Simple closed geodesics on regular tetrahedra in spherical space
SSimple closed geodesics on regular tetrahedra in spherical space
Alexander A. Borisenko, Darya D. Sukhorebska ∗ Abstract . On a regular tetrahedron in spherical space there exist the finite number ofsimple closed geodesics. For any pair of coprime integers ( p, q ) it was found the numbers α and α depending on p , q and satisfying the inequalities π/ < α < α < π/ such that ona regular tetrahedron in spherical space with the faces angle α ∈ ( π/ , α ) there exists unique,up to the rigid motion of the tetrahedron, simple closed geodesic of type ( p, q ) , and on a regulartetrahedron with the faces angle α ∈ ( α , π/ there is no simple closed geodesic of type ( p, q ) . Keywords : closed geodesic, regular tetrahedron, spherical space.
MSC : 53С22, 52B10
Working on the three-body problem, Poincare conjectured the existing of a simple (with-out points of self-intersection) closed geodesic on a smooth closed convex surface in three-dimensional Euclidean space. In 1929 Lyusternik and Shnirelman proved that on a Riemannianmanifold homeomorphic to a sphere there exist at least three simple closed geodesics (see [1], [2]).In 1898 Hadamard showed that on a closed surface of negative curvature any closed curve,that is not homotopic to zero, could be deformed into the convex curve of minimal length withinits free homotopy group. This minimal curve is unique and it is a closed geodesic (see [3]). Aninteresting problem is to find the asymptotic behavior of the number of simple closed geodesics,depending on the length of these geodesics, on a compact manifold of negative curvature Forinstance, Huber proved that on a complete closed two-dimensional manifold of constant negativecurvature the number of closed geodesics of length at most L has the order of growth e L /L as L → + ∞ (see. [4], [5]). In Rivin’s work [6], and later in Mirzakhani’s work [7], it’s proved thaton a surface of a constant negative curvature of genus g and with n cusps (points at infinity)the number of simple closed geodesics of length at most L is asymptotic to (positive) constanttimes L g − n as L → + ∞ .The substantive results about the behavior of geodesic lines on a convex two-dimensionalsurface was found by Cohn-Vossen [8], Alexandrov [9], Pogorelov [10]. In one of the earliestwork Pogorelov proved that a geodesic of length ≤ π/ √ k realized the shortest path betweenits endpoint on a closed convex surface of the Gaussian curvature ≤ k [11]. Toponogov provedthat on C -regular closed surface of curvature ≥ k > the length of a simple closed geodesicis ≤ π/ √ k [12]. Vaigant and Matukevich obtained that on this surface a geodesic of length ≥ π/ √ k has point of self-intersection [13].Geodesics have also been studied on non-smooth surfaces, including convex polyhedra(see [14] and [15]). D. Fuchs and E. Fuchs supplemented and systematized the results onclosed geodesics on regular polyhedra in three-dimensional Euclidean space (see [16] and [17]).Protasov obtained a condition for the existence of simple closed geodesics on a tetrahedron in ∗ The second author is supported by IMU Breakout Graduate Fellowship a r X i v : . [ m a t h . M G ] F e b uclidean space and evaluated the number of these geodesics in terms of the difference between π and the sum of the angles at a vertex of the tetrahedron [18].A simple closed geodesic is said to be of type ( p, q ) if it has p vertices on each of twoopposite edges of the tetrahedron, q vertices on each of other two opposite edges, and p + q vertices on each of the remaining two opposite edges. Geodesics are called equivalent if theyintersect the same edges of the tetrahedron in the same order.On a regular tetrahedron in Euclidean space, for each ordered pair of coprime integers ( p, q ) there exists a class of equivalent simple closed geodesics of type ( p, q ) , up to the isometryof the tetrahedron. Each of these classes contains an infinity many geodesics. Furthermore,into the class there is a simple close geodesic passing through the midpoints of two pairs ofopposite edges of the tetrahedron.In [19] we studied simple closed geodesics on a regular tetrahedra in Lobachevsky (hy-perbolic) three-dimensional space. In Euclidean space, the faces of a tetrahedron have zeroGaussian curvature, and the curvature of a tetrahedron is concentrated only at its vertices. InLobachevsky space, the Gaussian curvature of faces is -1, then the curvature of a tetrahedronis determined not only by its vertices, but also by its faces. Moreover, in hyperbolic spacethe value α of faces angle satisfies < α < π/ . The intrinsic geometry of such tetrahedrondepends on the value of its faces angle. It follows that the behavior of closed geodesics on aregular tetrahedron in Lobachevsky space differs from the Euclidean case.It is proved that on a regular tetrahedron in hyperbolic space for any coprime integers ( p, q ) , ≤ p < q , there exists unique, up to the rigid motion of the tetrahedron, simple closedgeodesic of type ( p, q ) , and it passes through the midpoints of two pairs of opposite edges of thetetrahedron. These geodesics exhaust all simple closed geodesics on a regular tetrahedron inhyperbolic space. The number of simple closed geodesics of length bounded by L is asymptoticto constant (depending on α ) times L , when L tends to infinity [19].In this work we considered simple closed geodesics on a regular tetrahedron in sphericalthree-dimensional space. In this space the curvature of a face equals 1, then the curvature of atetrahedron is also determined by its vertices and faces. The intrinsic geometry of a tetrahedrondepends on the value α of its faces angle, where α satisfies π/ < α ≤ π/ . If α = 2 π/ ,then the tetrahedron coincides with the unit two-dimensional sphere. Hence there are infinitelymany simple closed geodesics on it and they are great circles of the sphere. On a regular tetrahedron in spherical space there exists the finite number of simple closedgeodesics. The length of all these geodesics is less than π .For any coprime integers ( p, q ) we presented the numbers α and α , depending on p , q and satisfying the inequalities π/ < α < α < π/ , such that1) if π/ < α < α , then on a regular tetrahedron in spherical space with the faces angle α thereexists unique simple closed geodesic of type ( p, q ) , up to the rigid motion of this tetrahedron;2) if α < α < π/ , then on a regular tetrahedron with the faces angle α there is not simpleclosed geodesic of type ( p, q ) . A geodesic is locally the shortest curve. On a convex polyhedron, a geodesic has the followingproperties (see [9]):1) it consists of line segments on faces of the polyhedron;2) it forms equal angles with edges of adjacent faces;3) a geodesic cannot pass through a vertex of a convex polyhedron.Note, that by the ‘line segment’ we mean a geodesic segment in a space of constant cur-vature, where the polyhedron lies in.Let us take two tetrahedra in the spaces of constant curvature and consider a closed2eodesic on each of them. Construct a bijection between the vertices of the tetrahedra and givethe same labels to the corresponding vertices. Hence closed geodesics on these tetrahedra iscalled equivalent if they intersect the same-labeling edges in the same order [18].Fix the point of a geodesic on a tetrahedron’s edge and roll the tetrahedron along theplane in such way that the geodesic always touches the plane. The traces of the faces formthe development of the tetrahedron on a plane and the geodesic is a line segment inside thedevelopment.A spherical triangle is a convex polygon on a unit sphere bounded by three the shortestlines. A regular tetrahedron in three-dimensional spherical space S is a closed convex polyhe-dron such that all its faces are regular spherical triangles and all its vertices are regular trihedralangles. The value α of its faces angle satisfies the conditions π/ < α ≤ π/ . Note, than thereexist a unique (up to the rigid motion) tetrahedron in spherical space with a given value of afaces angle. The edge length is equal a = arccos (cid:18) cos α − cos α (cid:19) , (2.1) lim α → π a = 0; lim α → π a = π α → π a = π − arccos . (2.2)If α = 2 π/ , then a tetrahedron coincides with a unit two-dimensional sphere. Hencethere are infinitely many simple closed geodesics on it. In the following we consider α that π/ < α < π/ . Consider a regular tetrahedron A A A A in Euclidean space with the edge of length . Adevelopment of the tetrahedron is a part of the standard triangulation of the Euclidean plane.Denote the vertices of the triangulation in accordance with the vertices of the tetrahedron.Choose two identically oriented edges A A of the triangulation, which don’t belong to the sameline. Take two points X and X (cid:48) at equal distances from the vertex A such that the segment XX (cid:48) doesn’t contain any vertex of the triangulation. Hence the segment XX (cid:48) corresponds tothe closed geodesic on the tetrahedron A A A A . Any closed geodesic on a regular tetrahedronin Euclidean space can be constructed in this way (see Figure 1).Note, that the segments of geodesics lying on the same face of the tetrahedron are parallelto each other. It follows that any closed geodesic on a regular tetrahedron in Euclidean spacedoes not have points of self-intersection.We introduce a rectangular Cartesian coordinate system with the origin at A and the x -axis along the edge A A containing X . Then the vertices A and A has the coordinates (cid:0) l, k √ (cid:1) , and the coordinates of A and A are (cid:0) l + 1 / , k √ / (cid:1) , where k, l are integers.The coordinates of X and X (cid:48) equal ( µ, and ( µ + q + 2 p, q √ , where < µ < . The segment XX (cid:48) corresponds to the simple closed geodesic γ of type ( p, q ) on a regular tetrahedron inEuclidean space. If ( p, q ) are coprime integers then γ does not repeat itself. The length of γ isequal L = 2 (cid:112) p + pq + q . (3.1)Note, that for each coprime integers ( p, q ) there exist infinitely many simple closed geodesicsof type ( p, q ) , and all of them are parallel in the development and intersect the tetrahedron’sedges in the same order.If q = 0 and p = 1 , then geodesic consists of four segments that consecutively intersectfour edges of the tetrahedron, and doesn’t go through the one pair of opposite edges.3igure 1 Proposition 1. (see [19])
For each pair of coprime integers ( p, q ) there exists a simple closedgeodesic intersecting the midpoints of two pairs of opposite edges of the regular tetrahedron inEuclidean space. Proposition 2. (see [19])
The development of the tetrahedron obtained by unrolling along aclosed geodesic consists of four equal polygons, and any two adjacent polygons can be transformedinto each other by a rotation through an angle π around the midpoint of their common edge. Lemma 3.1.
Let γ be a simple closed geodesic of type ( p, q ) on a regular tetrahedron in Eu-clidean space such that γ intersects the midpoints of two pairs of opposite edges. Then thedistance h from the tetrahedron’s vertices to γ satisfies the inequality h ≥ √ (cid:112) p + pq + q . (3.2) Proof.
Let us take a regular tetrahedron A A A A in Euclidean space with the edge of length .Suppose γ intersects the edge A A at the midpoint X . Consider the development of thetetrahedron along γ starting from the point X and introduce a Cartesian coordinate sys-tem as described above. The geodesic γ is unrolled into the segment XX (cid:48) lying at the line y = q √ q +2 p ( x − ) (see Figure 1). The segment XX (cid:48) intersect the edges A A at the points ( x b , y b ) = (cid:16) q +2 p ) k + q q , k √ (cid:17) , where k ≤ q . Since γ does not pass through a vertex of the tilling,then x b couldn’t be an integer. Hence on the edge A A the distance from the vertices to thepoints of γ is not less than / q .In the same way on the edge A A the distance from the vertices to the points of γ is notless than / p .Choose the points B at the edge A A and B at the edge A A such that the length A B is / q and the length A B equals / p (see Figure 2). The distance h from the vertex A to γ is not less than the height A H of the triangle B A B . The length of B B equals √ p + pq + q pq . Then the length of A H is | A H | = √ (cid:112) p + pq + q . Hence the inequality (3.2) is proved.Introduce some definitions following [18]. A broken line on a tetrahedron is a curve consist-ing of the line segments, which connect points on the edges of this tetrahedron consecutively.4igure 2A generalized geodesic on a tetrahedron is a closed broken line with following properties:(1) it does not have points of self-intersection and adjacent segments of it lie on different faces;(2) it crosses more than three edges on the tetrahedron and doesn’t pass through tetrahedron’svertices.
Proposition 3. (V. Protasov [18])
For every generalized geodesic on a tetrahedron in Euclideanspace there exists a simple closed geodesic on a regular tetrahedron in Euclidean space that isequivalent to this generalized geodesic. (0 , and (1 , on a regular tetrahedronin S Let us remind that a simple closed geodesic γ has type ( p, q ) if it has p vertices on each oftwo opposite edges of the tetrahedron, q vertices on each of other two opposite edges, and p + q vertices on each of the remaining two opposite edges. If q = 0 and p = 1 , then geodesicconsists of four segments that consecutively intersect four edges of the tetrahedron, and doesn’tgo through the one pair of opposite edges. Lemma 4.1.
On a regular tetrahedron in spherical space there exist three different simple closedgeodesics of type (0 , . They coincide under isometries of the tetrahedron.Proof. Consider a regular tetrahedron A A A A in S with the faces angle α where π/ <α < π/ . Let X and X be the midpoints of A A and A A , and Y , Y be the midpointsof A A and A A respectively. Join these points consecutively with the segments trough thefaces. We obtain a closed broken line X Y X Y on the tetrahedron. Since the points X , Y , X and Y are midpoints, then the triangles X A Y , Y A X , X A Y and Y A X are equal.It follows that the broken line X Y X Y is a simple closed geodesic of type (0 , on a regulartetrahedron in spherical space (see Figure 3). Choosing the midpoints of other pairs of oppositeedges, we can construct other two geodesics of type (0 , on the tetrahedron. Lemma 4.2.
On a regular tetrahedron in spherical space with the faces angle α < π/ thereexist three simple closed geodesics of type (1 , .Proof. Consider a regular tetrahedron A A A A in S with the faces angle α where π/ < α <π/ . As above, the points X and X are the midpoints of A A and A A , and Y , Y are themidpoints of A A and A A .Develop two adjacent faces A A A and A A A into the plain and join the points X and Y with the line segment. Since α < π/ , then the segment X Y is contained inside thedevelopment and intersects the edge A A at right angle. Then develop another two adjacent5igure 3. Simple closed geodesic of type (0 , on a regular tetrahedron in S faces A A A and A A A and construct the segment Y X . In the same way join the points X and Y within the faces A A A and A A A , and join Y and X within A A A and A A A (see Figure 4). Since the points X , Y , X and Y are the midpoints, it follows, thatthe triangles X A Y , Y A X , X A Y и Y A X are equal. Hence, the segments X Y , Y X , X Y , Y X form a simple closed geodesic of type (1 , on the tetrahedron.Another two geodesics of type (1 , on a tetrahedron could be constructed in the sameway, if we choose the midpoints of other pairs of opposite edges.Figure 4. Simple closed geodesic of type (1 , on a regular tetrahedron in S Lemma 4.3.
On a regular tetrahedron in spherical space with the faces angle α ≥ π/ thereexists only three simple closed geodesics and all of them have type (0 , .Proof. Consider a regular tetrahedron in spherical space with the faces angle α ≥ π/ . Since ageodesic is a line segment inside the development of the tetrahedron, then it cannot intersectthree edges of the tetrahedron, coming out from the same vertex, in succession.If a simple closed geodesic on the tetrahedron is of type ( p, q ) , where p = q = 1 or < p < q ,then this geodesic intersect three tetrahedron’s edges, starting at the same vertex, in succession(see [18]). And only a simple closed geodesic of type (0 , intersects two tetrahedron’s edges,that have a common vertex, and doesn’t intersects the third. It follows that on a regulartetrahedron in spherical space with the faces angle α ∈ [ π/ , π/ there exist only threesimple closed geodesic of type (0 , and others don’t exist.In the next sections we will assume that α satisfying π/ < α < π/ .6 The length of a simple closed geodesic on a regular tetra-hedron in S Lemma 5.1.
The length of a simple closed geodesic on a regular tetrahedron in spherical spaceis less than π .Proof. Consider a regular tetrahedron A A A A in S with the faces angle of value α , where π/ < α < π/ . A spherical space S of curvature is realized as a unite tree-dimensionalsphere in four-dimensional Euclidean space. Hence the tetrahedron A A A A is situated inan open hemisphere. Consider a tangent Euclidean space to this hemisphere at the centerof the tetrahedron (by ‘center of the tetrahedron’ we mean a center of circumscribed sphereof the tetrahedron). A central projection of the hemisphere to this tangent space maps theregular tetrahedron in spherical space to the regular tetrahedron in Euclidean space. A simpleclosed geodesic γ on A A A A is mapped into generalized geodesic on a regular Euclideantetrahedron. From Proposition 3 we know that there exists a simple closed geodesic on aregular tetrahedron in Euclidean space equivalent to this generalized geodesic. It follows, that a simple closed geodesic on a regular tetrahedron in S is also uniquely determined with thepair of coprime integers ( p, q ) and has the same structure as a closed geodesic on a regulartetrahedron in Euclidean space. Investigate this structure following [18].A vertex of a geodesic is called a catching node if it and two adjacent vertices lie on thethree edges coming out from the same vertex A i of the tetrahedron and are the vertices of thegeodesic nearest to A i on these edges. By the ’vertex of a geodesic’ we mean a point of geodesicon an edge. Proposition 4. (V. Protasov [18])
Let γ and γ is a segments of a simple closed geodesic γ ,starting at a catching node on a regular tetrahedron, γ and γ is the segments following γ and γ and so on. Then for each i = 2 , . . . , p + 2 q − the segments γ i and γ i lie on the same faceof the tetrahedron, and there are no other geodesic points between them. The segments γ p +2 q и γ p +2 q meet at the second catching node of the geodesic. Figure 5Suppose γ has q points on the edges A A and A A , p points on A A and A A , and p + q points on A A and A A . Consider the catching node B of γ at the edge A A . The adjacentgeodesic vertices B and B are the nearest point of γ to A at the edges A A and A A respectively. The geodesic segments B B and B B correspond to γ and γ (see Figure 5).If we develop the faces A A A and A A A into the plain, then the segments γ and γ willform one line segment. The triangle B A B at the development is called catching triangle (seeFigure 6). 7he segments γ p +2 q and γ p +2 q meet at the second catching node B pq of the geodesic. Wewill assume, that B pq is the nearest to A geodesic vertex at the edge A A . The adjacentgeodesic vertices B pq and B pq are the nearest point of γ to A at A A and A A respectively.They form the second catching triangle B pq A B pq .From these catching triangles B A B and B pq A B pq it follows next inequalities | B B | < | B A | + | A B | ; | B pq B pq | < | B pq A | + | A B pq | . (5.1)Now let us develop the tetrahedron into two-dimensional sphere, starting from the face A A A and go along the segments γ i and γ i , i = 2 , . . . , p + 2 q − . The segments γ and γ start at the points B and B respectively, then they intersect the edge A A . Other segments γ i and γ i , i = 3 , . . . , p + 2 q − , are unrolled into two line segments, that intersect the sameedges in the same order and there are no other geodesic points between them. The last segments γ p +2 q − and γ p +2 q − intersect the edge A A , then go within A A A and end at the points B pq and B pq respectively (see Figure 7). It follows, that the tetrahedron’s vertices A and A lieinside the spherical lune, formed by two great arcs containing B B pq and B B pq . We obtaineda convex hexagon B A B B pq A B pq at the sphere.Figure 6 Figure 7From the inequalities (5.1) it follows that the length of the geodesic γ is less then theperimeter of the hexagon B A B B pq A B pq . Since the perimeter of convex polygon on a unitsphere is less than π , we get, that on a regular tetrahedron in Euclidean space with the face’sangle α < π/ the length of a simple closed geodesic is less than π .From Lemma 4.3 we know, that if the faces angle α satisfies the condition π/ ≤ α < π/ ,then on a regular tetrahedron there exist only three simple closed geodesics and they have type (0 , . The length of these geodesics equals L , = 4 arccos (cid:18) sin α α (cid:19) . (5.2)It is easy to check, that L , < π , when π/ ≤ α < π/ . Remark 1.
Lemma 5.1 could be considered as the particular case of the result [20] proved byfirst author about the generalization of V. Toponogov theorem [12] to the case of two-dimensionalAlexandrov space. Uniqueness of a simple closed geodesic on a regular tetra-hedron in S In a spherical space the analogical to Proposition 1 result is true.
Lemma 6.1.
On a regular tetrahedron in a spherical space a simple closed geodesic intersectsmidpoints of two pairs of opposite edges.Proof.
Let γ be a simple closed geodesic on a regular tetrahedron A A A A in S . In the part5 we showed, that there exists a simple closed geodesic (cid:101) γ on a regular tetrahedron in Euclideanspace such that (cid:101) γ is equivalent to γ . From Proposition 1 we assume (cid:101) γ intersects the midpoints (cid:101) X and (cid:101) X of the edges A A and A A on the tetrahedron. Denote by X and X the verticesof γ at the edges A A and A A on a spherical tetrahedron such that X and X are equivalentto the points (cid:101) X and (cid:101) X .Consider the development of the tetrahedron along γ starting from the point X on atwo-dimensional unite sphere. The geodesic γ is unrolled into the line segment X X (cid:48) of lengthless than π inside the development. Denote by T and T the parts of the development along X X and X X (cid:48) respectively.On the tetrahedron A A A A mark the midpoints M and M of the edges A A and A A respectively. Rotation by the angle π over the line M M is an isometry of the tetrahedron. Itfollows that the development of the tetrahedron is centrally symmetric with the center M .On the other hand symmetry over M replaces T and T . The point X (cid:48) at the edge A A of the part T is mapped into the point (cid:98) X (cid:48) at the edge A A containing X on T , and thelengths of A X and (cid:98) X (cid:48) A are equal.The image of the point X on T is a point (cid:98) X at the edge A A on T . Since M is amidpoint of A A , then the symmetry maps the point X at A A to the point (cid:98) X at the sameedge A A such that the lengths of A X and (cid:98) X A are equal. Thus, the segment X X (cid:48) ismapped into the segment (cid:98) X (cid:48) (cid:98) X inside the development.Figure 8 Figure 9Suppose the segments (cid:98) X (cid:48) (cid:98) X and X X intersect at the point Z inside T . Then thesegments (cid:98) X (cid:98) X and X X (cid:48) intersect at the point Z inside T , and the point Z is symmetricto Z over M (see Figure 8). We obtain two great arcs X X (cid:48) and (cid:98) X (cid:48) (cid:98) X intersecting in twopoints. It follows that Z and Z are antipodal points on a sphere and the length of the geodesicsegment Z X Z equals π .Now consider the development of the tetrahedron along γ starting from the point X . Thisdevelopment also consists of spherical polygons T and T , but in this case they are glued bythe edge A A and centrally symmetric with the center M (see Figure 9).By analogy with previous case apply the symmetry over M . The segment X X X (cid:48) ismapped into the segment (cid:98) X (cid:98) X (cid:98) X (cid:48) inside the development. Since the symmetries over M M correspond to the same isometry of the tetrahedron, then the arcs X X X (cid:48) and (cid:98) X (cid:98) X (cid:98) X (cid:48) also intersect at the points Z and Z (see Figure 9). It follows that the length ofgeodesic segment Z X Z is also equal π . Hence the length of the geodesic γ on a regulartetrahedron in spherical space equals π , that contradicts to Lemma 5.1. We get that thesegments (cid:98) X (cid:48) (cid:98) X and X X either don’t intersect or coincide at the development part T .Figure 10If the X X and (cid:98) X (cid:48) (cid:98) X at the development don’t intersect, then they form the quadrilateral X X (cid:98) X (cid:98) X (cid:48) on the unite sphere. Since γ is closed geodesic, then ∠ A X X + ∠ A (cid:98) X (cid:48) (cid:98) X (cid:48) = π .Furthermore, ∠ X X A + ∠ (cid:98) X (cid:48) (cid:98) X A = π . We obtain the convex quadrilateral on a spherewith the sum of inner angles π (see Figure 10). It follows that the integral of the Gaussiancurvature over the interior of X X (cid:98) X (cid:98) X (cid:48) on a sphere is equal zero. Hence, the segments X X and (cid:98) X (cid:48) (cid:98) X coincide under the symmetry of the development. We obtain that the points X and X of geodesic γ are the midpoints of the edges A A and A A respectively.Similarly it can be proved that γ intersect the midpoints of the second pair of the oppositeedges of the tetrahedron. Corollary 6.1.
If two simple closed geodesic on a regular tetrahedron in spherical space inter-sect the edges of the tetrahedron in the same order, then they coincide. S Lemma 7.1.
On a regular tetrahedron with the faces angle α in spherical space the length of asimple closed geodesic of type ( p, q ) satisfies the inequality L p,q > (cid:112) p + pq + q (cid:113) α − α . (7.1) Proof.
Consider a regular tetrahedron A A A A in S with the faces angle α and γ is a simpleclosed geodesic of type ( p, q ) on it.Each of the faces of this tetrahedron is a regular spherical triangle. Consider a two-dimensional unit sphere containing the face A A A . Construct the Euclidean plane Π passingthrough the points A , A and A . The intersection of the sphere with Π is a small circle. A raystarting at the sphere’s center O and going through a point at the triangle A A A intersectsthe plane Π . So we get the geodesic map between the sphere and the plane Π . The image ofthe spherical triangle A A A is the triangle (cid:52) A A A at the Euclidean plane Π . The edges10f (cid:52) A A A are the chords joining the vertices of the spherical triangle. From (2.1) it followsthat the length (cid:101) a of the plane triangle’s edge equals (cid:101) a = (cid:113) α − α . (7.2)The segments of the geodesic γ lying inside A A A are mapped into the straight line segmentsinside (cid:52) A A A (see Figure 11). Figure 11In the similar way the other tetrahedron faces A A A , A A A and A A A are mappedinto the plane triangles (cid:52) A A A , (cid:52) A A A and (cid:52) A A A respectively. Since the sphericaltetrahedron is regular, the constructed plane triangles are equal. We can glue them togetheridentifying the edges with the same labels. Hence we obtain the regular tetrahedron in theEuclidean space. Since the segments of γ are mapped into the straight line segments within theplane triangles, then they form the generalized geodesic (cid:101) γ on the regular Euclidean tetrahedron.Furthermore (cid:101) γ is equivalent to γ , so (cid:101) γ passes through the tetrahedron’s edges at the same orderas the simple closed geodesic of type ( p, q ) .Let us show that the length of γ is greater than the length of (cid:101) γ . Consider an arc M N ofthe geodesic γ within the face A A A . The rays OM and ON intersect the plane Π at thepoints (cid:102) M and (cid:101) N respectively. The line segment (cid:102) M и (cid:101) N lying into (cid:52) A A A is the image ofthe arc M N under the geodesic map (see Figure 11). Suppose that the length of the arc
M N is equal to φ , then the length of the segment (cid:102) M (cid:101) N equals φ . Thus the length of γ on aregular tetrahedron in spherical space is greater than the length of its image (cid:101) γ on a regulartetrahedron in Euclidean space.From Proposition 3 we know that on a regular tetrahedron in Euclidean space there existsa simple closed geodesic (cid:98) γ such that (cid:98) γ is equivalent to (cid:101) γ . Since on the development of thetetrahedron the geodesic (cid:98) γ is a line segment, and the generalized geodesic (cid:101) γ is a broken line,then the length of (cid:98) γ is less than the length of (cid:101) γ .Hence we get, that on a regular tetrahedron A A A A in S with the faces angle α thelength L p,q of a simple closed geodesic γ of type ( p, q ) is greater than the length of a simpleclosed geodesic (cid:98) γ of type ( p, q ) on a regular tetrahedron in Euclidean space with the edge length (cid:101) a . From the equations (3.1) and (7.2) we get, that L p,q > (cid:112) p + pq + q (cid:113) α − α . heorem 1. On a regular tetrahedron with the faces angle α in spherical space such that α > (cid:115) p + pq + q p + pq + q ) − π , (7.3) where ( p, q ) is a pair of coprime integers, there is no simple closed geodesic of type ( p, q ) .Proof. From Lemma 5.1 and the inequality (7.1) we get, that if α fulfills the inequality (cid:112) p + pq + q (cid:113) α − α > π, (7.4)then the necessary condition for the existence of a simple closed geodesic of type ( p, q ) on aregular tetrahedron with face’s angle α in spherical space is failed. Hence, after modifying (7.4),we get that, if α > (cid:115) p + pq + q p + pq + q ) − π , then there is no simple closed geodesics of type ( p, q ) on the tetrahedron with face’s angle α inspherical space. Corollary 7.1.
On a regular tetrahedron in spherical space there exist a finite number of simpleclosed geodesics.Proof.
If the integers ( p, q ) go to infinity, then lim p,q →∞ (cid:115) p + pq + q p + pq + q ) − π = 2 arcsin 12 = π . From the inequality (7.3) we get, that for the large numbers ( p, q ) a simple closed geodesic oftype ( p, q ) could exist on a regular tetrahedron with the faces angle α closed to π/ in sphericalspace.The pairs p = 0 , q = 1 and p = 1 , q = 1 don’t satisfy the condition (7.3). Geodesics of thistypes are considered in Section 4. S In previous sections we assumed that the Gaussian curvature of the tetrahedron’s faces isequal in spherical space. In this case faces of a tetrahedron were regular spherical triangleswith angles α on a unit two-dimensional sphere. The length a of the edges was the functiondepending on α according to (2.1). In current section we will assume that the tetrahedron’sfaces are spherical triangles with the angle α on a sphere of radius R = 1 /a . In this case thelength of the tetrahedron edges equals , and the faces curvature is a .Since α > π/ , then we can assume α = π/ ε , where ε > . Taking into accountLemma 4.3 we also expect ε < π/ . Let us proof some subsidiary results.12 emma 8.1. The length of the edges of a regular tetrahedron in spherical space of curvatureone satisfies inequality a < π (cid:114) π · √ ε, (8.1) where α = π + ε is a value of faces angles.Proof. From the equality (2.1) it follows, that sin a = (cid:113) α −
12 sin α . Substituting α = π + ε into this equality, we get sin a = (cid:113) sin ε cos (cid:0) π − ε (cid:1) sin (cid:0) π + ε (cid:1) . Since ε < π , then cos (cid:0) π − ε (cid:1) < cos π , sin (cid:0) π + ε (cid:1) > sin π and sin ε < ε . Using this estima-tions, we obtain sin a < (cid:114) π · √ ε. (8.2)Since a < π , then sin a > π a . Substituting this estimation into (8.2), we get the necessaryinequality (8.1).Consider a parametrization of a two-dimensional sphere S of radius R in three-dimensionalEuclidean case: x = R sin φ cos θy = R sin φ sin θz = − R cos φ , (8.3)where φ ∈ [0 , π ] , θ ∈ [0 , π ) . Let the point P have the coordinates φ = r/R, θ = 0 , where r/R < π/ , and the point X correspond to φ = 0 . Apply a central projection of the hemisphere φ ∈ [0 , π/ , θ ∈ [0 , π ) to the tangent plane at the point X . Lemma 8.2.
The central projection of the hemisphere of radius R = 1 /a to the tangent planeat the point X maps the angle α = π/ ε with the vertex P ( R sin rR , , − R cos rR ) at thishemisphere to the angle (cid:98) α r on a plane, which satisfies the inequality (cid:12)(cid:12)(cid:12)(cid:98) α r − π (cid:12)(cid:12)(cid:12) < π tan rR + ε. (8.4) Proof.
Construct two planes Π and Π passing through the center of a hemisphere and thepoint P ( R sin rR , , − R cos rR ) : Π : a cos rR x + (cid:113) − a y + a sin rR z = 0;Π : a cos rR x + (cid:113) − a y + a sin rR z = 0 , where | a | , | a | ≤ . (8.5)Let the angle between this two planes Π and Π equals α . Then cos α = a a + (cid:113) (1 − a )(1 − a ) . (8.6)13igure 12The equation of the tangent plane at the X to the sphere S is equal z = − R . Thistangent plane intersects the planes Π and Π along the lines, which form the angle (cid:98) α r (seeFigure 12), and cos (cid:98) α r = a a cos rR + (cid:112) (1 − a )(1 − a ) (cid:113) − a sin rR (cid:113) − a sin rR . (8.7)From the equations (8.6) and (8.7) we get | cos (cid:98) α r − cos α | < | a a sin rR | (cid:113) − a sin rR (cid:113) − a sin rR . (8.8)Applying the inequality (8.5) and the estimation (8.8) we get | cos (cid:98) α r − cos α | < tan rR . (8.9)From the equation | cos (cid:98) α r − cos α | = (cid:12)(cid:12)(cid:12) (cid:98) α r − α (cid:98) α r + α (cid:12)(cid:12)(cid:12) and inequalities (cid:12)(cid:12)(cid:12)(cid:12) sin (cid:98) α r + α (cid:12)(cid:12)(cid:12)(cid:12) > sin π , (cid:12)(cid:12)(cid:12)(cid:12) sin (cid:98) α r − α (cid:12)(cid:12)(cid:12)(cid:12) > π (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) α r − α (cid:12)(cid:12)(cid:12)(cid:12) , and since α > π and (cid:98) α r < π it follows π (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) α r − α (cid:12)(cid:12)(cid:12)(cid:12) < | cos (cid:98) α r − cos α | . From (8.10), (8.9) and equality α = π + ε we obtain (cid:12)(cid:12)(cid:12)(cid:98) α r − π (cid:12)(cid:12)(cid:12) < π tan rR + ε. Let us consider the arc of length one starting at the point P with the coordinates φ = r/R, θ = 0 , where r/R < π/ , on a sphere (8.3). Apply the central projection of this arc to theplane z = − R , which is tangent to the sphere at the point X ( φ = 0) .14 emma 8.3. The central projection of the hemisphere of radius R = 1 /a to the tangent planeat the point X maps the arc of the length one starting from the point P ( R sin rR , , − R cos rR ) into the segment of length (cid:98) l r satisfying the inequality (cid:98) l r − < cos π (4 + π (2 r + 1) ) (cid:0) − π a ( r + 1) (cid:1) · ε. (8.10) Proof.
The point P ( R sin rR , , − R cos rR ) on the sphere S is projected to the point (cid:98) P ( R tan rR , , − R ) on the tangent plane z = − R .Take the point Q ( Ra , Ra , Ra ) on a sphere such that the spherical distance P Q equals .Then ∠ P OQ = 1 /R , where O is a center of the sphere S (see Figure 13). Thus we get thefollowing conditions for the constants a , a , a : a sin rR − a cos rR = cos 1 R ; (8.11) a + a + a = 1 . (8.12)The central projection into the plane z = − R maps the point Q to the point (cid:98) Q (cid:16) − a a R, − a a R, − R (cid:17) . The length of (cid:98) P (cid:98) Q equals | (cid:98) P (cid:98) Q | = R (cid:115)(cid:18) a a − tan rR (cid:19) + a a (8.13)Applying the method of Lagrange multipliers for finding the local extremum of the length (cid:98) P (cid:98) Q , we get, that the minimum of | (cid:98) P (cid:98) Q | reaches when Q has the coordinates (cid:0) R sin r − R , , R cos r − R (cid:1) .Then | (cid:98) P (cid:98) Q | min = R (cid:12)(cid:12)(cid:12)(cid:12) tan rR − tan r − R (cid:12)(cid:12)(cid:12)(cid:12) = R sin R cos rR cos r − R . Note that | (cid:98) P (cid:98) Q | min > . Figure 13The maximum of | (cid:98) P (cid:98) Q | reaches at the point Q (cid:0) R sin r +1 R , , R cos r +1 R (cid:1) . This maximumvalue equals | (cid:98) P (cid:98) Q | max = R (cid:12)(cid:12)(cid:12)(cid:12) tan rR − tan r + 1 R (cid:12)(cid:12)(cid:12)(cid:12) = R sin R cos rR cos r +1 R . R = 1 /a , then the length (cid:98) l r of the projection of P Q satisfies (cid:98) l r < sin aa cos( ar ) cos (cid:0) a ( r + 1) (cid:1) . Applying the estimation sin a < a , we obtain (cid:98) l r − < − cos a − cos (cid:0) a (2 r + 1) (cid:1) ar ) cos (cid:0) a ( r + 1) (cid:1) . (8.14)From the equation (8.2) it follows − cos a = sin a a ≤ π ε. (8.15)In the same way from the inequality (8.1) we obtain − cos ( a (2 r + 1)) ≤ π cos π
12 (2 r + 1) ε ; (8.16)Estimate the denominator of the (8.14), using the inequality cos x > − π x where x < π/ .Applying the inequalities (8.15) и (8.16), we get (cid:98) l r − < π + π cos π (2 r + 1) (cid:0) − π a ( r + 1) (cid:1) · ε. Theorem 2.
Let ( p, q ) be a pair of coprime integers, ≤ p < q , and let ε satisfy ε < min √ c (cid:112) p + q + pq (cid:80) [ p + q ] +2 i =0 (cid:16) c l ( i ) + (cid:80) ij =0 c α ( j ) (cid:17) ; 18 cos π ( p + q ) , (8.17) where c = 3 − ( p + q +2) π cos π ( p + q ) − (cid:80) [ p + q ] +2 i =0 tan (cid:16) πi p + q ) (cid:17) − ( p + q +2)2 π cos π ( p + q ) − (cid:80) [ p + q ] +2 i =0 tan (cid:16) πi p + q ) (cid:17) ,c l ( i ) = cos π ( p + q ) (4 + π (2 i + 1) )( p + q − i − ,c α ( j ) = 4 (cid:18) π ( p + q ) cos π
12 tan πj p + q ) + 1 (cid:19) . Then on a regular tetrahedron in spherical space with the faces angle α = π/ ε there existsa unique, up to the rigid motion of the tetrahedron, simple closed geodesic of type ( p, q ) .Proof. Take the pair of coprime integers ( p, q ) , where < p < q . Consider a simple closedgeodesic (cid:101) γ of type ( p, q ) on a regular tetrahedron (cid:101) A (cid:101) A (cid:101) A (cid:101) A with the edge of the length one inEuclidean space. Suppose, that (cid:101) γ passes through the midpoints (cid:101) X , (cid:101) X and (cid:101) Y , (cid:101) Y of the edges (cid:101) A (cid:101) A и (cid:101) A (cid:101) A and (cid:101) A (cid:101) A , (cid:101) A (cid:101) A respectively.Consider the development (cid:101) T pq of the tetrahedron along (cid:101) γ starting from the point (cid:101) X . Thegeodesic unfolds to the segment (cid:101) X (cid:101) Y (cid:101) X (cid:101) Y (cid:102) X (cid:48) inside the development (cid:101) T pq . From Proposition 2we know, that the parts of the development along geodesic segments (cid:101) X (cid:101) Y , (cid:101) Y (cid:101) X , (cid:101) X (cid:101) Y and16igure 14 Figure 15 (cid:101) Y (cid:101) X (cid:48) are equal, and any two adjacent polygons can be transformed into each other by a rotationthrough an angle π around the midpoint of their common edge (see Figure 14).Now consider a two-dimensional sphere S of radius R = 1 /a , where a depends on α according to (2.1). On this sphere we take the several copies of regular spherical triangles withthe angle α at vertices, where π/ < α < π/ . Put this triangles up in the same order aswe develop the faces of the Euclidean tetrahedron along (cid:101) γ into the plane. In other words, weconstruct a polygon T pq on a sphere S formed by the same sequence of regular triangles as thepolygon (cid:101) T pq on Euclidean plane. Denote the vertices of T pq according to the vertices of (cid:101) T pq . Bythe construction the spherical polygon T pq has the same properties of the central symmetry asthe Euclidean (cid:101) T pq . Since the groups of isometries of regular tetrahedra in spherical space andin Euclidean space are equal, then T pq corresponds to the development of a regular tetrahedronwith the faces angle α in spherical space.Denote by X , X (cid:48) and X , Y , Y the midpoints of the edges A A , A A , A A , A A on T pq such that these midpoints correspond to the points (cid:101) X , (cid:101) X (cid:48) and (cid:101) X , (cid:101) Y , (cid:101) Y on the Euclideandevelopment (cid:101) T pq . Construct the great arcs X Y , Y X , X Y и Y X (cid:48) on a sphere. From theproperties of the central symmetry of T pq we obtain that these arcs form the one great arc X X (cid:48) on S (see Figure 15). Since the polygon T pq is not convex, we want to find α such that thepolygon T pq contains the arc X Y inside. Therefore the whole arc X X (cid:48) will be also containedinside T pq and X X (cid:48) will correspond to the simple closed geodesic of type ( p, q ) on a regulartetrahedron with the faces angle α in spherical space.In the following we will consider the part of the polygon T pq along X Y and will alsodenote it as T pq . This part consists of p + q regular spherical triangles with the edges of lengthone. Then if a satisfies the following inequality a ( p + q ) < π , (8.18)then the polygon T pq is contained inside the open hemisphere. Since α = π/ ε , then fromthe condition (8 . we get that the estimation (8.18) holds if ε <
18 cos π ( p + q ) . (8.19)In this case the length of the arc X Y is less than π/ a , so X Y satisfies the necessary conditionfrom Lemma 5.1.Apply a central projection of the T pq into the tangent plane T X S at the point X tothe sphere S . Since the central projection is a geodesic map, then the image of the sphericalpolygon T pq on T X S is a polygon (cid:98) T pq . 17enote by (cid:98) A i the vertex of (cid:98) T pq , which is an image of the vertex A i on T pq . The arc X Y maps into the line segment (cid:98) X (cid:98) Y on T X S , that joins the midpoints of the edges (cid:98) A (cid:98) A и (cid:98) A (cid:98) A .If for some α the segment (cid:98) X (cid:98) Y lies inside the polygon (cid:98) T pq , then the arc X Y is also containinginside T pq on the sphere.The vector (cid:98) X (cid:98) Y equals (cid:98) X (cid:98) Y = (cid:98) a + (cid:98) a + · · · + (cid:98) a s + (cid:98) a s +1 , (8.20)where (cid:98) a i are the sequential vectors of the (cid:98) T pq boundary, (cid:98) a = (cid:99) X (cid:99) A , (cid:98) a s +1 = (cid:99) A (cid:98) Y , and s = (cid:2) p + q (cid:3) + 1 (if we take the boundary of (cid:98) T pq from the other side of (cid:98) X (cid:98) Y , then s = (cid:2) p + q (cid:3) ) (seeFigure 16).On the other hand at the Euclidean plane T X S there exists a development (cid:101) T pq of a regularEuclidean tetrahedron (cid:101) A (cid:101) A (cid:101) A (cid:101) A with the edge of length one along a simple closed geodesic (cid:101) γ .The development (cid:101) T pq is equivalent to T pq , and then it’s equivalent to (cid:98) T pq . The segment (cid:101) X (cid:101) Y lies inside (cid:101) T pq and corresponds to the segment of (cid:101) γ .Let the development (cid:101) T pq be placed such that the point (cid:101) X coincides with (cid:98) X of (cid:98) T pq , andthe vector (cid:98) X (cid:98) A has the same direction with (cid:101) X (cid:101) A . Similarly the vector (cid:101) X (cid:101) Y equals (cid:101) X (cid:101) Y = (cid:101) a + (cid:101) a + · · · + (cid:101) a s + (cid:101) a s +1 , (8.21)where (cid:101) a i are the sequential vectors of the (cid:101) T pq boundary, s = (cid:2) p + q (cid:3) + 1 and (cid:101) a = (cid:101) X (cid:101) A , (cid:101) a s +1 = (cid:101) A (cid:101) Y (see Figure 16). Figure 16Suppose the minimal distance from the vertices of (cid:101) T pq to the segment (cid:101) X (cid:101) Y reaches at thevertex (cid:101) A k and equals (cid:101) h from the formula (3.2). Let us find the estimation of the distance (cid:98) h between the segment (cid:98) X (cid:98) Y and the corresponding vertex (cid:98) A k on (cid:98) T pq . A geodesic on a regulartetrahedron in Euclidean space intersects at most three edges starting from the same tetrahe-dron’s vertex. It follows, that the interior angles of the polygon (cid:101) T pq is not greater than π , andthe angles of the corresponding vertices on (cid:98) T pq is not greater than (cid:98) α i . Applying the formula(8.4) for ≤ i ≤ s we get, that the angle between (cid:98) a i and (cid:101) a i satisfies the inequality ∠ ( (cid:98) a i , (cid:101) a i ) < i (cid:88) j =0 (cid:18) π tan jR + ε (cid:19) . (8.22)18ince R = 1 /a , then using (8.1) we obtain tan jR < tan (cid:18) jπ (cid:114) π √ ε (cid:19) . (8.23)The inequality (8.18) holds if the following condition fulfills tan (cid:18) jπ (cid:114) π √ ε (cid:19) < tan πj p + q ) . (8.24)If tan x < tan x , then tan x < tan x x x . From (8.24) it follows tan (cid:18) jπ (cid:114) π √ ε (cid:19) < p + q ) tan πj p + q ) (cid:114) π √ ε. (8.25)Therefore from (8.23) and (8.25) we get tan jR < p + q ) tan πj p + q ) (cid:114) π √ ε. (8.26)Using (8.22) and (8.26) we obtain the final estimation for the angle between the vectors (cid:98) a i and (cid:101) a i : ∠ ( (cid:98) a i , (cid:101) a i ) < i (cid:88) j =0 (cid:18) π ( p + q ) cos π
12 tan πj p + q ) + 1 (cid:19) ε. (8.27)Figure 17 Figure 18Now estimate the length of the vector (cid:98) a i − (cid:101) a i (see Figure 17). The following inequalityholds | (cid:98) a i − (cid:101) a i | ≤ (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) a i | (cid:98) a i | − (cid:101) a i (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) a i − (cid:98) a i | (cid:98) a i | (cid:12)(cid:12)(cid:12)(cid:12) . (8.28)Since (cid:101) a i is a unite vector, then (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) a i | (cid:98) a i | − (cid:101) a i (cid:12)(cid:12)(cid:12)(cid:12) ≤ ∠ ( (cid:98) a i , (cid:101) a i ) и (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) a i − (cid:98) a i | (cid:98) a i | (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:98) l i − . (8.29)From the inequality (8.10) we get (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) a i − (cid:98) a i | (cid:98) a i | (cid:12)(cid:12)(cid:12)(cid:12) < cos π (4 + π (2 i + 1) ) (cid:0) − π a ( i + 1) (cid:1) · ε. (8.30)19sing the estimation (8.18), we obtain (cid:12)(cid:12)(cid:12)(cid:12) (cid:98) a i − (cid:98) a i | (cid:98) a i | (cid:12)(cid:12)(cid:12)(cid:12) < cos π ( p + q ) (4 + π (2 i + 1) )( p + q − i − · ε. (8.31)Therefore, from (8.28), (8.27) and (8.31) we get | (cid:98) a i − (cid:101) a i | ≤ (cid:32) c l ( i ) + i (cid:88) j =0 c α ( j ) (cid:33) ε, (8.32)where c l ( i ) = cos π ( p + q ) (4 + π (2 i + 1) )( p + q − i − , (8.33) c α ( j ) = 4 (cid:18) π ( p + q ) cos π
12 tan πj p + q ) + 1 (cid:19) . (8.34)Using (8.32) we estimate the length of (cid:98) Y (cid:101) Y | (cid:98) Y (cid:101) Y | < s +1 (cid:88) i =0 | (cid:98) a i − ˜ a i | < s +1 (cid:88) i =0 (cid:32) c l ( i ) + i (cid:88) j =0 c α ( j ) (cid:33) ε. (8.35)From (8.27) it follows that the angle ∠ (cid:98) Y (cid:98) X (cid:101) Y satisfies ∠ (cid:98) Y (cid:98) X (cid:101) Y < s +1 (cid:88) i =0 c α ( i ) ε. (8.36)The distance between the vertices (cid:98) A k and (cid:101) A k equals | (cid:98) A k (cid:101) A k | < k (cid:88) i =0 (cid:32) c l ( i ) + i (cid:88) j =0 c α ( j ) (cid:33) ε. (8.37)We drop a perpendicular (cid:98) A k (cid:98) H from the vertex (cid:98) A k into the segment (cid:98) X (cid:98) Y . The length of (cid:98) A k (cid:98) H equals (cid:98) h . Then we drop the perpendicular (cid:101) A k (cid:101) H into the segment (cid:101) X (cid:101) Y and the length of (cid:101) A k (cid:101) H equals (cid:101) h .Let the point F on (cid:101) X (cid:101) Y be such that the segment (cid:101) A k F intersect (cid:98) X (cid:98) Y at right angle.Then the length of (cid:101) A k F is not less than (cid:101) h . Let the point G on (cid:101) X (cid:101) Y lie at the extension ofthe segment (cid:98) A k (cid:98) H , and F K is perpendicular to (cid:98) HG (see Figure 18). Then the length of F K isnot greater than the length of (cid:98) A k (cid:101) A k , and ∠ KF G = ∠ (cid:98) Y (cid:98) X (cid:101) Y . From the triangle GF K we get,that | F G | = | F K | cos ∠ (cid:98) Y (cid:98) X (cid:101) Y . (8.38)Applying the inequality cos x > − π x , where x < π , to (8.38). We obtain | F G | < | (cid:98) A k (cid:101) A k | − π ∠ (cid:98) Y (cid:98) X (cid:101) Y . (8.39)So from (8.36), (8.37), and (8.39) it follows | F G | < (cid:80) ki =0 (cid:16) c l ( i ) + (cid:80) ij =0 c α ( j ) (cid:17) ε − (cid:80) si =0 (cid:16) π ( p + q ) cos π tan πi p + q ) + π (cid:17) ε . (8.40)20rom (8.19) and (8.40) we obtain | F G | < (cid:80) ki =0 (cid:16) c l ( i ) + (cid:80) ij =0 c α ( j ) (cid:17) ε − ( p + q +2)2 π cos π ( p + q ) − (cid:80) s +1 i =0 tan (cid:16) πi p + q ) (cid:17) . (8.41)From the our construction it follows (cid:101) h ≤ (cid:101) A k F ≤ (cid:98) h + | (cid:98) HG | + | (cid:98) A k (cid:101) A k | + | F G | ; (8.42)Note, that | (cid:98) HG | < | (cid:98) Y (cid:101) Y | . From Lemma 3.1 we know, that the distance (cid:101) h satisfies the inequality (cid:101) h > √ (cid:112) p + q + pq . Hence from (8.42) it follows, that (cid:98) h > √ (cid:112) p + q + pq − | (cid:98) Y (cid:101) Y | − | (cid:98) A k (cid:101) A k | − | F G | . (8.43)Applying the estimations (8.35), (8.37), (8.41) and the equality s = (cid:2) p + q (cid:3) + 1 , we get (cid:98) h > √ (cid:112) p + q + pq − c [ p + q ] +2 (cid:88) i =0 (cid:32) c l ( i ) + i (cid:88) j =0 c α ( j ) (cid:33) ε, (8.44)where c l ( i ) is from (8.33), and c α ( j ) is from (8.34) and c = 3 − ( p + q +2) π cos π ( p + q ) − (cid:80) [ p + q ] +2 i =0 tan (cid:16) πi p + q ) (cid:17) − ( p + q +2)2 π cos π ( p + q ) − (cid:80) [ p + q ] +2 i =0 tan (cid:16) πi p + q ) (cid:17) , From the inequality (8.44) we obtain, that if ε satisfies the condition ε < √ c (cid:112) p + q + pq (cid:80) [ p + q ] +2 i =0 (cid:16) c l ( i ) + (cid:80) ij =0 c α ( j ) (cid:17) , (8.45)then the distance from the vertices of the polygon (cid:98) T pq to (cid:98) X (cid:98) Y is greater than zero.Since we use the estimation (8.19), we get, that if ε < min √ c (cid:112) p + q + pq (cid:80) [ p + q ] +2 i =0 (cid:16) c l ( i ) + (cid:80) ij =0 c α ( j ) (cid:17) ; 18 cos π ( p + q ) , (8.46)then the segment (cid:98) X (cid:98) Y lies inside the polygon (cid:98) T pq . It follows that the arc X Y on a sphere isalso containing inside the polygon T pq . The arc X Y corresponds to a simple closed geodesic γ of type ( p, q ) on a regular tetrahedron with the faces angle α = π/ ε in spherical space. FromCorollary 6.1 we get, that this geodesic is unique, up to the rigid motion of the tetrahedron.Note, that the geodesic γ is invariant under the rotation of the tetrahedron of the angle π over the line passing through the midpoints of the opposite edges of the tetrahedron. Therotation of the tetrahedron of the angle π/ or π/ over the line passing through the tetra-hedron’s vertex and the center of its opposite face changes γ into another geodesic of type ( p, q ) .The rotations over others lines connected other tetrahedron’s vertex with the center of itsopposite face produce the existing geodesics. 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