On framed cobordism

Question

If two manifolds are h-cobordant then their homotopy groups agree. The notion of framed cobordism is supposedly a weaker notion. How much weaker than h-cobordism is it? What can be said about two framed cobordant manifolds if not that they are of same homotopy type?

Answer 1

Stupid example: The 0-dimensional manifold with two points, each with opposite framing, is framed-nullbordant (i.e. equivalent to the empty set), on the other hand, it is not homotopy equivalent to the empty set. In general taking two copies of a given manifold with opposite framings gives an example of this type.

More interesting: As a consequence of the recent resolution of the Kervaire invariant problem (except maybe dim 126), we now know that any stable framed manifold is framed cobordant to a sphere as long as the manifold has dimension different than 2, 6, 14, 30, and 62. As a result, one can quickly produce myriad examples by picking your favorite parallelizable manifold that is not homotopy equivalent to a sphere.

I should say that quoting HHR here is wayyyy overkill just to get a bunch of examples of the form you're interested in. The technique of starting with a manifold and then killing homology groups until we have a sphere is called "surgery theory" and you should Google it- it's pretty neat, and classical.

Simplest nontrivial example: A torus with its invariant framing is framed cobordant to a sphere because we can cut out one of the copies of $S^1$ and replace it with two copies of a disk (capping off the ends). This process corresponds to a framed cobordism if done properly. Alternatively you can just imagine the hole in the middle of the donut shrinking to a point and then place your hands on either side of the singularity and pop it up to inflate the beach ball.