# Why does the boundary of the Mandelbrot set contain a cardioid?

In a comment to a previous answer it has been mentioned that the boundary of the Mandelbrot set contains the cardioid $$c = e^{it} \, \frac{2 - e^{it}}{4}$$ but how can we prove this?

• One comment on the linked comment: I retract my statement that the boundary of the Mandelbrot set contains a line segment. Jun 27 '15 at 14:59

The cardioid is the boundary of the set of $c$s for which the iteration of $$z\mapsto z^2+c$$ has an attractive fixpoint.

Given $c$ we can find the fixpoints easily by solving a quadratic equation, and we get $$z_* = \frac{1\pm\sqrt{1-4c}}2$$ From the general theory of iterated systems, we know that the fixpoint is attractive if $\left|\frac d{dz}(z^2+c) \right| < 1$ at $z_*$, which works out to $|z_*|<\frac12$. Thus we ought to find the boundary of the area in question by equating $z_*$ to $\frac12 e^{it}$. From $$\frac12 e^{it} = \frac{1\pm\sqrt{1-4c}}{2}$$ a bit of simple algebra will give you $$c = e^{it}\frac{\pm2-e^{it}}4$$ as claimed. (And the solution with $\pm2=-2$ just corresponds to taking $e^{i(t+\pi)}$ instead).

Knowing that the boundary of "the area where the iteration has an attractive fixpoint" is included in the boundary of the entire Mandelbrot set is more involved, and I don't know how to prove that.

• Yes, this is essentially how I would have responded. The fact that this cardioid lies on the boundary of the Mandelbrot set can be proved using rational external rays. It turns out that every point of the form $e^{2\pi i t}(2-e^{2\pi i t})$ for rational $t$ is the base of a bulb off of the main cardioid and that some external ray lands exactly at that point. Jun 27 '15 at 14:29

As Henning Makholm mentioned, the interior of the main cardioid of the Mandelbrot set is precisely the set of $c$-values for which $z^2 + c$ has an attracting fixed point. All such points lie in the Mandelbrot set, since the corresponding filled Julia set must contain the basin of attraction of the attracting fixed point, and hence cannot be totally disconnected.

So the question is, how do we know that the cardioid itself is contained in the boundary of the Mandelbrot set?

One answer is that parabolic parameters are dense on the boundary of the cardioid. A fixed point $p$ for a holomorphic function $f$ is said to be parabolic if $f'(p)$ is a root of unity. The dynamics near such points are well understood -- see Chapter 10 of Milnor, Dynamics in One Complex Variable. A parameter value $c$ on the cardioid is called a parabolic parameter if $f(z) =z^2 +c$ has a parabolic fixed point. (More generally, a point in the Mandelbrot set is called a parabolic parameter if $z^2+c$ has a parabolic periodic point.)

Now, the cardioid is precisely the set of parameter values for which $f (z) = z^2 +c$ has a neutral fixed point, i.e. a fixed point $p$ with $|f'(p)|=1$. Indeed, if $\sqrt{-}$ denotes the branch of the square root function with a cut along the negative real axis, then the map $$c \;\mapsto\; 1-\sqrt{1-4c}$$ is a homeomorphism from the cardioid to the unit circle, and maps each parameter value $c$ to the derivative of $z^2+c$ at the neutral fixed point. Since the roots of unity are dense on the unit circle, it follows that the parabolic parameters are dense on the main cardioid. (You can actually see this in the picture of the Mandelbrot set -- it turns out that the parabolic parameter values on the cardioid are precisely the points at which the main cardioid touches the boundaries of the adjacent hyperbolic components.)

Now, it is known that each parabolic parameter in the Mandelbrot set is the landing point of exactly two external rays -- see this paper by Dierk Schleicher for a proof. Then the parabolic parameters must be in the closure of the complement, so each parabolic parameter value lies on the boundary of the Mandelbrot set. Since the parabolic parameters are dense in the cardioid and the boundary is a closed set, it follows that the entire cardioid is contained in the boundary.

A similar argument shows that, if $c$ is any parameter value for which $z^2+c$ has a neutral periodic point (i.e. a point $p$ of period $k$ for which $|(f^k)'(p)|=1$), then $c$ must lie on the boundary of the Mandelbrot set.

The Mandelbrot set is a compact set as the intersection of decreasing compact sets. If an open set is included into a compact set, so is its closure. Apply this result to the interior of the cardioid to show that its boundary is also included into the Mandelbrot set

• That shows just that the boundary of the cardioid is in the Mandelbrot set, not that it is in the boundary of the Mandelbrot set. Oct 3 '17 at 4:02