2
$\begingroup$

In Chap 8 of Stein's complex analysis, he proved that all the conformal maps $f$ from the upper half plane $\mathbb{H}$ to a given polygon $P$ is of the form $c_1S(z)+c_2$, where $c_1, c_2$ are complex numbers and S(z) is the Schwarz-Christoffel integral.

In the proof, he proposed that for any vertice $a_k$ of the polygon, assign a new function$h_k(z)=(f(z)-a_k)^{1/\alpha_k}$, where $\alpha_k$ is the inner angle at $a_k$, then $h_k$ maps the segment $[A_{k-1},A_{K+1}]$ to a line segment$L_k$, and then he applied the Schwarch reflection principle to see that $h_k$ is analytically continuable to a holomorphic function in the two way infinite strip $A_{k-1}<Re(z)<A_{k+1}$, and then "since $h_k$ is injective up to $L_k$ the symmetry in the Schwarz reflection principle guarantees that $h_k$ is injective in the whole disc centered at $x$" and later on "finally the Schwarz reflection principle shows that $f$ is continuable in the exterior of a disk $|z|\leq R$, for large $R$"

Can anyone help explain how the Schwarz reflection principle is used in the two quoted text? Thank you very much!

$\endgroup$

1 Answer 1

2
$\begingroup$

So the usual reflection principle is for reflection across the real axis, assuming the function maps the real axis to the real axis. In general, you can generalize the Schwarz reflection principle to mappings from line segments or circular arcs to other line segments or circular arcs, the idea is just to find a linear fractional transformation that transfers your problem to that of the real axis to real axis mapping.

Thus, an analytic function that is defined on one side of a segment/circular arc which maps that segment/circular arc to another segment/circular arc can be extended across the segment/circular arc while preserving symmetries. (The symmetric point of a point inside a circle lies on the same radial line but goes from being distance $d$ from the center to being radius $R/d$ from the center where $R$ is the radius of the circle).

PS circle reflection problem: The Schwarz Reflection Principle for a circle , relevant to the second set of quotes perhaps.

$\endgroup$
8
  • $\begingroup$ Thanks for this guideline. But how would the principle indicate that the function is injective on the whole disc after we extend it in the first quotes. $\endgroup$
    – Alex
    Aug 27, 2013 at 16:20
  • $\begingroup$ Yes, it says injective on the upper half of the circle gives you the same on the lower half of the circle. To justify, on the lower half, take two points, reflect them by symmetry to transfer the problem to the upper half, injectivity shows they map to two different points, and then reflect back to finish. I suppose it remains to show the boundary, but I think that is by construction (assume $F$ maps real line segment to angled segment, construct $h$ to flatten the angle, etc) $\endgroup$
    – Evan
    Aug 27, 2013 at 16:25
  • $\begingroup$ And if you wonder why can't a point on the upper half be mapped to the same thing as a point on the lower half, it's also because the image of the upper half is disjoint from the image of the lower half (symmetrically reflected) $\endgroup$
    – Evan
    Aug 27, 2013 at 16:27
  • $\begingroup$ I understand "the image of the upper half is disjoint from the image of the lower half "as follows: in the situation where the line segment is real axis and the image on it is real. If $f(z_1)=f(\overline{z_2})$,where $z_1,z_2 \in \mathbb{H}$, then since $f(\overline{z_2})=\overline{f(z_2)}$,we must have $f(z_1)=\overline{f(z_2)}$,therefore, any line segment between $z_1$ and $z_2$ must have a point $\omega$ on which $f$ is real, therefore contradicting the fact that $f$ is injective? $\endgroup$
    – Alex
    Aug 27, 2013 at 18:12
  • $\begingroup$ I mean $f$ as the function $h$. But the problem is $h(\omega)$ can be different from any value of $h$ on the line segment of the real axis that is also in the circle. Or is it due to other reasons, not injectivity that resulted in "the image of the upper half is disjoint from the image of the lower half "? @Evan $\endgroup$
    – Alex
    Aug 27, 2013 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.