Self map of the disk: does a large degree on the boundary imply a fixed point in the interior?

Question

Lets's suppose we have a continuous map of the disk

$$f: \mathbb D \to \mathbb D,$$

where $\mathbb D := \{z \in \mathbb C\,:\, |z|\le 1 \}$.

We know by the Brouwer fixed point theorem that $f$ has a fixed point in $\mathbb D$. But lets's suppose further that $f$ preserves the unit circle $S^1 = \{z\,:\,|z| = 1\}$ (i.e. the boundary of $\mathbb D$):

$$f(S^1) = S^1$$

and that $f$ restricted to this circle is a degree $n \ge 2$ map. That is, it's action on the fundamental group,

$$f_\ast: \pi_1(S^1) \to \pi_1(S^1)$$

takes $[\gamma] \longmapsto [\gamma]^n\ $ for $[\gamma]\in \pi_1(S^1)$.

Question:

Is it true that $f$ has a fixed point in the interior of $\mathbb D$?

Answer 1

This claim is simply false, here is an example. I will identify the circle $S^1$ with the unit circle in the complex plane. Then $D$ is the closed disk $\{|z|\le 1\}$.

First of all, for $n\ge 2$ consider the map $h: S^1\to S^1$ of degree $n$, $h(z)=z^n$.

I will define a map $f: D\to D$ by "coning off" from $1$: Consider the family of round circles $S_t$ (of radii $t\in [0,1]$) contained in $D$ and tangent to $S^1$ at $1$. For each $S_t$ let $g_t: S_t\to S^1$ denote the Euclidean dilation fixing $1$. Let $$ r(t)=t^2. $$ This map has no fixed points in the interval $(0,1)$.

For each $t\in [0,1]$ define the map $$ h_t= g_{r(t)}^{-1} \circ h \circ g_t: S_t\to S_{r(t)}. $$ Lastly, define $f: D\to D$ by combining the maps $h_t$: Every $z\in D$ belongs to a unique circle $S_t$. Then set $$ f(z)= h_t(z). $$ I will leave it to you to verify continuity to $f$. I claim that $f$ has no fixed points in the interior of $D$. Indeed, if $z\in S_t$ for $t\in (0,1)$ then $f(z)\in S_{r(t)}$. But $S_{r(t)}\cap S_t=\{1\}$ for $t\ne 1$. Hence, $z$ cannot be fixed by $f$ provided $|z|<1$.

Thus, all the fixed points of $f$ are on the boundary circle $S^1$. The restriction of $f$ to $S^1$ is the map $h$ which has degree $n$.

Answer 2

I believe I have an explicit counterexample which does the same thing: (a hyperbolic isometry squared)

$$f: z \longmapsto \left(\frac{3z+1}{3+z}\right)^2.$$

Maps the disk into the disk and fixes the boundary

Firstly, its a holomorphic map which preserves the circle and the disk, which can be seen by checking a few values:

$$(1,\ -1,\ i,\ 0) \longmapsto (1,\ 1,\ \frac7{25}-\frac{24}{25}i,\ \frac 19).$$

The first three mean it preserves the unit circle, the last means the interior of the disk is mapped to itself.

Has only one fixed point, which is on the boundary

$$f(z) - z = \frac{(1-z)^3}{(3+z)^2},$$

so the only fixed point of $f$ is $1 \in S^1$.

Restricts to a degree 2 map on the boundary

I'm not really sure how you check this in practical terms, besides a winding number computation:

$$\frac1{2\pi}\oint_{f \,\circ\, \gamma} \frac1z \mathrm dz = \int_0^{2\pi}\frac{f'(e^{2\pi i \theta})}{f(e^{2\pi i \theta})}e^{2\pi i \theta}\mathrm d \theta = \cdots =\int_0^{2\pi}\frac{8 }{3 \cos (2 \pi \theta )+5}\mathrm d \theta = 2$$

for $\gamma:[0,1]\to S^1$ is the "standard generator" $\gamma(t) = \exp (2\pi i t)$.

In other words, $f$ takes this curve of winding number 1 onto one of winding number 2; and is thus doubling.