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8 Surfaces Invariant by One-Parameter Group of Hyperbolic Isometries Having Constant Mean Curvature in ˜ PSL2(R,Τ)

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8 Surfaces Invariant by One-Parameter Group of Hyperbolic Isometries Having Constant Mean Curvature in ˜ PSL2(R,Τ) 8 Surfaces invariant by one-parameter group of hyperbolic isometries having constant mean curvature in PSL2(R,τ) g On (3) Ricardo Sa Earp gave explicit formulas for parabolic and hyper- bolic screw motions surfaces immersed in H2 R. There, they gave several × examples. In this chapter we only consider surfaces invariant by one-parameter group of hyperbolic isometries having constant mean curvature immersed in P] SL (R, τ). Since for τ 0 we are in H2 R then we have generalized the result 2 ≡ × obtained by Ricardo Sa Earp when the surface is invariant by one-parameter group of hyperbolic isometries having constant mean curvature. In this chapter we focus our attention on surfaces invariant by one- parameter group of hyperbolic isometries. To study this kind of surfaces, we take M 2 = H2 that is the half plane model for the hyperbolic space. Thus, the metric of M 2 is given by: 1 dσ2 = λ2(dx2 + dy2), λ = . y From Proposition 5.1.1, we know that to obtain a hyperbolic motion on ] P SL2(R, τ), it is necessary consider a hyperbolic motion (hyperbolic isometry) on H2. 8.1 Surfaces invariant by one-parameter group of hyperbolic isometries main lemma The idea to obtain a surface invariant by one-parameter group of hyper- bolic isometries is simple, we will take a curve in the xt plane and we will apply ] one-parameter group Γ of hyperbolic isometries on P SL2(R, τ). We denote by α(x)=(x, 1,u(x)) the curve in the xt plane and by S = Γ(α), the surface invariant by one-parameter group of hyperbolic isometries generate by α. Since the most simple hyperbolic isometry on H2 is the homotetia from the origin, we re-parameterized the hyperbolic plane whit coordinates R+ and Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 124 θ (0, π), so that ∈ x = R cos(θ) y = R sin(θ) where R R and θ (0, π). ∈ ∈ Thus, the surface S is parameterized by, ϕ(ρ, θ)=(R cos(θ), R sin(θ),u(θ)) Now, we rewrite all the expressions in this news coordinates, thus we obtain the next lemma. Lemma 8.1.1. By considering the above re-parametrization to the hyperbolic plane, we may rewrite all terms in the form, sin(θ) – ∂ = cos(θ)∂ ∂ x R − R θ cos(θ) – ∂ = sin(θ)∂ + ∂ y R R θ – Rx = cos(θ) – Ry = sin(θ) sin(θ) – θ = x − R cos(θ) – θ = y R 1 – λ = R sin(θ) 1 – dσ2 = (dR2 + R2dθ2) R2 sin2(θ) Proof. From, x = R cos(θ), (8-1) ( y = R sin(θ), we obtain, dx = cos(θ)dR R sin(θ)dθ, − (8-2) ( dy = sin(θ)dR + R cos(θ)dθ, since, dx2 + dy2 = dR2 + R2dθ2, 1 λ = R sin(θ) Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 125 1 then λ2 = , thus, R2 sin2(θ) dσ2 = λ2(dx2 + dy2) 1 = dR2 + R2dθ2 R2 sin2(θ) Furthermore, setting ∂x = a∂R + b∂θ and evaluating in equation (8-2). We obtain, 1 = cos(θ)a R sin(θ)b, − ( 0 = sin(θ)a + R cos(θ)b, hence, sin(θ) a = cos(θ), b = − R that is, sin(θ) ∂ = cos(θ)∂ ∂ . x R − R θ By considering, ∂y = a∂R + b∂θ and evaluating in equation (6-2). We obtain, 0 = cos(θ)a R sin(θ)b, − ( 1 = sin(θ)a + R cos(θ)b, so cos(θ) a = sin(θ), b = R that is, cos(θ) ∂ = sin(θ)∂ + ∂ . y R R θ From x2 + y2 = R2, we obtain, 2x = 2RRx 2y = 2RRy Considering the equation (8-1), we obtain, Rx = cos(θ) Ry = sin(θ) Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 126 Derivation of equation (8-1) with respect to x and y respectively gives, 1= R cos(θ) R sin(θ)θ x − x 1= Ry sin(θ)+ R cos(θ)θy Now, by using Rx and Ry, we obtain, sin(θ) θ = x − R cos(θ) θ = y R This completes the proof. The next Lemma is crucial for our study. We follow ideas of Appendix A of (11). Consider the graph t = u(θ) in the plane xt, and denote by S = grap(Γu). Supposing that S has constant mean curvature H, we have the lemma: Lemma 8.1.2. The function u satisfies (d 2H cot(θ)) 1+4τ 2 cos2(θ) u(θ)= − dθ 2τθ (8-3) 1 sin2(θ)(d 2H cot(θ))2 − Z − p− where d is a real number.p Proof. Since S has mean curvature H, then by lemma 5.2.1 the function u satisfies the equation α β 2H = divH2 e + e , (8-4) W 1 W 2 u u where W = 1+ α2 + β2, α = x 2τ, and β = y . λ − λ p By abuse of notation we consider u(x, y)= u(R(x, y), θ(x, y)), since S is a surface invariant by one-parameter group of hyperbolic isometries, then the function u does not depend of R, so, ux = uRRx + uθθx = uθθx, ( uy = uRRy + uθθy = uθθy, Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 127 By considering the lemma 8.1.1, this implies, u x = u sin2(θ) λ − θ u y = u sin(θ) cos(θ) λ θ Thus α = u sin2(θ) 2τ − θ − β = uθ sin(θ) cos(θ) this implies, 2 2 2 2 2 2 α + β = sin (θ)(uθ +2τ) +4τ cos (θ) 2 2 2 2 2 W = 1 + sin (θ)(uθ +2τ) +4τ cos (θ) α β 1 ∂ ∂ Setting X = e + e = α x + β y . u W 1 W 2 W λ λ We need express Xu in coordinates R and θ. Observe that, ∂ x = R sin θ cos θ∂ sin2 θ∂ λ R − θ and ∂ y = R sin2 θ∂ + sin θ cos θ∂ λ R θ Furthermore, ∂ ∂ α x + β y = (sin2 θu +2τ)(R sin θ cos θ∂ sin2 θ∂ )+ λ λ − θ R − θ 2 + sin θ cos θuθ(R sin θ∂R + sin θ cos θ∂θ) = [R sin3 θ cos θu +2τR sin θ cos θ]∂ + [sin4 θu +2τ sin2 θ]∂ − θ R θ θ 3 2 2 +R sin θ cos θuθ∂R + sin θ cos θuθ∂θ = 2τR sin θ cos θ∂ + sin2 θ(u +2τ)∂ − R θ θ So, we conclude that, 1 X = sin2 θ(u +2τ)∂ 2τR sin θ cos θ∂ u W θ θ − R and 2 2 2 2 W = 1+4τ cos θ + sin θ(uθ +2τ) q Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 128 Let θ , θ (0, π) with θ < θ and R , R R+ with R < R and 0 1 ∈ 0 1 0 1 ∈ 0 1 consider the domain Ω = [θ , θ ] [R , R ] in the plane R+θ. 0 1 × 0 1 η3 γ3 θ1 γ4 η2 Ω η4 γ 2 θ0 γ1 η1 R0 R1 By integrating the equation (8-4), we obtain X , η =2HArea([θ , θ ] [R , R ]) h u i 0 1 × 0 1 Z∂(Ω) where η is the outer co-normal. Since X , η = X , η + X , η + X , η + X , η h u i h u 1i h u 2i h u 3i h u 4i Z∂Ω Zγ1 Zγ2 Zγ3 Zγ4 we compute each integral: For the fist integral: Observe that, γ (s)=(s, θ ), R s R . 1 0 0 ≤ ≤ 1 This implies, γ1′ = ∂R, thus 1 γ′ = , | 1| R sin θ and η = sin θ∂ . 1 − θ Furthermore sin3 θ(u +2τ)(θ ) 1 sin θ(u +2τ) X , η = θ 0 = θ , h u 1i − W sin2 θ − W Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 129 hence R1 u +2τ X , η = θ (θ )dR h u 1i − RW 0 Zγ1 ZR0 For the third integral: Observe that, γ (s)=(R s, θ ), 0 s R R . 3 1 − 1 ≤ ≤ 1 − 0 This implies, γ′ = ∂ , 3 − R thus 1 γ′ = , | 3| R sin θ and η3 = sin θ∂θ. Furthermore sin3 θ(u +2τ)(θ ) 1 sin θ(u +2τ) X , η = θ 1 = θ , h u 3i W sin2 θ W hence, R1 R0 R1 − sin θ(u +2τ) 1 (u +2τ) X , η = θ ds = θ (θ )dR h u 3i W (R s) sin θ RW 1 Zγ3 Z0 1 − ZR0 For the second integral: Observe that, γ (s)=(R ,s), θ s θ . 2 1 0 ≤ ≤ 1 This implies, γ2′ = ∂θ, thus 1 γ′ = , | 2| sin θ and η2 = R sin θ∂R. Furthermore 2τR2 sin2 θ cos θ 1 2τ cos θ X , η = = , h u 2i − W R2 sin2 θ − W Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 130 hence, θ1 2τ cos θ X , η = dθ h u 2i − W sin θ Zγ2 Zθ0 For the four integral: Observe that, γ (s)=(R , θ s), 0 s θ θ . 4 0 1 − ≤ ≤ 1 − 0 This implies, γ′ = ∂ , 4 − θ thus 1 γ′ = , | 4| sin θ and η = R sin θ∂ . 4 − R Furthermore 2τR2 sin2 θ cos θ 1 2τ cos θ X , η = = , h u 4i W R2 sin2 θ W hence, θ1 θ0 θ1 − 2τ cos(θ s) 1 2τ cos θ X , η = 1 − ds = dθ h u 4i W sin(θ s) W sin θ Zγ4 Z0 1 − Zθ0 Taking into account this four integrals, we obtain R1 (u +2τ) dR R1 (u +2τ) dR X , η = θ (θ ) θ (θ ) h u i W 1 R − W 0 R Z∂Ω ZR0 ZR0 Observe that, θ1 R1 Area(Ω) = det(gij)dRdθ θ0 R0 Z Z q θ1 R1 1 = dRdθ R sin2 θ Zθ0 ZR0 So, we conclude that R1 (u +2τ) dR R1 (u +2τ) dR θ1 R1 1 θ (θ ) θ (θ ) = dRdθ W 1 R − W 0 R R sin2 θ ZR0 ZR0 Zθ0 ZR0 Surfaces of Constant Mean Curvature in Homogeneous Three Manifolds with ] Emphasis in PSL2(R, τ) 131 which we can write in the form: θ1 (u +2τ) θ1 1 ∂ θ dθ =2H dθ θ W sin2 θ Zθ0 Zθ0 By considering that Ω is any domain in the plane Rθ, and taking the derivative with respect to ρ we obtain: u +2τ 2H ∂ θ = θ W sin2 θ by integrating this expression: u +2τ θ = 2H cot θ + d 2 2 2 2 1+4τ cos θ + sin θ(uθ +2τ) − where d R.p This implies: ∈ (u +2τ)2[1 sin2 θ(d 2H cot θ)2]=(d 2H cot θ)2[1+4τ 2 cos2 θ] θ − − − so the function u satisfies: (d 2H cot θ)√1+4τ 2 cos θ u(θ)= − dθ 2τθ 1 sin2 θ(d 2H cot θ)2 − Z − − p 8.2 Examples of surfaces invariant by one-parameter group of hyperbolic isometries in PSL2(R, τ) Now, we explore the Lemma 8.1.2.
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