If I understand correctly, I can obtain the $O$-cobordism group of
$$
\Omega^{O}\_3(BO(3))=(\mathbb{Z}/2\mathbb{Z})^4,
$$
The 3d cobordism invariants have 4 generators of mod 2 classes, are generated by
$$g^3,$$
$$g w\_2'(V\_{SO(3)}),$$
$$w\_3'(V\_{SO(3)}),$$
$$g w\_1(T)^2.$$


Denote: 


The $T$ for the spacetime tangent bundle. 


The $O(3)=\mathbb{Z}\_2 \times SO(3)$.


The $V\_{SO(3)}$ for the vector bundle of $SO(3)$.


Here $g$ is related to the $\mathbb{Z}\_2$ generator of $g=H^1(B\mathbb{Z}\_2,\mathbb{Z}\_2)$.



> 
> Questions: What are the corresponding 3-manifold generators of O-co/bordism invariants $\Omega\_{O}^3(BO(3))$, for $g^3,$ $g w\_2'(V\_{SO(3)}),$
> $w\_3'(V\_{SO(3)}),$
> $g w\_1(T)^2.$?
> 
> 
> 


Hint: I think the manifold generator for $g^3$ is $\mathbb{RP}^3$.


The manifold generator for $g w\_1(T)^2$ is $S^1 \times \mathbb{RP}^2$?



> 
> What are manifold generators of $g w\_2'(V\_{SO(3)}),$ and $w\_3'(V\_{SO(3)})$? Can each manifold generator of 4 generators of mod 2 classes be unique and distinct?
> 
> 
> 



Manifold generators of O-bordism invariants
$\newcommand{\RP}{\mathbb{RP}}\newcommand{\ang}[1]{\langle #1\rangle}\newcommand{\R}{\mathbb R}$A representative of
a class in $\Omega\_3^O(BO\_3)$ is a 3-manifold $M$ together with a principal $O\_3$-bundle $P\to M$. Principal
$O\_3$-bundles are equivalent to rank-3 real vector bundles, so I will use 3-manifolds $M$ together with rank-3
real vector bundles $E\to M$. In this setting, your invariants correspond to products of Stiefel-Whitney numbers of
$M$ and $E$, evaluated on the mod 2 fundamental class of $M$.


Specifically, I will use the more standard notation $g^3 = \ang{w\_1(E)^3, [M]}$, $gw\_2'(V\_{SO(3)}) =
\ang{w\_1(E)w\_2(E), [M]}$, $w\_3'(V\_{SO(3)}) = \ang{w\_3(E), [M]}$, and $gw\_1(T)^2 = \ang{w\_1(E)w\_1(M)^2, [M]}$. For
brevity I'll write $w\_1(E)^3$ for $\ang{w\_1(E)^3, [M]}$, etc., and let $S(M,E)$ denote the tuple $(w\_1(E)^3,
w\_1(E)w\_2(E), w\_3(E), w\_1(E)w\_1(M)^2)$.


Let $\ell\_{\RP^n}\to\RP^n$ denote the tautological line bundle (I'll also do this for $\RP^1 = S^1$); if $x\in H^1(\RP^n;\mathbb Z/2)$ denotes the
generator, then $w(\ell\_{\RP^n}) = 1+x$. Then, a generating set for
$\Omega\_3^O(BO\_3)$ is


$$\{ (\RP^3, \ell\_{\RP^3}^{\oplus 3}), (\RP^3, \ell\_{\RP^3} \oplus\underline\R^2), (S^1\times\RP^2,
\ell\_{S^1}\oplus\underline\R^2), (S^1\times\RP^2, \ell\_{S^1}\oplus\ell\_{\RP^2}\oplus\underline\R)\}. $$


Specifically, using the Whitney sum formula one can calculate that


* $S(\RP^3, \ell\_{\RP^3}^{\oplus 3}) = (1, 1, 1, 0)$,
* $S(\RP^3, \ell\_{\RP^3}\oplus\underline\R^2) = (1, 0, 0, 0)$,
* $S(S^1\times\RP^2, \ell\_{S^1}\oplus\underline\R^2) = (0, 0, 0, 1)$, and
* $S(S^1\times\RP^2, \ell\_{S^1}\oplus\ell\_{\RP^2}\oplus\underline\R) = (1, 1, 0, 1)$.


These four vectors are linearly independent in $\mathbb F\_2^4$, so these manifolds generate the bordism group.


If you want manifolds equal to $1$ on one invariant and $0$ on the others, you can take disjoint unions of these
manifolds: for example, $(0, 1, 0, 0)$ is represented by the disjoint union of the last three generators. It may be
possible to find simpler representatives.


