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All: I know any two Cantor sets; "fat" , and "Standard"(middle-third) are homeomorphic to each other. Still, are they diffeomorphic to each other? I think yes, since they are both $0$-dimensional manifolds (###), and any two $0$-dimensional manifolds are diffeomorphic to each other. Still, I saw an argument somewhere where the claim is that the two are not diffeomorphic.

The argument is along the lines that, for $C$ the characteristic function of the standard Cantor set integrates to $0$ , since $C$ has (Lebesgue) measure zero, but , if $g$ where a diffeomorphism into a fat Cantor set $C'$, then: $ f(g(x))$ is the indicator function for $C'$, so its integral is positive.

And (my apologies, I don't remember the Tex for integral and I don't have enough points to look at someone else's edit ; if someone could please let me know )

By the chain rule, the change-of-variable $\int_0^1 f(g(x))g'(x)dx$ should equal $\int_a^b f(x)dx$ but $g'(x)>0$ and $f(g(x))>0$ . So the change-of-variable is contradicted by the assumption of the existence of the diffeomorphism $g$ between $C$ and $C'$.

Is this right?

(###)EDIT: I realized after posting --simultaneously with "Lost in Math"* , that the Cantor sets {C} are not 0-dimensional manifolds (for one thing, C has no isolated points). The problem then becomes, as someone posted in the comments, one of deciding if there is a differentiable map $f:[0,1]\rightarrow [0,1]$ taking C to C' with a differentiable inverse.

  • I mean, who isn't, right?
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    By definition a zero-dimensional manifold is locally homeomorphic to the space consisting of a single point. Thus a zero-dimensional manifold consists of isolated points. In the Cantor set (fat or standard) there is a nonisolated point so it's not a zero-dimensional manifold.2011-10-19
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    I just thought that none of the Cantor sets are manifolds, so it may not make sense to talk about diffeomorphisms between them.2011-10-19
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    Is there then, a definition of diffeomorphism that makes sense between non-manifolds?2011-10-19
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    There is. You can define a smooth function on any set C in R^n as a function that extends to a smooth one on some open neighbourhood U containing C. Then a diffemorphism between your sets is just a smooth function with a smooth inverse. Also, I believe the answer to be no, because your smooth function would take a zero-measure subset to a non-zero one.2011-10-19
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    \int_a^b f(x)\, dx is the TeX code for $\int_a^b f(x)\, dx $2011-10-19
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    Piotr: I think absolutely continuous functions preserve measure, but I don't know if this is true for smooth ones, and is definitely not true for continuous alone, with the function mapping the middle-third Cantor set into [0,1].2011-10-19
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    Jim: thanks; unfortunately, you took away my excuse for not editing, and now I am just lazy :).2011-10-19
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    @Gary: smooth functions are continuously differentiable, and hence (locally) Lipschitz-continuous, and hence absolutely continuous.2011-10-19
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    @Willie: I may have mixed up my terms; diffeomorphism here originally only meant that there is a differentiable map with a differentiable inverse, so I think the issue is more difficult. I think if we have at least $C^1$, then, by compactness, f would be Lipschitz, and , as you said, absolutely continuous. But I don't think differentiable with differentiable inverse is enough to conclude a.c.2011-10-26
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    But I maybe wrong....2011-10-26
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    @Gary: when people say smooth (with no other qualifiers), usually $C^\infty$ is meant. My comment was in response to what you wrote on Oct 19 at 19:22. If you actually meant only differentiable but not necessarily continuously differentiable, then you are right that the argument outlined in my comment won't apply.2011-10-29

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I'm pretty sure Hausdorff dimension is a diffeomorphism invariant. Hausdorff measure of course is not. The basic idea is that if you have a ball of radius $r$ and a diffeomorphism the image of the ball of radius $r$ contains a ball of radius $Mr$ where $M$ is the maximum of the norm of $(f^{-1})'$. Also, the image of the ball is contained in a ball of radius $Nr$, where $N$ is the maximum of the norm of $f'$. Basically you just have to worry about how diffeomorphisms distort the radius of balls (up to inclusion). Diffeomorphisms do so in a tame fashion, provided they're at least $C^1$.

So although all Cantor sets are homeomorphic, up to diffeomorphism you have at least the Hausdorff dimension that separates them -- I think likely there are many more invariants but I haven't given it much thought.

More generally speaking, given a homeomorphism between two metric spaces $f : X \to Y$ which is bi-lipschitz,

$$d(f(x),f(y)) \leq Md(x,y)$$

and

$$d(f^{-1}(x),f^{-1}(y)) \leq Nd(x,y)$$

where $N,M > 0$, the Hausdorff dimension of $A \subset X$ is equal to the Hausdorff dimension of $f(A)$.

A diffeomorphism has the property that it's bi-lipschitz with $M = max ||f'||$ and $N = max ||(f^{-1})'||$.

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    I guess we must be assuming that the Cantor set is embedded in a fixed $\mathbb R^n$. Otherwise ambient dimension is an invariant. In the case of $n=3$, I would think that Antoine's necklace is not diffeomorphic to a standard Cantor set since any neighborhood sees the local linking of the necklace.2011-10-19
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    Right, the notion of "differentiable structure" that I'm using is inherited as a subset of some ambient manifold. So a map $f : X \to Y$ is a diffeomorphism means $X$ and $Y$ are subsets of some manifold, $f$ is 1-1 and onto $Y$, and around every point of $X$ there is a local extension to an open neighbourhood in the manifold that this map is differentiable, similarly for the inverse $f^{-1} : Y \to X$.2011-10-19
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    I see, so diffeomorphisms are sort of like quasi-conformal maps which send small disks to ellipses with controlled eccentricity then, right?2011-10-19
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    Infinitesimally that's what they do, but macroscopic small discs get sent to relatively hard to describe sets, but they're *contained* (and respectively contain) discs whose radius are controlled linearly in terms of the domain disc radius and certain maxima of norms of $f'$ and $(f^{-1})'$. I first saw these details in the proof of the multi-variable change-of-variables theorem for integration -- at some step you have to prove that a diffeomorphism sends measure zero sets to measure zero sets. That's the key moment.2011-10-19
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I think I found an answer to my question, coinciding with the idea in Ryan's last paragraph: absolute continuity takes sets of measure zero to sets of measure zero. A diffeomorphism defined on [0,1] is Lipshitz continuous, since it has a bounded first derivative (by continuity of f' and compactness of [0,1]), so that it is absolutely continuous, so that it would take sets of measure zero to sets of measure zero.