I start with an automorphism $f$ of the complex unit disc $S^1 = \{ z \in \mathbb{C} : |z| \leq 1\}$. I assume that such a map is given by a Mobius transform, namely $$ f(z) = \frac{z - a}{-\overline{a}z + 1}, $$ for some $a \in S^1$ with $|a| < 1$. (I know the automorphisms of $S^1$ are slightly more general but for my purpose this is enough.) My question is how far does this map move points on the boundary? In other words, what is the most a point on the boundary can rotate by. Are there 'good' bounds on this by which I mean that these bounds should go to zero as $|a|$ does.
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$\begingroup$ It is usually much easierto do those computations in the Iwasawa coordinates of $PSL_2$, which in turn are more easier to handle in the upper-half-plane model. (You can transfer between the models by the Cayley transform). $\endgroup$– AsafMay 9, 2014 at 8:00
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You can assume $a$ is real. Then, $f(z) - z = (z-1)/(-\overline{z} a + 1),$ for $z$ on the unit circle. Since the modulus of this thing is the chordal distance, which is related to the angle in the obvious way, this reduces to a simple calculus optimization problem.
answered May 13, 2014 at 19:27
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$\begingroup$ I made a typo in writing down the automorphism of the unit disc. It should be $f(z) = (z-a)/(-\overline{a}z +1)$. I think this issue messes up the calculation that you did. Sorry about that. $\endgroup$– BenMay 13, 2014 at 19:39
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$\begingroup$ Doesn't mess it up too much, actually. The numerator is now $-a + \overline{a} z^2,$ which leads to an only slightly more complicated optimization (which I am too lazy to perform myself...) $\endgroup$ May 13, 2014 at 20:06