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Again, is there a conformal self-map that interchanges two points in the upper half-plane? I'm beginning to think this isn't so. Such a map would be a FLT $\frac{az+b}{cz+d}, ad-bc=1$, with real coefficients. If I construct a map that sends some $w \mapsto t$ and $t \mapsto w$, I can't find values for the coefficients that work, though I may not have done this with sufficient generality.

I'm also thinking that given two points and considering a circle that they describe, the self-map would have to reverse the orientation of the circle which I don't think is possible.

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Hm, you can certainly do it for some points. I'm a little rusty on my conformal mappings so I can't quite remember how much control you can get. But certainly you can map the half plane conformally to the disk and then do a rotation to swap any two points on opposite sides of the same line in the disk. Then pull back to the plane. –  JSchlather Nov 30 '12 at 20:13
    
Oh, right, it may be better to think about this on the unit disk since it and the upper half-plane are isomorphic. I'll see what I can cook up, thanks. –  Kannaguchi O. Nov 30 '12 at 20:18
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In the unit disk there is always the Blaschke factor (maybe with a minus sign) that swaps $0$ and a point you choose. –  user27126 Nov 30 '12 at 20:44
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And (also in the unit disk) $z \mapsto -z$ swaps everything in sight. :-) In the upper half plane, this corresponds to $f(z)=-1/z$. –  Hans Lundmark Nov 30 '12 at 21:10

1 Answer 1

Your fractional linear transformation $f(z) = \dfrac{az+b}{cz+d}$ interchanges $p$ and $f(p)$ if and only if $f(f(p)) = p$, which implies that either $a+d=0$ or $c p^2 + (d-a) p - b = 0$. In the second case, $p$ is actually a fixed point of $f$. But in the first case, for any $p = x+iy$ and $q = u+iv$ in the upper half plane, I get a solution with $f(p)=q, f(q)=p$ with $$ a=-d={\frac {c \left( xv+yu \right) }{y+v}},b=-{\frac { c\left( y{v}^{2}+v{x}^{2}+v{y}^{2}+y{u}^{2} \right) }{y+v}}$$

EDIT: There is a geometric interpretation. Let $r$ be the point where the line from $p$ to $\overline{q}$ intersects the real line. Then $p$ and $2r-\overline{q}$ are inverse points with respect to the circle with centre $r$ and radius $\rho = \sqrt{|(p-r)(q-r)|}$. The transformation is then $$z \to r - \rho^2/(z-r)$$

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