I Difference between diffeomorphism and homeomorphism

[Silviu]

Hello! I just started reading something on differential geometry and I am not sure I understand the Difference between diffeomorphism and homeomorphism. I understand that the homeomorphism means deforming the topological spaces from one to another into a continuous and bijective way (like a sphere into a cube). As far as I understand diffeomorphism means the same thing, but instead of going directly from a space to the other, you go through $R^n$ before. I can understand that mathematically diffeomorphism is a stronger requirement, but I am not sure I can visualize the difference. Can someone explain this to me? Thank you!

 

[fresh_42]

I just started reading something on differential geometry and I am not sure I understand the Difference between diffeomorphism and homeomorphism.

Both are bijective. Homeomorphisms (and their inverses) are continuous, diffeomorphisms (and their inverses) are continuously differentiable, which implies continuity of themselves and their derivatives, i.e. diffeomorph implies homeomorph.

So the difference is the same as with $\{(x,|x|)\,\vert \,x \in \mathbb{R})\}$ and $\{(x,x^2)\,\vert \,x \in \mathbb{R})\}$ at $x=0$. With homeomorphisms you can "smoothen" sharp edges, with diffeomorphism you can not. E.g. let's consider the everywhere favorable example of a mug and a torus: they are homeomorph but not diffeomorph, because the handle of the mug (normally) isn't clued differentiable. Or to be on the save side: a cube and a sphere are homeomorphic nut not diffeomorphic.

One may also shortly say:
Homeomorphisms are the bijective mappings in the category of topological spaces, whereas diffeomorphisms are the bijective mappings in the category of differentiable manifolds.

This also illustrates the difference: differentiable manifolds are also topological spaces, but not vice versa.

 

[Silviu]

Both are bijective. Homeomorphisms (and their inverses) are continuous, diffeomorphisms (and their inverses) are continuously differentiable, which implies continuity of themselves and their derivatives, i.e. diffeomorph implies homeomorph.

So the difference is the same as with $\{(x,|x|)\,\vert \,x \in \mathbb{R})\}$ and $\{(x,x^2)\,\vert \,x \in \mathbb{R})\}$ at $x=0$. With homeomorphisms you can "smoothen" sharp edges, with diffeomorphism you can not. E.g. let's consider the everywhere favorable example of a mug and a torus: they are homeomorph bur not diffeomorph, because the handle of the mug (normally) isn't clued differentiable. Or to be on the save side: a cube and a sphere are homeomorphic nut not diffeomorphic.

Thank you for your reply. It makes sense, however I read that diffeomorphically inequivalent homeomorphisms arise only for dim 4 or higher. So, based on this, shouldn't the sphere and the cube (which are 2D objects inside a 3D space) be diffeomorphic (as they are homeomorphic already)?

 

[fresh_42]

Thank you for your reply. It makes sense, however I read ...

It's hard to tell what you have read.

... that diffeomorphically inequivalent homeomorphisms arise only for dim 4 or higher.

Here's what Wikipedia says about it:

For differentiable (smooth) manifolds in dimension less than 4, homeomorphism always implies diffeomorphism: two differentiable (smooth) manifolds of the dimension less than or equal to 3, which are homeomorphic, are also diffeomorphic. That is, if there is a homeomorphism, then there is also a diffeomorphism. This does not mean that any homeomorphism would be a diffeomorphism.

So, based on this, shouldn't the sphere and the cube (which are 2D objects inside a 3D space) be diffeomorphic (as they are homeomorphic already)?

My examples with the absolute value function or the cube are of lower dimension, too. How do you calculate $\left.\frac{d}{dx}\right|_{x=0}|x|\,?$ Furthermore, we can have topological spaces (and with them homeomorphisms) which do not carry a differentiable structure at all, e.g. finite spaces. The crucial point is, that these are not differentiable manifolds, which is a strong additional requirement. And the above is only a statement about existence, not about inclusion, implication or equality.

An example, that not the same functions are meant by "existence" would be: A circle and an ellipse are both differential manifolds and homeomorphic as well as diffeomorphic. We can transform the circle homeomorphis into and onto a square and this square to the ellipse. Those functions will be a homeomorphism, but no diffeomorphism. Nevertheless, there is also a diffeomorphism which transforms the circle into and onto the ellipse - without having edges in between.

 

[Orodruin]

Isn't that just the embedding of the cube in $\mathbb R^3$ that is not smooth? That the embedding in $\mathbb R^3$ is not smooth does not mean that there cannot be a differentiable structure. You could define a differentiable structure on the cube just by projecting it onto the sphere. This structure would (naturally) be diffeomorphic to the sphere (it has to be, there is only one differentiable structure on the 2-sphere).

 

[lavinia]

Thank you for your reply. It makes sense, however I read that diffeomorphically inequivalent homeomorphisms arise only for dim 4 or higher. So, based on this, shouldn't the sphere and the cube (which are 2D objects inside a 3D space) be diffeomorphic (as they are homeomorphic already)?

Some topological manifolds can have more than one differentiable structure. This means that as smooth manifolds, they are homeomorphic but not diffeomorphic.

 

[fresh_42]

You could define a differentiable structure on the cube just by projecting it onto the sphere

Interesting question. It boils down to the question: What is an edge on a manifold? I would define it as continuous but not differentiable point, i.e. the (general) curves through it. However, I'm not sure if it is a reasonable definition this way. But to have a counterexample, we can live with embeddings, aka subspaces of $\mathbb{R}^n$. Manifold doesn't require not to be embedded carrying the induced topology. And as a counterexample, it has not to be true for all differentiable structures.

 

[Orodruin]

The point is that if you can put a differentiable structure on a topological manifold, you should be able to put one on all manifolds that are homeomorphic to it by using the homeomorphism.

You later have the fact lavinia mentions - a topological manifold might allow several distinct (under diffeomorphisms) differentiable structures. The 2-sphere only admits one though.

 

[fresh_42]

This means the topological structure, which is used to define the homeomorphism, is already given by the differentiable structure. This is quite a strong requirement which restricts possible topologies. Is this why topologies which don't have this property in dimensions 4 and higher are called exotic?

We have had discussions about the subject ealier (2014): https://www.physicsforums.com/threads/diffeomorphism-vs-homeomorphism.731632/
which is probably on A level, but Google found me more on PF, which I haven't checked. Would be interesting to find out about the difference between this Google search list and our own.

I've found these two articles which also might be of interest in this context (the second for whom is interested in mathematical history):
https://www3.nd.edu/~lnicolae/FYsem2003.pdf
http://www.mathunion.org/ICM/ICM1958/Main/icm1958.0433.0440.ocr.pdf

 

[lavinia]

This means the topological structure, which is used to define the homeomorphism, is already given by the differentiable structure. This is quite a strong requirement which restricts possible topologies. Is this why topologies which don't have this property in dimensions 4 and higher are called exotic?

I think "exotic" refers to a non-standard differentiable structure on a given topological manifold.

An exotic 7 sphere is a topological 7 sphere with a differentiable structure that is not diffeomorphic to the standard 7 sphere in Euclidean space. Milnor's original paper was called "On Manifolds Homeomorphic to the 7 Sphere."

I don't think any exotic 7 sphere can be embedded smoothly in $R^8$.

 

[WWGD]

A standard, kind of simple example is that of $(\mathbb R, Id.)$ and $(\mathbb R, x^3)$ , which are homeomorphic but not diffeomorphic, although $\mathbb R$ has just one differential structure. Meaning the coordinate maps are, resp. the Id and $x^3$; this last is not differentiable at $x=0$.EDIT: in 4-manifolds, the existence of differentiable structures is given by the mid-homology intersection form. EDIT2: I wrote this quickly last night; as @vanhees71 points out , the issue here is with the inverse, which is not differentiable at $x=0$.

 

[vanhees71]

Hm, the polynom $f(x)=x^3$ is differentiable everywhere. It's the inverse function $g(x)=x^{1/3}$ that is not differentiable at $x=0$. So indeed, $f$ is not a diffeomorphism at $x=0$.

 

[lavinia]

A standard, kind of simple example is that of $(\mathbb R, Id.)$ and $(\mathbb R, x^3)$ , which are homeomorphic but not diffeomorphic, although $\mathbb R$ has just one differential structure. Meaning the coordinate maps are, resp. the Id and $x^3$; this last is not differentiable at $x=0$.EDIT: in 4-manifolds, the existence of differentiable structures is given by the mid-homology intersection form.

Let $F: (\mathbb R, Id.) → (\mathbb R, x^3)$ by $F(x) = x^{1/3}$.

Isn't $F$ a diffeomorphism?

 

[fresh_42]

Let $F: (\mathbb R, Id.) → (\mathbb R, x^3)$ by $F(x) = x^{1/3}$.

Isn't $F$ a diffeomorphism?

I thought the derivative had to be continuous to establish a diffeomorphism.

 

[vanhees71]

$f(x)=x^3$ is differentiable on $\mathbb{R}$ and invertible, but the inverse is not differentiable. The inverse is $g(x)=f^{-1}(x)=x^{1/3}$, since $f'(0)=0$ the inverse $g$ is not differentiable in $x=0$ and thus $f$ is not a diffeomorphism.

 

[WWGD]

I thought the derivative had to be continuous to establish a diffeomorphism.

Maybe that is required in some cases, even $C^{\infty}$ , but the issue is, as Vanhees71 pointed out, that $f: \mathbb R \rightarrow \mathbb R : f(x)=x^3$ is differentiable, but its inverse is not, at $x=0$.

 

[WWGD]

Still, an interesting question, IMHO, is , which properties are preserved by diffeomorphisms that are not preserved by homeos., or, which properties do pairs of diffeomorphic spaces share that homeomorphic ones do not? Other than the collection of differentiable functions, I can't tell. Maybe DeRham Cohomology ( Since forms are pulled back by F )?

 

[fresh_42]

Maybe that is required in some cases, even $C^{\infty}$ , but the issue is, as Vanhees71 pointed out, that $f: \mathbb R \rightarrow \mathbb R : f(x)=x^3$ is differentiable, but its inverse is not, at $x=0$.

... which is the same as saying it is not continuous at $0$. It isn't defined, so whether you say not differentibale or not continuous doesn't make the difference here.

 

[lavinia]

What is wrong with this argument?

Given a function $F:M→N$ between two smooth manifolds then $F$ is smooth if for any coordinate chart $Φ: U⊂N→\mathbb R$, $Φ \circ F$ is smooth on $M$.

$(\mathbb R, x^3)$ has only one chart $Φ(x) = x^3$ and $Φ \circ F = Id_{(\mathbb R, Id.)}$ is smooth. Therefore $F$ is a smooth bijection.

$F^{-1}:(\mathbb R, x^3)→(\mathbb R, Id.)$ is $F^{-1}(x) = x^3$. The only chart on $(\mathbb R, Id.)$ is $Φ(x) = x$ so $Φ \circ F^{-1} = x^3$ and this is smooth on $(\mathbb R, x^3)$. So $F^{-1}$ is a smooth bijection.

 

[WWGD]

... which is the same as saying it is not continuous at $0$. It isn't defined, so whether you say not differentibale or not continuous doesn't make the difference here.

But the derivative being continuous is stronger than the inverse being differentiable. Yes, they do agree in this case, but not always.

 

[WWGD]

What is wrong with this argument?

Given a function $F:M→N$ between two smooth manifolds then $F$ is smooth if for any coordinate chart $Φ: U⊂N→R$, $Φ \circ F$ is smooth on $M$.

$(\mathbb R, x^3)$ has only one chart $Φ(x) = x^3$ and $Φ \circ F = Id_{(\mathbb R, Id.)$ is smooth. Therefore $F$ is a smooth bijection.

$F^{-1}:(\mathbb R, x^3)→(\mathbb R, Id.)$ is $F^{-1}(x) = x^3$. The only chart on $(\mathbb R, Id.)$ is $Φ(x) = x$ so $Φ \circ F^{-1} = x^3$ sand this is smooth on $(\mathbb R, x^3)$. So $f^{-1}$ is a smooth bijection.

I think you're showing that the differential structures are equivalent (by definition of their equivalence), but it is still the case that $f^{-1}(x)=x^{1/3}$ is not differentiable at $0$.

 

[lavinia]

I think you're showing that the differential structures are equivalent (by definition of their equivalence), but it is still the case that $f^{-1}(x)=x^{1/3}$ is not differentiable at $0$.

Didn't I show that the map is a diffeomorphism?

 

[WWGD]

Didn't I show that the map is a diffeomorphism?

But the original map is from $( \mathbb R,Id )$ to itself., and your argument does not hold for this case. I think so.

 

[lavinia]

But the original map is from $( \mathbb R,Id )$ ) to itself.

No it is from $( \mathbb R,Id )$ to $( \mathbb R,x^3)$

 

[WWGD]

No it is from $( \mathbb R,Id )$ to $( \mathbb R,x^3)$

But my argument is that my map $f(x)=x^3$ from $(\mathbb R, Id)$ to itself is a homeo but not a diffeo. As you showed, $f(x)=x^3$ is a diffeo. between $(\mathbb R, Id) , (\mathbb R, x^{-1/3})$ , but that is a different argument.

 

[fresh_42]

What is wrong with this argument?

Given a function $F:M→N$ between two smooth manifolds then $F$ is smooth if for any coordinate chart $Φ: U⊂N→\mathbb R$, $Φ \circ F$ is smooth on $M$.

$(\mathbb R, x^3)$ has only one chart $Φ(x) = x^3$ and $Φ \circ F = Id_{(\mathbb R, Id.)}$ is smooth. Therefore $F$ is a smooth bijection.

$F^{-1}:(\mathbb R, x^3)→(\mathbb R, Id.)$ is $F^{-1}(x) = x^3$. The only chart on $(\mathbb R, Id.)$ is $Φ(x) = x$ so $Φ \circ F^{-1} = x^3$ and this is smooth on $(\mathbb R, x^3)$. So $F^{-1}$ is a smooth bijection.

Don't we have to require the same condition on $F^{-1}:N→M$ and a chart $\Psi: V⊂M→\mathbb R\,$?

 

[lavinia]

Don't we have to require the same condition on $F^{-1}:N→M$ and a chart $\Psi: V⊂M→\mathbb R\,$?

Yes.

 

[fresh_42]

Yes.

I think you are right. The process of a pointwise deformation from $\{x\,\vert \,x \in \mathbb{R}\}$ to $\{x^3\,\vert \,x \in \mathbb{R}\}$ is smooth in both directions: $0$ is a fix point and everything around it is smooth. We do not need $x \mapsto \frac{1}{3}x^{-\frac{2}{3}}$ as we do not convert the function, but only the parametrizations between the two functions.

I guess if we simply would have drawn a circle (to be $\mathbb{R}$) and some kind of smooth loop (cluing the ends of $(x,x^3)$) and haven't mentioned the function, nobody would have asked whether they are diffeomorph.

 

[lavinia]

Here is something strange.

Two coordinate charts $φ:U →\mathbb {R^{n}}$ and $ψ:V → \mathbb {R^{n}}$ are "compatible" if both $φ \circ ψ^{-1}$ and $ψ \circ φ^{-1}$ are smooth functions on the inverse images of $U∩V$.

For the differentiable manifold $( \mathbb R,x^3)$ all charts must be compatible with $ψ(x) = x^3$. So $φ(x^{1/3})$ must be a smooth function on any open set $φ^{-1}(U)$. So in particular it seems that the identity function $φ(x) = x$ (considered as a coordinate map) is not smooth on $( \mathbb R,x^3)$.

 

[fresh_42]

Here is something strange.

Two coordinate charts $φ:U →\mathbb {R^{n}}$ and $ψ:V → \mathbb {R^{n}}$ are "compatible" if both $φ \circ ψ^{-1}$ and $ψ \circ φ^{-1}$ are smooth functions on the inverse images of $U∩V$.

For the differentiable manifold $( \mathbb R,x^3)$ all charts must be compatible with $ψ(x) = x^3$. So $φ(x^{1/3})$ must be a smooth function on any open set $U$. So in particular it seems that the identity function $φ(x) = x$ (considered as a coordinate map) is not smooth on $( \mathbb R,x^3)$.

This time I looked up definitions to contain my confusions. Say we have the two smooth manifolds $M=\{x^3\}$ and $N=\mathbb{R}$. Now here's the first trap. Do you consider the manifold $N$ or the Euclidean space $\mathbb{R}$ which provides the atlas for $M$, when you say compatible with $ψ(x) = x^3$, i.e. do we consider $\psi : N \longrightarrow M$ or $\psi$ as an inverse of a chart of $M\,$? Do you mean the compatibility of two charts $\chi_\alpha \; , \;\psi^{-1}$ of $M$, or what does compatible mean in case $\psi$ is considered a map between manifolds?

I don't see, why we cannot simply take $U_\alpha = (a^3,b^3)$ and $\chi_\alpha (p) = p$ as charts of $M$ and why the identity function $\varphi : M \longrightarrow M$ isn't simply $\varphi(p)=p \,$, because we do not need to switch over to roots in this case, except we consider mappings between manifolds, in which case the question might be, if the especially given $\psi : N \longrightarrow M$ is smooth, invertible and whether the inverse is smooth, too.

Sorry for my confusion caused by the triple role of $\mathbb{R}$: twice an atlas and once a manifold.

 

[lavinia]

This time I looked up definitions to contain my confusions. Say we have the two smooth manifolds $M=\{x^3\}$ and $N=\mathbb{R}$. Now here's the first trap. Do you consider the manifold $N$ or the Euclidean space $\mathbb{R}$ which provides the atlas for $M$, when you say compatible with $ψ(x) = x^3$, i.e. do we consider $\psi : N \longrightarrow M$ or $\psi$ as an inverse of a chart of $M\,$? Do you mean the compatibility of two charts $\chi_\alpha \; , \;\psi^{-1}$ of $M$, or what does compatible mean in case $\psi$ is considered a map between manifolds?

The definition of a smooth atlas that I have seen requires that $ψ \circ φ^{-1}$ and $φ \circ φ^{-1}$ are smooth maps from open sets in $\mathbb{R}^{n}$ into $\mathbb{R}^{n}$ with the usual definition of smooth in multivariable calculus.

 

[fresh_42]

So you consider an atlas of $M=\{x^3\}$ with one chart $\mathbb{R}$ and two chart mappings $\varphi$ and $\psi$. These should be functions from $M$ to $\mathbb{R}$. Say we define $\varphi(p)=x$ and $\psi(p)=x^3$ if $p=x^3$ and the question is, whether they both can be put in one atlas. Is this correct?

Edit: In this case the overlaps are $\varphi \psi^{-1}(x)=\varphi(x)=x$ and $\psi \varphi^{-1}(x)=\psi(x^3)=x^3$ which are both smooth as it should be.

 

[vanhees71]

... which is the same as saying it is not continuous at $0$. It isn't defined, so whether you say not differentibale or not continuous doesn't make the difference here.

Being continuous is not sufficient, it must be differentiable. Of course, if it's not continuous, it's also not differentiable...

 

[fresh_42]

Being continuous is not sufficient, it must be differentiable. Of course, if it's not continuous, it's also not differentiable...

The definitions I have found all require the $k-$th derivative to be continuous if the diffeomorphism is of $k-$th degree. So here $f'$ obviously wasn't continuous at $0$, which - to me - was the shortest way to say that the given $f$ is no diffeomorphism, because a $\mathcal{C}^1$ diffeomorphism requires a continuous derivative (acc. to a couple of sources I have checked to be sure). I did not refer to $f$ but to $f'$.

 

[vanhees71]

That's, of course, also a valid argument.

 

[lavinia]

So you consider an atlas of $M=\{x^3\}$ with one chart $\mathbb{R}$ and two chart mappings $\varphi$ and $\psi$. These should be functions from $M$ to $\mathbb{R}$. Say we define $\varphi(p)=x$ and $\psi(p)=x^3$ if $p=x^3$ and the question is, whether they both can be put in one atlas. Is this correct?

Edit: In this case the overlaps are $\varphi \psi^{-1}(x)=\varphi(x)=x$ and $\psi \varphi^{-1}(x)=\psi(x^3)=x^3$ which are both smooth as it should be.

Yes. In the manifold $(\mathbb R, x^{3})$ all charts must be compatible with $φ(x) = x^3$. The differentiable structure is a maximal atlas of mutually compatible charts. Compatible just means that the transition functions are smooth. https://en.wikipedia.org/wiki/Atlas_(topology)#Transition_maps
https://en.wikipedia.org/wiki/Smooth_structure

For any open set $U⊂(\mathbb R, x^3)$ the map $ψ(x) = x$ is not compatible with $φ(x) = x^3$ since if it were, the transition function $ψ \circ φ^{-1}$ would have to be a smooth map from $φ(U)⊂ \mathbb{R}$ into $\mathbb{R}$. But $ψ \circ φ^{-1}(t) = ψ(t^{1/3}) = t^{1/3}$ which is not even differentiable at zero.

 

[fresh_42]

Would you mind to explicitly define domain and codomain of $\psi$ and $\varphi$, because it looks to me as if you define the inverse of chart mappings here. The fact that there is $\mathbb{R}$ all around is quite confusing and a potential cause for misunderstandigs. The points in $U$ are already of the form $p=x^3$ so $x^3 \mapsto x$ won't be a problem for the overlaps or transitions or coordinate changes, whatever. To me it seems as if you started with an inverse function $\mathbb{R} \longrightarrow U$ of a map and I haven't found any statement about those inverses in the definition of an atlas. I first thought you might be talking about a smooth map between the manifolds $\{x^3\}$ and $\{x\}$ but my attempt to clarify this was in vain.

 

[lavinia]

$ψ$ and $φ$ are maps from the topological manifold $R^1$ endowed with the smooth structure $(\mathbb{R},x^3)$ into the Euclidean space $\mathbb{R}^1$. $U$ is an open set in the topological manifold $R^1$.

This refers to the case where I was showing that $ψ(x)=x$ is not a smooth chart on $(\mathbb{R},x^3)$ where the domain of $ψ$ is $(\mathbb{R},x^3)$ and its range is the Euclidean space $R^1$

 

[lavinia]

Would you mind to explicitly define domain and codomain of $\psi$ and $\varphi$, because it looks to me as if you define the inverse of chart mappings here. The fact that there is $\mathbb{R}$ all around is quite confusing and a potential cause for misunderstandigs. The points in $U$ are already of the form $p=x^3$ so $x^3 \mapsto x$ won't be a problem for the overlaps or transitions or coordinate changes, whatever. To me it seems as if you started with an inverse function $\mathbb{R} \longrightarrow U$ of a map and I haven't found any statement about those inverses in the definition of an atlas. I first thought you might be talking about a smooth map between the manifolds $\{x^3\}$ and $\{x\}$ but my attempt to clarify this was in vain.

$\mathbb{R}^1$ will denote the 1 dimensional Euclidean space. It has the usual Euclidean metric.

$\mathbb{R}$ will denote the underlying topological manifold. It has no metric relations and no differential structure.

Now give $\mathbb{R}$ two different differential structures to create two differentiable manifolds. We have called them $(\mathbb{R},x)$ and $(\mathbb{R},x^3)$.

This notation means that the maximal atlas for $(\mathbb{R},x)$ is all homeomorphisms $ψ$ from the open sets $U$ in the topological manifold $\mathbb{R}$ onto open sets in the Euclidean space $\mathbb{R}^1$ that are compatible with the function $φ(x) = x$, the map that maps the point $x$ in the topological manifold $\mathbb{R}$ to the point $x$ in the Euclidean space $\mathbb{R}^{1}$.

Similarly, the maximal atlas for $(\mathbb{R},x^3)$ is all homeomorphisms $ψ$ from the open sets $U$ in the topological manifold $\mathbb{R}$ onto open sets in the Euclidean space $\mathbb{R}^1$ that are compatible with the function $φ(x) = x^3$, the map that sends the point $x$ in the topological manifold $\mathbb{R}$ to the point $x^3$ in the Euclidean space $\mathbb{R}^1$.

Two homeomorphisms $ψ$ and $γ$ from open sets $U$ and $V$ in the topological manifold $\mathbb{R}$ into the Euclidean space $\mathbb{R}^1$ are compatible if $ψ \circ γ^{-1}$ and $γ \circ ψ^{-1}$ are smooth functions from open sets in the Eulcidean space $\mathbb{R}^1$ into itself.

- A function $f$ on either of the two manifolds into the Euclidean space $\mathbb{R}^1$ is smooth if for every coordinate chart in the atlas, $ψ:U→\mathbb{R}^1$ the function $f \circ ψ^{-1}$ from $ψ^{-1}(U)$ into $\mathbb{R}^1$ is smooth. This is a smooth map from the open set $ψ^{-1}(U)$ in the Euclidean space $\mathbb{R}^1$ into the Euclidean space $\mathbb{R}^1$.

Clearly $f(x) = x$ is not smooth on the manifold $(\mathbb{R},x^3)$.

- A homeomorphism between two differentiable manifolds $f:M→N$ is smooth if for every coordinate chart $ψ$ on $M$ and coordinate chart $γ$ on $N$ the map $γ \circ f \circ ψ^{-1}$ is a smooth map from an open set in Euclidean space into Euclidean space. $f$ is a diffeomorphism if it is bijective and its inverse is also smooth.

Consider the map $f:(\mathbb{R},x)→(\mathbb{R},x^3)$ defined by $f(x) = x^{1/3}$. Let $ψ$ be a chart on $(\mathbb{R},x)$ and $γ$ a chart on $(\mathbb{R},x^3)$. By definition $f$ is smooth if $γ \circ f \circ ψ^{-1}$ is a smooth map from an open set in the Euclidean space $\mathbb{R}^1$ into $\mathbb{R}^1$. Equivalently, $f$ is smooth if $(γ\circ x^{1/3}) \circ (x^3 \circ f \circ (x)^{-1} )\circ (x \circ ψ^{-1})$ is smooth. This is a composition of three maps from open sets in the Euclidean space $\mathbb{R}^1$ into $\mathbb{R}^1$. The left and right maps are smooth by compatibility. The middle map is the function $x→x$ and is also smooth. The composition of smooth maps is smooth so $γ \circ f \circ ψ^{-1}$ is smooth.

This shows that $f(x) = x^{1/3}$ is a smooth map from $(\mathbb{R},x)$ to $(\mathbb{R},x^3)$

 

[Ben Niehoff]

A standard, kind of simple example is that of $(\mathbb R, Id.)$ and $(\mathbb R, x^3)$ , which are homeomorphic but not diffeomorphic, although $\mathbb R$ has just one differential structure. Meaning the coordinate maps are, resp. the Id and $x^3$; this last is not differentiable at $x=0$.EDIT: in 4-manifolds, the existence of differentiable structures is given by the mid-homology intersection form. EDIT2: I wrote this quickly last night; as @vanhees71 points out , the issue here is with the inverse, which is not differentiable at $x=0$.

What did you intend this to be an example of, exactly? As you point out, $\mathbb{R}$ has only one differential structure. If I am understanding your notation correctly, I think the point is that $x^3$ only gives you a differential structure on $\mathbb{R}^+$, not on $\mathbb{R}$, since its inverse fails to be differentiable at 0.

Of course, $\mathbb{R}^+$ is homeomorphic to $\mathbb{R}$ via a map like, say, $\log x$. Under such a homeomorphism, you should see that the two differential structures agree (since $\mathbb{R}$ has only one differential structure!).

 

[WWGD]

What did you intend this to be an example of, exactly? As you point out, $\mathbb{R}$ has only one differential structure. If I am understanding your notation correctly, I think the point is that $x^3$ only gives you a differential structure on $\mathbb{R}^+$, not on $\mathbb{R}$, since its inverse fails to be differentiable at 0.

Of course, $\mathbb{R}^+$ is homeomorphic to $\mathbb{R}$ via a map like, say, $\log x$. Under such a homeomorphism, you should see that the two differential structures agree (since $\mathbb{R}$ has only one differential structure!).

It is a self-homeomorphism that is not a diffeomorphism; it is continuous with a continuous, non-differentiable inverse.

 

[Ben Niehoff]

It is a self-homeomorphism that is not a diffeomorphism; it is continuous with a continuous, non-differentiable inverse.

Ah, ok, then I agree.

 

[lavinia]

A standard, kind of simple example is that of $(\mathbb R, Id.)$ and $(\mathbb R, x^3)$ , which are homeomorphic but not diffeomorphic, although $\mathbb R$ has just one differential structure. Meaning the coordinate maps are, resp. the Id and $x^3$; this last is not differentiable at $x=0$.EDIT: in 4-manifolds, the existence of differentiable structures is given by the mid-homology intersection form. EDIT2: I wrote this quickly last night; as @vanhees71 points out , the issue here is with the inverse, which is not differentiable at $x=0$.

Here you say that the two differentiable structures are not diffeomorphic which is why I gave a proof that they are.

 

[WWGD]

Here you say that the two differentiable structures are not diffeomorphic which is why I gave a proof that they are.

Yes, I got mixed up with all the back and forth after the example of $f(x)=x^3$. I said the opposite in post #21, which is what I meant. But, ultimately, $f(x)=x^3$ is an example of a self-map of $(\mathbb R, id)$ which is a homeomorphism, but not a diffeomorphism. I made the post you refer to at the last minute before coffee shop closed down, should have waite dto make post more carefully.

 

[Ben Niehoff]

One can come up with more obvious examples, such as

$$f(x) = \begin{cases} x, & x<0 \\ \frac12 x, & x \geq 0 \end{cases}$$

 

[WWGD]

One can come up with more obvious examples, such as

$$f(x) = \begin{cases} x, & x<0 \\ \frac12 x, & x \geq 0 \end{cases}$$

One can come up with more obvious examples, such as

$$f(x) = \begin{cases} x, & x<0 \\ \frac12 x, & x \geq 0 \end{cases}$$

OR even more so: $f(x)=x , x<0 ; f(x)=x^{13} ,x \geq 0$. By parts you can just patch up " indiferentiabilities" ( equiv. to discontinuities) more clearly, as long as you have two increasing/decreasing pieces. And you can have as many problem points as you wish with piece-wise definitions. And you even have a cusp.

 

[WWGD]

A challenge:
Can anyone think of how to do homeomorphisms that are not diffeomorphisms for $(\mathbb R^2, Id )$ or higher?

 

[Ben Niehoff]

A challenge:
Can anyone think of how to do homeomorphisms that are not diffeomorphisms for $(\mathbb R^2, Id )$ or higher?

The piecewise-linear idea works just as well. You'll have to do some linear algebra to find the boundaries between the pieces, but I'm sure it isn't hard.

 

[WWGD]

It may be possible to just slightly perturb a diffeomorphism slightly to not be differentiable. I would assume that the self-diffeos are nowhere-dense in the space of self-maps.

 

[mathwonk]

as ben said, i would try something like (a,b)-->(a+b,b) on the upper half plane, and (a,b)-->(a,b) on the lower half plane. on the x-axis these agree since then b=0. i.e. this seems not differentiable along the x axis. i.e. ask yourself what the best linear approximation should be say in a neighborhood of (0,0)?

or, even easier, just take products of examples from R^1.