# Spin structure and bordism

I have some questions about bordism and spin structures on manifolds.

If you have any examples or references I would appreciate it.

1. Is there a 3-manifold $$M$$, orientable, which does not support 3 linearly independent sections in $$TM$$?

2. Is there a 4-manifold, with a non-zero signature, that is bordant (oriented) to the sphere $$\mathbb{S}^4$$?

3. Are there two spin varieties that are bordant, in the oriented sense, but are not in the spin sense?

4. Are there two 2-manifolds with $$w_1 (TM) = 0 = w_2 (M)$$ such that are not null-bordant?

Where $$w_i (X)$$ is the $$i$$-th class of Stiefel-Whitney of $$X$$.

1) As Peter points out in your comments, every closed orientable $$3$$-manifold is paralellizable. (You can use Wu's Formula (Theorem 11.4 in Milnor-Stasheff) to show all the Stiefel-Whitney numbers vanish, and then by work of Thom this is equivalent to being orientedly null-bordant since there are no Pontryagin classes.) In fact any counterexample would have to be non-compact but I don't know one.
2) No, the signature is an oriented bordism invariant, meaning that if $$M^{4k}$$ and $$N^{4k}$$ are orientedly bordant then $$\sigma(M) = \sigma(N)$$, in particular if $$M\sim_{SO} S^4$$ then $$\sigma(M)=0$$. The signature is one of the most computable ways of showing a manifold is not null-bordant, and in fact dimension $$4$$ is particularly nice because $$\sigma\colon \Omega^{SO}_4 \to \mathbb{Z}$$ is an isomorphism. Be careful however because there are many manifolds with vanishing signature which are not null-bordant, such as $$(\mathbb{C}P^2 \times \mathbb{C}P^2) \# \overline{\mathbb{C}P^4}$$.
3) Yes, it's known that the spin bordism group $$\Omega^{spin}_1$$ is $$\mathbb{Z}/2$$, where the two classes are represented by spin structures on $$S^1$$. $$S^1$$ is orientedly bordant to itself via $$S^1\times I$$ but there is no spin bordism between $$(S^1,P)$$ and $$(S^1, P')$$ where $$P$$ and $$P'$$ are the two different spin structures. I don't know if there is a proof of this result which is both clean and elementary, but I wrote something up as an asnwer to this question.
4) No, every orientable surface is null-bordant, by the classification of surfaces, i.e. $$w_1(TM) = 0$$ means that $$M$$ is null-bordant.