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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. –  LostInMath Oct 19 '11 at 18:56
I just thought that none of the Cantor sets are manifolds, so it may not make sense to talk about diffeomorphisms between them. –  gary Oct 19 '11 at 18:57
Is there then, a definition of diffeomorphism that makes sense between non-manifolds? –  gary Oct 19 '11 at 18:59
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. –  Piotr Pstrągowski Oct 19 '11 at 19:04
\int_a^b f(x)\, dx is the TeX code for $\int_a^b f(x)\, dx $ –  Grumpy Parsnip Oct 19 '11 at 19:15
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2 Answers 2

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,

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


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

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. –  Grumpy Parsnip Oct 19 '11 at 22:04
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 $f^{-1} : Y \to X$. –  Ryan Budney Oct 19 '11 at 22:06
I see, so diffeomorphisms are sort of like quasi-conformal maps which send small disks to ellipses with controlled eccentricity then, right? –  gary Oct 19 '11 at 22:09
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. –  Ryan Budney Oct 19 '11 at 22:12
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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.

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