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math185-s23:s:hexokinase:start [2023/04/07 20:34] hexokinase [Homework solutions] |
math185-s23:s:hexokinase:start [2026/02/21 14:41] (current) |
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| Similar argument but with Fourier inversion formula proves the $\impliedby$ direction. | Similar argument but with Fourier inversion formula proves the $\impliedby$ direction. | ||
| + | ====Fixed points of automorphisms of $\mathbb{D}$==== | ||
| + | Denote $B_a(z) = \frac{ a-z }{ 1-\overline{a}z }$ and recall SS ch 1 ex 7.\\ | ||
| + | **Definition: | ||
| + | **Definition: | ||
| + | |||
| + | **Claim:** Let $f \in \operatorname{Aut}(\mathbb{D})$. If $f$ is nontrivial, | ||
| + | then $f$ has at most one fixed point in $\mathbb{D}$.\\ | ||
| + | **Proof: | ||
| + | We prove the contrapositive; | ||
| + | The function $g = B_{z_0} \circ f \circ B_{z_0}$ fixes $0$ and $B_{z_0}(z_1)$\\ | ||
| + | so Schwarz lemma gives $g = (z \mapsto \lambda z)$ for some $\lambda \in \partial\mathbb{D}$, | ||
| + | and we must have $\lambda = 1$ since $g$ fixes $B_{z_0}(z_1)$, | ||
| + | hence $g = \operatorname{id}_\mathbb{D}$\\ | ||
| + | hence $f = B_{z_0} \circ g \circ B_{z_0} = \operatorname{id}_\mathbb{D}$.\\ | ||
| + | QED. | ||
| + | |||
| + | **Examples: | ||
| + | If $f = (z \mapsto \lambda z)$ with $1 \neq \lambda \in \partial\mathbb{D}$, | ||
| + | If $f = B_a$, with $a \in \mathbb{D}$, | ||
| + | If $f = -B_a$, with $0 \neq a \in \mathbb{D}$, | ||
| + | |||
| + | **Hyperbolic geometry**: every nontrivial isometry of the hyperbolic plane has at most one fixed point; this fails for the euclidean plane because of reflections. | ||
| + | |||
| + | **Fixed point in $\overline{\mathbb{D}}$: | ||
| + | |||
| + | **< | ||
| + | Maybe they are precisely those $f$ which have two fixed points in $\partial\mathbb{D}$. | ||
| + | |||
| + | **Thoughts: | ||
| + | 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' | ||
| + | |||
| + | **Solution: | ||
| + | $$ u \left(1 \pm \sqrt{ 1 - |u|^{-2} } \right) $$ | ||
| + | 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===== | =====Homework solutions===== | ||
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