Larry Riddle, Agnes Scott College

Construction

Animation

Let K = {f_{1}, f_{2}} be the two functions in the Golden Dragon IFS. Let G = {g_{1}, g_{2}} be the Z_{2} cyclic group. Then g_{1} is the identity element and g_{2} is rotation through 180°. We form a new IFS by applying each of the symmetries in G to each of the functions in K to form the set of affine transformations GK = {g_{1}f_{1}, g_{2}f_{1}, g_{1}f_{2}, g_{2}f_{2}}. The first iteration would transform the horizontal line segment into the following figure using the four functions in this set. The red line segments would be the results of the usual Golden dragon IFS and the blue line segments are the rotations by 180°.

Recall that \({\bf{r}} = \left( \dfrac{1}{\phi}\right)^{\frac{1}{\phi}} = 0.74274\) where \(\phi\) is the golden ratio, and that A = 32.893818° and B = 46.98598225°. Now continue to repeat the construction on each of the four new segments to get the Z_{2} Golden dragon.

Function

System

IFS

Animation

(with squares)

The IFS for the Golden Z_{2} dragon will consist of four functions: the original two functions from the Golden dragon IFS and the rotation of those two functions through 180°.

\({h_1}({\bf{x}}) = {g_1}{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{0.623653} & { - 0.403372} \\
{0.403372} & {0.623653} \\
\end{array}} \right]{\bf{x}}\) |
scale by r, rotate by 32.894° |

\({h_2}({\bf{x}}) ={g_2}{f_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{-0.623653} & { 0.403372} \\
{-0.403372} & {-0.623653} \\
\end{array}} \right]{\bf{x}}\) |
scale by r, rotate by −147.106° |

\({h_3}({\bf{x}}) ={g_1}{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ - 0.376347} & { - 0.403372} \\
{0.403372} & { - 0.376347} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
1 \\
0 \\
\end{array}} \right]\) |
scale by r^{2}, rotate by 133.014° |

\({h_4}({\bf{x}}) = {g_2}{f_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}}
{ 0.376347} & { 0.403372} \\
{-0.403372} & { 0.376347} \\
\end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}}
-1 \\
0 \\
\end{array}} \right]\) |
scale by r^{2}, rotate by −46.986° |

If H is the unique attractor for this IFS, then

\[H = h_1(H) \cup h_2(H) \cup h_3(H) \cup h_4(H) \]
Rotating H by 180° counterclockwise is the same as applying the function g_{2} from the cyclic group Z_{2} to the set H. Because g_{2}g_{2} = g_{1} (identity) and g_{2}g_{1} = g_{2}, we get

This shows that H has 180° rotational symmetry. Also notice that h_{1}(H) = h_{2}(H). That is because h_{2} does the same scaling to H as h_{1} but the two rotations differ by exactly 180°. Because of the 180° rotational symmetry of H, however, the result will be the same in both cases. You can see what happens by clicking on each of the buttons to the left to cover H with the four scaled and rotated copies.

- Michael Field and Martin Golubitsky, "Symmetric Fractals" (chapter 7) in
*Symmetry in Chaos: A Search for Pattern in Mathematics, Art and Nature*, Oxford University Press (Edition 1, 1992/1995) and Siam (Edition 2, 2009) [see Preview at Google Books].