- #1

Silviu

- 624

- 11

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Silviu
- Start date

- #1

Silviu

- 624

- 11

- #2

- 17,220

- 17,190

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

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.

- #3

Silviu

- 624

- 11

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)?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.

- #4

- 17,220

- 17,190

It's hard to tell what you have read.Thank you for your reply. It makes sense, however I read ...

Here's what Wikipedia says about it:... that diffeomorphically inequivalent homeomorphisms arise only for dim 4 or higher.

Fordifferentiable (smooth) manifoldsin dimension less than 4, homeomorphism always implies diffeomorphism: twodifferentiable (smooth) manifoldsof the dimension less than or equal to 3, which are homeomorphic, are also diffeomorphic. That is,if there isa homeomorphism,then there is alsoa diffeomorphism.This does not mean that any homeomorphism would be a diffeomorphism.

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.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)?

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.

Last edited:

- #5

- 19,539

- 9,906

- #6

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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.

- #7

- 17,220

- 17,190

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

- #8

- 19,539

- 9,906

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.

- #9

- 17,220

- 17,190

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

- #10

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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##.

Last edited:

- #11

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

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##.

Last edited:

- #12

- 21,130

- 12,011

- #13

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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?

- #14

- 17,220

- 17,190

I thought the derivative had to be continuous to establish a diffeomorphism.Let ##F: (\mathbb R, Id.) → (\mathbb R, x^3)## by ##F(x) = x^{1/3}##.

Isn't ##F## a diffeomorphism?

- #15

- 21,130

- 12,011

- #16

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

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

- #17

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

- #18

- 17,220

- 17,190

... 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.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##.

- #19

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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.

- #20

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

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

- #21

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

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 ##.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.

- #22

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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

- #23

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

But the original map is from ## ( \mathbb R,Id ) ## to itself., and your argument does not hold for this case. I think so.Didn't I show that the map is a diffeomorphism?

- #24

lavinia

Science Advisor

Gold Member

- 3,283

- 673

No it is from ## ( \mathbb R,Id ) ## to ## ( \mathbb R,x^3) ##But the original map is from ## ( \mathbb R,Id ) ## ) to itself.

- #25

WWGD

Science Advisor

Gold Member

- 6,037

- 7,349

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

- #26

- 17,220

- 17,190

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

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.

- #27

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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

Yes.

- #28

- 17,220

- 17,190

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.Yes.

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.

- #29

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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)##.

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)##.

Last edited:

- #30

- 17,220

- 17,190

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?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)##.

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.

Last edited:

- #31

lavinia

Science Advisor

Gold Member

- 3,283

- 673

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.

- #32

- 17,220

- 17,190

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.

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.

Last edited:

- #33

- 21,130

- 12,011

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

- #34

- 17,220

- 17,190

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

- #35

- 21,130

- 12,011

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

Share:

- Last Post

- Replies
- 11

- Views
- 4K

- Last Post

- Replies
- 13

- Views
- 16K

- Replies
- 1

- Views
- 4K

- Replies
- 10

- Views
- 4K

- Last Post

- Replies
- 12

- Views
- 7K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 4K

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 5K