The Generalized Mandelbrot Set

In this section, we want to give a version of the Mandelbrot set for the bicomplex numbers. First, we recall the definition of the Mandelbrot set for the complex plane:

Definition 2   Let $P_{c}(z)=z^2+c$ where $z,c\in\mathbb {C}$ and $P_{c}^{\circ n}:=(P_{c}^{\circ (n-1)}\circ P_{c})(z)$. Then the Mandelbrot set is defined as follows: $\mathcal{M}=\{c\in\mathbb {C}: P_{c}^{\circ n}(0) \mbox{ is bounded } \forall n\in\mathbb {N}\}$. When we take $z, c\in\mathbb {C}_{1}$, we denote the Mandelbrot set by $\mathcal{M}_{1}$.

Figure 1 gives an illustration of the Mandelbrot set with some of its ``filled-Julia" sets. In fact, our figure is a rotation by $90^{\circ}$ of the original Mandelbrot set. This rotation will give a better vantage point when we shall work on our version of the Mandelbrot set in $\mathbb {R}^{3}$. Also, the colours around the Mandelbrot set have been determined by the number of iterations needed before $\vert P_{c}^{\circ n}(0)\vert>2$. This is well justified by the fact that the Mandelbrot set can also be characterized as follows: $\mathcal{M}=\{c\in\mathbb {C}:\vert P_{c}^{\circ n}(0)\vert\leq 2\mbox{ }\forall n\in\mathbb {N}\}$ [1]. Then, the colours give information about the manner in which the algorithm for the Mandelbrot set diverges to infinity. This information will be almost the only possible one to approach our version of the Mandelbrot set in dimension three.

We also recall the following beautiful property of the Mandelbrot set [3]:

Theorem 1 (Douady and Hubard, 1982)   The Mandelbrot set $\mathcal{M}$ is connected.

Now, to give a version of the Mandelbrot set for the bicomplex numbers we have only to reproduce the algorithm of Definition 2 for the bicomplex numbers. This is the next definition.

Definition 3   Let $P_{c}(w)=w^2+c$ where $w,c\in\mathbb {C}_{2}$ and $P_{c}^{\circ n}(w):=(P_{c}^{\circ (n-1)}\circ P_{c})(w)$. Then the generalized Mandelbrot set for bicomplex numbers is defined as follows: $\mathcal{M}_{2}=\{c\in\mathbb {C}_{2}: P_{c}^{\circ n}(0) \mbox{ is bounded } \forall n\in\mathbb {N}\}$.

The next lemma is a characterization of $\mathcal{M}_{2}$ using only $\mathcal{M}_{1}$. This lemma will be useful to prove that $\mathcal{M}_{2}$ is also a connected set.

Lemma 1   $\mathcal{M}_{2}=\mathcal{M}_{1}\times_{e}\mathcal{M}_{1}$.

Proof. First, we prove that $\mathcal{M}_{2}\subseteq\mathcal{M}_{1}\times_{e}\mathcal{M}_{1}$. Let $c\in\mathbb {C}_{2}$ such that $P_{c}^{\circ n}(0)$ is bounded $\forall n\in\mathbb {N}$. We have

$$P_{c}(w)=w^{2}+c=[(z_1-z_2i_1)^{2}+(c_1-c_2i_1)]e_1+[(z_1+z_2i_1)^{2}+(c_1+c_2i_1)]e_2$$

where $w=(z_1-z_2i_1)e_1+(z_1+z_2i_1)e_2$ and $c=(c_1-c_2i_1)e_1+(c_1+c_2i_1)e_2$. Then,

$$P_{c}^{\circ n}(w)=P_{c_1-c_2i_1}^{\circ n}(z_1-z_2i_1)e_1+P_{c_1+c_2i_1}^{\circ n}(z_1+z_2i_1)e_2.$$

By hypothesis,

$$P_{c}^{\circ n}(0)=P_{c_1-c_2i_1}^{\circ n}(0)e_1+P_{c_1+c_2i_1}^{\circ n}(0)e_2\mbox{ is bounded } \forall n\in\mathbb {N}.$$

Hence, $P_{c_1-c_2i_1}^{\circ n}(0)$ and $P_{c_1+c_2i_1}^{\circ n}(0)$ are also bounded $\forall n\in\mathbb {N}$. Then $c_1-c_2i_1$, $c_1+c_2i_1\in\mathcal{M}_{1}$ and $c=(c_1-c_2i_1)e_1+(c_1+c_2i_1)e_2\in\mathcal{M}_{1}\times_{e}\mathcal{M}_{1}$.

Conversely, if we take $c\in\mathcal{M}_{1}\times_{e}\mathcal{M}_{1}$, we have $c=(c_1-c_2i_1)e_1+(c_1+c_2i_1)e_2$ with $c_1-c_2i_1$, $c_1+c_2i_1\in\mathcal{M}_{1}$. Hence, $P_{c_1-c_2i_1}^{\circ n}(0)$ and $P_{c_1+c_2i_1}^{\circ n}(0)$ are also bounded $\forall n\in\mathbb {N}$. Then $P_{c}^{\circ n}(0)$ is bounded $\forall n\in\mathbb {N}$, that is $c\in\mathcal{M}_{2}$. $\Box$

Theorem 2   The generalized Mandelbrot set $\mathcal{M}_{2}$ is connected.

Proof. Define a mapping $e$ as follows:

$$\mathbb {C}^{2}_1=\mathbb {C}_1\times \mathbb {C}_1\stackrel{... ...rrow} \mathbb {C}_{1}\times_{e} \mathbb {C}_{1}=\mathbb {C}_{2}$$

$$\hspace{0.6cm} (w_1,w_2)\longmapsto w_1e_1+w_2e_2.$$

The mapping $e$ is clearly a homeomorphism. Then, if $X_1$ and $X_{2}$ are connected subsets of $\mathbb {C}_{1}$ we have that $e(X_{1}\times X_{2})=X_{1}\times_{e} X_{2}$ is also connected. Now, by Lemma 1, $\mathcal{M}_{2}=\mathcal{M}_{1}\times_{e}\mathcal{M}_{1}$. Moreover, by Theorem 1, $\mathcal{M}_{1}$ is connected. It follows, if we let $X_1=X_2=\mathcal{M}_{1}$, that $\mathcal{M}_{2}$ is connected.$\Box$