answers from you are encouraged very much. – annie heart Oct 15 at 15:39
up vote 5 down vote accepted
+50

(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 – annie heart Oct 17 at 17:32
  • I think you mean it is $(1/2) P(x) $? – annie heart Oct 17 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? – annie heart Oct 17 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). – Johannes Nordström Oct 17 at 19:05
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    @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. – Mike Miller Oct 18 at 3:29

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