Can one give an immersion of exotic sphere $S^7$ in a standard sphere $S^8$ of radius $1$?
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$\begingroup$ Dear Li Yan. You might want to explain what you mean by "exhaustive", and also what you mean by the notation S^8(1). I don't know what these terms mean, and I think that other MO users will also not know. Please edit your question to add this information. $\endgroup$– André HenriquesCommented Jun 22, 2013 at 13:21
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$\begingroup$ Sorry,a serious mistake! I mean " John Milnor's exotic spheres"."S^(R)"means n-sphere of radius R. Li Yan $\endgroup$– Li YanCommented Jun 22, 2013 at 13:42
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1$\begingroup$ Note that the radius of the sphere is not important since $S^8$ of different radii are diffeomorphic. $\endgroup$– Lennart MeierCommented Jun 26, 2013 at 8:18
2 Answers
Yes. By Smale-Hirsch theory it is enough to find a bundle injection $T\Sigma \to \epsilon^8$, so it is enough to find a trivialisation of $T\Sigma \oplus \epsilon^1$. It is a theorem of Kervaire and Milnor that every exotic sphere is stably framable, so $T\Sigma \oplus \epsilon^N$ is trivial for some large N, and the connectivity of $BO(8) \to BO$ means you can destabilise this to trivialise $T\Sigma \oplus \epsilon^1$.
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$\begingroup$ Wow! Does this also allow one to say something meaningful about the complement, e.g. is it two standard open balls? $\endgroup$ Commented Jun 26, 2013 at 11:17
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10$\begingroup$ No--this is an immersion, not an embedding. If you had an embedding, then the allegedly exotic sphere would bound a contractible manifold, and hence be diffeomorphic to $S^7$ by the h-cobordism theorem. One probably could say something interesting about the topology of the double point set, though. $\endgroup$ Commented Jun 26, 2013 at 11:28
Slightly different: We computed the group of immersions of homotopy 4k-1 spheres into $R^{4k+1}$ and also to some other euclidean spaces here:
https://www.researchgate.net/publication/243028484_The_group_of_immersions_of_homotopy_4k-1-spheres