EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 17, No. 4, 2024, 3932-3944 ISSN 1307-5543 – ejpam.com Published by New York Business Global Some Rigidity Theorems of Closed Geodesic Polygons and Spherical Curves in Metric Spaces with Curvature Bounded Below Chanpen Phokaew1, Areeyuth Sama-ae1,∗ 1 Department of Mathematics and Computer Science, Faculty of Science and Technology, Prince of Songkla University, Pattani Campus, Pattani, 94000, Thailand Abstract. This paper examines characterizations of closed curves in a geodesic metric space with curvature bounded below, including closed geodesic polygons and closed spherical curves bounding surfaces isometric to convex polygons and circles in the model space. 2020 Mathematics Subject Classifications: 51K05, 54E35, 54E40 Key Words and Phrases: Spaces with curvature bounded below, convex hull, closed geodesic polygon, spherical curve, isometry 1. Introduction and preliminaries In this paper, we investigate the characterizations of closed curves in a geodesic metric space with curvature bound below in the sense of Alexandrov. We discuss the characteri- zations for closed geodesic polygons in the space that bound surfaces isometric to regions bounded by closed convex polygons in the model space RK , and we also look at the charac- terizations for closed spherical curves in the space that bound surfaces isometric to regions bounded by circles in the model space RK with the same perimeter. Alexandrov [1–5, 13] introduced lower and upper curvature bounds on metric spaces without Riemannian structure, which extended concepts to arbitrary spaces. This led to the theorems of Riemannian geometry, which defined bounded curvature as bounded sectional curvature. Examples include the Remannian manifolds with sectional curvature are not less than K and its convex subset, and Hilbert spaces. Let (X, d) be a metric space and γ : [a, b] → X a curve in X. The length ℓ(γ) of γ is defined by ℓ(γ) = sup k∑ i=1 d(γ(ti−1), γ(ti)), ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v17i4.5571 Email addresses: chunpen.t@psu.ac.th (C. Phokaew), areeyuth.s@psu.ac.th (A. Sama-ae) https://www.ejpam.com 3932 Copyright: © 2024 The Author(s). (CC BY-NC 4.0) C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3933 where the supremum is taken over all partitions a = t0 < t1 < · · · < tk = b of [a, b]. Hence, d∗(x, y) := inf{ℓ(γ)| γ is a curve from x to y}, for all x and y ∈ X, defines a metric on X with distance values in [0,∞]. If d = d∗, then (X, d) is called a length space. A geodesic in X is an isometry from R = (−∞,∞) into X. We may also refer to the image of this isometry as a geodesic. A geodesic path joining two points x and y is a map c : [0, a] ⊂ R → X such that c(0) = x and c(a) = y, and d(c(t), c(t′)) = |t − t′| for all t, t′ ∈ [0, a]. Usually, the image c([0, a]) is called a geodesic segment joining x and y, and if there is a unique geodesic segment joining two points x and y, then [x, y] is denoted the geodesic segment joining them. The metric space (X, d) is called a geodesic space if each pair of two points of X is joined by a geodesic segment. Definition 1. [5] Let K be a real number. The RK is one of the following spaces, de- pending on the sign of K: R2, if K = 0, the Euclidean sphere of radius 1/ √ K, if K > 0, and the hyperbolic plane with curvature K, if K < 0. We can learn more about the RK spaces in [4, 7–11]. A geodesic triangle△(p, q, r) in X is a triangle with points p, q, r as its vertices and three chosen geodesics [p, q], [q, r], [p, r] as its sides. A comparison triangle in RK for the geodesic triangle △(p, q, r) in X is a triangle △(p̃, q̃, r̃) in RK such that d(p, q) = dK(p̃, q̃), d(q, r) = d(q̃, r̃), and d(p, r) = d(p̃, r̃). Such a triangle △(p̃, q̃, r̃) always exists if d(p, q) + d(q, r) + d(p, r) < 2π√ K and it is unique up to isometries. Given a pair of a triangle △(p, q, r) in X and its comparison triangle △(p̃, q̃, r̃) in RK , the comparison point for a point x ∈ [q, r] is the point denoted by x̃ in [q̃, r̃] such that d(q, x) = d(q̃, x̃), and the comparison angle at q of the triangle △(p, q, r) is the angle at q̃ of triangle △(p̃, q̃, r̃). ∠p(q, r) denotes the angle at p of △(p, q, r) in X. We let ∠̃p(q, r) or ∠p̃(q̃, r̃) denote the angle at p̃ of a triangle △(q̃, p̃, r̃) in RK . Sometimes, for convenience we let a triangle △̃(p, q, r) in RK be a comparison triangle of △(p, q, r) in X. Definition 2. [6] Let X be a length space. A locally complete space X is a space with curvature bounded below by a real number K if every point x ∈ X has a neighborhood U(x) the following condition is satisfied: (A) for any four distinct points p, q, r, s ∈ U(x), ∠̃s(q, p) + ∠̃s(q, r) + ∠̃s(p, r) ≤ 2π. For spaces in which, locally, any two points are joined by a geodesic, in particular for locally compact spaces, the condition (A) in Definition 2 can be replaced by the condition: (B) for any triangle △(p, q, r) in U(x) and any point s on the side [q, r] the inequality d(p, s) ≥ d(p̃, s̃) is satisfied, where s̃ is the corresponding point of s on the side [q̃, r̃] of comparison triangle △̃(p, q, r). Let X be a space with curvature bounded below by K and α and β be two geodesics starting at a point p in X. The angle between α and β is defined by lim s→0 cos−1 ( d2(p, α(s)) + d2(p, β(s))− d2((α(s), β(s)) 2d(p, α(s))d(p, β(s)) ) . C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3934 The angle at p of a triangle △(q, p, r) is the angle between [p, q] and [p, r]. The condition (B) is equivalent to the following condition: (B̃) for any triangle △(p, q, r) in U(x), ∠p(q, r) ≥ ∠̃p(q, r), ∠q(p, r) ≥ ∠̃q(p, r), and ∠r(p, q) ≥ ∠̃r(p, q), where △̃(p, q, r) is a comparision triangle in RK of the triangle △(p, q, r). Spaces with curvature bounded below were defined above using local conditions. How- ever, for complete spaces, the global conditions may be deduced from the corresponding local ones. The metric space X considered in this work is complete. We then call X a metric space with curvature bounded below in the large. Theorem 1. [5] If X is a metric space with curvature bounded below by K in the large, where K > 0, then dim(X) ≤ π/ √ K and any triangle in X has perimeter no greater than 2π/ √ K. Theorem 2. [12] Let X be a metric space with curvature bounded below by K in the large, △(p, q, r) a triangle in X and △(p̃, q̃, r̃) a triangle in RK . If d(p, q) = d(p̃, q̃), d(p, r) = d(p̃, r̃), and ∠p(q, r) = ∠p̃(q̃, r̃), then d(q, r) ≤ d(q̃, r̃). 2. Closed geodesic polygons A closed curve in a metric space X is a continuous map of an oriented circle in the 2- dimensional Euclidean space. A chain V on a closed curve γ is a set of points corresponding to finitely many parameter values in order. The points in V are called the vertices of the chain. If γ consists of geodesic segments joining adjacent pairs in V , then γ and V form a closed geodesic polygon with a vertex chain in V . A subset A of a metric space (X, d) is defined as convex if, for any two points x, y ∈ A, the segment joining x and y is also included in A. C(A) represents the convex hull of a subset A, defined as the smallest convex set that contains A. An isometry between two metric spaces (X, d) and (Y, d∗) is a function i : X → Y such that d(x, y) = d∗(i(x), i(y)), for all points x, y ∈ X. We will start by describing a triangle in metric space with curvature bounded below whose convex hull is isometric to a comparison triangle’s convex hull in model space RK . We note that if p1, p2, p3, p4, p5 = p1 is a closed polygon in X, a metric space with curvature bounded below by K, and p′1, p ′ 2, p ′ 3, p ′ 4, p ′ 5 = p′1 is a closed polygon in RK with d(p1, p2) = d(p′1, p ′ 2), d(p2, p3) = d(p′2, p ′ 3), d(p3, p4) = d(p′3, p ′ 4), d(p4, p5) = d(p′4, p ′ 5) and d(p1, p4) = d(p′1, p ′ 4 then ∠p1(p2, p4) ≤ ∠p1(p2, p3) + ∠p1(p4, p3) ≥ ∠p′1 (p′2, p ′ 3) + ∠p′1 (p′4, p ′ 3) ≥ ∠p′1 (p′2, p ′ 4). Hence, it is not possible to compare the values of ∠p1(p2, p4) and ∠p′1 (p′2, p ′ 4). Therefore, we must suppose that the equation ∠p1(p2, p4) = ∠p1(p2, p3) +∠p1(p4, p3) is true in order to obtain ∠p1(p2, p4) ≥ ∠p′1 (p′2, p ′ 4). Lemma 1. Let X be a metric space with curvature bounded below by K in the large. Let △(p, q, r) be a triangle in X and △(p′, q′, r′) be its comparison triangle in RK , x be a point on [p, q], and x′ ∈ [p′, q′] be a comparison point of x. If ∠p(x, r) = ∠p′(x ′, r′), then d(x, r) = d(x′, r′). C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3935 Proof. Since X is a metric space with curvature bounded below by K in the large, we have that d(x, r) ≥ d(x′, r′). Using Theorem 2, we have d(x, r) ≤ d(x′, r′). Hence, we have the result. Theorem 3. Let X be a metric space with curvature bounded below by K in the large. Let △(p, q, r) be a triangle in X and △(p′, q′, r′) be its comparison triangle in RK . Suppose that the following statements hold: (i) ∠r(p, q) = ∠r′(p ′, q′); (ii) ∠p(x, r) = ∠p(q, r), ∠q(x, r) = ∠q(p, r) and ∠r(p, q) = ∠r(p, x) + ∠r(q, x) for any x ∈ [p, q]. Then the convex hull of △(p, q, r) is isometric to the convex hull of △(p′, q′, r′). Proof. By Lemma 1, we have that d(x, r) = d(x′, r′) if x ∈ [p, q] and x′ ∈ [p′, q′] such that d(x, p) = d(x′, p′). Let j be the map from the convex hull C(△(p′, q′, r′)) in RK to the convex hull C(△(p, q, r)) in X which, for every x′ ∈ [p′, q′], sends the geodesic segment [r′, x′] isometrically onto the geodesic segment [r, x]. We claim that j is an isometry onto its image; it then follows that the unique geodesic joining any two points of the image of j will be contained in the image, so j maps C(△(p′, q′, r′)) onto C(△(p, q, r)). Consider two points a′ ∈ [r′, x′] and b′ ∈ [r′, y′] in C(△(p′, q′, r′)), where x′, y′ ∈ [p′, q′] and x′ is between q′ and y′. Let x = j(x′), y = j(y′), a = j(a′) and b = j(b′). Since d(r, x) = d(r′, x′), d(r, y) = d(r′, y′) and d(x, y) = d(x′, y′), we have that △(r′, x′, y′) is a comparison triangle of △(r, x, y). Since X is a metric space with curvature bounded below by K in the large, we have d(a, b) ≥ d(a′, b′). As ∠r(a, b) ≤ ∠r(x, y) = ∠r′(x ′, y′) = ∠r′(a ′, b′), applying Theorem 2, we get that d(a, b) ≤ d(a′, b′). Therefore, d(a, b) = d(a′, b′), as required. We then describe that the convex hull of a closed geodesic polygon in a metric space with curvature bounded below is isometric to that of a polygon in the model space RK . Theorem 4. Let X be a metric space with curvature bounded below by K in the large. Let σ be a closed geodesic polygon with ordered vertices p1, p2, p3, p4, p1 with perimeter less than π/ √ K in X and let σ′ be a convex polygon with ordered vertices p′1, p ′ 2, p ′ 3, p ′ 4, p ′ 1 in the model space RK . Suppose the following statements hold: (i) C({p1, p2, p3}) and C({p1, p3, p4}) are isometric to C({p′1, p′2, p′3}) and C({p′1, p′3, p′4}), respectively; (ii) the geodesic [p1, p3] intersects the geodesic [p2, p4] at a point p; (iii) ∠p2(p̃, p3) = ∠p′2 (p′, p′3), ∠p2(p̃, p1) = ∠p′2 (p′, p′1), ∠p4(p̃, p1) = ∠p′4 (p′, p′1) and ∠p4(p̃, p3) = ∠p′4 (p′, p′3) for all p̃ ∈ [p2, p4] and p′ is the intersection of [p′1, p ′ 3] and [p′2, p ′ 4]; (iv) ∠p1(p2, p4) = ∠p1(p2, p̃) + ∠p1(p̃, p4) and ∠p3(p2, p4) = ∠p3(p2, p̃) + ∠p3(p̃, p4) for all p̃ ∈ [p2, p4]. C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3936 Then the convex hull of σ is isometric the convex hull of σ′. Proof. First, we shall prove that the point p is a corresponding point under both isometries to the point p′. Let p∗ be a point on the geodesic segment [p1, p3] such that d(p∗, p1) = d(p′, p′1) and let p′′ ∈ [p′1, p ′ 3] be a corresponding point of p under both isome- tries. Thus, d(p1, p3) = d(p1, p) + d(p, p3) ≤ d(p1, p ∗) + d(p∗, p3) = d(p′1, p ′) + d(p′, p′3) = d(p′1, p ′ 3), and d(p′1, p ′ 3) = d(p′1, p ′) + d(p′, p′3) ≤ d(p′1, p ′′) + d(p′′, p′3) = d(p1, p) + d(p, p3) = d(p1, p3), and hence, d(p1, p3) = d(p′1, p ′ 3). So we have p = p∗ and p′ = p′′, as required. By (iii) and (iv), we employ Theorem 3, C({p2, p3, p4}) is isometric to C({p′2, p′3, p′4}) and C({p1, p2, p4}) is isometric to C({p′1, p′2, p′4}). The next step is to confirm that C(σ) and C(σ′) are isometric to each other. By the definition of convex hull, C(σ) exists and is distinct, as we have noted. Let i1 : C({p1, p2, p3}) → C({p′1, p′2, p′3}) and i2 : C({p1, p3, p4}) → C({p′1, p′3, p′4}) be such that i1(pj) = p′j , j = 1, 2, 3 and i2(pk) = p′k, k = 1, 3, 4. Let i be a map from C(σ) to C(σ′) = C({p′1, p′2, p′3}) ∪ C({p′1, p′3, p′4}) such that i|C({p1,p2,p3}) = i1 and i|C({p1,p3,p4}) = i2. We must demonstrate that i is an isometry from C(σ) to C(σ′) by verifying that (∗) i is an isometry onto its image, and (∗∗) C(σ) = C({p1, p2, p3}) ∪ C({p1, p3, p4}). That i is surjective is obvious. Additionally, i is injective due to the circumstances of intersecting geodesic segments and isometric convex hulls. To prove (∗), let x1, x2 ∈ C(σ), x′1 = i(x1) and x′2 = i(x2). We shall verify that d(x1, x2) = d(x′1, x ′ 2). There is nothing to prove if x1, x2 ∈ C({p1, p2, p3}) or x1, x2 ∈ C({p1, p2, p4}) or x1, x2 ∈ C({p1, p3, p4}) or x1, x2 ∈ C({p2, p3, p4}). Without loss of generality, we assume that x1 is in C(p1, p2, p) and x2 is in C(p3, p4, p), where p is the point at which the geodesic segments [p1, p3] and [p2, p4] cross. Let x′1 and x′2 be corresponding points of x1 and x2, respectively. We suppose that the segment [x′1, x ′ 2] meets the segment [p′1, p ′ 3] at a point x′3 and meets the segment [p′2, p ′ 4] at a point x′4 such that x′3 ∈ [x′1, x ′ 4] (if x ′ 4 ∈ [x′1, x ′ 3] we can prove in the same manner). Let x3 and x4 be two points such that x′3 = i1(x3) and x′4 = i2(x4). In RK , we have [x′1, x ′ 4] = [x′1, x ′ 3] ∪ [x′3, x ′ 4] and [x′3, x ′ 2] = [x′3, x ′ 4] ∪ [x′4, x ′ 2]. Due to the fact that C({p1, p2, p4}) is isometric to C({p′1, p′2, p′4}) and [x′1, x ′ 4] is in C({p′1, p′2, p′4}), we have C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3937 that [x1, x4] = [x1, x3] ∪ [x3, x4] is in C({p1, p3, p4}) which such that d(x1, x3) = d(x′1, x ′ 3) and d(x3, x4) = d(x′3, x ′ 4), and then, d(x1, x4) = d(x1, x3) + d(x3, x4) = d(x′1, x ′ 3) + d(x′3, x ′ 4) = d(x′1, x ′ 4). Because C({p1, p3, p4}) is isometric to C({p′1, p′3, p′4}) and [x′3, x ′ 2] is in C({p′1, p′3, p′4}), we have [x3, x2] = [x3, x4]∪ [x4, x2] is in C({p1, p3, p4}), which such that d(x3, x4) = d(x′3, x ′ 4) and d(x4, x2) = d(x′4, x ′ 2), and thus, d(x3, x2) = d(x3, x4) + d(x4, x2) = d(x′3, x ′ 4) + d(x′4, x ′ 2) = d(x′3, x ′ 2). Hence, [x1, x2] = [x1, x3] ∪ [x3, x4] ∪ [x4, x2] forms a geodesic seqment, and therefore, d(x1, x2) = d(x1, x3) + d(x3, x4) + d(x4, x2) = d(x′1, x ′ 3) + d(x′3, x ′ 4) + d(x′4, x ′ 2) = d(x′1, x ′ 2). We now demonstrate (∗∗). For convenience, we set A = {p1, p2, p3} and B = {p1, p3, p4}. First, we must establish the convexity of C(A) ∪ C(B). Let x1, x2 be two points in C(A) ∪ C(B). The geodesic segment [x1, x2] must be declared to be in C(A) ∪ C(B). It makes no difference if x1, x2 are both in C(A) or C(B). We assume, without loss of generality, that x1 ∈ C(A) and x2 ∈ C(B). Let x′1 and x′2 be corresponding points to x1 and x2, respectively, and t′ the point of intersection of the two segments [x′1, x ′ 2] and [p′1, p ′ 3] and let t′ = i(t). Thus d(x1, x2) = d(x′1, x ′ 2) = d(x′1, t ′) + d(t′, x′2) = d(x1, t) + d(t, x2). This suggests that geodesic segments [x1, t] and [t, x2] form a geodesic segment connecting points x1 and x2. That is [x1, x2] = [x1, t] ∪ [t, x2] ⊂ C(A) ∪ C(B). Accordingly, C(A) ∪ C(B) is convex. We obtain that C(A ∪ B) ⊂ C(A) ∪ C(B), because C(A ∪ B) is the smallest convex set containing A ∪B. We also obtain C(A) ∪ C(B) ⊂ C(A ∪B) because both C(A) and C(B) are subsets of C(A ∪B). Therefore, C(A ∪B) = C(A) ∪ C(B). It is important to note that Theorem 4 depends on the intersection of two geodesics. We consider the space R3 with usual metric as a metric space with curvature bounded below by 0 in the large. Let X be a triangle with points (−1, 0, 0), (1, 0, 0) and (0, 1, 0) and Y be a triangle with points (−1, 0, 0), (1, 0, 0) and (0, 0, 1). In R0 space, we let X ′ be a triangle with points (0,−1), (0, 1) and (1, 0) and B′ be a triangle with points (0,−1), (0, 1) and (−1, 0). Thus, we have that X ′ and Y ′ are corresponding triangles of X and Y , respectively. It is evident that C(X ′) is isometric to C(X) and C(Y ′) is isometric to C(Y ) but that C(X ′∪Y ′) is not isometric to C(X∪Y ) since the segment connecting points (−1, 0, 0) and (1, 0, 0) do not meet the segment connecting points (0, 1, 0) and (0, 0, 1). The following theorem can be proven using the concept of the proof of Theorem 4 Theorem 5. Let X be a metric space with curvature bounded below by K in the large. Let σ be a closed geodesic polygon with ordered vertices p1, p2, p3, ..., pn, p1 with perimeter less than π/ √ K in X and σ′ be a convex polygon with ordered vertices p′1, p ′ 2, p ′ 3, ..., p ′ n, p ′ 1 in the model space RK , for n > 4. Suppose that the following statements hold: C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3938 (i) C({p1, p2, ..., pt}) is isometric to C({p′1, p′2, ..., p′t}) and C({pt, pt+1, ..., pn}) is iso- metric to C({p′t, p′t+1, ..., p ′ n}); (ii) the geodesic segment [p1, pt] intersects the geodesic segment [pi, pj ] at a point for some i ∈ {1, 2, ..., t− 1} and j ∈ {t+ 1, t+ 2, ..., n− 1}. (iii) ∠pi(p ∗, pt) = ∠p′i (p′, p′t), ∠pi(p ∗, p1) = ∠p′i (p′, p′1), ∠pj (p ∗, p1) = ∠p′j (p′, p′1) and ∠pj (p ∗, pt) = ∠p′j (p′, p′t) for all p∗ ∈ [pi, pj ]; (iv) ∠p1(pi, pj) = ∠p1(pi, p ∗) + ∠p1(p ∗, pj) and ∠pt(pi, pj) = ∠pt(pi, p ∗) + ∠pt(p ∗, pj) for all p∗ ∈ [pi, pj ] Then the convex hull of σ is isometric the convex hull of σ′, that is the totally geodesic surface bounded by σ and the region bounded by σ′ are isometric to each other. 3. Spherical curves If there is a point p in the metric space X and a positive real number r such that d(x, p) = r for all x in γ, the curve γ is said to be spherical. The radius of γ is the actual value r. For example, a circle of radius r > 0 in the model space RK is a closed spherical curve at a distance r from its center. In the subsequent discussion, we define γab as a spherical curve in a metric space with curvature bound below with endpoints a and b, and γ′a′b′ as a subarc of a circle in the model space RK with endpoints a′ and b′. We describe a closed spherical curve bounding a surface isometric to a region bounded by a circle in the model space RK . Lemma 2. Let X be a metric space with curvature bounded below by K in the large. Let γ be a spherical curve at a distance r < π 2 √ K from a point p with endpoints a, b in X and γ′ be a subarc of a circle of radius r centered at a point p′ in RK with endpoints a′, b′ such that d(a, b) = d(a′, b′). If that c ∈ γ is between a, b and c′ ∈ γ is between a′, b′ with conditions d(a, c) = d(a′, c′) and d(b, c) = d(b′, c′), then the geodesic segment [p, c] intersects the geodesic segment [a, b] at a point q which corresponds to the point q′ of intersection of [p′, c′] and [a′, b′]. Proof. We shall prove that the segment [a, b] intersects the segment [p, c] at a point. Suppose that q′ is the point of intersection of the segments [a′, b′] and [p′, c′]. Let q be a point on the segment [p, c] such that d(p, q) = d(p′, q′). Thus, q′ is a corresponding point of q in two triangles △(p, a, c) and △(p, b, c), and hence, d(a, q) = d(a′, q′) and d(b, q) = d(b′, q′). Consequently, d(a, b) ≤ d(a, q) + d(q, b) = d(a′, q′) + d(q′, b′) = d(a′, b′) = d(a, b), that is, the point q is the intersection of [a, b] and [p, c]. C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3939 Lemma 3. Let X be a metric space with curvature bounded below by K in the large. Let γ be a spherical curve at a distance r < π 2 √ K from a point p with endpoints a, b in X and γ′ be a subarc of a circle of radius r centered at a point p′ in RK with endpoints a′, b′. Suppose that c ∈ γ is between a, b and c′ ∈ γ is between a′, b′. Assume that the following statements hold: (i) ℓ(γ) = ℓ(γ′) ≤ π√ K ; (ii) d(a, b) = d(a′, b′); (iii) d(a, c) = d(a′, c′) and d(b, c) = d(b′, c′); (iv) ∠p(a, b) = ∠p′(a ′, b′) and ∠c(a, b) = ∠c′(a ′, b′); (v) for any triangle △(u, v, w) in X, ∠u(v, x) = ∠u(v, w) and ∠u(w, x) = ∠u(w, v) for all x ∈ [v, w]; (vi) for any triangle △(u, v, w) in X, ∠u(v, w) = ∠u(v, x) + ∠u(x,w) for all x ∈ [v, w]; Then C({p, a, c, b}) is isometric to C({p′, a′, c′, b′}). Proof. By (ii) and (iii), we have that triangles △(a′, b′, c′) and △(a′, b′, p′) are corre- sponding triangles of △(a, b, c) and △(a, b, p), respectively. By (iv) and (v), and applying Theorem 4, we get that the convex hulls of triangles △(a′, b′, c′) and △(a′, b′, p′) are iso- metric to the convex hulls of triangles △(a, b, c) and △(a, b, p), respectively. By (vi), we obtain ∠a(p, c) = ∠a(p, q) + ∠a(q, c) = ∠a′(p ′, q′) + ∠a′(q ′, c′) = ∠a′(p ′, c′) and ∠b(p, c) = ∠b(p, q) + ∠b(q, c) = ∠b′(p ′, q′) + ∠b′(q ′, c′) = ∠b′(p ′, c′), and using (iv) and Theorem 3, we have that the convex hulls of triangles △(a′, p′, c′) and △(b′, p′, c′) are isometric to the convex hulls of triangles △(a, p, c) and △(b, p, c), respectively. We prove, as the same proof of Theorem 4, that C({p, a, c, b}) is isometric to C({p′, a′, c′, b′}). Theorem 6. Let X be a metric space with curvature bounded below by K in the large. Let γ be a spherical curve at a distance r < π 2 √ K from a point p with endpoints a, b in X and γ′ be a subarc of a circle of radius r centered at a point p′ in RK with endpoints a′, b′. Suppose that c ∈ γ is between a, b and c′ ∈ γ is between a′, b′. Assume that the following statements hold: (i) ℓ(γ) = ℓ(γ′) ≤ π√ K ; (ii) d(a, b) = d(a′, b′); (iii) C({p, a, c}) is isometric to C({p′, a′, c′}) and C({p, b, c}) is isometric to C({p′, b′, c′}); C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3940 (iv) ∠c(a, b) = ∠c′(a ′, b′) and ∠p(a, b) = ∠p′(a ′, b′); (v) ∠a(x, c) = ∠a(b, c), ∠b(x, c) = ∠b(a, c) and ∠c(a, b) = ∠c(a, x) + ∠c(x, b) for any x ∈ [a, b]; (vi) ∠a(x, p) = ∠a(b, p), ∠b(x, p) = ∠b(a, p) and ∠p(a, b) = ∠p(a, x) + ∠p(x, b) for any x ∈ [a, b]. Then C({p, a, c, b}) is isometric to C({p′, a′, c′, b′}). Proof. By Lemma 2, the segment [a, b] intersects the segment [p, c] at a point q and using Theorem 4, we get that C({p, a, c, b}) is isometric to C({p′, a′, c′, b′}). By Theorem 6, we have the following corollary. Corollary 1. Let X be a metric space with curvature bounded below by K in the large. Let γ be a spherical curve at a distance r < π 2 √ K from a point p with endpoints a, b in X and γ′ be a subarc of a circle of radius r centered at a point p′ in RK with endpoints a′, b′. Suppose that a = c1, c2, ..., cn = b ∈ γ are consecutive points on γ and a′ = c′1, c ′ 2, ..., c ′ n = b′ ∈ γ are consecutive points on γ′. Assume that the following statements hold: (i) ℓ(γ) = ℓ(γ′) ≤ π√ K ; (ii) d(a, b) = d(a′, b′); (iii) C({p, c1, c2, ..., ct}) and C({p, ct, ct+1, ..., cn}) are isometric to C({p′, c′1, c′2, ..., c′t}) and C({p′, c′t, c′t+1, , ..., cn}), respectively, for some t ∈ {2, 3, ..., n− 1}; (iv) ∠ct(a, b) = ∠c′t (a′, b′) and ∠p(a, b) = ∠p′(a ′, b′); (v) ∠a(x, ct) = ∠a(b, ct), ∠b(x, ct) = ∠b(a, ct) and ∠ct(a, b) = ∠ct(a, x) + ∠ct(x, b) for any x ∈ [a, b]; (vi) ∠a(x, p) = ∠a(b, p), ∠b(x, p) = ∠b(a, p) and ∠p(a, b) = ∠p(a, x) + ∠p(x, b) for any x ∈ [a, b]. Then C({p, c1, c2, ..., cn}) is isometric to C({p′, c′1, c′2, ..., c′n}). Proof. First, we will demonstrate that the segments [a, b] and [p, c] cross at a specific location. Assume that the intersection of the segments [a′, b′] and [p′, c′] is at q′. A point on the segment [p, c] with the property d(p, q) = d(p′, q′) is called q. Because q′ is a point that corresponds to q, d(a, q) = d(a′, q′) and d(b, q) = d(b′, q′). Consequently, d(a, b) ≤ d(a, q) + d(q, b) = d(a′, q′) + d(q′, b′) = d(a′, b′) = d(a, b), that is the point q is the intersection of [a, b] and [p, c]. We can see from (iv), (v), and (vi) that △(a, b, c) and △(a, b, p) are isometric to △(a′, b′, c′) and △(a′, b′, p′), respectively. We proceed in the same way as Theorem 4, having established that C({p, a, c, b}) is isometric to C({p′, a′, c′, b′}). C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3941 Theorem 7. Let X be a metric space with curvature bounded below by K in the large, and γ be a spherical curve at a distance r < π 2 √ K from a point p with endpoints a, b in X. Let γ′ be a subarc of a circle of radius r centered at a point p′ in RK with endpoints a′, b′. Suppose that the following statements hold: (i) ℓ(γ) = ℓ(γ′) ≤ π√ K ; (ii) d(a, b) = d(a′, b′); (iii) ℓ(γxy) = ℓ(γ′x′y′) if and only if d(x, y) = d(x′, y′) for all x, y ∈ γ and x′, y′ ∈ γ′; (iv) ∠p(a, b) = ∠p′(a ′, b′) and ∠c(a, b) = ∠c′(a ′, b′); (v) for any triangle △(u, v, w) in X, ∠u(v, x) = ∠u(v, w) and ∠u(w, x) = ∠u(w, v) for all x ∈ [v, w]; (vi) for any triangle △(u, v, w) in X, ∠u(v, w) = ∠u(v, x) + ∠u(x,w) for all x ∈ [v, w]; Then ∪e∈γ [p, e] = C({p}∪γ) and C({p}∪γ) is isometric to C({p′}∪γ′), that is the totally geodesic surface bounded by γ and the region bounded by γ′ are isometric to each other. Proof. We will first demonstrate that ∪e∈γ [p, e] = C({p} ∪ γ). By the definition of C({p} ∪ γ), we have that ∪e∈γ [p, e] ⊂ C({p} ∪ γ). The next step is to confirm that ∪e∈γ [p, e] is convex . Let m1,m2 ∈ ∪e∈γ [p, e]. Therefore, m1 ∈ [p, e1] and m2 ∈ [p, e2] for some e1, e2 ∈ γ. If e1 = e2, there is nothing to prove; hence, we can assume that e1 ̸= e2 without losing generality. Let e′1, e ′ 2 be two points on γ such that ℓ(γ′a′e′1 ) = ℓ(γae1) and ℓ(γ′a′e′2 ) = ℓ(γae2). Assuming without loss of generality, that e′1 lies between a′ and e′2. Since ℓ(γa,e1) = ℓ(γa′,e′1) and ℓ(γa,e2) = ℓ(γa′,e′2), we have that ℓ(γe1e2) = ℓ(γe′1e′2). By (iii), d(a, e1) = d(a′, e′1), d(e1, e1) = d(e′1, e ′ 2) and d(e2, b) = d(e′2, b), and hence we have those triangles△(p′, a′, e′1), △(p′, e′1, e ′ 2) and △(p′, e′2, b ′) are comparison triangles of△(p, a, e1), △(p, e1, e2) and △(p, e2, b), respectively. As X is metric space with curvature bounded below, ∠p(a, e1) ≥ ∠p′(a ′, e′1), ∠p(e1, e2) ≥ ∠p′(e ′ 1, e ′ 2) and ∠p(e2, b) ≥ ∠p′(e ′ 2, b ′). Since ∠p′(a ′, b′) = ∠p′(a ′, e′1) + ∠p′(e ′ 1, e ′ 2) + ∠p′(e ′ 2, b ′) ≤ ∠p(a, e1) + ∠p(e1, e2) + ∠p(e2, b) = ∠p(a, b) = ∠p′(a ′, b′), we get ∠p(e1, e2) = ∠p′(e ′ 1, e ′ 2). Because d(p,m1) = d(p′,m′ 1), d(p,m2) = d(p′,m′ 2) and ∠p(e1, e2) = ∠p(m1,m2), by using Theorem 2, d(m1,m2) ≤ d(m′ 1,m ′ 2). As ℓ(γe1e2) = ℓ(γe′1e′2), by (iii), we have d(e1, e2) = d(e′1, e ′ 2). Now we have that a triangle △(p′, e′1, e ′ 2) is a comparison triangle of a triangle △(p, e1, e2). As a result, X is a metric space with curvature bounded below, d(m1,m2) ≥ d(m′ 1,m ′ 2). Consequently, d(m1,m2) = d(m′ 1,m ′ 2). That means that for any point on [m1,m2], lies on a segment [p, t], for some t ∈ γ. We may infer that, the geodesic segment [m1,m2] is contained in ∪e∈γ [p, e]. C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3942 We will then prove that C({p′} ∪ γ′) is isometric to C({p} ∪ γ) . Define a map i from C({p′} ∪ γ′) to C({p} ∪ γ) in such a way that every geodesic segment [p′, w′] from p′ to a point w′ on γ′ is sent isometrically onto the segment [p, w] where w is a point on γ with ℓ(γaw) = ℓ(γa′w′). The fact that i is a bijection is evident. Simply confirming that i maintains distances between points will demonstrate that i is an isometry from C({p′}∪γ′) onto C({p}∪γ). Let x′1 and x′2 be two points on segments [p′, y′1] and [p′, y′2], respectively, for some y′1, y ′ 2 ∈ γ′. On corresponding geodesic segments [p, y1] of [p ′, y′1] and [p, y2] of [p′, y′2], we let x1 and x2 be the points corresponding to x′1 and x′2, respectively. We can verify d(x1, x2) = d(x′1, x ′ 2) similarly as above, the result is completely proven. We describe characterizations of a closed spherical curve in a metric space with curva- ture bounded below by K in the large and having the same length as a circle in the model space RK in the last theorem. Theorem 8. Let X be a metric space with curvature bounded below by K in the large, and γ be a closed spherical curve at a distance r < π 2 √ K from a point p. Let γ′ be a circle of radius r centered at a point p′ in RK . Suppose that the following statements hold: (i) ℓ(γ) = ℓ(γ′); (ii) ℓ(γab) = ℓ(γ′a′b′) iff d(a, b) = d(a′, b′) iff ∠p(a, b) = ∠p′(a ′, b′), for all a, b ∈ γ and a′, b′ ∈ γ′; (iii) for any triangle △(u, v, w) in X, ∠u(v, x) = ∠u(v, w) and ∠u(w, x) = ∠u(w, v) for all x ∈ [v, w]; (iv) for any triangle △(u, v, w) in X, ∠u(v, w) = ∠u(v, x) + ∠u(x,w) for all x ∈ [v, w]. Then C(γ) is isometric to C(γ′), that is, the totally geodesic surface bounded by γ and the disk bounded by γ′ are isometric to each other. Proof. Let x, y ∈ γ and x′, y′ ∈ γ′ be such that ℓ(γxy) = ℓ(γ′x′y′) ≤ π√ K . We can conclude from (ii) that d(x, y) = d(x′, y′). We establish a map j1 from C({p′} ∪ γ′x′y′) to C({p} ∪ γxy) such that each segment [p′, z′] from p′ to z′ on γ′x′y′ is transferred on to the geodesic segment [p, z] from p to a point z on γxy, where z is the point such that ℓ(γxz) = ℓ(γ′x′z′) and a map j2 is defined from C({p′} ∪ γ′y′x′) to C({p} ∪ γyx) similar to j1. Lemma 3 indicates that j1 and j2 are isometries. We will now demonstrate that C(γ′) and C(γ) are isometric to each other. By the definition of convex hull, we observe that C(γ) exists and is unique. Let i be a map from C(γ′) = C(γ′x′y′) ∪ C(γ′y′x′) to C(γ) in such a way that the function i on C(γ′x′y′) is j1 and on C(γ′y′x′) is j2. We must demonstrate that i is an isometry from C(γ′) to C(γ), we must show that i is an isometry onto its image and C(γ) = C(γxy) ∪ C(γyx) = C(γxy ∪ γyx). It is clear that i is surjective. Additionally, as we shown in Lemma 3, i is injective as a result of the requirements of intersecting geodesics and isometric convex hulls. Let u′1 and u′2 be in C(γ′) and u1 = i(u′1) and u2 = i(u′2). We will demonstrate that d(u1, u2) = d(u′1, u ′ 2). If u′1, u ′ 2 ∈ C(γ′x′y′) or u′1, u ′ 2 ∈ C(γ′y′x′), neither case can be proven. We assume that C. Phokaew, A. Sama-ae, / Eur. J. Pure Appl. Math, 17 (4) (2024), 3932-3944 3943 u′1 ∈ C(γ′x′y′) and u′2 ∈ C(γ′y′x′). Let u′1 ∈ [p′, v′1] and u′2 ∈ [p′, v′2] for some v′1 ∈ C(γ′x′y′) and v′2 ∈ C(γ′y′x′). On X, we let [q, v1] be the geodesic segment containing u1 and let [q, v2] be the geodesic segment containing u2 where v1 ∈ C(γxy) and v2 ∈ C(γyx). If ℓ(γ′v′1v′2 ) ≤ ℓ(γ′)/2, we then have γ′v′1v′2 = γ′v′1y′ ∪ γ′y′v′2 . By (ii), C(γ′v′1y′ ) is isometric to C(γv1y) by j1 and C(γ′v′2y′ ) is isometric to C(γv2y) by j2, we thus get that C(γ′v′1v′2 ) is isometric to C(γv1v2) by i. Consequently, we get d(u1, u2) = d(u′1, u ′ 2). Additionally, we also have d(u1, u2) = d(u′1, u ′ 2) if ℓ(γ′v′2v′1 ) ≤ ℓ(γ′)/2. We will now demonstrate that C(γ) = C(γxy) ∪ C(γyx) = C(γxy ∪ γyx). It is necessary to demonstrate that the set C(γxy ∪ γyx) is convex. Without losing generality, we suppose that x1 is in C(γxy) and x2 is in C(γyx). Let [q, w1] and [q, w2] be the segments containing x1 and x2, respectively, where [q, w1] is the segment containing x1 and [q, w2] is the segment containing x2. Since j1 is the isometry from C(γ′x′y′) to C(γxy) and j2 is the isometry from C(γ′y′x′) to C(γyx), we let two points w′ 1 and w′ 2 in RK be the points corresponding to w1 and w2, respectively, and let two points x′1 and x′2 in RK be the points corresponding to x1 and x2, respectively. If ℓ(γ′w′ 1w ′ 2 ) ≤ ℓ(γ′)/2, then γ′w′ 1w ′ 2 = γ′w′ 1y ′ ∪ γ′y′w′ 2 is the result. As C(γ′w′ 1y ′) is isometric to C(γw1y) by j1 and C(γ′w′ 2y ′) is isometric to C(γw2y) by j2, we thus obtain that C(γ′w′ 1w ′ 2 ) is isometric to C(γw1w2) by i. Consequently, d(x1, x2) = d(x′1, x ′ 2) is what we have. Let x′′ be the point where [x′1, x ′ 2] and [x′, y′] intersect, and let x̂ = j1(x ′′) = j2(x ′′) = i(x′′). Hence, d(x1, x2) = d(x′1, x ′ 2) = d(x′1, x ′′) + d(x′′, x′2) = d(x′1, x̂) + d(x̂, x′2), so [x1, x2] = [x1, x̂] ∪ [x̂, x2] ⊂ C(γxy) ∪ C(γyx). Therefore, C(γxy) ∪ C(γyx) is a convex set. 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