# Every Manifold Cobordant to a Simply Connected Manifold

I am wondering if it is true that every compact, connected, oriented manifold is cobordant to a simply connected manifold.

I believe that some sort of surgery will do the trick. Roughly speaking, I want to add handles so that I can kill representative loops. However, I don't know if my surgery process builds a cobordism and it is hard to for me to see what the new boundary is. Another possibility is to build a Morse function that constructs the cobordism for free.

It is hard for me to get an intuition for what is going on, because all the compact, oriented 2-manifolds are boundaries and consequently cobordant to the empty set. CP^2 seems like the "easiest" test case, but it is already simply-connected for cellular reasons. Lens Spaces might be a nice candidate, but the 3 dimensional ones can be realized as boundaries of some disc bundle on S^2.

If possible, I'd prefer a constructive procedure, but any answer that helps elucidate the material is welcome.

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My favorite reference regarding surgeries is Milnor's book "Lectures on the h-cobordism theorem". It's hard to find, but I think there is something on gigapedia, and hopefully it's in your library. It's one of the best math books I've ever read but, unfortunately, it has a lot of typos, so be wary. (Also, some arguments are longer than would be absolutely necessary today, for example it reproves cellular homology). –  Ilya Grigoriev Feb 10 '10 at 17:52
@ Ilya, Milnor's book "Lectures on the h-cobordism theorem" can be found on Andrew Ranicki’s Homepage. But I think it is more about how to manipulate the Morse Function. –  Xiaolei Wu Apr 7 '11 at 18:41

Assume that $M^n$ has $\pi_1$ finitely generated (Edit: and n>3). Choose a generator. We will construct (using surgery) a cobordism to $M'$ which kills that generator, and by induction we can kill all of $\pi_1$. Choose an embedded loop which represents the generator, and choose a tubular neighborhood of the loop. We can view this as a (n-1)-dimensional vector bundle over $S^1$, the normal bundle. Since $M$ is oriented, this is a trivial vector bundle so we can identify this tubular neighborhood with $S^1 \times D^{n-1}$.

Now we build the cobordism. We take $M \times I$, which is a cobordism from $M$ to itself. To one end we glue $D^2 \times D^{n-1}$ along the boundary piece $S^1 \times D^{n-1}$ via its embedding into $M$. This is just attaching a handle to $M \times I$. This new manifold is a cobordism from $M$ to $M'$, where $M'$ is just $M$ where we've done surgery along the given loop.

A van Kampen theorem argument shows that we have exactly killed the given generator of $\pi_1$. Repeating this gives us a cobordism to a simply connected manifold.

Note that it is essential that our manifold was oriented. $\mathbb{RP}^2$ is a counter example in the non-oriented setting, as all simply connected 2-manifolds are null-cobordant, but $\mathbb{RP}^2$ is not.

[I was implicitly thinking high dimensions. Thanks to Tim Perutz for suggesting something was amiss when n=3]

If n=3 then this is "surgery in the middle dimension" and it is more subtle. First of all the normal bundle is an oriented 2-plane bundle over the sphere, so there are in fact $\mathbb{Z} = \pi_1(SO(2))$ many ways to trivialize the bundle (these are normal framings). Ignoring this, if you carry out the above construction, you will see that (up to homotopy) M' is the union of $M - (S^1 \times D^2)$ and $D^2 \times S^1$ along $S^1 \times S^1$. This can (and does) enlarge the fundamental group.

However a different argument works in dimensions n=1,2,3. The oriented bordism groups in those dimensions are all zero (see the Wikipedia entry on cobordism), so in fact every oriented 3-manifold is cobordant to the empty set (a simply connected manifold). The fastest way to see this is probably a direct calculation of the first few homotopy groups of the Thom spectrum MSO.

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Thanks! As I am just learning some surgery theory, it is a new vocabulary that requires some practice. Do you have references that you recommend? –  Justin Curry Feb 10 '10 at 15:16
btw, if you try to do this same thing but using framed manifolds you can't. Even in dimension 2 there is an obstruction known as the Arf invariant. This lead to a higher dimensional obstruction known as the Kervaire invariant, which has been the subject of some very exciting recent research. –  Chris Schommer-Pries Feb 10 '10 at 15:36
Chris, what happens if your generator happens to be a contractible loop in a 3-manifold? –  Tim Perutz Feb 10 '10 at 16:29
On the subject of references for surgery theory, there are a number of "goodies" at my colleague Andrew Ranicki's web page: maths.ed.ac.uk/~aar/surgery –  José Figueroa-O'Farrill Feb 10 '10 at 19:15
Maybe instead of saying that the empty set is simply-connected it is better to say that since a closed oriented 3-manifold bounds a compact 4-manifold, it is cobordant to the 3-sphere; the cobordism is obtained by removing an open disk from the interior of the 4-manifold. –  Igor Belegradek Feb 10 '10 at 19:58