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 $$g^3,$$ $$g w_2'(V_{SO(3)}),$$ $$w_3'(V_{SO(3)}),$$ $$g w_1(T)^2.$$


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?

  • $\begingroup$ Why do you write co/bordism? $\endgroup$ Oct 30, 2018 at 17:50
  • $\begingroup$ It means bordism OR cobordism $\endgroup$ Oct 30, 2018 at 17:51
  • $\begingroup$ Is there a difference? $\endgroup$ Oct 30, 2018 at 17:52
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    $\begingroup$ they are Potryagin dual. Bordism generators are the manifolds. Cobordism generators are the topological terms (characteristic classes, etc). $\endgroup$ Oct 30, 2018 at 17:58
  • $\begingroup$ I see, thanks. Somehow I always assumed that the two terms were basically synonymous. $\endgroup$ Oct 30, 2018 at 17:59

1 Answer 1


$\newcommand{\RP}{\mathbb{RP}}\newcommand{\ang}[1]{\langle #1\rangle}\newcommand{\R}{\mathbb R}$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.

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    $\begingroup$ I think the second generator should be $(\mathbb{RP}^3,l_{\mathbb{RP}^3}\oplus\underline{\mathbb{R}}^2)$ and $S(\mathbb{RP}^3,l_{\mathbb{RP}^3}\oplus\underline{\mathbb{R}}^2)=(1,0,0,0)$. $\endgroup$
    – Borromean
    Oct 31, 2018 at 8:02
  • $\begingroup$ @ZheyanWan you're absolutely right. Thanks for the correction! I'll fix it. $\endgroup$ Oct 31, 2018 at 12:36

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