) R 2. 2tR 2 R 2 +t 2 y = t2 R 2. x = R y, t 2 +R 2 R. 2uR2 R 2 +u 2 +v 2 2vR2. x = y = R z v = Ry, R z. z = u2 +v 2 R 2.

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1 Homework. Solutions 1 a Write own explicit formulae expressing stereographic coorinates for n-imensional sphere x x n+1 of raius via coorinates x 1,..., x n+1 an vice versa. For simplicit ou ma consier cases n, 3. b Check that for unit sphere S x + + z 1 all the points with rational cartesian coorinates x,, z have rational stereographic coorinates u, v an vice versa. a Write own the stereographic projection from the North pole of the sphere point N 0, 0,..., on the plane x n+1 0. Consier the segment ND which intersects the sphere at the point x 1,..., x n+1 x 1 + x x n+1. This segment intersects the plane x n+1 0 at the point D with coorinates x i u i for i 1,..., n. Then comparing similar triangles we have an ui xn+1 x i, i.e. x i ui i 1,..., n xn+1 x i ui x n+1, i 1,..., n. Using the fact that x x n+1 we come to x x n x n+1 n u i x n+1 + x n+1. Diviing b x n+1 x n+1 since North pole is remove we come to n x n+1 i1 ui n i1 ui +, u i xi n i 1,,... i1 ui + For projection with centre in South pole we have to change x n+1 x n+1. Write own these formulae for cases n 1,, 3, Case n 1: Circle x +. Stereographic coorinate t. Centre of projection 0, : i1 t x, x t +t t t + 1 Case n : Sphere x + + z. Stereographic coorinates u, v. Centre of projection 0, 0, : u x z v, z x u +u +v v +u +v z u +v u +v + Case n 3: 3-imensional sphere x + + z + t. Stereographic coorinates u, v, w. Centre of projection 0, 0, 0, : u x u x t +u +v +w v v t, +u +v +w w z w z t +u +v +w z u +v +w u +v +w + b We see that from explicit formulae. This is rational transformation of conic surfaces. Consier the iemannian metric on the circle of the raius inuce b the Eucliean metric on the ambient plane. 1

2 a Express it using polar angle as a coorinate on the circle. b Express the same metric using stereographic coorinate t obtaine b stereographic projection of the circle on the line, passing through its centre. iemannian metric of Eucliean space is G x +. x cos ϕ a using the angle: In this case parametric equation of circle is sin ϕ. Then G x + x cos ϕ, sin ϕ cos ϕ + sin ϕ ϕ. b In stereographic coorinate using 1 we have: G x + xxt,t t t + t + + t t + t 4t t tt + t + + t tt [ t t + t + t t + t + t tt ] + t t t + t + t + 44 t + t 44 t + t 3 Consier the iemannian metric on the sphere of the raius inuce b the Eucliean metric on the ambient 3-imensional space. a Express it using spherical coorinates on the sphere. b Express the same metric using stereographic coorinates u, v obtaine b stereographic projection of the sphere on the plane, passing through its centre. Then Solution iemannian metric of Eucliean space is G x + + z. a using the spherical coorinates: In this case parametric equation of sphere is x sin θ cos ϕ sin θ sin ϕ. z cos θ G x + +z x sin θ cos ϕ, sin θ sin ϕ,z cos θ sin θ cos ϕ + sin θ sin ϕ + cos θ cos θ cos ϕθ sin θ sin ϕϕ + cos θ sin ϕθ + sin θ cos ϕϕ + sin θθ θ + sin θϕ. b in stereographic coorinates using we have G x + + z xxu,v,u,v,zzu,v u + u + v + v + u + v u + v + + u + v 4 u uuu + vv + u + v + u + v + v vuu + vv + u + v + u + v + uu + vv + + u + v u + v uu + vv + u + v u + v 4 [ u + v u uvv ] + [ v + u v uvu ] + 4 uu + vv } u + v 4 + u + v } u + v 44 u + v + u + v

3 emark In the case of n-imensional sphere S n of raius in n + 1-imensional Eucliean space E n+1 it can be efine b the equation x x n+1 1 in cartesian coorinates x 1,..., x n, x n+1 iemannian metric on this sphere inuce b the Eucliean metric in the ambient space in stereographic coorinates has following appearance: G x x n+1 n x µ x i u i j1 u j + n i1 ui 44 n i1 ui + n i1 ui 4 Consier the surface L which is the upper sheet of two-sheete hperboloi in 3 : L: z x 1, z > 0. n i1 + ui + n i1 ui a Fin parametric equation of the surface L using hperbolic functions cosh, sinh following an analog with spherical coorinates on the sphere. The surface L sometimes is calle pseuo-sphere. b Consier the stereographic projection of the surface L on the plane OXY, i.e. the central projection on the plane z 0 with the centre at the point 0, 0, 1. Show that the image of projection of the surface L is the open isc x + < 1 in the plane OXY. x sinh θ cos ϕ a Parametric equation is sinh θ sin ϕ z cosh θθ cos ϕ We see that the conition z x 1 is fulfille. Compare with equation of sphere in spheric coorinates. b Calculations are ver similar to the case of stereographic coorinates for -sphere x + + z 1 of the raius 1. Stereographic coorinates u, v. Centre of projection 0, 0, 1: We have u x v 1. u x Hence v. Since x u1 + z, v1 + z then z 1 x + an z 1 u + v 1 + z, i.e. z 1+u +v 1 u v. We come to u x v The image of upper-sheet is an open isc u + v 1 since u + v x + sheet z > 1 then 0 z 1 z+1 < 1., u x 1 u v v 1 u v, u < 1, v < 1. 4 z u +v +1 1 u v z 1 5 Consier the pseuo-iemannian, pseuo-eucliean metric on 3 given b the formula s x + z. z 1 z+1. Since for upper Calculate the inuce metric on the surface L consiere in the Exercise 4, an show that it is a iemannian metric it is positive-efinite. Perform calculations in spherical-like coorinates see Exercise 4a above an in stereographic coorinates see exercise 4b above 3

4 emark The surface L sometimes is calle pseuosphere. The iemannian metric on this surface sometimes is calle Lobachevsk hperbolic metric. The surface L with this metric realises Lobachevsk hperbolic geometr, where Eucli s 5-th Axiom fails. This iemannian manifol manifol+iemannian metric we call Lobachevsk hperbolic plane. In stereographic coorinates we come to realisation of Lobachevsk plane on the isc in E. It is so calle Poincare moel of Lobachevsk geometr. Solution. The calculations will be ver similar to the calculations performe in the exercise 3 above. Just we nee consier cosh θ, sinh θ instea cos θ, sin θ an an sometimes changes the signs. First of all consier spherical-like coorinates: x sinh θ cos ϕ Equation of two-sheete hperboloi is sinh θ sin ϕ. Then z cosh θ G x + z xsinh θ cos ϕ,sinh θ sin ϕ,zcosh θ sinh θ cos ϕ + sinh θ sin ϕ cosh θ cosh θ cos ϕθ sinh θ sin ϕϕ + cosh θ sin ϕθ + sinh θ cos ϕϕ + sinh θθ θ + sinh θϕ. 1 0 matrix of iemannian metric is G 0 sinh. In the same wa as for sphere these coorinates are θ well-efine in all points except z ±1, where sin θ 0. Now express iemannian metric in stereographic coorinates 4: G x + z xxu,v,u,v,zzu,v u 1 u v + v 1 u v Compare with calculations for sphere x + + z 1. We have G x + z u uuu + vv v vuu + vv 1 u + v 1 u v + 1 u + v 1 u v uu + vv 1 u v + u + v + 1uu + vv 1 u v 4u + 4v 1 u v. u + v u v To perform these calculations it is convenient to enote b s 1 u v. esume: We come to the inuce iemannian metric on the surface from the pseuo-iemannian metric in ambient space. 6 Lobachevsk plane hperbolic plane L in stereographic coorinates can be consiere as an open isc u + v < 1 in the plane. In the previous exercise in particularl we calculate iemannian metric of L in these coorinates. Fin new coorinates x, such that in these coorinates Lobachevsk plane hperbolic plane can be consiere as an upper half plane > 0} an write own explicitl iemannian metric in these coorinates. Hint: You ma use complex coorinates: z x + i, z x i, w u + iv, w u iv an fin an holomorphic transformation w wz of the open isc w w < 1 onto the upper plane Imz > 0. Solution. ecall that in the previous exercise we calculate expression for Lobachevsk metric in stereographic coorinates u, v, u + v < 1. We come to the answer: G 4u +4v 1 u v see the previous exercise. It was 4

5 realisation of Lobachevsk plane on the Eucliean isc. Sometimes it calle Poincare moel of Lobachevsk hperbolic geometr. In complex coorinates w u + iv, w u iv the metric G 4u +4v obtaine in the exercise 8 can be rewritten G 4w w 1 w w. Inee G 1 u v 4w w 4u + ivu iv G 1 w w 1 u + ivu iv 4u + 4v 1 u v. It is a beautiful problem in complex analsis: fin Mobius transformation w az+b cz+ transformation which transforms the interior of circle w w 1 into upper half plane Imz > 0. One can see that w 1 + iz 1 iz, z i1 w 1 + w is the transformation which we nee Can ou fin all Mobius transformations which transform upper half plane to the interior of unit circle?. Now calculate G in coorinates z, z. i.e. in coorinates x, : We have Hence G w G 4u + 4v 4w w 1 u v 1 w w 1 + iz iz i z, w 1 iz 1 iz 1 + i z, 1 w w iz 1 i z 1 iz 1 + i z i z z 1 iz1 + i z 4w w 4 1 w w iz i z 1 iz 1+i z 4 z z 1 iz 1+i z 4 z z z x +, since z x + i an z z i. We come to the ver useful interpretation of hperbolic geometr: upper half plane in E with metric G x +. Later b efault we will call Lobachevsk hperbolic plane the realisation of Lobachevsk plane as an half-upper plane in E with these coorinates x, > 0 with metric G x +. 7 Consier the metric inuce on one-sheete hperboloi x + z 1 embee in 3 with the pseuo-eucliean metric x + z see the exercise 5. Show that this metric is not iemannian one. Solution. One can perform straightforwar calculations in spherical-like coorinates: Equation of onesheete hperboloi is cosh θ sin ϕ. Then x cosh θ cos ϕ z sinh θ G x + z xcosh θ cos ϕ,cosh θ sin ϕ,zsinh θ cosh θ cos ϕ + cosh θ sin ϕ sinh θ sinh θ cos ϕθ cosh θ sin ϕϕ + sinh θ sin ϕθ + cosh θ cos ϕϕ cosh θθ θ + cosh θϕ. 1 0 matrix is G 0 cosh. The conition of positive-efiniteness is not fulfille. This is not iemannian θ metric. Another solution Consier the vectors e an f z attache at the point 1, 0, 0. One can see that these vectors are tangent to the hperboloi, but the have the length of ifferent sign. One of these vectors is space-like vector, another time like vector. We have pseuoriemannian metric at the tangent space spanne b these two vectors. 5