Solutions of homework 1. 2 a) Using the stereographic projection from the north pole N = (0, 0, 1) introduce stereographic coordinates

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1 Solutions of homework 1 1 a) Using the stereographic projection from the north pole N (0, 1) introduce stereographic coordinate for the part of the circle S 1 ( + 1) without the north pole. b) Do the same but using the south pole S (0, 1) instead of the north pole. c) epress the stereographic coordinates obtained in a) and b) in terms of the angle ϕ (for polar coordinates in R, r cos ϕ, r sin ϕ). a) Consider the line which passes through the north pole (0, 1) the point (, ) on the circle and intersects the ais O at the point (u, 0); u u(, ) is a stereographic coordinate. Three points: north pole, the point (, ) on the circle and the point (u, 0) are on the same line. Hence u + 1. On the other hand + 1. Substitute 1 /u in the equation of the circle we come to quadratic equation: ( + 1 u) 1 This equation has two solution 0, ( 1) it is the North pole (0, 1) which belongs to the line and the circle. The second solution is u. Hence 1 /u 1 u and u 1. We come to the relations: u 1 u, u 1 (1.1) b) Consider the line which passes through the south pole (0, 1), the point (, ) on the circle and intersects the ais O at the point (u, 0). u u(, ) is a stereographic coordinate related with south pole. Under the reflection with respect to the ais O this line will go to the line which pass through the south pole (0, 1) and the point (, ) on the circle:, in the equation (). u 1+(u ), u (1.) (u ) (u ) The relation between coordinates u and u is uu 1. (1.3) c) If ϕ is an angle between ais O and the point (, ) on the circle, then cos ϕ, sin ϕ and u u 1 cos ϕ 1 sin ϕ sin( π ϕ) ( π 1 cos ( π ϕ) cotanh 4 ϕ ), 1 + cos ϕ 1 + sin ϕ sin( π ϕ) ( π 1 + cos ( π ϕ) tanh 4 ϕ ). (1.4) a) Using the stereographic projection from the north pole N (0, 0, 1) introduce stereographic coordinates (u, v) for the part of the sphere S ( + + z 1) without the north pole. b) Do the same but introduce stereographic coordinates (u, v ) using the south pole S (0, 0, 1) instead of the north pole. c) epress these stereographic coordinates via spherical coordinates θ, ϕ (for spherical coordinates in R 3, r sin θ cos ϕ, r sin θ sin ϕ, z r cos θ). a) In the same wa as in the previous eercise consider the line which passes through the North Pole (0, 0, 1) the point (,, z) on the sphere and intersects the plane O at the point with coordinates (u, v, 0). u u(,, z), v v(,, z) are stereographic coordinates of the sphere. We come to conditions + + z 1 and the condition that three points (0, 0, 1), (,, z) and (u, v, 0) belong to the same line, i.e. : : z 1 u : v : 1. We have the sstem of equations u v 1 z v + + z 1 1

2 Solving these equations we come to u +v v +v and z 1 u v +v u v 1 z 1 z (.1) b) Consider the line which passes through the South Pole (0, 0, 1) the point (,, z) on the sphere and intersects the plane O at the point with coordinates (u, v, 0). u u (,, z), v v (,, z) are stereographic coordinates of the sphere related with the South pole. Note that changing z z in the previous formulae lead to these stereographic coordinates related with South Pole (0, 0, 1): The relation between u, v and u, v are u +v v +v z u +v 1 u +v +1 and u v 1+z 1+z (.) u v u v uu + vv 1, u u u +v v v (.3) u +v Now epress stereographic coordinates via spherical coordinates: u sin θ cos ϕ 1 z 1 cos θ cotanθ cos ϕ i.e. v sin θ sin ϕ 1 z 1 cos θ cotanθ sin ϕ (.4) u + iv cotan θ eiϕ Respectivel for coordinates u, v u sin θ cos ϕ 1 + z 1 + cos θ tan θ cos ϕ i.e. v sin θ sin ϕ 1 + z 1 + cos θ tan θ sin ϕ (.4) u + iv tan θ eiϕ 3 Introduce stereographic coordinates on n-dimensional sphere S n. Let 1,..., n+1 be cartesian coordinates in R n+1. The point (0, 0,..., 1) is the North pole of the sphere ( 1 ) ( n+1 ) 1. The straight line passes through the point of the sphere and the point on the plane n+1 0 with the coordinates (u 1,..., u n, 0) then u 1 u 1 ( 1,..., n+1 ),..., u n u n ( 1,..., n+1 ) are stereographic coordinates. Performing the analogous calculations we come to the answer: k u k 1 + (u 1 ) + (u ) (u n ), k uk 1 n+1 for k 1,, 3,..., n; n+1 1 (u1 ) (u )... (u n ) 1 + (u 1 ) + (u ) (u n ) (3.1)

3 Respectivel for projective coordinates related with the South Pole we have: k u k 1 + (u 1 ) + (u ) (u n ), k u k 1 + n+1 for k 1,, 3,..., n; n+1 (u 1 ) + (u ) (u n ) 1 (u 1 ) + (u ) (u n ) + 1 (3.1) Coordinates (u 1,..., u n ) and (u 1,..., u n ) are related with relations u k u k 1 + (u 1 ) + (u ) (u n ) k 1,,..., n 4 Show that stereographic projection is the bijection between rational points on the unit sphere (the points on the unit sphere with rational coordinates) and rational points on the plane. Find all rational solutions of the equations + 1, + + z 1. Find all integer solutions of the equation + z (Pthagoreans triples. ) The fact that it is bijection between rational points follows from eplicit formulae above. On the other hand this is eas to understand without an formula. The relation is given b intersection of straight line with plane and points of the sphere. If coordinates of the sphere are rational then the line has rational coefficients. We come to quadratic equation with rational coefficients. One of its solution is rational, hencethe second is rational too. Hence the line intersects the plane at the point with rational coordinates. The formulae (1.3), (.3) give solutions:,, z Q u, v, Q In particular u, 1 u are rational iff u is rational. Hence taking u p q we come to Pthagorian triples (p, p q, p + q ). 5 Considering the natural bijection of the part of real projective space 1) RP n on the plane n+1 1 in R n+1 (( 1,..., n, n+1 ) are cartesian coordinates on R n+1 ) one can introduce inhomogeneous coordinates on the part of the RP n. a) do this eercise for n 1, ; b) do this eercise for an arbitrar n. 6 Find different inhomogeneous coordinates on RP considering natural bijections of the parts of RP on the plane 3 1 or on the plane 1 or on the plane 1 1. Find relations between these coordinates. Do the eercises 5 and 6 simultaneousl. Consider in RP n n + 1 the domains U 1, U,..., U n+1 such that U 1 is the set of points with homogeneous coordinates [ 1 : :... : n+1 ] such that 1 0 U is the set of points with homogeneous coordinates [ 1 : :... : n+1 ] such that 0... U n+1 is the set of points with homogeneous coordinates [ 1 : :... : n+1 ] such that n+1 0 We define on ever set U k the map ϕ k of U k on R n b dividing the coordinates on the coordinate k. Geometricall this means that we find the coordinates of the intersection of the line [ 1 : :... : n+1 ] with the plane k1 Write down detailed formulae for the cases n 1, 1) n 1, RP 1 is the set of equivalence classes [ 1 : ]} ϕ 1 : [ 1 : ] u (1), u (1) 1, (6.1) 1) Real projective space RP n set of straight lines in R n+1 which pass throug the origin}, or in other words RP n is the set of equivalence classes of non-zero vectors in R n+1, i.e. RP n [ 1 : :... : n+1 }] where [ 1 : :... : n+1 ] stands for the equivalence class of a non-zero vector ( 1,..., n+1 ) R n+1. Two non-zero vectors ( 1,..., n+1 ), ( 1,..., n+1 ) are considered equivalent if the are proportional: 1 λ 1, λ,..., n λ n, n+1 λ n+1. 3

4 ( i.e. the line [ 1 : ] intersects the plane 1 1 at the point 1, ). 1 ϕ : [ 1 : ] u (), u () 1 (6.) i.e. the line [ 1 : ] intersects the plane 1 at the point 1, 1 We define two charts on RP 1 : coordinate u (1) on the domain U 1 ( 1 0), and the coordinate u () on the domain U ( 0). These charts form the atlas. Transition function is: u () 1 u (1) ) n, RP n is the set of equivalence classes [ 1 : : 3 ]} ϕ 1 : [ 1 : : 3 ] (u 1 (1), u (1) ), u1 (1) 1, i.e. the line [ 1 : : 3 ] intersects the plane 1 1 at the point ϕ : [ 1 : : 3 ] (u 1 (), u () ), u1 () 1, i.e. the line [ 1 : : 3 ] intersects the plane 1 at the point ( 1, 1, 3 1 )., u (1) 3 1, (6.1), u () 3, (6.) 1, 1, 3 and ϕ 3 : [ 1 : : 3 ] (u 1 (3), u (3) ), u1 (3) 1 3,, u (3) 3, (6.3) i.e. the line [ 1 : : 3 ] intersects the plane 3 1 at the point 1, 3, 1 3. We define three charts on RP n : coordinates (u 1 (1), u (1) ) on the domain U 1 ( 1 0), coordinates (u 1 (), u () ) on the domain U ( 0) and coordinates (u 1 (3), u (3) ) on the domain U 3 ( 3 0). These charts form the atlas. Transition functions are the following: between (u 1 (1), u (1) ) and (u1 (), u () ): u1 () 1 u 1, u () u (1) (1) u 1 (1) between (u 1 (1), u (1) ) and (u1 (3), u (3) ): u1 (3) 1 u, u (3) u1 (1) (1) u (1) between (u 1 (), u () ) and (u1 (3), u (3) ): u1 (3) u1 () u, u (3) 1 () u () 7 Do the previous eercise for comple projective spaces CP 1 and CP. Our formulae formall have the same appearance: 1) CP 1 is a set of comple lines [z 1 : z ] (real planes: (z 1, z ) (λz 1, λz for comple λ). Consider two charts. First chart: the set U 1 of the equivalence classes [z 1 : z ] such that z 1 0 and the map ϕ 1 : U 1 C: u z, (u C) z1 Second chart: the set U of the equivalence classes [z 1 : z ] such that z 0 and the map ϕ : U 1 C: w z1, (w C) z 4

5 We see that u 1 w. In terms of real coordinates u + i 1 w 1 v + it, i.e.,, it is a composition of inversion and reflection with respect to ais. v v + t, t v + t ) CP is a set of comple lines [z 1 : z : z 3 ] (real planes: (z 1, z ) (λz 1, λz for comple λ). Consider three charts. First chart: the set U 1 of the equivalence classes [z 1 : z : z 3 ] such that z 1 0 and the map ϕ 1 : U 1 C : u z z 1, v z3 z 1 Second chart: the set U of the equivalence classes [z 1 : z : z 3 ] such that z 0 and the map ϕ : U 1 C : u z1 z, v z3 z and third chart: the set U 3 of the equivalence classes [z 1 : z : z 3 ] such that z 3 0 and the map ϕ 3 : U 1 C : ũ z1 z 3, ṽ z z 3 These charts form atlas. 5