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Obtaining Non-Euclidean Geometries

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Obtaining Non-Euclidean Geometries Obtaining Non-Euclidean Geometries David Mwakima Fall 2016 1 Introduction Hyperbolic geometry is the geometry obtained from the metric (first fundamental form) of the pseudosphere. It is a non-Euclidean geometry, which means that the models of hyperbolic geometry satisfy some but not all of the axioms of Euclidean geometry. In particular, in this paper I shall show that the so-called “parallel ax- iom/postulate” of Euclidean geometry does not hold in the model of hyperbolic geometry I shall be considering, which is the upper plane model. Moreover, I shall show that in hyperbolic geometry the sum of internal angles of a hyperbolic trian- gle is strictly less than π contrary to the well-known result of Euclidean geometry, viz. the sum of internal angles of a Euclidean triangle is equal to π. An interesting fact also emerges from hyperbolic geometry, which is that the angles of a hyperbolic polygon determine its area, which is a result we do not have in Euclidean geometry. (Cf. Walkden (2016) p. 42 for an interesting discussion of these facts.) In section 2 I derive the metric of the upper plane model of hyperbolic geometry and show, using this metric, covariant derivatives and Christoffel symbols; that the Gaussian Curvature K of the pseudosphere is in fact -1. In section 3 I identify the geodesics of the upper-plane model and show that the parallel axiom does not hold in hyperbolic geometry. Finally in section 4 I show how to compute distances and area in hyper- bolic geometry and show that the sum of interior angles of a hyperbolic triangle is less than π and the more general result relating the area of a hyperbolic polygon with the sum of its interior angles. 1 2 The upper plane model of hyperbolic geometry Following the discussion on surfaces of constant Gaussian curvature in Pressley (2010) section 8.3. pp. 196ff, we can obtain the pseudosphere given by the coordinate patch p σ(w; v) = (ewcos(v); ewsin(v); 1 − e2w − cosh−1(e−w)) We obtain this coordinate patch by considering the case where K < 0 of the surface of − f¨ revolution σ(u; v) = (f(u)cos(v); f(u)sin(v); g(u)), which has K = f and choosing K = −1 In his discussion on p. 270, Pressley (2010) does not say why we must have w > 1 for coordinate patch he considers to be smooth and well-defined. But by considering the following derivation we can see why this is the case. Given σ(w; v) and setting u = e−w gives the reparametrized surface r ! 1 1 1 σ~(u; v) = cosv; sinv; 1 − − cosh−1u u u u2 Writing u~ = v....(1) and v~ = u = e−w....(2) we can get the reparametrization map Φ given by: ( w = −ln(~v) by solving for w in (2) Φ = v =u ~ iff J(Φ) is invertible, where J(Φ) is the Jacobian matrix of the map Φ ( ) ( ) @w @w − 1 @u~ @v~ 0 v~ J(Φ) = @v @v = @u~ @v~ 1 0 detJ(Φ) =6 0 provided u =v ~ > 1. So, it is because of this that Pressley (2010) p. 270 says w > 1. For otherwise the Jacobian tells us that the reparametrization map is not invertible. 2 We know the first fundamental form g of surface of revolution is given by: g = du2 + f(u)2dv2 (Cf. Pressley (2010) p. 123) In our case with σ(w; v) it is: 2 2w 2 gw;v = dw + e dv T and by taking J (gw;v)J we get du2 + dv2 g = u;v u2 for the reparametrization σ~(u; v) The upper half-plane model of hyperbolic geometry is the surface H = f(v; u) 2 R2ju > 0g equipped with the metric gu;v: Following Pressley (2010) p. 270, we can identify the upper half of the vu-plane described above with the upper half of the complex plane. So that we have: H = fz 2 CjIm(z) > 0g Given our model of hyperbolic geometry, we can then compute distance, areas and angles on our surface. But first we need to know what the length minimizing curves i.e. geodesics of hyperbolic geometry are. The length minimizing curves are the analogues of the familiar straight lines from Euclidean geometry. Using the geodesics of our model of hyperbolic geometry, I will show that the “parallel axiom” does not hold in hyperbolic geometry. Before doing this, however, I show how using covariant derivatives and Christoffel symbols we can confirm that the Gaussian Curvature K is indeed −1 The covariant derivatives with respect to v and u of some vector a = (α; β) are given by: r v v u u va = (@vα + Γvvα + Γvuβ; @vβ + Γvvα + Γvuβ) r v v u u ua = (@uα + Γuvα + Γuuβ; @uβ + Γuvα + Γuuβ) 3 We know that the Gaussian Curvature K measures the failure of rv and ru to commute. So we obtain the expression. 2 1 ? (rvru − rurv)a = K(EG − F ) 2 a (1) · · · ? − where E =σ ~u σ~u, G =σ ~v σ~v and F =σ ~u σ~v are the entries of g(u;v) and a = (β; α) if a = (α; β) du2+dv2 Given our metric g(u;v) = u2 , we can compute the Christoffel symbols Γ. After doing so we find that the only non-zero Christoffel symbols are: 1 1 Γu = Γv = Γv = − and Γu = uu uv vu u vv u Picking a = (1; 0) we compute the left hand side of (1) as follows: ( ) ( ) 1 1 1 r v u − va = (@v(1) + Γvu(0);@v(0) + Γvv(1)) = 0 (0); 0 + (1) = 0; ( u u ) ( u ) 1 1 1 r v u − − − ua = (@u(1) + Γuv(1);@u(0) + Γuu(0) = 0 (1); 0 (0) = ; 0 ( ) ( ( ))u ( u ) u r r r − 1 − 1 1 − 1 − 1 v( ua) = v ; 0 = 0 (0); 0 + = 0; 2 ( u) ( u ( )u (u )) ( u ) r r r 1 − 1 1 − 1 1 − 2 u( va) = u 0; = 0 (0);@u = 0; 2 ( u) ( u ) ( u ) u u u 1 2 1 r (r a) − r (r a) = 0; − − 0; − = 0; v u u v u2 u2 u2 q 1 1 1 ? − − 2 2 − The right hand side of (1) is: K(0; u2 ) since (EG F ) = u4 and a = (0; 1) Therefore, comparing the left and right hand side we find that K = −1, which is what we were to show. 3 Geodesics of hyperbolic geometry and the “Par- allel Axiom” Let γ(t) = (x(t); (y(t)) be a unit speed curve in H 4 By proposition 9.2.3 we say γ(t) is a geodesic iff it satisfies the geodesic equations: u 2 u u 2 u¨ + Γuuu_ + 2Γuvu_v_ + Γvvv_ = 0 (2) v 2 v v 2 v¨ + Γuuu_ + 2Γuvu_v_ + Γvvv_ = 0 (3) du2+dv2 u v Given our metric g(u;v) = u2 , we see that the only non-zero Γ’s are: Γuu = Γuv = − 1 u 1 H u and Γvv = u . So, we demand that γ(t) satisfy the geodesic equations of given by: 1 1 =u ¨ − u_ 2 + v_ 2 = 0 (4) u u 2 =v ¨ − u_v_ = 0 (5) u But we also know by Theorem 9.2.1. and equation (9.2) in Pressley (2010) p. 220 and proposition 9.2.3 on p. 223 that (5) implies that if γ(t) is a geodesic d (F u_ + Gv_) = 0 dt d v_ v_ 2 So, dt ( u2 ) = 0 since F = 0 and u2 = c where c is a constant and v_ = cu :::(τ). There are two cases: Case 1: c = 0 Then v_ = 0 and v = k =) the geodesics are vertical straight lines in the vu - plane. Case 21: c =6 0 Here we use the fact that γ(t) is unit speed and so we write: u_ 2 +v _ 2 = 1 (6) u2 u_ 2 = u2 − v_ 2 (7) p u_ = u (1 − c2u2) plugging in v_ = cu2::::(τ) in (7) and simplying. (8) 1I thank Cliff Taubes for making this clear tome 5 We can divide (τ) and (8) to get: dv cu2 = p du u (1 − c2u2) cu = p (1 − c2u2) 1 d p = ∓ (1 − c2u2) c du Rearranging gives: ! d 1p v (1 − c2u2) = 0 du c which implies 1p v (1 − c2u2) = v where v is some constant c 0 0 1p v − v = ∓ (1 − c2u2) 0 c 1 (v − v )2 = (1 − c2u2) 0 c2 1 (v − v )2 + u2 = (9) 0 c2 But (9) is the equation of a semi-circle with center (v0; 0) in the vu - plane. Having exhausted all the cases, it follows that there are two length minimizing curves in H. The geodesics v = k for some constant k i.e. vertical half-lines orthogonal to the real axis (u = 0) and the semi-circles with centers on the real axis. These results explain Proposition 11.1.2 in Pressley (2010) p. 271, which we then use to state and explain Proposition 11.1.3 in (Ibid.). Proposition 11.1.3 (i) There is a unique geodesic passing through any two distinct points of H (ii) Suppose γ is a geodesic in H and that p is a point in the upper-half plane that is not in γ. Then there are infinitely many geodesics though p that do not interesect γ Proof 6 (i) Let α; β 2 H and α =6 β. If the line passing though α and β is parallel to the imaginary axis, then the unique geodesic passing through the points α, β is the half-vertical line containing them.
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