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EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 9, No. 1, 2016, 57-63

ISSN 1307-5543 – www.ejpam.com

On Convexity in Product of Riemannian Manifolds

Wedad Saleh, Adem Kiliçman ∗

Department of Mathematics and Institute for Mathematical Research, University Putra Malaysia,

43400 UPM, Serdang, Selangor, Malaysia

Abstract. In this paper, the concept of convexity and starshapedness in the cartesian product of two

complete, simple connected smooth Riemannian manifolds without conjugate points are studied in

terms of the same concepts in the components of product. We also discuss some of their properties

in the cartesian product of Riemannian manifolds without conjugate points. Results obtained in this

paper may inspire future research in convex analysis and related optimization fields.

2010 Mathematics Subject Classifications: 52A20, 52A30,53B20

Key Words and Phrases: Convex sets,Conjugate points,Kernel,Starshaped sets

1. Introduction

Convexity and starshapedness play an important role in optimization theory, convex analy-

sis , Minkowski space and fractal mathematics [4, 7–9, 11–14, 16]. In [15], Pandey introduced

an interesting form of a Riemannian metric g and connection ▽ on the cartesian product

M1 × M2 of two C∞ Riemannian manifolds M1 and M2. The main result in [2] is that the

product M1×M2 of two Riemannian manifolds is free from conjugate (rep.focal) points under

the metric and connection given in [15] if and only if both M1 and M2 are free from conjugate

(resp. focal) points under their own metrics and connections. In [3], there are some interest-

ing results in product of two C∞ complete, simple connected smooth Riemannian manifolds

without conjugate points. The main aim of this paper is studying the convexity and star-

shapedness in the cartesian product of two complete, simple connected smooth Riemannian

manifolds without conjugate points.

2. Preliminaries

In this section, we recall some definitions and properties , which are used further in this

paper. We refer to [18] for the standard material on differential geometry.

∗Corresponding author.

Email addresses: wed_10_777@hotmail.com (W. Saleh), akilic@upm.edu.my (A. Kiliçman)

http://www.ejpam.com 57 c© 2016 EJPAM All rights reserved.



W. Saleh, A. Kiliccman / Eur. J. Pure Appl. Math, 9 (2016), 57-63 58

Let N be a C∞ n-dimensional Riemannian manifold, and TzN be the tangent space to N

at z. Also, assume that µz(x1, x2) is a positive inner product on the tangent space

TzN ( x1, x2 ∈ TzN), which is given for each point of N . Then, a C∞ map µ: z −→ µz , which

assigns a positive inner product µz to TzN for each point z of N is called a Riemannian metric.

The length of a piecewise C1 curve η : [a1, a2] −→ N which is defined as follows:

L(η) =

∫ a2

a1

‖ή(x)‖d x .

We define d(z1, z2) = in f
�

L(η): η is a piecewise C1 curve joining z1 to z2

	

for any points

z1, z2 ∈ N . ▽X Y , X , Y ∈ N is a unique determined Riemannian connection which called Levi-

Civita connection on every Riemannian manifolds. Furthermore, a smooth path η is a geodesic

if and only if its tangent vector is a parallel vector field along the path η, i.e , η satisfies the

equation▽ή(t)ή(t) = 0. Every path η is joining z1, z2 ∈ N where L(η) = d(z1, z2) is a minimal

geodesic.

Finally, assume that (N ,η) is a complete n-dimensional Riemannian manifold with Rie-

mannian connection ▽. Let x1, x2 ∈ N and η : [0,1] −→ N be a geodesic joining the points

x1 and x2 , which means that ηx1,x2
(0) = x2 and ηx1,x2

(1) = x1.

Definition 1 (see[10]). A subset B in a Riemannian manifold N is convex if for each pair points

p,q ∈ N, there is a unique minimal geodesic segment from p to q and this segment is in B.

When dealing with a subset B ⊂ W , where W is a C∞ complete, simply connected n-

dimensional Riemannian manifold without conjugate points, the word "a unique minimal

geodesic segment" should be replaced by "the geodesic segment".

The following theorem was proved in [1] :

Theorem 1. Let A⊂W be an open convex subset. Then,

(i) The closure of A (Ā) is also convex.

(ii) The interior of A (Int(A)) is also convex.

The following theorem gives the relationship between global supporting and convexity:

Theorem 2 (see [5]). Let A⊂W be an open subset whose boundary A is a smooth hypersurface

of W. Then, A is convex if and only if A is globally supported at each boundary point .

Definition 2 (see [17]). A subset S in a Riemannian manifold N is starshaped if there is a point

p ∈ S such that for all q ∈ S there is a unique minimal geodesic segment γpq from p to q and this

segment is in S. In such a case, the set S is starshaped with respect to p or p sees S via S.

Remark 1. (i) The subset of S consisting of all points like p is called the kernel of S ( kerS).

(ii) In W, a subset S is starshaped if there is a point p ∈ Ssuch that for all q ∈ S, the geodesic

segment γpq joining p and q is contained in S.



W. Saleh, A. Kiliccman / Eur. J. Pure Appl. Math, 9 (2016), 57-63 59

Theorem 3 (see [6]). Let S ⊂W be an open starshaped subset with respect to some point p ∈ S,

then the closure S̄ is also starshaped with respect to the same point p .

Definition 3 (see [5]). Let A be an open subset of W whose boundary ∂ A is a smooth hypersurface

of W. A is called globally supported at p ∈ ∂ A if A is contained in one side of the tangent geodesic

hypersurface Sp at p ∈ ∂ A.

Let N1 and N2 be two complete Riemannian manifolds with Riemannian metrics g1 and g2

and Riemannian connections▽1 and▽2, respectively. A Riemannian metric g on N1×N2 was

defined as follows (see[19])

g(X , Y ) = g((X1, X2), (Y1, Y2)) = g1(X1, Y1) + g2(X2, Y2)

where X i , Yi ∈ ℑ(Ni) and ℑ denotes the set of all vector fields on Ni, i = 1,2. Similarly, a

Riemannian connection ▽ on N1 × N2 wwas given by [19]

▽X Y =▽(X1,X2)
(Y1, Y2) = (▽

1
X1

Y1,▽2
X2

Y2).

If γ : [0,λ]→ N1 × N2 is a smooth curve in N1 × N2, then the natural projections

γ1 : [0,λ]→ N1 and γ2 : [0,λ]→ N2 of γ on both N1 and N2, respectively, are smooth curves.

Moreover, γ is a geodesic in N1 × N2 if and only if both γ1 and γ2 are geodesics in N1 and N2

, respectively. Which means ▽γ̇γ =▽(γ̇1,γ̇2)
(γ̇1, γ̇2) = (▽

1
γ̇1
γ̇1,▽2

γ̇2
γ̇2), where γ̇ is the velocity

vector field along the curve γ. Consequently, ▽γ̇γ̇= 0 if and only if ▽i
γ̇i
γ̇i = 0 for i = 1,2, see

[3].

Let W1 and W2 be C∞ complete, simply connected Riemannian manifolds without con-

jugate points,then W1 ×W2 is also a C∞ complete, simply connected Riemannian manifold

without conjugate points. Notice that dim(W1 ×W2) = dim(W1) + dim(W2). Consequently,

each pair of different points (p1, p2) and (q1,q2) in W1×W2 are joined by a unique geodesic γ.

This segment when naturally projected on W1 and W2 yields two geodesic segments γi ⊂ Wi

joining pi and qi , i = 1,2 each one is unique in its own manifolds.The natural projection will

be denoted by ηi : W1 ×W2→Wi where ηi(p1, p2) = pi , i = 1,2 see[3].

The following propositions were proved in [3] :

Proposition 1. Let A1 ⊂W1 and A2 ⊂W2 be subsets of W1 and W2. Then, A1 ×A2 ⊂W1 ×W2 is

convex if and only if both A1 and A2 are convex.

Proposition 2. Let A1 ⊂W1 and A2 ⊂W2 be two subsets. Then,

(i) A1 × A2 ⊂W1 ×W2 is starshaped if and only if both A1 and A2 are starshaped.

(ii) ker(A1 × A2) = (kerA1)× (kerA2)

3. Convexity in Riemannian Manifolds Product

In this section, we study some properties of convexity in Riemannian manifolds product.

Proposition 3. The intersection of any number of product convex subsets is convex subset.



W. Saleh, A. Kiliccman / Eur. J. Pure Appl. Math, 9 (2016), 57-63 60

Proof. Let A1, A2, B1, and B2 be convex subsets, then A1 × A2, and B1 × B2 are convex

subsets. We know that (A1×A2)∩ (B1×B2) = (A1∩B1)× (A2∩B2). Since A1∩B1, and A2∩B2

are convex, then (A1 ∩ B1)× (A2 ∩ B2) is also convex. Therefore, the proof is complete.

Remark 2. The above proposition is not true in general for the union of subsets of W1 ×W2.

Theorem 4. Let A1 ⊂ W1 and A2 ⊂ W2 be an open convex subsets, then the closure A1 × A2 is

also convex.

Proof. Assume that both A1 ⊂ W1 and A2 ⊂ W2 are convex subsets. Then, Ā1 ⊂ W1 and

Ā2 ⊂W2 are convex, which means that Ā1 × Ā2 is convex. Then, A1 × A2 is convex.

Theorem 5. Let A1 ⊂W1 and A2 ⊂W2 be convex subsets, then the interior of

A1 × A2 (Int(A1)× Int(A2)) is also convex.

Proof. The proof is direct in the light of Theorem 1.

Notice that if Ai ⊂ Wi , i = 1,2 is an open subset such that A1 × A2 is convex, then Ai ,

i = 1,2 is not necessarily convex. The following example indicates this claim.

Example 1. Let A1 = S1 = {(x , y) : x2+ y2 ≤ 1}\{(0,0)} and A2 = [0,1]\{12}. Clearly A1 × A2

is a convex subset of R3 while A1 and A2 are non-convex.

The relationship between global supporting and convexity in the cartesian product of Rie-

mannian manifolds without conjugate points is given in the following theorem:

Theorem 6. Let A1 ⊂W1 and A2 ⊂W2 be open subsets whose boundary ∂ A1 and ∂ A2 are smooth

hypersurface, respectively. Then, A = A1 × A2 ⊂ W1 ×W2 is convex if and only if A1 and A2 are

globally supported at each boundary point.

Proof. Let A1 and A2 be globally supported at each boundary point, then by using Theorem 2

we have that A1 and A2 are convex which implies that A1×A2 is convex. Now, let A= A1×A2

be a convex, then A1 and A2 are convex, by using Theorem 2, A1 and A2 are globally supported

at each boundary point.

Corollary 1. Let A1 ⊂W1 and A2 ⊂W2 be open subsets whose boundary ∂ A1 and ∂ A2 are smooth

hypersurface of W1 and W2, respectively. Then, A= A1×A2 is convex if and only if every maximal

tangent geodesic of ∂ A1 and ∂ A2 have an empty intersection with A1 and A2.

4. Starshapedness in Riemannian Manifolds Product

In this section, we aim to establish some properties of starhapedness in Riemannian man-

ifolds product.

Theorem 7. Let A be a non-empty closed subset of W. If ∂ A is starshaped, then ker(∂ A) ⊂ kerA.



W. Saleh, A. Kiliccman / Eur. J. Pure Appl. Math, 9 (2016), 57-63 61

Proof. Let ∂ A be starshaped with respect to x , i.e., x ∈ ker(∂ A). Suppose that x is not

in kerA, i.e., there is a point y ∈ A such that γ[x y] is not contained in A. Since A is closed,

there is a point y1 ∈ ∂ A∩ γ[x y] such that γ(x y1)
∩A= φ. Thus, x does not see y1 via ∂ A. This

contradicts the fact that ∂ A is starshaped with respect to x .

Therefore, we can state the following result as well.

Theorem 8. Let A1 be a non-empty closed subset of W1, and A2 be a non-empty closed subset of

W2. If ∂ A1 and ∂ A2 are starshaped,then ker(∂ A1 × ∂ A2) ⊂ ker(A1 × A2).

Proof. Since

ker(∂ A1 × ∂ A2) =ker(∂ A1)× ker(∂ A2)

⊂ kerA1 × kerA2

=ker(A1 × A2).

Then, ker(∂ A1 × ∂ A2) ⊂ ker(A1 × A2).

Theorem 9. Let S = S1 × S2 be an open starshaped subset with respect to some point

p = (p1, p2) ∈ S, then the closure S̄ = S1 × S2 is also starshaped with respect to the same point

p = (p1, p2).

Proof. Suppose that S = S1 × S2 is starshaped with respect to the point p = (p1, p2), then

S1 and S2 are starshaped with respect to the points p1, and p2 , respectively. This implies, by

using Theorem 3 , to S̄1 and S̄1 are starshaped with respect to the points p1 and p2, respectively.

Then, S̄1 × S̄2 = S1 × S2 is starshaped with respect to the point p = (p1, p2).

Corollary 2. Let S = S1 × S2 ⊂W1 ×W2 be an open starshaped subset. Then, the kernel of S is

contained the kernel of S̄(ker(S1 × S2) ⊂ ker(S1 × S2)).

Proof. Let p = (p1, p2) ∈ kerS, i.e., p ∈ ker(S1 × S2), then S is starshaped with respect to

p = (p1, p2). By Theorem 9, S̄ = S1 × S2 is also starshaped with respect to p = (p1, p2), which

implies that p ∈ kerS̄. Hence, ker(S1 × S2) ⊂ ker(S1 × S2).

The relation between ker(S1 × S2) and ker(S1 × S2) for any arbitrary open starshaped

subset S = S1 × S2 ⊂W1 ×W2 is given in the following theorem:

Theorem 10. Let S = S1×S2 ⊂W1×W2 be an open starshaped subset such that ∂ S is a smooth

hypersurface. Then, ker(S1 × S2) = ker(S1 × S2).

Proof. Since

ker(S1 × S2) =ker(S̄1 × S̄2)

=ker(S̄1)× ker(S̄2)

=ker(S1)× ker(S2)

=kerS1 × kerS2



REFERENCES 62

=ker(S1 × S2).

Then, ker(S1 × S2) = ker(S1 × S2).

Corollary 3. Let S = S1×S2 ⊂W1×W2 be a closed starshaped such that S is smooth hypersurface.

Then, kerS is a closed subset.

Proof. The proof is direct from Theorem 10 since S = S1 × S2 is closed if and only if S = S̄

which implies that kerS = kerS.

5. Concluding Remarks

(i) All results in this paper are valid in the cartesian product of Euclidean as well as hyper-

bolic spaces as examples of manifold without conjugate points. Moreover, these results

are valid in the case of cartesian product of manifolds without focal points as every

manifold without focal points has no conjugate points.

(ii) The results will be more interesting in the cartesian product of general Riemannian man-

ifolds.

(iii) The study has been established in this paper could be considered as a base of a study of

other concepts such as local convexity and so on.

ACKNOWLEDGEMENTS The authors are exceptionally grateful to the anonymous referees

for their valuable suggestions and comments, which helped the authors to improve the work.

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