We recall that a point $z_0$ is

The point $z_0=\frac 1 2$ is fixed by $q_{{\frac 1 4}}$, and $q_{\frac 1 4}'(z_0)=1$, so this is parabolic. It has been proved that:

The situation is more complicated when $c>\frac 1 4$. The figures below show how the Julia sets $J_{\frac 1 4 + \epsilon}$ evolve as $\epsilon$ decreases to 0 (as we move from right to left). Julia sets $J_{\frac 1 4 + \epsilon}$: "Cauliflower" ($\epsilon=0$) on left; $\epsilon=.0005$ in center; $\epsilon=.01$ on right.

We can see that $J_{\frac1 4 +.0005}$ looks somewhat like $J_{\frac 1 4}$ from the "outside", but on the "inside" there are curlicues; pairs of them are vaguely reminiscent of "butterflies". As $t\to0$, these butterflies persist and remain uniformly large. We think of $t$ as representing time, which decreases to 0. The fact that they suddenly disappear for $t=0$ is the phenomenon called "implosion". Or, if we think of time starting at $t=0$, then the instantaneous appearance of large "butterflies" for $t>0$ may be thought of as "explosion".

The counterpart to ${\cal B}$ for negative time is the

A natural way to represent the basin ${\cal B}$ is to draw the intersection ${\cal B}\cap \Sigma$ in terms of the Fatou coordinate $H$. For instance, for $a = .15$ (and $c$ chosen according to (2)), we find the picture at left. By the Abel-Fatou property, the whole picture is invariant under translation to the right (adding $+1$). The gray region is the complement ${\Bbb C}^2-{\cal B}$, and the rest is ${\cal B}$. We can observe cusps in the boundary that are reminiscent of the cusps in the "Cauliflower" Julia set. The green lines are level sets of the imaginary part of $\Phi_{AF}$, and the pink/gray lines are level sets of the real part. These level sets form rectangles in much of the picture. Where they form octagons, there must be a critical point of $\Phi_{AF}$ inside. Also, the large green "X" indicates a critical point.

In order to have pictures in ${\Bbb C}^2$, which has real dimension 4, we cannot simply draw pictures like the "Cauliflower" above. Thus we draw pictures inside the slice by $\Sigma$.

As examples of the "curlicues" that arise in the "butterflies" we have pictures like the ones on the right. These pictures are invariant under horizontal translation $\zeta\mapsto\zeta+1$ and represent maps of the form (1) with $a$ and $c$ satisfying (2).

In the first two pictures, we have $a=.1$ The bottoms of these pictures show the boundary $(\partial{\cal B})\cap\Sigma$, which can be seen to be the same in both cases. The difference in the "curlicues" arises because $\epsilon\to0$ along different paths. The color pictures on the main page also show examples of imploded sets.

We have $a=.1\sqrt{-1}$ in the next two pictures. The alternating gray/white indicates the complement of ${\cal B}$. (The alternating white/gray was not displayed when the first two pictures were made.)

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