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Let $S^1=\mathbb R^1/\mathbb Z$. Consider a family $\varphi_t$ of pieceswise smooth injective maps $\varphi_t:S^1\to \mathbb C^1$ depending continuously on $t$. Then each curve $\varphi_t(S^1)$ is a simple closed curve in $\mathbb C^1$, i.e. it bounds an open complex disk. Using Riemann mapping theorem we can identify such a disk with the unit disk $|z|\le 1$. In particular, for each $t$, by identifying the disk bounded by $\varphi_t(S^1)$ with the disk $|z|\le 1$, we can define the cross-ratio of points $\varphi_t(0)$, $\varphi_t(\frac{1}{4})$, $\varphi_t(\frac{1}{2})$, $\varphi_t(\frac{3}{4})$ as points lying in $|z|\le 1$ . I am pretty sure that such a cross-ratio is a continuous function of $t$.

Question. Does this statement follow from some standard result? How to prove it?

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  • $\begingroup$ 1. What does the Riemann map have to do with this question? 2. You say $\phi_t$ continuously depends on $t$. Does not this mean that $\phi_t(0),\phi_t(1/2),...$ continuously depend on $t$ and since they are all distinct the cross ratio also continuously depend in $t$? $\endgroup$
    – Alexandre Eremenko
    Commented Sep 5, 2020 at 14:21
  • $\begingroup$ Sorry Alexander, it was a bit hard to phrase this question properly. I want to measure the cross ratio of these 4 points not inside $\mathbb C^1$ but inside the disk. This is why I mentioned Rimenann mapping theorem. Using this theorem we see these 4 points as points in the boundary of the disk $|z|\le 1$ and then measure the corss-ratio in the disk $|z|\le 1$. Is it more clear now? $\endgroup$
    – aglearner
    Commented Sep 5, 2020 at 14:29
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    $\begingroup$ Then set it up properly in the question text rather than in the comments. $\endgroup$
    – Alexandre Eremenko
    Commented Sep 5, 2020 at 14:31

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I realised that the answer to this question follows from a different question on Mathoverflow almost 10 years ago:

Does Riemann map depend continuously on the domain?

These were good times... People were not voting to close questions just because they don't understand them...

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