Why does every 7-manifold bound an 8-manifold?

Solution 1:

Milnor is referring to oriented bordism there, not un-oriented bordism. Not every 7-manifold bounds an 8-manifold as you've observed from Thom's computation of the unoriented bordism ring, being a polynomial ring and having generators in dimension 2 and 5 (and many other dimensions).

But I believe it goes back to Rene Thom, that the bordism group of oriented 7-manifolds is trivial.

Solution 2:

EDIT: The first paragraph is nonsense... but I'm keeping this here since the comments might be helpful to people looking at this question.

Looks like I figured out the answer to my own question! In the shower just now, I came up with a 7-manifold not bounded by an 8-manifold: $\mathbb{RP}^2 \times S\mathbb{RP}^4$ (Edit: this doesn't quite work yet; don't trust things you find out about in the shower. I just need a 5-manifold with odd Euler characteristic... any ideas?). Indeed, notice that its Euler characteristic is the product of the Euler characteristics of the factors, which are both odd. Boundaries of manifolds have even Euler characteristic, therefore this one is not a boundary, and it is clearly 7-dimensional.

So my confusion stemmed from the fact that the claim `every 7-manifold is a boundary' is false! After a more careful skimming of Milnor's paper, it looks like, even though he uses the unqualified phrase "every 7-manifold is a boundary" he really only uses that "every orientable 7-manifold is a boundary of an orientable 8-manifold."

I'm not entirely sure why this is true, but some evidence comes from the fact that if we invert $2$ in the oriented cobordism ring, then it becomes isomorphic to a polynomial algebra over $\mathbb{Z}[1/2]$ with generators in every 4th dimension... so at least we know that there's nothing appearing in dimension 7 except possibly some 2-torsion. Anyone know why there's no 2-torsion there?

Solution 3:

Because all its Stiefel-Whitney numbers are all zero.