Hyperbolic Geometry David Gu Yau Mathematics Science Center Tsinghua University Computer Science Department Stony Brook University [email protected] September 12, 2020 David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 1 / 65 Uniformization Figure: Closed surface uniformization. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 2 / 65 Hyperbolic Structure Fundamental Group Suppose (S; g) is a closed high genus surface g > 1. The fundamental group is π1(S; q), represented as −1 −1 −1 −1 π1(S; q) = a1; b1; a2; b2; ; ag ; bg a1b1a b ag bg a b : h ··· j 1 1 ··· g g i Universal Covering Space universal covering space of S is S~, the projection map is p : S~ S.A ! deck transformation is an automorphism of S~, ' : S~ S~, p ' = '. All the deck transformations form the Deck transformation! group◦ DeckS~. ' Deck(S~), choose a pointq ~ S~, andγ ~ S~ connectsq ~ and '(~q). The 2 2 ⊂ projection γ = p(~γ) is a loop on S, then we obtain an isomorphism: Deck(S~) π1(S; q);' [γ] ! 7! David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 3 / 65 Hyperbolic Structure Uniformization The uniformization metric is ¯g = e2ug, such that the K¯ 1 everywhere. ≡ − 2 Then (S~; ¯g) can be isometrically embedded on the hyperbolic plane H . The On the hyperbolic plane, all the Deck transformations are isometric transformations, Deck(S~) becomes the so-called Fuchsian group, −1 −1 −1 −1 Fuchs(S) = α1; β1; α2; β2; ; αg ; βg α1β1α β αg βg α β : h ··· j 1 1 ··· g g i The Fuchsian group generators are global conformal invariants, and form the coordinates in Teichm¨ullerspace. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 4 / 65 Hyperbolic Plane Definition (Poincar´eModel) 2 The upper half complex plane H = z C z = x + iy; y > 0 equipped with the hyperbolic metric f 2 j g dx2 + dy 2 dz 2 h = = j j ; y 2 y 2 is the Poincar´emodel of hyperbolic plane. We can see the hyperbolic metric is conformal to the Euclidean metric, hence the model is angle preserving. 2 2 2λ 1 1 1 2 @ @ e = ; K = ∆g log = y + log y = 1: y 2 − 2 y 2 @x2 @y 2 − David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 5 / 65 Hyperbolic Geometrics Consider γ(t) = (x(t); y(t))t [a; b] the hyperbolic length of the curve is 2 b b p Z Z x_ 2(t) +y _ 2(t) L(γ) = γ_ (t) hdt = dt; a j j a y(t) whereγ _ (t) = (x _(t); y_(t)). ib γ(t) ia Figure: Vertical axis is a geodesic. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 6 / 65 Hyperbolic Geodesics Lemma The positive imaginary axis is a geodesic b d(ia; ib) = ln : a Proof. Construct a smooth curve γ(a) = ia, γ(b) = ib, real number a < b, y(t) > 0 Z b Z b y_(t) y_(t) b L(γ) j jdt dt = ln ; ≥ a y(t) ≥ a y(t) a Equality holds, if and only ifx _(t) 0,y _(t) 0, equivalently γ([a; b]) is the imaginary axis, and y(t) is monotonously≡ ≥ increasing. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 7 / 65 Isometric Transformation Lemma 2 2 M¨obiustransformation f : H H ! az + b a b f (z) = ; SL(2; R); cz + d c d 2 is hyperbolic isometric, where SL(2; R) is the special linear transformation group, ad bc = 1. − Proof. f is the composition of the following maps 1 z z + b; b R; z λz; λ R; z : 7! 2 7! 2 7! −z 2 it is obvious that f (z) = z + b Iso(H ). 2 David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 8 / 65 Isometric Transformation Proof. Now, let w = λz, dw 2 λ2 dz 2 dz 2 w ∗(h) = j j = j j = j j : (w)2 λ2 (z)2 (z)2 = = = 2 1 1 1 1 1 hence f (z) = λz Iso(H ). Let w = ,dw = 2 dz, (w) = ( ), 2 − z z = 2i z¯ − z 2 1 2 1 2 2 dw jzj4 dz jzj4 dz dz w ∗(h) = j j = j j = j j = j j : (w)2 1 ( 1 1 )2 1 1 (z z¯)2 (z)2 = − 4 z¯ − z − 4 (¯zz)2 − = 1 2 hence f (z) = Iso(H ). − z 2 David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 9 / 65 Hyperbolic Geodesics Figure: All hyperbolic geodesics are lines or circular arcs orthogonal to the real axis. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 10 / 65 Hyperbolic Geodesics Corollary All hyperbolic geodesics are lines (z) = c or circular arcs < z a = ra R orthogonal to the real axis. j − j 2 Proof. 2 imaginary axis is a geodesic, f (z) = z + b Iso(H ), hence all vertical lines (z) = c are geodesics. 2 < David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 11 / 65 Hyperbolic Geodesics z 1 7→ −z Figure: Reflection transforms vertical lines to circular arcs orthogonal to the real axis. Isometry z 1 maps a vertical line (z) = c to a circular arc 7! − z < w + 1=(2c) = 1=(2 c ), j j j j 1 1 iy c 1 1 − + = − = : c + iy 2c iy + c 2c 2 c j j composed with isometry z z + b, we obtain circular arc z a = r, 7! j − j a R. Hence all z a = r are hyperbolic geodesics. 2 j − j David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 12 / 65 Hyperbolic Geodesic v v p p Figure: Unique geodesic through p and with tangent direction v. We show all geodesics are vertical lines or circular arcs. Through any point 2 p on H and with tangent direction v, there is a unique geodesic. We can construct a circular arc or vertical line τ through p with tangent direction v. By the uniqueness of geodesics, τ coincides with γ, γ = τ. Hence all geodesics are vertical lines or circular arcs orthogonal to the real axis. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 13 / 65 Hyperbolic Isometric Transformation Group Lemma Isometric transformation group on the hyperbolic plane is isomorphic to the M¨obiustransformation group: 2 PSL(2; R) = SL(2; R)= Id = Iso(H ): ± ∼ Proof. We define M¨obiustransformation cos θz + sin θ cos θ sin θ 'θ(z) = ; SL(2; R); sin θz + cos θ sin θ cos θ 2 − − We have 1 ' (i) = i;'0 (z) = ;'0 (i) = ei2θ: θ θ (cos θ sin θz)2 θ − David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 14 / 65 Hyperbolic Transformation Group 2 Choose arbitrarily a τ Iso(H ), assume τ(i) = ai + b. Construct 2 f PSL(2; R), f (z) = az + b, then f (i) = τ(i). Since both τ(z) and f (z) are2 isometric transformations, hence τ 0(i) = (τ(i)) = f 0(i) : j j = j j Construct a composition map g = f ' , satisfying ◦ θ 0 0 0 i2θ 0 g (i) = f (i)'θ(i) = e f (i): Let θ = 1 (arg τ 0(i) arg f 0(i)), then we get g(i) = τ(i) and g 0(i) = τ 0(i). 2 − Then we show g(z) = τ(z). David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 15 / 65 Hyperbolic Isometric Transformation Group Construct an isometric transformation η = τ g −1, then η(i) = i, and 0 2 ◦ η (i) = 1. Given a pont z H , z = i, there is a unique geodesic γ 2 6 connecting i and z, γ(0) = i, γ(s) = z, where the arc length parameter s equals to the hyperbolic distance between i and z. Isometric map η maps the geodesic γ to geodesicγ ~ = η(γ), preserving the arc length s, hence η(z) = η(γ(s)) =γ ~(s): (1) By construction γ~(0) = η(γ(0)) = η(i) = i = γ(0); furthermore γ~0(0) = η0(γ(0))γ0(0) = η0(i)γ0(0) = γ0(0): hence geodesics γ andγ ~ coincide, γ(s) =γ ~(s), combining Eqn.(1), we have η(z) =γ ~(s) = γ(s) = z: −1 hence η = τ g = id, τ = g, τ SL(R; 2). ◦ 2 David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 16 / 65 Hyperbolic Isometric Transformation Group By the definition of M¨obiustransformation, we have az + b a b a b f (z) = ; corresponds to and − − ; cz + d c d c d − − Hence M¨obiustransformation group is isomorphic to PSL(R; 2). David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 17 / 65 Cross Ration Definition (Cross Ratio) Suppose a; b; c; d C^ are distinct points on the extended complex plane, the cross ratio is defined2 as a c b c (a; b; c; d) = − : − : a d b d − − d c θ2 θ1 b a Figure: Four points are co-circular if and only if their cross ration is real. David Gu (Stony Brook University) Computational Conformal Geometry September 12, 2020 18 / 65 Cross Ratio Lemma Assume a; b; c; d C^, the four points are co-circular, if and only if their 2 cross ratio is real, (a; b; c; d) R. 2 Proof. As shown in figure (6), the surficient and necessary condition for the four ponts to be co-circular is θ1 equals to θ2, this is equivalent to (a; b; c; d) R.