Rich wrote:
Model the (unit) sphere as three unit vectors along the axes, arranged as a 3x3 matrix. Rolling a short distance multiplies this by another matrix I+A, where (the entries in) A are bounded by D, the distance rolled. The difference between (I+A)(I+B) and (I+B)(I+A) is AB-BA, which is bounded by twice the product of the two distances. Any change in contact point, or orientation, is also bounded by the product. For the snowflake, going down a level produces 4x as many edges of 1/3 the length. The difference between rolling along an edge, versus rolling along 4 smaller edges, is proportional to the square of the edge length. The effect of successive levels is multiplied by roughly 4/9 for each level, giving a convergent series.
On the other hand, consider rolling a ball along the 2D graph of a typical 1D Brownian motion. To make things analogous with the iterative Koch construction, we'll use Levy's iterative construction of Brownian motion on [0,1], which at the nth level specifies the function at the dyadic rationals with denominator 2^n (so you can think of the corresponding approximation to the fractal as being a piecewise-linear function with 2^n pieces). When we interpolate, the value of the function at the midpoint of an interval of width 1/2^n is defined as a Gaussian whose mean is the average of the already-chosen values of the function at the endpoints of the interval of width 1/2^n and whose standard deviation is sqrt(1/2^n) (up to a constant that we can ignore). The difference between rolling along an edge, versus rolling along two (not always smaller) edges, is proportional to the square of the edge length. But this edge length is roughly sqrt(1/2^n), so the squared edge-length is roughly 1/2^n, and when we add all 2^n of these contributions together, we find that the difference between successive levels of the construction is O(1), not O((4/9)^n). So we don't appear to get a convergent series. Does this seem right? Or am I missing something? Perhaps a finer analysis would show that you can roll a ball along the 2D graph of 1D Brownian motion (with probability 1, I mean). A related question is, can you roll a ball along the path traveled by a particle doing 2D Brownian motion? For that, matter, we could turn the question around: given a sphere executing a rotational Brownian dance, we could attempt to transfer those fractal jitters to a tangent plane. But I don't know how to define Brownian motion in the configuration space of a sphere. (Is there a general theorem that says that the heat equation, Brownian motion, etc. are all well-defined on a manifold that has properties X, Y, and Z?) Jim Propp