# A Note on Starshaped Sets in 2-Dimensional Manifolds without Conjugate Points

Authors: Adem Kılıcman; Wedad Saleh
Journal: Journal of Function Spaces
 (2014)

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## Abstract

LetWn be C∞ complete, simply connected n-dimensional Riemannian manifolds without conjugate points. Assume that S⊂W2 is starshaped where kerS≠S. For every point x∈S∖kerS, define A(x)={y: y lies on some geodesic segment in S from x to a point of kerS}. There is a finite collection A of all maximal A sets whose union is S. Further, kerS=∩{A:A in A}.


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## Body

## 1. Introduction

We begin with some definitions from [1–3]. A subset B in a Riemannian manifold M is convex if, for each pair point p,q∈B, there is a unique minimal geodesic segment from p to q and this segment is in B. When dealing with a subset B⊂Wn, the word “a unique minimal geodesic segment” should be replaced by “the geodesic.” A subset S in a Riemannian manifold M is starshaped if there is a point p∈S such that, for all q∈S, there is a unique minimal geodesic segment from p to q and this segment is in S. The subset of S consisting of all points like p is called the kernel of S  (ker⁡S). In Wn, a subset S is starshaped if there is a point p∈S such that, for all q∈S, the geodesic segment joining p and q is contained in S. The subset S is convex if and only if ker⁡S=S. Throughout the paper, Int(B) and ∂B will denote the interior and the boundary of the subset B, respectively. Let d(p,q) denote the distance between the two points p,q∈Wn. For the pair of points p,q∈Wn, γpq will denote the geodesic segment joining p, q. For x∈S and B⊂S, d(x,B)=inf⁡{d(x,b):b∈B}. The distance to B⊂S is locally maximal at x∈S if there is some neighborhood U of x such that d(y,B)≤d(x,B)  ∀y∈U∩S. Finally, all manifolds, maps, fields, and so forth are discussions to make sense. All curves are parametrized by arc-length.Readers may refer to [4] for discussions concerning visibility via geodesic segment and starshaped sets. The behavior geodesic in Wn has been discussed by many geometrers as Eberlein [2] and Goto [5]. For more properties of convexity and starshapedness in Wn, see [6–9].


## 2. Main Results

LetW2 be 2-dimensional manifolds without conjugate points, and assume that S⊂W2 is starshaped. For every x∈S∖ker⁡S, we define A(x)={y:y∈γxk⊂S,k∈ker⁡S}. Clearly S=∪{A(x):x∈S∖ker⁡S}.Lemma 1.
LetS be starshaped in 2-dimensional manifolds without conjugate point. Then (a)
the setker⁡S is convex,(b)
ifγ is any geodesic segment in S joining points of ker⁡S, then γ⊂ker⁡S,(c)
for every pointx∈S, there is a unique point k of ker⁡S closest to x; that is, there is a unique point k of ker⁡S such that d(x,k)=d(x,ker⁡S).Proof.
We have the following.(a)
SinceS is a starshaped subset of W2, by [10], ker⁡S is convex.(b)
Sinceker⁡S is convex, clearly any geodesic segment in S joining points of ker⁡S is in ker⁡S.(c)
Letx be an arbitrary point in S. If x∈ker⁡S, then the result is trivial. x∈S∖ker⁡S. Let there be two points k1 and k2 in ker⁡S closest to x, that is, d(x,k1)=d(x,k2)=d(x,ker⁡S), to show that k1=k2. Suppose on the contrary that k1≠k2. Since k1 and k2 are in ker⁡S, then γk1k2⊂ker⁡S. Let k∈γk1k2⊂ker⁡S, such that k≠k1 and k≠k2. However, k is closer to x than k1 and k2, contradicting the assumption that k1 and k2 are closest to x; then k1=k2.Proposition 2.
LetS be starshaped in 2-dimensional manifolds without conjugate points. For every point x∈S∖ker⁡S, one defines A(x)={y:y∈γxk⊂S,k∈ker⁡S}. For any pair of points x,z∈S∖ker⁡S, if x∈A(z), then A(x)⊆A(z).Proof.
Lets∈A(x); then s∈γxk where k∈ker⁡S. Since x∈A(z), then x∈γzk where k∈ker⁡S. This implies that γxk⊂γzk; then s∈γzk. We conclude that s∈A(z) and the proof is complete.Proposition 3.
LetS⊂W2 be a starshaped set. Let x∈S∖ker⁡S. If x∉∂S, then there is some point z∈∂S such that x∈A(z) and hence A(x)⊂A(z). Moreover, d(x,ker⁡S) is not locally maximal at x.Proof.
Letz∈∂S and x∉∂S. Let γxk⊂S be a geodesic segment from x to k where k∈ker⁡S. Then γzx∪γxk is a geodesic segment in S from z to x to k. Hence x∈A(z). Then, by Proposition 2, we have A(x)⊆A(z). Moreover, z∈A(z)∖A(x), because z∉A(x). Then A(z)⊈A(x), and A(x)⊂A(z). Finally, it is clear that, for all points s∈γzx, d(s,ker⁡S)>d(x,ker⁡S). Then the distance to ker⁡S is not locally maximal at x.Proposition 4.
Letx∈S∖ker⁡S. If A(x) is a maximal A set, then d(x,ker⁡S) is locally maximal at x.Proof.
Assume on the contrary that every neighborhood ofx contains some point z for which d(x,ker⁡S)<d(z,ker⁡S). By Proposition 3, let z∈∂S. Then x∈γzx∪γxk where k∈ker⁡S. Then x∈γzx, so x∈A(z). By Proposition 2, A(x)⊂A(z). Since A(x) is maximal, then A(z)⊂A(x) which means that z∈γxk where k∈ker⁡S, which contradicts d(x,ker⁡S)<d(z,ker⁡S)Proposition 5.
LetS⊂W2 be a starshaped set. Let x∈S such that d(x,ker⁡S) is locally maximal at x. Then (a)
the pointx is a vertex of S,(b)
for every pointz near x, z≠x, d(z,ker⁡S) is not locally maximal at z,(c)
A(x)⊈A(w) for any point w≠x.Proof.
We have the following.(a)
Sinced(x,ker⁡S) is locally maximal at x, then x∈S∖ker⁡S and every neighborhood of x contains some point z such that d(x,ker⁡S)>d(z,ker⁡S). By Proposition 3, x∈∂S. Then x must be a vertex of S.(b)
It is direct from (a).(c)
x∈A(x)∖A(w), because x∉A(w) for any w≠x. Then A(x)⊈A(w).Theorem 6.
LetS⊂W2 be a starshaped set. For every point x∈S, there is a finite subcollection A of all maximal A. Then, the following conditions hold:(a)
S=∪{A:AinA},(b)
for every pointz∈S∖ker⁡S, A(z) in A if and only if A(z) is a maximal A set.Proof.
Assume thatA={Ai(x):x∈S∖ker⁡S,Ai(x)⊈Aj(x)∀1≤i,j≤n}.(a)
Forz∈S, z belongs to at least one set Ai. Either Ai is in A or Ai⊂Aj where Aj is in A. This implies that z∈∪{A:AinA}; then S⊆∪{A:AinA}.The reverse inclusion is immediate, so the sets are equal.(b)
z∈S∖ker⁡S. Let A(z) be in A; then A(z) is a maximal set. To prove that A(z) is in A, assume on the contrary that A(z) is not in A. Since z∈S∖ker⁡S, by (a), there is Ai(w) in A such that A(z)⊂Ai(w), but A(z) is a maximal A set; this implies that A(z)=Ai(w). Then, A(z) is not subset of any set Ai, soA(z) is in A.Lemma 7.
LetS⊂W2 be a starshaped set. Each point x∈S∖ker⁡S. Then (a)
the setA(x)={y:y∈γxy⊂S,k∈ker⁡S} is convex,(b)
ifγ is any geodesic segment in S joining points of A(x), then γ⊂A(x).Proof.
We have the following.(a)
Letk1,k2∈ker⁡S; then γk1k2⊂ker⁡S, because ker⁡S is convex. Moreover γxk1∪γxk2∪γk1k2⊂S. Let z1,z2∈A(x), such that zi∈γxki,i=1,2, respectively. We will prove that γz1z2⊂A(x). Assume on the contrary that γz1z2 is not in A(x); then there is z∈γz1z2 and z∈S∖A(x). Let k be an arbitrary point of ker⁡S such that z∈γxk, but z∉A(x) which means that γxk is not in A(x), which is a contradiction.(b)
It is clear that any geodesic segment inS joining points of A(x) is in A(x), because A(x) is convex.Theorem 8.
LetS⊂W2 be a starshaped set. Then ker⁡S=∩{A:AinA}=∩{A(x):x∈S∖ker⁡S}.Proof.
Clearlyker⁡S⊆∩{A(x):x∈S∖ker⁡S}⊆∩{A:AinA}. Now, we will prove that ∩{A:AinA}⊆ker⁡S. Let z1∉ker⁡S; then there is a point z2∈S such that z1 does not see z2 via S. Since z2∈S, by Theorem 6, there is Ai(w) in A such that z2∈Ai(w); by Lemma 7, Ai(w) is convex; then z1∉Ai(w); this implies that z1∉∩{A:AinA}; then ∩{A:AinA}⊆ker⁡S and the proof is complete.


## 3. Conclusion

All results of the present work are valid in Euclidean spaceEn as manifolds without conjugate points [5], but the generalization of these results to Wn,n≥3, is more difficult and is left as open problem.



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*Source: 675735-2014-08-28.xml*