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Let $M^d$ be a nontrivial bordism generator for the bordism group $$ \Omega_d^G= \mathbb{Z}_n, $$ suppose $G$ (like O, SO, Spin, etc) specify the group structure of the boridsm group. The $\mathbb{Z}_n$ is a finite group. The $M^d$ is the bordism manifold generator genearting the bordism group $\Omega_d^G= \mathbb{Z}_n$.

Suppose some integral cohomology class of $M^d$ is an integral class $$ H^k(M^d, \mathbb{Z})=\mathbb{Z}. $$

I suspect that there exists NO extended group $G'$,

  1. Such that $$ G' \to G $$ is surjective by any group extension of $N$ (so $G=G'/N$ is a quotient group of $G'$)

  2. The bordism invariant $M^d$ which is a nonzero generator of $\Omega_d^G$, but is a zero element (trivial element) in $$ \Omega_d^{G'}. $$

My question: Is it true about my statement on None existences of any extended group $G'$ satisfy all the criteria 1 and 2 above? How can we prove it rigorously (or at least based on whatever claim we can convince the statement is true?) Or at least is there some argument for my statement to be true, or a counter example for my statement to be false?

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Your suspicion is correct: such a $G'$ cannot exist. The idea is that a $G'$-structure determines a $G$-structure, so if $M = \partial W$ as $G'$-manifolds, then $M = \partial W$ as $G$-manifolds.


In a bit more detail, suppose $\mathfrak U = \{(U, \phi_U\colon\mathbb R^n\to U)\}$ is an atlas for $M$ such that $TM|_U\cong U\times\mathbb R^n$ for all charts $U$. If $U$ and $V$ are intersecting charts, there are transition functions $g_{UV} := d\phi_V^{-1}\circ d\phi_U$ from the description of $TM$ in $U$-coordinates to the description in $V$-coordinates; they are invertible, so define maps $g_{UV}\colon U\cap V\to\mathrm{GL}_n(\mathbb R)$. Moreover, the transition functions satisfy a cocycle condition on triple intersections $U\cap V\cap W$:

$$ g_{WU}\circ g_{VW}\circ g_{UV} = 1.$$

Now suppose $\rho\colon G\to\mathrm{GL}_n(\mathbb R)$ is a homomorphism of topological groups. There are a few ways to define a $G$-structure on $M$; one hands-on definition is a lift of the transition functions across $\rho$. That is, a $G$-structure is data of functions $\widetilde g_{UV}\colon U\cap V\to G$ for all $U,V\in\mathfrak U$ such that $\rho\circ\widetilde g_{UV} = g_{UV}$ and the cocycle condition is satisfied: $\widetilde g_{WU}\circ\widetilde g_{VW}\circ \widetilde g_{UV} = 1$ on all triple intersections. (We also have to specify that $\widetilde g_{UU} = 1$.)

For example, a Riemannian metric determines an $\mathrm O_n$-structure; then an orientation is an $\mathrm{SO}_n$-structure, and a spin structure is a $\mathrm{Spin}_n$-structure.

The point of all this is that if $\psi\colon G'\to G$ is a homomorphism of topological groups, $\rho\circ\psi\colon G'\to\mathrm{GL}_n(\mathbb R)$ allows us to define $G'$-structures on manifolds. The key fact is that a $G'$-structure $\{\widetilde g_{UV}\}$ determines a $G$-structure $\{\psi\circ\widetilde g_{UV}\}$, and this is compatible with taking boundaries (a $G$-structure on $W$ induces a $G$-structure on $\partial W$ by restricting the transition maps to the boundary). If $[M] = 0$ in $\Omega_d^{G'}$, there is a compact $(d+1)$-dimensional $G'$-manifold $W$ such that $M = \partial W$ as $G'$-manifolds. Composing with $\psi$, we obtain $G$-structures on $M$ and $W$, and see that $M = \partial W$ as $G$-manifolds, which contradicts the initial hypothesis that $[M]\ne 0$ in $\Omega_d^G$.

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  • $\begingroup$ thanks very much +1, I will for a few days before accepting ur answer $\endgroup$ Nov 1, 2018 at 19:51
  • $\begingroup$ In your proof, I wonder where do you use this fact "some integral cohomology class of $M^d$ is an integral class $ H^k(M^d, \mathbb{Z})=\mathbb{Z}$" ??? I can accept your answer after your response on this $\endgroup$ Nov 16, 2018 at 23:03
  • $\begingroup$ Is that $[M]\ne 0$ in $\Omega_d^G$ and $[M]= 0$ in $\Omega_d^{G'}$? $\endgroup$ Nov 16, 2018 at 23:04
  • $\begingroup$ @annieheart I didn't use the fact about cohomology, because it turned out to not be necessary: the result is true whether or not that's the case. $\endgroup$ Nov 16, 2018 at 23:17
  • $\begingroup$ I think to me this fact "$H^k(M^d, \mathbb{Z})=\mathbb{Z}$" is crucial -- I dont know why you dont need it. What is the criteria you use then? $\endgroup$ Nov 16, 2018 at 23:20

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