4.II.24H

Differential Geometry | Part II, 2005

(i) Define what is meant by an isothermal parametrization. Let ϕ:UR3\phi: U \rightarrow \mathbb{R}^{3} be an isothermal parametrization. Prove that

ϕuu+ϕvv=2λ2H\phi_{u u}+\phi_{v v}=2 \lambda^{2} \mathbf{H}

where H\mathbf{H} is the mean curvature vector and λ2=ϕu,ϕu\lambda^{2}=\left\langle\phi_{u}, \phi_{u}\right\rangle.

Define what it means for ϕ\phi to be minimal, and deduce that ϕ\phi is minimal if and only if Δϕ=0\Delta \phi=0.

[You may assume that the mean curvature HH can be written as

H=eG2fF+gE2(EGF2).]\left.H=\frac{e G-2 f F+g E}{2\left(E G-F^{2}\right)} .\right]

(ii) Write ϕ(u,v)=(x(u,v),y(u,v),z(u,v))\phi(u, v)=(x(u, v), y(u, v), z(u, v)). Consider the complex valued functions

φ1=xuixv,φ2=yuiyv,φ3=zuizv\varphi_{1}=x_{u}-i x_{v}, \quad \varphi_{2}=y_{u}-i y_{v}, \quad \varphi_{3}=z_{u}-i z_{v}

Show that ϕ\phi is isothermal if and only if φ12+φ22+φ320\varphi_{1}^{2}+\varphi_{2}^{2}+\varphi_{3}^{2} \equiv 0.

Suppose now that ϕ\phi is isothermal. Prove that ϕ\phi is minimal if and only if φ1,φ2\varphi_{1}, \varphi_{2} and φ3\varphi_{3} are holomorphic functions.

(iii) Consider the immersion ϕ:R2R3\phi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} given by

ϕ(u,v)=(uu3/3+uv2,v+v3/3u2v,u2v2)\phi(u, v)=\left(u-u^{3} / 3+u v^{2},-v+v^{3} / 3-u^{2} v, u^{2}-v^{2}\right)

Find φ1,φ2\varphi_{1}, \varphi_{2} and φ3\varphi_{3}. Show that ϕ\phi is an isothermal parametrization of a minimal surface.

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