On the geodesic diameter of surfaces with involutive isometry
aa r X i v : . [ m a t h . DG ] N ov ON THE GEODESIC DIAMETER OF SURFACESWITH INVOLUTIVE ISOMETRY
YU.G. NIKONOROV
Abstract.
This is an English translation of the following paper, published sev-eral years ago:
Nikonorov Yu.G. On the geodesic diameter of surfaces with in-volutive isometry (Russian), Tr. Rubtsovsk. Ind. Inst., 2001, V. 9, 62–65, Zbl.1015.53041 . All inserted footnotes provide additional information related to thementioned problem.2010 Mathematical Subject Classification: 51M16, 53C45, 57M60.Key words and phrases: geodesic diameter, length metric space, intrinsic distance,surface, involutive isometry.
Let us consider a length metric space (
M, ρ ), i. e. the metric ρ on M is intrinsic [1].This means that for every two points x, y ∈ M , the distance ρ ( x, y ) is equal to theinfimum of the lengths of all paths γ ⊂ M , connected the points x and y . Fora curve γ , we denote by l ( γ ) the length of γ with respect to the metric ρ . Theshortest path is a curve γ ⊂ M homeomorphic to the interval [0 , ⊂ R and suchthat l ( γ ) = ρ ( x, y ), where the points x and y are the endpoints of the curve γ . Ifa length metric space ( M, ρ ) is compact, then for every points x, y ∈ M there is ashortest path γ , connected them [1]. The value diam( M ) = sup ρ ( x, y ), where x and y are taken in M , is called the geodesic diameter of the space ( M, ρ ). If (
M, ρ )is compact, then diam( M ) could be calculated as the maximal distance between twopoints x, y ∈ M .Note that we can consider the boundary of a convex body in Euclidean space withan intrinsic metric, induced by Euclidean one, as the space ( M, ρ ). Even due thistype of examples, it is clear that calculating of the geodetic diameter is an extremelydifficult problem. For example, if we consider the boundary (surface) of a rectan-gular three-dimensional parallelepiped, then the farthest point (in terms of intrinsicboundary metrics) from a given vertex of the parallelepiped is not necessarily theopposite vertex [2]. However, the presence of some symmetry in the space (
M, ρ )allows to simplify the calculation of the geodetic diameter.For a given metric space (
M, ρ ), a map I : M → M is called an isometry if ρ ( I ( x ) , I ( y )) = ρ ( x, y ) for all x, y ∈ M .An isometry I : M → M is called involutive if the equality I ◦ I = Id fulfilled,i. e. we get the identical map twice applying I . In the case when the isometry hasno fixed point ( I ( x ) = x for all x ∈ M ), points of the form x and I ( x ) will be calledantipodal. Recall that y = I ( x ) implies x = I ( y ). See also
Vyalyi M.N. The shortest paths along the parallelepiped surface (Russian) // Mat.Pros., Ser. 3, 9 (2005), 203–206; Miller S.M., Schaefer E.F. The distance from a point to itsopposite along the surface of a box // Pi Mu Epsilon J. 14(2) (2015), 143-154, arXiv:1502.01036.
Examples of metric spaces with involutive isometry are surfaces bounding cen-trally symmetric convex bodies in Euclidean space. Involutive isometry in this caseis the restriction of the central symmetry to the boundary of the body under con-sideration. It is natural to assume that the geodesic diameter of such spaces can becalculated as the maximum distance between pairs of antipodal points.We have the following
Theorem 1.
Let ( M, ρ ) be a length metric space that is homeomorphic to the two-dimensional sphere S and let I : M → M be an involutive isometry without fixedpoints. Then there is x ∈ M such that diam( M ) = ρ ( x, I ( x )) . Proof.
It is easy to see that the map I is a homeomorphism. Let us consider points x and y such that diam( M ) = ρ ( x, y ). Such points do exist since the space ( M, ρ ) iscompact. Suppose that y = I ( x ) and consider the shortest curve γ x,I ( x ) , connecting x and I ( x ). Note that the point y (as well as its image I ( y ) under the map I ) couldnot be situated on the considered shortest curve. Indeed, otherwise the inequality ρ ( x, y ) < ρ ( x, I ( x )) ≤ diam( M ) holds that is impossible. Let us consider on thecurve γ x,I ( x ) all pairs of points ( u, v ) such that v = I ( u ). The set of such pairs isnot empty, since the pair ( x, I ( x )) has a suitable property. Among all pairs withthis property, we chose the pair ( e u, e v ) with the minimal value of ρ ( u, v ). Such pairexists due to the compactness of γ x,I ( x ) .Let γ be a part of the curve γ x,I ( x ) between the chosen points e u and e v . It is easyto see that every points u and v on the curve γ with the property v = I ( u ) are theendpoints of γ . Indeed, the existence of some other such points contradicts to thechoice of the pair ( e u, e v ). Therefore, if we consider the curve e γ , the image of thecurve γ under the map I , then e γ ∩ γ = { e u, e v } . Hence, the curve η = e γ ∪ γ is homeomorphic to the circle S , and according to theJordan curve theorem, it separates M into two regions homeomorphic to the two-dimensional disc (open ball), i. e. M = η ∪ D ∪ D , where each D i is homeomorphicto D .It is clear that the curve η does not contain the points y and I ( y ) (this contradictsto the choice of the points x and y ). Without loss of generality, we may assumethat y ∈ D . Let us show that I ( y ) ∈ D . Indeed, since I ( η ) = η , then either I ( D ) = D , or I ( D ) = D . In the first case we get I ( η ∪ D ) = η ∪ D . Since η ∪ D is homeomorphic to the closed two-dimensional ball, then by Brouwer’s fixed-point theorem, there is a point z ∈ η ∪ D such that I ( z ) = z . The latter equalitycontradicts to the assumption that the map I has no fixed point. Therefore, wehave I ( D ) = D and I ( y ) ∈ D .Now, let γ y,I ( y ) be the shortest curve connected the points y and I ( y ). Since y ∈ D and I ( y ) ∈ D , then the curves η and γ y,I ( y ) have a non-empty intersection,i. e. there is a point t ∈ M such that t ∈ γ y,I ( y ) ∩ η . Without loss of generality we In the paper
Vˆılcu C. On two conjectures of Steinhaus // Geometriae Dedicata, 79 (2000),267–275 , the following version of this theorem is proved (see Proposition 6):
Let F be a convexcentrally symmetric surface in E , and diam( F ) its intrinsic diameter. If x, y ∈ F are such that diam( F ) is equal to the intrinsic distance between x and y on F , then the points x and y aresymmetric to each other under the central symmetry of the surface F . N THE GEODESIC DIAMETER OF SURFACES . . . 3 may assume that t ∈ γ x,I ( x ) (otherwise, one can consider the image of γ x,I ( x ) underthe map I ).It is clear that ρ ( I ( x ) , I ( y )) = ρ ( x, y ) = diam( M ). Let us consider the triples ofpoints ( x, t, y ) and ( I ( x ) , t, I ( y )). According to the triangle inequality, we get ρ ( x, y ) ≤ ρ ( x, t ) + ρ ( t, y ) ,ρ ( I ( x ) , I ( y )) ≤ ρ ( I ( x ) , t ) + ρ ( t, I ( y )) . Adding these inequalities we get the following:2 diam( M ) = ρ ( x, y ) + ρ ( I ( x ) , I ( y )) ≤ ρ ( x, t ) + ρ ( I ( x ) , t ) + ρ ( t, y ) + ρ ( t, I ( y )) . Further, using the fact that the point t is situated on two shortest curves γ x,I ( x ) and γ y,I ( y ) , we get ρ ( x, t ) + ρ ( I ( x ) , t ) = ρ ( x, I ( x )) = l ( γ x,I ( x ) ) ≤ diam( M ) ,ρ ( y, t ) + ρ ( I ( y ) , t ) = ρ ( y, I ( y )) = l ( γ y,I ( y ) ) ≤ diam( M ) . Consequently, three last inequalities are fulfilled if and only if they become equalities.In particular, ρ ( x, I ( x )) = ρ ( y, I ( y )) = diam( M ), q.e.d. (cid:3) It follows from the proved theorem that the geodesic diameter of the surfacebounding the convex centrally symmetric body in three-dimensional space, equal tothe maximum distance between pairs of symmetric points, which greatly simplifiesthe calculation of this value. It is also natural to raise the following question:
Is there an analogue of Theorem 1for length metric spaces homeomorphic to the sphere S q for some q ≥ ? In particular, this result was used for the computation of the geodesic diameter of the boundaryof a rectangular three-dimensional parallelepiped in the paper
Nikonorov Yu.G., Nikonorova Yu.V.The intrinsic diameter of the surface of a parallelepiped // Discrete and Computational Geometry,40(4) (2008), 504–527; see also
Hess R., Grinstead Ch., Grinstead M., Bergstrand D. Hermitpoints on a box // College Math. J. 39(1) (2008), 12–23. It follows from recent results of A.V. Podobryaev and Yu.L. Sachkov (
Podobryaev A.V.,Sachkov Yu.L. Cut locus of a left invariant Riemannian metric on SO in the axisymmetric case// Journal of Geometry and Physics, 110 (2016), 436–453; Podobryaev A.V. Diameter of theBerger Sphere // Mathematical Notes, 193(5) (2018), 846—851 ; see the formula (4) in the firstpaper and the proof of Theorem 1 in the second paper) that a three-dimensional analogue ofTheorem 1 is not true even for some of the Berger spheres (the three-dimensional sphere withstandard Riemannian metric compressed along the fibers of the Hopf fibration). The group SU (2)is diffeomorphic to S and could be considered as the group of unit quaternions ( a + a i + a j + a k , a i ∈ R , a + a + a + a = 1). The center of SU (2) consists of two elements ±
1. The set of left-invariant metrics on SU (2) is 3-parametrical, the corrresponding parameters are the eigenvalues I , I , I > SU (2). The Bergerspheres correspond to the case I = I . Due to the homogeneity, the diameter of any Bergersphere could be computed as the intrinsic distance between 1 and the farthest point from 1 in S .It should be noted that the farthest point from 1 is exactly the point − I = I < I ,whereas the set of the farthest points from 1 is the circle { a + a i + a j | a = I / ( I − I ) } for I = I > I (we thank Alexey Podobryaev for an interesting discussion). It is clear that forany left-invariant metric on SU (2) the map a
7→ − a (which is the multiplication by the centralelement −
1) is an involutive isometry (and a Clifford–Wolf translation, moving all points the samedistance). Therefore, a three-dimensional analogue of Theorem 1 is not true for this involutiveisometry on every Berger metric with I = I > I . YU.G. NIKONOROV