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This question is truly stupid, but is driving me crazy. I just need an outside viewpoint to sort out what's going on.

  1. In my textbook: "Show that every linear fractional (LF) transformation of $\hat{\mathbb{C}}$ is orientation preserving.

This is completely baffling. $f(z) = \frac{1}{z}$ is certainly LF, but it's equal to $\frac{\overline{z}}{|z|^2}$. I can't think of a more not orientation preserving map than complex conjugation followed by scaling.

  1. From Wikipedia: The set of LF transformations is isomorphic to the orientation-preserving isometries of $\mathbb{H}^3$.

What is the isomorphism here, explicitly? Realize $\mathbb{H}^3$ as the upper-half space in $E^3$. It can't be Poincaré extension, since we can just take the map $f$ from above and extend it to get reflection through a sphere, which is not orientation preserving. What orientation-preserving map of $\mathbb{H}^3$ will be the image of $f$?

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All analytic functions are orientation-preserving. –  Thomas Andrews Feb 23 '13 at 16:00
$f(z)=z$ is orientation-preserving, but $z=|z|^2/\bar z$ –  Thomas Andrews Feb 23 '13 at 16:03
Try it out: consider the two vectors centered at $1$ that point in the positive real axis and the positive imaginary axis. Which way do they point after applying $f(z)$? –  Hurkyl Feb 23 '13 at 16:04
Aha. I am an idiot, thank you. I was thinking about what it did to points in the plane and not the tangent spaces. –  Zach L. Feb 23 '13 at 16:11
Check how the reciprocal function looks like on the Sphere: It is a $180^\circ$ rotation around the axis through $-1$ and $1$, swapping $\infty\leftrightarrow 0$ and $i\leftrightarrow -i$. Such a rotation is of course orientation preserving. –  Hagen von Eitzen Feb 23 '13 at 16:17

1 Answer 1

up vote 4 down vote accepted

Holomorphic functions, including Möbius transformations, are locally orientation preserving. This means that a small circle $t\mapsto z_0+r e^{it}$ $\ (0\leq t\leq 2\pi)$ going counterclockwise around a point $z_0$ where $f'(z_0)\ne0$ is mapped onto a small circlelike closed curve going counterclockwise around the point $f(z_0)$. This has to do with the fact that the Jacobian determinant at $z_0$ of the underlying real map ${\bf f}:\ {\bf z}\to{\bf f}({\bf z})$ is given by $|f'(z_0)|^2>0$.

Draw a figure and convince yourself that this property is manifest also in the case of the map $f:\ z\mapsto {1\over z}$ wherever it is defined.

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Right, I wasn't thinking locally. I was looking at how the images of the points 1 and i and getting confused. Somehow I also inconveniently forgot that 0 was sent to $\infty$. If I had remembered this, I would have instantly seen that my "counterexample" was actually fine, since map still preserved the angles I wanted at $\infty$. Thanks all for helping me sort this all out. –  Zach L. Feb 23 '13 at 16:28

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