Notes on spherical geometry

Transcription

1 Notes on spherical geometry Math 130 Course web site: This handout covers some spherical trigonometry (following yan s exposition) and the topic of dual triangles. 1. Some tools Trig identities We will need to remember some trigonometric identities. In particular, we should recall cos(a + b) cos(a) cos(b) sin(a) sin(b) (1) cos(a b) cos(a) cos(b) + sin(a) sin(b) and the related formula ( X + Y cos(x ) cos(y ) 2 sin 2 ) ( ) X Y sin. (2) 2 Cross products For vectors A and B in 3, the cross-product A B can be characterized geometrically by the conditions: A B is zero if A and B are linearly dependent; otherwise: A B is orthogonal to both A and B;

2 A, B and A B form a right-handed basis; the length of A B is A B sin θ, where θ is the angle between the vectors. Algebraically, if then The inner product a 1 is the determinant of the matrix b 1 A a 2 and B b 2 a 3 b 3 a 2 b 3 a 3 b 2 A B a 3 b 1 a 1 b 3. a 1 b 2 a 2 b 1 A, B C a 1 b 1 c 1 a 2 b 2 c 2. a 3 b 3 c 3 From the familiar symmetries of the determinant, we can deduce A, B C B,C A. (3) The vector cross-product is not associative, so (A B) C and A (B C ) are usually different. There is a very useful formula involving these triple products: (A B) C A,C B B,C A. (4) 2. Spherical geometry Incidence geometry on the sphere We write S 2 for the unit sphere in 3 : S 2 { x 3 x 1 }.

3 By a line in spherical geometry, we will mean a great circle on the sphere S 2. In spherical geometry, any two distinct lines meet in exactly two (antipodal) points. If P and Q are points in S 2 that are neither equal nor antipodal, then there is a unique line containing both of them. To set down some formulae for these things, we first note that we can described a great circle in S 2 as the set of unit vectors that are orthogonal to a given unit vector ξ in 3. That is, every line l in spherical geometry can be written as l { x S 2 x, ξ 0 }, for some ξ in S 2. The point ξ is called a pole for the line l. The line l is called the polar line for the point ξ. For each line l, there are two possible choices for the pole: if ξ is one, then ξ is the other. To compute the line that goes through two given points on S 2, we proceed as follows. If P and Q are in S 2 and are neither equal nor parallel, then they are linearly independent vectors in 3 ; and P Q is therefore non-zero. The point ξ P Q P Q is therefore a point on the unit sphere which is orthogonal to the vectors P and Q. It is the pole corresponding to the unique line through P and Q. To compute the intersection of two distinct lines l and m in S 2, we take ξ and η to be poles for the two lines; then the points of l m are the points on S 2 that are orthogonal to both ξ and η. Again, the cross-product allows us to construct these points: the two points of intersection of l and m are ± ξ η ξ η. Distances and angles on the sphere Let P and Q be distinct points on the sphere S 2. If P and Q are not antipodal, then there is exactly one line (great circle) passing through both points. The points themselves divide the great circle into two segments. The shorter of these is called the minor segment. The distance from P to Q is the length of this minor segment. If

4 P and Q are antipodal, then all great-circle segments from P to Q have length π, and in this case d(p,q ) is π. All this can be captured more simply by the formula d(p,q ) cos 1 P,Q. Now let P, Q and R are three points on S 2. Suppose that Q ±P and Q ±R. This means that there are unique minor segments joining Q to P and Q to R. How should we understand the angle between these to segments, where they meet at the point Q? The great circles through QP and QR are the intersection of the sphere with two planes in 3, and the angle we seek is the angle between these two planes. This is the same as the angle between vectors orthogonal to these two planes. So we must measure the angle between the poles corresponding to these great circles. To get the internal angle θ PQR at Q, we must make the correct choice for the poles. The right choice is to take the poles for these two lines to be Q P Q P and Q R Q R.

5 This leads to the formula for the angle, Q P θ cos 1 Q P, Q R. (5) Q R As a check of the signs, observe that when P R, the inner product on the right is 1 (not 1), so that the angle θ comes out as 0, which is what we want. (The wrong choice of poles might have led to the inner product being and the angle being π; this would have been the external angle at Q, instead of the internal angle.) Spherical triangles By a spherical triangle we mean a triple A, B, C of non-collinear points on S 2 together with the minor segments joining them. (The fact that A, B and C are not collinear implies in particular that no one can be the antipode of the another.) We write the side-lengths of the triangle as a d(b,c ) b d(c, A) c d(a, B).

6 We write the angles of the triangle as α C AB β ABC γ BC A. The cosine rule The cosine rule in Euclidean geometry expresses a relation between four quantities associated to a Euclidean triangle: the lengths of the three sides, and one of the angles. There is a version of the cosine rule in spherical geometry. Let ABC be as above. We shall express the cos(α) in terms of the lengths of the sides. We start with (5), which gives cos α A B A B, A C. A C Next we use the fact that A B is sin of the angle between the unit vectors A and B, which is the side-length c of the spherical triangle. Treating A C similarly, we get 1 cos α A B, A C. Now we use (3) and (4), obtaining 1 cos α A, B (A C ) 1 A, (C A) B 1 A, C, B A A, B C 1 ( ) B,C A, A A, B A,C 1 ( ) B,C A,C A, B Finally we remember that A,C is cos of the spherical distance d(a,c ), etcetera, and we end up with: cos a cosb cosc cos α. (6)

7 This is the spherical cosine rule. We can rewrite it as: cos a cosb cosc + cos α. Let us see what happens when the side-lengths of the triangle are all small. In this case, we approximate cos a by 1 a 2 /2 and sinb by b. We get or more simply, 1 2 a2 /2 1 2 (b2 + c 2 ) + cos αbc, a 2 b 2 + c 2 2ab cos α. We recognize this as the cosine rule from Euclidean geometry: the formulae from spherical geometry coincide with the Euclidean formulae to leading order when the triangle is small. The sine rule Having obtained a formula for cos α, we can obtain a formula now for sin α. We start with sin α (1 cos α)(1 + cos α) M N. We then calculate (using the cosine rule (6) above), M 1 cos α ( + cos a cosb cos a )/ () ( + cosb cosc cos a )/ (). Next we apply our cosine formulae (1) and (2) to get M + cosb cosc cos a cos(b c) cos a sinb ( sinc ) ( ) 2 sin b c+a 2 sin b c a 2 2 sin(s c) sin(s b).

8 where s (a + b + c)/2. Similarly, cosb cosc + cos a N cos(b + c) + cos a sinb ( sinc) ( ) 2 sin a+b+c 2 sin a b c 2 2 sin(s) sin(s a). Putting it all together and diving by sin 2 a, we get sin 2 α sin 2 a 4sin(s) sin(s a) sin(s b) sin(s c) (sin a ) 2. The key point now is that the expression on the right is symmetric in a, b and c. We therefore deduce: sin α sin a sin β sinb sin γ sinc. (7) This is the spherical sine rule. As for the cosine rule, when a, b and c are all small, this formula coincides to leading order with the Euclidean sine rule: sin α a sin β b sin γ c. 3. Duality The dual triangle Let ABC again be a spherical triangle. To each of the vertices A, B and C of our spherical triangle, we can associate its polar line: three great circles l A, l B and l C. And to each side of the triangle, BC, C A and AB, we can associate a point: the pole for the corresponding great circle (though we have a choice of two). In order to nail down the choices a little, we proceed as follows. We will suppose that the vertices of our triangle ABC are labelled so that the unit vectors A, B, C

9 in 3 form a right-handed basis. This is equivalent to asking that As pole for the line AB, we take the point A B,C > 0. (8) A B A B. (This is one of the two possible poles, but we choose this one.) For the other two lines BC and C A, we use the same recipe, permuting the letters cyclically. Thus the three chosen poles for the three sides of the triangle are the points ξ a B C B C ξ b C A C A ξ c A B A B. The dual triangle to the triangle ABC is the triangle ξ a ξ b ξ c whose vertices are these poles corresponding to the sides of the original triangle. Thus, to every spherical triangle, we have assigned a dual triangle. The dual of the dual The relationship between a triangle and its dual is a symmetrical one. That is, if we form the dual of the triangle ξ a ξ b ξ c, we will end up with the original triangle ABC again. We can verify this as follows. If we form the dual of the triangle ξ a ξ b ξ c, we will get a triangle A B C where (for example) C ξ a ξ b ξ a ξ b. What we must show is that C C. We observe that C is a positive multiple of ξ a ξ b, which in turn is a positive multiple of (B C ) (C A).

10 We use the identity (4) to write this last vector as B,C A C C,C A B B,C A C 0 A B,C C The coefficient A B,C is positive, because of our condition that A, B, C is a right-handed basis (8). Putting it all together, we see that C is a positive multiple of C; and since both are unit vectors, it follows that C C. Thus the dual of the dual triangle is the original triangle. Left or right handed? In showing that the dual of the dual is the original triangle, we used the fact that (8) was a positive quantity: the right-handed condition. What would have happened if the original triple A, B, C had been a left-handed basis? It turns out that ξ a, ξ b, ξ c (the vertices of the dual triangle) will be a right-handed basis of 3 no matter what. When we take the dual of the dual, we will end up with A, B and C that are again right-handed: these points will be the antipodal points to the original triangle A, B, C. Side-lengths and angles of the dual triangle Let us compute the side-lenghts of the dual triangle. The distance between the vertices ξ b, ξ c of the dual triangle is given by the usual formula for spherical distance: d(ξ b, ξ c ) cos 1 ξ b, ξ c. From the formulae defining ξ b and ξ c, we see that this becomes C A d(ξ b, ξ c ) cos 1 C A, A B A B ) A C cos ( 1 A C, A B A B A C π cos 1 A C, A B A B π BAC π α.

11 from (5). Thus the side-lengths of the dual triangle are π α, π β, π γ, where α, β and γ are the angles of the original triangle. What about the angles of the dual triangle? We could do another calculation; or we could exploit the fact that the dual of the dual is the original. That is, we reverse the role of the original triangle and its dual in the above statement, and we see that the side-lengths of the original triangle are π θ, π φ, π ψ, where θ, φ and ψ are the angles of the dual triangle. estating this: the angles of the dual triangle are π a, π b, π c, where a, b and c are the side-lengths of the original triangle. The dual cosine rule We can apply the cosine rule (6) to the dual triangle ξ a ξ b ξ c. The role of α is now played by the angle of the dual triangle at the vertex ξ a, which is π a, and so on. Thus we obtain cos(π a) cos(π α) cos(π β) cos(π γ ), sin(π β) sin(π γ ) which simplifies to cos a cos α + cos β cos γ sin β sin γ, (9) This is the dual cosine rule. It expresses the side-lengths of the triangle ABC (e.g. the side-length a) in terms of the angles of the triangle. Note that it is a special feature of spherical geometry (not shared with Euclidean geometry) that the side-lengths of a triangle are determined by the angles. An example Consider a regular pentagon on the sphere. Specifically, let us think of five points P 1,..., P 5 in the northern hemisphere, all the same distance from the north pole; these are the vertices of our pentagon. The edges are arcs of great circles, and all five edges have the same length. Suppose that the internal angles of the pentagon are all 120 (that is, 2π/3). How long are the edges?

12 Let N be the north pole and let us look at the triangle P 1 P 2 N. The line from N to P 1 bisects the internal angle of the pentagon at P 1. So N P 1 P 2 π/3. Similarly P 1 P 2 N π/3. At N the rays from all the P i meet at equal angles, so P 1 N P 2 2π/5. If x d(p 1, P 2 ), then from the dual cosine rule above we get cos x cos(2π/5) + cos2 (π/3) sin 2 (π/3) cos(2π/5) + 1/4 3/4 4 cos(2π/5) So x cos 1 ( 5/3), which is about 0.73.