# Contractability of Exotic R^4s

Notation: $\mathbf{R}^4$ is a smooth manifold with underlying topology $(\mathbb{R})^4$; ${\mathbb{R}}^4$ is the standard smooth structure.

The two things I know best about $\mathbf{R}^4$ is that it is locally diffeomorphic to $\mathbb{R} ^4$, and that it's contractible. It's easy to see that the contraction can be acheived by a smooth map ${\mathbb{R}}^4\times I\rightarrow{\mathbb{R}}^4$.

• Do I suppose correctly that the same contraction is not smooth as a map ${\mathbf{R}}^4\times I\rightarrow{\mathbf{R}}^4$?
• Do the exotic smooth structures have any smooth contractions?
• If not, are there continuous contractions $\mathbf{R}^4\times I\rightarrow\mathbf{R}^4$ within the smooth maps $\mathbf{R}^4\rightarrow\mathbf{R}^4$?
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What's a "contraction"? Is it a homotopy between the identity and a constant map? Any continuous map can be approximated by a smooth one, and nearby maps are homotopic, so it seems you can find a smooth homotopy between the identity and a constant map. –  Igor Belegradek Apr 14 '10 at 22:17
@Igor: yes, a contraction is a homotopy as you describe. You can certainly smooth any continuous map of smooth spaces; but whether the result will still be a contraction isn't clear to me. A related issue is the failure in the smooth category of "Mazur's Swindle", that shows all spheres to be topologically irreducible w.r.t. connect-sum. –  some guy on the street Apr 15 '10 at 2:38
some guy on the street: The identity map and a constant map are both smooth, and it's an easy exercise fom Igor's statement to show that in fact all homotopic smooth maps are smoothly homotopic. It is an exercise in milnor's Topology from a Differentiable viewpoint. –  Harry Gindi Apr 15 '10 at 3:20
Perhaps this question could be interpreted as follows: topologically, there is no difference between exotic-R^4 and R^4 but as manifolds, they are different. This leads one to wonder at what point it is possible to distinguish between the two. In particular, is it possible to distinguish exotic-R^4 and R^4 using smooth homotopy theory (i.e. homotopy theory but everything has to be smooth)? –  Andrew Stacey Apr 15 '10 at 12:26
$R^4\times I\rightarrow R^4$ grrr! preview comments? author edit comments?? don't mind me! –  some guy on the street Apr 15 '10 at 16:21
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