Can someone give an example of a non-differentiable manifold?

Question

A topological space $M$ is a manifold of dimension $n\geq 1$ iff it is a second countable space that is locally homeomorphic to the Euclidean space $R^n$. So if $M$ is a manifold there exists a map $f: M \rightarrow R^n$ such that $f$ is continuous and bijective in both directions($f$ and $f^{-1}$). Now lemme point out that while continuity is a necessary condition for differentiability, it not a sufficient. That is,even if $f$ is continuous it *need not be differentiable *.A good example of a function that is continuous everywhere but differentiable nowhere is the Weierstrass function. So by "non-differentiable" manifold I mean one for which every chart in its atlas is continuous but nowhere differentiable. Or in particularly, there exists a bijective map $\Phi : M \rightarrow R^n$ such that $\Phi$ is continuous everywhere on $M$ but differentiable nowhere(on $M$). So clearly such a thing can exist. Any examples of such a manifold?

Answer 1

Your question is not quite correctly formulated. If $M$ is a general, abstract manifold, defining what it means for a map $M \rightarrow \mathbb{R}^n$ to be differentiable is not immediate and requires actually more structure than $M$ itself (it is the notion of differential structure).

A differential structure is an (equivalence class of) atlas in which all transition maps are differentiable. This is well defined since transition maps are defined on open sets of $\mathbb{R}^n$ and not on the manifold $M$. Once you have such an atlas, it makes sense to say that a function is differentiable if written in any charts of the atlas it is differentiable.

When I say that it is more information than just $M$, I mean that there are examples of manifolds $M$ which have several different differentiable structures, i.e. different atlas such that a map from say $M$ to $\mathbb{R}$ may be differentiable when $M$ is equipped with a certain atlas and not differentiable when $M$ is equipped with another atlas. It is possible to prove that all topological manifolds of dimension up to 3 are in fact $C^\infty$ manifolds, and that any $C^1$ manifold is in fact a $C^\infty$ manifold. However these results are hard.

So to sum things up : this is a tricky topic. Your last question is in fact not a special case of your first question, which should be reformulated as "are there topological manifolds that are not smooth manifolds?". Also note that you won't find any homeomorphism from $\mathbb{R}$ to $\mathbb{R}$ nowhere differentiable, as such a homeomorphism must be monotone and monotone maps can be shown to be almost everywhere differentiable.

ADDENDUM : this is all tricks and subtleties. If you are interested in submanifolds rather than manifolds (which is probable if this is the first time you encounter such questions) then defining the regularity of a map defined on $M$ is much easier and natural (say that such a map is $C^k$ if it is the restriction of a $C^k$ ambient map). Then your question makes sense and the answer of Tomas is correct. Indeed, if a map $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ maps homeomorphically the graph of the Weierstrass function to the real axis, then this function can not be differentiable anywhere on the graph.