Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
We give a complete list of left-invariant unit vector fields on three-dimensional Lie groups equipped with a left-invariant metric that generate a totally geodesic submanifold in the unit tangent bundle of a group equipped with the Sasaki metric. As a result we obtain that each three-dimensional Lie...
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irk-123456789-1064492016-09-29T03:02:18Z Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups Yampolsky, A. We give a complete list of left-invariant unit vector fields on three-dimensional Lie groups equipped with a left-invariant metric that generate a totally geodesic submanifold in the unit tangent bundle of a group equipped with the Sasaki metric. As a result we obtain that each three-dimensional Lie group admits totally geodesic unit vector eld under some conditions on structural constants. From a geometrical viewpoint, the field is either parallel or a characteristic vector field of a natural almost contact structure on the group. 2007 Article Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups / A. Yampolsky // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 2. — С. 253-276. — Бібліогр.: 9 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106449 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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We give a complete list of left-invariant unit vector fields on three-dimensional Lie groups equipped with a left-invariant metric that generate a totally geodesic submanifold in the unit tangent bundle of a group equipped with the Sasaki metric. As a result we obtain that each three-dimensional Lie group admits totally geodesic unit vector eld under some conditions on structural constants. From a geometrical viewpoint, the field is either parallel or a characteristic vector field of a natural almost contact structure on the group. |
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Article |
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Yampolsky, A. |
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Yampolsky, A. Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups Журнал математической физики, анализа, геометрии |
| author_facet |
Yampolsky, A. |
| author_sort |
Yampolsky, A. |
| title |
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups |
| title_short |
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups |
| title_full |
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups |
| title_fullStr |
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups |
| title_full_unstemmed |
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups |
| title_sort |
invariant totally geodesic unit vector fields on three-dimensional lie groups |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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http://dspace.nbuv.gov.ua/handle/123456789/106449 |
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Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups / A. Yampolsky // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 2. — С. 253-276. — Бібліогр.: 9 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT yampolskya invarianttotallygeodesicunitvectorfieldsonthreedimensionalliegroups |
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2025-07-07T18:30:30Z |
| last_indexed |
2025-07-07T18:30:30Z |
| _version_ |
1837013959090110464 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2007, vol. 3, No. 2, pp. 253�276
Invariant Totally Geodesic Unit Vector Fields
on Three-Dimensional Lie Groups
A. Yampolsky
Department of Mechanics and Mathematics, V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:AlexYmp@gmail.com
Received March 22, 2006
We give a complete list of left-invariant unit vector �elds on three-
dimensional Lie groups equipped with a left-invariant metric that generate
a totally geodesic submanifold in the unit tangent bundle of a group equipped
with the Sasaki metric. As a result we obtain that each three-dimensional
Lie group admits totally geodesic unit vector �eld under some conditions
on structural constants. From a geometrical viewpoint, the �eld is either
parallel or a characteristic vector �eld of a natural almost contact structure
on the group.
Key words: Sasaki metric, totally geodesic unit vector �eld, almost con-
tact structure, Sasakian structure.
Mathematics Subject Classi�cation 2000: 53B20, 53B25 (primary); 53C25
(secondary).
Introduction
The problem on description of all totally geodesic submanifolds in tangent
and unit tangent bundle of space forms was formulated by A. Borisenko in [2,
Probl. 1]. In general setting the problem is unsolved up to now. More progress is
achieved for a special class of submanifolds in the unit tangent bundle formed by
unit vector �elds on the base manifold. We begin with a de�nition.
Let (Mn; g) be a Riemannian manifold and (T1M
n; gs) its unit tangent bundle
with Sasaki metric. Consider a unit vector �eld � as a mapping
� : Mn ! T1M
n:
De�nition 1. A unit vector �eld � on the Riemannian manifold Mn is called
totally geodesic if the image of the (local) imbedding � : Mn ! T1M
n is a totally
geodesic submanifold in the unit tangent bundle T1M
n with the Sasaki metric.
c
A. Yampolsky, 2007
A. Yampolsky
In the two-dimensional case the problem is solved [13]. In the case of higher di-
mensions only partial results are known. A. Borisenko conjectured that the Hopf
unit vector �eld on each odd-dimensional sphere is totally geodesic. The conjec-
ture was approved in a more general case. If M2m+1 is a Sasakian manifold and �
is a characteristic vector �eld of the Sasakian structure, then �(M2m+1) is totally
geodesic in T1M
2m+1 [12].
Note that the Hopf vector �eld belongs to the class of left-invariant unit vector
�elds on S3 as a Lie group with the left-invariant Riemannian metric. In this
paper, we give a full description of three-dimensional Lie groups with the left-
invariant metric which admit a totally geodesic left-invariant unit vector �eld and
the �elds themselves. As a consequence, we show that, in nontrivial cases, for
each totally geodesic left-invariant unit vector �eld � the structure
�
� = �r�, �,
� = g(�; �)
�
is an almost contact structure on the corresponding Lie group and �
is a characteristic vector �eld of this structure. If � is a Killing unit vector �eld,
then the structure is Sasakian.
It is worthwhile to note that in a similar way one can de�ne a locally minimal
unit vector �eld as a �eld of zero mean curvature. A number of examples of locally
minimal unit vector �elds were found recently [5, 6]. In particular, K. Tsukada
and L. Vanhecke [9] described all minimal left-invariant unit vector �elds on three-
dimensional Lie groups with the left-invariant metric. While the totally geodesic
unit vector �elds form a subclass in a class of minimal unit vector �elds, no method
to distinguish minimal and totally geodesic �elds was proposed in [9].
The paper is organized as follows. In Section 1, we give some preliminaries
and formulate the results. In Section 2, we consider the unimodular Lie groups.
We prove that if a totally geodesic unit vector �eld exists on a given group, then
it is an eigenvector of the Ricci tensor which corresponds to the Ricci principal
curvature � = 2. Moreover, we give a complete list of totally geodesic unit vector
�elds on a corresponding Lie group as well as the conditions on the structure
constants of the group. In a series of Props. 2.2�2.6, we give a description of
totally geodesic unit vector �elds in unimodular case from the contact geometry
viewpoint. In Section 3 we consider the nonunimodular case. We give an explicit
expression for the totally geodesic unit vector �eld as well as the conditions on the
structure constants of the corresponding group. Finally, Prop. 3.1 gives a geo-
metrical characterization of the totally geodesic unit vector �eld and clari�es
a structure of the corresponding nonunimodular Lie group.
1. Preliminaries and Results
Let (M; g) be an n-dimensional Riemannian manifold with metric g and TM
be its tangent bundle. Denote by � : TM ! M the bundle projection. Denote
by Q a point on TM . Then Q = (q; �), where q 2 M and � 2 TqM . Let
254 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
~X; ~Y 2 TQTM . A natural (Sasaki) Riemannian metric ~g on the tangent bundle
is de�ned by the following scalar product
~g( ~X; ~Y )
��
Q
= ~g(�� ~X;�� ~Y )
��
q
+ ~g(K ~X;K ~Y )
��
q
;
where �� and K are the di�erentials of the bundle projection and the connection
map [3] respectively. A unit tangent bundle T1M is a subbundle in TM and
a hypersurface in (TM; ~g) with a pull-back metric.
Suppose that u := (u1; : : : ; un) are local coordinates onM . Denote by (u; �) :=
(u1; : : : ; un; �1; : : : ; �n) the natural local coordinates in the tangent bundle TM .
If �(u) is a unit vector �eld on M , then it de�nes a mapping � : M ! T1M; given
by �(u) = (u; �(u)): The image �(M) is a submanifold in T1M with a pull-back
metric.
Denote by r the Levi�Civita connection on M . Introduce a pointwise linear
operator A� : TqM
n ! �?q by
A�X = �rX�:
From the de�nition of the connection map it follows that the pull-back metric on
�(M) is de�ned by
~g(��X; ��Y
�
)
��
(u;�(u))
= g(X;Y )
��
q
+ g(A�X;A�Y )
��
q
:
From intrinsic viewpoint, this metric can be considered as a metric on M addi-
tively deformed by the �eld �.
When �? is an integrable distribution, the unit vector �eld � is called holo-
nomic, otherwise it is called nonholonomic. In holonomic case the operator A�
is symmetric (w.r. to metric g) and is known as Weingarten or the shape opera-
tor for each hypersurface of the foliation. In general (nonholonomic) case, A� is
not symmetric but formally satis�es the Codazzi equation. Namely, a covariant
derivative of A� is de�ned by
(rXA�)Y = �rXrY � +rrXY �:
Then for the curvature operator of M we have
R(X;Y )� = (rYA�)X � (rXA�)Y;
which gives a Codazzi-type equation. From this viewpoint, it is natural to call
the operator A� a nonholonomic shape operator.
Introduce a symmetric tensor �eld
Hess�(X;Y ) =
1
2
�
(rYA�)X + (rXA�)Y
�
; (1)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 255
A. Yampolsky
which is a symmetric part of the covariant derivative of A�. The trace
�
P
n
i=1Hess�(ei; ei) := ��; where e1; : : : ; en is an orthonormal frame, known as
the rough Laplacian [1] of the �eld �. Therefore, one can treat the tensor �eld (1)
as a rough Hessian of the �eld.
A unit vector �eld is called harmonic, if it is a critical point of the energy
functional of mapping � : Mn ! T1M
n. This de�nition presumes the variation
within the class of unit vector �elds. A unit vector �eld is harmonic if and only
if �� = �jr�j2� (see [10]). There exist harmonic unit vector �elds that fail to
be critical within a wider class of all mappings f : Mn ! T1M
n [4]. Introduce
a tensor �eld
Hm�(X;Y ) =
1
2
�
R(�;A�X)Y +R(�;A�Y )X
�
:
A harmonic unit vector �eld � de�nes a harmonic mapping � : Mn ! T1M
n
if and only if
P
n
i=1 Hm�(ei; ei) = 0 (see [4]). The following lemma [14] gives
the condition on � to be totally geodesic in terms of Hess� and Hm�.
Lemma 1.1. A unit vector �eld � on a given Riemannian manifold Mn is
totally geodesic if and only if
Hess�(X;Y ) +A�Hm�(X;Y )� g(A�X;A�Y ) � = 0
for all vector �elds X;Y on Mn.
For the sake of brevity, denote
TG�(X;Y ) := Hess�(X;Y ) +A�Hm�(X;Y )� g(A�X;A�Y ) �: (2)
The treatment of three-dimensional Lie groups with the left-invariant met-
rics is based on J. Milnor's description of three-dimensional Lie groups via the
structure constants [8] and splits into two natural cases.
The unimodular case. In this case, there is an orthonormal frame e1; e2; e3
of its Lie algebra such that the bracket operations are de�ned by
[e2; e3] = �1e1; [e3; e1] = �2e2; [e1; e2] = �3e3: (3)
The constants �1; �2; �3 completely determine a topological structure of the cor-
responding Lie group as in the following table:
Signs of �1; �2; �3 Associated Lie group
+;+;+ SU(2) or SO(3)
+;+;� SL(2;R) or O(1; 2)
+;+; 0 E(2)
+;�; 0 E(1; 1)
+; 0; 0 Nil3 (Heisenberg group)
0; 0; 0 R � R � R
:
The main result for this case is the following.
256 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
Theorem 1.1. Let G be a three-dimensional unimodular Lie group with the
left-invariant metric and let fei; i = 1; 2; 3g be an orthonormal basis for the Lie
algebra satisfying (3). Moreover, assume that �1 � �2 � �3. Then the left-
invariant totally geodesic unit vector �elds on G are given as follows:
G Conditions on �1; �2; �3 �
SU(2) �1 = �2 = �3 = 2 arbitrary
left-invariant
�1 = �2 = � > �3 = 2 �e3
�1 = �2 = � > 2 > �3 = ��
p
�2 � 4 cos t e1 + sin t e2
�1 = 2 > �2 = �3 = � > 0 �e1
�1 = �+
p
�2 � 4 > � = �2 = �3 > 2 cos t e2 + sin t e3
�1 > �2 > �3 > 0, �2m � (�i � �k)
2 = 4 �em (i,k,m=1,2,3)
SL(2,R) �23 � (�1 � �2)
2 = 4 �e3
�21 � (�2 � �3)
2 = 4 �e1
E(2) �1 = �2 > 0; �3 = 0 �e3, cos t e1+sin t e2
�21 � �22 = 4, �1 > �2 > 0, �3 = 0 �e1
E(1,1) �21 � �22 = �4, �1 > 0; �2 < 0; �3 = 0 �e2
�21 � �22 = 4, �1 > 0; �2 < 0; �3 = 0 �e1
Heisenberg
group
�1 = 2; �2 = 0; �3 = 0 �e1
R�R�R �1 = �2 = �3 = 0 arbitrary
left-invariant
.
The case of nonunimodular groups. Let e1 be a unit vector orthogonal
to the unimodular kernel U and choose an orthonormal basis fe2; e3g of U which
diagonalizes the symmetric part of ade1
��
U
. Then the bracket operation can be
expressed as
[e1; e2] = �e2 + � e3; [e1; e3] = �� e2 + Æ e3; [e2; e3] = 0: (4)
If necessary, by changing e1 to �e1, we can assume � + Æ > 0 and by possibly
alternating e2 and e3, we may also suppose � � Æ [9].
The main result in this case is the following one.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 257
A. Yampolsky
Theorem 1.2. Let G be a nonunimodular Lie group with the basis satisfying
(4) of its Lie algebra. In assumption � + Æ > 0 and � � Æ , the left-invariant
totally geodesic unit vector �elds on G are given as follows:
Conditions on �; �; Æ Left-invariant totally geodesic
unit vector �eld
Geometrical
structure of G
� = Æ = 0 �e3 L2(��)�E1
� = �1; �Æ = �1 �
�
�
1p
1 + � 2
e2 +
�p
1 + � 2
e3
�
Sasakian
manifold
.
2. Unimodular Case
Choose the orthonormal frame as in (3). De�ne connection numbers by
�i =
1
2
(�1 + �2 + �3)� �i:
Then the Levi�Civita covariant derivatives can be expressed via the cross-products
as follows rei
ek = �i ei � ek: For any left-invariant unit vector �eld � = x1e1 +
x2e2 + x3e3 we have
rei
� = �i ei � �:
Denote Ni = ei � �. Then
rei
� = �i ei � � = �iNi:
As a consequence, the matrix of the Weingarten operator takes the form
A� =
0
@ 0 ��2x3 �3x2
�1x3 0 ��3x1
��1x2 �2x1 0
1
A : (5)
The following technical lemma can be checked by direct computation.
Lemma 2.1. Let G be a three-dimensional unimodular Lie group with the
left-invariant metric g and let fei; i = 1; 2; 3g be an orthonormal basis for the Lie
algebra satisfying (3). Then for any left-invariant unit vector �eld � = x1e1 +
x2e2 + x3e3 we have
A�ei = ��i ei � � = ��iNi;
(rei
A�)ei = �2
i
(� � xiei);
(rei
A�)ek = "ikm�i�mNm � �i�kxiek; i 6= k;
R(ei; ek)� = �"ikm�ikNm;
where �ik = �ki = �i�m + �k�m � �i�k and "ikm = g(ei � ek; em) .
258 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
Remark that the chosen frame diagonalizes the Ricci tensor [8]. Moreover,
2�i�k = �m; where �m is the principal Ricci curvature and i 6= k 6= m. It also
worthwhile to mention that �ik = 1
2
(�k + �i � �m) is a sectional curvature of the
left-invariant metric in the direction of ei ^ ek.
The following Lemma is also a result of direct computations.
Lemma 2.2. Let G be a three-dimensional unimodular Lie group with the
left-invariant metric g and let fei; i = 1; 2; 3g be an orthonormal basis for the Lie
algebra satisfying (3). Then the left-invariant unit vector �eld � = x1e1 + x2e2 +
x3e3 is totally geodesic if and only if for any i 6= k 6= m
TG(ei; ei) = xi�i
n
xm(�ik�k � �i)Nk � xk(�im�m � �i)Nm
o
= 0;
2TG(ei; ek) = "ikm
n
� xixm�i(�ik�i � �k)Ni + xkxm�k(�ik�k � �i)Nk
+
�
�i�m(1� �km)� �k�m(1� �im) + �i(�km�m � �k)x
2
i
��k(�im�m � �i)x
2
k
�
Nm
o
= 0;
where �ik = �ki = �i�m + �k�m � �i�k and "ikm = g(ei � ek; em) .
Now we can prove the main Lemma.
Lemma 2.3. Let G be a three-dimensional unimodular Lie group with the
left-invariant metric and let fei; i = 1; 2; 3g be an orthonormal basis for the Lie
algebra satisfying (3). Denote by �1; �2; �3 the principal Ricci curvatures of the
given group. Then the set of left-invariant totally geodesic unit vector �elds can
be described as follows:
�1 �2 �3 �1 �2 �3 �
0 0 0 0 0 0 arbitrary left-invariant
0 0 0 6= 0 0 0 �e1, cos t e2 + sin t e3
0 0 0 0 6= 0 0 �e2, cos t e1 + sin t e3
0 0 0 0 0 6= 0 �e3, cos t e1 + sin t e2
2 �e1
2 �e2
2 �e3
2 2 cos t e1 + sin t e2
2 2 cos t e1 + sin t e3
2 2 cos t e2 + sin t e3
2 2 2 arbitrary left-invariant
.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 259
A. Yampolsky
P r o o f. Rewrite the result of Lem. 2.2 for various combinations of indices
to get
(1; 1) x1�1
n
x3(�12�2 � �1)N2 � x2(�13�3 � �1)N3
o
= 0;
(2; 2) x2�2
n
x3(�21�1 � �2)N1 � x1(�23�3 � �2)N3
o
= 0;
(3; 3) x3�3
n
x2(�31�1 � �3)N1 � x1(�32�2 � �3)N2
o
= 0;
(1; 2) � x1x3�1(�12�1 � �2)N1 + x2x3�2(�12�2 � �1)N2 +
�
�1�3(1� �23)
� �2�3(1� �13) + �1(�23�3 � �2)x
2
1 � �2(�13�3 � �1)x
2
2
�
N3 = 0;
(2; 3) � x2x1�2(�23�2 � �3)N2 + x3x1�3(�23�3 � �2)N3 +
�
�2�1(1� �31)
� �3�1(1� �21) + �2(�31�1 � �3)x
2
2 � �3(�21�1 � �2)x
2
3
�
N1 = 0;
(3; 1) � x3x2�3(�13�3 � �1)N3 + x1x2�1(�13�1 � �3)N1 +
�
�3�2(1� �12)
� �1�2(1� �32) + �3(�12�2 � �1)x
2
3 � �1(�32�2 � �3)x
2
1
�
N2 = 0:
The vectors N1; N2 and N3 are linearly dependent:
x1N1 + x2N2 + x3N3 = 0;
but linearly independent in the pairs for general (not speci�c) �elds �.
The case x1 6= 0; x2 6= 0; x3 6= 0.
The subcase 1: �1 = 0; �2 = 0; �3 = 0. All equations are ful�lled evidently.
Therefore, the arbitrary left-invariant vector �eld is totally geodesic in this case,
and we get the �rst row in the table.
The subcase 2: �1 = 0, �2 6= 0 or �3 6= 0. Then from (2,2) and (3,3) we see,
that �2 = 0; �3 = 0, which gives a contradiction. In a similar way we exclude
the cases when �i = 0, but �2
k
+ �2m 6= 0 for arbitrary triple of di�erent indices
(i; k;m).
The subcase 3: �1 6= 0; �2 6= 0; �3 6= 0. Since N1; N2 and N3 are linearly
independent in pairs, from (1,1), (2,2) and (3,3) we conclude:�
�12�2 � �1 = 0;
�12�1 � �2 = 0;
�
�13�3 � �1 = 0;
�13�1 � �3 = 0;
�
�23�2 � �3 = 0;
�23�3 � �2 = 0:
(6)
As a consequence, we get
(�12 � 1)(�1 + �2) = 0; (�13 � 1)(�1 + �3) = 0; (�23 � 1)(�2 + �3) = 0:
260 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
Taking into account (6), the rest of the equations yields8<
:
�1�3(1� �23)� �2�3(1� �13) = 0;
�1�2(1� �13)� �1�3(1� �12) = 0;
�2�3(1� �12)� �1�2(1� �23) = 0:
Since �i 6= 0, i = 1; 2; 3, we conclude �ik = 1, i; k = 1; 2; 3, and therefore �i = 2,
i = 1; 2; 3. This is the case of the last row in the table.
The case x1 6= 0, x2 6= 0, x3 = 0. In this case x1N1 + x2N2 = 0, but N1, N3
and N2, N3 are linearly independent in pairs. Rewrite the system for this case
as follows:
(1; 1) �1(�13�3 � �1) = 0; (2; 2) �2(�23�3 � �2) = 0; (3; 3) � 0;
(1; 2) �1�3(1� �23)� �2�3(1� �13) + �1(�23�3 � �2)x
2
1
��2(�13�3 � �1)x
2
2 = 0;
(2; 3) x21�2(�23�2 � �3) + �1�2(1� �31 � �1�3(1� �21)
+�2(�13�1 � �3)x
2
2 = 0;
(3; 1) �x22�1(�13�1 � �3 + �2�3(1� �12)� �1�2(1� �32)
��1(�23�2 � �3)x
2
1 = 0:
Set �1 = �2 = 0. Then the system is ful�lled for the arbitrary �3. The case
�3 = 0 has already been considered. The case �3 6= 0 gives the cos t e1 + sin t e2
in the 4-th row of the table.
Set �1 = 0; �2 6= 0. Then �12 = �2�3; �13 = �2�3; �23 = ��2�3. The equation
(2,2) yields ��22(�23 + 1) = 0, which gives a contradiction.
Set �1 6= 0; �2 = 0. Then �12 = �1�3; �13 = ��1�3; �23 = �1�3. The equation
(1,1) yields ��21(�23 + 1) = 0, which gives a contradiction.
Set �1 6= 0; �2 6= 0. Then �1 = �13�3; �2 = �23�3 and the substitution into (1,2)
yields �33(�2 � �1) = 0: The case �3 = 0 contradicts �1 6= 0; �2 6= 0, as one can
see from (1,1) and (2,2). Thus, set �1 = �2 = � 6= 0. Then �13 = �23 = �2 and
from (1,1) and (2,2) we conclude
��3 � 1 = 0: (7)
In this case we have
�12 = 2� �2; �13 = �2; �23 = �2: (8)
If we plug (7) and (8) into the system, then we get an identity. Since ��3 = 1
means that �1 = �2 = 2, we get the 8-th row of the table.
The case x1 6= 0; x2 = 0; x3 6= 0 after similar computations results cos t e1 +
sin t e3 in the 3-rd and in the 9-th rows of the table.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 261
A. Yampolsky
The case x1 = 0; x2 6= 0; x3 6= 0 results cos t e2 + sin t e3 in the 2-nd and in
the 10-th rows of the table.
The case x1 = 1, x2 = 0; x3 = 0. In this case N1 = 0 and the equations
(1,1), (2,2), (3,3) and (2,3) are ful�lled regardless the geometry of the group.
The equations (1,2) and (1,3) take the forms
(1; 2) �1�3(1� �23)� �2�3(1� �13) + �1(�23�3 � �2) = 0;
(1; 3) �2�3(1� �12)� �1�2(1� �23)� �1(�23�2 � �3) = 0:
After simpli�cations, we get
(1; 2) �13(�2�3 � 1) = 0; (1; 3) �12(�2�3 � 1) = 0:
The case �2�3 = 1means �1 = 2, and we have the 5-th row of the table. Consider
the case �12 = 0; �13 = 0 which is equivalent to �2�3 = 0 and �1(�2 � �3) = 0:
We have four possible solutions:
(i) �1 = 0; �2 = 0; �3 = 0; (ii) �1 = 0; �2 = 0; �3 6= 0;
(iii) �1 = 0; �2 6= 0; �3 = 0; (iv) �1 6= 0; �2 = 0; �3 = 0:
The case (i) is already included into the 1-st row of the table; the case (ii) is
already included into cos t e1 + sin t e2 case in the 4-st one of the table; the case
(iii) is already included into cos t e1 + sin t e3 case in the 3-rd row of the table.
The case (iv) is a new one and yields e1 in the 2-nd row of the table.
The case x1 = 0; x2 = 1; x3 = 0 yields e2 in the 3-rd and the 6-th rows of
the table. The case x1 = 0; x2 = 0; x3 = 1 yields e3 in the 4-th and the 7-th
rows of the table.
If we specify the result of Lem. 2.3 to each unimodular group, then we get the
result of Th. 1.1.
2.1. Geometrical Characterization of Totally Geodesic Unit Vector
Fields on Unimodular Groups
Let M be an odd-dimensional smooth manifold. Denote by �, �, � a (1; 1)-
tensor �eld, a vector �eld and a 1-form on M respectively. A triple (�; �; �) is
called an almost contact structure on M if
�2X = �X + �(X)�; �� = 0; �(�) = 1; (9)
for any vector �eld X onM . The manifold M with an almost contact structure is
called an almost contact manifold. If M is endowed with the Riemannian metric
g(�; �) such that
g(�X; �Y ) = g(X;Y ) � �(X)�(Y ); �(X) = g(�;X) (10)
262 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
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for all vector �elds X and Y onM , then a quadruple (�; �; �; g) is called an almost
contact metric structure and the manifold is called an almost contact metric
manifold. The �rst of the conditions above is called a compatibility condition for
� and g. If the 2-form d�, given by d�(X;Y ) = 1
2
�
X�(Y )� Y �(X)� �([X;Y ])
�
;
satis�es
d�(X;Y ) = g(X;�Y ) ;
then the structure (�; �; �; g) is called a contact metric structure and the man-
ifold with a contact metric structure is called a contact metric manifold. A
contact metric manifold is called K-contact if � is a Killing vector �eld. The Ni-
jenhuis torsion of tensor �eld T of type (1; 1) is given by
[T; T ](X;Y ) = T 2[X;Y ] + [TX; TY ]� T [TX; Y ]� T [X;TY ]
and de�nes a (1; 2)-tensor �eld on M . An almost contact structure (�; �; �) is
called normal if
[�; �](X;Y ) + 2d�(X;Y ) � = 0: (11)
Finally, a contact metric structure (�; �; �; g) is called Sasakian, if it is normal.
A manifold with the Sasakian structure is called a Sasakian manifold. In the
Sasakian manifold necessarily � = A� and � = g(�; �) . The unit vector �eld � is
called a characteristic vector �eld of the Sasakian structure and is a Killing one.
This vector �eld is always totally geodesic [12].
In the three-dimensional case we have a stronger result.
Theorem 2.1. [12]. Let � be a unit Killing vector �eld on a three-dimensional
Riemannian manifold M3. If �(M3) is totally geodesic in T1M
3, then either�
� = A�; �; � = g(�; � )
�
is a Sasakian structure on M3, or M3 = M2 � E1 metrically and � is a unit
vector �eld of the Euclidean factor.
Now we can give a geometrical description of totally geodesic unit vector �elds.
Proposition 2.1. Let � be a left-invariant totally geodesic unit vector �eld on
SU(2) with the left-invariant metric g and let fei; i = 1; 2; 3g be an orthonormal
basis for the Lie algebra satisfying (3). Assume in addition that �1 � �2 � �3.
Then �
� = A�; �; � = g(�; � )
�
(12)
is an almost contact structure on SU(2). Moreover,
� if �1 = �2 = �3 = 2 or �1 = �2 > �3 = 2 or �1 = 2 > �2 = �3, then the
structure is Sasakian;
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 263
A. Yampolsky
� if �1 = �2 = � > 2 > �3 = ��
p
�2 � 4 or �1 = � +
p
�2 � 4 > � = �2 =
�3 > 2, then the structure is neither normal nor metric;
� if �1 > �2 > �3, then the structure is normal only for
� = e1; �1 = �2 +
1
�2
; �3 =
1
�2
; �2 > 1:
P r o o f. Consider the cases from Th. 1.1.
� In the case of �1 = �2 = �3 = 2 we have �1 = �2 = �3 = 1 and hence
A� =
0
@ 0 �x3 x2
x3 0 �x1
�x2 x1 0
1
A :
Therefore, the �eld � is the Killing one. By Theorem 2.1, the structure (12) is
Sasakian.
For �1 = �2 = � > �3 = 2 we have �1 = 1; �2 = 1; �3 = � � 1 and � = �e3.
For � = + e3 we �nd
A� =
0
@ 0 �1 0
1 0 0
0 0 0
1
A
and so � is again a Killing unit vector �eld and the structure (12) is Sasakian.
For �1 = 2 > �2 = �3 = � > 0, we have �1 = �1 + �; �2 = 1; �3 = 1 and
� = �e1. For � = + e1 we �nd
A� =
0
@ 0 0 0
0 0 �1
0 1 0
1
A
and see that again � is the Killing unit vector �eld. Therefore, the structure (12)
is Sasakian.
� Consider the case �1 = �2 = � > 2 > �3 = ��
p
�2 � 4 and � = x1e1+x2e2.
We have
�1 =
1
2
(��
p
�2 � 4); �2 =
1
2
(��
p
�2 � 4); �3 =
1
2
(�+
p
�2 � 4):
For brevity, put � = 1
2
(��
p
�2 � 4) and �� = 1
2
(�+
p
�2 � 4). Then
�1 = �; �2 = �; �3 = ��; ��� = 1; � 6= 1; �� 6= 1;
and for this case we have
� = A� =
0
@ 0 0 ��x2
0 0 ���x1
��x2 �x1 0
1
A :
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Since � 6= ��, the �eld � is never a Killing one but it is geodesic, since A�� = 0.
The structure (12) is an almost contact one on SU(2). Indeed,
�2 = A2
� =
0
@ �x22 x1x2 0
x1x2 �x21 0
0 0 �1
1
A and �2Z = �Z + g(�; Z) �:
This structure is not metric. For the compatibility condition (10), we have
g(�Z; �W ) = �2(z1w1 + z2w2) + ��2x3w3 � g(�; Z) g(�;W )
6= g(Z;W ) � g(�; Z) g(�;W ) :
This structure is not normal. To prove this, check the normality condition (9).
We have
[�; �](e1; e2) = �(�2 � 1)e3 6= 2d�(e1; e2)�:
In a similar way we can analyze the case �1 = � +
p
�2 � 4 > � = �2 = �3 > 2
with the same result.
� Consider the case �1 > �2 > �3, � = �ei. We have
�1 =
1
2
(��1 + �2 + �3); �2 =
1
2
(�1 � �2 + �3); �3 =
1
2
(�1 + �2 � �3):
Set � = e1. The condition �21 � (�2 � �3)
2 = 4 means that �2�3 = 1. The matrix
A� takes the form
A� =
0
@ 0 0 0
0 0 ��3
0 �2 0
1
A = �:
Since �2 6= �3, the �eld � is not a Killing one, but it is geodesic. The structure
(12) is almost contact. Indeed,
�2 =
0
@ 0 0 0
0 ��2�3 0
0 ��3�2
1
A =
0
@ 0 0 0
0 �1 0
0 0 �1
1
A
and hence
�2Z = �Z + g(�; Z) �:
The structure is normal if and only if
�1 = �2 +
1
�2
; �3 =
1
�2
; �2 > 1: (13)
Indeed, note that �e1 = 0; �e2 = �2e3; �e3 = ��3e2: Now put Z = e1;W = e2.
Then we have
[�; �][e1; e2] = (��3 + �22�2) e3; d�(e1; e2) = 0:
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 265
A. Yampolsky
Therefore, the �rst necessary condition of normality is �3 = �22�2. Since �2�3 = 1,
we can rewrite this condition as
�3�3 = �2�2: (14)
Put Z = e1;W = e3. Then we have
[�; �][e1; e3] = (�2 � �22�3) e2; d�(e1; e3) = 0:
The second necessary condition of normality is �2 = �23�3, which is equivalent
to (14).
Finally, put Z = e2;W = e3. Then we have
[�; �][e2; e3] = �1 e1; d�(e2; e3) = �1
2
e1
and (11) is ful�lled. The equation (14) can be simpli�ed to
(�3 � �2)(�1 � (�2 + �3)) = 0:
Since �2 6= �3, we get �1 = �2 + �3: Then �1 = 0, �2 = �3, �3 = �2 and, from
the condition �2�3 = 1, we �nd �2�3 = 1: Since �1 > �2 > �3, we get (13).
The structure is not metric, since
g(�Z; �W ) = �23z3w3 + �22z2w2 6= g(Z;W ) � g(�; Z) g(�;W ) = z2w2 + z3w3:
Making similar computations for � = e2, we get the normality condition of the
form �2 = �1 + �3 which contradicts the condition �1 > �2 > �3. The structure
is not metric. Finally, for � = e3, we get the normality condition of the form
�3 = �1 + �2 which contradicts again the condition �1 > �2 > �3 and the
structure is not metric again.
In a similar way we prove the following propositions.
Proposition 2.2. Let � be a left-invariant totally geodesic unit vector �eld on
SL(2; R) with the left-invariant metric g and let fei; i = 1; 2; 3g be an orthonormal
basis for the Lie algebra satisfying (3). Assume in addition that �1 � �2 > 0,
�3 < 0. Then �
� = A�; �; � = g(�; � )
�
is the almost contact structure on SL(2; R), where g(� ; �) is the scalar product
with respect to g. Moreover, if
� �1 = �2; �3 = �2, then the structure is Sasakian;
� �3 = �
p
4 + (�1 � �2)2 < �2 or �1 =
p
4 + (�2 � �3)2 , then the structure
is neither normal nor metric.
266 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
Proposition 2.3. Let � be a left-invariant totally geodesic unit vector �eld on
E(2) with the left-invariant metric g and let fei; i = 1; 2; 3g be an orthonormal
basis for the Lie algebra satisfying (3). Assume in addition that �1 � �2 > 0,
�3 = 0.
If �1 = �2 = � > 0, then the group is �at. Moreover,
� if � = e3, then � is a parallel vector �eld on E(2);
� if � = x1e1 + x2e2, then � moves along e3 with the constant angle speed �.
If �1 > �2 > 0, then
�
� = A�; �; � = g(�; � )
�
is an almost contact structure on
E(2). This structure is neither metric nor normal.
Proposition 2.4. Let � be a left-invariant totally geodesic unit vector �eld on
E(1; 1) with the left-invariant metric and let fei; i = 1; 2; 3g be an orthonormal
basis for the Lie algebra satisfying (3). Assume in addition that �1 > 0, �2 < 0,
�3 = 0. Then �
� = A�; �; � = g(�; � )
�
is an almost contact structure on E(1; 1). This structure is neither metric nor
normal.
Proposition 2.5. Let � be a left-invariant totally geodesic unit vector �eld
on the Heisenberg group with the left-invariant metric and let fei; i = 1; 2; 3g be
an orthonormal basis for the Lie algebra satisfying (3). Moreover, assume that
�1 > 0; �2 = 0; �3 = 0. Then�
� = A�; �; � = g(�; � )
�
is a Sasakian structure.
3. Nonunimodular Case
Choose the orthonormal frame e1; e2; e3 as in (4). Then the Levi�Civita con-
nection is given by the following table:
r e1 e2 e3
e1 0 � e3 �� e2
e2 ��e2 � e1 0
e3 �Æ e3 0 Æ e1
:
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 267
A. Yampolsky
For any left-invariant unit vector �eld � = x1e1 + x2e2 + x3e3 we have
re1� = � e1 � �; re2� = �� e3 � �; re3� = Æ e2 � �:
Denote
N1 = e1 � � = �x3 e2 + x2 e3;
N2 = e3 � � = �x2 e1 + x1 e2;
N3 = e2 � � = x3 e1 � x1 e3:
(15)
Then the matrix of A� takes the form
A� =
0
B@
0 ��x2 �Æ x3
� x3 �x1 0
�� x2 0 Æ x1
1
CA : (16)
A direct computation gives the following result.
Lemma 3.1. The derivatives (rei
A�)ek of the Weingarten operator A� for
the left-invariant unit vector �eld are as in the following table:
e1 e2 e3
e1 �� 2(x1e1 � �) � Æ N3 + � � x1e3 � �N2 � � Æ x1e2
e2 � 2N2 + � �x3e1 � �N1 � � 2(x3e3 � �) � Æ x3e2
e3 �Æ 2N3 � � Æ x2e1 � Æ x2e3 � Æ N1 � Æ 2(x2e2 � �)
:
By a straightforward application of the Codazzi equation and Lem. 3.1, we
can easily prove the following.
Lemma 3.2. The curvature operator of the nonunimodular group with respect
to the chosen frame takes the form
R(e1; e2)� = � 2N2 + � (� � Æ )N3;
R(e1; e3)� = �Æ 2N3 � � (� � Æ )N2;
R(e2; e3)� = � Æ N1:
The following Lemma gives the components of (2).
Lemma 3.3. Let G be a nonunimodular Lie group with the basis satisfying
(4). Then the left-invariant unit vector �eld � = x1e1 + x2e2 + x3e3 is totally
geodesic if and only if it satis�es the following equations:
(1; 1) � x1
n�
� [1 + � (� � Æ )]x2 + � 3x3
�
N2
�
�
� [1� Æ (� � Æ )]x3 � Æ 3x2
�
N3
o
= 0;
268 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
(2; 2) �
nh
� [1 + � 2(1� x23)]� [� + � 2(� � Æ )]x2x3
�i
N1
+ �
h
1 + Æ 2
i
x1x3N3
o
= 0;
(3; 3) Æ
nh
� [1 + Æ 2(1� x22)] + [Æ � � 2(� � Æ )]x2x3
�i
N1
� Æ
h
1 + � 2
i
x1x2N2
o
= 0;
(1; 2) � x1
h
[� + � 2(� � Æ )]x2 + � � 2x3
i
N1
+ �
h
� [1 + � 2(1� x23)]� � [1 + � (� � Æ )]x2x3
i
N2
+
h
� Æ
�
� Æ (1�x21)� Æ 2x2x3+� (� � Æ )(1�x23)
�
+� � (x23�x21)+� Æ
i
N3 = 0;
(1; 3) � x1
h
[Æ � � 2(� � Æ )]x3 � � Æ 2x2
i
N1
�
h
� Æ
�
�� (1� x21) + � 2x2x3 � � (� � Æ )(1� x22)
�
+ � � + � Æ (x22 � x21)
i
N2
+ Æ
h
� [�1 + Æ (� � Æ )]x2x3 � Æ [1 + Æ 2(1� x22)]
i
N3 = 0;
(2; 3)
h
�
�
� Æ (� + Æ )x2x3 � � (� � Æ )(� (1� x23) + Æ (1� x22))
�
+ � Æ (x22 � x23)
i
N1 + � Æ
h
1 + � 2
i
x1x3N2 � � Æ
h
1 + Æ 2
i
x1x2N3 = 0:
The proof consists of rather long computations of the corresponding compo-
nents TG�(ei; ek) for various combinations of (i; k) similar to those in the uni-
modular case.
P r o o f of the Theorem 1.2. Set � = x1e1 + x2e2 + x3e3 and suppose
x1 6= 0. From (15) it follows that N2 6= 0, N3 6= 0 and they are always linearly
independent. Moreover, the vectors N1 and N3 are linearly dependent if and only
if x3 = 0. If x3 6= 0, then the equation (2; 2) implies x3 = 0 and we come to
a contradiction.
Put x3 = 0. If x2 6= 0, then N1 and N2 are linearly independent and (3; 3)
implies Æ = 0. In this case we can rewrite (1; 1) as � 2x1x2(1 + � 2)N2 = 0 and
we have � = 0. In this case the equation (1; 2) takes the form � 2(1 +� 2)N2 = 0
and we have a contradiction.
Put x3 = x2 = 0. In this case � = e1, N1 = 0, N2 = e2 and N3 = �e3.
The equation (1; 2) takes the form � 2(1+� 2)N2 = 0. This gives a contradiction.
Suppose � = x2e2 + x3e3. Since x1 = 0, we have N1 6= 0 and N1 is linearly
independent with either N2 or N3.
Suppose � = 0. Then (2; 2) implies �� 2x2x3 = 0 and we have the following
cases:
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 269
A. Yampolsky
� Case x3 = 0. Then N1 = � e3, N2 = � e1, N3 = 0 and the equation (1; 2)
takes the form � 2(1 + � 2)N2 = 0, which is a contradiction.
� Case x2 = 0. Then N1 = � e2, N2 = 0, N3 = � e1. The equation (1; 3)
then takes the form �Æ 2(1 + Æ 2)N3 = 0 and we should set Æ = 0. It is easy
to check that if � = Æ = 0, then all equations are ful�lled. Moreover, the
�eld � = �e3 becomes a parallel vector �eld, since A� � 0.
Suppose � 6= 0, Æ = 0. Then (1; 3) implies � �N2 = 0 and we have x2 = 0.
In this case x23 = 1 and (2; 2) yields �� N1 = 0. This gives a contradiction.
Suppose � 6= 0; Æ 6= 0. In this case we apply a di�erent method based on the
explicit expression for the second fundamental form of �(Mn) � T1M
n [11].
Lemma 3.4. Let � be a unit vector �eld on a Riemannian manifold Mn+1.
The components of second fundamental form of �(M) � T1M
n+1 are given by
~
�jij = 1
2
��ij
n
�
(rei
A�)ej + (rej
A�)ei; f�
�
+��
�
�j
R(e�; ei)�; fj
�
+ �i
R(e�; ej)�; fi
��o
;
where ��ij = [(1 + �2�)(1 + �2
i
)(1 + �2
j
)]�1=2, �0 = 0; �1; : : : ; �n are the singular
values of the matrix A� and e0; e1; : : : ; en; f1; : : : ; fn are the orthonormal frames
of singular vectors (i; j = 0; 1; : : : ; n; � = 1; : : : ; n).
Since x1 = 0, the matrix (16) takes the form
A� =
0
B@
0 ��x2 �Æ x3
� x3 0 0
�� x2 0 0
1
CA :
Denote by ~e0; ~e1; ~e2; ~f1; ~f2 the orthonormal singular frames of A�. The matrix
At
�
A� takes the form
At
�A� =
0
B@
� 2 0 0
0 � 2x22 � Æ x2x3
0 � Æ x2x3 Æ 2x23
1
CA : (17)
The eigenvalues are
�
0; � 2; � 2x22+ Æ 2x23
�
. Denote m =
q
� 2x22 + Æ 2x23. Then the
singular values are
�0 = 0; �1 = j� j; �2 = m:
270 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
The singular frame ~e0; ~e1; ~e2 consists of the eigenvectors of the matrix (17),
namely,
~e0 =
1
m
�
� Æ x3 e2 + �x2 e3
�
; ~e1 = e1; ~e2 =
1
m
�
�x2 e2 + Æ x3 e3
�
:
To �nd ~f1 and ~f2, compute
A� ~e1 = �
�
x3 e2 � x2 e3
�
; A� ~e2 = �me1:
Denote " = sign(� ). Then
~f1 = "
�
x3 e2 � x2 e3
�
; ~f2 = �e1:
Now we have
~
�j00 = � 1p
1 + �2�
g((r ~e0A�) ~e0; ~f�) :
If � is totally geodesic, then � satis�es
0 = (r ~e0A�) ~e0 = r ~e0(A� ~e0)�A�r ~e0 ~e0 = A�A ~e0 ~e0:
Since (16) is applicable to any left-invariant unit vector �eld, we easily calculate
A ~e0 ~e0 = � 1
m2
� Æ (Æ x23 + �x22) ~e1:
Therefore,
A�A ~e0 ~e0 = �"� � Æ (Æ x23 + �x22)
~f1:
Since � 6= 0; � 6= 0 and Æ 6= 0, we have(
�x22 + Æ x23 = 0;
x22 + x23 = 1:
Solving the system, we get
x22 =
�Æ
� � Æ
; x23 =
�
� � Æ
:
Remind that � + Æ > 0; � � Æ by the choice of the frame. Therefore, the
solution exists if Æ < 0 and, as a consequence, � > 0. Thus,
� = �
r
�Æ
� � Æ
e2 �
r
�
� � Æ
e3:
Denote � = � 1. Without loss of generality, we can put
� = �
r
�Æ
� � Æ
e2 +
r
�
� � Æ
e3:
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 271
A. Yampolsky
As a consequence, we get m =
p
�� Æ : Moreover,
�
m
x2 = �
�p
�� Æ
r
�Æ
� � Æ
= � x3;
Æ
m
x3 =
Æp
�� Æ
r
�
� � Æ
=
�
p
(�Æ )2p
�� Æ
r
�
� � Æ
= �� x2
and we have
~e0 = 1
m
�
� Æ x3 e2 + �x2 e3
�
= � �; ~e1 = e1 = � ~f2;
~e2 = 1
m
�
�x2 e2 + Æ x3 e3
�
= �(x3 e2 � x2 e3) = �" ~f1:
With respect to this frame the matrix A� takes the form
A� =
0
@ 0 0 0
0 0 �m
0 �� 0
1
A :
A simple calculation yields
r ~e0 ~e1 ~e2
~e0 0 ��m ~e2 �m ~e1
~e1 �� ~e2 0 � ~e0
~e2 �m ~e1 ��m ~e0 � (� + Æ ) ~e2 (� + Æ ) ~e1
: (18)
With respect to a new frame, the derivatives (r ~eiA�) ~ek form the following table:
~e0 ~e1 ~e2
~e0 0 �m(�m� � ) ~e1 m(�m� � ) ~e2
~e1 �m� ~e1 �� 2 ~e0 0
~e2 �m� ~e2 �(� + Æ )(m� �� ) ~e1 �m2 ~e0 + (� + Æ )(m� �� ) ~e2
.
Finally, the necessary components of the curvature operator can be found from
the latter table and take the form
R( ~e0; ~e1)� = m(�m� 2� ) ~e1; R( ~e0; ~e2)� = ��m2 ~e2;
R( ~e1; ~e2)� = �(� + Æ )(m� �� ) ~e1:
(19)
272 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
Now, we can compute all the entries of the matrices ~
�. As a result, we have
~
1 =
0
BBBBB@
0 0
1
2
"�m(m� 2�� )(m� � �)p
(1 + � 2)(1 +m2)
0 0 0
1
2
"�m(m� 2�� )(m� � �)p
(1 + � 2)(1 +m2)
0
"�(� + Æ )(�m� � )(m� � �)p
(1 + � 2)(1 +m2)
1
CCCCCA ;
~
2 =
0
BBBBBBB@
0
m2(m� � �)
2
p
(1 + � 2)(1 +m2)
0
m2(m� � �)
2
p
(1 + � 2)(1 +m2)
0
(� + Æ )(�� �m)
2
p
(1 + � 2)
0
(� + Æ )(�� �m)
2
p
(1 + � 2)
0
1
CCCCCCCA
:
Thus, for a totally geodesic �eld �, we have a unique possible solution � = �m,
m� = �. It follows that �� Æ = m2 = 1, � = �(= �1). As a consequence,
� � = �
1p
1 + � 2
e2 +
�p
1 + � 2
e3
is the corresponding totally geodesic unit vector �eld.
Now we give a geometrical description of totally geodesic unit vector �eld and
the group.
Proposition 3.1. Let G be a nonunimodular three-dimensional Lie group with
the left-invariant metric. Suppose that G admits a left-invariant totally geodesic
unit vector �eld �. Then either
� G = L2(��2) � E1, where L2(��2) is the Lobachevski plane of curvature
�� 2, and � is a parallel unit vector �eld on G tangent to the Euclidean
factor, or
� G admits a Sasakian structure; moreover, G admits two hyperfoliations
L1;L2 such that:
(i) the foliations L1 and L2 are intrinsically �at, mutually orthogonal and
have constant extrinsic curvature;
(ii) one of them, say L2, is minimal;
(iii) the integral trajectories of the �eld � are L1 \ L2.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2 273
A. Yampolsky
P r o o f. Suppose � is as in the hypothesis. Consider the case � = Æ = 0
and � = e3 of Th. 1.2. The bracket operations take the form
[e1; e2] = � e2; [e1; e3] = 0; [e2; e3] = 0;
and we conclude that the group admits three integrable distributions, namely,
e1^e2, e1^e3 and e2^e3. The table of the Levi�Civita connection takes the form
r e1 e2 e3
e1 0 0 0
e2 �� e2 �� e1 0
e3 0 0 0
.
The only nonzero component of the curvature tensor of the group is of the form
R(e1; e2)e2 = �� 2e1: Thus, G = L2(�� ) � R1 and the �eld � = e3 is a parallel
unit vector �eld on G tangent to the Euclidean factor.
Consider the second case of Th. 1.2. If � = �, m =
p
�� Æ = 1, then with
respect to the singular frame, the matrix A� takes the form A� =
0
@ 0 0 0
0 0 �1
0 1 0
1
A
and hence, � = � ~e0 is a Killing unit vector �eld. Therefore, by Th. 2.1, the
structure
�
� = A�; �; � = g(�; �)
�
is Sasakian.
We can also say more about this Sasakian structure. The table (18) in the
case under consideration takes the form
r ~e0 ~e1 ~e2
~e0 0 �� ~e2 � ~e1
~e1 �� ~e2 0 � ~e0
~e2 � ~e1 �� ~e0 � (� + Æ ) ~e2 (� + Æ ) ~e1
; (20)
and hence, for the brackets we have
[ ~e0; ~e1] = 0; [ ~e0; ~e2] = 0; [ ~e1; ~e2] = 2� ~e0 + (� + Æ ) ~e2: (21)
From (21) we see that the distributions ~e0^ ~e2 and ~e0^ ~e1 are integrable. Denote
by L1 and L2 the corresponding foliations generated by these distributions. Then
the integral trajectories of the �eld � are exactly L1 \ L2.
274 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 2
Invariant Totally Geodesic Unit Vector Fields on Three-Dimensional Lie Groups
Denote by
(1) and
(2) the second fundamental forms of L1 and L2 res-
pectively. Since ~e1 and ~e2 are unit normal vector �elds for the corresponding
foliations, from (20) we can easily �nd
(1) =
�
0 1
1 � + Æ
�
;
(2) =
�
0 ��
� 0
�
and see that L2 is a minimal foliation.
Putting � = � ~e0, we �nd from (19) the corresponding curvature components:
R( ~e0; ~e2) ~e0 = � ~e2; R( ~e0; ~e1) ~e0 = � ~e1:
Denote by K
(i)
int
and K
(i)
ext
the intrinsic and extrinsic curvatures of the correspond-
ing foliations (i = 1; 2). Then K
(i)
ext
=
R( ~e0; ~ei) ~ei; ~e0
�
= 1. The Gauss equation
implies
K
(i)
int
= K
(i)
ext
+ det
(i) = 0:
Therefore, both foliations are intrinsically �at and have a constant extrinsic cur-
vature K
(i)
ext
= 1.
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