# Any cobordism invariant made of “characteristic classes”, on unorientable manifolds, must be a mod 2 class?

For any cobordism invariant (or simply bordism invariant) quantity $$\omega$$ that satisfy the conditions:

• $$\omega$$ can be fully decomposed from the cup product of characteristic classes (such as Stiefel Whitney class, Chern class, etc.).

• $$\omega$$ can be defined on an unorientable smooth manifold.

then question:

Could we show that this $$\omega$$ has to be $$\mathbb{Z}/2\mathbb{Z} : = \mathbb{Z}_2$$ valued (a mod 2 class)? Is this true or false?

From the recent post: Cobordism invariants: topological v.s. geometric, I recall that:

(1). The 2d $$Pin^-$$ bordism invariant (defined on unorientable manifolds) from $$\Omega_2^{Pin^-}=\mathbb{Z}_8$$, where any 2-manifold $$M$$ always admits a $$Pin^-$$ structure. The $$\mathbb{Z}_8$$ is a Arf-Brown-Kervaire (ABK) invariant. $$Pin^-$$ structures are in one-to-one correspondence with quadratic enhancements $$q: H^1(M,\mathbb{Z}_2)\to\mathbb{Z}_4$$ such that $$x,y \in \mathbb{Z}_2$$, we have $$q(x+y)-q(x)-q(y)=2\int_M x\cup y\mod4.$$ In particular, $$q(x)=\int_M x\cup x\mod2.$$ The $$x,y$$ counts as a cohomology class, but the ABK invariant seems not be decomposed from decomposed from a cup product of characteristic classes only? But does this count as a counter example of the above statement?

(2) The 4d $$\Omega_4^{Pin^+}(B^2 \mathbb{Z}_2)=\mathbb{Z}_4 \times \mathbb{Z}_{16}$$ contains an $$\eta$$ invariant of $$\mathbb{Z}_{16}$$, and a quadratic refinement of $$\mathbb{Z}_4$$ (from $$B \in H^2(M,\mathbb{Z}_2)$$ to $$H^4(M,\mathbb{Z}_4)$$. (It is probably similar to Pontryagin square $$P(B)$$ of $$B \in H^2(M,\mathbb{Z}_2)$$; but Pontryagin square $$P(B)$$ is defined on an orientable $$SO$$ structure, here I am looking at the unorientable $$Pin^+$$ structure. But does this count as a counter-example of the above statement?

If the above statement is false, can one give counter examples?