On Toponogov s Theorem

Transcription

1 On Toponogov s Theorem Viktor Schroeder 1 Trigonometry of constant curvature spaces Let κ R be given. Let M κ be the twodimensional simply connected standard space of constant curvature κ. Thus M κ is the round sphere of radius 1/ π for κ > 0, the euclidean plane for κ = 0 and the hyperbolic space with curvature κ for κ < 0. Let { π/ κ for κ > 0, D κ := diam M κ = for κ 0. To describe the trigonometry we use the functions sn κ, cs κ, which are the solutions of the differential equation f + κf = 0 with sn κ (0) = 0 sn κ (0) = 1 cs κ (0) = 1 cs κ(0) = 0 Thus we have 1 κ sin( κ t) for κ > 0, sn κ (t) = t for κ = 0, 1 κ sinh( κ t) for κ < 0. cos( κ t) for κ > 0, cs κ (t) = 1 for κ = 0, cosh( κ t) for κ < 0. We have the following formulae sn κ = cs κ, cs κ = κsn κ, 1 = cs 2 κ +κsn 2 κ sn κ (a + b) = sn κ (a)cs κ (b) + cs κ (a)sn κ (b) cs κ (a + b) = cs κ (a)cs κ (b) κsn κ (a)sn κ (b) 1

2 for the formulation of the law of cosine also the following function is useful: { t 1 md κ (t) := sn κ (τ)dτ = κ (1 cs κ(t)) for κ 0, t2 for κ = 0. Now the law of cosine in M κ can be written as md κ (c) = md κ (a b) + sn κ (a)sn κ (b)(1 cos(γ)) where a,b,c are the lengths of the sides and α,β,γ the opposite angles of a triangle. In the case κ = 0,1, 1 this corresponds to the classical formulae c 2 = a 2 + b 2 2abcos(γ) cos(c) = cos(a) cos(b) + sin(a) sin(b) cos(γ) cosh(c) = cosh(a) cosh(b) sinh(a) sinh(b) cos(γ) In the following we derive some consequences of this formula. Lemma 1.1. Let a,b,c,c be positive numbers with a+b+c, a+b+c < 2D k. Let be the triangle in M 2 κ with sides a,b,c and opposite angles α,β,γ. Let be the triangle with sides a,b,c and angles α,β,γ. Then: (a) c c if and only if γ γ. (b) If α,β π 2 and c c, then α α and β β. Proof. (a) is clear from the law of cosine. We prove (b) for simplicity only in the case κ = 1. We consider α in dependence of c, where a and b are fixed: cos(b) cos(a)cos(c) cos(α) = = f(c). sin(a) sin(c) We show that f (c) 0 as long as α,β π/2. This proves the desired monotonicity in α (and by symmetry of the argument also in β). Since α π/2 and β π/2 we see from this formula in particular and by symmetry Now the nominator of f (c) is cos(b) cos(a) cos(c) cos(a) cos(b) cos(c). sin(a)[cos(a) cos(c) cos(b)] 0, which shows that α (and by symmetry also β) is decreasing in dependence of c. 2

3 c c b a Figure 1: blabla a γ Lemma 1.2. Consider two triangles in M κ, one with sides a,b,c and angles α,β,γ and one with sides a,b,c and angles α,β.γ. Assume that γ π 2, b D k /2, a + b D κ and a a. Then c c. Proof. The case κ 0 is quite clear, so we prove the statement for κ = 1. We consider c in dependence of a, where b and γ are fixed and have: cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(γ) = f(a). We will show that f (a) 0 as long as 0 a,a + b π, 0 b π/2 and γ π/2. This then shows the desired monotonicity. We compute f (a) = cos(γ)sin(a + b) (1 cos(γ))sin(a)cos(b) 0 since by assumption cos(γ) 0, sin(a), sin(a + b) 0, cos(b) Alexandrov Lemma... 2 The Theorem of Toponogov Let M be a geodesic metric space. For p,q M let pq the distance between p and q. We denote by pq a shortest geodesic from p to q. This is an abuse of notation, since this shortest curve is in general not unique. Our general convention with this notation is the following: if in a given context (say a proof) we introduce the notation pq, it stands for an arbitrary shortest geodesic between p and q with the restriction: if we have later (in this context) points x pq and y pq, then xy denotes the geodesic which is (considered as a set) a subset of pq. A point x pq is called an interior point, if p x q. Given geodesics pq, ps, we denote with p (q,s) the Alexandrov angle. Definition 2.1. The space M satisfies the (Aπ)-property (angles sum up to π), if for an interior point x pq, and s x we have x (p,s) + x (s,q) = π. 3

4 For example every Riemannian manifold satisfies the (Aπ)-property. Given three point p,s,q M, we denote by psq a triangle with this vertices, where ps, sq and qp are edges of this triangle. If for some κ the perimeter Per(psq) := ps + sq + qp < 2D κ, then there exists a (up to congruence) unique comparison triangle p,s,q. We denote with κ p(q,s) := p κ (q,s) the comparison angle. Here p κ(q,s) is the usual angle in M κ. Definition 2.2. M is called CBB(κ)-space, if every point p M has a neighborhood U, such that for all triangles xyx contained in U we have y (x,z) κ y(x,z). It is not difficult to show that a CBB(κ)-space satisfies the (Aπ)- property. We will prove the following local to global theorem, which is called Theorem of Toponogov. Toponogov proved it first in the context of Riemannian manifolds. In the present form it is due to Burago-Gromov-Perelman. Theorem 2.3. Let M be a complete geodesic space which is CBB(κ). If psq is a triangle in M with Per(psq) < 2D κ, then p (s,q) κ p (s,q). We will use the following notation. Given a triangle psq in M. We say: Top(κ) holds in psq at the vertex p if p (s,q) κ p(s,q). We say Top(κ) holds in psq, if Top(κ) hold at every vertex of the triangle. The essential step in the proof is the following result Proposition 2.4. Let M be a geodesic space which satisfies the (Aπ)- property. Let psq be a triangle in M with Per(psq) l < 2D κ. Assume that Top(κ) holds for all triangles in M with Per( ) 3 4l and a vertex on one of the sides of psq. Then Top(κ) holds for the triangle psq. From Proposition 2.4 Theorem 2.3 follows easily: Proof. (of Theorem 2.3) Assume that the theorem does not hold. Then there exists a triangle 0 = (p 0,s 0,q 0 ) such that Top(κ) does not hold. Let l = Per( 0 ) < 2D κ. By Proposition 2.4 there exists p 1 on one of the sides of 0 and a triangle 1 = (p 1,s 1,q 1 ) with Per( 1 ) 3 4l such that Top(κ) does not hold for 1. We have p 0 p l 3 4l. Inductively there exists a sequence of triangles k = (p k,s k,q k ) with Per( k ) ( 3 4 )k l and p k 1 p k ( 3 4 )k l such that Top(κ) does not hold for k. Then p k is a Cauchy sequence which converges to p M, since M is complete. Since M is CBB(κ), there exists a neighborhood U such that Top(κ) holds for all triangles contained in U. For k sufficiently large, k U. This is a contradiction. 4

5 In the proof of Proposition 2.4 we will use the following (well known) Lemma 2.5. Let xyz be a triangle in M with perimeter < 2D κ. Let y = p 0,...,p k = z consecutive points on yz and assume that Top(κ) holds in every triangle xp i 1 p i at the vertices p i 1 and p i. Then Top(κ) holds in xyz at the vertices y and z....proof used Alexandrov lemma... The following definition turns out to be very useful. Definition 2.6. A triangle xyz in a geodesic space M is called κ-straddled, if Per(xyz) < 2D κ and κ x(y,z), κ z(y,x) π 2, κ y(x,z) π 2. Thus a κ-straddled triangle has two acute and one obtuse comparison angle. x π 2 π 2 z We derive the following π 2 y Figure 2: A straddled triangle Lemma 2.7. Let M be a geodesic metric space and xyz a κ-straddled triangle in M. Assume further that Top(κ) holds for xyz. Let xy z be a configuration in M κ with Per(x y z) < 2D κ and xy = x y, yz = y z, y (x,z) κ y (x,z). Then xz xz and x (y,z) κ x (y,z), y(x,z) κ y (x,z) Proof. Let ˆxŷ ẑ the comparison triangle for xyz in M κ. Since Top(κ) holds for xyz, we have κ ŷ (ˆx,ẑ) y(x,z) κ y (x,z), where the last inequality is just our assumption. Thus to obtain the configuration xy z from ˆxŷ ẑ, we have to increase the angle at ŷ of the hinge (ˆxŷ),(ŷẑ) to obtain the triangle xy z. By Lemma 1.1 (a) we obtain xz = ˆxẑ xz. By Lemma 1.1 (b) we have κ x (y,z) κˆx (ŷ,ẑ) x(y,z), where the last inequality is just Top(κ) for xyz at the vertex x. In the same way we see κ z (y,x) z(y,x). Proof. (of Proposition 2.4) We show that α := s (p,q) κ s(p,q). Let a = ps, δ = sq. Thus 2a Per(psq) 2a + 2δ. We first make an additional assumption and assume that the triangle is thin (δ << a) in the following sense: any triangle xyz with a 3 δ xy, yz a 3 + δ and xy + yz xz + 2δ is κ-straddled. It is easy to show that for fixed a with 0 < a < D(κ) there exists a δ = δ(a) > 0 with this property. In addition we assume that δ < a 12. 5

6 We construct now sequences x i,y i M in the following way: Take x 1 ps to be the point with px 1 = a 3 and choose a shortest x 1q. Let y 1 x 1 q with y 1 q = a 3. Assume that x i,y i are already defined, then let x i+1 py i such that px i+1 = a 3 and y i+1 x i+1 q with y i+1 q = a 3. q σ 1 y 1 τ x 1 2 p α 1 β 1 x 1 Figure 3: The picture in M α s By construction (and obvious triangle inequality arguments) we have: (i) px i = y i q = a 3 (ii) a 3 + δ x iy i x i+1 y i x i+1 y i+1 a 3 δ We define β 1 = x1 (s,q), β i = xi (y i 1,q) for i 2, α i = xi (p,q), τ i = yi (x i,p), σ i = yi (p,q). By the (Aπ)-property of M we have α i + β i = π and σ i + τ i = π. Thus we have the following picture: We have a first triangle (s) = x 1 sq, and then a sequence of triangle (x i ) = px i y i and (y i ) = x i+1 y i q. Observe that the relations (i), (ii) together with our additional assumption imply that the triangles (x i ) and (y i ) are κ-straddled. Since the perimeter of all triangle (s), (x i ) and (y i ) is 4 3 a + 2δ 3 2 a 3 4l, and one vertex of these triangles coincides with p or q, we have by assumption that T op(κ) holds for (s), (x i ), (y i ). Since the triangles (x i ) and (y i ) are κ-straddled and Top(κ) holds for these triangles, we have α i π/2 σ i π/2 (1) Note that by Top(κ) for (x i ), we have α i κ x i (p,y i ) = γ κ ( px i, x i y i, y i p ) π, where γ κ (a,b,c) is the value of the angle γ in the triangle of sides a,b,c in M κ given by the law of cosine. The convergence π easily follows from px i = a/3, a/3 δ x i y i a/2 + δ, and px i + x i y i y i p = x i+1 y i x i y i 0, because of (ii). Thus we have α i π. (2) Now we look to a comparison situation in Mκ 2. Let (p s,sq) the comparison hinge for ps,sq in M κ with p s = ps, s q = sq and κ s (p,q) = κ s (p,q) = α. Now our claim α κ s(p,q) is,by Lemma 1.1 (a), equivalent to pq p q. (3) 6

7 y 2 σ 1 y τ 1 1 x 2 α q q 1 2 q p 2 p 1 α 1 β 1 x 1 α s Figure 4: The picture in M κ To show this, we define p 1 := p. Let x 1 p 1 s be the point with p 1 x 1 = a 3. Then choose q 1 such that x 1 s q 1 is the comparison triangle for (s) = x 1 sq. Let y 1 x 1 q 1 with y 1 q 1 = a 3. Let α = κ s (p,q), and β 1 = κ x 1 (q,s). Since Top(κ) holds for (s) we have α α, β 1 β 1. The first inequality gives p 1 q 1 pq, (4) the second inequality α 1 α 1, (5) here α 1 = κ x 1 (s,q 1 ). Let us recollect the following comparison data. We have the hinge p,x 1,y 1,q in M with y 1 x 1 q and a comparison hinge p 1,x 1,y 1,q 1 in M κ with y 1 x 1 q 1 such that px 1 = p 1 x 1, x 1 y 1 = x 1 y 1, y 1 q = y 1 q 1, α 1 α 1. (6) Since (x 1 ) is a κ-straddled triangle for which Top(κ) holds, we have from Lemma 2.7, that p 1 y 1 py 1 and τ 1 := y κ 1 (x 1,p 1 ) τ 1. The first inequality shows that there exists a point p 2 p y 1 with p 2 y 1 = py 1 and a point x 2 p 2 y 1 with x 2 p 2 = a 3. The second inequality shows that σ 1 := y κ 1 (p 2,q 1 ) σ 1. Again Lemma 2.7 now applied to (y 1 ) shows that x 2 q 1 x 2 q and β 2 := x κ 2 (y 1,q 1 ) β 2. The first inequality shows again that we find points q 2 x 2 q 1 with x 2 q 2 = x 2 q 2 and y 2 x 2 q 2 with x 2 q 2 = a 3. The second inequality gives α 2 := κ x 2 (p 2,q 2 ) α 2. Thus we now have a hinge p,x 2,y 2,q in M and a comparison hinge p 2,x 2,y 2,q 2 in M κ with px 2 = p 2 x 2, x 2 y 2 = x 2 y 2, y 2 q = y 2 q 2, α 2 α 2. (7) Note further that we have σ 1 σ 1 π/2 and α 2 α 2 π/2 and hence by Lemma 1.2 we conclude p 1 q 2 p 1 q 1 and q 2 p 2 q 2 p 1, and thus p 2 q 2 p 1 q 1. (8) 7

8 Now we can repeat the avove agument recursively and obtain to the hinge p,x k,y k,q in M a comparison hinge p k,x k,y k,q k in M κ with with px k = p k x k, x k y k = x k y k, y k q = y k q k, α k α k. (9) p k q k p k 1 q k 1. (10) Using the fact that p i x i q i spans a triangle i M κ and i+1 i it is not difficult to prove, that these sequences converge, i.e. p i p, x i x, y i y, q i q and since α i α i π,here we use (3), these points lie on a line, i.e. p q = p x + x y + p q. We now conclude pq lim inf[ px i + x i y i + y i q ] = lim inf[ p i x i + x i y i + y i q i ] = [ p x + x y + p q ] = p q pq where the last inequality comes from the fact that p i q i is monotonous by (10) and p 1 q 1 pq by (4). This proves our claim (3). Finally we have to get rid of the additional assumption on δ. Assume that psq is as in the assumption of the proposition. In the case p sq, the comparison triangle is degenerated and T op(κ) holds for trivial reasons. Thus we can assume that px ã > 0 for all x sq. Choose 0 < δ << ã such that our additional assumption holds for ã instead of a. Then subdivide sq with points s = t 0,...,t k = q, such that t i t i+1 δ. We can now apply the first part of the proof to the triangle pt i t i+1. Note that the triangles of diameter 3 4l which we used in the proof and for which Top(κ) holds have still one vertex either in p or in t m, thus one vertex on a side of the triangle psq. Thus by the assumption, Top(κ) holds for the triangles pt i t i+1 at the vertices t i and t i+1. By Lemma 2.5 Top(κ) holds in psq at s and q. References [BGP] Y. Burago, M. Gromov, G. Perelman... [M] W. Meyer, Toponogov s theorem and Applications, notes, 52p. [T] V. Toponogov Riemannian spaces having their curvature bounded below by a positive number Am. Math. Soc. Transl. 37 (1964),