It can be found that there are the following [bordism group](https://en.wikipedia.org/wiki/Cobordism) $\Omega\_0^{G}$ at $d=0$ and 1 dimensions by requiring $G$ structure for $d$-manifolds:


$$
\Omega\_0^{SO} = \mathbb{Z} , \quad \Omega\_1^{SO} = 0.
$$
$$
\Omega\_0^{Spin} = \mathbb{Z} , \quad \Omega\_1^{Spin} = \mathbb{Z}/2 .
$$
$$
\Omega\_0^{O} = \mathbb{Z}/2 , \quad \Omega\_1^{O} = 0.
$$


The group classification as the cobordism class is obtained by comparing any two sets of $d$-manifolds $M\_1$ and $M\_2$ and by seeking whether there are $d+1$-manifolds $\Sigma$ making $M\_1$ and $M\_2$ as $\Sigma$'s boundary. If yes, then $M\_1$ and $M\_2$ are in the same cobordism class or correspond to the same group element in the bordism group. All $M\_1, M\_2, \Sigma$ should have equipped with $G$ structure.


So generally we look at:
$$
\Omega\_0^{G} = ? , \quad \Omega\_1^{G} = ?.
$$


1. The manifold generates $n \in \mathbb{Z}$ class for $d=0$ in $\Omega\_0^{SO}=\Omega\_0^{Spin} = \mathbb{Z}$ must be an integer $\mathbb{Z}$ number of $d=0$ points (as manifold generators). Is this true that we cannot regard the $n\_1,n\_2 \in \mathbb{Z}$ points as the boundary of 1d lines, if $n\_1 \neq n\_2$? But only if $n\_1 = n\_2=n$, we can use $n$ 1d lines to connect the two sets of $n$ points as boundaries of $n$ 1d lines?
2. Since $\Omega\_0^{O} = \mathbb{Z}/2$, this means two $d=0$ points can be cobordant to nothing via an unoriented $d=1$ line segment. I may conjecture that:



> 
> For any $G$ structure, its bordism group must be either $\Omega\_0^{G} =\mathbb{Z}$ or $\Omega\_0^{G} =\mathbb{Z}\_2$. Namely, it cannot be $\Omega\_0^{G} =\mathbb{Z}/N$ for $N\neq 2$ or
> or $\Omega\_0^{G} =0$ for any $G$. Is this correct?
> 
> 
> 


3. I may conjecture that:



> 
> there exists no $G$ structure with its bordism group $\Omega\_1^{G} =\mathbb{Z}$. Namely, we only have either
> $\Omega\_1^{G} =\mathbb{Z}/N$
> or
> $\Omega\_1^{G} =0$ for any $G$. Is this correct?
> (Is this true that we must have $\Omega\_1^{G} =\mathbb{Z}/N$ with $N =1$ as 0 and $N =2$?)
> 
> 
> 


If so:
This means that for some finite $N$, the $N$ sets of $1$-manifolds ($S^1$) must be regarded as some bundary of a $G$ structure $2$-manifold. Thus, the $N$ sets of $S^1$ must be cobordant to nothing via this $G$ structure $2$-manifold. Can we prove this? Is this true that $N\leq 2$?



Prove or disprove that there exists no $G$ structure with its bordism group $\Omega_1^{G} =\mathbb{Z}/N$ for $N>2$
$\newcommand{\Z}{\mathbb Z}$Part 2 is correct: for any $G$-structure, $\Omega\_0^G$ is either isomorphic to $\Z$ or
$\Z/2$. Part 3 isn't correct, and I'll give a counterexample.


First, part 2: let $\rho\colon G\to O$ be a $G$-structure. $\Omega\_0^G$ is isomorphic to $\pi\_0(MG)$, where $MG$ is
the Thom spectrum of the pullback of the tautological bundle $V\to BO$ across $B\rho\colon BG\to BO$. Since $MG$ is
connective, the Hurewicz map $\pi\_0(MG)\to \tilde H\_0(MG)$ is an isomorphism. The Thom isomorphism implies
$\tilde H\_0(MG)\cong H\_0(BG; \Z\_{\sigma})$, where $\Z\_\sigma$ is the orientation local system for $B\rho^\*V$.


For a general space $X$, the zeroth twisted homology with coefficients in a $\Z[\pi\_1(X)]$-module $V$ is the
coinvariants $V/\{v - \gamma\cdot v\mid \gamma\in \pi\_1(X)\}$ (See
[Davis-Kirk](https://jfdmath.sitehost.iu.edu/teaching/m623/book.pdf), Proposition 5.14.) In our case, only two
things can happen.


* If $B\rho^\*V$ is orientable, $\Z\_\sigma$ is a trivial local system, and we obtain $H\_0(BG;\Z\_\sigma) \cong \Z/0 =
 \Z$.
* If $B\rho^\*V$ is unorientable, $\Z\_\sigma$ is a nontrivial local system. This means there is an element of
$\pi\_1(BG)$ which sends $1\mapsto -1$, so we're quotienting by a subgroup containing $2\Z$. However, as
$\pi\_1(BG)$ is a group, its action cannot send $1$ to $0$, so we're quotienting out by exactly $2\Z$, and we
obtain $H\_0(BG;\Z\_\sigma)\cong \Z/2$.


For part 3: for a $G'$-structure with $\Omega\_1^{G'}\cong\Z$, consider $G' = \mathrm{SO}\times\Z$. This encodes the notion
of oriented manifolds with a principal $\Z$-bundle, or equivalently oriented manifolds with a map to $S^1$. The
Atiyah-Hirzebruch spectral sequence
$$E\_2^{p,q} = H\_p(S^1; \Omega\_q^{\mathrm{SO}}) \Longrightarrow \Omega\_{p+q}^{\mathrm{SO}}(S^1) = \Omega\_{p+q}^{G'}$$ collapses in
degrees $3$ and below, producing an isomorphism $\Omega\_1^{G'} \cong H\_1(S^1)\cong\Z$. One can take as a generator
the $1$-manifold $S^1$ with principal $\Z$-bundle $\mathbb R\to \mathbb R/\Z\cong S^1$.


