# Examples of smooth fractals

A classic example of a fractal curve is the Koch Snowflake. This is a topological manifold (as opposed to many other fractals which are not), but it also clearly not smooth.

Question: Are there any curve-type fractals that are actually smooth? Or does the infinite self-similarity eventually pose an insurmountable barrier to smoothness?

Technically speaking, $\mathbb{R}$ is a smooth fractal too, so for the above question, I'd only introduce the caveat that the curve be 'interesting' as a fractal (or at least non-trivial).

Intuitively, I see no reason for such objects to not exist, but this is far from any area of math I'm familiar with.

• Do you have a precise definition of a fractal in mind? If you're just after self-similarity, a line would do.
– mrf
Jul 28, 2016 at 14:38
• Well the Koch snowflake is not smooth at any iteration, but you could smooth it out at each iteration by replacing the points of the triangles with semi-circles or something. That would be $C^1$ anyway, of not $C^{\infty}$. Do you think in the limit it loses its roundedness and becomes non-smooth? Hard to picture... Jul 28, 2016 at 14:40
• @mrf Nope, just self-similarity. That's why I put the 'interesting' caveat in at the end! Jul 28, 2016 at 14:45
• @GregoryGrant I would, purely speculatively, imagine it would lose its smoothness in the limit. That's basically exactly what prompted this question. Jul 28, 2016 at 14:45
• @PeteCaradonna Maybe it goes the other way, maybe the Koch Snoflake smooths out in the limit even though each iteration is pointy. How do you know for sure it is not smooth? Jul 28, 2016 at 14:48

Nash's isometric embedding theorem gives an affirmative answer to your question. Visualisation of such an example of smooth fractal can be made on google images typing "isometric embedding gnash torus".

Here is a link to a picture of the sphereflake fractal.

https://en.wikipedia.org/wiki/Koch_snowflake#/media/File:Sf6.jpg

Here is a link to a page from a textbook:

• If I understand the construction correctly, the sphere-flake is not even a manifold, let alone smooth, at all such points at which the iterative construction affixes a sphere. It's a step in the direction I'm looking for though because it's clearly smooth somewhere unlike the snowflake, but I'm really looking for something smooth everywhere. Jul 28, 2016 at 18:41
• Upon further consideration it seems possible to inflate those points and get something smoothed out. Thanks! Jul 28, 2016 at 18:57
• This clearly does not answer the question. It is not smooth and any attempt on smoothing will destroy self similarity. Jul 29, 2016 at 1:53

I think, it depends on the precise notion of self-similarity and smoothness in your question. In my answer I consider planar curves which are self-similar with respect to affine transformations. De Rham curves provide examples of self-similar curves in this sense. Parabola is self-similar and smooth. More interestingly, there are other de Rham curves which are merely $C^1$-smooth, see the discussion in:

V.Protasov, On the regularity of de Rham curves, Izvestia Math., 2007

freely available here.

My guess is that the only $C^2$-smooth affine self-similar planar curves are algebraic of degree $\le 2$.

Update. It took me awhile to find proper references, but here it goes. First of all, by a self-similar subset of $E^n$ I will mean the limit set $\Lambda$ of an "iterated functional system" of a collection of contracting maps $S_1,...,S_k$ of $E^n$, which belong to some group $G$ of transformations of the Euclidean space or the extended Euclidean space. The most commonly considered groups are:

1. The group $Sim(E^n)$ os Euclidean similitudes (compositions of Euclidean rigid motions and dilations).

2. The group $Aff(E^n)$ of affine transformations.

3. The group $Mob(S^n)$ of Moebius transformations.

I will restrict to the case $n=2$, just for simplicity, much of what I will say holds in higher dimensions as well.

1. In the case $G=Sim(E^2)$, if the self-similar set $\Lambda$ is a differentiable curve (or, even weaker is a curve of Hausdorff dimension 1) then $\Lambda$ is a subset of a straight line. See

V. Mayer, M. Urbański, Finer geometric rigidity of limit sets of conformal IFS. Proc. Amer. Math. Soc. 131 (2003), no. 12, 3695–3702.

1. In the case $G=Aff(E^2)$, the self-similar set can be $C^1$-smooth (see above), but if it is $C^2$ then it has to be contained in a straight line or a parabola, see

C. Bandt, A. Kravchenko, Differentiability of fractal curves. Nonlinearity 24 (2011), no. 10, 2717–2728.

1. In the case $G=Mob(S^2)$ if a self-similar set is a differentiable curve (or, even weaker is a curve of Hausdorff dimension 1) then $\Lambda$ is a subset of a straight line or of a round circle. See the reference in 1.