Disc Retraction

Thursday, Aug 24, 2023

We sketch here some more details of the construction of a retraction $\rho: D^2 \to S^1$ from the disc $D^2$ to the circle $S^1$ given a map $f: D^2 \to D^2$ without fixed points.

Geometric construction

1. For a point $x\in D^2$, let $\lambda_x: [0, \infty) \to \mathbb{R}^2$ be the ray from $f(x)$ through $x$, with $\lambda_x(1)= 1$.
2. By convexity, for $\lambda_x(s)\in (0, 1)$, $\lambda(s)$ is in interior of the disc $D^2$.
3. Hence, there is a unique $\sigma_x\geq 1$ so that$\lambda_x(\sigma_x)$ lies on $S^1$.
4. Define $\rho(x) = \lambda_x(\sigma_x)$.

Algebraic construction

1. Fix $x\in D^2$.

2. Define $\lambda_x: [0, \infty) \to \mathbb{R}^2$ by $\lambda_x(s) = (1-s)f(x) + sx$.

3. By convexity, for $\lambda_x(s)\in (0, 1)$, $\Vert\lambda(s)\Vert< 1$.

4. Consider the equation $\Vert\lambda_x(s)\Vert^2 = 1$, i.e., $\Vert (1-s)f(x) + sx\Vert^2=1$.

5. This can be written as $$\Vert x - f(x)\Vert^2 s^2 + 2\langle x - f(x), f(x)\rangle s - (1 - \Vert f(x)\Vert^2) = 0.$$ which is of the form $a_xs^2 + b_xs + c_x$ = 0 with $a_x = \Vert x - f(x)\Vert^2$, $b_x = 2\langle x - f(x), f(x)\rangle$, $c_x = -(1 - \Vert f(x)\Vert^2)$.

6. As $f$ has no fixed points, $a = \Vert x - f(x)\Vert^2 > 0$ for all $x$. Thus, the equation is a non-degenerate quadratic equation. Further, by compactness there exists $\epsilon > 0$ such that $\Vert x - f(x)\Vert^2 \geq \epsilon$ for all $x$.

7. We will consider the larger real root of this equation. First we establish that there are distinct real roots.

8. The discriminant of the equation is $$D = D_x = 4\langle x - f(x), f(x)\rangle^2 + 4\Vert x - f(x)\Vert^2(1 - \Vert f(x)\Vert^2)$$.

9. As $\Vert f(x)\Vert^2 \leq 1 1$, both the terms of $D$ are non-negative. Hence, $D\geq 0$.

10. Further, if $D=0$, then $\langle x - f(x), f(x)\rangle = 0$ and $\Vert f(x)\Vert^2 = 1$. We deduce (by Pythagoras theorem or properties of inner products) that $\Vert x\Vert^2 = \Vert f(x)\Vert^2 + \Vert x - f(x)\Vert^2 > 1 + \epsilon$, which is a contradiction as $x \in D^2$.

11. We can thus define $\sigma_x$ to be the larger real root of the equation $\Vert\lambda_x(s)\Vert^2 = 1$ using the quadratic equation solution formula $$\sigma_x = \frac{-b_x + \sqrt{D_x}}{2a_x}.$$

12. This is continuous as both the denominator and the term under the square root are positive.

13. Define $\rho(x) = \lambda_x(\sigma_x)$. This is a composition of continuous functions and hence continuous.

14. By construction of $\sigma_x$, $\rho(x)\in S^1$ for all $x\in D^2$.

15. Finally, if $x\in S^1$, then $\sigma_x = 1$ as $1$ is a root of the equation, and hence the unique root that is at least $1$. Hence, $\rho(x) = x$ for all $x\in S^1$.