a characterization of spaces of constant curvature by minimum covering radius of triangles

10
Available online at www.sciencedirect.com ScienceDirect Indagationes Mathematicae 25 (2014) 608–617 www.elsevier.com/locate/indag A characterization of spaces of constant curvature by minimum covering radius of triangles Bal´ azs Csik ´ os , M´ arton Horv´ ath otv¨ os Lor´ and University, Budapest, Hungary Received 29 November 2012; received in revised form 30 August 2013; accepted 18 February 2014 Communicated by M. Crainic Abstract In this paper, we prove that if in a Riemannian manifold, the minimum covering radius of a point triple of small diameter depends only on the geodesic distances between the points, then the manifold must be of constant curvature. This implies that if in a complete connected Riemannian manifold, the volume of the intersection of three small geodesic balls of equal radii depends only on the distances between the centers and the radius, then it is one of the simply connected spaces of constant curvature. This generalizes an earlier result of the first author and D. Kunszenti-Kov´ acs (2010). c 2014 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. Keywords: Kneser–Poulsen conjecture; Constant curvature spaces; Minimum covering radius; Comparison theorems 1. Introduction The results presented in this paper were motivated by the question of the extendability of the Kneser–Poulsen conjecture to Riemannian manifolds. The long-standing Kneser–Poulsen conjecture [16,14] predicts that if a finite collection of congruent balls in the Euclidean space is rearranged in such a way that the distances between the centers do not increase, then the volume of the union of the balls does not grow. The conjecture has hitherto been proved only in the plane by K. Bezdek and R. Connelly [2]. Furthermore, there are many known results supporting the Corresponding author. E-mail addresses: [email protected] (B. Csik´ os), [email protected] (M. Horv´ ath). http://dx.doi.org/10.1016/j.indag.2014.02.005 0019-3577/ c 2014 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved.

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Available online at www.sciencedirect.com

ScienceDirect

Indagationes Mathematicae 25 (2014) 608–617www.elsevier.com/locate/indag

A characterization of spaces of constant curvature byminimum covering radius of triangles

Balazs Csikos∗, Marton Horvath

Eotvos Lorand University, Budapest, Hungary

Received 29 November 2012; received in revised form 30 August 2013; accepted 18 February 2014

Communicated by M. Crainic

Abstract

In this paper, we prove that if in a Riemannian manifold, the minimum covering radius of a point tripleof small diameter depends only on the geodesic distances between the points, then the manifold must be ofconstant curvature. This implies that if in a complete connected Riemannian manifold, the volume of theintersection of three small geodesic balls of equal radii depends only on the distances between the centersand the radius, then it is one of the simply connected spaces of constant curvature. This generalizes anearlier result of the first author and D. Kunszenti-Kovacs (2010).c⃝ 2014 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved.

Keywords: Kneser–Poulsen conjecture; Constant curvature spaces; Minimum covering radius; Comparison theorems

1. Introduction

The results presented in this paper were motivated by the question of the extendability ofthe Kneser–Poulsen conjecture to Riemannian manifolds. The long-standing Kneser–Poulsenconjecture [16,14] predicts that if a finite collection of congruent balls in the Euclidean space isrearranged in such a way that the distances between the centers do not increase, then the volumeof the union of the balls does not grow. The conjecture has hitherto been proved only in the planeby K. Bezdek and R. Connelly [2]. Furthermore, there are many known results supporting the

∗ Corresponding author.E-mail addresses: [email protected] (B. Csikos), [email protected] (M. Horvath).

http://dx.doi.org/10.1016/j.indag.2014.02.0050019-3577/ c⃝ 2014 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved.

B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617 609

conjecture in higher dimensions (see e.g. [6–8], or Chapter 5 in [1]). Most of these results holdalso for balls of different radii, so it is reasonable to extend the conjecture without the conditionthat the balls have the same radius. As many of the partial results were proved also in the sphericaland hyperbolic spaces, and the conjecture itself can be formulated for any connected Riemannianmanifold, it is natural to ask whether the Kneser–Poulsen conjecture can be true in Riemannianmanifolds more general than the spaces of constant curvature.

If the Kneser–Poulsen conjecture holds for geodesic balls of arbitrary radii in a Riemannianmanifold, then by the inclusion–exclusion principle, the manifold must have the property thatthe volume of the intersection of k geodesic balls can depend only on the distances betweenthe centers and the radii of the balls (see [11]). For brevity, call the latter property the K Pkproperty. Similarly, if the Kneser–Poulsen conjecture holds for geodesic balls of equal radii in aRiemannian manifold, then the volume of the intersection of k geodesic balls of the same radiusdepends only on the distances between the centers and the common value of the radius of theballs. This property will be called the K P=

k property.The K P1 = K P=

1 property is quite closely related to the notion of ball-homogeneity, in-troduced by O. Kowalski and L. Vanhecke [15]. Recall that a Riemannian manifold is calledball-homogeneous if the volume of small geodesic balls depends only on the radius. Ball-homogeneous spaces have been studied extensively, see e.g. [3,4,13] and the references therein.K P1 manifolds are obviously ball-homogeneous and ball-homogeneous spaces are of constantscalar curvature.

It is clear that the K P2 property holds for all 2-point homogeneous spaces which are not all ofconstant curvature. In the subsequent papers [9,10], the authors proved that every K P=

2 manifoldis harmonic. Combining this with a result of Z. I. Szabo (Corollary 2.1 [17]), we obtain that acomplete, connected and simply connected Riemannian manifold is K P=

2 if and only if it isharmonic, and both properties are equivalent to the K P2 property.

The first author and D. Kunszenti-Kovacs [11] proved that if a complete connected Rieman-nian manifold has the K P3 property, then it is one of the simply connected spaces of constantcurvature. Of course, all complete, connected and simply connected spaces of constant curvaturesatisfy all the K Pk conditions.

In this paper, as an accomplishment of the subject, we study geometrical consequencesof the K P=

3 property. We shall see that the minimum covering radius of a point triple in aK P=

3 manifold depends only on the geodesic distances between the points. Applying the lattercondition to triangles of some special form and using a version of Rauch’s comparison theoremwe prove that such a space has constant sectional curvature. Consequently, the Kneser–Poulsenconjecture for balls with equal radii cannot be generalized for spaces of non-constant curvature.

For a connected Riemannian manifold (M, g), the following notations will be used throughoutthis paper. The exponential map at the point P ∈ M is denoted by expP . We use ⟨v, w⟩ for theinner product of the tangent vectors v and w, and ∥v∥ for the length of the tangent vector vwith respect to the Riemannian metric g. The geodesic distance of the points A and B will bedenoted by d(A, B), and ℓ(c) will denote the length of a curve c. Let γAB : [0, d(A, B)] → Mbe the minimal unit speed geodesic arc joining A and B such that γAB(0) = A provided that itexists uniquely. When γAB is defined, [A, B] denotes the image of γAB . If we consider pointsA, B in a submanifold N ⊆ M , we use notations γ N

AB and γ MAB to mark in which manifold

γAB is taken. Let B(P, r) stand for the geodesic open ball centered at P with radius r andB(P, r) stand for its closure. For a linear space V , denote by Gr(2, V ) the Grassmann manifold of2-dimensional linear subspaces of V . Let K : Gr2(T M) → R be the sectional curvature function,where Gr2(T M) is the manifold of tangential 2-planes of M .

610 B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617

Fig. 1. Construction of the triangle ABC .

2. Minimum covering radius of a triangle

By a triangle in a manifold, we mean an arbitrary triple of points in the manifold. We definethe minimum covering radius rABC of a triangle ABC in a connected Riemannian manifold Mas the infimum of the radii r such that the intersection of the geodesic balls centered at A, B, andC with radius r is nonempty:

rABC = inf {r | B(A, r) ∩ B(B, r) ∩ B(C, r) = ∅} .

If M is complete or one of the ball closures B(A, ρ), B(B, ρ), B(C, ρ) is compact for a certainρ > rABC , then we also have

rABC = minr | B(A, r) ∩ B(B, r) ∩ B(C, r) = ∅

= min

r | ∃P ∈ M such that A, B, C ∈ B(P, r)

.

In a K P=

3 manifold, the volume of the intersection B(A, r) ∩ B(B, r) ∩ B(C, r), consequentlyits emptiness or non-emptiness depend only on the lengths of the sides of the triangle, i.e., thegeodesic distances between the vertices. Thus, the following proposition is obvious.

Proposition 1. In a K P=

3 manifold, the minimum covering radius of a triangle depends only onthe lengths of the sides of the triangle. �

For a given connected Riemannian manifold M , choose a function r: M → (0, ∞) such that forevery P ∈ M, expP is defined on the ball of radius r(P) centered at the origin of TP M , and theball B(P, r) is a geodesically convex set for any r ≤ r(P). We call a subset D ⊆ M geodesicallyconvex if for any two points A, B ∈ D, the geodesic arc γ M

AB exists uniquely, and [A, B] ⊆ D.We shall call a triangle small if it is contained in one of the balls B(P, r(P)), where P ∈ M .

Consider a 2-plane σ < TP M , and a tangent vector v ∈ σ of length r = ∥v∥ < r(P), seeFig. 1. Set A = expP (v) and B = expP (−v). Let w ∈ σ be one of the two tangent vectors oflength r perpendicular to v. Define the map u: [0, π] → TP M by u(ϕ) = v cos(ϕ) + w sin(ϕ).Obviously, we have u(0) = v, u(π) = −v, and ∥u(ϕ)∥ = r for all ϕ ∈ [0, π]. The distancebetween A and expP (u(ϕ)) changes from 0 to 2r continuously and the distance between B andexpP (u(ϕ)) changes from 2r to 0 continuously, so there is a value ϕ0, where these distances areequal. Set C = expP (u(ϕ0)). Then d(A, C) = d(B, C) by the construction. The points A, B and

B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617 611

Fig. 2. Moving C away from P preserving the condition d(A, C) = d(B, C).

C depend on the choices of P, σ, r, v, w, ϕ0. We shall call the ordered triple ABC the triangleconstructed from the data (P, σ, r, v, w, ϕ0). For a given 2-plane σ < TP M and r > 0, denoteby ∆(P, σ, r) the set of all triangles that can be constructed from some data (P, σ, r, v, w, ϕ0).

Proposition 2. If M has the property that the minimum covering radius of small trianglesdepends only on the lengths of the sides, then there is a function a: [0, ∞) → R such thatd(A, C) = a(r) for any triangle ABC that can be constructed from some data (P, σ, r, v, w, ϕ0)

with r <r(P)

3 .

Proof. Suppose to the contrary that there are two triangles ABC and A′ B ′C ′ in the sets∆(P, σ, r) and ∆(P ′, σ ′, r), where 3r is less than min

r(P), r(P ′)

, such that the sides

a = d(A, C) = d(B, C) and a′= d(A′, C ′) = d(B ′, C ′) are not equal, say a < a′. By triangle

inequalities, r < a < a′ < 2r . For a point S ∈ B(P, 3r), we define the radial unit vector fieldX S on B(P, 3r) \ {S} by X S(T ) = −γ ′

T S(0). Create the smooth vector field X =X A+X B

1+⟨X A,X B ⟩on

B(P, 3r)\[A, B]. Let γ : [0, b) → B(P, 3r)\[A, B] be the integral curve of X starting at γ (0) =

C , where 0 < b ≤ +∞ is such that γ cannot be extended as an integral curve of X beyond b.Consider a point Q such that r ≤ d(P, Q) = r < 3r , see Fig. 2. As B(P, r) is geodesicallyconvex, and A, B ∈ B(P, r), we have ⟨X P (Q), X A(Q)⟩ > 0 and ⟨X P (Q), X B(Q)⟩ > 0 by theGauss lemma. Consequently, ⟨X P (Q), X (Q)⟩ > 0 as 1 + ⟨X A(Q), X B(Q)⟩ > 0. Thus,

ddt

d(P, γ (t)) = ⟨X P (γ (t)), γ ′(t)⟩ = ⟨X P (γ (t)), X (γ (t))⟩ > 0,

if r ≤ d(P, γ (t)) < 3r. (1)

We know that d(P, γ (0)) = r and (1) implies that d(P, γ (t)) increases in a neighborhood oft whenever d(P, γ (t)) ≥ r , therefore, d(P, γ (t)) must be an increasing function. Consider thederivative

ddt

d(A, γ (t)) = ⟨X A(γ (t)), γ ′(t)⟩ =

X A(γ (t)),

X A(γ (t)) + X B(γ (t))

1 + ⟨X A(γ (t)), X B(γ (t))⟩

= 1.

Similarly, ddt d(B, γ (t)) = 1. These equations imply that

d(A, γ (t)) = d(A, γ (0)) + t = a + t = d(B, γ (t)) (2)

612 B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617

for 0 ≤ t < b. Thus, d(P, γ (t)) ≤ d(P, A)+d(A, γ (t)) = r +a + t < 3r if t < 2r −a, so b ≥

2r−a > a′−a. Let D be equal to γ (a′

−a). Then we have d(A, D) = d(B, D) = a+a′−a = a′

by (2). Since d(P, D) > r and the closed balls B(A, r) and B(B, r) intersect one another only atthe point P , the intersection B(A, r)∩B(B, r)∩B(D, r) is empty, therefore, rAB D > r = rA′ B′C ′ .This contradicts our assumption as the sides of the triangles AB D and A′ B ′C ′ are equal. �

3. Application of Rauch’s comparison theorem

We will use a sharpened version of Rauch’s comparison theorem.

Theorem 1. Let M, M0 be Riemannian manifolds, and let γ : [0, l] → M, γ0: [0, l] → M0 beunit-speed geodesics, and set γ ′

= T, γ ′

0 = T0. Assume that for no t ∈ [0, l] is γ0(t) conjugateto γ0(0) along γ0. Let V, V0 be Jacobi fields along γ, γ0 such that V (0), V0(0) are tangent toγ, γ0 and

⟨T (0), V (0)⟩ = ⟨T0(0), V0(0)⟩, ⟨T (0), V ′(0)⟩ = ⟨T0(0), V ′

0(0)⟩,

∥V ′(0)∥ = ∥V ′

0(0)∥.

Assume that for each t ∈ [0, l] and any X0 ∈ Tγ0(t)M0, the sections σ0 spanned by X0, T0(t)and σ spanned by V (t), T (t) satisfy K (σ0) ≥ K (σ ). Then for all t ∈ [0, l], we have ∥V (t)∥ ≥

∥V0(t)∥.

Rauch’s first comparison theorem as formulated in 1.28, Section 10, Chapter 1 [5] assumesstronger comparison conditions on the curvatures, but the proof given there works for Theorem 1as well. Now we also modify its corollary (1.30, Section 10, Chapter 1 [5]).

Theorem 2. Let M, M0 be Riemannian manifolds with dim M0 ≥ dim M, and let P ∈ M, P0 ∈

M0. Let r be chosen such that expP |B(0,r) is an embedding and expP0|B(0,r) is nonsingular. Let

I : TP M → TP0 M0 be a linear map preserving inner products. Let c: [a, b] → expP (B(0, r))

be a curve, and letc: [a, b] → B(0, r) be the unique curve in B(0, r) such that expPc(t) = c(t).Set c0(t) = expP0

(I (c(t))). Consider the singular rectangle αP : [0, 1]×[a, b] → M, αP (s, t) =

expP (sc(t)). Assume K (σ ) ≤ K (σ0) for all σ0 < T M0 and σ ∈ SP (c), where

SP (c) =im T(s,t)αP | s ∈ [0, 1], t ∈ [a, b] and dim im T(s,t)αP = 2

.

Then we have ℓ(c) ≥ ℓ(c0).

Theorem 2 can be deduced from Theorem 1 in the same way as Corollary 1.30 is proved fromTheorem 1.28 in [5].

We will need a lemma.

Lemma 1. For any open neighborhood U ⊆ Gr2(T M) of a 2-plane σ < TP M, there exists aρ > 0 such that for any r < ρ and for any triangle ABC in ∆(P, σ, r), we have SP (γAC ) ⊆ U.

Remark. Though the proof of the lemma is quite technical, one has to be cautious, because thelemma would not be true if we replaced γAC with an arbitrary curve c connecting A to C inB(P, ρ).

Proof. Let ϕ = (ϕ1, . . . , ϕn): V → Rn be a normal chart around P . For a point Q ∈ V , wewill use notation Qϕ for the image ϕ(Q) of Q under ϕ. Denote (gi j )

ni, j=1 the matrix of the

B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617 613

Riemannian metric, and Γ ki j the Christoffel symbols with respect to ϕ. As ϕ is normal, we have

gi j (P) = δi j and Γ ki j (P) = 0. For any ε > 0, say for ε =

12 , there exists a ball B(P, δ) ⊆ V

such that

12

I = (1 − ε)I < g(Q) < (1 + ε)I =32

I andΓ i

jk(Q)

< ε =12

(3)

for every 1 ≤ i, j, k ≤ n and Q ∈ B(P, δ), where we write A < B for n × n matrices A and Bif and only if (B − A) is positive definite.

Consider a unit-speed geodesic γ : [0, T ] → B(P, δ). Its coordinate functions γ i= ϕi

◦ γ

(1 ≤ i ≤ n) fulfill the differential equation

γ i ′′(t) +

nj,k=1

Γ ijk(γ (t)) · γ j ′(t) · γ k ′

(t) = 0.

From this, we obtain

γ i ′(t) − γ i ′(0) =

t

0−

nj,k=1

Γ ijk(γ (τ )) · γ j ′(τ ) · γ k ′

(τ ) dτ. (4)

We give an upper bound for this integral. As

12

ni=1 γ i ′2(t) ≤

γ ′(t) = 1 by (3), we haveγ i ′(t)

≤√

2, so (3) and (4) yield

γ i ′(t) − γ i ′(0)

t

0n2

·12

·√

2 ·√

2 dτ = n2t. (5)

Integrating (5), we getγ i (t) − γ i (0)

− tγ i ′(0)

=

t

0γ i ′(τ ) − γ i ′(0) dτ

t

0

γ i ′(τ ) − γ i ′(0)

dτ ≤12

n2t2.

This inequality gives an upper bound on the Euclidean norm (also denoted by ∥·∥) of thedifference between the image γ ϕ of γ under ϕ and the linear curve t → γ ϕ(0) + tγ ϕ ′(0), t ∈

[0, T ]:γ ϕ(t) −γ ϕ(0) + tγ ϕ ′

(0) ≤

12

n5/2t2. (6)

Assume that 0 < ρ < minδ,

r(P)3

. For an r < ρ, consider a triangle ABC from ∆(P, σ, r).

Applying inequality (6) for the geodesic γAC and t = T = d(A, C), we obtainCϕ−

+ T γϕAC

′(0) ≤

12

n5/2T 2,

hence γ ϕAC

′(0) −

Cϕ− Aϕ

T

≤12

n5/2T . (7)

614 B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617

Applying the triangle inequality, inequality (6) to γ = γAC , and inequality (7) multiplied by twe obtainγ ϕ

AC (t) −

+ tCϕ

− Aϕ

T

γ ϕAC (t) −

+ tγ ϕAC

′(0)

+

tγ ϕAC

′(0) − t

Cϕ− Aϕ

T

12

n5/2t2+

12

n5/2tT ≤ n5/2T 2. (8)

From (5), we can deduceγ ϕ

AC′(t) − γ

ϕAC

′(0)

≤ n5/2t , which yieldsγ ϕAC

′(t) −

Cϕ− Aϕ

T

≤32

n5/2T (9)

together with (7). Using inequalities (3) and (9), we obtain

T =

T

0

γ ′

AC (t) dt ≤

T

0

32

γ ϕAC

′(t) dt

32

T

0

Cϕ− Aϕ

T

+32

n5/2T

dt

=

32

Cϕ− Aϕ

+32

n5/2T 2

.

Rearranging gives23

T −32

n5/2T 2≤Cϕ

− Aϕ . (10)

Similarly, we can get23

T −32

n5/2T 2≤Cϕ

− Bϕ . (11)

By Thales’ theorem, the angle AϕCϕ Bϕ equals π2 , so the altitude of the triangle AϕCϕ Pϕ from

Pϕ(=0) is equal to ∥Cϕ−Bϕ∥

2 , thus

∥Cϕ− Bϕ∥

2≤

Aϕ+ t

Cϕ− Aϕ

T

. (12)

Inequalities (8), (11) and (12) yield

T√

6−

74

n5/2T 2≤γ ϕ

AC (t) .

This inequality and (8) giveγ ϕAC (t) −

+ t Cϕ−Aϕ

T

γ ϕAC (t)

≤n5/2T 2

T√

6−

74 n5/2T 2

. (13)

B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617 615

SP (γAC ) consists of 2-planes

im T(s,t)αP = spanTsγAC (t) expP (sγAC

′(t)), TsγAC (t) expP (γAC (t)),

where s ∈ [0, 1], t ∈ [0, T ] and dim im T(s,t)αP = 2. As ρ tends to 0, sγAC (t) converges to theconstant 0 function uniformly, so TsγAC (t) expP tends to T0 expP , which is identity with the usualidentification T0TP M ∼= TP M . Thus, it is enough to prove that span

sγAC

′(t), γAC (t)

is closeto σ . In terms of normal coordinates, using the Plucker embedding Gr(2, Rn) ↩→ P(Rn

∧Rn), we

want to show thatγ

ϕAC

′(t) ∧ γ

ϕAC (t)

∈ P(Rn

∧ Rn) is close to [Cϕ∧ Aϕ] ∈ P(Rn

∧ Rn) when

ρ is small enough. We can represent the lines [Cϕ∧ Aϕ] and

γ

ϕAC

′(t) ∧ γ

ϕAC (t)

by the rescaled

vectors Cϕ∧Aϕ

T∥γϕAC (t)∥

and γϕAC

′(t)∧

γϕAC (t)

∥γϕAC (t)∥

, 0 ≤ t ≤ T , respectively. The advantage of considering

these rescaled direction vectors is that they are separated from 0 if ρ is small. Indeed, if α denotesthe angle Aϕ PϕCϕ , then Bϕ PϕCϕ = π − α. As r < T < 2r and γAC (t) ∈ B(P, r), usinginequalities (10) and (11) we get Cϕ

∧ Aϕ

Tγ ϕ

AC (t) =

r2 sin(Aϕ PϕCϕ )

Tγ ϕ

AC (t) >

sin(α)

2

= sinα

2

cos

α

2

= sin

α

2

sin

π − α

2

=∥Cϕ

− Aϕ∥

2r·∥Cϕ

− Bϕ∥

2r≥

23 T −

32 n5/2T 2

2r

2

>

23 −

32 n5/2T

2

2

≥19

if 2ρ(> T ) is small enough.We can give an upper bound on the distance between the above rescaled wedge products with

the help of inequalities (9) and (13):γ ϕAC

′(t) ∧

γϕAC (t)γ ϕAC (t)

−Cϕ

∧ Aϕ

Tγ ϕ

AC (t)

=

γ ϕAC

′(t) ∧

γϕAC (t)γ ϕAC (t)

−Cϕ

− Aϕ

T∧

Aϕ+ t Cϕ

−Aϕ

Tγ ϕAC (t)

=

γϕAC

′(t) −

Cϕ− Aϕ

T

γϕAC (t)γ ϕAC (t)

+

Cϕ− Aϕ

T∧

γ

ϕAC (t)γ ϕAC (t)

−Aϕ

+ t Cϕ−Aϕ

Tγ ϕAC (t)

<32

n5/2T · 1 + 2 ·n5/2T 2

T√

6−

74 n5/2T 2

.

As this upper bound tends to 0 as 2ρ(> T ) tends to 0, the lemma is proved. �

616 B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617

4. Main results

Now we are ready to prove our main theorem.

Theorem 3. If M has the property that the minimum covering radius of small triangles dependsonly on the lengths of the sides, then M has constant sectional curvature. In particular, everyK P=

3 manifold is of constant sectional curvature.

Proof. Assume that there are two 2-planes σ < TP M and σ ′ < TP ′ M such that K (σ ) = K (σ ′).Set κ = K (σ ) and κ ′

= K (σ ′), and suppose that κ < κ ′. Let κ1 =2κ+κ ′

3 and κ2 =κ+2κ ′

3 .Choose an open neighborhood U of σ in Gr2(T M) such that K (σ ) < κ1 for every σ ∈ U .By Lemma 1, we can choose ρ such that SP (γAC ) ⊆ U for every triangle ABC in ∆(P, σ, r)

with r < ρ. For an arbitrary ρ′ > 0, the sectional curvature of the surface expP ′(σ ′∩ B(0, ρ′))

at P ′ is κ ′. Choose a small ρ′ such that the sectional curvature of this surface is larger than κ2

everywhere. Set ρ = minρ, ρ′,

r(P)3 ,

r(P ′)3

.

Let ABC and A′ B ′C ′ be triangles in the sets ∆(P, σ, r) and ∆(P ′, σ ′, r) with r < ρ.Set a = d(A, C) = d(B, C) and a′

= d(A′, C ′) = d(B ′, C ′). Our goal is to show thata > a′, which will contradict Proposition 2. For i = 1, 2, consider the connected, simplyconnected, complete Riemannian manifold Mκi of constant curvature κi , choose an arbitrarypoint Pi ∈ Mκi and a 2-plane σi < TPi Mκi . Take a triangle Ai Bi Ci from ∆(Pi , σi , r) and setai = d(Ai , Ci ) = d(Bi , Ci ). As κ1 < κ2, we have a1 > a2.

Suppose that the angle APC is at least π2 . Let A′

1, P ′

1, C ′

1 be points in Mκ1 such thatd(P ′

1, A′

1) = d(P ′

1, C ′

1) = r and A′

1 P ′

1C ′

1 = APC . We can apply Theorem 2 to manifolds

Mκ1 and M , and the curve c = γ MAC because the curvature condition is fulfilled by the choice of

ρ. Then we get that there is a curve c0 in Mκ1 joining A′

1 and C ′

1 such that a = ℓγ M

AC

≥ ℓ(c0).

Obviously ℓ(c0) ≥ ℓγ

Mκ1A′

1C ′

1

. As we have A′

1 P ′

1C ′

1 =

π2 ≥ A1 P1C1 , the SAS inequality

yields ℓγ

Mκ1A′

1C ′

1

≥ ℓ

γ

Mκ1A1C1

= a1. From this we get a ≥ a1.

Consider the submanifolds N ′= expP ′(σ ′

∩ B(0, ρ)) and Nκ2 = expP2(σ2 ∩ B(0, ρ)).

Suppose that the angle A′ P ′C ′ is at most π2 . Let A′

2, P ′

2, C ′

2 be points in Nκ2 such thatd(P ′

2, A′

2) = d(P ′

2, C ′

2) = r and A′

2 P ′

2C ′

2 = A′ P ′C ′ . Applying Theorem 2 to manifolds

Nκ2 and N ′, and c = γNκ2A′

2C ′

2, we get that ℓ

γ

Nκ2A′

2C ′

2

≥ ℓ(c0) holds for a certain curve c0 joining

A′ and C ′ in N ′. Obviously ℓ(c0) ≥ ℓγ N ′

A′C ′

≥ ℓ

γ M

A′C ′

= a′. Since Nκ2 is a totally geodesic

submanifold in Mκ2 , ℓγ

Mκ2A′

2C ′

2

= ℓ

γ

Nκ2A′

2C ′

2

. As the angle A′

2 P ′

2C ′

2 is at most π2 , we have

a2 = ℓγ

Mκ2A2C2

≥ ℓ

γ

Mκ2A′

2C ′

2

. So we get a2 ≥ a′.

If we combine our inequalities, we get a ≥ a1 > a2 ≥ a′, that is a > a′, which contradictsProposition 2. �

The following theorem is formulated for K P2 manifolds in [11], but the proof given thereuses only the K P=

2 property.

Theorem 4 ([11]). Suppose that M is a complete connected n-dimensional Riemannian K P=

2manifold with constant sectional curvature κ . Then M is either simply connected or it is theelliptic space RPn

κ with a metric of constant sectional curvature κ > 0.

B. Csikos, M. Horvath / Indagationes Mathematicae 25 (2014) 608–617 617

If we combine the above theorems, we can get the following theorem easily.

Theorem 5. Suppose that M is a complete connected K P=

3 manifold. Then M is a simplyconnected space of constant curvature.

Proof. By Theorems 3 and 4, it is enough to prove that RPnκ is not a K P=

3 manifold. Thoughthere is a proof in [12], we give a simpler one. We assume κ = 1 for simplicity. Considerthree points A1, B1, C1 on a great circle of Sn with pairwise spherical distances 2π

3 , andthree other points A2, B2, C2 in an open hemisphere with pairwise spherical distances π

3 .Let A′

1, B ′

1, C ′

1, A′

2, B ′

2, C ′

2 ∈ RPn be the images of these points via the factorization mapSn

→ RPn . Then all the sides of the triangles A′

1 B ′

1C ′

1 and A′

2 B ′

2C ′

2 are π3 . The minimum

covering radius of A′

1 B ′

1C ′

1 is π3 , while the minimum covering radius of A′

2 B ′

2C ′

2 is less than π3 ,

so RPn is not a K P=

3 manifold by Proposition 1. �

Acknowledgment

Both authors were supported by the Hungarian National Science and Research FoundationOTKA K72537.

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