# Bordism invariants vanishes in a lifted twisted $Pin^- \times Spin$-structure

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.

## 1 Answer

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.