问题

施瓦茨导数(Schwarzian derivative)

对于解析函数 $f(z)$, 称下面的表达式

$$\{f,z\}:=\frac{f^{‴}(z)}{f^{\prime }(z)}-\frac{3}{2}(\frac{f^{″}(z)}{f^{\prime }(z)}{)}^2$$

为施瓦茨导数(Schwarzian derivative). 

由于

$$(\frac{f^{″}(z)}{f^{\prime }(z)}{)}^{\prime }=\frac{f^{‴}(z)f^{\prime }(z)-f^{″}(z)f^{″}(z)}{(f^{\prime }(z){)}^2}=\frac{f^{‴}(z)}{f^{\prime }(z)}-(\frac{f^{″}(z)}{f^{\prime }(z)}{)}^2,$$

施瓦茨导数 $\{f,z\}$ 又可写为

$$\{f,z\}=(\frac{f^{″}(z)}{f^{\prime }(z)}{)}^{\prime }-\frac{1}{2}(\frac{f^{″}(z)}{f^{\prime }(z)}{)}^2.$$

若令

$$g(z)=\frac{af(z)+b}{cf(z)+d},$$

则有

$$\frac{g^{‴}(z)}{g^{\prime }(z)}-\frac{3}{2}(\frac{g^{″}(z)}{g^{\prime }(z)}{)}^2=\frac{f^{‴}(z)}{f^{\prime }(z)}-\frac{3}{2}(\frac{f^{″}(z)}{f^{\prime }(z)}{)}^2.$$

即 $\{g,z\}=\{f,z\}$, 即得到如下命题.

命题.  施瓦茨导数(Schwarzian derivative)在莫比乌斯变换下不变.

对于解析函数 $f(z)$, 若 $s$ 是另一参数, 则根据 M. Barner, 可证明

$$\{f,t\}={\phi }^{-2}\cdot (\{f,z\}+2A^2-2A^{\prime })$$

这里 $\phi =\frac{\mathrm{d}t}{\mathrm{d}z}$, $A=\frac{1}{2}\frac{{\phi }^{\prime }}{\phi }$.

References:

[1] Schwarzian derivative - Encyclopedia of Mathematics