For a locally univalent holomorphic map \(f\), define \[ S_f=N_f'-\frac12N_f^2, \qquad N_f=\frac{f''}{f'}. \] Prove the composition law \[ S_{f\circ g}=S_g+(S_f\circ g)(g')^2. \]
Solution
Use the nonlinearity chain rule \[ N_{f\circ g}=N_g+(N_f\circ g)g'. \] Differentiating gives \[ N_{f\circ g}' =N_g' +(N_f'\circ g)(g')^2 +(N_f\circ g)g''. \] Also \[ \begin{aligned} \frac12N_{f\circ g}^2 &=\frac12N_g^2 +N_g(N_f\circ g)g'\\ &\qquad +\frac12(N_f\circ g)^2(g')^2. \end{aligned} \] Since \[ N_g=\frac{g''}{g'}, \] the two mixed terms cancel: \[ (N_f\circ g)g'' =N_g(N_f\circ g)g'. \] Therefore \[ \begin{aligned} S_{f\circ g} &=N_{f\circ g}'-\frac12N_{f\circ g}^2\\ &=\left(N_g'-\frac12N_g^2\right) +\left(N_f'-\frac12N_f^2\right)\!\circ g\,(g')^2\\ &=S_g+(S_f\circ g)(g')^2. \end{aligned} \] Equivalently, for the quadratic differential \(S_f(z)\,dz^2\), \[ S(f\circ g)=S(g)+g^*S(f). \]