From what I understand, an exotic n-sphere is a manifold which is homeomorphic to the n-sphere but not diffeomorphic to it. Now I have read that there are no exotic 2-spheres. 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 ?
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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. $\endgroup$– PetyaMar 12, 2010 at 12:33
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2$\begingroup$ I believe you're confusing properties of an embedding of a sphere in 3-space with its intrinsic differentiable structure $\endgroup$– j.c.Mar 12, 2010 at 12:33
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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 2-sphere-. $\endgroup$– HenrikRüpingMar 12, 2010 at 13:10
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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 "well-known" 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 YuryevMar 12, 2010 at 13:47
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1$\begingroup$ @HerinkRüping: You should repost your comment as an answer so that Cosmonut can accept it. $\endgroup$– Omar Antolín-CamarenaMar 13, 2010 at 1:36
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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 2-sphere.
