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Schwarzian differential operator and Möbius transformations

The Schwarzian derivative is given by the formula \begin{displaymath}
\mathcal{S}(f) = \left( \frac{f''(z)}{f'(z)} \right)' - \frac12 \left(
\frac{f''(z)}{f'(z)} \right)^2
\end{displaymath} It's miraculous properties are that for any Möbius transformation $m(z)$ we have \begin{displaymath}
\mathcal{S}(m\circ f) =\mathcal{S}(f)
\end{displaymath} and \begin{displaymath}
\mathcal{S}(f\circ m) (z) = \mathcal{S}(f)(m(z)) \, m'(z)^2
\end{displaymath} It takes some very careful application of the chain rule just to verify these formulas. It is more of a miracle that if two functions $f$, $g$ have the same Schwarzian, then there must be a Möbius transformation $m$ such that $f=g\circ m$. This comes from the theory of linear differential equations. A basic reference is [Ford, 1951], which has many other wonderful things in it as well.



David Wright
2004-11-24