# $W$ spin implies $\partial W$ spin

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 $$M$$.

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.