2
$\begingroup$

Schwarz reflection across a real analytic arc $C$ in $\mathbb{R}^2$ is usually defined analytically. Thinking of the arc as the image of an interval in $\mathbb{R}$ under an invertible holomorphic map $T$ of a vicinity of $C$ in $\mathbb{C}=\mathbb{R}^2$ the reflection is defined as $R_T=T\circ R\circ T^{-1}$, where $R$ is the usual reflection across $\mathbb{R}$. For lines and circles we get the usual reflection and the circle inversion. A point and its Schwarz reflection are called symmetric with respect to the arc. Of course, analogous construction can be done in $\mathbb{C}^2$. In On the Conformal Geometry of Analytic Arcs Pfeiffer gives a geometric description of it, which I do not understand:

"The definition given above of two real points which are symmetric with respect to a real analytic arc is extended to apply to the complex plane in that the arc $C$ may be a complex analytic arc and the transformation $T$ a general non-singular conformal transformation. In all that follows, it is assumed that the slope of the arc at the point about which the power series which represents the arc is developed is not minimal. It may now readily be shown that two points which are symmetric with respect to an analytic arc are such that the two minimal lines on each point intersect on the analytic arc and, conversely, two points which are such that the two minimal lines on each intersect on the analytic arc are symmetric with respect to the analytic arc." He further mentions a theorem that "under any general non-singular conformal transformation minimal lines are transformed into minimal lines".

Minimal lines? Schwarz reflection across analytic arcs is discussed e.g. by Caratheodory in Conformal Representation (§ 139), but he mentions no "minimal lines". Pfeiffer and his teacher Kasner are the only people I saw use this term, so perhaps they are called differently now. I am not sure what "minimal slope" means either.

Can someone explain what "two minimal lines on each point" mean in this context, and how it can "readily be shown" that the condition on them is equivalent to Schwarz reflection symmetry.

$\endgroup$

0

You must log in to answer this question.