Minimal surfaces in the 3-dimensional Heisenberg group

166 Jun-ichi InoguchiProposition 2.1 ([4]-II) Let {ω 1 , ω 2 , ω 3 } be a solution to(5)(6)(7)¯∂ω 1 =λ2 (ω2 ∧ ω 3 − ω 2 ∧ ω 3 );¯∂ω 2 = − λ 2 (ω1 ∧ ω 3 − ω 1 ∧ ω 3 );¯∂ω 3 = − λ 2 (ω1 ∧ ω 2 + ω 1 ∧ ω 2 )on a simply connected coordinate region D ⊂ C. Then∫ z(ϕ(z, ¯z) = 2 Re ω 1 , ω 2 , ω 3 − λ )z 02 (x2 · ω 1 − x 1 · ω 2 )is a harmonic map into Nil 3 . Moreover if in addition, {ω 1 , ω 2 , ω 3 } satisfiesandω 1 ⊗ ω 1 + ω 2 ⊗ ω 2 + ω 3 ⊗ ω 3 = 0,ω 1 ⊗ ω 1 + ω 2 ⊗ ω 2 + ω 3 ⊗ ω 3 ≠ 0.Then ϕ is an minimal immersion.Conversely, any harmonic map of D into Nil 3 can be represented in this form.3 The normal Gauss mapLet ϕ : M → G(λ) be a conformal immersion. Take a unit normal vector field Nalong ϕ. Then, by the left translation we obtain the following smooth map:ψ := ϕ −1 · N : M → S 2 ⊂ g(λ).The resulting map ψ takes value in the unit sphere in the Lie algebra g(λ). Here,via the orthonormal basis {E 1 , E 2 , E 3 }, we identify g(λ) with Euclidean 3-spaceE 3 (u 1 , u 2 , u 3 ).The smooth map ψ is called the normal Gauss map of ϕ ([6], [3, p. 370]).Express the data as ω i = φ i dz. First we consider the case “φ 3 ≡ 0”. In this case,we have(x 1 z) 2 + (x 2 z) 2 = 0, |x 1 z| 2 + |x 2 z| 2 > 0,x 3 z + λ 2 (x1 zx 2 − x 1 zx 2 ) = 0.These equation imply that the minimal surface determined by the condition φ 3 ≡ 0is an integral surface of the distribution ω = 0. The distribution ω = 0 is integrableif and only if λ = 0. When λ = 0, then the surface is a vertical plane x 3 = constant.Hereafter we assume that φ 3 ≢ 0 and introduce the mapping f and g by(1) f := φ 1 − √ −1φ 2 , g :=φ 3φ 1 − √ −1φ 2 .By definition, f and g take values in the extended complex plane C = C∪{∞}. Usingthese functions f and g, we obtain the following formula.