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Cobordism is a fundamental equivalence relation on the class of compact manifolds of the same dimension, set up using the concept of the boundary of a manifold.

14 votes
1 answer
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What is known about exotic spheres up to stable diffeomorphism?

In even dimensions $n=2k$ we can define two smooth manifolds $M$ and $N$ to be stably diffeomorphic if they become diffeomorphic after the connect sum with $r$ many copies of $S^k \times S^k$ for some …
28 votes
Accepted

Nilpotence of the stable Hopf map via framed cobordism

I am probably going to conflate these framed manifolds with their classes in frame cobordism. I hope you forgive me. … that $24 \nu = 0$ and I will start there. 24 Torsion in the 3rd stable stem, via geometry The 3rd stable stem ( = the 3rd stable homotopy group of spheres = group of stably framed 3-manifolds up to cobordism
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16 votes
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Super-cobordisms

There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. I am going to ignore …
Chris Schommer-Pries's user avatar
5 votes

An orthogonal factorization system on 1-Cob?

[Note: This answer pertains to the orignal version of the question, and no longer applies given the subsequent edits]. To be a weak factorization system you need to factor morphisms as composites $ …
Chris Schommer-Pries's user avatar
19 votes

Twistings for other cohomology theories

Twisted forms exist for all multiplicative generalized cohomology theories. A nice paper which discusses a modern point of view for twists of homology, K-theory, and TMF is the following paper Twists …
Chris Schommer-Pries's user avatar
20 votes
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Every Manifold Cobordant to a Simply Connected Manifold

We will construct (using surgery) a cobordism to $M'$ which kills that generator, and by induction we can kill all of $\pi_1$. … Now we build the cobordism. We take $M \times I$, which is a cobordism from $M$ to itself. …
Chris Schommer-Pries's user avatar