问题及解答

施瓦茨导数(Schwarzian derivative)

对于解析函数 $f(z)$, 称下面的表达式

\[
\{f,z\}:=\frac{f'''(z)}{f'(z)}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2
\]

为施瓦茨导数(Schwarzian derivative). 

由于

\[
\biggl(\frac{f''(z)}{f'(z)}\biggr)'=\frac{f'''(z)f'(z)-f''(z)f''(z)}{(f'(z))^2}=\frac{f'''(z)}{f'(z)}-\biggl(\frac{f''(z)}{f'(z)}\biggr)^2,
\]

施瓦茨导数 $\{f,z\}$ 又可写为

\[
\{f,z\}=\biggl(\frac{f''(z)}{f'(z)}\biggr)'-\frac{1}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2.
\]

若令

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

则有

\[
\frac{g'''(z)}{g'(z)}-\frac{3}{2}\biggl(\frac{g''(z)}{g'(z)}\biggr)^2=\frac{f'''(z)}{f'(z)}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2.
\]

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

\[
\{f,t\}=\varphi^{-2}\cdot\Bigl(\{f,z\}+2A^2-2A'\Bigr)
\]

这里 $\varphi=\frac{\mathrm{d}t}{\mathrm{d}z}$, $A=\frac{1}{2}\frac{\varphi'}{\varphi}$.

References:

[1] Schwarzian derivative - Encyclopedia of Mathematics

证明.  令

\[
g(z)=\frac{af(z)+b}{cf(z)+d},\quad ad-bc\neq 0.
\]

则

\[
g'(z)=\frac{af'(z)(cf(z)+d)-(af(z)+b)cf'(z)}{(cf(z)+d)^2}=\frac{(ad-bc)f'(z)}{(cf(z)+d)^2}.
\]

为简洁, 记 $f'$ 为 $f'(z)$ 等等.

\[
\begin{split}
g''&=\biggl[\frac{(ad-bc)f'}{(cf+d)^2}\biggr]'=(ad-bc)\cdot\frac{f''\cdot(cf+d)^2-f'\cdot 2(cf+d)cf'}{(cf+d)^4}\\
&=(ad-bc)\cdot\frac{(cf+d)f''-2c(f')^2}{(cf+d)^3},
\end{split}
\]

\[
\begin{split}
g'''&=(ad-bc)\cdot\biggl[\frac{(cf+d)f''-2c(f')^2}{(cf+d)^3}\biggr]'\\
&=(ad-bc)\cdot\frac{\bigl[(cf+d)f''-2c(f')^2\bigr]'\cdot(cf+d)^3-\bigl[(cf+d)f''-2c(f')^2\bigr]\cdot 3(cf+d)^2\cdot cf'}{(cf+d)^6}\\
&=\frac{ad-bc}{(cf+d)^4}\biggl[\Bigl(cf'\cdot f''+(cf+d)f'''-4cf' f''\Bigr)(cf+d)-\Bigl((cf+d)f''-2c(f')^2\Bigr)\cdot 3cf'\biggr]\\
&=\frac{ad-bc}{(cf+d)^4}\biggl[\Bigl((cf+d)f'''-3cf' f''\Bigr)(cf+d)-3c(cf+d)f' f''+6c^2(f')^3\biggr]\\
&=\frac{ad-bc}{(cf+d)^4}\biggl[(cf+d)^2 f'''-6cf' f''\cdot(cf+d)+6c^2(f')^3\biggr]
\end{split}
\]

于是

\[
\begin{split}
\frac{g'''}{g'}&=\frac{ad-bc}{(cf+d)^4}\biggl[(cf+d)^2 f'''-6cf' f''\cdot(cf+d)+6c^2(f')^3\biggr]\cdot\frac{(cf+d)^2}{(ad-bc)f'}\\
&=\frac{1}{(cf+d)^2}\biggl[(cf+d)^2\frac{f'''}{f'}-6c(cf+d)f''+6c^2(f')^2\biggr]\\
&=\frac{f'''}{f'}-\frac{6cf''}{cf+d}+\frac{6c^2(f')^2}{(cf+d)^2}
\end{split}
\]

\[
\begin{split}
\frac{3}{2}\biggl(\frac{g''}{g'}\biggr)^2&=\frac{3}{2}\biggl[(ad-bc)\cdot\frac{(cf+d)f''-2c(f')^2}{(cf+d)^3}\cdot\frac{(cf+d)^2}{(ad-bc)f'}\biggr]^2\\
&=\frac{3}{2}\biggl[\frac{(cf+d)f''-2c(f')^2}{(cf+d)f'}\biggr]^2\\
&=\frac{3}{2}\biggl[\frac{f''}{f'}-\frac{2cf'}{cf+d}\biggr]^2\\
&=\frac{3}{2}\biggl[\biggl(\frac{f''}{f'}\biggr)^2-\frac{4cf''}{cf+d}+\frac{4c(f')^2}{(cf+d)^2}\biggr]\\
&=\frac{3}{2}\biggl(\frac{f''}{f'}\biggr)^2-\frac{6cf''}{cf+d}+\frac{6c(f')^2}{(cf+d)^2},
\end{split}
\]

因此,

\[
\frac{g'''}{g'}-\frac{3}{2}\biggl(\frac{g''}{g'}\biggr)^2=\frac{f'''}{f'}-\frac{3}{2}\biggl(\frac{f''}{f'}\biggr)^2.
\]

注意 $\varphi(z)=\frac{\mathrm{d}t}{\mathrm{d}z}=t'(z)$, 故 $\varphi^{-2}=\dfrac{1}{(t'(z))^2}$.

\[
A(z)=\frac{1}{2}\frac{\varphi'(z)}{\varphi(z)},
\]

于是

\[
A'(z)=\frac{1}{2}\frac{\varphi''(z)\varphi(z)-(\varphi'(z))^2}{\varphi^2(z)}.
\]

因此

\[
\begin{split}
2(A^2-A')&=2\biggl[\frac{1}{4}\frac{\varphi'^2}{\varphi^2}-\frac{1}{2}\frac{\varphi''\varphi-\varphi'^2}{\varphi^2}\biggr]\\
&=\frac{1}{2}\frac{\varphi'^2}{\varphi^2}-\frac{\varphi''\varphi-\varphi'^2}{\varphi^2}\\
&=\frac{3\varphi'^2-2\varphi''\varphi}{2\varphi^2}\\
&=\frac{3}{2}\biggl(\frac{\varphi'}{\varphi}\biggr)^2-\frac{\varphi''}{\varphi}\\
&=\frac{3}{2}\biggl(\frac{t''(z)}{t'(z)}\biggr)^2-\frac{t'''(z)}{t'(z)}.
\end{split}
\]

因此,

\[
\begin{split}
&\varphi^{-2}\Bigl[\{f,z\}+2A^2-2A'\Bigr]\\
=&\frac{1}{(t'(z))^2}\biggl[\frac{f'''(z)}{f'(z)}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2+\frac{3}{2}\biggl(\frac{t''(z)}{t'(z)}\biggr)^2-\frac{t'''(z)}{t'(z)}\biggr]\\
=&\frac{f'''(z)}{f'(z)}\frac{1}{(t'(z))^2}-\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2\frac{1}{(t'(z))^2}+\frac{3}{2}\biggl(\frac{t''(z)}{t'(z)}\biggr)^2\frac{1}{(t'(z))^2}-\frac{t'''(z)}{t'(z)}\frac{1}{(t'(z))^2}
\end{split}\tag{*}
\]

若 $f$ 是 $t$ 的函数, $t=t(z)$, 则

\[
f'(z)=f'(t)\cdot t'(z)
\]

\[
f''(z)=(f'(t)\cdot t'(z))'_z=f''(t)\cdot (t'(z))^2+f'(t)\cdot t''(z).
\]

\[
\begin{split}
f'''(z)&=\bigl[f''(t)\cdot (t'(z))^2+f'(t)\cdot t''(z)\bigr]'_z\\
&=f'''(t)\cdot(t'(z))^3+f''(t)\cdot 2t'(z)t''(z)+f''(t)\cdot t'(z)t''(z)+f'(t)\cdot t'''(z).
\end{split}
\]

于是

\[
\begin{split}
\frac{f'''(z)}{f'(z)}\frac{1}{(t'(z))^2}&=\frac{f'''(t)\cdot(t'(z))^3+f''(t)\cdot 2t'(z)t''(z)+f''(t)\cdot t'(z)t''(z)+f'(t)\cdot t'''(z)}{f'(t)\cdot (t'(z))^3}\\
&=\frac{f'''(t)}{f(t)}+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{t'''(z)}{(t'(z))^3},
\end{split}
\]

\[
\begin{split}
\frac{3}{2}\biggl(\frac{f''(z)}{f'(z)}\biggr)^2\frac{1}{(t'(z))^2}&=\frac{3}{2}\biggl(\frac{f''(t)\cdot (t'(z))^2+f'(t)\cdot t''(z)}{f'(t)\cdot t'(z)}\biggr)^2\cdot\frac{1}{(t'(z))^2}\\
&=\frac{3}{2}\frac{1}{(t'(z))^2}\biggl(\frac{f''(t)}{f'(t)}t'(z)+\frac{t''(z)}{t'(z)}\biggr)^2\\
&=\frac{3}{2}\frac{1}{(t'(z))^2}\biggl[\biggl(\frac{f''(t)}{f'(t)}\biggr)^2(t'(z))^2+2\frac{f''(t)}{f'(t)}t''(z)+\biggl(\frac{t''(z)}{t'(z)}\biggr)^2\biggr]\\
&=\frac{3}{2}\biggl(\frac{f''(t)}{f'(t)}\biggr)^2+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{3}{2}\frac{(t''(z))^2}{(t'(z))^4}
\end{split}
\]

代入 (*) 式, 得

\[
\begin{split}
&\varphi^{-2}\Bigl[\{f,z\}+2A^2-2A'\Bigr]\\
=&\biggl(\frac{f'''(t)}{f(t)}+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{t'''(z)}{(t'(z))^3}\biggr)-\biggl[\frac{3}{2}\biggl(\frac{f''(t)}{f'(t)}\biggr)^2+3\frac{f''(t)}{f'(t)}\frac{t''(z)}{(t'(z))^2}+\frac{3}{2}\frac{(t''(z))^2}{(t'(z))^4}\biggr]+\frac{3}{2}\frac{(t''(z))^2}{(t'(z))^4}-\frac{t'''(z)}{(t'(z))^3}\\
&=\frac{f'''(t)}{f'(t)}-\frac{3}{2}\biggl(\frac{f''(t)}{f'(t)}\biggr)^2
\end{split}
\]