For a locally univalent holomorphic map \(f\), define its nonlinearity by \[ N_f=\frac{f''}{f'}. \] Prove the composition law \[ N_{f\circ g} =N_g+(N_f\circ g)g'. \]
Solution
By the ordinary chain rule, \[ (f\circ g)'=(f'\circ g)g'. \] Differentiating once more gives \[ (f\circ g)'' =(f''\circ g)(g')^2+(f'\circ g)g''. \] Because \(f\) and \(g\) are locally univalent, the relevant first derivatives do not vanish, so division by \((f\circ g)'\) is legitimate. Therefore \[ \begin{aligned} N_{f\circ g} &=\frac{(f\circ g)''}{(f\circ g)'}\\ &=\frac{(f''\circ g)(g')^2+(f'\circ g)g''} {(f'\circ g)g'}\\ &=\left(\frac{f''}{f'}\circ g\right)g' +\frac{g''}{g'}\\ &=(N_f\circ g)g'+N_g. \end{aligned} \] This is exactly the asserted cocycle law. In differential-form notation it reads \[ N(f\circ g)=N(g)+g^*N(f). \]