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Let $M$ and $N$ be topological manifolds that admit differential structures and let $f:M\to N$ be continuous. Can $M$ and $N$ always be given differential structures to become differentiable manifolds $\widetilde M$ and $\widetilde N$ such that $f:\widetilde M\to\widetilde N$ is differentiable? What if we impose further restrictions, such as $\widetilde M$ and $\widetilde N$ being smooth manifolds, or $f$ becoming smooth? Are there certain differential structures which we can't do this with (i.e. if we want to make $f$ differentiable, we can never make $N$ diffeomorphic to $\overline N$ where $\overline N$ is some differential structure on $N$)? What if $M$ already has a fixed differential structure?

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  • $\begingroup$ Shouldn't the 1-dimensional case yield examples? I mean there are continuous $f : \mathbb{R} \to \mathbb{R}$ which are not differentiable and I somehow think that there shouldn't be more differentiable structures on $\mathbb{R}$ then the natural one? $\endgroup$
    – user301452
    Commented Jan 1, 2021 at 13:33
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    $\begingroup$ There are multiple differential structures on $\mathbb{R}$. They're all diffeomorphic, but they are distinct. For a maximal atlas on $\mathbb{R}$ and a homeomorphism $h:\mathbb{R}\to\mathbb{R}$ we can take a chart $\varphi$ in the atlas and obtain a new chart $\varphi\circ h$. We can define a new atlas consisting of all the charts in the previous one composed with $h$. Clearly the transition maps are differentiable. So if $h$ wasn't itself differentiable, we have a new differential structure, since $h\circ\varphi$ and $\varphi$ clearly aren't compatible. $\endgroup$ Commented Jan 1, 2021 at 14:07
  • $\begingroup$ Sorry, the last part doesn't work. I confused $\varphi\circ h$ and $h\circ\varphi$. However, if $\varphi$ commutes with $h$, then it's fine, so if, e.g. the identity is a chart (which of course it is in some maximal atlas), then we have a new differential structure. $\endgroup$ Commented Jan 2, 2021 at 0:38

1 Answer 1

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There is not necessarily a way to make a map smooth. For example, suppose $M=\mathbb{R}$, and suppose $N$ is any topological manifold of dimension $2$ or more. Let $f:M\rightarrow N$ be any continuous function whose image contains a non-empty open subset of $N$ (e.g., take a space filling curve onto $\mathbb{R}^n$ and then think of this $\mathbb{R}^n$ as a chart).

Then there are no smooth structures on $M$ and $N$ which makes $f$ smooth. In fact, you cannot even make $f$ continuously differentiable. One way to see this is to use Sard's Theorem: if you could make $f$ continuously differentiable, then the set of regular values would be open and dense. Because $M$ has a lower dimension than $N$, regular values are points of $N$ which are not in the image of $f$. But then the rest that the image of $f$ contains an open subset of $N$ means the set of regular values of $f$ is not dense.

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    $\begingroup$ This is a great answer. $\endgroup$
    – Tom Ariel
    Commented Jan 1, 2021 at 14:29
  • $\begingroup$ I just updated the answer to change "smooth" to "continuously differentiable". $\endgroup$ Commented Jan 1, 2021 at 16:12

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