Spin structure and bordism

Question

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 $.

Answer 1

A good reference for a lot of this is "Characteristic Classes" by Milnor and Stasheff. I can definitely answer 2, 3, 4 and partially answer 1.

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.