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Let $P$ be a point where three arcs of circle meet at equal angles (120 degrees). Suppose that the sum of the curvatures (with sign given by orientation) of the three arcs is zero.

Is it true that this property is preserved under a moebius transformation? Angles are preserved, but what about the sum of curvatures?

What if we don't know anything about the angles? Is the sum of curvatures anyway preserved (if it is zero)?


What seems to be true is that given three arcs coming out from a point $O$ with equal angles (120 degrees) and with curvatures whose sum is zero, then the arcs meet (all three) at another point $P$. This can be found in Lemma 8.1 of Cox, Harrison, Hutchings "The shortest enclosure of three connected areas in $\mathbb R^2$" (Real Anal. Exch.)". The proof is by direct computation, left to the reader.

It seems natural that also the converse is true i.e.: if the arcs all meet in another single point, then the sum of curvatures is zero. This would solve the problem, since moebius transform preserve this condition.

It seems to me that these properties should be known, because they seem quite important... don't they?

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Any Moebius transformation can be written as a composition of maps of the form $$z\mapsto z+c\>; \qquad z\mapsto c\>z, \quad c\in{\mathbb C}^*; \qquad z\mapsto{1\over z}\ .$$ For the first two of these it is obvious what happens to the curvatures of any smooth curves, so that it remains to investigate the map $z\mapsto{1\over z}$, and it is even sufficient to consider curves through the point $z=1$.

When a curve $\gamma$ is written in terms of arc length as parameter its parametrization looks like $$z(t)=z_0+\left( t+{i t^2\over2}\kappa\right)e^{i\theta}+?t^3\ .\tag{1}$$ Here $\theta=\arg\bigl(\dot z(0)\bigr)$ is the initial argument of $\gamma$, and $\kappa={d\arg\bigl(z(t)\bigr)\over dt}\Biggr|_{t=0}$ is the signed curvature of $\gamma$ at $z_0$. One can read off that $$\kappa={\rm Im}\left({\ddot z\over \dot z}\Biggr|_{t=0}\right)\ .$$ When we have a curve $\hat\gamma$ parametrized as $t\mapsto w(t)$, where $t$ is an arbitrary parameter then the derivative of $w$ with respect to arc length is given by $$\dot w={w'\over\sigma'}\tag{2}$$ where $$\sigma'^2=w'\bar w'$$ and therefore $$2\sigma'\sigma''=w''\bar w'+w'\bar w''\ .$$ It follows that $$\ddot w(t):={1\over\sigma'}\left({w'\over\sigma'}\right)'={w'\over\sigma'^4}\>{w''\bar w'-w'\bar w''\over2}\ ,$$ so that $(2)$ implies $${\ddot w\over \dot w}={1\over\sigma'^3}{w''\bar w'-w'\bar w''\over2}\ .\tag{3}$$ In our case the curve $\hat\gamma$ is the image of the curve $(1)$ under the map $z\mapsto w:={1\over z}$. From $$\eqalign{w(t)={1\over z(t)}&={1\over 1 +(t + i t^2\kappa/2)e^{i\theta}+?t^3}\cr &=1-t e^{i\theta}+{t^2\over2}(i\kappa e^{i\theta}+2e^{2i\theta})+?t^3\cr}$$ From this we read off $\sigma'(0)=1$ and compute, using $(3)$, $${\ddot w\over\dot w}\Biggr|_{t=0}=- i(\kappa + 2\sin\theta)\ .$$ It follows that the curvature of the image curve $\hat\gamma$ at $1$ is given by $$\hat\kappa=-(\kappa+2\sin\theta)\ .$$ From this formula we can immediately see that when the curvatures of $n$ "equally spaced" curves through $1$ sum to zero then the curvatures of the image curves sume to zero as well.

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I think you are right, the point is to reduce the problem to a single inversion and to use isometric transformation to reduce the problem. I would reduce to $z=0$ instead of $z=1$, what do you think? The point $z=0$ will go to infinity, but I only have to measure the curvature of the three circles... I will try the computation. – Emanuele Paolini May 5 '14 at 16:04
It is important that you look at an ordinary point, like $z=1$, because at $\infty$ all information about curvatures goes lost. – Christian Blatter May 5 '14 at 16:17
You are right... but at least I can suppose one of the three circles is $|z|=1$, don't I? – Emanuele Paolini May 5 '14 at 16:58
I don't know. You have to play with this problem, e.g., considering circles with center on the positive $x$-axis, their angles of intersection with the unit circle, and what happens to their curvature under inversion. This is not "homework", but a "seminar problem". – Christian Blatter May 5 '14 at 19:16
please look at the update... have you some other hint? – Emanuele Paolini May 6 '14 at 8:35

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