We return to fractals again, this time to examine them in more detail at a basic level. Here I want to compare two simple fractal spaces based on an object known as "the Sierpinski gasket". The most elemental form is depicted in Fig. 1 above and can be compared with Fig 2 which is generated from it.
For Fig. 1, we regard the simple (1/a) space as the generator of more complex fractal spaces F that are essentially infinite. To undertake the cyclic generation process we require that the specification of the vertices (a1, a2, a3) be non-degenerate, i.e. no more than one eigenvalue: a1, a2, or a3 can be assigned for any triangular space. The fundamental space depicted in Fig. 1 we shall call a "Planckian gasket" and note that its fractal mass density can be found as well as its fractal dimension.
The fractal mass density can be computed from:
rho(f) = {N(s) - N(h)}/ N(s)
or the number of scale elements minus "hole" elements divided by scale elements. In this case: N(s) = 1, and N(h) = 0, so:
rho(f) = {1 - 0}/ 1 = 1
The fractal dimension D_f is the inverse, or: D_f = 1/ rho(f)
In this case, D_f = 1
Now, examining the larger space in Fig. 2 defined by {2/a}, we again obtain the vertex designation (a1, a2, a3).
However, we further note that to get from the primitive space to the space {2/a} we require the transposition (see the blog entry on transpositions from two years ago):
(a1, a2, a3) -> (a1, a3, a2)
That is, a1 remains fixed, and a2 - a3 are "mirrored" through a bsiector axis identificed above from a1-a1. Note that the "hole" (in black) represents the inversion space (or negative space: -a1a2a3) a result of the inversion of the positive space of the primitive {1/a}.
The fractal mass density can be computed from:
rho(f) ={4 - 1}/ 4 = 3/4 = 0.75
And the fractal dimension is:
D_f = 1/ rho(f) = 1/ 0.75 = 1.333
Exercise for the adventurous reader: Draw the space for {3/a} with three symmetric holes of oriented vertices a1, a2, and a3 inside it. Then work out the fractal mass density and fractal dimension. We will look at the solution in the next instalment, and also see how algebraic homology can be applied to these triangular spaces!
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