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--- math185-s23:s:hexokinase:start [2023/04/14 04:25]
hexokinase
+++ math185-s23:s:hexokinase:start [2026/02/21 14:41] (current)
@@ Line 25: / Line 25: @@
 **Claim:** Let $f \in \operatorname{Aut}(\mathbb{D})$. If $f$ is nontrivial,
-then $f$ has at most one fixed point.\\
+then $f$ has at most one fixed point in $\mathbb{D}$.\\
 **Proof:**\\
 We prove the contrapositive; assume $f$ has two distinct fixed points $z_0,z_1 \in \mathbb{D}$.\\
@@ Line 37: / Line 37: @@
 **Examples:**\\
 If $f = (z \mapsto \lambda z)$ with $1 \neq \lambda \in \partial\mathbb{D}$, then $0$ is the unique fixed point of $f$ because $f$ is a nontrivial rotation.\\
-If $f = B_a$ (with $a \in \mathbb{D}$), then $f$ has a unique fixed point in $\mathbb{D}$ because the equation $B_a(z) = z$ has a unique solution in $\mathbb{D}$.
+If $f = B_a$, with $a \in \mathbb{D}$, then $f$ has a unique fixed point in $\mathbb{D}$ because the equation $B_a(z) = z$ has a unique solution in $\mathbb{D}$.\\
+If $f = -B_a$, with $0 \neq a \in \mathbb{D}$, then $f$ has no fixed points in $\mathbb{D}$; its fixed points are $\pm \frac{ a }{ |a| } \in \partial\mathbb{D}$, since those are the solutions of $-B_a(z) = z$.
 **Hyperbolic geometry**: every nontrivial isometry of the hyperbolic plane has at most one fixed point; this fails for the euclidean plane because of reflections.
-I suspect that every nontrivial automorphism has a fixed point.\\
+**Fixed point in $\overline{\mathbb{D}}$:** the Brouwer fixed point theorem implies that every $f \in \operatorname{Aut}(\mathbb{D})$ has a fixed point in $\overline{\mathbb{D}}$ (once $f$ has been extended holomorphically to an open superset of $\overline{\mathbb{D}}$). It would be interesting to see a complex-analytic proof, especially since I don't know how to prove the Brouwer fixed point theorem.
-Two ideas for how to prove this:\\
-• Showing that the equation $\lambda B_a(z) = z$ has a solution in $\mathbb{D}$ for $a\in\mathbb{D}, \lambda\in\partial\mathbb{D}$.\\
-• Rouché's theorem.\\
+**<del>Un</del>resolved question:** which $f \in \operatorname{Aut}(\mathbb{D})$ have no fixed point in $\mathbb{D}$?\\
+Maybe they are precisely those $f$ which have two fixed points in $\partial\mathbb{D}$.
+**Thoughts:** The function $f = \lambda B_a$ is nonconstant whenever $a \notin \partial\mathbb{D}, \lambda\neq0$. It's meromorphic on $\hat{\mathbb{C}}$. Let $a \neq 0$; then the quadratic formula shows that $f$ has exactly two fixed points with multiplicity. They may be equal: let $\lambda \in \partial\mathbb{D}$; then $f$ has a double fixed point iff $ \frac{ |1+\lambda| }{ 2|a| } = 1$, and the double fixed point is $\frac{ 1+\lambda }{ 2\overline{a} }$. (Note that $\arg \lambda = 2 \arg (1+\lambda)$; this has a straightedge-and-compass proof.) Let $\lambda \neq -1,1$; those $a$ solving the equation form a circle centered at $0$ with radius $\tfrac{1}{2} | 1+\lambda | < 1$.\\
+From this we can see that for any $a \in \mathbb{D}$ there exists a unique rotation $(z \mapsto \lambda z)$ such that $\lambda B_a$ has a double fixed point on the unit circle (and no other fixed points).\\
+Rouché's theorem might give a little more info.
+**Solution:** Just realized that, whenever $\lambda \in \partial\mathbb{D}$ and $0 \neq a \notin \partial\mathbb{D}$, the fixed points of $\lambda B_a$ are
+$$ u \left(1 \pm \sqrt{ 1 - |u|^{-2} } \right) $$
+where $u = \frac{ 1+\lambda }{ 2\overline{a} }$.\\
+This completely resolves the unresolved question;
+I might use this to write up a cleaner version of this post at some point.\\
 =====Homework solutions=====
 {{ :math185-s23:s:hexokinase:1.pdf |}}\\
@@ Line 53: / Line 63: @@
 {{ :math185-s23:s:hexokinase:7.pdf |}}\\
 {{ :math185-s23:s:hexokinase:8.pdf |}}\\
-{{ :math185-s23:s:hexokinase:9.pdf |}}
+{{ :math185-s23:s:hexokinase:9.pdf |}}\\
+{{ :math185-s23:s:hexokinase:10.pdf |}}

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