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In his "Classification of (n-1)-connected 2n-dimensional manifolds and the discovery of exotic spheres", Milnor observes that since his exotic 7-spheres admit a Morse function with only two critical points, they are diffeomorphic to two 7-disks glued along a diffeomorphism $g : S^6 \to S^6$, which can't be isotopic to the identity due to the fact the result can't be diffeomorphic to $S^7$ (here I suppose Milnor forgot to comment that $g$ preserves orientation).

How far does this relationship go in general? If I take an orientation-preserving diffeomorphism $g : S^n \to S^n$ which is not isotopic to the identity, and construct a topological sphere by gluing two copies of $D^{n + 1}$ along $g$, will that sphere be exotic? Can every exotic sphere be obtained in this way?

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Look under "Twisted Spheres" in the Wikipedia article: https://en.wikipedia.org/wiki/Exotic_sphere. For n>6, all diffeomorphisms not isotopic to the identity give an exotic sphere. Edit: As Mike Miller points out and the article as well, all exotic spheres are thus obtainable.

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    $\begingroup$ I don't understand. The h-cobordism theorem implies that all exotic spheres are obtainable in this way, at least for $n \geq 6$; delete two discs, and relative to say the leftmost boundary component this is diffeomorphic to $S^{n-1} \times [0,1]$. Because we can do this relatively we glue in the leftmost disc to see that the sphere is diffeomorphic to the union of two discs along their boundary by some diffeomorphism. $\endgroup$ – user98602 Jun 1 '16 at 0:47
  • $\begingroup$ @MikeMiller: of course you are right. I was misled by the fact that $\Gamma_n$ was given a different name than $\Theta_n$ in the article. But it then goes on to say they are isomorphic. Oops. $\endgroup$ – Cheerful Parsnip Jun 1 '16 at 0:50
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    $\begingroup$ The fact that this is then isomorphic to $\pi_0 \text{Diff}^+(S^{n-1})$ is such a fantastic theorem, and the pseudoisotopy theorem and its various improvements absolutely insane; I think it's very easy to believe that none of these various groups will be the same :) Of course, because (Cerf again) $\Gamma_3 = 0$, if 4D Poincare is false the counterexamples couldn't possibly be twisted, and it's entirely possible that $\pi_0 \text{Diff}^+(S^4)$ is a silly group, compared to the rather simple $\Theta_5 = 0$. $\endgroup$ – user98602 Jun 1 '16 at 1:05
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Grumpy Parsnip answered the question, but let me summarize the situation and add some other aspects of the story.

Denote by

  • $Diff^\partial(D^n)$ the group of diffeomorphisms of the $n$-disc that are the identity on a neighborhood of the boundary, by
  • $Diff^+(S^n)$ the group of orientation preserving diffeomorphisms and by
  • $\Theta_n$ the group of exotic spheres (which coincides with the group of h-cobordism classes of homotopy spheres provided $n\neq 4$ by the h-cobordism theorem).

We have maps $$\pi_0(Diff^\partial(D^n))\rightarrow \pi_0(Diff^+(S^n))\rightarrow\mu(Diff^+(S^n))\rightarrow \Theta_{n+1},$$

where

  • $\mu(Diff(M))$ denotes the group of pseudoisotopy classes of diffeomorphisms of a manifold $M$,
  • the first map is induced by the map $Diff^\partial(D^n)\rightarrow Diff^+(S^n)$ that crushes the boundary of the $n$-disc,
  • the second map is provided as isotopy implies pseudoisotopy and
  • the last map is given by the twisted spheres construction (gluing two $n+1$ discs via the given diffeomorphism to obtain a, possibly exotic, spheres).

What can we say about the quality of these maps?

  • The last map is an isomorphism for $n\neq 4 $ by Smale's h-cobordism theorem.
  • The middle map is an isomorphism for $n\ge 5$ by the pseudoistopy theorem of Cerf which says that $\mu(Diff(M))\cong \pi_0(Diff(M))$ for simply connected manifolds of dimension $n\ge 5$.

So we are left with the question in which dimensions the first map is an isomorphism as well. We have a map $Diff^+(S^n)\rightarrow SO(n+1)$ given by taking the derivative at a fixed point $x$ and identifying the frame bundle of $S^n$ with $SO(n+1)$. This is a fibration with fiber the group of diffeomorphism that fix $x$ and the tangent space of $x$, which is homotopy equivalent to the group of diffeomorphisms fixing $x$ and a neighborhood thereof, which is $Diff^\partial(D^n)$. We hence get a fibration sequence $$Diff^\partial(D^n)\rightarrow Diff^+(S^n)\rightarrow SO(n+1),$$ which splits up to homotopy (a splitting is given by the inclusion $SO(n+1)\subseteq Diff^+(S^n)$), so we get $$Diff^+(S^n)\simeq SO(n+1)\times Diff^\partial(D^n).$$ As $\pi_0(SO(n+1))$ is trivial, the first map in the above chain is always an isomorphism.

Now the question arises whether actually $SO(n+1)$ is equivalent to the path component of the identity of $Diff^+(S^n)$. This is

  • true in dimensions 1-3,
  • unknown in dimension 4 and
  • wrong in dimensions 5 and above.
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