# Arf-Brown-Kervaire invariant and a surjective map $G \to Pin^-$

We know that the Arf-Brown-Kervaire (abk) invariant is a bordism invariant of $$\Omega_2^{Pin^-}(pt)=\mathbb{Z}/(8\mathbb{Z}),$$ where the $$\mathbb{Z}/(8\mathbb{Z})$$ is generated by a 2-manifold $$M^2$$ generator, such as $$\mathbb{RP}^2,$$ or the invariant $$\exp( 2 \pi i (k/8) \int_{M^2} \; \text{(abk)}),$$ with $$k \in \mathbb Z_8$$

• My question: Does there exist any surjective $$G \to Pin^-,$$ such that it is a surjective map $$G \to Pin^-$$, and the Arf-Brown-Kervaire invariant $$\exp( 2 \pi i (k/8) \int_{M^2} \; \text{(abk)})$$ at the even integer $$k$$ of $$\Omega_2^{Pin^-}(pt)$$ becomes a trivial bordism invariant, under pullback, evaluated at $$\Omega_2^{G}?$$ Namely the corresponding invariant for $$\Omega_2^{G}$$ on $$\mathbb{RP}^2$$ (at the even integer $$k$$) becomes 1. (i.e. a trivial group element.)

p.s. It is obviously possible to find $$G$$, if the $$G \to Pin^-$$ is an injective instead of surjective map. Say, we can take $$G=Spin \to Pin^-,$$ then $$\Omega_2^{Spin}(pt)=\mathbb{Z}/(2\mathbb{Z})$$ is obviously trivial, when we map back from any even integer $$k$$ in $$\Omega_2^{Pin^-}(pt)=\mathbb{Z}/(8\mathbb{Z})$$ to $$k \mod 2$$ as $$0$$ in $$\Omega_2^{Spin}(pt)=\mathbb{Z}/(2\mathbb{Z}).$$

• The notation $\Omega_3^G$ is almost always used to denote cobordism classes of 3-manifolds with $G$-structure, rather than 2-manifolds (so it's more common to write $\Omega_2^{\mathrm{Pin}^-}$, $\Omega_2^{\mathrm{Spin}}$, etc.). – Arun Debray Nov 20 '18 at 2:03
• What restrictions do you place on the G? For example let $G = Spin \times_{Pin^-} P(Pin^-)$ where $P(H)$ is the free path space on $H$ with pointwise multiplication. The fiber product uses restriction to one end of a path. Restriction to the other end of the path gives a surjective homomorphism from $G$ to $Pin^-$. But $G \simeq Spin$ and so $G$-bordism is the same as $Spin$-bordism. – Chris Schommer-Pries Nov 20 '18 at 20:29