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I have found the following derivation of the Schwarzian derivatives in the book of Ovsienko and Tabachnikov:

For a diffeomorphism $\gamma$ which acts on 4 points $t_1,t_2,t_3,t_4 \in \mathbb{RP}_1$, We assume that the 4 points are spread so that $t_2,t_3,t_4$ can be defined by their distances to $t_1$ as a function of $\epsilon\in\mathbb{R}$: $t_2 = t_1 + \epsilon$, $t_3 = t_1 + 2\epsilon$ and $t_4 = t_1 + 3\epsilon$. So, the 4 points become $t_1, t_1 + \epsilon, t_1 + 2\epsilon, t_1 + 3\epsilon$ and they are related by the variable $\epsilon$. The Schwarzian derivative measures the effect of $\gamma$ on the cross-ratio as $\epsilon$ tends to zero. In other words, the Schwarzian derivative measures the cross-ratio of the points when they are infinitesimally close. To obtain the Schwarzian derivative one forms the Taylor expansion of $\Phi$ when $\epsilon$ goes to zero and keeps the first non-zero term of the expansion:

$$\Phi(\gamma(t_1),\gamma(t_2),\gamma(t_3),\gamma(t_4))= \Phi(t_1,t_2,t_3,t_4) - \epsilon^2 S[\gamma](t_1) + O(\epsilon^3) \tag 1$$

In the above equation $S[\gamma]$ is the Schwarzian derivative for $\mathbb{RP}_1$, defined by: $ S[\gamma]=\frac{\gamma'''}{\gamma'}-\frac{3}{2} \left (\frac{\gamma''}{\gamma'}\right )^2$

By construction when $\gamma$ is a linear-fractional function, $S[\gamma]$ and all the higher order terms are zero as linear-fractional exactly preserves the cross-ratio.

According to the book, the converse is also true i.e. if $S[\gamma]=0$, $\gamma$ is a linear-fractional. But I cannot see why, because in equation (1), there are higher order terms of $\epsilon$. Could anyone please tell know what I am missing here?

Michael Hardy
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odvut
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  • Not an answer, but an analogy: A functions has derivative 0 everywhere if $f(x+h)=f(x)+o(h)$ for all $x$. If a function has derivative 0, it's constant. But why's it constant if there's an $o(h)$ term? Answer: because you need a theorem like Rolle's theorem to say that if the infinitesimal condition is satisfied everywhere, then the error term is actually 0. – Anthony Quas Jul 03 '14 at 11:29
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    Check out http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deri – Igor Rivin Jul 03 '14 at 14:46

2 Answers2

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$S[f]=0$ is a third order ordinary differential equation, and all fractional-linear functions are solutions. This is a $3$-parametric family. You can match any initial conditions with a fractional-linear function, then uniqueness theorem says that all solutions are fractional-linear.

Alexandre Eremenko
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    This is wrong: Only Moebius transformations are solutions. The equation $Sf=2q$ is equivalent to the equation $u" +qu=0$. The space of solutions of the latter is 2-dimensional, ratios of solutions $f=u_1/u_2$ are exactly the solutions of $Sf=2q$. If you set $q=0$ you get that all solutions are ratios of linear functions, hence, are linear-fractional. – Misha Jul 03 '14 at 21:24
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    Sorry. I corrected. – Alexandre Eremenko Jul 04 '14 at 06:27
  • Sorry for my delayed reply and thank you for your answers. I don't understand how your proof, using ratios of solutions of $u''+qu=0$, extends to general functions where it may not be possible to express them as the ratio of two solutions of $u''+qu=0$. – odvut Jul 18 '14 at 12:22
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The Schwatzian derivative measures the deviation of $\phi$ (which I take here as an orientation preserving diffeomorphism of $\mathbb R$, for simplicity's sake) from being a Moebius transformation; here is a direct computational proof. Of course, Alexandre Eremenko's proof is more elegant. $$ S(\phi)=0 \iff \phi(x)=\frac{ax+b}{cx+d}\text{ for } \begin{pmatrix} a & b \\ c & d\end{pmatrix} \in SL(2,\mathbb R). $$ Indeed, $S(\phi)=0$ if and only if $g=\log(\phi')'=\frac{\phi''}{\phi'}$ satisfies the differential equation $g'= g^2/2$, so that $\frac{2\,dg}{g^2}=dx$ or $\frac{-2}{g}=x+\frac dc$ which means $\log(\phi')'(x)=g(x)=\frac{-2}{x+d/c}$ or again $$\log(\phi'(x))=\int\frac{-2dx}{x+d/c}=-2\log(x+d/c)-2\log(c)=\log(\frac 1{(cx+d)^{2}}).$$ Therefore, $\phi'(x)= \frac 1{(cx+d)^{2}}=\partial_x\frac{ax+b}{cx+d}$.

Peter Michor
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