3
$\begingroup$

Take a branch of the function $f(t) = \sqrt{1-t^2}$ on the closed upper-half plane ($\bar{H}$) so that $f(i) = \sqrt{2} > 0$. Then, we can define

$\phi(z) = \int_0^{z} \frac{dt}{f(t)}$, where the path is taken in $\bar{H}$.

This function maps $\bar{H}$ to a half-infinite rectangular strip $[-\pi/2,\pi/2] \times [0,\infty)$.

I don't completely understand why this is the case.

Wouldn't

$\psi(z) = \int_0^{z} f(t) dt$ map $\bar{H}$ onto $[-\pi/4,\pi/4] \times (-\infty,0]$?

The argument for this is the same for that of $\phi$: The function $\psi$ maps $[-1,1]$ to $[-\pi/4,\pi/4]$. Every time $z$ passes a branch point along the real axis, the argument changes by $\pm\pi$, so this results in a $\pm\frac{\pi}{2}$ turn.

So it appears $\psi$ essentially does what $\phi$ does. Then is there any reason to look at $\phi$ rather than $\psi$? (More generally, why does the Scwarz Christoffel mapping need all those things appear in the denominator rather than in the numerator?)

p.s. I don't really understand why the the turns happen in the directions that they do rather than making like a stair-case like shape in the case of $\phi$ or $\psi$.

EDIT: Here are some references that helped me clarify this:

  1. http://www.mth.kcl.ac.uk/~shaww/web_page/books/complex/Chapter21Excerpt.pdf
  2. http://books.google.com/books/about/Complex_Variables.html?id=uYYpdrbKJTEC
  3. Ahlfors
$\endgroup$
3
$\begingroup$

why does the Schwarz-Christoffel mapping need all those things appear in the denominator rather than in the numerator?

It does not; this is just a traditional way of writing the formula. You can put all those factors in the numerator, with opposite sign of the exponent: $$f(\zeta) = \int^\zeta K (w-a)^{(\alpha/\pi)-1}(w-b)^{ (\beta/\pi)-1}(w-c)^{ (\gamma/\pi)-1} \cdots \,{d}w $$ Indeed, this makes the formula more transparent: we recognize $(w-a)^{ (\alpha/\pi)-1}$ as the derivative of $(w-a)^{ (\alpha/\pi)}$ (up to a constant factor), and the latter map creates an interior angle of $\alpha$ from a halfplane.

So it appears $\psi$ essentially does what $\phi$ does. Then is there any reason to look at $\phi$ rather than $\psi$?

There is a difference, masked in your text by $\pm$ signs. Namely, $\phi$ creates interior angles of size $\pi/2$, while $\psi $ creates interior angles of size $3\pi/2$. The boundary of image may stay essentially the same, but the image is on a different side of the boundary.


how to pick the right sign

Follow the real line from left to right. When you encounter a point where the integrand is $0$ or $\infty$ (such as $1$ or $-1$ here), make a small detour along a half-circle in the upper half-plane (which is the domain of our function). How much does the argument of the integrand change during this detour? Say, we go around $1$ in this fashion: then the argument of $z-1$ changes by $-\pi$ because we move in half-circle clockwise. So:

  • if the integrand has $(z-1)^{1/2}$, its argument changes by $-\pi/2$. Thus, the image makes a turn by 90 degrees to the right.
  • if the integrand has $(z-1)^{-1/2}$, its argument changes by $ \pi/2$. The image makes a turn by 90 degrees to the left.
  • generally, if the integrand has $(z-1)^{\beta}$, its argument changes by $-\pi \beta$
$\endgroup$
  • $\begingroup$ The deep problem I was having with the understanding was how the different branches of the logarithm was being chosen, and how the $\frac{1}{\sqrt{1-t^2}}$ is implicitly thought as a product of $i(t-1)^{-1/2}(t+1)^{-1/2}$. You said "There is a difference, masked in your text by ± signs," but I wasn't getting how to pick the right sign. $\endgroup$ – Braindead Jan 4 '14 at 5:49
  • $\begingroup$ @Braindead I edited in an explanation. $\endgroup$ – Post No Bulls Jan 4 '14 at 6:17
  • $\begingroup$ Thank you! I'm still a bit fuzzy about choosing the right branches, but I think I more or less understand this now. $\endgroup$ – Braindead Jan 4 '14 at 6:48
  • 1
    $\begingroup$ @Braindead It's okay to be fuzzy about branches; all the choice of a branch does is contribute a unimodular constant to the map. The change of argument of $(z-a)^\beta$ along a curve does not depend on any branch choices. $\endgroup$ – Post No Bulls Jan 4 '14 at 6:52

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.