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I'm glancing over Milnor's paper on exotic 7-spheres, and one of the first few lines says, `every closed 7-manifold $M^7$ is the boundary of an 8-manifold $B^8$.

Here's what I don't understand: The unoriented cobordism ring is isomorphic to a polynomial ring over $\mathbb{Z}_2$ with a generator in every degree except $2^m -1$ for any $m$. In particular, there are generators $x_2$ and $x_5$ in degrees $2$ and $5$, respectively. So: why does their product not represent a nontrivial element of the group of cobordism classes of 7-manifolds?

Apologies if this is a silly question... I'm new to all of this.

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Is your point that you think that a product of two non-trivial classes must be non-trivial? That's not true, as your reasoning shows. –  Rasmus Jan 19 '11 at 8:36
If you don't get a good answer within a couple of days, feel free to post this to MathOverflow.This is an appropriate level of question for that forum. –  Grumpy Parsnip Jan 19 '11 at 14:09
@Rasmus: No I don't think that it is always the case that the product of nontrivial classes is nontrivial. But in this case we're in a polynomial ring, which is a free algebra... so the product of two `indterminates' shouldn't be zero, right? Otherwise we would be imposing some extraneous relation... –  Dylan Wilson Jan 19 '11 at 15:05
Apologies for my earlier ignorant response. I didn't appreciate the intricate nature of the question. I would also like to second Jim's suggestion: it might be profitable to post this at MathOverflow. –  Glen Wheeler Jan 19 '11 at 16:20
Since the question has been resurrected, I'll add a small tidbit (4 years late!) Practically, Milnor only considers $S^3$ bundles over $S^4$ with structure group $SO(4)$. The linear action of $SO(4)$ on $S^3$ extends to a disc, and so each of his manifolds is the boundary of the associated disc bundle $D^4$ over $S^4$. In particular, you don't need Thom's computation of the oriented bordism ring for Milnor's paper to work –  Jason DeVito Apr 1 at 5:11

3 Answers 3

up vote 13 down vote accepted

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.

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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?

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This answer needs a slight tweaking and then it will be correct... I'll think about finding manifolds with odd Euler characteristic, and then check back here later today. –  Dylan Wilson Jan 19 '11 at 15:40
You probably know this, but just to be sure: a compact orientable manifold of odd dimension has vanishing Euler characteristic, by Poincare duality. –  Matt E Jan 19 '11 at 16:00
And so every compact 5-manifold has Euler characteristic zero, since non-orientable manifolds are finitely covered by orientable ones. This applies to all odd-dimensional manifolds. –  Ryan Budney Jan 19 '11 at 16:09
Ah, whoops. Well then can you think of an explicit example of a non-orientable 7-manifold that is not a boundary? –  Dylan Wilson Jan 19 '11 at 17:22
Take the Cartesian product of representatives of the generators $x_2$ and $x_5$ in the unoriented cobordism ring! We know $x_2=[\mathbb{R}P^2]$, but this only works for $x_{2i}$. Apparently Dold gives representatives for $x_{2i+1}$ (for $2i+1\not= 2^j-1$, of course!) in "Erzeugende der Thomschen Algebra $N$", 1956. –  Aaron Mazel-Gee Jan 20 '11 at 5:39

Because all its Stiefel-Whitney numbers are all zero.

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This is false in general, as other answers show. –  Mike Miller Apr 2 at 20:52

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