Let $M$ be a compact orientable manifold with the first two Stieffel-Whitney numbers equal to zero (this is my definition of SPIN manifold). Let $B$ be the boundary of $M$; I want to show that $B$ is spin (in the sense that I've already stated).
I know that there is a theorem of Pontryagin that holds in a more general context and gives a stronger result (if $M$ is a compact manifold with boundary, than the Stieffel-Whitney numbers $w_i$ with $i \geq 1$ of $\partial M$ vanish) but I was wandering if in this case there exists a simpler proof of this fact.
I know for example that if $TM$ is orientable (as bundle, since as manifold is always orientable) than $M$ is spin iff the restriction of the tangent bundle to the 2-skeleton of $M$ is trivial (given a cellularization of M as CW-complex). The problem is that I don't know how to use this, since I don't know how to relate the $2$-skeleton of $\partial M$ to the $2$-skeleton of $M$.
Best Answer
There are some confusions here I should clear up before answering.
1) This is a definition of spinnable manifold, not of a spin manifold. (Similar to the difference between oriented manifold and orientable manifold.) When spin structures exist, there are $H^1(W;\Bbb Z/2)$-many of them, and a spin manifold requires one such choice.
2) If $M$ is the boundary of another manifold $W$, then it is not necessarily true that the Stiefel-Whitney classes $w_i(M) \in H^i(M;\Bbb Z/2)$ vanish. For instance, if $M$ has nontrivial $w_i(M)$ for $i < \dim M$, then so does $M \# M$, but the latter is always null-bordant. What instead is the case is that the Stiefel-Whitney numbers vanish. These are the results of all products of $w_i(M)$ that land in $H^{\dim M}(M; \Bbb Z/2) \cong \Bbb Z/2$; they are labeled by partitions of $\dim M$.
3) A vector bundle $E$ of rank at least 3 is spinnable if and only if it is trivializable over the 2-skeleton. This is not true for bundles of rank 2: there are spinnable bundles which are not trivial over the 2-skeleton. Think $TS^2$.
As for your actual question, the point is that we have a natural isomorphism $T(\partial W) \oplus \Bbb R \cong TW\big|_{\partial W}$. The easy claim is that if $W$ is spinnable, then $\partial W$ is spinnable: naturality of Stiefel-Whitney classes implies that $j^*(w_i(W)) = w_i(\partial W)$, where $j: \partial W \to W$ is the inclusion. If you already know that $w_1(W) = w_2(W) = 0$, you hence know that as well for $\partial W$. This argument applies for any condition defined by the vanishing of some set of Stiefel-Whitney classes.
To actually pin down a specific spin structure, we need to be able to argue that a spin structure on $E \oplus \Bbb R$ induces a natural spin structure on $E$. (This is true for orientations!) This amounts to the inverse claim that the map sending spin structures on $E$ to spin structures on $E \oplus \Bbb R$ (via the natural map $\text{Spin}(n) \to \text{Spin}(n+1)$) is a bijection; and this may be verified using that spin structures are affine over the group of isomorphism classes of real line bundles, aka $H^1(W;\Bbb Z/2)$.
This argument doesn't work for arbitrary sorts of structure on the tangent bundle. For instance, one that famously doesn't work is "$W$ parallelizable implies $\partial W$ parallelizable". Every disc $D^n$ is parallelizable, but the only spheres which are are $S^0, S^1, S^3$, and $S^7$. In the argument above, the way that showed up was in our descriptions of spin structures as affine over $H^1(W;\Bbb Z/2)$, which is true for spin structures on rank $n$ bundles for all $n$; trivializations are affine over $[W, SO(n)]$, which depends on $n$ until $n$ is quite large.