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## Smooth representatives for elements of $\pi_7(\text{exotic$S^7$})$

Let $M$ be $S^7$ with an exotic smooth structure. Since one can smoothen maps, there exist smooth maps $f:S^7\to M$ which are homotopic to the identity (relative to a base point, if you want).

Can one make explicit one such map? Can such a map be an homeomorphism?

Little addendum. The smooth homeomorphism constructed in Ryan's answer below is of course not a diffeomorphism.

Does one have some control on the non-smooth locus of the inverse of smooth homeomorphims, or on the type of their non-smoothness there?

The inverse of Ryan's map is non-smooth only at the bad pole and I guess the initial map $h$ arises as the "conical differential" of the map there, so the singularity there is pretty bad. Maybe one can find other smooth homeomorphisms whose inverse has a larger non-smooth locus but with tamer non-smoothness there?

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Duràn wrote down an explicit formula for such map in "Pointed Wiedersehen Metrics on Exotic Spheres and Diffeomorphisms of $S^6$". That is he wrote an explicit formula for an exotic diffeomorphism from $S^6$ to $S^6$ which is homotopic but not isotopic to the identity. This the produces an explicit homeomorphism from $S^7$ to an exotic sphere by glueing as described by Ryan in his answer.

Geometric properties of that particular map were later studied by various people. For example, it's written down explicitly on page 1 in "Bootstrapping $Ad$-Equivariant Maps, Diffeomorphisms and Involutions" by Duràn and Rigas and Sperança (this link is freely accessible unlike the first one).

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 Thank you very much for the references. – Mariano Suárez-Alvarez Apr 8 2012 at 22:51

I'll answer your question in two steps. (1) You can make a degree one map $f : S^7 \to M$ a homeomorphism, and $C^\infty$-smooth on the complement of a point.

The idea is that you can construct $M$ as the union of two $D^7$'s

$$M = D^7 \sqcup_{h} D^7$$

where the gluing map $h : S^6 \to S^6$ is some diffeomorphisms of $S^6$ that does not extend to a diffeomorphism of $D^7$.

To get a map $S^7 \to M$, just write $S^7 = D^7 \sqcup_{Id_{S^6}} D^7$, the map on the first $D^7$ factor $S^7 \to M$ is just the identity, and on the 2nd $D^7$ factor you're coning-off $h$, i.e. the extension of $h : S^6 \to S^6$ to a homeomorphism of $D^7$ is given by $\tilde h(tv) = th(v)$ provided $v \in S^6$ and $t \in [0,1]$. This is usually called the Alexander trick, at least in the PL or topological categories. This map is smooth everywhere except $0$ in the 2nd $D^7$ factor.

(2) To fix this argument and get a smooth homeomorphism, replace $\tilde h(tv) = th(v)$ with $\tilde h(tv)=\beta(t)h(v)$ where $\beta : [0,1] \to [0,1]$ is a $C^\infty$-smooth homeomorphism with all derivatives of $\beta$ zero at $0$, but otherwise $\beta'(t) > 0$ for $t \in (0,1]$. You can cook up such functions readily using bump functions.

I think these arguments go back to Milnor's first papers on exotic smooth structures. You can make every representative of $\pi_7 M$ smooth, by iteratively suspending the degree $k$ map $S^1 \to S^1$ and appropriately "dampening" at the cone-points, such as the above construction.

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 For some reason I expected this would not work, but surely it does. Thanks! – Mariano Suárez-Alvarez Apr 8 2012 at 22:57