Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold

Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold

We give a full description of totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold of constant curvature and present a new class of a cylinder-type totally geodesic submanifolds in the general case.

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Datum:2005
1. Verfasser: Yampolsky, A.
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Veröffentlicht: Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України 2005
Schriftenreihe:Журнал математической физики, анализа, геометрии
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Zitieren:Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold / A. Yampolsky // Журнал математической физики, анализа, геометрии. — 2005. — Т. 1, № 1. — С. 116-139. — Бібліогр.: 14 назв. — англ.

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spelling irk-123456789-1065682016-10-01T03:02:05Z Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold Yampolsky, A. We give a full description of totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold of constant curvature and present a new class of a cylinder-type totally geodesic submanifolds in the general case. 2005 Article Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold / A. Yampolsky // Журнал математической физики, анализа, геометрии. — 2005. — Т. 1, № 1. — С. 116-139. — Бібліогр.: 14 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106568 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
language English
description We give a full description of totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold of constant curvature and present a new class of a cylinder-type totally geodesic submanifolds in the general case.
format Article
author Yampolsky, A.
spellingShingle Yampolsky, A.
Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold
Журнал математической физики, анализа, геометрии
author_facet Yampolsky, A.
author_sort Yampolsky, A.
title Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold
title_short Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold
title_full Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold
title_fullStr Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold
title_full_unstemmed Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold
title_sort totally geodesic submanifolds in the tangent bundle of a riemannian 2-manifold
publisher Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
publishDate 2005
url http://dspace.nbuv.gov.ua/handle/123456789/106568
citation_txt Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold / A. Yampolsky // Журнал математической физики, анализа, геометрии. — 2005. — Т. 1, № 1. — С. 116-139. — Бібліогр.: 14 назв. — англ.
series Журнал математической физики, анализа, геометрии
work_keys_str_mv AT yampolskya totallygeodesicsubmanifoldsinthetangentbundleofariemannian2manifold
first_indexed 2025-07-07T18:39:10Z
last_indexed 2025-07-07T18:39:10Z
_version_ 1837014503671201792
fulltext Journal of Mathematical Physics, Analysis, Geometry 2005, v. 1, No. 1, p. 116�139 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold Alexander Yampolsky Department of Mechanics and Mathematics, V.N. Karazin Kharkov National University 4 Svobody Sq., Kharkov, 61077, Ukraine E-mail:yamp@univer.kharkov.ua Received December 7, 2004 We give a full description of totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold of constant curvature and present a new class of a cylinder-type totally geodesic submanifolds in the general case. Introduction Let (Mn; g) be a Riemannian manifold with metric g and TMn its tangent bundle. S. Sasaki [7] introduced on TMn a natural Riemannian metric Tg. With respect to this metric, all the �bers are totally geodesic and intrinsically �at submanifolds. Probably M.-S. Liu [5] was the �rst who noticed that the base manifold embedded into TMn by the zero section is totally geodesic, as well. Soon afterwards, K. Sato [9] described geodesics (the totally geodesic submanifolds of dimension 1) in the tangent bundle over space forms. The next step was made by P. Walczack [10] who tried to �nd a nonzero section � : Mn ! TMn such that the image �(Mn) is a totally geodesic submanifold. He proved that if � is of constant length and �(Mn) is totally geodesic, then � is a parallel vector �eld. As a consequence, the base manifold should be reducible. The irreducible case stays out of considerations up to now. A general conjecture stated by A. Borisenko claims that, in irreducible case, the zero vector �eld is the unique one which generates a totally geodesic submanifold �(Mn) or, equivalently, the base manifold is the unique totally geodesic submanifold of dimension n in TMn transversal to �bers. A dimensional restriction is essential. M.T.K. Abbassi and the author [1] treated the case of �ber transversal submanifolds in TMn of dimension l < n and have found some examples of totally geodesic submanifolds of this type. Earlier this problem had been considered in [11]. Mathematics Subject Classi�cation 2000: 53B25 (primary); 53B20 (secondary). Key words: Sasaki metric, totally geodesic submanifolds in the tangent bundle. c Alexander Yampolsky, 2005 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold It is also worthwhile to mention that in the case of tangent sphere bundle the situation is di�erent. S. Sasaki [8] described geodesics in the tangent sphere bundle over space forms and P. Nagy [6] described geodesics in the tangent sphere bundle over symmetric spaces. The author has given a full description of totally geodesic vector �elds on 2-dimensional manifolds of constant curvature [12] and an example of a totally geodesic unit vector �eld on positively/negatively curved manifolds of nonconstant curvature [13]. A full description of 2-manifolds which admit a totally geodesic unit vector �eld was given in [14]. In this paper we consider a more general problem concerning the description of all possible totally geodesic submanifolds in the tangent bundle of Riemannian 2-manifold with a sign-preserving curvature. For the spaces of constant curvature this problem was posed by A. Borisenko in [2]. In Section 2 we prove the following theorems. Theorem 1. Let M2 be Riemannian manifold of constant curvature K 6= 0. Suppose that ~F 2 � TM2 is a totally geodesic submanifold. Then locally ~F 2 is one of the following submanifolds: (a) a single �ber TqM 2; (b) a cylinder-type surface based on a geodesic in M2 with elements generated by a parallel unit vector �eld along ; (c) the base manifold embedded into TM2 by zero vector �eld. Remark that the item (b) of Theorem 1 is a consequence of more general result. Theorem 2. Let M2 be a Riemannian manifold of sign-preserving curvature. Suppose that ~F 2 � TM2 is a totally geodesic submanifold having nontransversal intersection with the �bers. Then locally ~F 2 is a cylinder-type surface based on a geodesic in M2 with elements generated by a parallel unit vector �eld along . Moreover, a general Riemannian manifold Mn admits this class of totally geodesic surfaces in TMn (see Prop. 2.4). In Section 3 we prove the following general result. Theorem 3. Let M2 be a Riemannian manifold with sign-preserving curva- ture. Then TM2 does not admit a totally geodesic 3-manifold even locally. Acknowledgement. The author expresses his thanks to Professor E. Boeckx (Leuven, Belgium) for useful remarks in discussing the results. 1. Necessary facts about the Sasaki metric Let (Mn; g) be an n-dimensional Riemannian manifold with metric g. Denote by � ; � � the scalar product with respect to g. The Sasaki metric on TMn is de�ned Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 117 Alexander Yampolsky by the following scalar product: if ~X; ~Y are tangent vector �elds on TMn, then ~X; ~Y �� := �� ~X;�� ~Y � + K ~X;K ~Y � ; (1) where �� : TTMn ! TMn is the di�erential of the projection � : TMn ! Mn and K : TTMn ! TMn is the connection map [3]. The local representations for �� and K are the following ones. Let (x1; : : : ; xn) be a local coordinate system on Mn. Denote by @i := @=@xi the natural tangent coordinate frame. Then, at each point q 2 Mn, any tangent vector � can be decomposed as � = �i @ijq. The set of parameters fx1; : : : ; xn; �1; : : : ; �ng forms the natural induced coordinate system in TMn, i.e., for a point Q = (q; �) 2 TMn, with q 2 Mn; � 2 TqM n, we have q = (x1; : : : ; xn); � = �i @ijq. The natural frame in TQTM n is formed by ~@i := @ @xi jQ; ~@n+i := @ @�i jQ; and for any ~X 2 TQTM n we have the decomposition ~X = ~Xi ~@i + ~Xn+i ~@n+i : Now locally, the horizontal and vertical projections of ~X are given by �� ~X jQ = ~Xi @ijq; K ~XjQ = ( ~Xn+i + �i jk (q) �j ~Xk) @ijq; (2) where �i jk are the Christo�el symbols of the metric g. The inverse operations are called lifts . If X = Xi @i is a vector �eld on Mn then the vector �elds on TM given by Xh = Xi ~@i � �i jk �jXk ~@n+i; Xv = Xi ~@n+i are called the horizontal and vertical lifts of X respectively. Remark that for any vector �eld X on Mn it holds ��X h = X; KXh = 0; ��X v = 0; KXv = X: (3) There is a natural decomposition TQ(TM n) = HQ(TM n)� VQ(TMn); 118 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold where HQ(TM n) = kerK is called the horizontal distribution and VQ(TMn) = ker �� is called the vertical distribution on TMn. With respect to the Sasaki metric, these distributions are mutually orthogonal. The vertical distribution is integrable and the �bers are precisely its integral submanifolds. The horizontal distribution is never integrable except the case of a �at base manifold. For any vector �elds X;Y on Mn, the covariant derivatives of various com- binations of lifts to the point Q = (q; �) 2 TMn can be found by the formulas [4] ~rXhY hjQ = (rXY jq)h � 1 2 (Rq(X;Y )�)v; ~rXvY hjQ = 1 2 (Rq(�;X)Y )h; ~rXhY vjQ = (rXY jq)v + 1 2 (Rq(�; Y )X)h; ~rXvY vjQ = 0; (4) where r and R are the Levi�Civita connection and the curvature tensor of Mn respectively. R e m a r k. The formulas (4) are applicable to the lifts of vector �elds only. A formal application to a general �eld on tangent bundle may lead to wrong result. For example, ~rXv(�i(@i) h) = Xv(�i) @h i + �i ~rXv@h i = Xi@h i + �i 1 2 � R(�;X)@i �h = Xh + 1 2 � R(�;X)� �h and we have an additional term in the formulas. We will use this rule in our calculations without special comments. 2. Local description of 2-dimensional totally geodesic submanifolds in TM2 In this section we prove Theorem 1. The proof is given in a series of subsec- tions. Namely, in Subsection 2.1 we prove the item (a), in Subsection 2.2 we prove the item (c) and �nally, in Subsection 2.3 we prove Theorem 2 and therefore, the item (b) of Theorem 1. 2.1. Preliminary considerations Let ~F 2 be a submanifold in TM2. Let (x1; x2; �1; �2) be a local chart on TM2. Then locally ~F 2 can be given by mapping f of the form f : ( x1 = x1(u1; u2); x2 = x2(u1; u2); �1 = �1(u1; u2); �2 = �1(u1; u2); Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 119 Alexander Yampolsky where u1; u2 are the local parameters on ~F 2. The Jacobian matrix f� of the mapping f is of the form f� = 0 BBBBBB@ @x 1 @u1 @x 1 @u2 @x2 @u1 @x2 @u2 @� 1 @u1 @� 1 @u2 @�2 @u1 @�2 @u2 1 CCCCCCA : Since rank f� = 2, we have three geometrically di�erent possibilities to achieve the rank, namely (a) det 0 @ @x1 @u1 @x1 @u2 @x 2 @u1 @x 2 @u2 1 A 6= 0; (b) det 0 @ @x 1 @u1 @x 1 @u2 @�1 @u1 @�1 @u2 1 A 6= 0; (c) det 0 @ @�1 @u1 @�1 @u2 @�2 @u1 @�2 @u2 1 A 6= 0: Without loss of generality we can consider these possibilities in a way that (b) excludes (a), and (c) excludes (a) and (b) restricting the considerations to a smaller neighbourhood or even to an open and dense subset. Case (a). In this case one can locally parameterize the submanifold under consideration as f : ( x1 = u1; x2 = u1; �1 = �1(u1; u2); �2 = �2(u1; u2); and we can consider the submanifold ~F 2 as an image of the vector �eld �(u1; u2) on the base manifold. Denote ~F 2 in this case by �(M2). We analyze this case in subsection 2.2. Case (b). In this case one can parameterize the submanifold F 2 as f : ( x1 = u1; x2 = x2(u1; u2); �1 = u2; �2 = �2(u1; u2): Taking into account that we exclude the case (a) in considerations of the case (b), we should set det 0 @ @x1 @u1 @x1 @u2 @x 2 @u1 @x 2 @u2 1 A = det 0 @ 1 0 @x2 @u1 @x2 @u2 1 A = @x2 @u2 = 0: 120 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold Therefore, x2(u1; u2) does not depend on u2 and the local representation takes the form f : ( x1 = u1; x2 = x2(u1); �1 = u2; �2 = �2(u1; u2): Remark that �( ~F 2) = (u1; x2(u1) is a regular curve on M2. If we denote this projection by (s) parameterized by the arc-length parameter and set u2 := t, the local parametrization of ~F 2 takes the form (s) : ( x1 = x1(s); x2 = x2(s); �(t; s) : ( �1 = t; �2 = �2(t; s): (5) We can interpret this kind of submanifolds in TM2 as a one-parametric family of smooth vector �elds over a regular curve on the base manifold. We will refer to this kind of submanifolds as ruled submanifolds in TM2 and analyze their totally geodesic property in subsection 2.3. Case (c). It this case a local parametrization of ~F 2 can be given as f : ( x1 = x1(u1; u2); x2 = x2(u1; u2); �1 = u1; �2 = u2: Taking into account that we exclude the case (b) considering the case (c), we should suppose det 0 @ @x 1 @u1 @x 1 @u2 @�1 @u1 @�1 @u2 1 A = det 0 @ @x 1 @u1 @x 1 @u2 1 0 1 A = �@x 1 @u2 = 0; det 0 @ @x1 @u1 @x1 @u2 @�2 @u1 @�2 @u2 1 A = det 0 @ @x 1 @u1 @x 1 @u2 0 1 1 A = @x1 @u1 = 0: Thus, we conclude x1 = const. In the same way, we get x2 = const. Therefore, a submanifold of this kind is nothing else but the �ber, which is evidently totally geodesic and there is nothing to prove. 2.2. Totally geodesic vector �elds In [1] the author has found the conditions on a vector �eld to generate a totally geodesic submanifold in the tangent bundle. Namely, let � be a vector �eld on Mn. The submanifold �(Mn) is totally geodesic in TMn if and only if for any vector �elds X;Y on Mn the following equation holds: r(X;Y )� + r(Y;X)� �rh�(X;Y )� = 0; (6) Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 121 Alexander Yampolsky where r(X;Y )� = rXrY � �rrXY � is "half" the Riemannian curvature tensor and h�(X;Y ) = R(�;rX�)Y +R(�;rY �)X. It is natural to rewrite this equations in terms of � and e� where e� is a unit vector �eld and � is the length function of �. Lemma 2.1. Let � = � e� be a vector �eld on a Riemannian manifold Mn. Then �(Mn) is totally geodesic in TMn if and only if for any vector �eld X the following equations hold8< : Hess�(X;X) � �2 � R(e�;rXe�)X � (�)� � jrXe�j2 = 0; �3rR(e�;rXe�)Xe� � 2X(�)rXe� � �(r(X;X)e� + jrXe�j2 e�) = 0; (7) where Hess�(X;X) is the Hessian of the function �. P r o o f. Indeed, the equation (6) is equivalent to r(X;X)� = rR(�;rX�)X�; (8) where X is an arbitrary vector �eld. Setting � = � e�, where e� is a unit vector �eld, we have r(X;X)� = rXrX(� e�)�rrXX(� e�) = rX(X(�)e� + �rXe�)� (rXX)(�)e� � �rrXXe� = � X(X(�)) � (rXX)(�) � e� + 2X(�)rXe� + � r(X;X)e� and rR(�;rX�)X� = �2 � R(e� ;rXe�)X � (�) e� + �3rR(e�;rXe�)Xe�: If we remark that X(X(�)) � (rXX)(�) def = Hess�(X;X) and for a unit vector �eld e� r(X;X)e� ; e� � = �jrXe�j2; then we can easily decompose the equation (8) into components, parallel to and orthogonal to e�, which gives the equations (7). Corollary 2.1. Suppose thatMn admits a totally geodesic vector �eld � = � e�. Then (a) the function � has no strong maximums; (b) there is a bivector �eld e0 ^re0e0 such that e� is parallel along it. Particulary, if n = 2 then either M2 is �at or e0 is a geodesic vector �eld and � is linear with respect to the natural parameter along each e0 geodesic line. Moreover, the �eld � makes a constant angle with each e0 geodesic line. 122 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold P r o o f. Indeed, for any unit vector �eld � consider the linear mapping rZ�jq : TqM n ! �?q , where �?q is an orthogonal complement to � in TqM n. For dimensional reasons it follows that the kernel of this mapping is not empty. In other words, there exists a (unit) vector �eld e0 such that re0� = 0. Let e0 be a unit vector �eld such that re0e� = 0. Then from (7)1 we conclude Hess�(e0; e0) = 0 at each point of Mn. Therefore, the Hessian of � can not be positively de�nite. Moreover, from (7)2 we see that r(e0; e0)e� = 0, which gives re0re0e� � rre0 e0e� = �rre0 e0e� = 0. Setting Z = e0 ^re0e0, we get rZe� = 0: Suppose now that n = 2. If Z 6= 0 then e� is a parallel vector �eld on M2 which means that M2 is �at. If Z = 0 then evidently e0 is a geodesic vector �eld. Since in this case Hess�(e0; e0) = e0(e0(�)) = 0, we conclude that � is linear with respect to the natural parameter along each e0 geodesic line. As concerns the angle function e0; e� � , we have e0 e0; e� � = re0e0; e� � + e0;re0e� � = 0: Taking into account the Corollary 2.1, introduce on M2 a semi-geodesic coor- dinate system (u; v) such that e� is parallel along u-geodesics. Let ds2 = du2 + b2(u; v) dv2 (9) be the �rst fundamental form of M2 with respect to this coordinate system. De- note by @1 and @2 the corresponding coordinate vector �elds. Then the following equations should be satis�ed: r@1 e� = 0; @21(�) = 0: Introduce the unit vector �elds e1 = @1; e2 = 1 b @2: Then the following rules of covariant derivation are valid: re1e1 = 0; re1e2 = 0; re2e1 = �k e2 re2e2 = k e1; (10) where k is a (signed) geodesic curvature of v-curves. Remark that k = �@1b b : Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 123 Alexander Yampolsky With respect to chosen coordinate system, the �eld � can be expressed as � = � (cos! e1 + sin! e2); (11) where ! = !(u; v) is an angle function, i.e., e� = cos! e1 + sin! e2: Introduce a unit vector �eld �� by �� = � sin! e1 + cos! e2: Then we can easily �nd re1e� = @1! ��; re2e� = (e2(!)� k) �� : Since e� is parallel along u-curves, we conclude that @1! = 0, so that ! = !(v). Now the problem can be formulated as On a Riemannian 2-manifold with the metric (9), �nd a vector �eld of the form (11) with @21� = 0 and ! = !(v) (12) satisfying the equation (8). Lemma 2.2. Let M2 be a Riemannian 2-manifold with the metric (9) and � be a local vector �eld on M2 satisfying (12). Then � is totally geodesic if and only if re2re2� � (k + cK)re1� = 0; re1re2� +re2re1� + (k + cK)re2� = 0; (13) or in a scalar form ( e2(e2(�))� (k + cK) e1(�) = ��2; e2(c) = 0;( 2e1(e2(�)) + cK e2(�) = 0; e1(c) + c (k + cK) = 0; (14) where � := re2e�; �� � = e2(!)�k, c := �2� = �j �^re2�j and K is the Gaussian curvature of M2. 124 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold P r o o f. Indeed, re1� = e1(�) e� ; re2� = e2(�) e� + ����: So, taking into account (10) and (12), we have r(e1; e1)� = re1re1� �rre1 e1� = e1(e1(�)) e� = @21� e� = 0; r(e1; e2)� = re1re2� �rre1 e2� = re1re2�; r(e2; e1)� = re2re1� �rre2 e1� = re2re1� + kre2�; r(e2; e2)� = re2re2� �rre2e2 � = re2re2� � kre1�: As concerns the right-hand side of (8), we have R(�;re1�)e1 = 0; R(�;re1�)e2 = 0; R(�;re2�)e1 = �2�R(e� ; ��)e1 = ��2�K e2; R(�;re2�)e2 = �2�R(e� ; ��)e2 = �2�K e1: Therefore, setting X = e1 in (8), we obtain an identity. Setting X = e2, we have re2re2� � kre1� = �2�Kre1�: Setting X = e1 + e2, we obtain r(e1; e2)� + r(e2; e1)� = ��2�Kre2�; which can be reduced to re1re2� +re2re1� + kre2� = ��2�Kre2�: It remains to mention that j � ^re2�j2 = j�j2 jre2�j2 � �;re2� �2 = �2(e2(�) 2 + �2�2)� (e2(�)�) 2 = �4�2: So, if we set c = �2�, we evidently obtain (13). Moreover, continuing calculations, we see that re2re2� = h e2(e2(�)) � ��2 i e� + h e2(�)�+ e2(��) i �� = h e2(e2(�)) � ��2 i e� + 1 � e2(c) �� ; re1re2� +re2re1� = h e2(e1(�)) + e1(e2(�)) i e� + h e1(�)�+ e1(��) i �� = h e2(e1(�)) + e1(e2(�)) i e� + 1 � e1(c)�� : Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 125 Alexander Yampolsky Taking into account that e1(e2(�)) � e2(e1(�)) = k e2(�), the equations (13) can be written ash e2(e2(�))� ��2 i e� + 1 � e2(c) �� � (k + cK)e1(�) e� = 0;h 2 e1(e2(�)) � k e2(�) i e� + 1 � e1(c) �� + (k + cK) h e2(�) e� + �� �� i = 0; and after evident simpli�cations we obtain the equations (14). Proposition 2.1. Let M2 be a Riemannian manifold of constant curvature. Suppose � is a nonzero local vector �eld on M2 such that �(M2) is totally geodesic in TM2. Then M2 is �at. P r o o f. Let M2 a Riemannian manifold of constant curvature K 6= 0. Then the function b in (9) should satisfy the equation �@11b b = K: The general solution of this equation can be expressed in 3 forms: (a) b(u; v) = A(v) cos(u=r + �(v)) or b(u; v) = A(v) sin(u=r + �(v)) for K = 1=r2 > 0; (b) b(u; v) = A(v) cosh(u=r + �(v)) or b(u; v) = A(v) sinh(u=r + �(v)) for K = �1=r2 < 0; (c) b(u; v) = A(v)eu=r for K = �1=r2 < 0. Evidently, we may set A(v) � 1 (making a v-parameter change) in each of these cases. The equation (14)2 means that c does not depend on v. Since K is constant, the equation (14)4 implies e2(k) = 0: If we remark that k = �@1b b then one can easily �nd �(v) = const in cases (a) and (b). After a u-parameter change, the function b takes one of the forms: (a) b(u; v) = cos(u=r) or b(u; v) = sin(u=r) for K = 1=r2 > 0; (b) b(u; v) = cosh(u=r) or b(u; v) = sinh(u=r) for K = �1=r2 < 0; (c) b(u; v) = eu=r for K = �1=r2 < 0. 126 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold From the equation (14)4 we �nd cK = �e1(c) c � k = �e1(c) c + e1(b) b = e1(ln b=c): Suppose �rst that e2(�) 6= 0. Multiplying (14)3 by e2(�), we can easily solve this equation with respect to e2(�) by a chain of simple transformations: 2e2(�) � e1(e2(�)) + e1(ln b=c) � [e2(�)2] = 0; e1[e2(�) 2] + e1(ln b=c) � [e2(�)2] = 0; e1[e2(�) 2] e2(�)2 + e1(ln b=c) = 0; e1[ln e2(�) 2] + e1(ln b=c) = 0; e1(ln[e2(�) 2 b=c]) = 0 and therefore, e2(�) 2 b=c = h(v)2 or @2� = h(v) p c b: Since � is linear with respect to the u-parameter, say � = a1(v)u + a2(v), then @2� = a01u + a02 and therefore p cb is also linear with respect to u, namelyp cb = m1(v)u+m2(v) = a 0 1 h u+ a 0 2 h . But the functions c and b do not depend on v. Therefore m1 and m2 are constants, so a1 = m1 R h(v) dv; a2 = m2 R h(v) dv. Thus p cb = m1u+m2: Now the function c takes the form c(u) = (m1u+m2) 2 b and therefore e1(c) = 2m1(m1u+m2) b � (m1u+m2) 2@1b b2 : Substitution into (14)4 gives 2m1(m1u+m2) b � 2(m1u+m2) 2@1b b2 + (m1u+m2) 4 b2 K = 0 or (m1u+m2) b2 h 2m1b� 2(m1u+m2)@1b+ (m1u+m2) 3K i = 0: The expression in brackets is an algebraic one and can not be identically zero if K 6= 0. Therefore m1 = m2 = 0 and hence �2� := c = 0. But this identity implies Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 127 Alexander Yampolsky � = 0 or � = 0. If � = 0 then e� is a parallel unit vector �eld and therefore, M2 is �at and we come to a contradiction. Therefore � = 0. Re m a r k. If K = 0, we can not conclude that c = 0. In this case the expression in brackets can be identically zero for m1 = 0 and b = const. And we have c = m2 = const. Suppose now that e2(�) = 0. Then � = a1u+ a2; where a1; a2 are constants and we obtain the following system: �(k + cK)@1� = ��2; @2 c = 0; @1c+ c(k + cK) = 0: (15) If @1� = 0 then immediately � = 0 or � = 0. The identity � = 0 implies K = 0 as above. Therefore, � = 0. Suppose @1� 6= 0 or equivalently a1 6= 0. Then from (15)1 we get (k + cK) = ��� 2 a1 : (16) Since c = �2�, from (15)2 we see that @2� = 0 or @2 h @2!+@1b b i = 0. Since b does not depend on v, we have @22! = 0 or equivalently @2! = � = const. Thus, � = �+@1b b . Now we can �nd @1c in two ways. First, from (15)3 using (16) and keeping in mind that c = �2�: @1c = c ��2 a1 = �3�3 a1 : Second, directly: @1c = 2�@1��+ �2@1�: It is easy to see that @1� = k��K, and hence we get @1c = 2a1��+ �2(k��K): Equalizing, we have 2a1��+ �2(k��K)� �3�3 a1 = 0 or � a1 � 2a21�+ a1�(k��K)� �2�3 � = 0: 128 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold The expression in brackets is an algebraic one and can not be identically zero for K 6= 0. Since � 6= 0, we obtain a contradiction. Re m a r k. We do not obtain a contradiction if K = 0, since we have another solution � = 0 which gives @1b+ � = 0 and hence b = ��u+m. We have achieved the result by putting a restriction on the geometry of the base manifold. Putting a restriction on the vector �eld, we are able to achieve a similar result. Recall that a totally geodesic vector �eld necessarily makes a constant angle with some family of geodesics on the base manifold (see Cor. 2.1). It is not parallel along this family and this fact is essential for its totally geodesic property. Namely, Proposition 2.2. Let M2 be a Riemannian manifold. Suppose � is a nonzero local vector �eld on M2 which is parallel along some family of geodesics of M2. If �(M2) is totally geodesic in TM2 then M2 is �at. R e m a r k. Geometrically, this assertion means that if �(M2) is not transver- sal to the horizontal distribution on TM2 then �(M2) is never totally geodesic in TM2 except when M2 is �at. P r o o f. Let M2 be a non�at Riemannian manifold and suppose that the hypothesis of the theorem is ful�lled. Then, choosing a coordinate system as in Lemma 2.2, we have re1� = 0; and we can reduce (13) to re2re2� = 0; re1re2� + (k + cK)re2� = 0: (17) Now make a simple computation. R(e2; e1)re2� = re2re1re2� �re1re2re2� �r[e2;e1]re2� = re2re1re2� � kre2re2� = re2re1re2�: On the other hand, di�erentiating (17)2, we �nd re2re1re2� = �e2(k + cK)re2�: So we have R(e2; e1)re2� = �e2(k + cK)re2�: Therefore, either re2� = 0 or e2(k + cK) = 0. If we accept the �rst case we see that � is a parallel vector �eld on M2 and we get a contradiction. Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 129 Alexander Yampolsky If we accept the second case, we obtain R(e2; e1)re2� = 0; which means that re2� belongs to a kernel of the curvature operator of M2. In dimension 2 this means that M2 is �at or, equivalently, � is a parallel vector �eld and we obtain a contradiction, as well. 2.3. Ruled totally geodesic submanifolds in TM2 Proposition 2.3. Let M2 be a Riemannian manifold of sign-preserving cur- vature. Consider a ruled submanifold ~F 2 in TM2 given locally by (s) : ( x1 = x1(s); x2 = x2(s); �(t; s) : ( �1 = t; �2 = �2(t; s): Then ~F 2 is totally geodesic in TM2 if (s) is a geodesic in M2, �(t; s) = t �(s) e(s); where e(s) is a unit vector �eld which is parallel along and �(s) is an arbitrary smooth function. R e m a r k. Geometrically, ~F 2 is a cylinder-type surface based on geodesic (s) with elements directed by a unit vector �eld e(s) parallel along (s). P r o o f. Fixing s = s0, we see that F 2 meets the �ber over x1(s0); x 2(s0) by a curve �(t; s0). If F 2 is supposed to be totally geodesic, then this curve is a straight line on the �ber. Therefore, the family �(t; s) should be of the form �(t; s) : ( �1 = t; �2 = �(s)t+ �(s): Introduce two vector �elds given along (s) by a = @1 + �(s)@2; b = �(s)@2: (18) Then we can represent �(t; s) as �(t; s) = a(s) t+ b(s): 130 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold Denote by � and � the vectors of the Frenet frame of the curve (s). Denote also by ( 0) the covariant derivative of vector �elds with respect to the arc-length parameter on (s). Then � � 0 = k �; � 0 = �k �: Denote by ~@1; ~@2 the s and t coordinate vector �elds on F 2 respectively. A sim- ple calculation yields ~@1 = �h + (�0)v; ~@2 = av: One of the unit normal vector �elds can be found immediately, namely ~N1 = � h. Consider the conditions on F 2 to be totally geodesic with respect to the normal vector �eld ~N1. Using formulas (4), ~r~@1 ~N1 = ~r� h+(�0)v� h = �k� h � 1 2 h R(�; �)� iv + 1 2 h R(�; �0)� ih : Therefore, ~r~@1 ~N1; ~@2 �� = �1 2 R(�; �)�; a � = �1 2 R(�; �)b; a � = 0: Since M2 is supposed to be non�at, it follows b ^ a = 0. From (18) we conclude b = 0. Thus, �(t; s) = a(s) t. Moreover, ~r~@1 ~N1; ~@1 �� = �k � 1 2 R(�; �)�; �0 � + 1 2 R(�; �0)�; � � = �k + R(�; �0)�; � � = �k + t2 R(a; a0)�; � � = 0 identically with respect to parameter t. Therefore, k = 0 and a ^ a0 = 0. Thus, (s) is a geodesic line on M2. In addition, (a ^ a0 = 0) � (a0 = �a). Set a = �(s) e(s), where � = ja(s)j. Then (a0 = �a) � (�0 e + � e0 = �� e), which means that e0 = 0. From this we conclude �(t; s) = t�(s) e(s); where �(s) is arbitrary function and e(s) is a unit vector �eld, parallel along (s). Therefore, ~@1 = �h + t� 0 ev; ~@2 = � ev; and we can �nd another unit normal vector �eld ~N2 = (e?)v , where e?(s) is a unit vector �eld also parallel along (s) and orthogonal to e(s). For this vector �eld we have ~r~@1 ~N2 = ~r� h+(�0)v (e ?)v = [(e?)0]v + 1 2 h R(�; e?)� ih = 1 2 t� h R(e; e?)� ih ; ~r~@2 ~N2 = 0: Evidently, ~r~@i ~N2; ~@k �� = 0 for all i; k = 1; 2. Thus, the submanifold is totally geodesic. Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 131 Alexander Yampolsky The converse statement is true in general. Proposition 2.4. Let Mn be a Riemannian manifold. Consider a cylinder type surface ~F 2 � TMn parameterized as� (s); t �(s) e(s) ; where (s) is a geodesic in Mn, e(s) is a unit vector �eld, parallel along and �(s) is an arbitrary smooth function. Then ~F 2 is totally geodesic in TMn and intrinsically �at. P r o o f. Indeed, the tangent basis of ~F 2 is consisted of ~@1 = 0h + t� 0ev; ~@2 = � ev : By formulas (4), ~r~@1 ~@1 = (r 0 0)h + 1 2 � R( 0; 0)� �v = 0 ; ~r~@1 ~@2 = (r 0� e)v + 1 2 � R(�; � e) 0 �h = � 0ev � ~@2 ; ~r~@2 ~@1 = 1 2 � R(�; � e) 0 �h = 1 2 � R(t � e; � e) 0 �h = 0; ~r~@2 ~@2 = �ev(�) ev = 0: It is easy to �nd the Gaussian curvature of this submanifold, since it is equal to the sectional curvature of TM2 along the ~@1 ^ ~@2- plane. Using the curvature tensor expressions [4], we �nd Gauss( ~F 2) = ~R(� h; ev)ev; � h �� = 1 4 jR(�; e)� j2 = 0: 3. Local description of 3-dimensional totally geodesic submanifolds in TM2 Theorem 3.1. Let M2 be Riemannian manifold with Gaussian curvature K. A totally geodesic submanifold ~F 3 � TM2 locally is either a) a 3-plane in TM2 = E4 if K = 0, or b) a restriction of the tangent bundle to a geodesic 2M2 such that Kj = 0 if K 6� 0. If M2 does not contain such a geodesic, then TM2 does not admit 3-dimensional totally geodesic submanifolds. 132 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold P r o o f. Let ~F 3 be a submanifold it TM2. Let (x1; x2; �1; �2) be a local chart on TM2. Then locally ~F 3 can be given mapping f of the form f : 8>>>>>< >>>>>: x1 = x1(u1; u2; u3); x2 = x2(u1; u2; u3); �1 = �1(u1; u2; u3); �2 = �2(u1; u2; u3); where u1; u2; u3 are the local parameters on ~F 3. The Jacobian matrix f� of the mapping f is of the form f� = 0 BBBBBB@ @x1 @u1 @x1 @u2 @x1 @u3 @x 2 @u1 @x 2 @u2 @x 2 @u3 @�1 @u1 @�1 @u2 @�1 @u3 @�2 @u1 @�2 @u2 @�2 @u3 1 CCCCCCA : Since rank f� = 3, we have two geometrically di�erent possibilities to achieve the rank, namely (a) det 0 BBB@ @x 1 @u1 @x 1 @u2 @x 1 @u3 @x2 @u1 @x2 @u2 @x2 @u3 @� 1 @u1 @� 1 @u2 @� 1 @u3 1 CCCA 6= 0; (b) det 0 BBB@ @x1 @u1 @x1 @u2 @x1 @u3 @� 1 @u1 @� 1 @u2 @� 1 @u3 @� 2 @u1 @� 2 @u2 @� 2 @u3 1 CCCA 6= 0: Without loss of generality we can consider this possibilities in such a way that (b) excludes (a). Consider the case (a). In this case we can locally parameterize the submanifold F 3 as f : 8>>>>< >>>>: x1 = u1; x2 = u2; �1 = u3; �2 = �2(u1; u2; u3): By hypothesis, the submanifold ~F 3 is totally geodesic in TM2. Therefore, it intersects each �ber of TM2 by a vertical geodesic, i.e., by a straight line. Fix u0 = (u10; u 2 0). Then the parametric equation of ~F 3 \Tu0M2 with respect to �ber parameters is ( �1 = u3; �2 = �2(u10; u 2 0; u 3): Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 133 Alexander Yampolsky On the other hand, this equation should be the equation of a straight line and hence ( �1 = u3; �2 = �(u10; u 2 0)u 3 + �(u10; u 2 0); where �(u) = �(u1; u2) and �(u) = �(u1; u2) some smooth functions on M2. From this viewpoint, after setting u3 = t the submanifold under consideration can be locally represented as a one-parametric family of smooth vector �elds �t on M2 of the form �t(u) = t @1 + � �(u)t+ �(u) � @2 with respect to the coordinate frame @1 = @=@u1; @2 = @=@u2. Introduce the vector �elds a(u) = @1 + �(u) @2; b(u) = �(u) @2: (19) Then �t can be expressed as �t(u) = t a(u) + b(u): It is natural to denote by �t(M 2) a submanifold ~F 3 � TM2 of this kind. Denote by ~@i (i = 1; : : : ; 3) the coordinate vector �elds of �t(M 2). Then ~@1 = � 1; 0; 0; t @1�+ @1� ; ~@2 = � 0; 1; 0; t @2�+ @2� ; ~@3 = � 0; 0; 1; � : A direct calculation shows that these �elds can be represented as ~@1 = @h1 + t(r@1 a)v + (r@1 b)v; ~@2 = @h2 + t(r@2 a)v + (r@2 b)v; ~@3 = av: Denote by ~N a normal vector �eld of �t(M 2). Then ~N = (a?)v + Zh t ; where a?; a � = 0 and the �eld Zt = Z1 t @1 + Z2 t @2 can be found easily from the equations ~@i; ~N �� = Zt; @i � + t r@i a; a? � + r@i b; a? � = 0; i = 1; 2: 134 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold Using the formulas (4), one can �nd ~r~@i av = ~r @h i +t(r@i a)v+(r@i b)va v = (r@i a)v + 1 2 h R(�t; a)@i ih = (r@i a)v + 1 2 h R(b; a)@i ih : If the submanifold �t(M 2) is totally geodesic, then the following equations should be satis�ed identically: ~r~@i ~@3; ~N �� = r@i a; a? � + 1 2 R(b; a)@i; Zt � = 0 with respect to the parameter t. To simplify the further calculations, suppose that the coordinate system on M2 is the orthogonal one, so that @1; @2 � = 0 and R(b; a)@2 = g11K jb ^ aj@1; R(b; a)@1 = �g22K jb ^ aj@2; where K is the Gaussian curvature ofM2 and g11; g22 are the contravariant metric coe�cients. Then we have R(b; a)@1; Zt � = �g22K jb ^ aj Zt; @2 � = g22K jb ^ aj � t r@2 a; a? � + r@2 b; a? �� ; R(b; a)@2; Zt � = g11K jb ^ aj Zt; @1 � = �g11K jb ^ aj � t r@1 a; a? � + r@1 b; a? �� : Thus we get the system8< : g22Kjb ^ aj r@2 a; a? � t+ r@1 a; a? � + g22Kjb ^ aj r@2 b; a? � = 0; g11Kjb ^ aj r@1 a; a? � t� r@2 a; a? � + g11Kjb ^ aj r@1 b; a? � = 0; which should be satis�ed identically with respect to t. As a consequence, we have 3 cases: (i) K = 0; ( r@1 a; a? � = 0 r@2 a; a? � = 0 ; (ii) K 6= 0, jb ^ aj = 0; ( r@1 a; a? � = 0 r@2 a; a? � = 0 ; (iii) K 6= 0, jb ^ aj 6= 0, ( r@1 a; a? � = 0 r@2 a; a? � = 0 ; ( r@1 b; a? � = 0 r@2 b; a? � = 0 : Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 135 Alexander Yampolsky Case (i). In this case the base manifold is �at and we can choose a Cartesian coordinate system, so that the covariant derivation becomes a usual one and we have ( r@i a = � 0; @i� ; i = 1; 2; a? = � � �; 1 : From r@i a; a? � = 0 it follows that � = const, i.e., a is a parallel vector �eld. Moreover, in this case ~@1 = � 1; 0; 0; @1� = @h1 + (@1 b) v ; ~@2 = � 0; 1; 0; @2� = @h1 + (@1 b) v ; ~@3 = � 0; 0; 1; � ; ~N = � � @1�;�@2�;��; 1 : Now we can �nd ~r~@i ~@k = (r@i @kb) v = � 0; 0; 0; @ik� and the conditions ~r~@i ~@k; ~N �� = 0 imply @ik� = 0. Thus, � = m1u 1 +m2u 2 +m0, where m1;m2;m0 are arbitrary constants. As a consequence, the submanifold �t(M 2) is described by parametric equations of the form 8>>>>< >>>>: x1 = u1; x2 = u2; �1 = t; �2 = �t+m1u 1 +m2u 2 +m0 and we have a hyperplane in TM2 = E4. Case(ii). Keeping in mind (19), the condition b ^ a = 0 implies b = 0. The conditions ( r@1 a; a? � = 0; r@2 a; a? � = 0 imply r@1 a = �1(u) a; r@2 a = �2(u) a. As a consequence, we have �t = t a; ~@1 = @h1 + t(r@1 a)v = @h1 + t�1 a v; ~@2 = @h2 + t(r@2 a)v = @h2 + t�2 a v; ~@3 = av; ~N = (a?)v : 136 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold Using formulas (4), ~r~@i ~@k = ~r @hi +t�i a v � @h k + t�k a v � = ~r @hi @h k + t�i ~rav@ h k + ~r @hi (t�ka v) + t2�i�k ~rava v = (r@i @k) h � 1 2 h R(@i; @k)�t iv + t�i 1 2 h R(�t; a)@k ih + t@i(�k)a v + t�k(r@i a)v + t�k 1 2 h R(�t; a)@i ih = (r@i @k) h � t1 2 h R(@i; @k)a iv + t@i(�k)a v + t�k�i a v: Evidently, for i 6= k ~r~@i ~@k; ~N �� = �t1 2 R(@i; @k)a; a ? � 6= 0; since M2 is non�at and a 6= 0. Contradiction. Case (iii). The conditions imply ria = �i(u) a; rib = �i(u) a; i = 1; 2; and we have �t = t a+ b; ~@1 = @h1 + (t�1 + �1) a v ; ~@2 = @h1 + (t�2 + �2) a v ; ~@3 = av; ~N = (a?)v: A calculation as above leads to the identity ~r~@i ~@k; ~N �� = �1 2 R(@i; @k)�t; a ? � = �t1 2 R(@i; @k)a; a ? � � 1 2 R(@i; @k)b; a ? � = 0; which can be true if and only if( R(@i; @k)a; a ? � = 0; R(@i; @k)b; a ? � = 0: The �rst condition contradicts K 6= 0. Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 137 Alexander Yampolsky Consider the case (b). In this case the submanifold ~F 3 can be locally para- metrized by 8>>>>< >>>>: x1 = u1; x2 = x2(u1; u2; u3); �1 = u2; �2 = u3: Since we exclude the case (a), we should suppose det 0 BBB@ @x 1 @u1 @x 1 @u2 @x 1 @u3 @x 2 @u1 @x 2 @u2 @x 2 @u3 @�1 @u1 @�1 @u2 @�1 @u3 1 CCCA = det 0 BB@ 1 0 0 @x 2 @u1 @x 2 @u2 @x 2 @u3 0 1 0 1 CCA = �@x 2 @u3 = 0; det 0 BBB@ @x1 @u1 @x1 @u2 @x1 @u3 @x 2 @u1 @x 2 @u2 @x 2 @u3 @�2 @u1 @�2 @u2 @�2 @u3 1 CCCA = det 0 BB@ 1 0 0 @x2 @u1 @x2 @u2 @x2 @u3 0 0 1 1 CCA = @x2 @u2 = 0: Therefore, in this case we have a submanifold, which can be parametrized by8>>>>< >>>>: x1 = x1(s); x2 = x2(s); �1 = u2; �2 = u3; where s is a natural parameter of the regular curve (s) = � x1(s); x2(s) on M2. Geometrically, a submanifold of this class is nothing else but the restriction of TM2 to the curve (s). Denote by � and � the Frenet frame of (s). It is easy to verify that ~@1 = � h; ~@2 = @v1 ; ~@3 = @v2 ; ~N = � h: By formulas (4), for i = 1; 2 ~r~@1+i ~N; ~@1 �� = ~r@vi �h; �h �� = 1 2 R(�; @i)�; � � = 1 2 R(�; �)@i; � � = 0 for arbitrary �. Evidently, M2 must be �at along (s). 138 Journal of Mathematical Physics, Analysis, Geometry , 2005, v. 1, No. 1 Totally geodesic submanifolds in the tangent bundle of a Riemannian 2-manifold References [1] M.T.K. Abbassi and A. Yampolsky, Totally geodesic submanifolds transverse to �bers of the tangent bundle of a Riemannian manifold with Sasaki metric. � Math. Publ. Debrecen 64 (2004), 129�154. [2] A. Borisenko and A. Yampolsky, Riemannian greometry of bundles. � Uspechi Mat. Nauk (1991), v. 46, No. 6, p. 51�95 (Russian). Engl. transl.: Russian Math. Surveys 46 (1991), No. 6, 55�106. [3] P. Dombrowski, On the geometry of tangent bundle. � J. Reine Angew. Math. 210 (1962), No. 1�2, 73�88. [4] O. 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