I Difference between diffeomorphism and homeomorphism

[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.

 

[WWGD]

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.

But it seems to restrict to PL manifolds. Can you do it without linearity? And, is it easy to do this as n grows, say for n=17, or so? .How would you do it with, say, an S^n which is not a linear space? EDIT: I know I asked about $\mathbb R^n$ , but I am curious about the more general case. And, I don't know if I unde understood you correctly, but isn't a product function (fxg)(x,y) defined as f(x)g(y), so that the codomain is the Reals, and not $\mathbb R^n$? And if you consider pairs f,g of homeomorphism and
defining h: $\mathbb R \times \mathbb R : h(x,y)=(f(x), g(y))$, is that necessarily a homeomorphism?EDIT 2: Ah, nevermind, we do the same thing for any $\mathbb R^n,$ we just partition into "North and South"

 

[mathwonk]

yes, as you understood, i meant the product functor, which takes a sequence of n functions R-->R and produces one function R^n-->R^n. and of course like all functors it takes isomorphisms to isomorphisms, in any category, since the inverse is just the product of the inverses.

since we are thinking functorially, how aboiut isong the fact that there is a functor from R^n to S^n, namely one point compactification. that should easily make a homeo of R^n into one of S^n. is this ok ?

 

[WWGD]

yes, as you understood, i meant the product functor, which takes a sequence of n functions R-->R and produces one function R^n-->R^n. and of course like all functors it takes isomorphisms to isomorphisms, in any category, since the inverse is just the product of the inverses.

since we are thinking functorially, how aboiut isong the fact that there is a functor from R^n to S^n, namely one point compactification. that should easily make a homeo of R^n into one of S^n. is this ok ?

I am noit sure I have heard of that functor, but, e.g., f(x)=x and f(x)=2x are homeomorphisms of the Reals to themselves, yet the pair g(x,y)=(x,2x) is not a homeomorphim from $\mathbb R^2$ to itself.

 

[mathwonk]

I think the functor would take (x,y) to (x, 2y). No?

to be more precise, the functor takes a sequence of maps (f1,...,fn) each from R->R, to a single map

(f1,...,fn):R^n-->.R^n, defined by (f1,...,fn)(x1,...,xn) = (f1(x1), ..., fn(xn)).

How does that seem?

 

[WWGD]

I think the functor would take (x,y) to (x, 2y). No?

to be more precise, the functor takes a sequence of maps (f1,...,fn) each from R->R, to a single map

(f1,...,fn):R^n-->.R^n, defined by (f1,...,fn)(x1,...,xn) = (f1(x1), ..., fn(xn)).

How does that seem?

Ok, thanks, I see, that makes sense and works.

 

[mathwonk]

let me try to refresh my memory about functors. A functor takes objects of one sort to objects of another sort, as well as taking maps of one sort to maps of the other sort. They also have to satisfy two axioms, namely they take the identity map to the corresponding identity map, and they take compositions to compositions. As a consequence they always take isomorphisms to isomorphisms. here we have a functor taking a sequence of isomorphisms, such as x and 2x, to the isomorphism taking (x,y) to (x,2y). this is forced by the axioms, e.g. since the first axiom says that the identity, i.e. the sequence x-->x and x-->x must go to the identity, i.e. (x,y)-->(x,y). anyway, thanks for checking me.

 

[WWGD]

let me try to refresh my memory about functors. A functor takes objects of one sort to objects of another sort, as well as taking maps of one sort to maps of the other sort. They also have to satisfy two axioms, namely they take the identity map to the corresponding identity map, and they take compositions to compositions. As a consequence they always take isomorphisms to isomorphisms. here we have a functor taking a sequence of isomorphisms, such as x and 2x, to the isomorphism taking (x,y) to (x,2y). this is forced by the axioms, e.g. since the first axiom says that the identity, i.e. the sequence x-->x and x-->x must go to the identity, i.e. (x,y)-->(x,y). anyway, thanks for checking me.

Nice job, Wonk, now maybe we can try to see the domain and codomain categories where the functor is doing its mapping.

 

[mathwonk]

I suppose we could let C be the category of topological spaces and continuous maps, and C^n the category whose objects are sequences of n topological spaces, and sequences of n continuous maps. Then the product functor should go from C^n to C, i.e. it takes a sequence of topological spaces and maps, to a single topological space, and takes a sequence of continuous maps to a single continuous map.

i.e. (X1,...,Xn) goes to X1x...xXn and (f1,...,fn) goes to f1x...xfn.

how's that?

 

[cianfa72]

- 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 $\mathbb{R}^1$.

Great explanation !
Just to make sure I got it right, I believe the open set in the Euclidean space $\mathbb{R}^1$ should read as $ψ(U)$ since $ψ$ is a coordinate chart in the atlas from the open set $U$ in the topological manifold $\mathbb{R}$ into $\mathbb{R}^1$

 

[zinq]

If two smooth manifolds are diffeomorphic, that just means that the functions defining a homeomorphism between them can be chosen so as to be differentiable. Differentiability in differential geometry is usually taken to be "smooth" — which means infinitely differentiable. But also, a diffeomorphism is required to be of maximal rank at each point. (So as I think someone mentioned, the function x ⟼ x³ from the reals to the reals is a homeomorphism that is infinitely differentiable, but it is not a diffeomorphism.)

Up until 1956 it was generally believed that any two smooth manifolds that are homemorphic must also be diffeomorphic. But in that year John Milnor discovered the first examples of homeomorphic smooth manifolds that are not diffeomorphic; these were seven-dimensional spheres S⁷. It was eventually discovered that there are 15 distinct equivalence classes of topological 7-dimensional spheres. (The figure usually cited is 28; this refers to when orientation is also taken into account.) In fact, smooth manifolds of dimensions 1, 2, 3, 5, and 6 always have no alternative ("exotic") differentiable structure. (There are also some topological manifolds that have no differentiable structure at all.) In fact, for all dimensions n ≠ 4, Euclidean space Rⁿ has a unique differentiable structure. But it was discovered in the early 1980s that R⁴ actually has not only infinitely many non-diffeomorphic differentiable structures, but in fact an uncountable infinity of them!