# Manifold generators of O-bordism invariants

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?

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

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

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