It looks to me that the bordism group
$$\Omega\_3^{SO} (B(O(2) \times SO(3))) \tag{1}$$ 
(whose Pontryagin dual for the manifold generator) contains at least a nontrivial invariant: 
$$
w\_1(O(2))\big(w\_1(O(2))^2 +w\_2(SO(3))\big). \tag{2}
$$



> 
> Question: Is it true that if we lift this invariant respect to a new bordism group
>  $$
> \Omega^{SO}\_3 (B((Pin^-(2) \times Spin(3))/ \mathbb{Z}\_2)), \tag{3}
> $$
>  the invariant eq.(2) becomes zero in bordism group (2)?
> 
> 
> 




---


My trial/attempt: The lifting is simply that the $O(2)$ bundle in eq(1) is lifted to $Pin^-(2)$ bundle in eq(3), and the $SO(3)$ bundle in eq(1) is lifted to $Spin(3)$ bundle in eq(3). There is an overall constraint:
$$
w\_2(O(2))+w\_1(O(2))^2=w\_2(SO(3)). \tag{4}
$$


It also looks that, after I look it up, the $w\_3(O(2))=w\_1(O(2))w\_2(O(2))+\frac{d w\_2(O(2))}{2}=0$. But I am not sure that $\frac{d w\_2(O(2))}{2}$ is well-defined. By using the fact that the O(2) bundle can be lifted to $Pin^-(2)$ bundle, we shall have this constraint $w\_2(O(2))+w\_1(O(2))^2=0$. But here we may need to apply eq.(4) instead.



Bordism invariants vanishes in a lifted twisted $Pin^- \times Spin$-structure
Since you have the constraint 
$$w\_1(O(2))^2+w\_2(O(2))=w\_2(SO(3)),$$
then
$$w\_1(O(2))(w\_1(O(2))^2+w\_2(SO(3)))=w\_1(O(2))w\_2(O(2))=Sq^1(w\_2(O(2)))=w\_1(TM)w\_2(O(2))=0$$
by Wu formula. ($w\_1(TM)=0$ since you are considering oriented bordism.)


$\frac{dw\_2(O(2))}{2}=Sq^1(w\_2(O(2)))$ by the definition of Bockstein homomorphism.


