# Twisted spin-bordism invariant and a possible Postnikov square from $d=2$ to $d=5$

This is a follow up more advanced question following Twisted spin bordism invariants in 5 dimensions. We follow the definitions in the earlier post.

I had discussed my computation of

$$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2},$$

My specific question is how to pin down the $$\mathbb Z_{32}$$ generator as a topological invariant? (like characteristic class or eta invariant).

My current proposal is that the $$\mathbb Z_{32}$$ whether 1/2 of the $$\mathbb Z_{32}$$ is just the Postnikov square of $$b \in H^2(B \mathbb Z_8, \mathbb Z_8)$$.

$$\text{Postnikov square from b \in H^2(B \mathbb Z_8, \mathbb Z_8) to H^5(B \mathbb Z_8, \mathbb Z_{16})}$$

Can this be proved or disproved, so falsified?

Let $$X$$ is any $$\mathbb Z_{32}$$ topological term in the dimension $$d=5$$,

then consider $$X'$$ be the pullback of $$X$$ from $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}$$ to $$\Omega_5^{Spin \times_{} \mathbb Z_{8}}$$,

we find:

$$X' \mod 2 =(a \mod 2) (b \mod 2)^2= Sq^2\big((a \mod 2) (b \mod 2)\big)$$ $$=(w_2(TM)+w_1(TM)^2)\big((a \mod 2) (b \mod 2)\big) =0,$$ the last equality is based on the Wu formula on a 5d spin manifold.

My question is $$\frac{X'}{2}=\text{Postnikov square of } b?$$ where $$a \in H^1(B \mathbb Z_8,\mathbb Z_8),\quad b \in H^2(B \mathbb Z_8, \mathbb Z_8).$$

PS. Recall, we find the Adams $$\mathcal A$$ module structure for $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{4}} =\mathbb Z_{16}$$ and $$\Omega_5^{Spin \times_{\mathbb Z_2} \mathbb Z_{8}}= \mathbb Z_{32} \times \mathbb Z_{2}$$, are provided in my earlier post Twisted spin bordism invariants in 5 dimensions .