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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 $$\Omega_2^{Spin}(pt)=\mathbb{Z}/(2\mathbb{Z}).$$

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    $\begingroup$ 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.). $\endgroup$ – Arun Debray Nov 20 '18 at 2:03
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    $\begingroup$ 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. $\endgroup$ – Chris Schommer-Pries Nov 20 '18 at 20:29

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