Details

Autor(en) / Beteiligte

• Airault, Hélène

Titel

From Diff (S1) to Univalent Functions. Cases of Degeneracy

Ist Teil von

• Analysis and Mathematical Physics, p.1-19

Ort / Verlag

Basel: Birkhäuser Basel

Link zum Volltext

Quelle

Alma/SFX Local Collection

Beschreibungen/Notizen

• We explain in detail how to obtain the Kirillov vector fields (Lk)k∈Z on the space of univalent functions inside the unit disk. Following Kirillov, they can be produced from perturbations by vectors (eikθ)k∈Z of diffeomorphisms of the circle. We give a second approach to the construction of the vector fields. In our approach, the Lagrange series for the inverse function plays an important part. We relate the polynomial coefficients in these series to the polynomial coefficients in Kirillov vector fields. By investigation of degenerate cases, we look for the functions \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ f(z) = z + \sum\nolimits_{n \geqslant 1} {a_n z^{n + 1} } $$\end{document} such that Lkf = L-kf for k ≥1. We find that f(z) must satisfy the differential equation: (*)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \left[ {\frac{{z^2 w}} {{1 - zw}} + \frac{{zw}} {{w - z}}} \right]f'(z) - f(z) - \frac{{w^2 f'(w)^2 }} {{f(w)^2 }} \times \frac{{f(z)^2 }} {{f(w) - f(z)}} = 0. $$\end{document} We prove that the only solutions of (*) are Koebe functions. On the other hand, we show that the vector fields (Tk)k∈Z image of the (Lk)k∈Z through the map \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ g(z) = \frac{1} {{f(\tfrac{1} {z})}} $$\end{document} can be obtained directly as the (Lk) from perturbations of diffeomorphisms of the circle.