From what I understand, an exotic nsphere is a manifold which is homeomorphic to the nsphere but not diffeomorphic to it. Now I have read that there are no exotic 2spheres. But isn't something like a tetrahedron an example of a manifold which is homeomorphic to the sphere, but not diffeomorphic ? (Because of the corners and edges.) What am I missing ?

3$\begingroup$ You are missing the definition of a manifold, I guess. Tetrahedron is not a manifold if you don't specify atlases on it. $\endgroup$ – Petya Mar 12 '10 at 12:33

2$\begingroup$ I believe you're confusing properties of an embedding of a sphere in 3space with its intrinsic differentiable structure $\endgroup$ – j.c. Mar 12 '10 at 12:33

9$\begingroup$ A exotic sphere is (by definition) a differentiable manifold. So if you want to consider the tetrahedron, you have to specify, what differentiability at one of the edges means. As soon as you specified this, you will just get the 2sphere. $\endgroup$ – HenrikRüping Mar 12 '10 at 13:10

1$\begingroup$ On the other hand, some subsets of $\mathbb{R}^n$ (e.g. a sphere, a torus) are accepted as manifolds w/o specifying an atlas. I think the situation is that if there is a "wellknown" submersion such with this subset is one of the fibers, then this subset is accepted by community as a manifold. (A neccesary condition then is for this subset to have a unique tangent space at each point, so tetrahedron fails to satisfy this condition and you don't need to try to find a relevant submersion.) $\endgroup$ – Paul Yuryev Mar 12 '10 at 13:47

1$\begingroup$ @HerinkRüping: You should repost your comment as an answer so that Cosmonut can accept it. $\endgroup$ – Omar AntolínCamarena Mar 13 '10 at 1:36
A exotic sphere is (by definition) a differentiable manifold. So if you want to consider the tetrahedron, you have to specify, what differentiability at one of the edges means. As soon as you specified this, you will just get the 2sphere.