Larry Riddle, Agnes Scott College

## Rotating a Fractal

Let $$\{ f_i : i = 1,2,\dots,n \}$$ be an IFS where each affine transformation $$f_i : R^2 \to R^2$$ is of the form $$f_i (x) = \lambda_i M_ix + b_i$$ with $$0 < \lambda_i < 1$$, $$M_i$$ a rotation matrix, and $$b_i$$ a translation vector. Let $$R$$ be a (counterclockwise) rotation matrix through an angle $$\beta$$ and let $$\alpha > 0$$. Form a new IFS consisting of the functions $$g_i (x) = \lambda_i M_ix + \alpha Rb_i$$. In other words, the linear part of each transformation in the new IFS has not changed, but each translation vector in the original IFS has been rotated counterclockwise through the angle $$\beta$$ and scaled by the factor $$\alpha$$.

Let $$A$$ be the unique attractor for the IFS $$\{ f_i : i = 1,2,\dots,n \}$$. We claim that $$\alpha R(A)$$ is the attractor for the IFS $$\{ g_i : i = 1,2,\dots,n \}$$. First we note that $A = \bigcup_{i=1}^{n} f_i(A) = \bigcup_{i=1}^{n} \left( \lambda_i M_i(A) + b_i \right)$ Since rotation matrices commute, we get that \begin{align} \alpha R(A) &= \alpha R\left( \bigcup_{i=1}^{n} \left( \lambda_i M_i(A) + b_i \right) \right) \\ \\ &= \bigcup_{i=1}^{n} \left( \alpha \lambda_i RM_i(A) + \alpha Rb_i \right) \\ \\ &= \bigcup_{i=1}^{n} \left( \lambda_i M_i \left( \alpha R(A) \right) + \alpha Rb_i \right) \\ \\ &= \bigcup_{i=1}^{n} g_i \left( \alpha R(A) \right) \end{align}

This shows that $$\alpha R(A)$$ is the fixed point for the $$\{g_i\}$$ IFS and thus is the attractor for that IFS.

For example, if we take the IFS for the Sierpinski gasket

$${f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}}$$

$${f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ 0 \\ \end{array}} \right]$$

$${f_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} {1/4} \\ {\sqrt{3}/4} \\ \end{array}} \right]$$

and we want to rotate that through an angle of 135° counterclockwise and scale by $$\frac{1}{\sqrt 2}$$, the new translation vectors would be

$$\frac{1}{{\sqrt 2 }}\left[ {\begin{array}{*{20}{c}} { - \frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}}\\ {\frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\\ 0 \end{array}} \right]$$

$$\frac{1}{{\sqrt 2 }}\left[ {\begin{array}{*{20}{c}} { - \frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}}\\ {\frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {1/2}\\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{4}}\\ {\frac{1}{4}} \end{array}} \right]$$

$$\frac{1}{{\sqrt 2 }}\left[ {\begin{array}{*{20}{c}} { - \frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}}\\ {\frac{{\sqrt 2 }}{2}}&{ - \frac{{\sqrt 2 }}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {1/4}\\ {\sqrt 3 /4} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{1}{8} - \frac{{\sqrt 3 }}{8}}\\ {\frac{1}{8} - \frac{{\sqrt 3 }}{8}} \end{array}} \right]$$

The figure below shows the original Sierpinski gasket in red and the attractor for the new IFS in blue after the rotation and scaling.