Rolling a sphere on a fractal curve

Question

Given a rectifiable curve C : [0, 1] → R² in the plane, it makes sense to roll a unit sphere S² on the plane along that curve and ask what is its net rotation in SO(3).

I wonder if this also makes sense for a wider class of curves than just the rectifiable ones.

For simplicity let C : [0, 1] → R² be a homeomorphism onto its image C([0, 1]), and assume C([0, 1]) has constant Hausdorff dimension d with 1 < d < ∞. (I.e., every neighborhood of every point of C([0, 1]) has the same Hausdorff dimension d.)

Is it possible that, although C([0, 1]) is not rectifiable, it nevertheless makes sense to ask what the net rotation is of a unit sphere rolled along it from beginning to end?

For definiteness, we can consider the "Koch curve" (2nd paragraph of https://mathworld.wolfram.com/KochSnowflake.html) that is one-third of the Koch snowflake curve.

Call this Koch curve W. W is the limit of a sequence of standard stages W_n as n → ∞.

Its Hausdorff dimension is easily computed to be log₃4 = 1.2618595+. Since this exceeds 1, it cannot be rectifiable.

If net rotation makes sense for this curve, what element of SO(3) is it? (Assume that Stage 1, the "witch's hat" of four equal line segments, has euclidean length = 4/3.)

I would be happy to define this concept as the limit as n → ∞ of the net rotation of S² along the nth stage W_n if that limit exists.

Answer 1

Numerically on the Koch curve, it seems to quickly converge. Symbolically, it is tedious to figure out any closed formula. At stage $n$, rolling the unit sphere on the $(x,y)$-plane along $[0,\alpha]\times\{0\}\times\{0\}$ with $\alpha=3^{-n}$ rotates it around the $y$-axis with the matrix $A := A(\alpha):=\left(\begin{smallmatrix}\cos\alpha&0&\sin\alpha\\0&\;1\;&0\\-\sin\alpha&0&\cos\alpha\end{smallmatrix}\right)$. We only need to compose it with the $\beta:=\pi/3$ rotation around the $z$-axis whose matrix is $B:=\left(\begin{smallmatrix}\cos\beta&\sin\beta&0\\-\sin\beta&\cos\beta&0\\0&0&\;1\;\end{smallmatrix}\right)$ according to the recurrence $$R_{n,0} := A(3^{-n})\;,\;\;R_{n,k+1} := R_{n,k}BR_{n,k}B^{-2}R_{n,k}BR_{n,k}\,.$$ The net rotation is $R_{n,n}$. In PARI/GP with 512-bit precision we instantaneously get $R_{100,100} $. Its first 20 decimals digits stabilize starting with $R_{56,56}$. Here they are : $$\begin{pmatrix} \;\;\;0.54420\,52646\,18151\,51477 & \;\;\;0.07439\,24346\,27331\,14793 & \;\;\;0.83564\,72914\,04756\,45330\\ -0.07439\,24346\,27331\,14793 & \;\;\;0.99641\,61277\,92991\,15095 & -0.04025\,74955\,03816\,29478\\ -0.83564\,72914\,04756\,45330 & -0.04025\,74955\,03816\,29478 & \;\;\;0.54778\,91368\,25160\,36382\end{pmatrix}$$ Does anybody recognize something there?

EDITS:

1. I prefixed most transforms by a rotation $B^m$ around the $z$-axis and suffixed them with the inverse rotation $B^{-m}$ so as to keep rolling along the $x$-axis and benefit from repeated calculations. It amounts to look at the oriention of the next straight segment we want to roll the sphere over, rotate the whole plane, curve and sphere together so as the straight segment now looks in the unchanged $x$-direction and orientation, roll, and rotate the whole back.

2. What if the limit net rotation depended on the sequence of approximating curves we used? For example the Koch curve can also be approximated "from above" by carving in a roof starting as two line segments of length $1/\sqrt{3}$ at angle $\pm\pi/6$ from the $x$-axis. Fortunately, in that specific example we get the same numerical limit as with the previous calculation. This hints at celebrating the net rotation as a universal property of fractals!

3. Here is my PARI-GP code :

default(realbitprecision, 512);
R(n,k) = if(!k, A, my(M=R(n,k-1)); M*B*M*B^-2*M*B*M);
f(n) = { \\ Standard approximation by growing a unit segment
  my(a = 3^-n, b = Pi/3); 
  A = [cos(a), 0, sin(a); 0, 1, 0; -sin(a), 0, cos(a)];
  B = [cos(b), sin(b), 0; -sin(b), cos(b), 0; 0, 0, 1];
  R(n,n)
};
g(n) = { \\ Alternative approximation by carving in a roof
  my(a = 3^(-1/2-n), b = -Pi/3, c=Pi/6); 
  A = [cos(a), 0, sin(a); 0, 1, 0; -sin(a), 0, cos(a)];
  B = [cos(b), sin(b), 0; -sin(b), cos(b), 0; 0, 0, 1];
  C = [cos(c), sin(c), 0; -sin(c), cos(c), 0; 0, 0, 1];
  C*R(n,n)*C^-2*R(n,n)*C
};
f(100)
g(100)

4. To get a closed formula, we may try to describe $R = \lim\limits_{n \to \infty}R_{n,n}$ as a fixed point of some transform. Or see how $R$ behaves as the initial unit segment is replaced by a very small segment. Maybe we'll recognize the coefficients in the matrix as a known taylor series in the segment length.